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Sixth Edition

Website www.mhhe.com/grifﬁth The text-speciﬁc website provides students with useful study tools designed to help improve their understanding of the material presented in the text and class. For the instructor, the website is designed to help ease the time burdens of the course by providing valuable presentation and preparation tools.

For Instructors

• Student study guide

• All student content

– – – – –

Mastery quiz Know Understand Study hints Practice problems

• Animations

MD DALIM #971965 7/14/08 CYAN MAG YELO BLK

• Answers to selected questions

from the text

• Instructor’s manual • PowerPoint lectures • PowerPoint ﬁles of the text’s

images • Sample syllabi, including an

energy-themed course • Clicker questions • Formula summaries

of Everyday Phenomena

For Students

Physics

This imposing wave is seen breaking on the Oregon Coast in the United States’ Paciﬁc Northwest. How are waves created? How do waves travel? What affects their speed and frequency? What do water waves have in common with light and sound waves? Can we use the energy of ocean waves to generate electric power? How can knowledge of wavelengths and frequencies be applied to examination of the galaxies and the current hypothesis that the universe is expanding? The answers to these questions and many more can all be learned through your study of The Physics of Everyday Phenomena.

The

On the cover

Griffith Brosing

Physics y

The

of Everyday Phenomena A Conceptual Introduction to Physics Sixth Edition

W. Thomas Griffith Juliet W. Brosing

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Sixth Edition

of

THE

PHYSICS Everyday Phenomena

A Conceptual Introduction to Physics W. Thomas Griffith Pacific University

Juliet W. Brosing Pacific University

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THE PHYSICS OF EVERYDAY PHENOMENA: A CONCEPTUAL INTRODUCTION TO PHYSICS, SIXTH EDITION Published by McGraw-Hill, a business unit of The McGraw-Hill Companies, Inc., 1221 Avenue of the Americas, New York, NY 10020. Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved. Previous editions © 2007, 2004, 2001, 1998, and 1992. No part of this publication may be reproduced or distributed in any form or by any means, or stored in a database or retrieval system, without the prior written consent of The McGraw-Hill Companies, Inc., including, but not limited to, in any network or other electronic storage or transmission, or broadcast for distance learning. Some ancillaries, including electronic and print components, may not be available to customers outside the United States. This book is printed on acid-free paper. 1 2 3 4 5 6 7 8 9 0 DOW/DOW 0 9 8 ISBN 978–0–07–351211–2 MHID 0–07–351211–7 Publisher: Thomas Timp Sponsoring Editor: Debra B. Hash Director of Development: Kristine Tibbetts Senior Developmental Editor: Mary E. Hurley Senior Marketing Manager: Lisa Nicks Project Manager: Joyce Watters Senior Production Supervisor: Kara Kudronowicz Lead Media Project Manager: Stacy A. Patch Associate Design Coordinator: Brenda A. Rolwes Cover/Interior Designer: Studio Montage, St. Louis, Missouri (USE) Cover Image: © Images Ideas/PictureQuest Lead Photo Research Coordinator: Carrie K. Burger Photo Research: Pam Carley/Sound Reach Compositor: Laserwords Private Limited Typeface: 10/12 Times Printer: R. R. Donnelley Willard, OH The credits section for this book begins on page 503 and is considered an extension of the copyright page. Library of Congress Cataloging-in-Publication Data Griffith, W. Thomas. The physics of everyday phenomena : a conceptual introduction to physics. — 6th ed. / W. Thomas Griffith, Juliet W. Brosing. p. cm. Includes index. ISBN 978–0–07–351211–2 — ISBN 0–07–351211–7 (hard copy : alk. paper) 1. Physics. I. Brosing, Juliet Wain, 1953- II. Title. QC23.2.G75 2009 530—dc22

www.mhhe.com

2008018203

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brief contents 1

Physics, the Fundamental Science

Unit One

2 3 4 5 6 7 8

12 13 14

The Newtonian Revolution

Describing Motion

18

Falling Objects and Projectile Motion Newton’s Laws: Explaining Motion

38

Energy and Oscillations

15 16 17

79

102

Momentum and Impulse

124

Rotational Motion of Solid Objects

145

The Behavior of Fluids

170

Temperature and Heat

191

Electric Circuits

18 19 20 21 iii

257

281

Wave Motion and Optics

Making Waves

306

Light Waves and Color

330

Light and Image Formation

355

The Atom and Its Nucleus

The Structure of the Atom

382

The Nucleus and Nuclear Energy

Unit Six

Heat Engines and the Second Law of Thermodynamics 212

236

Magnets and Electromagnetism

Unit Five

Fluids and Heat

Electricity and Magnetism

Electrostatic Phenomena

Unit Four

58

Circular Motion, the Planets, and Gravity

Unit Two

9 10 11

Unit Three

1

408

Relativity and Beyond

Relativity

434

Beyond Everyday Phenomena

457

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detailed contents 3

Preface viii Acknowledgments xiv Secrets to Success in Studying Physics xvi

1

Physics, the Fundamental Science 1.1 The Scientific Enterprise 1.2 The Scope of Physics

1

3.4 Projectile Motion 3.5 Hitting a Target

49

Summary 53, Key Terms 53, Questions 54, Exercises 55, Synthesis Problems 56, Home Experiments and Observations 57

10

Summary 13, Key Terms 13, Questions 14, Exercises 14, Synthesis Problems 15, Home Experiments and Observations 15

2

46

everyday phenomenon box 3.1 Shooting a Basketball 50

1.3 The Role of Measurement and Mathematics in Physics 7

4

The Newtonian Revolution

Newton’s Laws: Explaining Motion 4.1 A Brief History

59

4.2 Newton’s First and Second Laws 4.3 Mass and Weight

64

18 19

4.5 Applications of Newton’s Laws

22

5

25

2.4 Graphing Motion

27

everyday phenomenon box 2.2 The 100-m Dash 30 2.5 Uniform Acceleration

70

Summary 74, Key Terms 74, Questions 75, Exercises 76, Synthesis Problems 77, Home Experiments and Observations 78

everyday phenomenon box 2.1 Transitions in Traffic Flow 22 2.3 Acceleration

67

everyday phenomenon box 4.2 Riding an Elevator 69

2.1 Average and Instantaneous Speed 2.2 Velocity

61

everyday phenomenon box 4.1 The Tablecloth Trick 64 4.4 Newton’s Third Law

Describing Motion

39

42

3.3 Beyond Free Fall: Throwing a Ball Upward 44

4

everyday phenomenon box 1.1 The Case of the Malfunctioning Coffee Pot 5

Unit One

3.1 Acceleration Due to Gravity 3.2 Tracking a Falling Object

2

1.4 Physics and Everyday Phenomena

Falling Objects and Projectile Motion 38

31

Summary 33, Key Terms 34, Questions 34, Exercises 35, Synthesis Problems 37, Home Experiments and Observations 37

Circular Motion, the Planets, and Gravity 79 5.1 Centripetal Acceleration 5.2 Centripetal Forces

80

83

everyday phenomenon box 5.1 Seat Belts, Air Bags, and Accident Dynamics 84 5.3 Planetary Motion

iv

87

58

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5.4 Newton’s Law of Universal Gravitation 5.5 The Moon and Other Satellites everyday phenomenon box 5.2 Explaining the Tides 94

6 7

Energy and Oscillations

Unit Two

9

Summary 97, Key Terms 97, Questions 98, Exercises 99, Synthesis Problems 100, Home Experiments and Observations 101

6.1 Simple Machines, Work, and Power

103

The Behavior of Fluids

170

9.1 Pressure and Pascal’s Principle

171

9.2 Atmospheric Pressure and the Behavior of Gases 173

9.3 Archimedes’ Principle

6.2 Kinetic Energy 106 6.3 Potential Energy 108 6.4 Conservation of Energy 110 everyday phenomenon box 6.1 Conservation of Energy 111 everyday phenomenon box 6.2 Energy and the Pole Vault 114

9.4 Fluids in Motion

178

181

9.5 Bernoulli’s Principle

183

everyday phenomenon box 9.2 Throwing a Curveball 186 Summary 187, Key Terms 188, Questions 188, Exercises 189, Synthesis Problems 190, Home Experiments and Observations 190

6.5 Springs and Simple Harmonic Motion 115 Summary 118, Key Terms 119, Questions 119, Exercises 121, Synthesis Problems 122, Home Experiments and Observations 123

Momentum and Impulse

Fluids and Heat

everyday phenomenon box 9.1 Measuring Blood Pressure 175

102

10

124

191

10.2 Heat and Specific Heat Capacity

10.4 Gas Behavior and the First Law 10.5 The Flow of Heat

192

195

201

204

everyday phenomenon box 10.1 Solar Collectors and the Greenhouse Effect 206

133

Summary 208, Key Terms 208, Questions 209, Exercises 210, Synthesis Problems 210, Home Experiments and Observations 211

Summary 139, Key Terms 140, Questions 140, Exercises 142, Synthesis Problems 143, Home Experiments and Observations 144

Rotational Motion of Solid Objects 145 8.1 What is Rotational Motion? 146 8.2 Torque and Balance 149 8.3 Rotational Inertia and Newton’s Second Law 152 8.4 Conservation of Angular Momentum

Temperature and Heat

10.1 Temperature and Its Measurement 10.3 Joule’s Experiment and the First Law of Thermodynamics 198

7.1 Momentum and Impulse 125 7.2 Conservation of Momentum 128 everyday phenomenon box 7.1 The Egg Toss 129 7.3 Recoil 131 7.4 Elastic and Inelastic Collisions 7.5 Collisions at an Angle 135 everyday phenomenon box 7.2 An Automobile Collision 137

8

90

93

11

Heat Engines and the Second Law of Thermodynamics 212 11.1 Heat Engines

213

everyday phenomenon box 11.1 Hybrid Automobile Engines 216 11.2 The Second Law of Thermodynamics

216

11.3 Refrigerators, Heat Pumps, and Entropy 220 155

everyday phenomenon box 8.1 Achieving the State of Yo 158 8.5 Riding a Bicycle and Other Amazing Feats 159 everyday phenomenon box 8.2 Bicycle Gears 162 Summary 163, Key Terms 164, Questions 164, Exercises 166, Synthesis Problems 167, Home Experiments and Observations 168

11.4 Thermal Power Plants and Energy Resources 223 11.5 Perpetual Motion and Energy Frauds

226

everyday phenomenon box 11.2 A Productive Pond 228 Summary 229, Key Terms 230, Questions 230, Exercises 232, Synthesis Problems 232, Home Experiments and Observations 233

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Unit Three

12

Unit Four

Electricity and Magnetism

Electrostatic Phenomena 12.1 Effects of Electric Charge 12.2 Conductors and Insulators

15

236 237 240

246

12.5 Electric Potential

248

13

Summary 253, Key Terms 253, Questions 254, Exercises 255, Synthesis Problems 255, Home Experiments and Observations 256

257

13.1 Electric Circuits and Electric Current

16

258

263

13.3 Series and Parallel Circuits

264

13.4 Electric Energy and Power

268

everyday phenomenon box 13.2 The Hidden Switch in Your Toaster

Light Waves and Color

330

16.1 Electromagnetic Waves

331

16.2 Wavelength and Color

335

16.3 Interference of Light Waves 16.4 Diffraction and Gratings

338

342

everyday phenomenon box 16.2 Antireflection Coatings on Eyeglasses 16.5 Polarized Light

272

Summary 275, Key Terms 276, Questions 276, Exercises 278, Synthesis Problems 279, Home Experiments and Observations 280

Magnets and Electromagnetism 14.1 Magnets and the Magnetic Force

281 282

14.2 Magnetic Effects of Electric Currents 14.3 Magnetic Effects of Current Loops

285 288

everyday phenomenon box 14.1 Direct-Current Motors 291

17

343

346

Summary 350, Key Terms 351, Questions 351, Exercises 352, Synthesis Problems 353, Home Experiments and Observations 353

Light and Image Formation

355

17.1 Reflection and Image Formation 17.2 Refraction of Light

356

359

everyday phenomenon box 17.1 Rainbows 362 17.3 Lenses and Image Formation

364

17.4 Focusing Light with Curved Mirrors

367

17.5 Eyeglasses, Microscopes, and Telescopes 370

14.4 Faraday’s Law: Electromagnetic Induction 292

everyday phenomenon box 17.2 Laser Refractive Surgery 373

everyday phenomenon box 14.2 Vehicle Sensors at Traffic Lights 295 14.5 Generators and Transformers

321

Summary 325, Key Terms 326, Questions 326, Exercises 327, Synthesis Problems 328, Home Experiments and Observations 329

everyday phenomenon box 16.1 Why Is the Sky Blue? 338

260

13.2 Ohm’s Law and Resistance

314

317

15.5 The Physics of Music

everyday phenomenon box 13.1 Electrical Impulses in Nerve Cells

307

everyday phenomenon box 15.2 A Moving Car Horn and the Doppler Effect 320

13.5 Alternating Current and Household Circuits 271

14

306

everyday phenomenon box 15.1 Electric Power from Waves 308 15.2 Waves on a Rope 311 15.4 Sound Waves

everyday phenomenon box 12.2 Lightning 252

Electric Circuits

Making Waves

15.1 Wave Pulses and Periodic Waves

15.3 Interference and Standing Waves

everyday phenomenon box 12.1 Cleaning Up the Smoke 243 12.3 The Electrostatic Force: Coulomb’s Law 244 12.4 The Electric Field

Wave Motion and Optics

296

Summary 299, Key Terms 299, Questions 300, Exercises 301, Synthesis Problems 302, Home Experiments and Observations 303

Summary 376, Key Terms 377, Questions 377, Exercises 378, Synthesis Problems 379, Home Experiments and Observations 380

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Unit Five

18

everyday phenomenon box 20.1 The Twin Paradox 446

The Structure of the Atom

20.5 General Relativity

382

18.1 The Existence of Atoms: Evidence from Chemistry 383 everyday phenomenon box 18.1 Fuel Cells and the Hydrogen Economy

386

18.2 Cathode Rays, Electrons, and X Rays

389

everyday phenomenon box 18.2 Electrons and Television 390

21

Beyond Everyday Phenomena

457

21.1 Quarks and Other Elementary Particles 458 21.2 Cosmology and the Beginning of Time 462

18.4 Atomic Spectra and the Bohr Model of the Atom 396

21.3 Semiconductors and Microelectronics

Summary 404, Key Terms 404, Questions 405, Exercises 406, Synthesis Problems 406, Home Experiments and Observations 407

The Nucleus and Nuclear Energy 19.1 The Structure of the Nucleus 19.2 Radioactive Decay

408

409

everyday phenomenon box 21.1 Holograms 471 Summary 472, Key Terms 473, Questions 473, Exercises 474, Synthesis Problems 474, Home Experiments and Observations 474

Appendix A Using Simple Algebra

412

Decimal Fractions, Percentages, and Scientific Notation 478

420

Appendix C

everyday phenomenon box 19.2 What Happened at Chernobyl? 424

Vectors and Vector Addition

19.5 Nuclear Weapons and Nuclear Fusion 424

Answers to Selected Questions, Exercises, and Synthesis Problems 486 Glossary 493 Photo Credits 503 Index 504

Relativity and Beyond

Relativity

434

20.1 Relative Motion in Classical Physics

482

Appendix D

Summary 429, Key Terms 429, Questions 430, Exercises 431, Synthesis Problems 431, Home Experiments and Observations 432

Unit Six

475

Appendix B

19.3 Nuclear Reactions and Nuclear Fission 417 19.4 Nuclear Reactors

464

21.4 Superconductors and Other New Materials 468

everyday phenomenon box 19.1 Smoke Detectors 414

20

449

Summary 453, Key Terms 454, Questions 454, Exercises 455, Synthesis Problems 455, Home Experiments and Observations 456

18.3 Radioactivity and the Discovery of the Nucleus 393

18.5 Particle Waves and Quantum Mechanics 400

19

20.4 Newton’s Laws and Mass-Energy Equivalence 445

The Atom and Its Nucleus

435

20.2 The Speed of Light and Einstein’s Postulates 438 20.3 Time Dilation and Length Contraction

442

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preface T

essential by many instructors, even in a one-semester course. If included in such a course, we recommend curtailing coverage in other areas to avoid student overload. Sample syllabi for these different types of courses can be found on the Instructor Center of the Online Learning Center. Some instructors would prefer to put chapter 20 on relativity at the end of the mechanics section or just prior to the modern physics material. Relativity has little to do with everyday phenomena, of course, but is included because of the high interest that it generally holds for students. The final chapter (21) introduces a variety of topics in modern physics—including particle physics, cosmology, semiconductors, and superconductivity—that could be used to stimulate interest at various points in a course. One plea to instructors, as well as to students using this book: Don’t try to cram too much material into too short a time! We have worked diligently to keep this book to a reasonable length while still covering the core concepts usually found in an introduction to physics. These ideas are most enjoyable when enough time is spent in lively discussion and in consideration of questions so that a real understanding develops. Trying to cover material too quickly defeats the conceptual learning and leaves students in a dense haze of words and definitions. Less can be more if a good understanding results.

he satisfaction of understanding how rainbows are formed, how ice skaters spin, or why ocean tides roll in and out—phenomena that we have all seen or experienced— is one of the best motivators available for building scientific literacy. This book attempts to make that sense of satisfaction accessible to non-science majors. Intended for use in a one-semester or two-quarter course in conceptual physics, this book is written in a narrative style, frequently using questions designed to draw the reader into a dialogue about the ideas of physics. This inclusive style allows the book to be used by anyone interested in exploring the nature of physics and explanations of everyday physical phenomena.

“Griffith has done a very respectable job in presenting his conceptual physics course in a clear, useable fashion. It is a fine work that is evidently quickly evolving into a top-notch textbook.” —Michael Bretz, University of Michigan

How This Book Is Organized With the exception of the reorganization of chapters 15, 16, and 17 introduced in the fourth edition, we have retained the same order of topics as in the previous editions. It is traditional with some minor variations. The chapter on energy (chapter 6) appears prior to that on momentum (chapter 7) so that energy ideas can be used in the discussion of collisions. Wave motion is found in chapter 15, following electricity and magnetism and prior to chapters 16 and 17 on optics. The chapter on fluids (chapter 9) follows mechanics and leads into the chapters on thermodynamics. The first 17 chapters are designed to introduce students to the major ideas of classical physics and can be covered in a one-semester course with some judicious paring. The complete 21 chapters could easily support a twoquarter course, and even a two-semester course in which the ideas are treated thoroughly and carefully. Chapters 18 and 19 on atomic and nuclear phenomena, are considered

Mathematics in a Conceptual Physics Course The use of mathematics in a physics course is a formidable block for many students, particularly non-science majors. Although there have been attempts to teach conceptual physics without any mathematics, these attempts miss an opportunity to help students gain confidence in using and manipulating simple quantitative relationships. Clearly mathematics is a powerful tool for expressing the quantitative relationships of physics. The use of mathematics can be carefully limited, however, and subordinated to the physical concepts being addressed. Many users of the first edition of this text felt that mathematical viii

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ix

expressions appeared too frequently for the comfort of some students. In response, we substantially reduced the use of mathematics in the body of the text in the second edition. Most users have indicated that the current level is about right, so we have not changed the mathematics level in subsequent editions. “The level of presentation is pitch-perfect for a college physics course. I happen to have a need for a book at just this level, compromising between a math-free conceptual book and one that goes for the full college-level (but not university-level) treatment. The brevity of presentation also lends itself well to a one-semester survey course format.” —Brent Royuk, Concordia University

Logical coherence is a strong feature of this book. Formulas are introduced carefully after conceptual arguments are provided, and statements in words of these relationships generally accompany their introduction. We have continued to fine tune the example boxes that present sample exercises and questions. Most of these provide simple numerical illustrations of the ideas discussed. No mathematics prerequisite beyond high school algebra should be necessary. A discussion of the basic ideas of very simple algebra is found in appendix A, together with some practice exercises, for students who need help with these ideas.

New to This Edition We have made several significant changes to the sixth edition. As the book has evolved, however, we have tried to remain faithful to the principles that have guided the writing of the book from the outset. One of these has been to keep the book to a manageable length, both in the number of chapters and the overall content. Many books become bloated as users and reviewers request more and more pet topics, We have tried to add material judiciously and have pared material elsewhere so that the overall length of the book has not changed. The changes include the following: 1. New Everyday Phenomenon Boxes. We have added five new everyday phenomenon boxes to this edition. Three of these are related to energy issues designed to better support instructors interested in building an energy emphasis into their courses. The new boxes are: everyday phenomenon box 6.1 Conservation of Energy everyday phenomenon box 7.1 The Egg Toss everyday phenomenon box 12.1 Cleaning Up the Smoke everyday phenomenon box 15.1 Electric Power from Waves everyday phenomenon box 19.1 Smoke Detectors 2. New Sample Exercises. Many users have pointed out a need for more sample exercises in some chapters. We

have added several new sample exercises in places where the need was apparent. Except for the first and the last chapter, most chapters now have three or four sample exercises. 3. Building an Energy Emphasis. Although this book remains a basic conceptual physics text, we are working to make the book better serve instructors who want to teach a conceptual physics course with an energy emphasis. This is reflected in the new everyday phenomenon boxes, but also in other places within the body of the text. In the past few editions, we have added everyday phenomenon boxes on fuel cells and hybrid automobiles, and boxes on solar collectors and nuclear reactors were already included. We have enhanced the discussion of the greenhouse effect in everyday phenomenon box 10.1. A syllabus for instructors wishing to teach a course with an energy emphasis can be found on the Instructor Center of the text website. We plan to continue building this emphasis in future editions. 4. New Home Experiments. We have added several new home experiments and have also added a few new synthesis problems. Many users have found these features to be very useful. 5. Continued Refinements in Artwork and Textual Clarity. Although the textual clarity of this text has been extensively praised by many reviewers and users, it can always be improved. Reviewers continue to point out places where either the art or text can be improved, and we have responded to many of these suggestions. To this end, we have made many changes, often subtle, to both the art and text. More noticeable changes include an improved and simplified discussion of planetory motion and Kepler’s Laws in chapter 5 and on updated discussion of integrated circuits in chapter 21.

Learning Aids The overriding theme of this book is to introduce physical concepts by appealing to everyday phenomena whenever possible. To achieve this goal, this text includes a variety of features to make the study of The Physics of Everyday Phenomena more effective and enjoyable. A few key concepts form the basis for understanding physics, and the textual features described here reinforce this structure so that the reader will not be lost in a flurry of definitions and formulas.

“The presentation is outstanding: Clear, concise, not too complicated, not trivial either. The style is refreshing. Students are invited to think; they are not overwhelmed by complicated explanations. . . .” —Klaus Rossberg, Oklahoma City University

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Chapter Openers Each chapter begins with an illustration from everyday experience and then proceeds to use it as a theme for introducing relevant physical concepts. Physics can seem abstract to many students, but using everyday phenomena and concrete examples reduces that abstractness. The chapter overview previews the chapter’s contents and what students can expect to learn from reading the chapter. The overview introduces the concepts to be covered, facilitating the integration of topics, and helping students to stay focused and organized while reading the chapter for the first time. The chapter outline includes all the major topic headings within the body of the chapter. It also contains questions that provide students with a guide of what they will be expected to know in order to comprehend the major concepts of the chapter. (These questions are then correlated to the end-ofchapter summaries.)

chapter

7

Momentum and Impulse

chapter overview In this chapter, we explore momentum and impulse and examine the use of these concepts in analyzing events such as a collision between a fullback and defensive back. The principle of conservation of momentum is introduced and its limits explained. A number of examples will shed light on how these ideas are used, particularly conservation of momentum. Momentum is central to all of these topics—it is a powerful tool for understanding a lot of life’s sudden changes.

chapter outline

1

2 3 4

Conservation of momentum. What is the principle of conservation of momentum, and when is it valid? How does this principle follow from Newton’s laws of motion? Recoil. How can we explain the recoil of a rifle or shotgun using momentum? How is this similar to what happens in firing a rocket? Elastic and inelastic collisions. How can collisions be analyzed using conservation of momentum? What is the difference between an elastic and an inelastic collision? Collisions at an angle. How can we extend momentum ideas to two dimensions? How does the game of pool resemble automobile collisions?

Other Text Features Running summary paragraphs are found at the end of each chapter section to supplement the more general summary at the end of the chapter.

Rotational displacement, rotational velocity, and rotational acceleration are the quantities that we need to fully describe the motion of a rotating object. They describe how far the object has rotated (rotational displacement), how fast it is rotating (rotational velocity), and the rate at which the rotation may be changing (rotational acceleration). These definitions are analogous to similar quantities used to describe linear motion. They tell us how the object is rotating, but not why. Causes of rotation are considered next.

“I found the liberal use of questions such as “Do you believe in atoms? And, if so, why?” to motivate the discussion to be outstanding. I also found the interwoven history used to guide the discussion to be excellent. I often use that approach myself. It usually leads to a natural flow of concepts and also informs the student how we know what we know, as well as giving them training in scientific thinking and showing them how science is done in real life. . . . Only someone who actively resisted understanding could fail to understand Griffith’s text. He writes clearly, logically, and interestingly.” —Charles W. Rogers, Southwestern Oklahoma State University

unit one

5

Momentum and impulse. How can rapid changes in motion be described using the ideas of momentum and impulse? How do these ideas relate to Newton’s second law of motion?

The chapter outlines, questions, and summaries provide a clear framework for the ideas discussed in each chapter. One of the difficulties that students have in learning physics (or any subject) is that they fail to construct the big picture of how things fit together. A consistent chapter framework can be a powerful tool in helping students see how ideas mesh.

124

“Very good chapter overview and chapter outline for each chapter and for each unit. Very clear introduction and illustration of physics phenomena, concepts, and principles, and excellent exercises, problems, and home experiments/observations at the end of each chapter.” —Hai-Sheng Wu, Minnesota State University, Mankato

Subsection headings are often cast in the form of questions to motivate the reader and pique curiosity.

What is the difference between speed and velocity? Imagine that you are driving a car around a curve (as illustrated in figure 2.5) and that you maintain a constant speed of 60 km/h. Is your velocity also constant in this case? The answer is no, because velocity involves the direction of motion as well as how fast the object is going. The direction of motion is changing as the car goes around the curve. To simply state this distinction speed as we have de

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Study hints and study suggestions provide students with pointers on their use of the textbook, tips on applying the principles of physical concepts, and suggestions for home experiments. Everyday phenomenon boxes relate physical concepts discussed in the text to real-world topics, societal issues, and modern technology, underscoring the relevance of physics and how it relates to our day-to-day lives. The list of topics includes: The Case of the Malfunctioning Coffee Pot (chapter 1) Transitions in Traffic Flow (chapter 2) The 100-m Dash (chapter 2) Shooting a Basketball (chapter 3) The Tablecloth Trick (chapter 4) Riding an Elevator (chapter 4) Seat Belts, Air Bags, and Accident Dynamics (chapter 5) Explaining the Tides (chapter 5) Conservation of Energy (chapter 6) Energy and the Pole Vault (chapter 6) The Egg Toss (chapter 7) An Automobile Collision (chapter 7) Achieving the State of Yo (chapter 8) Bicycle Gears (chapter 8) Measuring Blood Pressure (chapter 9) Throwing a Curveball (chapter 9) Solar Collectors and the Greenhouse Effect (chapter 10) Hybrid Automobile Engines (chapter 11) A Productive Pond (chapter 11) Cleaning Up the Smoke (chapter 12) Lightning (chapter 12) Electrical Impulses in Nerve Cells (chapter 13) The Hidden Switch in Your Toaster (chapter 13) Direct-Current Motors (chapter 14) Vehicle Sensors at Traffic Lights (chapter 14) Electric Power from Waves (chapter 15) A Moving Car Horn and the Doppler Effect (chapter 15) Why Is the Sky Blue? (chapter 16) Antireflection Coatings on Eyeglasses (chapter 16) Rainbows (chapter 17) Laser Refractive Surgery (chapter 17) Fuel Cells and the Hydrogen Economy (chapter 18) Electrons and Television (chapter 18) Smoke Detectors (chapter 19) What Happened at Chernobyl? (chapter 19) The Twin Paradox (chapter 20) Holograms (chapter 21)

study hint Visualizing these angular momentum vectors and their changes can be an abstract and difficult task. The effect will seem much more real if you can directly experience it. If a bicycle wheel mounted on a hand-held axle (such as that pictured in figure 8.23) is available, try the tilt effect yourself. Grasp the wheel with both hands by the handles on each side and have someone give it a good spin with the wheel in a vertical plane. Then try tilting the wheel downward to the left to simulate a fall. The wheel will seem to have a mind of its own and will turn to the left as suggested by figure 8.22.

everyday phenomenon

box 9.1

Measuring Blood Pressure The Situation. When you visit your doctor’s office, the nurse will almost always take your blood pressure before the doctor spends time with you. A cuff is placed around your upper arm (as shown in the photograph) and air is pumped into the cuff, producing a feeling of tightness in your arm. Then the air is slowly released while the nurse listens to something with a stethoscope and records some numbers, such as 125 over 80.

through the compressed artery at the peak of the heart’s cycle. The lower reading, the diastolic pressure, is taken when blood flow occurs even at the low point in the cycle. There are distinctive sounds picked up by the stethoscope at these two points. The pressure recorded is actually the pressure in the air cuff for these two conditions. It is a gauge pressure, meaning that it is the pressure difference between the pressure being measured and atmospheric pressure. It is recorded in the units mm of mercury, which is the common way of recording atmospheric pressure. Thus a reading of 125 means that the pressure in the cuff is 125 mm of mercury above atmospheric pressure. A mercury manometer that is open to the air on one side (see the drawing) will measure gauge pressure directly.

Open end

Cuff Release valve

Having your blood pressure measured is a standard procedure for most visits to a doctor’s office. How does this process work?

What is the significance of these two numbers? What is blood pressure and how is it measured? Why are these readings an important factor, along with your weight, temperature, and medical history, in assessing your health? Stethoscope

The Analysis. Your blood flows through an elaborate system of arteries and veins in your body. As we all know, this flow is driven by your heart, which is basically a pump. More accurately, the heart is a double pump. One-half pumps blood through your lungs, where the blood cells pick up oxygen and discard carbon dioxide. The other half of the heart pumps blood through the rest of your body to deliver oxygen and nutrients. Arteries carry blood away from the heart into small capillaries that interface with other cells in muscles and organs. The veins collect blood from the capillaries and carry it back to the heart. We measure the blood pressure in a major artery in your upper arm at about the same height as your heart. When air is pumped into the cuff around your upper arm, it compresses this artery so that the blood flow stops. The nurse places the stethoscope, a listening device, near this same artery at a lower point in the arm and listens for the blood flow to restart as the air in the cuff is released. The heart is a pulsating pump that pumps blood most strongly when the heart muscle is most fully compressed. The pressure therefore fluctuates between high and low values. The higher reading in the blood pressure measurement, the systolic pressure, is taken when the blood just begins to spurt

An open-ended manometer can be used to measure the gauge pressure of the cuff. The stethoscope is used to listen for sounds indicating the restart of blood flow.

High blood pressure can be a symptom of many health problems, but most specifically, it is a warning sign for heart attacks and strokes. When arteries become constricted from the buildup of plaque deposits inside, the heart must work harder to pump blood through the body. Over time this can weaken the heart muscle. The other danger is that blood vessels might burst in the brain, causing a stroke, or blood clots might break loose and block smaller arteries in the heart or brain. In any case, high blood pressure is an important indicator of a potential problem. Low blood pressure can also be a sign of problems. It can cause dizziness when not enough blood is reaching the brain. When you stand up quickly, you sometimes experience a feeling of “light-headedness” because it takes a brief time for the heart to adjust to the new condition where your head is higher. Giraffes have a blood pressure about three times higher than humans (in gauge pressure terms). Why do you suppose this is so?

“This book compared to others is simply interesting. Topics like physics of music and color perception really engaged me, even as I read most of the chapters in one sitting. It indeed does a good job at getting at everyday phenomena.” —Tim Bolton, Kansas State University

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example box 2.4 Sample Exercise: Uniform Acceleration A car traveling due east with an initial velocity of 10 m/s accelerates for 6 seconds at a constant rate of 4 m/s2. a. What is its velocity at the end of this time? b. How far does it travel during this time?

Example boxes are included within the chapter and contain one or more concrete, worked examples of a problem and its solution as it applies to the topic at hand. Through careful study of these examples, students can better appreciate the many uses of problem solving in physics.

a. v0 10 m/s a 4

v v0 at 10 m/s (4 m/s2)(6 s)

m/s2

t 6s

10 m/s 24 m/s

v ?

34 m/s v 34 m/s due east

End-of-Chapter Features • The summary highlights the key elements of the chapter and correlates to the questions asked about the chapter’s major concepts on the chapter opener.

Summary

summary In this chapter, we recast Newton’s second law in terms of impulse and momentum to describe interactions between objects, such as collisions, that involve strong interaction forces acting over brief time intervals. The principle of conservation of momentum, which follows from Newton’s second and third laws, plays a central role.

Conservation of momentum, 126 Recoil, 129 Perfectly inelastic collision, 130

Elastic collision, 130 Partially inelastic collision, 130

Fnet∆t = ∆p,

4

p = mv

Conservation of momentum. Newton’s second and 2third laws combine to yield the principle of conservation of mo-

Elastic and inelastic collisions. A perfectly inelastic collision is one in which the objects stick together after the collision. If external forces can be ignored, the total momentum is conserved. An elastic collision is one in which the total kinetic energy is also conserved.

mentum: if the net external force acting on a system is zero, the total momentum of the system is a constant.

study hint Except for the examples involving impulse, most of the situations described in this chapter highlight the principle of conservation of momentum. The basic ideas used in applying conservation of momentum are:

3. Equality of momentum before and after the event can be used to obtain other information about the motion of the objects.

vi

vi

After

Before P

For review, look back at how these three points are used in each of the examples in this chapter. The total momentum of the system before and after the event is always found by adding the momentum values of the individual objects as vectors. You should be able to describe the magnitude and direction of this total momentum for each of the examples.

1. External forces are assumed to be much smaller than the very strong forces of interaction in a collision or other brief event. If external forces acting on the system can be ignored, momentum is conserved. 2. The total momentum of the system before the collision or other brief interaction pinitial is equal to the momentum after the event pfinal. Momentum is conserved and does not change.

vf

P

Elastic

Perfectly inelastic

5 Collisions at an angle. Conservation of momentum is not restricted to one-dimensional motion. When objects collide

If Fexternal = 0

at an angle, the total momentum of the system before and after the collision is found by adding the momentum vectors of the individual objects.

Ptotal = constant

pf

Before

After

questions pf = pi * more open-ended questions, requiring lengthier responses, suitable for group discussion Q sample responses are available in appendix D Q sample responses are available on the website

Q2. Two forces produce equal impulses, but the second force acts for a time twice that of the first force. Which force, if either, is larger? Explain. Q3. Is it possible for a baseball to have as large a momentum as a much more massive bowling ball? Explain. Q4. Are impulse and force the same thing? Explain. Q5. Are impulse and momentum the same thing? Explain. Q6. If a ball bounces off a wall so that its velocity coming back has the same magnitude that it had prior to bouncing: a. Is there a change in the momentum of the ball? Explain. b. Is there an impulse acting on the ball during its collision with the wall? Explain. Q7. Is there an advantage to following through when hitting a baseball with a bat, thereby maintaining a longer contact between the bat and the ball? Explain.

• Exercises and synthesis problems are intended to help students test their grasp of problem-solving. The odd-numbered exercises have answers in appendix D. By working through the odd-numbered exercises and checking the answer in appendix D, students can gain confidence in tackling the even-numbered exercises, and thus reinforce their problem-solving skills.

p1

p2 = –p1 ∆p

Impulse

Impulse, 124 Momentum, 124 Impulse-momentum principle, 124

Q1. Does the length of time that a force acts on an object have any effect on the strength of the impulse produced? Explain.

Exercises

p2

ment that the net impulse acting on an object equals the change in momentum of the object. Impulse is defined as the average force acting on an object multiplied by the time interval during which the force acts. Momentum is defined as the mass of an object times its velocity.

key terms

Questions

total momentum after the event must still be zero if there is no net external force. The final momentum vectors of the two objects are equal in size but opposite in direction.

and impulse. Newton’s second law can 1be recastMomentum in terms of momentum and impulse, yielding the state-

Key Terms • Key terms are page-referenced to where students can find the terms defined in context. • Questions are designed to challenge students to demonstrate their understanding of the key concepts. Selected answers are provided in appendix D to assist students with their study of more difficult concepts.

Recoil. If an explosion or push occurs between two ob3jects initially at rest, conservation of momentum dictates that the

Q8. What is the advantage of a padded dashboard compared to a rigid dashboard in reducing injuries during collisions? Explain using momentum and impulse ideas.

Q9. What is the advantage of an air bag in reducing injuries during collisions? Explain using impulse and momentum ideas. *Q10. If an air bag inflates too rapidly and firmly during a collision, it can sometimes do more harm than good in lowvelocity collisions. Explain using impulse and momentum ideas. Q11. If you catch a baseball or softball with your bare hand, will the force exerted on your hand by the ball be reduced if you pull your arm back during the catch? Explain.

Q13. Is the principle of conservation of momentum always valid, or are there special conditions necessary for it to be valid? Explain. Q14. A ball is accelerated down a fixed inclined plane under the influence of the force of gravity. Is the momentum of the ball conserved in this process? Explain.

synthesis problems

Q15. Two objects collide under conditions where momentum is SP1. A fast ball thrown with a velocity of 40 m/s (approximately conserved. Is the momentum of each object conserved in 90 MPH) is struck by a baseball bat, and a line drive the collision? Explain.

comes back toward the pitcher with a velocity of 60 m/s.

Q16. Which of Newton’s laws of motion are involved in justifyThe ball is in contact with the bat for a time of just 0.04 s. ing the principle of conservation of momentum? Explain.

The baseball has a mass of 120 g (0.120 kg). a. What is the change in momentum of the baseball during this process? b. Is the change in momentum greater than the final momentum? Explain. c. What is the magnitude of the impulse required to produce this change in momentum? d. What is the magnitude of the average force that acts on the baseball to produce this impulse?

exercises E1. An average force of 300 N acts for a time interval of 0.04 s on a golf ball. a. What is the magnitude of the impulse acting on the golf ball? b. What is the change in the golf ball’s momentum? E2. What is the momentum of a 1200-kg car traveling with a speed of 27 m/s (60 MPH)? E3. A bowling ball has a mass of 6 kg and a speed of 1.5 m/s. A baseball has a mass of 0.12 kg and a speed of 40 m/s. Which ball has the larger momentum? E4. A force of 45 N acts on a ball for 0.2 s. If the ball is initially at rest: a. What is the impulse on the ball? b. What is the final momentum of the ball? E5. A 0.12-kg ball traveling with a speed of 40 m/s is brought to rest in a catcher’s mitt. What is the size of the impulse exerted by the mitt on the ball? E6. A ball experiences a change in momentum of 24 kg·m/s. a. What is the impulse acting on the ball? b. If the time of interaction is 0.15 s, what is the magnitude of the average force acting on the ball? A 60 k f

i

i

Synthesis Problems

Q12. A truck and a bicycle are moving side by side with the same velocity. Which, if either, will require the larger impulse to bring it to a halt? Explain.

i i i ll

ih

SP2. A bullet is fired into a block of wood sitting on a block of E10. A fullback with a mass of 100 kg and a velocity of 3.5 m/s ice. with The bullet has an back initialwith velocity due west collides head-on a defensive a of 500 m/s and a mass of 0.005 mass of 80 kg and a velocity of 6kg. m/sThe duewooden east. block has a mass of 1.2 kg and is initially atofrest. bullet remains embedded in the block a. What is the initial momentum eachThe player? of wood afterward. b. What is the total momentum of the system before the a. Assuming that momentum is conserved, find the veloccollision? of the forces block can of wood and bullet after the collision. c. If they stick together andity external be ignored, b. What is the magnitude the impulse that acts on the what direction will they be traveling immediatelyofafter block of wood in this process? they collide? c. Does the change in momentum of the bullet equal that E11. An ice skater with a mass of 80 kg pushes off against a of the block of wood? Explain. second skater with a mass of 32 kg. Both skaters are iniSP3. Consider two cases in which the same ball is thrown against tially at rest. a wall withofthe initial velocity. a. What is the total momentum thesame system after they In case A, the ball sticks to the wall and does not bounce. In case B, the ball bounces push off? b. If the larger skater moves off the withsame a speed 3 m/s, back with speedofthat it came in with. what is the corresponding smaller skater?is the change in momentum a. Inspeed whichofofthe these two cases th fires l a t? E12. A rifle with a mass of 1.2 kg bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 600 m/s after the rifle is fired. a. What is the momentum of the bullet after the rifle is fired? b. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle?

SP4. A car traveling at a speed of 18 m/s (approximately 40 MPH) crashes into a solid concrete wall. The driver has a mass of 90 kg. a. What is the change in momentum of the driver as he comes to a stop? b. What impulse is required in order to produce this change in momentum? c. How does the application and magnitude of this force differ in two cases: the first, in which the driver is wearing a seat belt, and the second, in which he is not wearing a seat belt and is stopped instead by contact with the windshield and steering column? Will the time of action of the stopping force change? Explain. SP5. A 1500-kg car traveling due north with a speed of 25 m/s collides head-on with a 4500-kg truck traveling due south with a speed of 15 m/s. The two vehicles stick together after the collision. a. What is the total momentum of the system prior to the collision? b. What is the velocity of the two vehicles just after the collision? c. What is the total kinetic energy of the system before the collision? d. What is the total kinetic energy just after the collision? e. Is the collision elastic? Explain.

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• Because many courses for non-science majors do not have a laboratory component, home experiments and observations are found at the end of each chapter. The spirit of these home experiments is to enable students to explore the behavior of physical phenomena using easily available rulers, string, paper clips, balls, toy cars, flashlight batteries, and so on. Many instructors have found them useful for putting students into the exploratory and observational frame of mind that is important to scientific thinking. This is certainly one of our objectives in developing scientific literacy.

Home Experiments and Observations home experiments and observations HE1. Take two marbles or steel balls of the same size and practice shooting one into the other. Make these observations: a. If you produce a head-on collision with the second marble initially at rest, does the first marble come to a complete stop after the collision? b. If the collision with a second marble occurs at an angle, is the angle between the paths of the two marbles after the collision a right angle (90°)? c. If marbles of different sizes and masses are used, how do the results of parts a and b differ from those obtained with marbles of the same mass? HE2. If you have access to a pool table, try parts a and b of the observations in home experiment 1 on the pool table. What effect does putting spin on the first ball have on the collisions? HE3. If you have both a basketball and a tennis ball, try dropping the two of them onto a floor with a hard surface, first individually and then with the tennis ball placed on top of the basketball before the two are dropped together. a. Compare the height of the bounce of each ball in these different cases. The case where the two are dropped together may surprise you.

b. Can you devise an explanation for these results using impulse and Newton’s third law? (Consider the force between the basketball and the floor as well as that between the tennis ball and the basketball for the case where they are dropped together.) HE4. Place a cardboard box on a smooth tile or wood floor. Practice rolling a basketball or soccer ball at different speeds and allowing the ball to collide with the box. Observe the motion of both the box and the ball just after the collision. a. How do the results of the collision vary for different speeds of the ball (slow, medium, fast)? b. If we increase the weight of the box by placing books inside, how do the results of the collision change for the cases in part a? c. Can you explain your results using conservation of momentum?

“The selection of problems and questions at the end of each chapter is excellent. They provide students with a comprehensive review of the chapters and at the same time present challenges to reinforce the concepts. . . . Many students taking an introductory physics course do not have a chance to take a lab component with the course. The home experiments can go a long way toward addressing this deficiency.” —Farhang Amiri, Weber State University

Supplements Text Website A text-specific website that provides students with useful study tools designed to help improve their understanding of the material presented in the text and class. For the instructor, the website is designed to help ease the time burdens of the course by providing valuable presentation and preparation tools.

For Students Student Study Guide Integration • Mastery Quiz • Know • Understand • Study Hints • Practice Problems • Answers to Selected Questions Animations Crossword Puzzles Links Library Chapter Summary Chapter Objectives For Instructors All Student Content PowerPoint Lectures Instructor’s Manual Sample Syllabi CPS eInstruction Questions for Personal Response Systems Powerpoints of Art and Photos from the Text Test Bank Formula Summaries

Personal Response Systems Personal Response Systems (clickers) can bring interactivity into the classroom or lecture hall. Wireless response systems give the instructor and students immediate feedback from the entire class. The wireless response pads are essentially remotes that are easy to use and engage students, allowing instructors to motivate student preparation, interactivity, and active learning. Instructors receive immediate feedback to gauge which concepts students understand. Questions covering the content of The Physics of Everyday Phenomena text are formatted in PowerPoint and are available on the text website.

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Computerized Test Bank Online

Electronic Books

A comprehensive bank of test questions is provided on the text website within a computerized test bank powered by McGraw-Hill's flexible electronic testing program EZ Test Online (www.eztestonline.com). EZ Test Online allows you to create paper and online tests or quizzes in this easy to use program!

If you or your students are ready for an alternative version of the traditional textbook, McGraw-Hill brings you innovative and inexpensive electronic textbooks. By purchasing e-books from McGraw-Hill, students can save as much as 50% on selected titles delivered on the most advanced ebook platforms available.

Imagine being able to create and access your test or quiz anywhere, at any time, without installing the testing software. Now, with EZ Test Online, instructors can select questions from multiple McGraw-Hill test banks, or author their own, and then either print the test for paper distribution or give it online.

E-books from McGraw-Hill are smart, interactive, searchable, and portable, with such powerful tools as detailed searching, highlighting, note taking, and student-to-student or instructorto-student note sharing. E-books from McGraw-Hill will help students to study smarter and quickly find the information they need. Students will also save money. Contact your McGraw-Hill sales representative to discuss e-book packaging options.

Test Creation • Author/edit questions online using the 14 different question type templates • Create printed tests or deliver online to get instant scoring and feedback. • Create question pools to offer multiple versions online— great for practice • Export your tests for use in WebCT, Blackboard, PageOut, and Apple's iQuiz • Compatible with EZ Test Desktop tests you have already created • Sharing tests with collegues, adjuncts, TAs is easy Online Test Management • Set availability dates and time limits for your quiz or test • Control how your test will be presented • Assign points by question or question type with dropdown menu • Provide immediate feedback to students or delay until all finish the test • Create practice tests online to enable student mastery • Your roster can be uploaded to enable student selfregistration Online Scoring and Reporting • Automated scoring for most of EZ Test's numerous question types • Allows manual scoring for essay and other open response questions • Manual re-scoring and feedback is also available • EZ Test's grade book is designed to easily export to your grade book • View basic statistical reports Support and Help • • • • •

User's Guide and built-in page-specific help Flash tutorials for getting started on the support site Support Website - www.mhhe.com/eztest Product specialist available at 1-800-331-5094 Online Training: http://auth.mhhe.com/mpss/workshops/

Acknowledgments A large number of people have contributed to this sixth edition, either directly or indirectly. We extend particular thanks to those who participated in reviews of the fifth edition. Their thoughtful suggestions have had direct impact upon the clarity and accuracy of this edition, even when it was not possible to fully incorporate all of their ideas due to space limitations or other constraints. We would especially like to thank David H. Kaplan for his many useful suggestions on revising chapter 5 for the sixth edition. We also thank the contributors of the sixth edition supplements: website, Joseph Schaeffer; the Instructor’s Manual, Farhang Amiri; the Test Bank, Thomas Brueckner; and the PowerPoint Lectures, Daniel Bubb, Art Braundmeier, Virgil Stubblefield, and Mikolaj Sawicki.

List of Reviewers Farhang Amiri, Weber State University Tim Bolton, Kansas State University Etan Bourkoff, Baurch College-City Universtiy of New York Lois Breur Krause, Clemson University Zachary T. Bruno, Pittsburgh Technical Institute S.H Chung, William Paterson University of New Jersey Francis Cobbina, Colombus State Commnity College Orville W. Day, East Carolina University Alejandro L. Garcia, San Jose State University John D. Hopkins, The Pennsylvania State University Brian Oetiker, Sam Houston State University Steven T. Poindexter, Pittsburgh Technical Institute Kent Reinhard, Southeast Community College Andrew Robinson, University of Sasketchewan William K. Rule, SUNY Oswego Joseph J. Scanio, The University of Cincinnati

Reviewers of Previous Editions Murty A. Akundi, Xavier University of Louisiana Farhang Amiti, Weber State University

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Charles Ardary, Edmonds Community College James W. Arrison, Villanova University Richard A. Atneosen, Western Washington University Dr. Jean-Claude Ba, Columbus State Community College Maria Bautista, University of Hawaii Richard L. Bobst, LaSierra University Tim Bolton, Kansas State University Ferdinando Borsa, Iowa State University Art Braundmeier, Southern Illinois State University, Edwardsville Michael Bretz, University of Michigan Ron Brown, Mercyhurst College Dan Bruton, Stephen F. Austin State University Daniel M. Bubb, Seton Hall University Lou Cadwell, Providence College Richard A. Cannon, Southeast Missouri State University Richard Cardenas, St. Mary’s University Edward H. Carlson, Michigan State University Cary Caruso, Fayetteville State University Jerry Clifford, Seton Hall University Rory Coker, University of Texas, Austin Doug Davis, Eastern Illinois University Renee D. Diehl, Penn State University David Donnelly, Sam Houston State University Matt Evans, University of Wisconsin, Eau Claire Abbas M. Faridi, Orange Coast College Clarence W. Fette, Penn State University Lyle Ford, University of Wisconsin, Eau Claire Marco Fornari, Central Michigan University Bernard Gilpin, Golden West College Igor Glozman, Highline Community College Laura Greene, University of Illinois at Urbana Robert Grubel, Stephen F. Austin State University James Gundlach, John A. Logan College Kenneth D. Hahn, Truman State University John M. Hauptman, Iowa State University H. James Harmon, Oklahoma State University Lionel D. Hewett, Texas A & M University, Kingsville Robert C. Hudson, Roanoke College Scott Johnson, Idaho State University Stanley T. Jones, University of Alabama David H. Kaplan, Southern Illinois University, Edwardsville Kara Jagne Keeter, Idaho State University Sanford Kern, Colorado State University James Kernohan, Milton Academy John B. Laird, Bowling Green State University Paul L. Lee, California State University, Northridge

Joel M. Levine, Orange Coast College Ian M. Littlewood, California State University, Stanislaus Michael C. LoPresto, Henry Ford Community College John Lowenstein, New York University Michael J. Lysak, San Bernardino Valley College Robert R. Marchini, University of Memphis Paul Middents, Olympic College Joseph Mottillo, Henry Ford Community College William J. Mullin, University of Massachusetts, Amherst Jeffrey S. Olafsen, University of Kansas Mellvyn J. Oremland, Pace University Dr. Arnold Pagnamenta, University of Illinois at Chicago Russell L. Palma, Sam Houston State University Eric Peterson, Highland Community College T. A. K. Pillai, University of Wisconsin La Crosse Ervin Poduska, Kirkwood Community College Steven T. Poindexter, Pittsburgh Technical Institute Brian A. Raue, Florida International University Michael Read, College of the Siskiyous Charles W. Rogers, Southwestern Oklahoma State University Brent Royuk, Concordia University Mohammad Samiullah, Truman State University Joseph A. Schaefer, Loras College Rahim Setoodeh, Milwaukee Area Technical College Elwood Shapanasky, Santa Barbara City College Lawrence C. Shepley, University of Texas, Austin Bradley M. Sherrill, Michigan State University Cecil G. Shugart, University of Memphis John W. Snyder, Southern Connecticut State University Steve Storm, Arkansas State University, Herber Springs Thor F. Stromberg, New Mexico State University Charles R. Taylor, Western Oregon State College Oswald Tekyi-Mensah, Alabama State University Fiorella Terenzi, Pace University Fred Thomas, Sinclair Community College Jeffrey S. Thompson, University of Nevada, Reno Michael Thorensen, University of Northern Iowa Paul Varlashkin, East Carolina University Douglas Wendel, Snow College Scott W. Wissink, Indiana University Hai-Sheng Wu, Minnesota State University, Mankato BonnieWylo, Eastern Michigan University John Yelton, University of Florida Mike Young, Santa Barbara City College Raymond L. Zich, Illinois State University Jens Zorn, University of Michigan

We also wish to acknowledge the contributions of the editorial staff and book team members at McGraw-Hill Higher Education. Their commitment of time and enthusiasm for this work has helped enormously in pushing this project forward. We also owe a huge debt of thanks to our colleagues at Pacific University for helpful suggestions as well as for their forbearance when this project limited our time for other activities. Many other users have also provided constructive criticisms or suggestions, such as Jerry Clifford, Seton Hall University, Mikolaj Sawicki, John A. Logan College, and Mike Crivello, San Diego Mesa College. Last, but certainly not least, we would both like to acknowledge the support of our families, friends, and colleagues. Their encouragement has been essential and has allowed us to enjoy the pleasure of this endeavor.

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Secrets to Success in Studying Physics First of all, we should admit that there are no secrets. Conscientious work and follow-through with reading, problem assignments, and class participation will reap the rewards that students can expect from such efforts in other courses. Failing to do so will also lead to expected results. There are some ways, however, in which studying physics is different from your studies in biology, history, or many other courses. Physics is not an area of study that can be mastered by memorizing discrete facts or by cramming before tests. Students sometimes bring study strategies to physics that have worked in other courses and are disappointed when they fail to work in their physics class. The suggestions that follow are sure-fire steps to getting the most out of your physics course and this textbook.

provide practice with simple numerical applications of physics concepts. They are only useful if you do them yourself and write out the solution steps in such a way that you can follow your work. Copying answers and steps from classmates or other sources may gain points on the assignment but provides no benefit in understanding. As in sports and many other activities, success on physics exams will come to those who practice. 5. Be there. College students set their own priorities for use of time, and sometimes class attendance is not at the top of the list. In some classes, this may be justified by the nature of the benefit of class activities, but that is seldom the case in physics. The demonstrations, explanations, working of exercises, and class discussions that are usually part of what occurs during a physics class provide an invaluable aid to grasping the big picture and filling in holes in your understanding. The demonstrations alone are often worth the price of admission. (You do pay—it’s called tuition.)

1. Experiment. Experiments play a key role in the development of physics but also in the growth of understanding for anyone approaching physics concepts. We often suggest in the text that you try simple experiments that might involve throwing a ball, walking across a room, or other very rudimentary activities. Do them right away as they arise in the text. Not only will you gain the benefit of increased blood flow to various parts of the body including the brain, but what follows in your reading will make more sense. Experience with everyday phenomena cannot be gained passively.

6. Ask questions. If the explanations of demonstrations or other issues are not clear, ask questions. If you are confused, chances are good that many other students are likewise befuddled. They will love you for raising the flag. Unless the instructor is unusually insecure, he or she will also love you for providing the opportunity to achieve better clarity. Physics instructors already know this stuff, so they sometimes have difficulty seeing where student hang-ups may lie. Questions provide the lubrication for moving things forward.

2. Get the big picture. Physics is a big-picture subject. Your understanding of Newton’s laws of motion, for example, cannot be encapsulated by a formula or by memorizing the laws themselves. You need to see the entire context, understand the definitions, and work with how the laws are applied. The outlines and summaries provided at the beginning and end of each chapter can help to provide the context. They cannot stand alone, however. You need to place the examples and descriptions provided in the classroom and text in the framework provided by the outlines and summaries. If you grasp the big picture, the details will often follow.

7. Review understanding. Preparing for tests should not be a matter of last-minute cramming and memorization. Instead, you should review your understanding of the big picture and question yourself on why we did what we did in answering questions and working exercises done previously. Memorization is usually pointless because many physics instructors provide or permit formula sheets that may include definitions and other information. Late-night cramming is counterproductive because it detracts from getting a good night’s sleep. Sleep can be critical to having a clear head the next day to meet the challenges provided by the test.

3. Explore questions. The textbook provides a list of conceptual questions at the end of each chapter, but also raises questions in the body of the text. The greatest benefit is gained by attacking these questions first on your own and then by discussion with classmates. Write out answers to these questions using full sentences, not just short-answer phrases. Compare your answers with those provided at the back of the text for selected questions, but only after having a good crack at answering the questions yourself.

Although there is an element of common sense in most of these suggestions, you will probably not be surprised to learn that many students do not approach things following these guidelines. Old habits are hard to break and peer pressure can also be a negative influence at times. Students fall into patterns that they know are ineffective, but are unable to climb out of the rut. We have done our duty in disclosing these secrets. You are on your own if you choose a different path. Let us know if it works.

4. Try the exercises. The textbook also provides exercises and synthesis problems at the end of each chapter. Their purpose is to

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about the authors Tom Griffith is now Distinguished University Professor Emeritus at Pacific University in Forest Grove, Oregon, having recently retired after 36 years of teaching physics at Pacific. His continued interests in teaching and research mean that he can still be spotted about the halls of the science building or library, and he makes occasional guest appearances with his guitar in physics courses. Over the years he has also enjoyed hiking, bicycling, singing, reading, and performing in stage plays and musical comedies. During his years at Pacific, he has served as Physics Department Chair, Science Division Chair, Interim Dean of Enrollment Management, and Director of Institutional Research among other things, but his primary focus has always been teaching. He has been active in the Oregon Section of the American Association of Physics Teachers (AAPT) and the Pacific Northwest Association for College Physics (PNACP).

Juliet Brosing is a Professor of Physics at Pacific University in Forest Grove, Oregon, and has taught there for the past 20 years. Her research interests include nuclear physics, medical physics, and the application of teaching methods grounded in physics educational research. She helped run a summer science camp for 7th and 8th grade girls for ten years (as director for six). She is on the Reactor Operations Committee for the Reed Nuclear Reactor in Portland and often takes her students there for field trips. She is also the proud owner of three potato guns; therefore, parties with students at her home usually involve projectiles and noise. She remains active in both the state and national American Association of Physics Teachers (AAPT) and the Pacific Northwest Association for College Physics (PNACP). Above all, Dr. Brosing is dedicated to teaching pyhsics with a positive outlook and the use of methods that encourage and benefit her students, regardless of their chosen field of study.

The author and his wife, Adelia, hiking in the mountains of Oregon.

The author, Juliet Brosing, and her husband Keith LeComte at the Tualatin River near their home in Cherry Grove, Oregon.

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Physics, the Fundamental Science

1

chapter

chapter overview The main objective of this chapter is to help you understand what physics is and where it fits in the broader scheme of the sciences. A secondary purpose is to acquaint you with the metric system of units and the advantages of the use of simple mathematics.

chapter outline

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The scientific enterprise. What is the scientific method? How do scientific explanations differ from other types of explanation? The scope of physics. What is physics? How is it related to the other sciences and to technology? What are the major subfields of physics? The role of measurement and mathematics in physics. Why are measurements so important? Why is mathematics so extensively used in science? What are the advantages of the metric system of units? Physics and everyday phenomena. How is physics related to everyday experience and common sense? What are the advantages of using physics to understand common experience?

unit one

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magine that you are riding your bike on a country road on an Indian-summer afternoon. The sun has come out after a brief shower, and as the rain clouds move on, a rainbow appears in the east (fig. 1.1). A leaf flutters to the ground, and an acorn, shaken loose by a squirrel, misses your head by only a few inches. The sun is warm on your back, and you are at peace with the world around you. No knowledge of physics is needed to savor the moment, but your curiosity may bring some questions to mind. Why does the rainbow appear in the east rather than in the west, where it may also be raining? What causes the colors to appear? Why does the acorn fall more rapidly than the leaf? Why is it easier to keep your bicycle upright while you are moving than when you are standing still? Your curiosity about questions like these is similar to what motivates scientists. Learning to devise and apply theories or models that can be used to understand, explain, and predict such phenomena can be a rewarding intellectual game. Crafting an explanation and testing it with simple experiments or observations is fun. That enjoyment is often missed when the focus of a science course is on accumulating facts. This book can enhance your ability to enjoy the phenomena that are part of everyday experience. Learning to produce your own explanations and to perform simple experimental tests can be gratifying. The questions posed here lie in the realm of physics, but the spirit of inquiry and explanation is found throughout science

study hint If you have a clear idea of what you want to accomplish before you begin to read a chapter, your reading will be more effective. The questions in the chapter outline—as well as those in the subheadings of each section—can serve as a checklist for measuring your progress as you read. A clear picture of what questions are going to be addressed and where the answers will be found forms a mental road map to guide you through the chapter. Take a few minutes to study the outline and fix this road map in your mind. It will be time well spent.

1.1 The Scientific Enterprise How do scientists go about explaining something like the rainbow described in the introduction to this chapter? How do scientific explanations differ from other types of explanations? Can we count on the scientific method to explain almost anything? It is important to understand what science can and cannot do. Philosophers have devoted countless hours and pages to questions about the nature of knowledge, and of scientific knowledge in particular. Many issues are still being

and in many other areas of human activity. The greatest rewards of scientific study are the fun and excitement that come from understanding something that has not been understood before. This is true whether we are talking about a physicist making a major scientific breakthrough or about a bike rider understanding how rainbows are formed.

figure 1.1

A rainbow appears to the east in the Columbia River Gorge in Oregon. How can this phenomenon be explained? (See pages 354–355.)

refined and debated. Science grew rapidly during the twentieth century and has had a tremendous impact on our lives. Innovations in medicine, communications, transportation, and computer technology all have resulted from advances in science. What is it about science that explains its impressive advances and steady expansion?

Science and rainbows Let’s consider a specific example of how a scientific explanation comes to be. Where would you turn for an explanation of how rainbows are formed? If you returned from your bike ride with that question on your mind, you might turn to an encyclopedia or a textbook on physics, look up rainbow in the index, and read the explanation found there. Are you behaving like a scientist? The answer is both yes and no. Many scientists would do the same if they were unfamiliar with the explanation. When we do this, we appeal to the authority of the textbook author and to those who preceded the author in inventing the explanation. Appeal to authority is one way of gaining knowledge, but you are at the mercy of your source for the validity of your explanation. You are also hoping that someone has already raised the same question and done the work to create and test an explanation.

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Suppose you go back three hundred years or more and try the same approach. One book might tell you that a rainbow is a painting of the angels. Another might speculate on the nature of light and its interactions with raindrops but be quite tentative in its conclusions. Either of these books might have seemed authoritative in its day. Where, then, do you turn? Which explanation will you accept? If you are behaving like a scientist, you might begin by reading the ideas of other scientists about light and then test these ideas against your own observations of rainbows. You would carefully note the conditions when rainbows appear, the position of the sun relative to you and the rainbow, and the position of the rain shower. What is the order of the colors in the rainbow? Have you observed that order in other phenomena? You would then invent an explanation or hypothesis using current ideas on light and your own guess about what happens as light passes through a raindrop. You could devise experiments with water drops or glass beads to test your hypothesis. (See chapter 17 for a modern view of how rainbows are formed.) If your explanation is consistent with your observations and experiments, you could report it by giving a paper or talk to scientific colleagues. They may criticize your explanation, suggest modifications, and perform their own experiments to confirm or refute your claims. If others confirm your results, your explanation will gain support and eventually become part of a broader theory* about phenomena involving light. The experiments that you and others do may also lead to the discovery of new phenomena, which will call for refined explanations and theories. What is critical to the process just described? First is the importance of careful observation. Another aspect is the idea of testability. An acceptable scientific explanation should suggest some means to test its predictions by observations or experiment. Saying that rainbows are the paintings of angels may be poetic, but it certainly is not testable by mere humans. It is not a scientific explanation. Another important part of the process is a social one, the communication of your theory and experiments to colleagues (fig. 1.2). Submitting your ideas to the criticism (at times blunt) of your peers is crucial to the advancement of science. Communication is also important in assuring your own care in performing the experiments and interpreting the results. A scathing attack by someone who has found an important error or omission in your work is a strong incentive for being more careful in the future. One person working alone cannot hope to think of all of the possible ramifications, alternative explanations, or potential mistakes

*The concept of a theory, as used in science, is often misunderstood. It is much more than a simple hypothesis. A theory consists of a set of basic principles from which many predictions can be deduced. The basic principles involved in the theory are often widely accepted by scientists working in the field.

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figure 1.2

A scientific meeting. Communication and debate are important to the development of scientific explanations. The speaker is Albert Einstein.

in an argument or theory. The explosive growth of science has depended heavily on cooperation and communication.

What is the scientific method? Is there something we could call scientific method within this description, and if so, what is it? The process just described is a sketch of how the scientific method works. Although there are variations on the theme, this method is often described as shown in table 1.1. The steps in table 1.1 are all involved in our description of how to develop an explanation of rainbows. Careful observation may lead to empirical laws for when and where rainbows appear. An empirical law is a generalization derived from experiments or observations. An example of an empirical law is the statement that we see rainbows with the sun at our backs as we look at the rainbow. This is an

table 1.1 Steps in the Scientific Method 1. Careful observation of natural phenomena. 2. Formulation of rules or empirical laws based on generalizations from these observations and experiences. 3. Development of hypotheses to explain the observations and empirical laws, and the refinement of hypotheses into theories. 4. Testing of the hypotheses or theories by further experiment or observation.

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important clue for developing our hypothesis, which must be consistent with this rule. The hypothesis, in turn, suggests ways of producing rainbows artificially that could lead to experimental tests and, eventually, to a broader theory. This description of the scientific method is not bad, although it ignores the critical process of communication. Few scientists are engaged in the full cycle that these steps suggest. Theoretical physicists, for example, may spend all of their time with step 3. Although they have some interest in experimental results, they may never do any experimental work themselves. Today, little science is done simply by observing, as implied by step 1. Most experiments and observations take place to test a hypothesis or existing theory. Although the scientific method is presented here as a stepwise process, in reality these steps often happen simultaneously with much cycling back and forth between steps (fig. 1.3).

Observation or experiments

Generalization

Hypothesis or theory

figure 1.3

The scientific method cycles back to observations or experiments as we seek to test our hypotheses or theories.

The scientific method is a way of testing and refining ideas. Note that the method only applies when experimental tests or other consistent observations of phenomena are feasible. Testing is crucial for weeding out unproductive hypotheses; without tests, rival theories may compete endlessly for acceptance. Example box 1.1 provides a sample question and response illustrating these ideas.

example box 1.1 Sample Question: How Reliable Is Astrology? Question: Astrologers claim that many events in our lives are determined by the positions of the planets relative to the stars. Is this a testable hypothesis? Answer: Yes, it could be tested if astrologers were willing to make explicit predictions about future events that could be verified by independent observers. In fact, astrologers usually carefully avoid doing this, preferring to cast their predictions as vague statements subject to broad interpretation. This prevents clean tests. Astrology is not a science!

How should science be presented? Traditional science courses focus on presenting the results of the scientific process rather than the story of how scientists arrived at these results. This is why the general public often sees science as a collection of facts and established theories. To some extent, that charge could be made against this book since it describes theories that have resulted from the work of others without giving the full picture of their development. Building on the work of others, without needing to repeat their mistakes and unproductive approaches, is a necessary condition for human and scientific progress. This book attempts to engage you in the process of making your own observations and developing and testing your own explanations of everyday phenomena. By doing home experiments or observations, constructing explanations of the results, and debating your interpretations with your friends, you will appreciate the give-and-take that is the essence of science. Whether or not we are aware of it, we all use the scientific method in our everyday activities. The case of the malfunctioning coffee pot described in everyday phenomenon box 1.1 provides an example of scientific reasoning applied to ordinary troubleshooting. The process of science begins with, and returns to, observations of or experiments on natural phenomena. Observations may suggest empirical laws, and these generalizations may be incorporated into a more comprehensive hypothesis. The hypothesis is then tested against more observations or by controlled experiments to form a theory. Working scientists are engaged in one or more of these activities, and we all use the scientific method on everyday problems.

1.2 The Scope of Physics Where does physics fit within the sciences? Since this book is about physics, rather than biology, chemistry, geology, or some other science, it is reasonable to ask where we draw the lines between the disciplines. It is not possible, however, to make sharp distinctions among the disciplines or to provide a definition of physics that will satisfy everyone. The easiest way to give a sense of what physics is and does is by example, that is, by listing some of its subfields and exploring their content. First, let’s consider a definition, however incomplete.

How is physics defined? Physics can be defined as the study of the basic nature of matter and the interactions that govern its behavior. It is the most fundamental of the sciences. The principles and theories of physics can be used to explain the fundamental interactions involved in chemistry, biology, and other sciences at

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everyday phenomenon

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box 1.1

The Case of the Malfunctioning Coffee Pot The Situation. It is Monday morning, and you are, as usual, only half-awake and feeling at odds with the world. You are looking forward to reviving yourself with a freshly brewed cup of coffee when you discover that your coffeemaker refuses to function. Which of these alternatives is most likely to work? 1. Pound on the appliance with the heel of your hand. 2. Search desperately for the instruction manual that you probably threw away two years ago. 3. Call a friend who knows about these things. 4. Apply the scientific method to troubleshoot the problem.

Fixing a malfunctioning coffee pot—alternative 1.

The Analysis. All of these alternatives have some chance of success. The sometimes positive response of electrical or mechanical appliances to physical abuse is well documented. The second two alternatives are both forms of appeal to authority that could produce results. The fourth alternative, however, may be the most productive and quickest, barring success with alternative 1. How would we apply the scientific method as outlined in table 1.1 to this problem? Step 1 involves calmly observing the symptoms of the malfunction. Suppose that the coffeemaker

the atomic or molecular level. Modern chemistry, for example, uses the physical theory of quantum mechanics to explain how atoms combine to form molecules. Quantum mechanics was developed primarily by physicists in the early part of this century, but chemists and chemical knowledge also played important roles. Ideas about energy that

simply refuses to heat up. When the switch is turned on, no sounds of warming water are heard. You notice that no matter how many times you turn the switch on or off, no heat results. This is the kind of simple generalization called for in step 2. We can now generate some hypotheses about the cause of the malfunction, as suggested in step 3. Here are some candidates: a. b. c. d. e.

The coffee pot is not plugged in. The external circuit breaker or fuse has tripped. The power is off in the entire house or neighborhood. An internal fuse in the coffee pot has blown. A wire has come loose or burned through inside the coffeemaker. f. The internal thermostat of the coffeemaker is broken. No detailed knowledge of electrical circuits is needed to check these possibilities, although the last three call for more sophistication (and are more trouble to check) than the first three. The first three possibilities are the easiest to check and should be tested first (step 4 in our method). A simple remedy such as plugging in the pot or flipping on a circuit breaker may put you back in business. If the power is off in the building, other appliances (lights, clocks, and so on) will not work either, which provides an easy test. There may be little that you can do in this case, but at least you have identified the problem. Abusing the coffee pot will not help. The pot may or may not have an internal fuse. If it is blown, a trip to the hardware store may be necessary. A problem like a loose wire or a burnt-out connection often becomes obvious by looking inside after you remove the bottom of the pot or the panel where the power cord comes in. (You must unplug the pot before making such an inspection!) If one of these alternatives is the case, you have identified the problem, but the repair is likely to take more time or expertise. The same is true of the last alternative. Regardless of what you find, this systematic (and calm) approach to the problem is likely to be more productive and satisfying than the other approaches. Troubleshooting, if done this way, is an example of applying the scientific method on a small scale to an ordinary problem. We are all scientists if we approach problems in this manner.

arose initially in physics are now used extensively in chemistry, biology, and other sciences. The general realm of science is often divided into the life sciences and the physical sciences. The life sciences include the various subfields of biology and the healthrelated disciplines that deal with living organisms. The

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physical sciences deal with the behavior of matter in both living and nonliving systems. In addition to physics, the physical sciences include chemistry, geology, astronomy, oceanography, and meteorology (the study of weather). Physics underlies all of them. Physics is also generally regarded as the most quantitative of the sciences. It makes heavy use of mathematics and numerical measurements to develop and test its theories. This aspect of physics has often made it seem less accessible to students, even though the models and ideas of physics can be described more simply and cleanly than those of other sciences. As we will discuss in section 1.3, mathematics serves as a compact language, allowing briefer and more precise statements than would be possible without its use.

What are the major subfields of physics? The primary subfields of physics are listed and identified in table 1.2. Mechanics, which deals with the motion (or lack of motion) of objects under the influence of forces, was the first subfield to be explained with a comprehensive theory. Newton’s theory of mechanics, which he developed in the last half of the seventeenth century, was the first full-fledged physical theory that made extensive use of mathematics. It became a prototype for subsequent theories in physics. The first four subfields listed in table 1.2 were well developed by the beginning of the twentieth century, although all have continued to advance since then. These subfields— mechanics, thermodynamics, electricity and magnetism, and optics—are sometimes grouped as classical physics. The last four subfields—atomic physics, nuclear physics, particle physics, and condensed-matter physics—are often under the heading of modern physics, even though all of the subfields are part of the modern practice of physics. The distinction is made because the last four subfields all emerged during the twentieth century and only existed in rudimentary forms before the turn of that century. In addition to the subfields listed in table 1.2, many physicists work in

table 1.2 The Major Subfields of Physics Mechanics. The study of forces and motion. Thermodynamics. The study of temperature, heat, and energy. Electricity and Magnetism. The study of electric and magnetic forces and electric current. Optics. The study of light. Atomic Physics. The study of the structure and behavior of atoms. Nuclear Physics. The study of the nucleus of the atom. Particle Physics. The study of subatomic particles (quarks, etc.). Condensed-Matter Physics. The study of the properties of matter in the solid and liquid states.

figure 1.4

An optics experiment

using a laser.

figure 1.5

An infrared photograph showing patterns of heat loss from a house is an application of thermodynamics.

interdisciplinary fields such as biophysics, geophysics, or astrophysics. The photographs in this section (figs. 1.4–1.7) illustrate characteristic activities or applications of the subfields. The invention of the laser has been an extremely important factor in the rapid advances now taking place in optics (fig. 1.4). The development of the infrared camera has provided a tool for the study of heat flow from buildings, which involves thermodynamics (fig. 1.5). The rapid growth in consumer electronics, as seen in the availability of home computers, pocket calculators, and many other gadgets, has been made possible by developments in condensedmatter physics (fig. 1.6). Particle physicists use particle accelerators (fig. 1.7) to study the interactions of subatomic particles in high-energy collisions.

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Physics is the study of the basic characteristics of matter and its interactions. It is the most fundamental of the sciences; many other sciences build on ideas from physics. The major subfields of physics are mechanics, electricity and magnetism, optics, thermodynamics, atomic and nuclear physics, particle physics, and condensed-matter physics. Physics plays an important role in engineering and technology, but the real fun of physics comes from understanding how the universe works.

1.3 The Role of Measurement and Mathematics in Physics

figure 1.6

An integrated circuit employing semiconductor devices developed from knowledge of condensed-matter physics. Magnification: 50.

figure 1.7

A Super–Proton–Synchrotron (SPS) particle accelerator used to study interactions of subatomic particles at high energies. It is located at CERN, the European particle-physics laboratory in Switzerland.

Science and technology depend on each other for progress. Physics plays an important role in the education and work of engineers, whether they specialize in electrical, mechanical, nuclear, or other engineering fields. In fact, people with physics degrees often work as engineers when they are employed in industry. The lines between physics and engineering, or research and development, often blur. Physicists are generally concerned with developing a fundamental understanding of phenomena, and engineers with applying that understanding to practical tasks or products, but these functions often overlap. One final point: physics is fun. Understanding how a bicycle works or how a rainbow is formed has an appeal that anyone can appreciate. The thrill of gaining insight into the workings of the universe can be experienced at any level. In this sense, we can all be physicists.

If you go into your college library, find a volume of Physical Review or some other major physics journal, and open it at random, you are likely to find a page with many mathematical symbols and formulas. It would probably be incomprehensible to you. In fact, even many physicists who are not specialists in the particular subfield covered by the article might have difficulty making sense of that page, because they would not be familiar with the particular symbols and definitions. Why do physicists make such extensive use of mathematics in their work? Is knowledge of mathematics essential to understanding the ideas being discussed? Mathematics is a compact language for representing the ideas of physics that makes it easier to precisely state and manipulate the relationships between the quantities that we measure in physics. Once you are familiar with the language, its mystery disappears and its usefulness becomes more obvious. Still, this book uses mathematics in a very limited manner, because most ideas of physics can be discussed without extensive use of mathematics.

Why are measurements so important? How do we test theories in physics? Without careful measurements, vague predictions and explanations may seem reasonable, and making choices between competing explanations may not be possible. A quantitative prediction, on the other hand, can be tested against reality, and an explanation or theory can be accepted or rejected based on the results of measurements. If, for example, one hypothesis predicts that a cannonball will land 100 meters from us and another predicts a distance of 200 meters under the same conditions, firing the cannon and measuring the actual distance provides persuasive evidence for one hypothesis or the other (fig. 1.8). The rapid growth and successes of physics began when the idea of making precise measurements as a test was accepted. Everyday life is full of situations in which measurements, as well as the ability to express relationships between measurements, are important. Suppose, for example, that you normally prepare pancakes on Sunday morning for three people, but on a particular Sunday there is an

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figure 1.8

Cannonballs and a measuring tape: the proof lies in the measurement.

extra mouth to feed. What will you do—double the recipe and feed the rest to the dog? Or will you figure out just how much the quantities in the recipe should be increased to come out right? Let’s say that the normal recipe calls for 1 cup of milk. How much milk will you use if you are increasing the recipe to feed four people instead of three? Perhaps you can solve this problem in your head, but some might find that process dangerous. (Let’s see, 1 cup is enough for three people, so 1⁄ 3 cup is needed for each person, and 4 times 1⁄ 3 equals 4⁄ 3 or 11⁄ 3 cups. See figure 1.9.) If you had to describe this operation to someone else, for the milk and all the other ingredients, you might find yourself using a lot of words. If you looked closely at the person you were talking to, you might also notice his eyes glazing over and confusion setting in.

How can mathematics help? You can reduce the confusion by creating a statement that works for all of the ingredients in the recipe, thus avoiding the need to repeat yourself. You could say, “The quantity of each ingredient needed for four people is related to the quantity needed for three people as 4 is to 3.” That still takes quite a few words and might not be clear unless the person you were talking to was familiar with this way of stating a proportion. If a piece of paper was handy, you might communicate this statement in writing as: Quantity for four : Quantity for three 4 : 3. To make the statement even briefer, you could use the symbol Q4 to represent the quantity of any given ingredient needed to feed four people, and the symbol Q3 to represent

2 cups

2 cups

2 cups

2 cups

1 cup

1 cup

1 cup

1 cup

the quantity needed for three people. Then the statement can be expressed as a mathematical equation, Q4

4 . Q3 3

Using symbols is simply a compact way of saying the same thing that we expressed in words earlier. This compact statement also has the advantage of making manipulations of the relationship easier. For example, if you multiply both sides of this equation by Q3, it takes the form Q4

4 Q, 3 3

which in words says that the quantity needed for four people is 4⁄ 3 times what is needed for three people. If you are comfortable with fractions, you could use this relationship to find the proper amount for any ingredient quickly. There are two points to this example. The first is that making measurements is both a routine and important part of everyday experience. The second is that using symbols to represent quantities in a mathematical statement is a shorter way of expressing an idea involving numbers than the same statement in words would be. Using mathematics also makes it easier to manipulate relationships to construct concise arguments. These are the reasons that physicists (and many other people) find mathematical statements useful. Despite the brevity and apparent clarity of mathematical statements, many people are still more comfortable with words. This is a matter of personal choice and experience, although some fear of mathematics may also be involved. For this reason, word statements are provided in this book with most of the simple mathematical expressions that we will use. Together with the mathematical statement and the drawings, these word statements will help you to understand the concepts we will be discussing.

Why are metric units used?

figure 1.9

Two measuring cups, one containing enough milk to make pancakes to feed three people and the other enough for four people.

Units of measurement are an essential part of any measurement. We do not communicate clearly if we just state a number. If you just talked about adding 11⁄ 3 of milk, for example, your statement would be incomplete. You need to indicate whether you are talking about cups, pints, or milliliters.

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The liter and milliliter are metric units of volume. Cups, pints, quarts, and gallons are holdovers from the older English system of units. Most countries have now adopted the metric system, which has several advantages over the English system still used in the United States. The main advantage of the metric system is its use of standard prefixes to represent multiples of 10, making unit conversion within the system quite easy. The fact that a kilometer (km) is 1000 meters and a centimeter (cm) is 1⁄ 100 of a meter, and that the prefixes kilo and centi always mean 1000 and 1⁄ 100, makes these conversions easy to remember (see table 1.3). To convert 30 centimeters to meters, all we have to do is move the decimal point two places to the left to get 0.30 meter. Moving the decimal point two places to the left is equivalent to dividing by 100. Table 1.3 is a list of the common prefixes used in the metric system. (See appendix B for a discussion of the powers of 10 or scientific notation used for describing very large and very small numbers.) The basic unit of volume in the metric system is the liter (L), which is slightly larger than a quart (1 liter 1.057 quarts). A milliliter (mL) is 1⁄ 1000 of a liter, a convenient size for quantities in recipes. One milliliter is also equal to 1 cm3, or 1 cubic centimeter, so there is a simple relationship between the length and volume measurements in the metric system. Such simple relationships are hard to find in the English system, where 1 cup is 1⁄ 4 of quart, and a quart is 67.2 cubic inches. The metric system predominates in this book. English units will be used occasionally because they are familiar and can help in learning new concepts. Most of us still relate more readily to distances in miles than in kilometers, for example. That there are 5280 feet in a mile is a nuisance, however, compared to the tidy 1000 meters in 1 kilometer. Becoming familiar with the metric system is a worthy objective. Your ability to participate in international trade (for business or pleasure) will be enhanced if you are familiar with the system of units used in most

example box 1.2 Sample Exercise: Length Conversions If you are told that there are 2.54 cm in 1 inch, a. How many centimeters are there in 1 foot (12 inches)? b. How many meters does 1 foot represent? a. 1 inch 2.54 cm 1 foot 12 inches 1 foot ? (in cm) (1 ft) a

12 in 2.54 cm ba b 30.5 cm 1 foot 30.5 cm 1 ft 1 in

b. 1 foot 30.5 cm 1 m 100 cm 1 foot ? (in m) (1 ft) a

30.5 cm 1m ba b 0.305 cm 1 ft 100 cm 1 foot 0.305m

Lines drawn through the units indicate cancellation.

of the world. Example boxes 1.2 and 1.3 provide unit conversion exercises involving metric units. Stating a result or prediction in numbers lends precision to otherwise vague claims. Measurement is an essential part of science and of everyday life. Using mathematical symbols and statements is an efficient way of stating the results of measurements and eases manipulating the relationships between quantities. Units of measurement are an essential part of any measurement, and the metric system of units used in most of the world has a number of advantages over the older English system.

table 1.3

example box 1.3

Commonly Used Metric Prefixes Meaning in scientific notation

Sample Exercise: Rate Conversions

Prefix

in figures

tera

1 000 000 000 000

1012

1 trillion

giga

1 000 000 000

109

1 billion

mega

1 000 000

1 million

kilo

1000

103

centi

1⁄ 100

0.01

milli

1⁄ 1000

0.001 103

micro

1⁄ 1 000 000

1⁄ 106

106

1 millionth

nano

1⁄ 1 000 000 000

1⁄ 109

109

1 billionth

1⁄ 1012

pico

106 102

1012

in words

1 thousand 1 hundredth 1 thousandth

1 trillionth

9

If the rate of flow in an automatic watering system is 2 gallons/hour, how many milliliters per minute is this? 1 gallon 3.786 liters 1 liter 1000 ml 2 gal/hr ? (in ml/min) a

2 gallons 3.786 liter 1000 ml 1 hour b ba ba ba hour 1 gallon 1 liter 60 min 126.2 ml/min

2 gallons/hour 126.2 ml/min Lines drawn through the units indicate cancellation.

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figure 1.10

+

A bicycle wheel, a model of an atom, and a galaxy all involve the concept of angular momentum.

1.4 Physics and Everyday Phenomena Studying physics can and will lead us to ideas as earthshaking as the fundamental nature of matter and the structure of the universe. With ideas like these available, why spend time on more mundane matters like explaining how a bicycle stays upright or how a flashlight works? Why not just plunge into far-reaching discussions of the fundamental nature of reality?

Why study everyday phenomena? Our understanding of the fundamental nature of the universe is based on concepts such as mass, energy, and electric charge that are abstract and not directly accessible to our senses. It is possible to learn some of the words associated with these concepts and to read and discuss ideas involving them without ever acquiring a good understanding of their meaning. This is one risk of playing with the grand ideas without laying the proper foundation. Using everyday experience to raise questions, introduce concepts, and practice devising physical explanations has the advantage of dealing with examples that are familiar and concrete. These examples also appeal to your natural curiosity about how things work, which, in turn, can motivate you to understand the underlying concepts. If you can clearly describe and explain common events, you gain confidence in dealing with more abstract concepts. With familiar examples, the concepts are set on firmer ground, and their meaning becomes more real. For example, why a bicycle (or a top) stays upright while moving but falls over when at rest involves the concept of angular momentum, which is discussed in chapter 8. Angular momentum also plays a role in our understanding of atoms and the atomic nucleus—both in the realm of the very small—and the structure of galaxies at the opposite end of the scale (fig. 1.10). You are more likely to under-

stand angular momentum, though, by discussing it first in the context of bicycle wheels or tops. The principles explaining falling bodies, such as the acorn mentioned in the chapter introduction, involve the concepts of velocity, acceleration, force, and mass, which are discussed in chapters 2, 3, and 4. Like angular momentum, these concepts are also important to our understanding of atoms and the universe. Understanding how rainbows are formed involves the behavior of light, discussed in chapter 17. The behavior of light also plays a major role in how we think about atoms and the universe. Our “common sense” sometimes misleads us in our understanding of everyday phenomena. Adjusting common sense to incorporate well-established physical principles is one of the challenges we face in dealing with everyday experience. By performing simple experiments, either at home (as is often suggested in this book) or in laboratories and demonstrations associated with your course in physics, you can take an active part in building your own scientific worldview. Although it may seem like an oxymoron, everyday experience is extraordinary. A bright rainbow is an incredible sight. Understanding how rainbows originate does not detract from the experience. It adds excitement to explain such a beautiful display with just a few elegant concepts. In fact, people who understand these ideas see more rainbows because they know where to look. This excitement, and the added appreciation of nature that is a part of it, is accessible to all of us. Studying everyday phenomena can make abstract ideas more accessible. These ideas are needed to understand the fundamental nature of matter and the universe, but they are best encountered first in familiar examples. Being able to explain common phenomena builds confidence in using the ideas and enhances our appreciation of what happens around us..

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Key Terms

chapter Physics, the Fundamental Science chapter overview

1

The main objective of this chapter is to help you understand what physics is and where it fits in the broader scheme of the sciences. A secondary purpose is to acquaint you with the metric system of units and the advantages of the use of simple mathematics.

chapter outline

1 2 3

4

The scientific enterprise. What is the scientific method? How do scientific explanations differ from other types of explanation? The scope of physics. What is physics? How is it related to the other sciences and to technology? What are the major subfields of physics? The role of measurement and mathematics in physics. Why are measurements so important? Why is mathematics so extensively used in science? What are the advantages of the metric system of units? Physics and everyday phenomena. How is physics related to everyday experience and common sense? What are the advantages of using physics to understand common experience?

summary This first chapter introduces the scientific enterprise and its methods, the scope of physics, and the use of mathematics and measurement in physics. It also gives pointers on how to use the features of this book most effectively. The key points include:

The role of measurement and mathematics in 3physics. Much of the progress in physics can be attributed to its use of quantitative models, which yield precise predictions that

1

The scientific enterprise. Scientific explanations are developed by generalizing from observations of nature, forming hypotheses or theories, and then testing these theories by further experiments or observations. This process is often called the scientific method, but actual practice may depart in various ways from this model.

90 80 70 60 50 40

Observation or experiments

Generalization

30 20 10

Hypothesis or theory

scope of physics. Physics is the most fundamen2tal of theThenatural sciences because physical theories often underlie explanations in the other sciences. Its major subfields include mechanics, thermodynamics, electricity and magnetism, optics, atomic physics, nuclear physics, condensed-matter physics, and particle physics.

can be tested by making physical measurements. Mathematics is a compact language for describing and manipulating these results. The basic concepts of physics can often be described and understood with a minimum of mathematics.

Physics and everyday phenomena. Many of the 4basic concepts of physics become clearer if applied to everyday phenomena. Being able to understand and explain familiar phenomena makes the concepts more vivid. This adds to the enjoyment of studying physics.

key terms Hypothesis, 3 Theory, 3 Scientific method, 3 Empirical law, 3 Classical physics, 6 Modern physics, 6

Mechanics, 6 Thermodynamics, 6 Electricity and magnetism, 6 Optics, 6 Atomic physics, 6 Nuclear physics, 6

Particle physics, 6 Condensed-matter physics, 6 Proportion, 8 Metric system, 9 Powers of 10, 9 Scientific notation, 9

1

The chapter outline and chapter summary provide related frameworks for organizing concepts.

study hint: How to Use the Features of This Book This book has a number of features designed to make it easier for you to organize and grasp the concepts that we will explore. These features include the chapter overview and outline at the beginning of each chapter and the summary at the end of each chapter, as well as the structure of individual sections of the chapters. The questions, exercises, and synthesis problems at the end of each chapter also play an important role. How can these features be used to the best advantage?

Chapter outlines and summaries Knowing where you are heading before you set out on a journey can be the key to the success of your mission. Students get a better grasp of concepts if they have some structure or framework to help them to organize the ideas. Both the chapter overview and outline at the beginning of each chapter and the summary at the end are designed to provide such a framework. Having a clear idea of what you are trying

vvvvvvvvv

to accomplish before you invest time in reading a chapter will make your reading more effective and enjoyable. The list of topics and questions in the chapter outline can be used as a checklist for measuring your progress as you read. Each numbered topic in the outline, with its associated questions, pertains to a section of the chapter. The outline is designed to stimulate your curiosity by providing some blanks (unanswered questions) to be filled in by your reading. Without the blanks, your mind has no organizational structure to store the information. Without structure, recall is more difficult. You can use the questions in the outline to check the effectiveness of your reading. Can you answer all of the questions when you are done? Each section of a chapter also begins with questions, and the section subheadings are likewise often cast as questions. At the end of each section there is also an indented summary paragraph designed to help you tie the ideas in that section together. The end-of-chapter summary gives a short description of the key ideas in each section, often cast in the form of answers to the questions raised in the outline (see diagram). Summaries provide a quick review, but they are no substitute for a careful reading of the main text. By following

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Chapter 1 Physics, the Fundamental Science

the same organizational structure as the outline, the summary reminds you where to find a more complete discussion of these ideas. The purpose of both the outlines and the summaries is to make your reading more organized and effective. Studying any new discipline requires forming new patterns of thought that can take time to gel. The summaries at the end of each section, as well as at the end of the chapter, can help this gelling to take place. A structure is often built layer by layer, and the later layers will be shaky if the base is unstable.

How should the questions and exercises be used? At the end of each chapter you will find a group of questions, followed by a group of exercises, and, finally, by a small number of synthesis problems. Your grasp of the chapter will improve if you write out answers to the questions and exercises, either as assigned by your instructor or in independent study. The ideas contained in each chapter cannot be thoroughly mastered without this kind of practice. The questions are crucial to helping you fix the important concepts and distinctions in your mind. Most of the questions call for a short answer as well as an explanation. A few of the questions, marked with asterisks, are more open-ended and call for lengthier responses. It is a good idea to write out the explanations in clear sentences when you answer these questions, because it is only through reinforcement that ideas become a part of you. Also, if you can explain something clearly to someone else, you understand it. A sample question and answer appears in example box 1.1. The exercises are designed to give you practice in using the ideas and the related formulas to do simple computations. The exercises also help to solidify your understanding of concepts by giving you a sense of the units and the sizes of the quantities involved. Even though many of the exercises are straightforward enough to work in your head without writing much down, we recommend writing out the information given, the information sought, and the solution in the manner shown in example boxes 1.2 and 1.3 in section 1.3. This develops careful work habits that will help you

avoid careless mistakes. Most students find the exercises easier than the questions. The sample exercises scattered through each chapter can help you get started. The synthesis problems are more wide-ranging than the questions or exercises. They often involve features of both. Although not necessarily harder than the questions or exercises, they do take more time and are sometimes used to extend ideas beyond what was discussed in the chapter. Doing one or two of these in each chapter should build your confidence. They are particularly recommended for those students who have worked the exercises and want to explore the topic in more depth. Answers to the odd-numbered exercises, odd-numbered synthesis problems, and selected questions are found in the back of the book in appendix D. Looking up the answer before attempting the problem is self-defeating. It deprives you of practice in thinking things through on your own. Checking answers after you have worked an exercise can be a confidence builder. Answers should be used only to confirm or improve your own thinking.

Home experiments and everyday phenomenon boxes Reading or talking about physical ideas is useful, but there is no substitute for hands-on experience with the phenomena. You already have a wealth of experience with many of these phenomena, but you probably have not related it to the physical concepts you will be learning. Seeing things in new ways will make you a more astute observer. In addition to the home experiments at the end of each chapter, we often suggest some simple experiments in the main text or in the study hints. We strongly recommend making these observations and doing the experiments. Lecture demonstrations can help, but doing something yourself imprints it vividly on your mind. There is excitement in discovering things yourself and seeing them in a new light. The boxes that discuss everyday phenomena also give you practice in applying physical concepts. Most of the phenomena discussed in these boxes are familiar. The boxes allow us to explore these examples more thoroughly. Participating in these investigations of everyday phenomena can help bring the ideas home.

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Key Terms

13

summary This first chapter introduces the scientific enterprise and its methods, the scope of physics, and the use of mathematics and measurement in physics. It also gives pointers on how to use the features of this book most effectively. The key points include:

The role of measurement and mathematics in 3physics. Much of the progress in physics can be attributed to its use of quantitative models, which yield precise predictions that

1

The scientific enterprise. Scientific explanations are developed by generalizing from observations of nature, forming hypotheses or theories, and then testing these theories by further experiments or observations. This process is often called the scientific method, but actual practice may depart in various ways from this model.

90 80 70 60 50 40

Observation or experiments

Generalization

30 20 10

Hypothesis or theory

2tal of the natural sciences because physical theories often underlie

The scope of physics. Physics is the most fundamen-

explanations in the other sciences. Its major subfields include mechanics, thermodynamics, electricity and magnetism, optics, atomic physics, nuclear physics, condensed-matter physics, and particle physics.

can be tested by making physical measurements. Mathematics is a compact language for describing and manipulating these results. The basic concepts of physics can often be described and understood with a minimum of mathematics.

Physics and everyday phenomena. Many of the 4basic concepts of physics become clearer if applied to everyday phenomena. Being able to understand and explain familiar phenomena makes the concepts more vivid. This adds to the enjoyment of studying physics.

key terms Hypothesis, 3 Theory, 3 Scientific method, 3 Empirical law, 3 Classical physics, 6 Modern physics, 6

Mechanics, 6 Thermodynamics, 6 Electricity and magnetism, 6 Optics, 6 Atomic physics, 6 Nuclear physics, 6

Particle physics, 6 Condensed-matter physics, 6 Proportion, 8 Metric system, 9 Powers of 10, 9 Scientific notation, 9

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questions * more open-ended questions, requiring lengthier responses, suitable for group discussion Q sample responses are available in appendix D Q sample responses are available on the website

Q9. Based upon the brief descriptions provided in table 1.2, which subfield of physics would you say is involved in the explanation of rainbows? Which subfield is involved in describing how an acorn falls? Explain.

*Q1. Which of these criteria best distinguish between explanations provided by science and those provided by religion: truth, testability, or appeal to authority? How do religious explanations differ from scientific explanations?

Q10. Based upon the descriptions provided in table 1.2, which subfields of physics are involved in explaining why an ice cube melts? Which subfields are involved in explaining how an airplane flies? Explain.

Q2. A person claiming to have paranormal powers states that she can predict which card will come up next in a shuffled deck of cards simply by exercising her mental powers. Is this a testable claim? Explain.

Q11. Suppose that you are told that speed is defined by the relationship s d/t, where s represents speed, d represents distance, and t represents time. State this relationship in words, using no mathematical symbols.

Q3. Historians sometimes develop theories to explain observed patterns in the history of different countries. Are these theories testable in the same sense as a theory in physics? Explain.

Q12. Impulse is defined as the average force acting on an object multiplied by the time the force acts. If we let I represent impulse, F the average force, and t the time, is I F/t a correct way of expressing this definition? Explain.

*Q4. Over the years, there have been several credible claims by experienced observers of sightings of Unidentified Flying Objects (UFOs). Despite this, scientists have shied away from taking up serious study of UFOs, although there are ongoing searches for signals from extraterrestrial intelligent beings. Can you think of reasons why scientists have not taken UFOs seriously? What problems can you see in trying to study UFOs?

Q13. The distance that an object travels when it starts from rest and undergoes constant acceleration is one-half the acceleration multiplied by the square of the time. Invent your own symbols and express this statement in symbolic form.

Q5. Suppose that your car will not start and you form the hypothesis that the battery is dead. How would you test this hypothesis? Explain. Q6. Suppose that your phone has not rung in several days, but a friend tells you he has tried to call. Develop two hypotheses that could explain why the phone has not rung and state how you would test these hypotheses. *Q7. Suppose that a friend states the hypothesis that the color of socks that he wears on a given day, brown or black, will determine whether the stock market will go up or down. He can cite several instances in which this hypothesis has been apparently verified. How would you go about evaluating this hypothesis? Q8. Which of the three science fields: biology, chemistry, or physics, would you say is the most fundamental? Explain by describing in what sense one of these fields may be more fundamental than the others.

Q14. What are the primary advantages of the metric system of units over the older English system of units? Explain. Q15. What are the advantages, if any, of continuing to use the English system of units in industry and commerce rather than converting to the metric system? Explain. Q16. Which system of units, the metric system or English system, is used more widely throughout the world? Explain. Q17. The width of a man’s hand was used as a common unit of length several hundred years ago. What are the advantages and disadvantages of using such a unit? Explain. Q18. A pirate map indicates that a treasure is buried 50 paces due east and 120 paces due north of a big rock. Will you know where to dig? Explain. Q19. List the following volumes in descending order: gallon, quart, liter, milliliter. The conversion factors given on the inside front cover may be useful. Q20. List the following lengths in descending order: kilometer, feet, mile, centimeter, inch. The conversion factors given on the inside front cover may be useful.

exercises E1. Suppose that a pancake recipe designed to feed three people calls for 600 mL of flour. How many milliliters of flour would you use if you wanted to extend the recipe to feed five people? E2. Suppose that a cupcake recipe designed to produce twelve cupcakes calls for 900 mL of flour. How many milliliters of flour would you use if you wanted to make only eight cupcakes?

E3. It is estimated that six large pizzas are about right to serve a physics club meeting of 30 students. How many pizzas would be required if the group grows to 50 students? E4. A man uses his hand to measure the width of a tabletop. If his hand has a width of 12 cm at its widest point, and he finds the tabletop to be 10.5 hands wide, what is the width of the tabletop in cm? In meters?

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E5. A woman’s foot is 9 inches long. If she steps off the length of a room by placing one foot directly in front of the other, and finds the room to be 15 foot-lengths long, what is the length of the room in inches? In feet? E6. A book is 220 mm in width. What is this width in centimeters? In meters? E7. A crate has a mass of 8.60 kg (kilograms). What is this mass in grams? In milligrams? E8. A tank holds 2.18 kL (kiloliters) of water. How many liters is this? How many milliliters? E9. A mile is 5280 ft long. The sample exercise in example box 1.2 shows that 1 foot is approximately 0.305 m. How many meters are there in a mile? How many kilometers (km) are there in a mile?

15

E10. If a mile is 5280 ft long and a yard contains 3 ft, how many yards are there in a mile? E11. Area is found by multiplying the length of a surface times the width. If a floor measures 6.25 m2, how many square centimeters does this represent? How many square centimeters are there in 1 m2? E12. A common speed limit in Vancouver, British Columbia, is 80 km/hr. If you are going 55 MPH, are you speeding? Show by converting 55 MPH to km/hr using the conversion factors on the inside front cover. E13. If gas costs 80¢ a liter, how much does a gallon of gas cost? Show by converting gallons to liters using the conversion factors on the inside front cover.

synthesis problems SP1. Astrologers claim that they can predict important events in your life by the configuration of the planets and the astrological sign under which you were born. Astrological predictions, called horoscopes, can be found in most daily newspapers. Find these predictions in a newspaper and address the questions: a. Are the astrological predictions testable? b. Choosing the prediction for your own sign, how would you go about testing its accuracy over the next month or so? c. Why do newspapers print these readings? What is their appeal? SP2. In the United States a common quantity of hard liquor was historically a fifth, which represents a fifth of a US gallon. However, since the US wants to market its alcohol globally, and everyone else uses the metric system, it has retooled its packaging, so a common quantity is now 750 ml. a. How many liters are in a fifth? b. How many milliliters are in a fifth? c. Which is larger, 750 ml or a fifth of a gallon?

home experiments and observations HE1. Look around your house, car, or dormitory room to see what measuring tools (rulers, measuring cups, speedometers, etc.) you have handy. Which of these tools, if any, provides both English and metric units? For those that do, determine the conversion factor needed to convert the English units to metric units.

SP3. An energy-efficient bulb claims to have the brightness of a 75W bulb but only uses 15W of electrical power. a. If you have this light bulb on for 5 hours a day, for 350 days during a year, how many hours is it on? b. A kilowatt is 1000 watts. The kilowatt-hour is a common unit for energy, obtained by multiplying the power in kilowatts by the time used in hours. How many kilowatt-hours (kWh) will you use when burning the 75W bulb for the year? c. How many kilowatt-hours (kWh) will you use when burning the 15W bulb for the year? d. Assuming that the cost of electricity is 15¢ per kWh, what is the cost of using the 75W bulb for the year? e. Assuming this same cost, what is the cost of using the 15W bulb for the year? f. How much do you save by using the 15W bulb? g. How much would you save if you replaced 20 of the 75W bulbs with the 15W bulbs?

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unit

One The Newtonian Revolution

I

Although Newton’s theory of motion does not accurately describe the motion of very fast objects (which are now described using Einstein’s theory of relativity) and very small objects (where quantum mechanics must be used), it is still used extensively in physics and engineering to explain motion and to analyze structures. Newton’s theory has had enormous influence over the last three hundred years in realms of thought that extend well beyond the natural sciences and deserves to be understood by anyone claiming to be well educated. Central to Newton’s theory is his second law of motion. It states that the acceleration of an object is proportional to the net force acting on the object and inversely proportional to the mass of the object. Push an object and that object accelerates in the direction of the applied force. Contrary to intuition and to Aristotle’s teachings, acceleration, not velocity, is proportional to the applied force. To understand this idea, we will thoroughly examine acceleration, which involves a change in the motion of an object. Rather than plunging into Newton’s theory, we begin this unit by studying Galileo’s insights into motion and free fall. This provides the necessary foundation to tackle Newton’s ideas. To see well, we need to stand on the shoulders of these giants.

n 1687, Isaac Newton published his Philosophiae Naturalis Principia Mathematica or Mathematical Principles of Natural Philosophy. This treatise, often called simply Newton’s Principia, presented his theory of motion, which included his three laws of motion and his law of universal gravitation. Together these laws explain most of what was then known about the motion of ordinary objects near Earth’s surface (terrestrial mechanics) as well as the motion of the planets around the sun (celestial mechanics). Along the way, Newton had to invent the mathematical techniques that we call calculus. Newton’s theory of mechanics described in the Principia was an incredible intellectual achievement that revolutionized both science and philosophy. The revolution did not begin with Newton, though. The true rebel was the Italian scientist, Galileo Galilei, who died just a few months after Newton was born in 1642. Galileo championed the sun-centered view of the solar system proposed a hundred years earlier by Nicolaus Copernicus and stood trial under the Inquisition for his pains. Galileo also challenged the conventional wisdom, based on Aristotle’s teachings, about the motion of ordinary objects. In the process, he developed many of the principles of terrestrial mechanics that Newton later incorporated into his theory.

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Describing Motion

chapter overview The main purpose of this chapter is to provide clear definitions and illustrations of the terms used in physics to describe motion, such as the motion of the car described in this chapter’s opening example. Speed, velocity, and acceleration are crucial concepts for the analysis of motion in later chapters. Precise description is the first step to understanding. Without it, we remain awash in vague ideas that are not defined well enough to test our explanations. Each numbered topic in this chapter builds on the previous section, so it is important to obtain a clear understanding of each topic before going on. The distinctions between speed and velocity and velocity and acceleration are particularly important.

chapter outline

1

2 3 4 unit one

5

Average and instantaneous speed. How do we describe how fast an object is moving? How does instantaneous speed differ from average speed? Velocity. How do we introduce direction into descriptions of motion? What is the distinction between speed and velocity? Acceleration. How do we describe changes in motion? What is the relationship between velocity and acceleration? Graphing motion. How can graphs be used to describe motion? How can the use of graphs help us gain a clearer understanding of speed, velocity, and acceleration? Uniform acceleration. What happens when an object accelerates at a steady rate? How do the velocity and distance traveled vary with time when an object is uniformly accelerating?

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2.1 Average and Instantaneous Speed

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magine that you are in your car stopped at an intersection. After waiting for cross traffic, you pull away from the stop sign, accelerating eventually to a speed of 56 kilometers per hour (35 miles per hour). You maintain that speed until a dog runs in front of your car and you hit the brakes, reducing your speed rapidly to 10 km/h (fig. 2.1). Having missed the dog, you speed up again to 56 km/h. After another block, you come to another stop sign and reduce your speed gradually to zero. We can all relate to this description. Measuring speed in miles per hour (MPH) may be more familiar than the use of kilometers per hour (km/h), but speedometers in cars now show both. The use of the term acceleration to describe an increase in speed is also common. In physics, however, these concepts take on more precise and specialized meanings that make them even more useful in describing exactly what is happening. These meanings

figure 2.1

19

are sometimes different from those in everyday use. The term acceleration, for example, is used by physicists to describe any situation in which velocity is changing, even when the speed may be decreasing or the direction of the motion may be changing. How would you define the term speed if you were explaining the idea to a younger brother or sister? Does velocity mean the same thing? What about acceleration— is the notion vague or does it have a precise meaning? Is it the same thing as velocity? Clear definitions are essential to developing clear explanations. The language used by physicists differs from our everyday language, even though the ideas are related and the same words are used. What are the exact meanings that physicists attach to these concepts, and how can they help us to understand motion?

As the car brakes for the dog, there is a sudden change in speed.

2.1 Average and Instantaneous Speed Since driving or riding in cars is a common activity in our daily lives, we are familiar with the concept of speed. Most of us have had experience in reading a speedometer (or perhaps failing to read it carefully enough to avoid the attention of law enforcement). If you describe how fast something is moving, as we did in our example in the introduction, you are talking about speed.

example, that we travel a distance of 260 miles in a time of 5 hours, as shown on the road map of figure 2.2. The average speed is then 260 miles divided by 5 hours, which is equal to 52 MPH. This type of computation is familiar to most of us. We can also express the definition of average speed in a word equation as Average speed equals the distance traveled divided by the time of travel.

How is average speed defined? What does it mean to say that we are traveling at a speed of 55 MPH? It means that we would cover a distance of 55 miles in a time of 1 hour if we traveled steadily at that speed. Carefully note the structure of this description: there is a number, 55, and some units or dimensions, miles per hour. Numbers and units are both essential parts of a description of speed. The term miles per hour implies that miles are divided by hours in arriving at the speed. This is exactly how we would compute the average speed for a trip: suppose, for

or Average speed

distance traveled . time of travel

We can represent this same definition with symbols by writing s

d , t

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120 mi

Flagstaff

2.4 h

example box 2.1 Sample Exercise: Speed Conversions Convert 90 kilometers per hour to (a) miles per hour and (b) meters per second.

h 2. 6

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m

i

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a. 1 km 0.6214 miles 90 km/hr ? (in MPH) a

90 km 0.6214 miles b 55.9 MPH ba hr km

90 km/hr 55.9 MPH b. 1 km 1000 m

Phoenix

figure 2.2

A road map showing a trip of 260 miles, with driving times for the two legs of the trip.

where the letter s represents the speed, d represents distance, and t represents the time. As noted in chapter 1, letters or symbols are a compact way of saying what could be said with a little more effort and space with words. Judge for yourself which is the more efficient way of expressing this definition of average speed. Most people find the symbolic expression easier to remember and use. The average speed that we have just defined is the rate at which distance is covered over time. Rates always represent one quantity divided by another. Gallons per minute, pesos per dollar, and points per game are all examples of rates. If we are considering time rates, the quantity that we divide by is time, which is the case with average speed. Many other quantities that we will be considering involve time rates.

What are the units of speed? Units are an essential part of the description of speed. Suppose you say that you were doing 70—without stating the units. In the United States, that would probably be understood as 70 MPH, since that is the unit most frequently used. In Europe, on the other hand, people would probably assume that you are talking about the considerably slower speed of 70 km/h. If you do not state the units, you will not communicate effectively. It is easy to convert from one unit to another if the conversion factors are known. For example, if we want to convert kilometers per hour to miles per hour, we need to know the relationship between miles and kilometers. A kilometer is roughly 6⁄ 10 of a mile (0.6214, to be more precise). As shown in example box 2.1, 70 km/h is equal to 43.5 MPH. The process involves multiplication or division by the appropriate conversion factor.

a

90 km 1000 m ba b 90,000 m/hr hr km

However (1 hr) a a

60 min 60 sec ba b 3600 sec hr min

90,000 m 1 hr b 25.0m/sec ba hr 3600 sec

90 km/hr 25.0 m/sec Part b can also be done using the conversion factors for speed on the inside front cover: 1 km/hr 0.278 m/sec (90 km/hr) a

0.278 m/sec b 25.0 m/sec 1 km/hr

90 km/hr 25.0 m/sec Lines drawn through the units indicate cancellation.

Units of speed will always be a distance divided by a time. In the metric system, the fundamental unit of speed is meters per second (m/s). Example box 2.1 also shows the conversion of kilometers per hour to meters per second, done as a two-step process. As you can see, 70 km/h can also be expressed as 19.4 m/s or roughly 20 m/s. This is a convenient size for discussing the speeds of ordinary objects. (As shown in example box 2.2, the convenient unit for measuring the growth of grass has a very different size.) Table 2.1 shows some familiar speeds expressed in miles

table 2.1 Familiar Speeds in Different Units 20 MPH

32 km/h

40 MPH

64 km/h 18 m/s

60 MPH

97 km/h 27 m/s

9 m/s

80 MPH 130 km/h 36 m/s 100 MPH 160 km/h 45 m/s

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example box 2.2 Passing the truck

120

Sample Question: Watching Grass Grow

Answer: When grass is well fertilized and watered, it is not unusual for it to grow 3 to 6 centimeters in the course of a week. This can be seen by measuring the length of the clippings after mowing. If we measured the speed in m/s, we would obtain an extremely small number that would not provide a good intuitive sense of the rate of growth. The units of cm/week or mm/day would provide a better indication of this speed.

90 Speed (km/h)

Question: The units km/h or m/s have an appropriate size for moving cars or people. Many other processes move much more slowly, though. What units would have an appropriate size for measuring the average speed with which a blade of grass grows?

60 Behind a slow truck

30 Turn onto highway Stoplight

5

10 15 Time (minutes)

20

per hour, kilometers per hour, and meters per second to give you a sense of their relationships.

figure 2.4

What is instantaneous speed?

how the speed of a car varies during a portion of a trip could be provided by a graph such as that shown in figure 2.4. Each point on this graph represents the instantaneous speed at the time indicated on the horizontal axis. Even though we all have some intuitive sense of what instantaneous speed means from our experience in reading speedometers, computing this quantity presents some problems that we did not encounter in defining average speed. We could say that instantaneous speed is the rate that distance is being covered at a given instant in time, but how do we compute this rate? What time interval should we use? What is an instant in time? Our solution to this problem is simply to choose a very short interval of time during which a very short distance is covered and the speed does not change drastically. If we know, for example, that in 1 second a distance of 20 meters was covered, dividing 20 meters by 1 second to obtain a speed of 20 m/s would give us a good estimate of the instantaneous speed, provided that the speed did not change much during that single second. If the speed was changing rapidly, we would have to choose an even shorter interval of time. In principle, we can choose time intervals as small as we wish, but in practice, it can be hard to measure such small quantities. If we put these ideas into a word definition of instantaneous speed, we could state it as

If we travel a distance of 260 miles in 5 hours, as in our earlier example, is it likely that the entire trip takes place at a speed of 52 MPH? Of course not; the speed goes up and down as the road goes up and down, when we overtake slower vehicles, when rest breaks occur, or when the highway patrol looms on the horizon. If we want to know how fast we are going at a given instant in time, we read the speedometer, which displays the instantaneous speed (fig. 2.3). How does instantaneous speed differ from average speed? The instantaneous speed tells us how fast we are going at a given instant but tells us little about how long it will take to travel several miles, unless the speed is held constant. The average speed, on the other hand, allows us to compute how long a trip might take but says little about the variation in speed during the trip. A more complete description of

45

55

65 75

35 25 15

60 40

80 100 120 140

20 5

0 0

figure 2.3

85 95

Variations in instantaneous speed for a portion of a trip on a local highway.

160

km/h

180

105

MPH

A speedometer with two scales for measuring instantaneous speed, MPH and km/h.

Instantaneous speed is the rate at which distance is being covered at a given instant in time. It is found by computing the average speed for a very short time interval in which the speed does not change appreciably.

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everyday phenomenon

box 2.1

Transitions in Traffic Flow The Situation. Jennifer commutes into the city on a freeway every day for work. As she approaches the city, the same patterns in traffic flow seem to show up in the same places each day. She will be moving with the flow of traffic at a speed of approximately 60 MPH when suddenly things will come to a screeching halt. The traffic will be stop-and-go briefly and then will settle into a wavelike mode with speeds varying between 10 and 30 MPH. Unless there is an accident, this will continue for the rest of the way into the city.

The traffic in the upper lanes is flowing freely with adequate spacing to allow higher speeds. The higher-density traffic in the lower lanes moves much more slowly.

What causes these patterns? Why does the traffic stop when there is no apparent reason such as an accident? Why do ramp traffic lights seem to help the situation? Questions like these are the concern of the growing field of traffic engineering. The Analysis. Although a full analysis of traffic flow is complex, there are some simple ideas that can explain many of the patterns that Jennifer observes. The density of vehicles, measured in vehicles per mile, is a key factor. Adding vehicles at entrance ramps increases this vehicle density. When Jennifer and other commuters are traveling at 60 MPH, they need to keep a spacing of several car lengths between vehicles. Most drivers do this without thinking about it, although there are always some who follow too closely or tailgate. Tailgating runs the risk of rear-end collisions when the traffic suddenly slows. When more vehicles are added at an entrance ramp, the density of vehicles increases, reducing the distance between vehicles. As the distance between vehicles decreases, drivers should reduce their speed to maintain a safe stopping distance. If this occurred uniformly, there would be a gradual decrease in the average speed of the traffic to accommodate the greater density. This is not what usually happens, however.

(continued) Instantaneous speed is closely related to the concept of average speed but involves very short time intervals. When discussing traffic flow, average speed is the critical issue, as shown in everyday phenomenon box 2.1. We find an average speed by dividing the distance traveled by the time required to cover that distance. Average speed is therefore the average rate at which distance is being covered. Instantaneous speed is the rate that distance is being covered at a given instant in time and is found by considering very small time intervals or by reading a speedometer. Average speed is useful for estimating how long a trip will take, but instantaneous speed is of more interest to the highway patrol.

2.2 Velocity Do the words speed and velocity mean the same thing? They are often used interchangeably in everyday language,

but physicists make an important distinction between the two terms. The distinction has to do with direction: which way is the object moving? This distinction turns out to be essential to understanding Newton’s theory of motion (introduced in chapter 4), so it is not just a matter of whim or jargon.

What is the difference between speed and velocity? Imagine that you are driving a car around a curve (as illustrated in figure 2.5) and that you maintain a constant speed of 60 km/h. Is your velocity also constant in this case? The answer is no, because velocity involves the direction of motion as well as how fast the object is going. The direction of motion is changing as the car goes around the curve. To simply state this distinction, speed as we have defined it tells us how fast an object is moving but says nothing about the direction of the motion. Velocity includes the

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2.2 Velocity

A significant proportion of drivers will attempt to maintain their speed at 50 to 60 MPH even when densities have increased beyond the point where this is advisable. This creates an unstable situation. At some point, usually near an entrance ramp, the vehicle density becomes too large to sustain these speeds. At this point there is a sudden drop in average speed and a large increase in the local density. As shown in the drawing, cars can be separated by less than a car length when they are stopped or moving very slowly. Once the average speed of a few vehicles has slowed to less than 10 MPH, vehicles moving at 50 to 60 MPH begin to pile up behind this slower moving jam. Because this does not happen smoothly, some vehicles must come to a complete stop, further slowing the flow. At the front end of the jam, on the other hand, the density is reduced due to the slower flow behind. Cars can then start moving at a speed consistent with the new density, perhaps around 30 MPH. If every vehicle moved with the appropriate speed, flow would be smooth and the increased density could be safely accommodated. More often, however, overanxious drivers exceed the appropriate speed, causing fluctuations in the average speed as vehicles begin to pile up again. Notice that we are using average speed with two different meanings in this discussion. One is the average speed of an individual vehicle as its instantaneous speed increases and

N v2

E

B

v1

A

figure 2.5

The direction of the velocity changes as the car moves around the curve, so that the velocity v2 is not the same as the velocity v1 even though the speed has not changed.

23

decreases. The other is the average speed of the overall traffic flow involving many vehicles. When the traffic is flowing freely, the average speed of different vehicles may differ. When the traffic is in a slowly moving jam, the average speeds of different vehicles are essentially the same, at least within a given lane. Traffic lights at entrance ramps that permit vehicles to enter one-at-a-time at appropriate intervals can help to smoothly integrate the added vehicles to the existing flow. This reduces the sudden changes in speed caused by a rapid increase in density. Once the density increases beyond the certain level, however, a slowing of traffic is inevitable. The abrupt change from low-density, high-speed flow to higherdensity, slow flow is analogous to a phase transition from a gas to a liquid. (Phase transitions are discussed in chapter 10.) Traffic engineers have used this analogy to better understand the process. If we could automatically control and coordinate the speeds of all the vehicles on the highway, the highway might carry a much greater volume of traffic at a smooth rate of flow. Speeds could be adjusted to accommodate changes in density and smaller vehicle separations could be maintained at higher speeds because the vehicles would all be moving in a synchronized fashion. Better technology may someday achieve this dream.

idea of direction. To specify a velocity, we must give both its size or magnitude (how fast) and its direction (north, south, east, up, down, or somewhere in between). If you tell me that an object is moving 15 m/s, you have told me its speed. If you tell me that it is moving due west at 15 m/s, you have told me its velocity. At point A on the diagram in figure 2.5, the car is traveling due north at 60 km/h. At point B, because the road curves, the car is traveling northwest at 60 km/h. Its velocity at point B is different from its velocity at point A (because the directions are different). The speeds at points A and B are the same. Direction is irrelevant in specifying the speed of the object. It has no effect on the reading on your speedometer. Changes in velocity are produced by forces acting upon the car, as we will discuss further in chapter 4. The most important force involved in changing the velocity of a car is the frictional force exerted on the tires of the car by the road surface. A force is required to change either the size or the direction of the velocity. If no net force were acting on the car, it would continue to move at constant speed in a straight line. This happens sometimes when there is ice or oil on the road surface, which can reduce the frictional force to almost zero.

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Describing Motion

10 m/s

study hint Science has always relied on pictures and charts to get points across. Throughout the book, a number of concepts will be introduced and illustrated. In the illustrations, the same color will be used for certain phenomena. Blue arrows are velocity vectors. Green arrows depict acceleration vectors. Red arrows depict force vectors. Purple arrows show momentum, a concept we will explore in chapter 7.

What is a vector? Velocity is a quantity for which both the size and direction are important. We call such quantities vectors. To describe these quantities fully, we need to state both the size and the direction. Velocity is a vector that describes how fast an object is moving and in what direction it is moving. Many of the quantities used in describing motion (and in physics more generally) are vector quantities. These include velocity, acceleration, force, and momentum, to name a few. Think about what happens when you throw a rubber ball against a wall, as shown in figure 2.6. The speed of the ball may be about the same after the collision with the wall as it was before the ball hit the wall. The velocity has clearly changed in the process, though, because the ball is moving in a different direction after the collision. Something has happened to the motion of the ball. A strong force had to be exerted on the ball by the wall to produce this change in velocity. The velocity vectors in figures 2.5 and 2.6 are represented by arrows. This is a natural choice for depicting vectors, since the direction of the arrow clearly shows the

20 m/s

figure 2.7

The length of the arrow shows the size of the

velocity vector.

direction of the vector, and the length can be drawn proportional to the size. In other words, the larger the velocity, the longer the arrow (fig. 2.7). In the text, we will represent vectors by printing their symbols in boldface and larger than other symbols: v is thus the symbol for velocity. A fuller description of vectors can be found in appendix C.

How do we define instantaneous velocity? In considering automobile trips, average speed is the most useful quantity. We do not really care about the direction of motion in this case. Instantaneous speed is the quantity of interest to the highway patrol. Instantaneous velocity, however, is most useful in considering physical theories of motion. We can define instantaneous velocity by drawing on our earlier definition of instantaneous speed. Instantaneous velocity is a vector quantity having a size equal to the instantaneous speed at a given instant in time and having a direction corresponding to that of the object’s motion at that instant.

Instantaneous velocity and instantaneous speed are closely related, but velocity includes direction as well as size. It is changes in instantaneous velocity that require the intervention of forces. These changes will be emphasized when we explore Newton’s theory of mechanics in chapter 4. We can also define the concept of average velocity, but that is a much less useful quantity for our purposes than either instantaneous velocity or average speed.*

vfinal

vinitial

figure 2.6

The direction of the velocity changes when a ball bounces from a wall. The wall exerts a force on the ball in order to produce this change.

To specify the velocity of an object, we need to state both how fast and in what direction the object is moving; velocity is a vector quantity. Instantaneous velocity has a magnitude equal to the instantaneous speed and points in the direction that the object is moving. Changes in instantaneous velocity are where the action is, so to speak, and we will consider these in more detail when we discuss acceleration in section 2.3.

*Strictly speaking, velocity is the change in displacement divided by time, where displacement is a vector representing the change in position of an object. See Appendix C and figure C.2 for a discussion of displacement vectors. In one-dimensional motion when an object does not change direction, the distance traveled is equal to the magnitude of the displacement.

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2.3 Acceleration

25

2.3 Acceleration Acceleration is a familiar idea. We use the term in speaking of the acceleration of a car away from a stop sign or the acceleration of a running back in football. We feel the effects of acceleration on our bodies when a car’s velocity changes rapidly and even more strikingly when an elevator lurches downward, leaving our stomachs slightly behind (fig. 2.8). These are all accelerations. You can think of your stomach as an acceleration detector—a roller-coaster gives it a real workout! Understanding acceleration is crucial to our study of motion. Acceleration is the rate at which velocity changes. (Note that we said velocity, not speed.) It plays a central role in Newton’s theory of motion. How do we go about finding a value of an acceleration, though? As with speed, it is convenient to start with a definition of average acceleration and then extend it to the idea of instantaneous acceleration.

How is average acceleration defined? How would we go about providing a quantitative description of an acceleration? Suppose that your car, pointing due east, starts from a full stop at a stop sign, and its velocity increases from zero to 20 m/s as shown in figure 2.9. The change in velocity is found simply by subtracting the initial velocity from the final velocity (20 m/s 0 m/s 20 m/s).

v=0

v = 20 m/s

t=0

t=5s

figure 2.9

A car, starting from rest, accelerates to a velocity of 20 m/s due east in a time of 5 s.

To find its rate of change, however, we also need to know the time needed to produce this change. If it took just 5 seconds for the velocity to change, the rate of change would be larger than if it took 30 seconds. Suppose that a time of 5 seconds was required to produce this change in velocity. The rate of change in velocity could then be found by dividing the size of the change in velocity by the time required to produce that change. Thus the size of the average acceleration, a, is found by dividing the change in velocity of 20 m/s by the time of 5 seconds, a

20 m/s 4 m/s/s. 5s

The unit m/s/s is usually written m/s2 and is read as meters per second squared. It is easier to understand it, however, as meters per second per second. The car’s velocity (measured in m/s) is changing at a rate of 4 m/s every second. Other units could be used for acceleration, but they will all have this same form: distance per unit of time per unit of time. In discussing the acceleration of a car on a drag strip, for example, the unit miles per hour per second is sometimes used. The quantity that we have just computed is the size of the average acceleration of the car. The average acceleration is found by dividing the total change in velocity for some time interval by that time interval, ignoring possible differences in the rate of change of velocity that might be occurring within the time interval. Its definition can be stated in words as

Acceleration detector

Average acceleration is the change in velocity divided by the time required to produce that change.

We can restate it in symbols as Acceleration a

figure 2.8

Your acceleration detector senses the downward acceleration of the elevator.

or a

¢v . t

change in velocity elapsed time

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Because change is so important in this definition, we have used the special symbol (the Greek letter delta) to mean a change in a quantity. Thus v is a compact way of writing the change in velocity, which otherwise could be expressed as vf vi, where vf is the final velocity and vi is the initial velocity. Because the concept of change is critical, this delta () notation will appear often. The idea of change is all-important. Acceleration is not velocity over time. It is the change in velocity divided by time. It is common for people to associate large accelerations with large velocities, when in fact the opposite is often true. The acceleration of a car may be largest, for example, when it is just starting up and its velocity is near zero. The rate of change of velocity is greatest then. On the other hand, a car can be traveling at 100 MPH but still have a zero acceleration if its velocity is not changing.

What is instantaneous acceleration? Instantaneous acceleration is similar to average acceleration with an important exception. Just as with instantaneous speed or velocity, we are now concerned with the rate of change at a given instant in time. It is instantaneous acceleration that our stomachs respond to. It can be defined as Instantaneous acceleration is the rate at which velocity is changing at a given instant in time. It is computed by finding the average acceleration for a very short time interval during which the acceleration does not change appreciably.

If the acceleration is changing with time, choosing a very short time interval guarantees that the acceleration computed for that time interval will not differ too much from the instantaneous acceleration at any time within the interval. This is the same idea used in finding an instantaneous speed or instantaneous velocity.

vf = 20 m/s

∆v = 12 m/s

vi = 8 m/s vi

vf

vi

∆v

+

vf

=

a

figure

2.10 The acceleration vector is in the same direction as the velocity vectors when the velocity is increasing.

vectors, as shown in figure 2.11. Because the initial velocity vi is larger than the final velocity vf, the change in velocity must point in the opposite direction to produce a shorter vf arrow. The acceleration is also in the opposite direction to the velocity, since it is in the direction of the change in velocity. In Newton’s theory of motion, the force required to produce this acceleration would also be opposite in direction to the velocity. It must push backward on the car to slow it down. The term acceleration describes the rate of any change in an object’s velocity. The change could be an increase (as in our initial example), a decrease, or a change in direction. The term applies even to decreases in velocity (decelerations). To a physicist these are simply accelerations with a direction opposite that of the velocity. If a car is braking while traveling in a straight line, its velocity is decreasing and its acceleration is negative if the velocity is positive. This situation is illustrated in the sample exercise in example box 2.3. The minus sign is an important part of the result in the example in example box 2.3 because it indicates that the change in velocity is negative. The velocity is getting smaller. We can call it a deceleration if we like, but it is

What is the direction of an acceleration? Like velocity, acceleration is a vector quantity. Its direction is important. The direction of the acceleration vector is that of the change in velocity v. If, for example, a car is moving in a straight line and its velocity is increasing, the change in velocity is in the same direction as the velocity itself, as shown in figure 2.10. The change in velocity v must be added to the initial velocity vi to obtain the final velocity vf. All three vectors point forward. The process of adding vectors can be readily seen when we represent the vectors as arrows on a graph. (More information on vector addition can be found in appendix C.) If the velocity is decreasing, however, the change in velocity v points in the opposite direction to the two velocity

vi = 20 m/s

∆v = –12 m/s

vf = 8 m/s

vi

vi

vf

∆v

+

=

vf

a

figure

2.11 The velocity and acceleration vectors for decreasing velocity: v and a are now opposite in direction to the velocity. The acceleration a is proportional to v.

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example box 2.3

vi

Sample Exercise: Negative Accelerations

vf

The driver of a car steps on the brakes, and the velocity drops from 20 m/s due east to 10 m/s due east in a time of 2.0 seconds. What is the acceleration? vi 20 m/s due east vf 10 m/s due east t 2.0 s a ?

27

a

¢v t

t

10 m/s 20 m/s 2.0 s

10 m/s 2.0 s

5

a

vf vi

m/s2

a 5.0 m/s2 due west Notice that when we are dealing just with the magnitude of a vector quantity, we do not use the boldface notation. The sign can indicate direction, however, in a problem involving straight-line motion.

the same thing as a negative acceleration. One word, acceleration, covers all situations in which the velocity is changing.

Can a car be accelerating when its speed is constant? What happens when a car goes around a curve at constant speed? Is it accelerating? The answer is yes, because the direction of its velocity is changing. If the direction of the velocity vector is changing, the velocity is changing. This means that there must be an acceleration. This situation is illustrated in figure 2.12. The arrows in this drawing show the direction of the velocity vector at different points in the motion. The change in velocity v is the vector that must be added to the initial velocity vi to obtain the final velocity vf. The vector representing the change in velocity points toward the center of the curve, and therefore, the acceleration vector also points in that direction. The size of the change is represented by the length of the arrow v. From this we can find the acceleration. Acceleration is involved whenever there is a change in velocity, regardless of the nature of that change. Cases like figure 2.12 will be considered more fully in chapter 5 where circular motion is discussed.

vi ∆v vf

figure

2.12 A change in the direction of the velocity vector also involves an acceleration, even though the speed may be constant.

Acceleration is the rate of change of velocity and is found by dividing the change in the velocity by the time required to produce that change. Any change in velocity involves an acceleration, whether an increase or a decrease in speed, or a change in direction. Acceleration is a vector having a direction corresponding to the direction of the change in velocity, which is not necessarily the same direction as the instantaneous velocity itself. The concept of change is crucial. The graphical representations in section 2.4 will help you visualize changes in velocity as well as in other quantities.

2.4 Graphing Motion It is often said that a picture is worth a thousand words, and the same can be said of graphs. Imagine trying to describe the motion depicted in figure 2.4 precisely in words and numbers. The graph provides a quick overview of what took place. A description in words would be much less efficient. In this section, we will show how graphs can also help us to understand velocity and acceleration.

What can a graph tell us? How can we produce and use graphs to help us describe motion? Imagine that you are watching a battery-powered toy car moving along a meter stick (fig. 2.13). If the car is moving slowly enough, you could record the car’s position while also recording the elapsed time using a digital watch. At regular time intervals (say, every 5 seconds), you would note the value of the position of the front of the car on the meter stick and write these values down. The results might be something like those shown in table 2.2.

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figure

2.13 A toy car moving along a meter stick. Its position can be recorded at different times.

How do we graph these data? First, we create evenly spaced intervals on each of two perpendicular axes, one for distance traveled (or position) and the other for time. To show how distance varies with time, we usually put time on the horizontal axis and distance on the vertical axis. Such a graph is shown in figure 2.14, where each data point from table 2.2 is plotted and a line is drawn through the points. To make sure that you understand this process, choose different points from table 2.2 and find where they are located on the graph. Where would the point go if the car was at 21 centimeters at 25 seconds? The graph summarizes the information presented in the table in a visual format that makes it easier to grasp at a

table 2.2 Position of the Toy Car along the Meter Stick at Different Times Time Position 0s

0 cm

5s

4.1 cm

10 s

7.9 cm

15 s

12.1 cm

20 s

16.0 cm

25 s

16.0 cm

30 s

16.0 cm

35 s

18.0 cm

40 s

20.1 cm

45 s

21.9 cm

50 s

24.0 cm

55 s

22.1 cm

60 s

20.0 cm

glance. The graph also contains information on the velocity and acceleration of the car, although that is less obvious. For example, what can we say about the average velocity of the car between 20 and 30 seconds? Is the car moving during this time? A glance at the graph shows us that the distance is not changing during that time interval, so the car is not moving. The velocity is zero during that time, which is represented by a horizontal line on our graph of distance versus time. What about the velocity at other points in the motion? The car is moving more rapidly between 0 and 20 seconds than it is between 30 and 50 seconds. The distance curve is rising more rapidly between 0 and 20 seconds than between 30 and 50 seconds. Since more distance is covered in the same time, the car must be moving faster there. A steeper slope to the curve is associated with a larger speed. In fact, the slope of the distance-versus-time curve at any point on the graph is equal to the instantaneous velocity of the car.* The slope indicates how rapidly the distance is changing with time at any instant in time. The rate of change of distance with time is the instantaneous speed according to the definition given in section 2.1. Since the motion takes place along a straight line, we can then represent the direction of the velocity with plus or minus signs. There are only two possibilities, forward or backward. We then have the instantaneous velocity, which includes both the size (speed) and direction of the motion. When the car travels backward, its distance from the starting point decreases. The curve goes down, as it does between 50 and 60 seconds. We refer to this downwardsloping portion of the curve as having a negative slope and also say that the velocity is negative during this portion of the motion. A large upward slope represents a large instantaneous velocity, a zero slope (horizontal line) a zero velocity, and a downward slope a negative (backward) velocity. Looking at the slope of the graph tells us all we need to know about the velocity of the car.

Velocity and acceleration graphs These ideas about velocity can be best summarized by plotting a graph of velocity against time for the car (fig. 2.15). The velocity is constant wherever the slope of the distanceversus-time graph of figure 2.14 is constant. Any straightline segment of a graph has a constant slope, so the velocity changes only where the slope of the graph in figure 2.14 changes. If you compare the graph in figure 2.15 to the graph in figure 2.14 carefully, these ideas should become clear.

*Since the mathematical definition of slope is the change in the vertical coordinate d divided by the change in the horizontal coordinate t, the slope, d/t, is equal to the instantaneous velocity, provided that t is sufficiently small. It is possible to grasp the concept of slope, however, without appealing to the mathematical definition.

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Distance (cm)

30

20

10

20

40

60

Time (s)

figure

2.14 Distance plotted against time for the motion of the toy car. The data points are those listed in table 2.2.

Velocity (cm/s)

1.5 1.0 0.5 0

20

40

60

–0.5 –1.0 Time (s)

figure

2.15 Instantaneous velocity plotted against time for the motion of the toy car. The velocity is greatest when distance traveled is changing most rapidly.

29

What can we say about the acceleration from these graphs? Since acceleration is the rate of change of velocity with time, the velocity graph (fig. 2.15) also provides information about the acceleration. In fact, the instantaneous acceleration is equal to the slope of the velocity-versustime graph. A steep slope represents a rapid change in velocity and thus a large acceleration. A horizontal line has zero slope and represents zero acceleration. The acceleration turns out to be zero for most of the motion described by our data. The velocity changes at only a few points in the motion. The acceleration would be large at these points and zero everywhere else. Since our data do not indicate how rapidly the changes in velocity actually occur, we do not have enough information to say just how large the acceleration is at those few points where it is not zero. We would need measurements of distance or velocity every tenth of a second or so to get a clear idea of how rapid these changes are. As we will see in chapter 4, we know that these changes in velocity cannot occur instantly. Some time is required. So we can sketch an approximate graph of acceleration versus time, as shown in figure 2.16. The spikes in figure 2.16 occur when the velocity is changing. At 20 seconds, there is a rapid decrease in the velocity represented by a downward spike or negative acceleration. At 30 seconds, the velocity increases rapidly from zero to a constant value, and this is represented by an upward spike or positive acceleration. At 50 seconds, there is another negative acceleration as the velocity changes from a positive to a negative value. If you could put yourself inside the toy car, you would definitely feel these accelerations. (Everyday phenomenon box 2.2 provides another example of how a graph is useful for analyzing motion.)

Acceleration (cm/s 2 )

Can we find the distance traveled from the velocity graph?

20

40 60

Time (s)

figure

2.16 An approximate sketch of acceleration plotted against time for the toy-car data. The acceleration is non-zero only when the velocity is changing.

What other information can be gleaned from the velocityversus-time graph of figure 2.15? Think for a moment about how you would go about finding the distance traveled if you knew the velocity. For a constant velocity, you can get the distance simply by multiplying the velocity by the time, d vt. In the first 20 seconds of the motion, for example, the velocity is 0.8 cm/s and the distance traveled is 0.8 cm/s times 20 seconds, which is 16 cm. This is just the reverse of what we used in determining the velocity in the first place. We found the velocity by dividing the distance traveled by the time. How would this distance be represented on the velocity graph? If you recall formulas for computing areas, you may recognize that the distance d is the area of the shaded rectangle on figure 2.15. The area of a rectangle is found by multiplying the height times the width, just what we have done here. The velocity, 0.8 cm/s, is the height and the

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everyday phenomenon

box 2.2

The 100-m Dash The Situation. A world-class sprinter can run 100 m in a time of a little under 10 s. The race begins with the runners in a crouched position in the starting blocks, waiting for the sound of the starter’s pistol. The race ends with the runners lunging across the finish line, where their times are recorded by stopwatches or automatic timers.

runner’s maximum speed must be somewhat larger than this value, since we know that the instantaneous speed will be less than 10 m/s while the runner is accelerating. These ideas are easiest to visualize by sketching a graph of speed plotted against time, as shown. Since the runner travels in a straight line, the magnitude of the instantaneous velocity is equal to the instantaneous speed. The runner reaches top speed at approximately 2 to 3 s into the race.

s (m/s)

30

7/7/08

12 10 8 6 4 2

Zero acceleration Decreasing acceleration Constant acceleration 2

Runners in the starting blocks, waiting for the starter’s pistol to fire.

What happens between the start and finish of the race? How do the velocity and acceleration of the runners vary during the race? Can we make reasonable assumptions about what the velocity-versus-time graph looks like for a typical runner? Can we estimate the maximum velocity of a good sprinter? Most importantly for improving performance, what factors affect the success of a runner in the dash? The Analysis. Let’s assume that the runner covers the 100-m distance in a time of exactly 10 s. We can compute the average speed of the runner from the definition s d/t: s

100 m 10 m/s. 10 s

Clearly, this is not the runner’s instantaneous speed throughout the course of the race, since the runner’s speed at the beginning of the race is zero and it takes some time to accelerate to the maximum speed. The objective in the race is to reach a maximum speed as quickly as possible and to sustain that speed for the rest of the race. Success is determined by two things: how quickly the runner can accelerate to this maximum speed and the value of this maximum speed. A smaller runner often has better acceleration but a smaller maximum speed, while a larger runner sometimes takes longer to reach top speed but has a larger maximum speed. The typical runner does not reach top speed before traveling at least 10 to 20 m. If the average speed is 10 m/s, the

4

6 t (s)

8

10

A graph of speed versus time for a hypothetical runner in the 100-m dash.

The average speed (or velocity) during the time that the runner is accelerating is approximately half of its maximum value if the runner’s acceleration is more or less constant during the first 2 s. If we assume that the runner’s average speed during this time is about 5.5 m/s (half of 11 m/s), then the speed through the remainder of the race would have to be about 11.1 m/s to give an average speed of 10 m/s for the entire race. This can be seen by computing the distance from these values: d (5.5 m/s)(2 s) (11.1 m/s)(8 s) 11 m 89 m 100 m. What we have done here is to make some reasonable guesses for these values that will make the average speed come out to 10 m/s; we then checked these guesses by computing the total distance. This suggests that the maximum speed of a good sprinter must be about 11 m/s (25 MPH). For sake of comparison, a distance runner who can run a 4-min mile has an average speed of about 15 MPH, or 6.7 m/s. The runner’s strategy should be to get a good jump out of the blocks, keeping the body low initially and leaning forward to minimize air resistance and maximize leg drive. To maintain top speed during the remainder of the race, the runner needs good endurance. A runner who fades near the end needs more conditioning drills. For a given runner with a fixed maximum speed, the average speed depends on how quickly the runner can reach top speed. This ability to accelerate rapidly depends upon leg strength (which can be improved by working with weights and other training exercises) and natural quickness.

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2.5 Uniform Acceleration

time, 20 seconds, is the width of this rectangle on the graph. It turns out that we can find the distance this way even when the areas involved on the graph are not rectangles, although the process is more difficult when the curves are more complicated. The general rule is that the distance traveled is equal to the area under the velocity-versus-time curve. When the velocity is negative (below the time axis on the graph), the object is traveling backward and its distance from the starting point is decreasing. Even without computing the area precisely, it is possible to get a rough idea of the distance traveled by studying the velocity graph. A large area represents a large distance. Quick visual comparisons give a good picture of what is happening without the need for lengthy calculations. This is the beauty of a graph. A good graph can present a picture of motion that is rich in insight. Distance traveled plotted against time tells us not only where the object is at any time, but its slope also indicates how fast it was moving. The graph of velocity plotted against time also contains information on acceleration and on the distance traveled. Producing and studying such graphs can give us a more general picture of the motion and the relationships between distance, velocity, and acceleration.

change as the motion proceeds. It has the same value at any time, which produces a horizontal-line graph. The graph of velocity plotted against time for this same situation tells a more interesting story. From our discussion in section 2.4, we know that the slope of a velocity-versustime graph is equal to the acceleration. For a uniform positive acceleration, the velocity graph should have a constant upward slope; the velocity increases at a steady rate. A constant slope produces a straight line, which slopes upward if the acceleration is positive as shown in figure 2.18. In plotting this graph, we assumed that the initial velocity is zero. This graph can also be represented by a formula. The velocity at any time t is equal to the original velocity plus the velocity that has been gained because the car is accelerating. The change in velocity v is equal to the acceleration times the time, v at since acceleration is defined as v/t. These ideas result in the relationship v v0 at. The first term on the right, v0, is the original velocity (assumed to be zero in figure 2.18), and the second term,

a

2.5 Uniform Acceleration If you drop a rock, it falls toward the ground with a constant acceleration, as we will see in chapter 3. An unchanging or uniform acceleration is the simplest form of accelerated motion. It occurs whenever there is a constant force acting on an object, which is the case for a falling rock as well as for many other situations. How do we describe the resulting motion? The importance of this question was first recognized by Galileo, who studied the motion of balls rolling down inclined planes as well as objects in free fall. In his famous work, Dialogues Concerning Two New Sciences, published in 1638 near the end of his life, Galileo developed the graphs and formulas that are introduced in this section and that have been studied by students of physics ever since. His work provided the foundation for much of Newton’s thinking a few decades later.

t

figure

2.17 The acceleration graph for uniform acceleration is a horizontal line. The acceleration does not change with time.

v v

v– = 1–2 v

How does velocity vary in uniform acceleration? Suppose a car is moving along a straight road and accelerating at a constant rate. We have plotted the acceleration against time for this situation in figure 2.17. The graph is very simple, but it illustrates what we mean by uniform acceleration. A uniform acceleration is one that does not

t

figure

2.18 Velocity plotted against time for uniform acceleration, starting from rest. For this special case, the average velocity is equal to one-half the final velocity.

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at, represents the change in velocity due to the acceleration. Adding these two terms together yields the velocity at any later time t. A numerical example applying these ideas to an accelerating car is found in part a of example box 2.4. The car could not keep on accelerating indefinitely at a constant rate because the velocity would soon reach incredible values. Not only is this dangerous, but physical limits imposed by air resistance and other factors prevent this from happening. What happens if the acceleration is negative? Velocity would decrease rather than increase, and the slope of the velocity graph would slope downward rather than upward. Because the acceleration is then negative, the second term in the formula for v would subtract from the first term, causing the velocity to decrease from its initial value. The velocity then decreases at a steady rate.

How does distance traveled vary with time? If the velocity is increasing at a steady rate, what effect does this have on the distance traveled? As the car moves faster and faster, the distance covered grows more and more rapidly. Galileo showed how to find the distance for this situation. We find distance by multiplying velocity by time, but in this case we must use an average velocity since the velocity is changing. By appealing to the graph in figure 2.18, we can see that the average velocity should be just half the final velocity, v. If the initial velocity is zero, the final

example box 2.4 Sample Exercise: Uniform Acceleration A car traveling due east with an initial velocity of 10 m/s accelerates for 6 seconds at a constant rate of 4 m/s2. a. What is its velocity at the end of this time? b. How far does it travel during this time? a. v0 10 m/s

v v0 at

a 4 m/s2

10 m/s (4 m/s2)(6 s)

t 6s

10 m/s 24 m/s

v ?

34 m/s v 34 m/s due east 1

d

t

figure

2.19 As the car accelerates uniformly, the distance covered grows more and more rapidly with time because the velocity is increasing.

velocity is at, so multiplying the average velocity by the time yields d

1 2

at2.

The time t enters twice, once in finding the average velocity and then again when we multiply the velocity by time to find the distance.* The graph in figure 2.19 illustrates this relationship; the distance curve slopes upward at an ever-increasing rate as the velocity increases. This formula and graph are only valid if the object starts from rest as shown in figure 2.18. Since distance traveled is equal to the area under the velocity-versus-time curve (as discussed in section 2.4), this expression for distance can also be thought of as the area under the triangle in figure 2.18. The area of a triangle is equal to one-half its base times its height, which produces the same result. If the car is already moving before it begins to accelerate, the velocity graph can be redrawn as pictured in figure 2.20. The total area under the velocity curve can then be split in two pieces, a triangle and a rectangle, as shown. The total distance traveled is the sum of these two areas, d v0 t

1 2

at 2.

The first term in this formula represents the distance the object would travel if it moved with constant velocity v0, and the second term is the additional distance traveled because the object is accelerating (the area of the triangle in figure 2.20). If the acceleration is negative, meaning that the object is slowing down, this second term will subtract from the first.

b. d v0t 2 at2 1

(10 m/s)(6 s) 2 (4 m/s2)(6 s)2 60 m (2 m/s2)(36 s2) 60 m 72 m 132 m

*Expressing this argument in symbolic form, it becomes 1

1

The average velocity v 2 v 2 at

d vt

112 at2t 12 at2.

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v

v – v0 = ∆v

d = 1–2 at 2 v0

v0

d = v0 t

t

33

manner, and it is not difficult to add them together. The two portions of the graph in figure 2.20 represent these two contributions. The sample exercise in example box 2.4 provides a numerical example of these ideas. The car in this example accelerates uniformly from an initial velocity of 10 m/s due east to a final velocity of 34 m/s due east and covers a distance of 132 meters while this acceleration is taking place. Had it not been accelerating, it would have gone only 60 meters in the same time. The additional 72 meters comes from the acceleration of the car.

figure

2.20 The velocity-versus-time graph redrawn for an initial velocity different from zero. The area under the curve is divided into two portions, a rectangle and a triangle.

Acceleration involves change, and uniform acceleration involves a steady rate of change. It therefore represents the simplest kind of accelerated motion that we can imagine. Uniform acceleration is essential to an understanding of free fall, discussed in chapter 3, as well as to many other phenomena. Such motion can be represented by either the graphs or the formulas introduced in this section. Looking at both and seeing how they are related will reinforce these ideas.

This more general expression for distance may seem complex, but the trick to understanding it is to break it down into its parts, as just suggested. We are merely adding two terms representing different contributions to the total distance. Each one can be computed in a straightforward

summary The main purpose of this chapter is to introduce concepts that are crucial to a precise description of motion. To understand acceleration, you must first grasp the concept of velocity, which in turn builds on the idea of speed. The distinctions between speed and velocity, and between velocity and acceleration, are particularly important.

The instantaneous velocity of an object is a 2vector Velocity. quantity that includes both direction and size. The size of the velocity vector is equal to the instantaneous speed, and the direction is that of the object’s motion.

1

Average and instantaneous speed. Average speed is defined as the distance traveled divided by the time. It is the average rate at which distance is covered. Instantaneous speed is the rate at which distance is being covered at a given instant in time and requires that we use very short time intervals for computation.

v v = speed and direction

3

Acceleration. Acceleration is defined as the time rate of change of velocity and is found by dividing the change in velocity by the time. Acceleration is also a vector quantity. It can be computed as either an average or an instantaneous value. A change in the direction of the velocity can be as important as a change in magnitude. Both involve acceleration. d s=– t

vi

+

∆v

= ∆v a = _ t

vf

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Graphing motion. Graphs of distance, speed, velocity, 4and acceleration plotted against time can illustrate relationships

Uniform acceleration. When an object accelerates at 5a constant rate producing a constant-slope graph of velocity versus

between these quantities. Instantaneous velocity is equal to the slope of the distance-time graph. Instantaneous acceleration is equal to the slope of the velocity-time graph. The distance traveled is equal to the area under the velocity-time graph.

time, we say that it is uniformly accelerated. Graphs help us to understand the two formulas describing how velocity and distance traveled vary with time for this important special case. v

v

d

t

t

v0

t v = v0 + at d = v0t +

1 – at 2 2

key terms Speed, 19 Average speed, 19 Rate, 20 Instantaneous speed, 21 Velocity, 22

Magnitude, 23 Vector, 24 Vector quantity, 24 Instantaneous velocity, 24 Acceleration, 25

Average acceleration, 25 Instantaneous acceleration, 26 Slope, 28 Uniform acceleration, 31

questions * = more open-ended questions, requiring lengthier responses, suitable for group discussion Q = sample responses are available in appendix D Q = sample responses are available on the website Q1. Suppose that critters are discovered on Mars who measure distance in boogles and time in bops. a. What would the units of speed be in this system? Explain. b. What would the units of velocity be? Explain. c. What would the units of acceleration be? Explain. Q2. Suppose that we choose inches as our basic unit of distance and days as our basic unit of time. a. What would the units of velocity and acceleration be in this system? Explain. b. Would this be a good choice of units for measuring the acceleration of an automobile? Explain. Q3. What units would have an appropriate size for measuring the rate at which fingernails grow? Explain. Q4. A tortoise and a hare cover the same distance in a race. The hare goes very fast for brief intervals, but stops frequently,

whereas the tortoise plods along steadily and finishes the race ahead of the hare. a. Which of the two racers has the greater average speed over the duration of the race? Explain. b. Which of the two racers is likely to reach the greatest instantaneous speed during the race? Explain. Q5. A driver states that she was doing 80 when stopped by the police. Is that a clear statement? Would this be interpreted differently in England than it would be in the United States? Explain. Q6. Does the speedometer on a car measure average speed or instantaneous speed? Explain. Q7. Is the average speed over several minutes more likely to be close to the instantaneous speed at anytime for a car traveling in freely flowing, low-density traffic or for one traveling in high-density traffic? Explain. *Q8. The highway patrol sometimes uses radar guns to identify possible speeders and at other times uses associates in airplanes who note the time taken for a car to pass between two marks some distance apart on the highway. What do each of these methods measure, average speed or instantaneous

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Questions

speed? Can you think of situations in which either one of these methods might unfairly penalize a driver? Explain. Q9. A ball is thrown against a wall and bounces back toward the thrower with the same speed as it had before hitting the wall. Does the velocity of the ball change in this process? Explain.

Q19. A car moves along a straight line so that its position (distance from some starting point) varies with time as described by the graph shown here. a. Does the car ever go backward? Explain. b. Is the instantaneous velocity at point A greater or less than that at point B? Explain. d

Q10. A ball attached to a string is whirled in a horizontal circle such that it moves with constant speed. a. Does the velocity of the ball change in this process? Explain. b. Is the acceleration of the ball equal to zero? Explain.

B

*Q11. A ball tied to a string fastened at the other end to a rigid support forms a pendulum. If we pull the ball to one side and release it, the ball moves back and forth along an arc determined by the string length. a. Is the velocity constant in this process? Explain. b. Is the speed likely to be constant in this process? What happens to the speed when the ball reverses direction? Q12. A dropped ball gains speed as it falls. Can the velocity of the ball be constant in this process? Explain. Q13. A driver of a car steps on the brakes, causing the velocity of the car to decrease. According to the definition of acceleration provided in this chapter, does the car accelerate in this process? Explain. Q14. At a given instant in time, two cars are traveling at different velocities, one twice as large as the other. Based upon this information is it possible to say which of these two cars has the larger acceleration at this instant in time? Explain.

A t

Q19 Diagram Q20. For the car whose distance is plotted against time in question 19, is the velocity constant during any time interval shown in the graph? Explain. Q21. A car moves along a straight section of road so that its velocity varies with time as shown in the graph. a. Does the car ever go backward? Explain. b. At which of the labeled points on the graph, A, B, or C, is the magnitude of the acceleration the greatest? Explain. v

Q15. A car just starting up from a stop sign has zero velocity at the instant that it starts. Must the acceleration of the car also be zero at this instant? Explain. Q16. A car traveling with constant speed rounds a curve in the highway. Is the acceleration of the car equal to zero in this situation? Explain. Q17. A racing sports car traveling with a constant velocity of 100 MPH due west startles a turtle by the side of the road who begins to move out of the way. Which of these two objects is likely to have the larger acceleration at that instant? Explain. Q18. In the graph shown here, velocity is plotted as a function of time for an object traveling in a straight line. a. Is the velocity constant for any time interval shown? Explain. b. During which time interval shown does the object have the greatest acceleration? Explain. v

35

B C A

2

4

6

t (s)

Q21 Diagram Q22. For the car whose velocity is plotted in question 21, in which of the equal time segments 0–2 seconds, 2–4 seconds, or 4–6 seconds, is the distance traveled by the car the greatest? Explain. Q23. Look again at the velocity-versus-time graph for the toy car shown in figure 2.15. a. Is the instantaneous speed greater at any time during this motion than the average speed for the entire trip? Explain. b. Is the car accelerated when the direction of the car is reversed at t 50 s? Explain. Q24. Suppose that the acceleration of a car increases with time. Could we use the relationship v v0 at in this situation? Explain.

2

4

6 t (s)

Q18 Diagram

8

Q25. When a car accelerates uniformly from rest, which of these quantities increases with time: acceleration, velocity, and/or distance traveled? Explain.

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Q26. The velocity-versus-time graph of an object curves as shown in the diagram. Is the acceleration of the object constant? Explain. v

t

Q26 Diagram Q27. For a uniformly accelerated car, is the average acceleration equal to the instantaneous acceleration? Explain. Q28. A car traveling in the forward direction experiences a negative uniform acceleration for 10 seconds. Is the distance covered during the first 5 seconds equal to, greater than, or

less than the distance covered during the second 5 seconds? Explain. Q29. A car starts from rest, accelerates uniformly for 5 seconds, travels at constant velocity for 5 seconds, and finally decelerates uniformly for 5 seconds. Sketch graphs of velocity versus time and acceleration versus time for this situation. Q30. Suppose that two runners run a 100-meter dash, but the first runner reaches maximum speed more quickly than the second runner. Both runners maintain constant speed once they have reached their maximum speed and cross the finish line at the same time. Which runner has the larger maximum speed? Explain. Q31. Sketch a graph showing velocity-versus-time curves for the two runners described in question 30. (Sketch both curves on the same graph, so that the differences are apparent.) *Q32. A physics instructor walks with increasing speed across the front of the room then suddenly reverses direction and walks backward with constant speed. Sketch graphs of velocity and acceleration consistent with this description.

exercises E1. A traveler covers a distance of 460 miles in a time of 8 hours. What is the average speed for this trip? E2. A walker covers a distance of 1.8 km in a time of 30 minutes. What is the average speed of the walker for this distance in km/h? E3. Grass clippings are found to have an average length of 4.8 cm when a lawn is mowed 12 days after the previous mowing. What is the average speed of growth of this grass in cm/day? E4. A driver drives for 2.5 hours at an average speed of 54 MPH. What distance does she travel in this time? E5. A woman walks a distance of 240 m with an average speed of 1.2 m/s. What time was required to walk this distance? E6. A person in a hurry averages 62 MPH on a trip covering a distance of 300 miles. What time was required to travel that distance? E7. A hiker walks with an average speed of 1.2 m/s. What distance in kilometers does the hiker travel in a time of 1 hour? E8. A car travels with an average speed of 22 m/s. a. What is this speed in km/s? b. What is this speed in km/h? E9. A car travels with an average speed of 58 MPH. What is this speed in km/h? (See example box 2.1.) E10. Starting from rest and moving in a straight line, a runner achieves a velocity of 7 m/s in a time of 2 s. What is the average acceleration of the runner? E11. Starting from rest, a car accelerates at a rate of 4.2 m/s2 for a time of 5 seconds. What is its velocity at the end of this time?

E12. The velocity of a car decreases from 30 m/s to 18 m/s in a time of 4 seconds. What is the average acceleration of the car in this process? E13. A car traveling with an initial velocity of 12 m/s accelerates at a constant rate of 2.5 m/s2 for a time of 2 seconds. a. What is its velocity at the end of this time? b. What distance does the car travel during this process? E14. A runner traveling with an initial velocity of 2.0 m/s accelerates at a constant rate of 1.2 m/s2 for a time of 2 seconds. a. What is his velocity at the end of this time? b. What distance does the runner cover during this process? E15. A car moving with an initial velocity of 30 m/s slows down at a constant rate of 3 m/s2. a. What is its velocity after 3 seconds of deceleration? b. What distance does the car cover in this time? E16. A runner moving with an initial velocity of 4.0 m/s slows down at a constant rate of 1.5 m/s2 over a period of 2 seconds. a. What is her velocity at the end of this time? b. What distance does she travel during this process? E17. If a world-class sprinter ran a distance of 100 meters starting at his top speed of 11 m/s and running with constant speed throughout, how long would it take him to cover the distance? E18. Starting from rest, a car accelerates at a constant rate of 3.0 m/s2 for a time of 5 seconds. a. Compute the velocity of the car at 1 s, 2 s, 3 s, 4 s, and 5 s and plot these velocity values against time. b. Compute the distance traveled by the car for these same times and plot the distance values against time.

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synthesis problems SP1. A railroad engine moves forward along a straight section of track for a distance of 80 m due west at a constant speed of 5 m/s. It then reverses its direction and travels 20 m due east at a constant speed of 4 m/s. The time required for this deceleration and reversal is very short due to the small speeds involved. a. What is the time required for the entire process? b. Sketch a graph of average speed versus time for this process. Show the deceleration and reacceleration upon reversal as occurring over a very short time interval. c. Using negative values of velocity to represent reversed motion, sketch a graph of velocity versus time for the engine. d. Sketch a graph of acceleration versus time for the engine. SP2. The velocity of a car increases with time as shown in the graph. a. What is the average acceleration between 0 seconds and 4 seconds? b. What is the average acceleration between 4 seconds and 8 seconds? c. What is the average acceleration between 0 seconds and 8 seconds? d. Is the result in part c equal to the average of the two values in parts a and b? Compare and explain.

v (m/s)

12 8

SP3. A car traveling due west on a straight road accelerates at a constant rate for 10 seconds increasing its velocity from 0 to 24 m/s. It then travels at constant speed for 10 seconds and then decelerates at a steady rate for the next 5 seconds to a velocity of 10 m/s. It travels at this velocity for 5 seconds and then decelerates rapidly to a stop in a time of 2 seconds. a. Sketch a graph of the car’s velocity versus time for the entire motion just described. Label the axes of your graph with the appropriate velocities and times. b. Sketch a graph of acceleration versus time for the car. c. Does the distance traveled by the car continually increase in the motion described? Explain. SP4. A car traveling in a straight line with an initial velocity of 14 m/s accelerates at a rate of 2.0 m/s2 to a velocity of 24 m/s. a. How much time does it take for the car to reach the velocity of 24 m/s? b. What is the distance covered by the car in this process? c. Compute values of the distance traveled at 1-second intervals and carefully draw a graph of distance plotted against time for this motion. SP5. Just as car A is starting up, it is passed by car B. Car B travels with a constant velocity of 10 m/s, while car A accelerates with a constant acceleration of 4.5 m/s2, starting from rest. a. Compute the distance traveled by each car for times of 1 s, 2 s, 3 s, and 4 s. b. At what time, approximately, does car A overtake car B? c. How might you go about finding this time exactly? Explain.

4

2

4

6

8

10

t (s)

SP2 Diagram

home experiments and observations HE1. How fast do you normally walk? Using a meter stick or a string of known length, lay out a straight course of 40 or 50 meters. Then use a watch with a second hand or a stopwatch to determine: a. Your normal walking speed in m/s. b. Your walking speed for a brisk walk. c. Your jogging speed for this same distance. d. Your sprinting speed for this distance. Record and compare the results for these different cases. Is your sprinting speed more than twice your speed for a brisk walk? HE2. The speed with which hair or fingernails grow provides some interesting measurement challenges. Using a millimeter

rule, estimate the speed of growth for one or more of: fingernails, toenails, facial hair if you shave regularly, or hair near your face (such as sideburns) that will provide an easy reference point. Measure the average size of clippings or of growth at regular time intervals. a. What is the average speed of growth? What units are most appropriate for describing this speed? b. Does the speed appear to be constant with time? Does the speed appear to be the same for different nails (thumb versus fingers, fingernails versus toenails), or in the case of hair, for different positions on your face?

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Falling Objects and Projectile Motion

chapter overview Our main purpose in this chapter is to explore how objects move under the influence of the gravitational acceleration near the Earth’s surface. Uniform acceleration, introduced in chapter 2, plays a prominent role. We begin by considering carefully the acceleration of a dropped object, and then we will extend these ideas to thrown objects or objects projected at an angle to the ground.

chapter outline

1

2 3 4

Tracking a falling object. How do velocity and distance traveled vary with time for a falling object? How can we quickly estimate these values knowing the gravitational acceleration? Beyond free fall: Throwing a ball upward. What changes when a ball is thrown upward rather than being dropped? Why does the ball appear to hover near the top of its flight? Projectile motion. What determines the motion of an object that is fired horizontally? How do the velocity and position of the object change with time in this case? Hitting a target. What factors determine the trajectory of a rifle bullet or football that has been launched at some angle to the horizontal to hit a target?

unit one

5

Acceleration due to gravity. How does a dropped object move under the influence of the Earth’s gravitational pull? How is its acceleration measured, and in what sense is it constant?

38

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3.1 Acceleration Due to Gravity

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H

ave you ever watched a leaf or a ball fall to the ground? At times during your first few years of life, you probably amused yourself by dropping an object repeatedly and watching it fall. As we grow older, that experience becomes so common that we usually do not stop to think about it or to ask why objects fall as they do. Yet this question has intrigued scientists and philosophers for centuries. To understand nature, we must first carefully observe it. If we control the conditions under which we make our observations, we are doing an experiment. The observations of falling objects that you performed as a young child were a simple form of experiment, and we would like to rekindle that interest in experimentation here. Progress in science has depended on carefully controlled experiments, and your own progress in understanding nature will depend on your active testing of ideas through experiments. You may be amazed at what you discover. Look around for some small, compact objects. A short pencil, a rubber eraser, a paper clip, or a small ball will all do nicely. Holding two objects at arm’s length, release them simultaneously and watch them fall to the floor (fig. 3.1). Be careful to release them from the same height above the floor without giving either one an upward or downward push. How would you describe the motion of these falling objects? Is their motion accelerated? Do they reach the floor at the same time? Does the motion depend on the shape and composition of the object? To explore this last question, you might take a small piece of paper

3.1 Acceleration Due to Gravity If you dropped a few objects as suggested in the introduction, you already know the answer to one of the questions posed there. Are the falling objects accelerated? Think for a moment about whether the velocity is changing. Before you release an object, its velocity is zero, but an instant after the object is released, the velocity has some value different from zero. There has been a change in velocity. If the velocity is changing, there is an acceleration. Things happen so rapidly that it is difficult, just from watching the fall, to say much about the acceleration. It does appear to be large, because the velocity increases rapidly. Does the object reach a large velocity instantly, or does the acceleration occur more uniformly? To answer this question, we must slow the motion down somehow so that our eyes and brains can keep up with what is happening.

How can we measure the gravitational acceleration? There are several ways to slow down the action. One was pioneered by the Italian scientist, Galileo Galilei (1564–1642), who was the first to accurately describe the acceleration due to gravity. Galileo’s method was to roll or slide objects down a slightly

figure 3.1

An experimenter dropping objects of different mass. Do they reach the ground at the same time?

and drop it at the same time as an eraser or a ball. First, drop the paper unfolded. Then, try folding it or crumpling it into a ball. What difference does this make? From these simple experiments, we can draw some general conclusions about the motion of falling objects. We can also try throwing or projecting objects at different angles to study the motion of a projectile. We will find that a constant downward gravitational acceleration is involved in all of these cases. This acceleration affects virtually everything that we do when we move or play on the surface of this Earth.

inclined plane. This allows only a small portion of the gravitational acceleration to come into play, just that part in the direction of motion along the plane. Thus a smaller acceleration results. Other methods (not available to Galileo) use time-lapse photography, ultrasonic motion detectors, or video recording to locate the position of the falling object at different times. If you happen to have a grooved ruler and a small ball or marble handy, you can make an inclined plane yourself. Lift one end of the ruler slightly by placing a pencil under one end, and let the ball or marble roll down the ruler under the influence of gravity (fig. 3.2). Can you see it gradually pick up speed as it rolls? Is it clearly moving faster at the bottom of the incline than it was halfway down? Galileo was handicapped by a lack of accurate timing devices. He often had to use his own pulse as a timer. Despite this limitation, he was able to establish that the acceleration was uniform, or constant, with time and to estimate its value using inclined planes. We are more fortunate. We have devices that allow us to study the motion of a falling object more directly. One such device is a stroboscope, a rapidly blinking light whose flashes occur at regular intervals in time. Figure 3.3 is a photograph taken using a stroboscope to illuminate an object as it falls. The position of the object is pinpointed every time the light flashes.

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Computing values of the average velocity for each time interval will make this even clearer. The computation can be done if we know the time interval between flashes and can measure the position of the ball from the photograph, knowing the distance between the grid marks. Table 3.1 displays data obtained in this manner. It shows the position of a ball at intervals of 1⁄ 20 of a second (0.05 second). To see that the velocity is indeed increasing, we compute the average velocity for each successive time interval. For example, between the second and third flashes, the ball traveled a distance of 3.6 centimeters, which is found by subtracting 1.2 centimeters from 4.8 centimeters. Dividing this distance by the time interval of 0.05 second yields the average size of the velocity: v

figure 3.2

A marble rolling down a ruler serving as an inclined plane. Does the velocity of the marble increase as it rolls down the incline?

If you look closely at figure 3.3, you will notice that the distance covered in successive time intervals increases regularly. The time intervals between successive positions of the ball are all equal. (If the stroboscope light flashes every 1⁄ 20 of a second, you are seeing the position of the ball every 1⁄ 20 of a second.) Since the distance covered by the ball in equal time intervals is increasing, the velocity must be increasing. Figure 3.3 shows a ball whose velocity is steadily increasing in the downward direction.

3.6 cm 72 cm/s. 0.05 s

You could verify the other values shown in the third column of table 3.1 by doing similar computations. It is clear in table 3.1 that the velocity values steadily increase. To see that velocity is increasing at a constant rate, we can plot velocity against time (fig. 3.4). Notice that each velocity data point is plotted at the midpoint between the two times (or flashes) from which it was computed. This is because these values represent the average velocity for the short time intervals between flashes. For constant acceleration, the average velocity for any time interval is equal to the instantaneous velocity at the midpoint of that interval. Did you notice that the slope of the line is constant in figure 3.4? The velocity values all fall approximately on a constant-slope straight line. Since acceleration is the slope

table 3.1 Distance and Velocity Values for a Falling Ball Time

Distance

Velocity 24 cm/s

0.05 s

1.2 cm

0.10 s

4.8 cm

0.15 s

11.0 cm

0.20 s

19.7 cm

0.25 s

30.6 cm

0.30 s

44.0 cm

0.35 s

60.0 cm

0.40 s

78.4 cm

0.45 s

99.2 cm

0.50 s

122.4 cm

72 cm/s 124 cm/s 174 cm/s 218 cm/s 268 cm/s 320 cm/s 368 cm/s

figure 3.3

A falling ball is illuminated by a rapidly blinking stroboscope. The stroboscope blinks at regular time intervals.

416 cm/s 464 cm/s

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3.1 Acceleration Due to Gravity

500

Velocity (cm/s)

400 300 200 100 0

0.1

0.2 0.3 Time (s)

0.4

0.5

figure 3.4

Velocity plotted against time for the falling ball. The velocity values are those shown in table 3.1.

of the velocity-versus-time graph, the acceleration must also be constant. The velocity increases uniformly with time. To find the value of the acceleration, we choose two velocity values that lie on the straight line and calculate how rapidly the velocity is changing. For example, the last velocity value, 464 cm/s, and the second value, 72 cm/s, are separated by a time interval corresponding to 8 flashes or 0.40 second. The increase in velocity v is found by subtracting 72 cm/s from 464 cm/s, obtaining 392 cm/s. To find the acceleration, we divide this change in velocity by the time interval (a v/t), a

392 cm/s 980 cm/s2 9.8 m/s2. 0.4 s

This result gives us the acceleration due to gravity for objects falling near the Earth’s surface. Its value actually varies slightly from point to point on the Earth’s surface because of differences in altitude and other effects. This acceleration is used so often that it is given its own symbol g where

gravitational acceleration does not depend on the weight of the object. Galileo used similar experiments to prove this point. His experiments contradicted Aristotle’s view that heavier objects fall more rapidly. How could Aristotle’s idea have been accepted for so long when simple experiments can disprove it? Experimentation was not part of the intellectual outlook of Aristotle and his followers; they valued pure thought and logic more highly. Galileo and other scientists of his time broke new ground by using experiments as an aid to thinking. A new tradition was emerging. On the other hand, Aristotle’s view agrees with our intuition that heavy objects do fall more rapidly than some lighter objects. If, for example, we drop a brick together with a feather or unfolded piece of paper (fig. 3.5), the brick will reach the floor first. The paper or feather will not fall in a straight line but instead will flutter to the floor much as a leaf falls from a tree. What is happening here? You will probably recognize that the effects of air resistance impede the fall of the feather or paper much more than the fall of the brick, a steel ball, or a paper clip. When we crumple the piece of paper into a ball and drop it simultaneously with a brick or other heavy object, the two objects reach the floor at approximately the same time. We live at the bottom of a sea of air, and the effects of air resistance can be substantial for objects like leaves, feathers, or pieces of paper. These effects produce a slower and less regular flight for light objects that have a large surface area. If we drop a feather and a brick simultaneously in a vacuum or in the very thin atmosphere of the moon, they do reach the ground at the same time. Moonlike conditions are not part of our everyday experience, however, so we are used to seeing feathers fall more slowly than rocks or bricks. Galileo’s insight was that the gravitational acceleration is the same for all objects, regardless of their weight, provided that the effects of air resistance are not significant. Aristotle did not separate the effect of air resistance from that of gravity in his observations.

g 9.8 m/s2. Called the gravitational acceleration or acceleration due to gravity, it is valid only near the Earth’s surface and thus is not a fundamental constant.

How did Galileo’s ideas on falling objects differ from Aristotle’s? There is another sense in which the gravitational acceleration is constant, which takes us back to the experiments suggested in the chapter opener, p. 39. When you drop objects of different sizes and weights, do they reach the floor at the same time? Except for an unfolded piece of paper, it is likely that all of the objects that you test, regardless of their weight, reach the floor at the same time when released simultaneously. This finding suggests that the

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figure 3.5

The brick reaches the floor first when a brick and a feather are dropped at the same time.

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The gravitational acceleration for objects near the Earth’s surface is uniform and has the value of 9.8 m/s2. It can be measured by using stroboscopes or similar techniques to record the position of a falling object at regular, very small time intervals. This acceleration is constant in time. Contrary to Aristotle’s belief, it also has the same value for objects of different weight.

3.2 Tracking a Falling Object Imagine yourself dropping a ball from a sixth-story window, as in figure 3.6. How long does it take for the ball to reach the ground below? How fast is it traveling when it gets there? Things happen quickly, so the answers to these questions are not obvious. If we assume that air-resistance effects are small for the object we are tracking, we know that it accelerates toward the ground at the constant rate of 9.8 m/s2. Let’s make some quick estimates of how these values change with time without doing detailed computations.

How does the velocity vary with time? In making estimates of velocity and distance for a falling object, we often take advantage of the fact that the gravitationalacceleration value of 9.8 m/s2 is almost 10 m/s2 and round it up. (Here we are choosing the downward direction as positive.) This makes the numerical values easier to calculate without sacrificing much in accuracy. Multiplying by 10 is quicker than multiplying by 9.8. How fast is our dropped ball moving after 1 second? An acceleration of 10 m/s2 means that the velocity is increasing by 10 m/s each second. If its original velocity is zero, then after 1 second its velocity has increased to 10 m/s, in 2 seconds to 20 m/s, and in 3 seconds to 30 m/s. For each additional second, the ball gains 10 m/s in velocity.* To help you appreciate these values, look back at table 2.1, which shows unit comparisons for familiar speeds. A velocity of 30 m/s is roughly 70 MPH, so after 3 seconds the ball is moving quickly. After just 1 second, it is moving with a downward velocity of 10 m/s, which is over 20 MPH. The ball gains velocity at a faster rate than is possible for a high-powered automobile on a level surface.

How far does the ball fall in different times? The high velocities are more meaningful if we examine how far the ball falls during these times. As the ball falls, it gains speed, so it travels farther in each successive time interval, as in the photograph in figure 3.3. Because of uniform

figure 3.6

A ball is dropped from a sixth-story window. How long does it take to reach the ground?

acceleration, the distance increases at an ever-increasing rate. During the first second of motion, the velocity of the ball increases from zero to 10 m/s. Its average velocity during that first second is 5 m/s, and it travels a distance of 5 meters in that second. This can also be found by using the relationship between distance, acceleration, and time in section 2.5. If the starting velocity is zero, we found that 1 d 2 at2. After 1 second, the ball has fallen a distance 1

*In section 2.5, we noted that the velocity of an object moving with uniform acceleration is v v0 at, where v0 is the original velocity and the second term is the change in velocity, v at. When a ball is dropped, v0 0, so v is just at, the change in velocity.

d 2 (10 m/s2)(1 s)2 5 m. Since the height of a typical story of a multistory building is less than 4 meters, the ball falls more than one story in just a second.

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3.2 Tracking a Falling Object

During the next second of motion, the velocity increases from 10 m/s to 20 m/s, yielding an average velocity of 15 m/s for that interval. The ball travels 15 meters in that second, which, when added to the 5 meters covered in the first second, yields a total of 20 meters. After 2 seconds, the distance fallen is four times as large as the 5 meters traveled after 1 second.* Since 20 meters is roughly five stories in height, the ball dropped from the sixth story will be near the ground after 2 seconds. Figure 3.7 gives the velocity and distance fallen at halfsecond time intervals for a ball dropped from a six-story building. Notice that in just half a second, the ball falls 1.25 meters. An object dropped to the floor from an outstretched arm therefore hits the floor in roughly half a second. This makes it difficult to time with a stopwatch. (See example box 3.1.) The change in velocity is proportional to the size of the time interval selected. In 1 second the change in velocity is 10 m/s, so in half a second the change in velocity is 5 m/s. In each half-second the ball gains approximately 5 m/s in velocity, illustrated in figure 3.7. As the velocity gets larger, the arrows representing the velocity vectors grow. If we plotted these velocity values against time, we would get a simple upward-sloping straight-line graph as in figure 3.4. What does the graph of the distance values look like? The distance values increase in proportion to the square of the time, which means that they increase more and more rapidly as time elapses. Instead of being a straight-line graph, the graph of the distance values curves upward as in figure 3.8. The rate of change of distance with time is itself increasing with time.

Throwing a ball downward Suppose that instead of just dropping the ball, we throw it straight down, giving it a starting velocity v0 different from zero. How does this affect the results? Will the ball reach the ground more rapidly and with a larger velocity? You would probably guess correctly that the answer is yes. In the case of the velocity values, the effect of the starting velocity is not difficult to see. The ball is still being accelerated by gravity so that the change in velocity for each second of motion is still v 10 m/s, or for a halfsecond, 5 m/s. If the initial downward velocity is 20 m/s, after half a second, the velocity is 25 m/s, and after 1 second, it is 30 m/s. We simply add the change in velocity to the initial velocity as indicated by the formula v v0 at. In the case of distance, however, the values increase more rapidly. The full expression for distance traveled by a uniformly accelerated object (introduced in section 2.5) is 1

d v0t 2 at2. *This is a result of the time being squared in the formula for distance. 1 Putting 2 s in place of 1 s in the formula d 2 at2 multiplies the result by 2 a factor of 4 (2 4), yielding a distance of 20 m.

t = 0.5 s v = 5 m/s

d = 1.25 m a = 10 m/s2

t = 1.0 s d = 5.0 m v = 10 m/s a = 10 m/s2

t = 1.5 s d = 11.3 m v = 15 m/s a = 10 m/s2

t = 2.0 s d = 20 m v = 20 m/s a = 10 m/s2

figure 3.7

Velocity and distance values for the dropped ball shown at half-second time intervals.

example box 3.1 Sample Question: Using a Pulse Rate to Time a Falling Object Question: Suppose that Galileo’s resting pulse rate was 60 beats per minute. Would his pulse be a useful timer for getting position-versus-time data for an object dropped from the height of 2 to 3 meters? Answer: A pulse rate of 60 beats per minute corresponds to 1 beat per second. In the time of 1 second, a dropped object falls a distance of approximately 5 m. (It falls 1.22 m in just half a second as seen in table 3.1.) Thus this pulse rate (or most pulse rates) would not be an adequate timer for an object dropped from a height of a few meters. It could be slightly more effective for an object dropped from a tower several stories in height.

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After just 1 second the ball has already traveled 25 meters, which means that it would be near the ground if thrown from our sixth-story window. Keep in mind, though, that we have ignored the effects of air resistance in arriving at these results. For a compact object falling just a few meters, the effects of air resistance are very small. These effects increase as the velocity increases, however, so that the farther the object falls, the greater the effects of air resistance. In chapter 4, we will discuss the role of air resistance in more depth in the context of sky diving.

Distance (m)

20

10

0 0

1.0

2.0 Time (s)

figure 3.8

A plot of distance versus time for the dropped

ball.

The first term is the distance that the ball would travel if it continued to move with just its original velocity. This distance also increases with time. The second term is due to the acceleration and has the same values as shown in figures 3.7 and 3.8. In the sample exercise in example box 3.2, we calculate velocity and distance traveled during the first 2 seconds of motion for a ball thrown downward. Notice that after 2 seconds the ball has traveled a distance of 60 meters, much larger than the 20 meters when the ball is simply dropped.

example box 3.2

When an object is dropped, its velocity increases by approximately 10 m/s every second due to the gravitational acceleration. The distance traveled increases at an ever-increasing rate because the velocity is increasing. In just a few seconds, the object is moving very rapidly and has fallen a large distance. In section 3.3, we will explore the effects of gravitational acceleration on an object thrown upward.

3.3 Beyond Free Fall: Throwing a Ball Upward In section 3.2, we discussed what happens when a ball is dropped or thrown downward. In both of these cases, the ball gains velocity as it falls due to the gravitation acceleration. What if the ball is thrown upward instead, as in figure 3.9? How does gravitational acceleration affect the ball’s motion? What goes up must come down—but when and how fast are interesting questions with everyday applications.

Sample Exercise: Throwing a Ball Downward A ball is thrown downward with an initial velocity of 20 m/s. Using the value 10 m/s2 for the gravitational acceleration, find (a) the velocity and (b) the distance traveled at 1-s time intervals for the first 2 s of motion. a. v0 20 m/s a 10 m/s2 v ?

t 2s b. d ? t 1s

v v0 at for t 1 s v 20 m/s (10 m/s2)(1 s) 20 m/s 10 m/s 30 m/s

v 20 m/s (10 m/s2)(2 s) 20 m/s 20 m/s 40 m/s 1

d v0t 2 at2 d (20 m/s)(1 s)

1 2

(10 m/s2)(1 s)2

20 m 5 m 25 m t 2s

d (20 m/s)(2 s)

1 2

(10 m/s2)(2 s)2

40 m 20 m 60 m

figure 3.9

A ball thrown upward returns to the ground. What are the magnitude and direction of the velocity at different points in the flight?

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3.3 Beyond Free Fall: Throwing a Ball Upward

The directions of the acceleration and velocity vectors merit our close attention. The gravitational acceleration is always directed downward toward the center of the Earth, because that is the direction of the gravitational force that produces this acceleration. This means that the acceleration is in the opposite direction to the original upward velocity.

How does the ball’s velocity change? Suppose that we throw a ball straight up with an original velocity of 20 m/s. Many of us can throw a ball at this velocity: it is approximately 45 MPH. This is a lot less than a 90-MPH fastball, but throwing a ball upward with good velocity is harder than throwing it horizontally. Once the ball leaves our hand, the primary force acting on it is gravity, which produces a downward acceleration of 9.8 m/s2 or approximately 10 m/s2. (If we now choose the upward direction as positive, this acceleration is negative because it is downward.) Every second, there is a change in velocity of 10 m/s. This change in velocity is directed downward, however, opposite to the direction of the original velocity. It subtracts from the original velocity rather than adding to it. Once you are aware of how important direction is in observing the ball thrown upward, finding the velocity at different times is not hard. After 1 second, the velocity of the ball has decreased by 10 m/s, so if it started at 20 m/s (choosing the positive direction to be upward in this case), it is now moving upward with a velocity of just 10 m/s. After 2 seconds, it loses another 10 m/s, so its velocity is then zero. It does not stop there, of course. In another second (3 seconds from the start), its velocity decreases by another 10 m/s, and it is then moving downward at 10 m/s. The sign of the velocity indicates its direction. All of these values can be found from the relationship v v0 at, where v0 20 m/s and a 10 m/s2. Clearly, the ball has changed direction, as you might expect. Just as before, the velocity changes steadily at 10 m/s each second, due to the constant downward acceleration. After 4 seconds, the ball is moving downward with a velocity of 20 m/s and is back at its starting position. These results are illustrated in figure 3.10. The high point in the motion occurs at a time 2 seconds after the ball is thrown, where the velocity is zero. If the velocity is zero, the ball is moving neither upward nor downward, so this is the turnaround point. An interesting question, a favorite on physics tests (and often missed by students), asks for the value of acceleration at the high point in the motion. If the velocity is zero at this point, what is the value of the acceleration? The quick, but incorrect, response given by many people is that the acceleration must also be zero at that point. The correct answer is that the acceleration is still 10 m/s2. The gravitational acceleration is constant and does not change. The velocity of the ball is still changing at that instant, from a positive to a negative value, even though the instantaneous

t d v a t d v a

= = = =

= = = =

45

2s 20 m 0 –10 m/s2

1s 15 m +10 m/s –10 m/s2

t = 0 d = 0 v = +20 m/s

t d v a

= = = =

3s 15 m –10 m/s –10 m/s2

t d v a

= = = =

4s 0 –20 m/s –10 m/s2

figure 3.10

The changing velocity is indicated by the blue velocity vectors at different points in the flight of a ball thrown upward with a starting velocity of 20 m/s. The constant downward acceleration is shown as a green vector at each point.

velocity is zero. Acceleration is the rate of change of velocity and is unrelated to the size of the velocity. What would a graph of velocity plotted against time look like for the motion just described? If we make the upward direction of motion positive, the velocity starts with a value of 20 m/s and changes at a steady rate, decreasing by 10 m/s each second. This is a straight-line graph, sloping downward as in figure 3.11. The positive values of velocity represent upward motion, where the size of the velocity is decreasing, and the negative values of velocity represent downward motion. If the ball did not hit the ground, but was thrown from the edge of a cliff, it would continue to gain negative velocity as it moved downward.

How high does the ball go? The position or height of the ball at different times can be computed using the methods in section 3.2. These distance computations involve the formula for uniform acceleration developed in section 2.5. In the sample exercise in example box 3.3, we compute the height or distance traveled at 1-second intervals for the ball thrown upward at 20 m/s, using 10 m/s2 for the gravitational acceleration.

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+20

v (m/s)

+10 0 1

2

3

4

t (s)

–10 –20

figure

3.11 A plot of the velocity versus time for a ball thrown upward with an initial velocity of 20 m/s. The negative values of velocity represent downward motion.

What should you notice about these results? First, the high point of the motion is 20 meters above the starting point. The high point is reached when the velocity is zero, and we determined earlier that this occurs at a time of 2 seconds. This time depends on how fast the ball is thrown initially. The larger the original velocity, the greater the time to reach the high point. Knowing this time, we can use the distance formula to find the height. You should also notice that after just 1 second, the ball has reached a height of 15 meters. It covers just five additional meters in the next second of motion, and then falls back to 15 meters in the following second. The ball spends a full 2 seconds above the height of 15 meters, even though it only reaches a height of 20 meters. The ball is moving more slowly near the top of its flight than it is at lower points—this is why the ball appears to “hang” near the top of its flight.

Finally, the time taken for the ball to fall back to its starting point from the high point is equal to the time taken for the ball to reach the high point in the first place. It takes 2 seconds to reach the high point and another 2 seconds for it to return to the starting point. The total time of flight is just twice the time needed to reach the high point, in this case, 4 seconds. A larger starting velocity would produce a higher turnaround point and a greater “hang time” for the ball. A ball thrown upward is slowed by the downward gravitational acceleration until its velocity is reduced to zero at the high point. The ball then falls from that high point accelerating downward at the same constant rate as when it was rising. The ball travels more slowly near the top of its flight, so it appears to “hang” there. It spends more time in the top few meters than it does in the rest of the flight. We will find that these features are also present when a ball is projected at an angle to the horizontal, as discussed in section 3.5.

3.4 Projectile Motion Suppose that instead of throwing a ball straight up or down, you throw it horizontally from some distance above the ground. What happens? Does the ball go straight out until it loses all of its horizontal velocity and then starts to fall like the perplexed coyote in the Roadrunner cartoons (fig. 3.12)? What does the real path, or trajectory, look like? Cartoons give us a misleading impression. In fact, two different things are happening at the same time: (1) the ball is accelerating downward under the influence of gravity, and (2) the ball is also moving sideways with an approximately constant horizontal velocity. Combining these two motions gives the overall trajectory or path.

example box 3.3 Sample Exercise: Throwing a Ball Upward A ball is thrown upward with an initial velocity of 20 m/s. Find its height at 1-s intervals for the first 4 s of its flight. d ?

d v0 t

1 2

at2 1

t 1s

(20 m/s)(1 s) 2 (10 m/s2)(1 s)2 20 m 5 m 15 m

t 2s

d (20 m/s)(2 s) 2 (10 m/s2)(2 s)2 40 m 20 m 20 m

t 3s

d (20 m/s)(3 s) 2 (10 m/s2)(3 s)2 60 m 45 m 15 m

t 4s

d (20 m/s)(4 s) 2 (10 m/s2)(4 s)2 80 m 80 m 0 m

1

1

1

figure

3.12 A cartoon coyote falling off a cliff. Is this a realistic picture of what happens?

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3.4 Projectile Motion

What does the trajectory look like? You can perform a simple experiment to help you visualize the path that the projectile follows. Take a marble or small ball, roll it along the top of a desk or table, and let it roll off the edge. What does the path of the ball look like as it travels through the air to the floor? Is it like the coyote in figure 3.12? Roll the ball at different velocities and see how the path changes. Try to sketch the path after making these observations. How do we go about analyzing this motion? The key lies in thinking about the horizontal and vertical components of the motion separately and then combining them to get the actual path (fig. 3.13). The acceleration of the horizontal motion is zero, provided that air resistance is small enough to be ignored. This implies that the ball moves with a constant horizontal velocity once it has rolled off the table or has left the hand. The ball travels equal horizontal distances in equal time intervals, as shown across the top of figure 3.13. In constructing this diagram, we assumed an initial horizontal velocity of 2 m/s for the ball. Every tenth of a second, then, the ball travels a horizontal distance of 0.2 meter. At the same time that the ball travels with constant horizontal velocity, it accelerates downward with the constant gravitational acceleration g. Its vertical velocity increases exactly like that of the falling ball photographed for figure 3.3. This motion is depicted along the left side of figure 3.13. In each successive time interval, the ball falls a greater distance than in the time interval before, because the vertical velocity increases with time. Combining the horizontal and vertical motions, we get the path shown curving downward in figure 3.13. For each time shown, we draw a horizontal dashed line locating the vertical position of the ball, and a vertical dashed line for the horizontal position. The position of the ball at any time is the point where these lines intersect. The resulting t = 0.1 s

x t = 0.2 s

trajectory (the solid curve) should look familiar if you have performed the simple experiments suggested in the first paragraph on this page. If you understand how we obtained the path of the ball, you are well on your way to understanding projectile motion. The total velocity of the ball at each position pictured is in the direction of the path at that point, since this is the actual direction of the ball’s motion. This total velocity is a vector sum of the horizontal and vertical components of the velocity (fig. 3.14). (See appendix C for a discussion of vector components.) The horizontal velocity remains constant, because there is no acceleration in that direction. The downward (vertical) velocity gets larger and larger.

study hint If you are not familiar with vectors, you should take the time to read and work the exercises in appendix C. Appendix C describes what vectors are, how they are added using simple graphical procedures, and how vector components are defined. In this section, we use the ideas that a vector quantity such as velocity can have both horizontal and vertical components and that these components add to give the total velocity. These concepts are critical to your understanding of projectile motion. Vector addition and vector components are also used in many other situations that we will encounter in later chapters.

The actual shape of the path followed by the ball depends on the original horizontal velocity given the ball by throwing it or rolling it from the tabletop. If this initial horizontal velocity is small, the ball does not travel very far horizontally. Its trajectory will then be like the smallest starting velocity v1 in figure 3.15. The three trajectories shown in figure 3.15 have three different starting velocities. As you would expect, the ball

t = 0.3 s

v horizontal

v vertical

t = 0.4 s

v total

y

figure

3.13 The horizontal and vertical motions combine to produce the trajectory of the projected ball. The vertical and horizontal positions are shown at regular time intervals.

47

figure

3.14 The total velocity at any point is found by adding the vertical component of the velocity to the horizontal component.

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example box 3.4 Sample Exercise: Projectile Motion

v1

v2

v3

A ball rolls off a tabletop with an initial velocity of 3 m/s. If the tabletop is 1.25 m above the floor, a. How long does it take for the ball to hit the floor? b. How far does the ball travel horizontally? a. In figure 3.7, we saw that a ball will fall a distance of 1.25 m in approximately half a second. This could be found directly from

figure 3.15

Trajectories for different initial velocities of a ball rolling off a table: v3 is larger than v2, which in turn is larger than v1. The positions are shown at equal time intervals.

1 2

dvertical 1.25 m

dvertical

a g 10 m/s2

Solving for t 2:

t ?

t2

travels greater horizontal distances when projected with a larger initial horizontal velocity.

at 2

d 1 2

a

1.25 m 5 m/s2

0.25 s2 Taking the square root to get t:

What determines the time of flight? Which of the three balls in figure 3.15 would hit the floor first if all three left the tabletop at the same time? Does the time taken for the ball to hit the floor depend on its horizontal velocity? There is a natural tendency to think that the ball that travels farther takes a longer time to reach the floor. In fact, the three balls should all reach the floor at the same time. The reason is that they are all accelerating downward at the same rate of 9.8 m/s2. This downward acceleration is not affected by how fast the ball travels horizontally. The time taken to reach the floor for the three balls in figure 3.15 is determined strictly by how high above the floor the tabletop is. The vertical motion is independent of the horizontal velocity. This fact often surprises people. It contradicts our intuitive sense of what is going on but can be confirmed by doing simple experiments using two similar balls (fig. 3.16). If you throw one ball horizontally at the same time that you simply drop the second ball from the same height, the two balls should reach the floor at roughly the same time. They may fail to hit at the same time, most likely because it is hard to throw the first ball completely horizontally and to release both balls at the same time. A special spring gun, often used in demonstrations, will do this more precisely. If we know how far the ball falls, we can compute the time of flight. This can then be used to determine the horizontal distance that the ball will travel, if we know the initial horizontal velocity. The sample exercise in example box 3.4 shows this type of analysis. Notice that the horizontal distance traveled is determined by two factors: the time of flight and the initial velocity.

t 0.5 s b. Knowing the time of flight t, we can now compute the horizontal distance traveled: v0 3 m/s t 0.5 s dhorizontal ?

figure

dhorizontal v0t (3.0 m/s)(0.50 s) 1.5 m

3.16 A ball is dropped at the same time that a second ball is projected horizontally from the same height. Which ball reaches the floor first?

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figure 3.17

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A target shooter fires at a distant target. The bullet falls as it travels to the target.

Treating the vertical motion independently of the horizontal motion and then combining them to find the trajectory is the secret to understanding projectile motion. A horizontal glide combines with a vertical plunge to produce a graceful curve. The downward gravitational acceleration behaves the same as for any falling object, but there is no acceleration in the horizontal direction if air resistance can be ignored. The projectile moves with constant horizontal velocity while it is accelerating downward.

3.5 Hitting a Target As long as humans have been hunters or warriors, they have wanted to predict where a projectile such as a cannonball will land after it is fired. Being able to hit a target such as a bird in a tree or a ship at sea has obvious implications for survival. Being able to hit a catcher’s mitt with a baseball thrown from center field is also a highly valued skill.

Does the bullet fall when a rifle is fired? Imagine that you are firing a rifle at a small target some distance away, with the rifle and target at exactly the same distance above the ground (fig. 3.17). If the rifle is fired directly at the target in a horizontal direction, will the bullet hit the center of the target? If you think of the ball rolling off the table in section 3.4, you should conclude that the bullet will strike the target slightly below the center. Why? The bullet will be accelerated downward by Earth’s gravitational pull and will fall slightly as it travels to the target. Since the time of flight is small, the bullet does not fall very far, but it falls far enough to miss the center of the target. How do you compensate for the fall of the bullet? You aim a little high. You correct your aim either through trial and error or by adjusting your rifle sight so that your aim is automatically a little above center. Rifle sights are often adjusted for some average distance to the target. For longer distances you must aim high, for shorter distances a little low. If you aim a little high, the bullet no longer starts out in a completely horizontal direction. The bullet travels up slightly during the first part of its flight and then comes down to meet the target. This also happens when you fire a cannon or throw a ball at a distant target. A frequent demonstration to illustrate the independence of the vertical and horizontal motions of projectiles is

often referred to as “Shoot the Monkey” or “Monkey in a Tree.” A projectile is aimed directly at a toy monkey (or other suitable target) hanging from the ceiling. An electronic trigger allows the target to drop at the same time the projectile is launched. The target falls straight down at a rate governed by the acceleration of gravity. The projectile starts to move toward the initial position of the target, but also starts to fall at a rate governed by the acceleration of gravity. Due to the fact that both the projectile and target begin falling in the vertical direction at the exact same time and with the same downward acceleration, the projectile will always hit the target (fig. 3.18). It is crucial to recognize that the projectile hits below where it was aimed by an amount equal to the vertical distance the target drops since the acceleration of gravity has the same effect on both the projectile and the target. It is also important that the target is released and the projectile is fired at the exact same time. If the target was stationary, the projectile would have to be aimed above the target to compensate for the vertical drop due to the acceleration of gravity.

The flight of a football Whenever you throw a ball such as a football at a somewhat distant target, the ball must be launched at an angle above the horizontal so that the ball does not fall to the ground too soon. A good athlete does this automatically as a result of

line of sight to target

trajectory of projectile

figure

3.18 If the projectile is launched at the same time the target is dropped, will it hit the target?

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30°

figure

3.19 The flight of a football launched at an angle of 30° to the horizontal. The vertical and horizontal positions of the ball are shown at regular time intervals. practice. The harder you throw, the less you need to direct the ball upward, because a larger initial velocity causes the ball to reach the target more quickly, giving it less time to fall. Figure 3.19 shows the flight of a football thrown at an angle of 30° above the horizontal. The vertical position of the ball is plotted on the left side of the diagram, as in figure 3.13 for the horizontally projected ball. The horizontal position of the ball is shown across the bottom of the diagram. We have assumed that air resistance is small, so the ball travels with a constant horizontal velocity. Combining these two motions yields the overall path. As the football climbs, the vertical component of its velocity decreases because of the constant downward gravitational acceleration. At the high point, this vertical component of the velocity is zero, just as it is for a ball thrown straight upward. The velocity of the ball is completely horizontal at this high point. The ball then begins to fall, gaining downward velocity as it accelerates. Unlike the ball thrown straight upward, however, there is a constant horizontal component to the velocity throughout the flight. We need to add this horizontal motion to the up-and-down motion that we described in section 3.3.

everyday phenomenon

In throwing a ball, you can vary two quantities to help you hit your target. One is the initial velocity, which is determined by how hard you throw the ball. The other is the launch angle, which can be varied to fit the circumstances. A ball thrown with a large initial velocity does not have to be aimed as high and will reach the target more quickly. It may not clear the onrushing linemen, however, and it might be difficult to catch because of its large velocity. There is no time like the present to test these ideas. Take a page of scrap paper and crumple it into a compact ball. Then take your wastebasket and put it on your chair or desk. Throwing underhand, experiment with different throwing speeds and launch angles to see which is most effective in making a basket. Try to get a sense of how the launch angle and throwing speed interact to produce a successful shot. A low, flat-trajectory shot should require a greater throwing speed than a higher, arching shot. The flatter shot must also be aimed more accurately, since the effective area of the opening in the basket is smaller when the ball approaches at a flat angle. The ball “sees” a smaller opening. (This effect is discussed in everyday phenomenon box 3.1.)

box 3.1

Shooting a Basketball The Situation. Whenever you shoot a basketball, you unconsciously select a trajectory for the ball that you believe will have the greatest likelihood of getting the ball to pass through the basket. Your target is above the launch point (with the exception of dunk shots and sky hooks), but the ball must be on the way down for the basket to count. What factors determine the best trajectory? When is a high, arching shot desirable, and when might a flatter trajectory be more effective? Will these factors be different for a free throw than for a shot taken when you are guarded by another player? How can our understanding of projectile motion help us to answer these questions?

The Analysis. The diameter of the basketball and the diameter of the basket opening limit the angle at which the basketball can pass cleanly through the hoop. The second drawing shows the range of possible paths for a ball coming straight down and for one coming in at a 45° angle to the basket. The shaded area in each case shows how much the center of the ball can vary from the center line if the ball is to pass through the hoop. As you can see, a wider range of paths is available when the ball is coming straight down. The diameter of the basketball is a little more than half the diameter of the basket.

(continued)

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This second drawing illustrates the advantage of an arched shot. There is a larger margin of error in the path that the ball can take and still pass through the hoop cleanly. For the dimensions of a regulation basketball and basket, the angle must be at least 32° for a clean shot. As the angle gets larger, the range of possible paths increases. At smaller

angles, appropriate spin on the basketball will sometimes cause the ball to rattle through, but the smaller the angle, the less the likelihood of that happening. The disadvantage of the arched shot is less obvious. As you get farther away from the basket, launching conditions for an arched shot must be more precise for the ball to travel the horizontal distance to the basket. If an arched shot is launched from 30 ft, it must travel a much higher path than a shot launched at the same angle closer to the basket, as shown in the third drawing. Since the ball stays in the air for a longer time, small variations in either the release speed or angle can cause large errors in the distance traveled. This distance depends on both the time of flight and the horizontal component of the velocity.

Different possible trajectories for a basketball free throw. Which has the greatest chance of success?

An arched shot launched from a large distance stays in the air longer than one launched at the same angle from much closer to the basket.

45°

Possible paths for a basketball coming straight down and for one coming in at a 45° angle. The ball coming straight down has a wider range of possible paths.

A highly arched shot is more effective when you are close to the basket. You can then take advantage of the greater range of paths available to the arched shot without suffering much from the uncertainty in the horizontal distance. Away from the basket, the desirable trajectories gradually become flatter, permitting more accurate control of the shot. An arched shot is sometimes necessary from anywhere on the court, however, to avoid having the shot blocked. The spin of the basketball, the height of the release, and other factors all play a role in the success of a shot. A fuller analysis can be found in an article by Peter J. Brancazio in the American Journal of Physics (April 1981) entitled “Physics of Basketball.” A good understanding of projectile motion might improve the game of even an experienced player.

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v

v0

Chapter 3 Falling Objects and Projectile Motion

v0 70°

45° 20°

figure

3.20 Cannonball paths for different launch angles but the same initial launch speed.

How can we achieve maximum distance? In firing a rifle or cannon, the initial velocity of the projectile is usually set by the amount of gunpowder in the shell. The launch angle is then the only variable we can change in attempting to hit a target. Figure 3.20 shows three possible paths, or trajectories, for a cannonball fired at different launch angles for the same initial speed. For different launch angles, we tilt the cannon barrel by different amounts from the position shown. Note that the greatest distance is achieved using an intermediate angle, an angle of 45° if the effects of air resistance are negligible. The same considerations are involved in the shot put in track-and-field events. The launch angle is very important and, for the greatest distance, will be near 45°. Air resistance and the fact that the shot hits the ground below the launch point are also factors, so the most effective angle is somewhat less than 45° in the shot put. Thinking about what happens to the horizontal and vertical components of the initial velocity at different launch angles will show us why the angle for maximum distance is approximately 45°. (See figure 3.21.) Velocity is a vector, and its horizontal and vertical components can be found by drawing the vector to scale and adding dashed lines to the horizontal and vertical directions (fig. 3.21). This process is described more fully in appendix C. For the lowest launch angle 20°, we see that the horizontal component of the velocity is much larger than the vertical. Since the initial upward velocity is small, the ball does not go very high. Its time of flight is short, and it hits the ground sooner than in the other two cases shown. The ball gets there quickly because of its large horizontal velocity and short travel time, but it does not travel very far before hitting the ground. The high launch angle of 70° produces a vertical component much larger than the horizontal component. The ball thus travels much higher and stays in the air for a longer time than at 20°. It does not travel very far horizontally, however, because of its small horizontal velocity. The ball travels the same horizontal distance as for the 20°

20°

45°

70°

figure

3.21 Vector diagrams showing the horizontal and vertical components of the initial velocity for the three cases illustrated in figure 3.20.

launch, but it takes longer getting there.* (If we shot it straight up, the horizontal distance covered would be zero, of course.) The intermediate angle of 45° splits the initial velocity into equal-sized horizontal and vertical components. The ball therefore stays in the air longer than in the low-angle launch but also travels with a greater horizontal velocity than in the high-angle launch. In other words, with relatively large values for both the vertical and horizontal components of velocity, the vertical motion keeps the ball in the air long enough for the horizontal velocity to be effective. This produces the greatest distance of travel. The time of flight and the horizontal distance traveled can be found if the launch angle and the size of the initial velocity are known. It is first necessary to find the horizontal and vertical components of the velocity to do these computations, however, and this makes the problem more complex than those discussed earlier. The ideas can be understood without doing the computations. The key is to think about the vertical and horizontal motions separately and then combine them.

For a projectile launched at an angle, the initial velocity can be broken down into vertical and horizontal components. The vertical component determines how high the object will go and how long it stays in the air, while the horizontal component determines how far it will go in that time. The launch angle and the initial speed interact to dictate where the object will land. Through the entire flight, the constant downward gravitational acceleration is at work, but it changes only the vertical component of the velocity. Producing or viewing such trajectories is a common part of our everyday experience.

*The angles 20° and 70° are complementary because their sum is 90°. Any pair of complementary launch angles (30° and 60°, for example) yield the same horizontal range as one another.

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key terms

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summary The primary aim in this chapter has been to introduce you to the gravitational acceleration for objects near the Earth’s surface and to show how that acceleration affects the motion of objects dropped or launched in various ways.

1

Acceleration due to gravity. To find the accelera-

tion due to gravity, we use measurements of the position of a dropped object at different times. The gravitational acceleration is 9.8 m/s2. It does not vary with time as the object falls, and it has the same value for different objects regardless of their weight.

point, and then increases as the object falls. The object spends more time near the top of its flight because it is moving more slowly there. v

t

motion. If an object is launched horizon4tally, itProjectile moves with a constant horizontal velocity at the same time that it accelerates downward due to gravity. These two motions combine to produce the object’s curved trajectory.

2

Tracking a falling object. The velocity of a falling object increases by approximately 10 m/s every second of its fall. Distance traveled increases in proportion to the square of the time, so that it increases at an ever-increasing rate. In just 1 second, a dropped ball is moving with a velocity of 10 m/s and has traveled 5 meters.

path of an object launched at an angle to the horizontal. Once again, the horizontal and vertical motions combine to produce the overall motion as the projectile moves toward a target.

d

v

t v = v0 + a t

t d = 1–2 at 2

3

Beyond free fall: Throwing a ball upward. The speed of an object thrown upward first decreases due to the downward gravitational acceleration, passes through zero at the high

key terms Acceleration due to gravity, 41 Air resistance, 41

a target. There are two factors, the launch 5speed Hitting and the launch angle, that can be varied to determine the

Trajectory, 46 Projectile motion, 47

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* more open-ended questions, requiring lengthier responses, suitable for group discussion Q sample responses are available in appendix D Q sample responses are available on the website

Q1. A small piece of paper is dropped and flutters to the floor. Is the piece of paper accelerating at any time during this motion? Explain. Q2. The diagram shows the positions at intervals of 0.10 seconds of a ball moving from left to right (as in a photograph taken with a stroboscope that flashes every tenth of a second). Is the ball accelerated? Explain.

Q2 Diagram Q3. The diagram shows the positions at intervals of 0.05 seconds of two balls moving from left to right. Are either or both of these balls accelerated? Explain. A B

Q3 Diagram Q4. A lead ball and an aluminum ball, each 1 in. in diameter, are released simultaneously and allowed to fall to the ground. Due to its greater density, the lead ball has a substantially larger mass than the aluminum ball. Which of these balls, if either, has the greater acceleration due to gravity? Explain.

velocity

questions

time

Q9 Diagram Q10. A ball is thrown downward with a large starting velocity. a. Will this ball reach the ground sooner than one that is just dropped at the same time from the same height? Explain. b. Will this ball accelerate more rapidly than one that is dropped with no initial velocity? Explain. Q11. A ball thrown straight upward moves initially with a decreasing upward velocity. What are the directions of the velocity and acceleration vectors during this part of the motion? Does the acceleration decrease also? Explain. Q12. A rock is thrown straight upward reaching a height of 20 meters. On its way up, does the rock spend more time in the top 5 meters of its flight than in its first 5 meters of its flight? Explain. Q13. A ball is thrown straight upward and then returns to the Earth. Choosing the positive direction to be upward, sketch a graph of the velocity of this ball against time. Where does the velocity change direction? Explain. Q14. A ball is thrown straight upward. At the very top of its flight, the velocity of the ball is zero. Is its acceleration at this point also zero? Explain.

Q5. Two identical pieces of paper, one crumpled into a ball and the other left uncrumpled, are released simultaneously from the same height above the floor. Which one, if either, do you expect to reach the floor first? Explain.

Q15. A ball is thrown straight upward and then returns to the Earth. Does the acceleration change direction during this motion? Explain.

Q6. Two identical pieces of paper, one crumpled into a ball and the other left uncrumpled, are released simultaneously from inside the top of a large evacuated tube. Which one, if either, do you expect will reach the bottom of the tube first? Explain.

*Q16. A ball rolls up an inclined plane, slows to a stop, and then rolls back down. Do you expect the acceleration to be constant during this process? Is the velocity constant? Is the acceleration equal to zero at any point during this motion? Explain.

*Q7. Aristotle stated that heavier objects fall faster than lighter objects. Was Aristotle wrong? In what sense could Aristotle’s view be considered correct?

Q17. A ball rolling rapidly along a tabletop rolls off the edge and falls to the floor. At the exact instant that the first ball rolls off the edge, a second ball is dropped from the same height. Which ball, if either, reaches the floor first? Explain.

Q8. A rock is dropped from the top of a diving platform into the swimming pool below. Will the distance traveled by the rock in a 0.1-second interval near the top of its flight be the same as the distance covered in a 0.1-second interval just before it hits the water? Explain. Q9. The graph shows the velocity plotted against time for a certain falling object. Is the acceleration of this object constant? Explain.

Q18. For the two balls in question 17, which, if either, has the larger total velocity when it hits the floor? Explain. Q19. Is it possible for an object to have a horizontal component of velocity that is constant at the same time that the object is accelerating in the vertical direction? Explain by giving an example, if possible.

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Exercises

Q20. A ball rolls off a table with a large horizontal velocity. Does the direction of the velocity vector change as the ball moves through the air? Explain. Q21. A ball rolls off a table with a horizontal velocity of 5 m/s. Is this velocity an important factor in determining the time that it takes for the ball to hit the floor? Explain. Q22. An expert marksman aims a high-speed rifle directly at the center of a nearby target. Assuming that the rifle sight has been accurately adjusted for more distant targets, will the bullet hit the near target above or below the center? Explain. Q23. In the diagram, two different trajectories are shown for a ball thrown by a center fielder to home plate in a baseball game. Which of the two trajectories (if either), the higher one or the lower one, will result in a longer time for the ball to reach home plate? Explain.

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cannon. Will the 70° shot travel a greater horizontal distance than the 45° shot? Explain. Q27. Will a shot fired from a cannon at a 20° launch angle travel a longer horizontal distance than a 45° shot? Explain. Q28. The diagram shows a wastebasket placed behind a chair. Three different directions are indicated for the velocity of a ball thrown by the kneeling woman. Which of the three directions—A, B, or C—is most likely to result in the ball landing in the basket? Explain. A B C

Q28 Diagram Q29. In the situation pictured in question 28, is the magnitude of the velocity important to the success of the shot? Explain.

Q23 Diagram Q24. For either of the trajectories shown in the diagram for question 23, is the velocity of the ball equal to zero at the high point in the trajectory? Explain. Q25. Assuming that the two trajectories in the diagram for question 23 represent throws by two different center fielders, which of the two is likely to have been thrown by the player with the stronger arm? Explain.

Q30. In shooting a free throw in basketball, what is the primary advantage that a high, arching shot has over one with a flatter trajectory? Explain. Q31. In shooting a basketball from greater than free-throw range, what is the primary disadvantage of a high, arching shot? Explain. *Q32. A football quarterback must hit a moving target while eluding onrushing linemen. Discuss the advantages and disadvantages of a hard low-trajectory throw to a higherlofted throw.

Q26. A cannonball fired at an angle of 70° to the horizontal stays in the air longer than one fired at 45° from the same

exercises E1. A steel ball is dropped from a diving platform (with an initial velocity of zero). Using the approximate value of g 10 m/s2, a. What is the velocity of the ball 0.8 seconds after its release? b. What is its velocity 1.6 seconds after its release? E2. For the ball in exercise 1: a. Through what distance does the ball fall in the first 0.8 seconds of its flight? (Assume g 10 m/s2.) b. How far does it fall in the first 1.6 seconds of its flight? E3. A large rock is dropped from the top of a high cliff. Assuming that air resistance can be ignored and that the acceleration has the constant value of 10 m/s2, how fast would the rock be traveling 5 seconds after it is dropped? What is this speed in MPH? (See inside front cover for conversion factors.) E4. Suppose Galileo’s pulse rate was 80 beats per minute. a. What is the time in seconds between consecutive pulse beats?

b. How far does an object fall in this time when dropped from rest? E5. A ball is thrown downward with an initial velocity of 12 m/s. Using the approximate value of g 10 m/s2, what is the velocity of the ball 1.0 seconds after it is released? E6. A ball is dropped from a high building. Using the approximate value of g 10 m/s2, find the change in velocity between the first and fourth second of its flight. E7. A ball is thrown upward with an initial velocity of 15 m/s. Using the approximate value of g 10 m/s2, what are the magnitude and direction of the ball’s velocity: a. 1 second after it is thrown? b. 2 seconds after it is thrown? E8. How high above the ground is the ball in exercise 7: a. 1 second after it is thrown? b. 2 seconds after it is thrown?

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E9. At what time does the ball in exercise 7 reach the high point in its flight? (Use the approximate value of g 10 m/s2, and remember that the velocity is equal to zero at the high point.) E10. Suppose that the gravitational acceleration on a certain planet is only 3.0 m/s2. A space explorer standing on this planet throws a ball straight upward with an initial velocity of 18 m/s. a. What is the velocity of the ball 4 seconds after it is thrown? b. How much time elapses before the ball reaches the high point in its flight? E11. A bullet is fired horizontally with an initial velocity of 900 m/s at a target located 150 m from the rifle. a. How much time is required for the bullet to reach the target? b. Using the approximate value of g 10 m/s2, how far does the bullet fall in this time? E12. A ball rolls off a shelf with a horizontal velocity of 6 m/s. At what horizontal distance from the shelf does the ball land if it takes 0.4 s to reach the floor?

E13. A ball rolls off a table with a horizontal velocity of 4 m/s. If it takes 0.5 seconds for the ball to reach the floor, how high above the floor is the tabletop? (Use g 10 m/s2.) E14. A ball rolls off a table with a horizontal velocity of 5 m/s. If it takes 0.6 seconds for it to reach the floor: a. What is the vertical component of the ball’s velocity just before it hits the floor? (Use g 10 m/s2.) b. What is the horizontal component of the ball’s velocity just before it hits the floor? E15. A ball rolls off a platform that is 5 meters above the ground. The ball’s horizontal velocity as it leaves the platform is 6 m/s. a. How much time does it take for the ball to hit the ground? (See example box 3.3, use g 10 m/s2.) b. How far from the base of the platform does the ball hit the ground? E16. A projectile is fired at an angle such that the vertical component of its velocity and the horizontal component of its velocity are both equal to 30 m/s. a. Using the approximate value of g 10 m/s2, how long does it take for the ball to reach its high point? b. What horizontal distance does the ball travel in this time?

synthesis problems SP1. A ball is thrown straight upward with an initial velocity of 16 m/s. Use g 10 m/s2 for computations listed here. a. What is its velocity at the high point in its motion? b. How much time is required to reach the high point? c. How high above its starting point is the ball at its high point? d. How high above its starting point is the ball 2 seconds after it is released? e. Is the ball moving up or down 2 seconds after it is released? SP2. Two balls are released simultaneously from the top of a tall building. Ball A is simply dropped with no initial velocity, and ball B is thrown downward with an initial velocity of 12 m/s. a. What are the velocities of the two balls 1.5 seconds after they are released? b. How far has each ball dropped in 1.5 seconds? c. Does the difference in the velocities of the two balls change at any time after their release? Explain. SP3. Two balls are rolled off a tabletop that is 0.8 m above the floor. Ball A has a horizontal velocity of 3 m/s and that of ball B is 5 m/s. a. Assuming g 10 m/s2, how long does it take each ball to reach the floor after it rolls off the edge? b. How far does each ball travel horizontally before hitting the floor? c. If the two balls started rolling at the same time at a point 1.2 m behind the edge of the table, will they reach the floor at the same time? Explain. SP4. A cannon is fired over level ground at an angle of 30° to the horizontal. The initial velocity of the cannonball is 400 m/s,

but because the cannon is fired at an angle, the vertical component of the velocity is 200 m/s and the horizontal component is 346 m/s. a. How long is the cannonball in the air? (Use g 10 m/s2 and the fact that the total time of flight is twice the time required to reach the high point.) b. How far does the cannonball travel horizontally? c. Repeat these calculations, assuming that the cannon was fired at a 60° angle to the horizontal, resulting in a vertical component of velocity of 346 m/s and a horizontal component of 200 m/s. How does the distance traveled compare to the earlier result? SP5. A good pitcher can throw a baseball at a speed of 90 MPH. The pitcher’s mound is approximately 60 ft from home plate. a. What is the speed in m/s? b. What is the distance to home plate in meters? c. How much time is required for the ball to reach home plate? d. If the ball is launched horizontally, how far does the ball drop in this time, ignoring the effects of spin? SP6. An archeologist is running at 7 m/s with her hands outstretched above her head (1.95 m from feet to fingertips) while being chased by a tiger. She runs exactly horizontally off of a chasm and attempts to grab onto the opposite side. a. If the chasm is 4.55 meters wide, how long does she take to cover this distance? b. During this time, what distance has she fallen vertically (use g 10 m/s2)? c. How far above or below the edge of the opposite side do her fingertips fall? (Use to indicate distances above the edge and to indicate distances below the edge.)

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home experiments and observations HE1. Gather numerous small objects and drop them from equal heights, two at a time. Record which objects fall significantly more slowly than a compact dense object such as a marble or similar object. Rank order these slower objects by their time of descent. What factors seem to be important in determining this time? HE2. Working with a partner, you can get an estimate of your reaction time by catching a falling meter stick. Have your partner hold the meter stick from a point near the top while you place the finger and thumb of your catching hand about an inch apart on either side of the 50 cm mark. Without giving any cues, your partner then drops the meter stick, and when you see it move, you react to catch it by closing your finger and thumb. Record the distance that the meter stick moves between the time that your partner releases it and you catch it. a. Repeat this process several times for each partner and compute the average distance the meter stick traveled for each partner. Tabulate your results. b. Since the distance traveled in the time t that it takes for you 2d to react is d 1/2 gt2, the time of travel is t . Use Bg a calculator to compute the reaction time t for each partner from the average distance d (expressed in meters). Use g 10 m/s2. How does your average reaction time compare to your partner’s? c. A ‘normal’ reaction time is between 0.2 and 0.25 sec. Is your reaction time close to this? If not, explain why you think your reaction time is different. HE3. Try dropping a ball from one hand at the same time that you throw a second ball with your other hand. At first, try to throw the second ball horizontally, with no upward or downward component to its initial velocity. (It may take some practice.) a. Do the balls reach the floor at the same time? (It helps to enlist a friend for making this judgment.) b. If the second ball is thrown slightly upward from the horizontal, which ball reaches the ground first?

c. If the second ball is thrown slightly downward from the horizontal, which ball reaches the ground first? HE4. Take a ball outside and throw it straight up in the air as hard as you can. By counting seconds yourself, or by enlisting a friend with a watch, estimate the time that the ball remains in the air. From this information, can you find the initial velocity that you gave to the ball? (The time required for the ball to reach the high point is just half the total time of flight.) HE5. Take a stopwatch to a football game and estimate the hang time of several punts. Also note how far (in yards) each punt travels horizontally. Do the highest punts have the longest hang times? Do they travel the greatest distances horizontally? HE6. Using rubber bands and a plastic rule or other suitable support, design and build a marble launcher. By pulling the rubber band back by the same amount each time, you should be able to launch the marble with approximately the same speed each time. (Warning: Leave yourself ample room free of breakable objects!) a. Produce a careful drawing of your launcher and note the design features that you used. (Prizes may be available for the best design.) b. Placing your launcher at a number of different angles to the horizontal, launch marbles over a level surface and measure the distance that they travel from the point of launch. Which angle yields the greatest distance? c. Fire the marbles at different angles from the edge of a desk or table. Which angle yields the greatest horizontal distance? HE7. Try throwing a ball or a wadded piece of paper into a wastebasket placed a few meters from your launch point. a. Which is most effective, an overhanded or underhanded throw? (Five practice shots followed by ten attempts for each might produce a fair test.) b. Repeat this process with a barrier such as a chair placed near the wastebasket.

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chapter overview The primary purpose of this chapter is to explain Newton’s three laws of motion and how they apply in familiar situations. We begin with a historical sketch of their development and then proceed to a careful discussion of each law. The concepts of force, mass, and weight play critical roles in this discussion. We conclude the chapter by applying Newton’s theory to several familiar examples.

chapter outline

1 2

3 4

Newton’s first and second laws. How do forces affect the motion of an object? What do Newton’s first and second laws of motion tell us, and how are they related to one another? Mass and weight. How can we define mass? What is the distinction between mass and weight? Newton’s third law. Where do forces come from? How does Newton’s third law of motion help us to define force, and how is the third law applied? Applications of Newton’s laws. How can Newton’s laws be applied in different situations such as pushing a chair, sky diving, throwing a ball, and pulling two connected carts across the floor?

unit one

5

A brief history. Where do our ideas and theories about motion come from? What roles were played by Aristotle, Galileo, and Newton?

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A

large person gives you a shove, and you move in the direction of that push. A child pulls a toy wagon with a string, and the wagon lurches along. An athlete kicks a football or a soccer ball, and the ball is launched toward the goal. These are familiar examples involving forces in the form of pushes or pulls that cause changes in motion. To pick a less complex example, imagine yourself pushing a chair across a wood or tile floor (fig. 4.1). Why does the chair move? Will it continue its motion if you stop pushing? What factors determine the velocity of the chair? If you push harder, will the chair’s velocity increase? Up to this point, we have introduced ideas useful in describing motion, but we have not talked much about what causes changes in motion. Explaining motion is more challenging than describing it. You already have some intuitive notions about what causes the chair to move. Certainly, the push that you exert on the chair has something to do with it. But is the strength of that push more directly related to the velocity of the chair or to its acceleration? At this point, intuition often serves us poorly. Over two thousand years ago, the Greek philosopher Aristotle (384–322 B.C.) attempted to provide answers to some of these questions. Many of us would find that his explanations match our intuition for the case of the moving chair, but they are less satisfactory in the case of a thrown object where the push is not sustained. Aristotle’s ideas were widely accepted until they were replaced by a theory introduced by Isaac Newton in the seventeenth century. Newton’s theory of motion has proved to be a much more complete and satisfactory explanation of motion, and it permits quantitative predictions that were largely lacking in Aristotle’s ideas. Newton’s three laws of motion form the foundation of his theory. What are these laws and how are they used in explaining motion? How do Newton’s ideas differ from those of Aristotle, and why do Aristotle’s ideas often seem to fit our commonsense notions of what is

figure 4.1

Moving a chair. Will the chair continue to move when the person stops pushing?

happening? A good understanding of Newton’s laws will permit you to analyze and explain almost any simple motion. This understanding will provide you with insights useful in driving a car, moving heavy objects, and many other everyday activities.

4.1 A Brief History

Aristotle’s view of the cause of motion

Did some genius, sitting under an apple tree, concoct a full-blown theory of motion in a sudden, blinding flash of inspiration? Not quite. The story of how theories are developed and gain acceptance involves many players over long periods of time. Let’s highlight the roles of a few key people whose insights produced major advances. A glimpse of this history can help you appreciate the physical concepts we will discuss by showing when and how the theories emerged. It is important, for example, to know whether a theory was just proposed yesterday or has been tried and tested over a long time. Not all theories carry equal weight in their acceptance and use by scientists. Aristotle, Galileo, and Newton were major players in shaping our views of the causes of motion.

Questions about the causes of motion and changes in motion had perplexed philosophers and other observers of nature for centuries. For over a thousand years, Aristotle’s views prevailed. Aristotle was a careful and astute philospher of nature. Aristotle investigated an incredible range of subjects, and he (or perhaps his students) produced extensive writings on topics such as logic, metaphysics, politics, literary criticism, rhetoric, psychology, biology, and physics. In his discussions of motion, Aristotle conceived of force much as we have talked about it to this point: as a push or pull acting on an object. He believed that a force had to act for an object to move and that the velocity of the object was proportional to the strength of the force. A heavy object would fall more quickly toward the Earth than a lighter object, because there was a larger force pulling the object

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to the Earth. The strength of this force could be appreciated simply by holding the object in your hand. Aristotle was also aware of the resistance that a medium offers the motion of an object. A rock falls more rapidly through air than through water. Water provides greater resistance to motion than air, as you surely know from trying to walk through waist-deep water at the beach. Aristotle thus saw the velocity of the object as being proportional to the force acting on it and inversely related to the resistance, but he never defined the concept of resistance quantitatively. He did not distinguish acceleration from velocity, and he spoke of velocity by stating the time required to cover a fixed distance. Aristotle was an observer of nature rather than an experimenter. He did not make quantitative predictions that he checked by experiment. Even without such tests, however, some problems with his basic ideas of motion troubled Aristotle himself, as well as later thinkers. For example, in the case of a thrown ball or rock, the force that initially propels the object no longer acts once the ball leaves the hand. What keeps the ball moving? Since the ball does keep moving for some time after leaving the hand that throws it, a force was necessary, according to Aristotle’s theory. He suggested that the force that maintains the motion once the ball leaves the hand is provided by air rushing around to fill the vacuum in the spot where the ball has just been (fig. 4.2). This flow of air then pushes the ball from behind. Does this seem reasonable? Following the decline of the Roman Empire, only fragments of Aristotle’s writings were known to European thinkers for several centuries. His complete works, which had been preserved by Arab scholars, did not resurface in Europe until the twelfth century. Along with the work of other Greek thinkers, Aristotle’s works were translated into Latin during the twelfth and thirteenth centuries.

How did Galileo challenge Aristotle’s views? By the time that the Italian scientist Galileo Galilei (1564– 1642) came on the scene, Aristotle’s ideas were well established at European universities, including the universities of

Pisa and Padua where Galileo studied and taught. In fact, education at the universities was organized around the disciplines defined by Aristotle, and much of Aristotle’s natural philosophy had been incorporated into the teaching of the Roman Catholic Church. The Italian theologian Thomas Aquinas had carefully interwoven Aristotle’s thinking with the theology of the church. To challenge Aristotle was equivalent to challenging the authority of the church and could carry heavy consequences. Galileo was not alone in questioning Aristotle’s ideas on motion; others had noted that dropped objects of similar form but radically different weights fall at virtually the same rate, contrary to Aristotle’s theory. Although Galileo may never have dropped objects from the Leaning Tower of Pisa, he did perform careful experiments with dropped objects and actively publicized his results. Galileo’s primary problems with the church came from advocating the ideas of Copernicus. Copernicus had proposed a sun-centered (heliocentric) model of the solar system (discussed in chapter 5), which opposed the prevailing Earth-centered models of Aristotle and others. Galileo was an activist on several fronts in challenging Aristotle and the traditional thinking. This placed him in conflict with many of his university colleagues and with members of the church hierarchy. He was eventually tried by the Inquisition and found guilty of heresy. He was placed under house arrest and forced to retract some of his teachings. In addition to his work on falling objects, Galileo developed new ideas on motion that contradicted Aristotle’s theory. Galileo argued that the natural tendency of a moving object is to continue moving: no force is required to maintain this motion. (Think about the pushed chair again. Does this statement make sense in that situation?) Building on the work of others, Galileo also developed a mathematical description of motion that included acceleration. The relationship d 12 at 2 for the distance covered by a uniformly accelerating object was carefully demonstrated by Galileo. He published many of these ideas near the end of his life in his famous Dialogues Concerning Two New Sciences.

What did Newton accomplish?

v

figure 4.2

Aristotle pictured air rushing around a thrown object to continue pushing the object forward. Does this picture seem reasonable?

Isaac Newton (1642–1727; fig. 4.3) was born in England within a year of Galileo’s death in Italy. Building on the work of Galileo, he proposed a theory of the causes of motion that could explain the motion of any object—the motion of ordinary objects such as a ball or chair as well as the motion of heavenly bodies such as the moon and the planets. In the Greek tradition, celestial motions were thought of as an entirely different realm from Earthbound motions, thus requiring different explanations. Newton abolished this distinction by explaining both terrestrial and celestial mechanics with one theory. The central ideas in Newton’s theory are his three laws of motion (discussed in sections 4.2 and 4.4) and his law

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figure 4.3

A portrait of Isaac Newton.

of universal gravitation (discussed in chapter 5). Newton’s theory provided successful explanations of aspects of motion already known and offered a framework for many new studies in physics and astronomy. Some of these studies led to predictions of phenomena not previously observed. For example, calculations applying Newton’s theory to irregularities in the orbits of the known planets led to the prediction of the existence of Neptune, which was quickly confirmed by observation. Confirmed predictions are one of the marks of a successful theory. Newton’s theory served as the basic theory of mechanics for over two hundred years and is still used extensively in physics and engineering. Newton developed the basic ideas of his theory around 1665, when he was still a young man. To avoid the plague, he had returned to his family’s farm in the countryside where he had time to engage in serious thought with little interruption. The story has it that seeing an apple fall led to his insight that the moon also falls toward the Earth and that the force of gravity is involved in both cases. (See chapter 5.) Flashes of insight or inspiration were surely a part of the process. Although Newton developed much of his theory and its details in 1665, he did not formally publish his ideas until 1687. One reason for this delay was his need to develop some of the mathematical techniques required to calculate the effects of the proposed gravitational force on objects such as planets. (He is generally credited with being the coinventor of what we now call calculus.) The English title of Newton’s 1687 treatise is The Mathematical Principles of Natural Philosophy (Philosophiae Naturalis Principia Mathematica in Latin), which is often referred to as Newton’s Principia. Scientific theories like Newton’s do not just emerge in an intellectual vacuum. They are products of their time and

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the state of knowledge and worldview current then. They usually replace earlier and often cruder theories. The accepted theory of motion in Newton’s day was still that of Aristotle, although it had come under attack by Galileo and others. Its shortcomings were generally recognized. Newton provided the capstone for a revolution in thought that was already well under way. Although Aristotle’s ideas on motion are now considered unsatisfactory and are worthless for making quantitative predictions, they do have an intuitive appeal much like our own untrained thinking about motion. For this reason, we often speak of the need to replace Aristotelian ideas about motion with Newtonian concepts as we learn mechanics. Even though our own naive ideas about motion are not usually as fully developed as those of Aristotle, you may find that some of your commonsense notions will require modification. Newton’s theory, in turn, has been partially superseded by more sophisticated theories that provide more accurate descriptions of motion. These include Einstein’s theory of relativity as well as the theory of quantum mechanics, both of which arose early in the twentieth century. Although the predictions of these theories differ substantially from Newton’s theory in the realm of the very fast (in the case of relativity) and the very small (quantum mechanics), they differ insignificantly for the motion of ordinary objects traveling at speeds much less than that of light. Newton’s theory was a tremendous step forward and is still used extensively to analyze motion of ordinary objects. Aristotle’s ideas on motion, although not capable of making quantitative predictions, provided explanations that were widely accepted for many centuries and that fit well with some of our own commonsense thinking. Galileo challenged Aristotle’s ideas on free fall as well as his general assumption that a force was required to keep an object in motion. Building on Galileo’s work, Newton developed a more comprehensive theory of motion that replaced Aristotle’s ideas. Newton’s theory is still widely used to explain the motion of ordinary objects.

4.2 Newton’s First and Second Laws If we push a chair across the floor, what causes the chair to move or to stop moving? Newton’s first two laws of motion address these questions and, in the process, provide part of a definition of force. The first law tells us what happens in the absence of a force, and the second describes the effects of applying a force to an object. We discuss the first and second laws of motion together because the first law is closely related to the more general second law. Newton felt the need to state the first law separately, however, to counter strongly held Aristotelian ideas

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about motion. In doing so, Newton was following the lead of Galileo, who had stated a principle similar to Newton’s first law several years earlier.

Newton’s first law of motion In language not too different from his own, Newton’s first law of motion can be stated as An object remains at rest, or in uniform motion in a straight line, unless it is compelled to change by an externally imposed force.

In other words, unless there is a force acting on the object, its velocity will not change. If it is initially at rest, it will remain at rest; if it is moving, it will continue to do so with constant velocity (fig 4.4). Notice that, in paraphrasing Newton’s first law, we have used the term velocity rather than the term speed. Constant velocity implies that neither the direction nor the magnitude of the velocity changes. When the object is at rest, its velocity is zero, and that value remains constant in the absence of a force. If there is no force acting on the object, the acceleration of the object is zero. The velocity does not change. Although this law seems simple enough, it directly contradicts Aristotle’s ideas (and perhaps your own intuition as well). Aristotle believed that a force is required to keep an object moving. His views make intuitive sense if we are talking about moving a heavy object such as the chair mentioned in our introduction. If you stop pushing, the chair stops moving. This view encounters problems, however, if we consider the motion of a thrown ball, or even a chair moving on a slippery surface. These objects continue to move after the initial push. Newton (and Galileo) made the

If F = 0

v remains equal to 0 (at rest)

or v

v

v remains constant (uniform motion in a straight line)

figure 4.4

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Newton’s first law: In the absence of a force, an object remains at rest or moves with constant velocity.

strong statement that no force is needed to keep an object moving. How can Aristotle’s ideas be so different from those of Newton and Galileo and yet seem so reasonable in some situations? The key to answering that question involves the existence of resistive or frictional forces. The chair does not move far after you stop pushing because the frictional forces of the floor acting on the chair cause the velocity to quickly decrease to zero. A thrown ball would eventually stop moving, even if it did not fall to the ground, because the force of air resistance is pushing against it. It is really quite difficult to find a situation in which there are no forces acting upon an object. Aristotle recognized the presence of air resistance and similar effects but did not treat them as forces in his theory.

How is force related to acceleration? Newton’s second law of motion is a more complete statement about the effect of an imposed force on the motion of an object. Stated in terms of acceleration, it says The acceleration of an object is directly proportional to the magnitude of the imposed force and inversely proportional to the mass of the object. The acceleration is in the same direction as that of the imposed force.

This statement is most easily grasped in symbolic form. By choosing appropriate units for force, we can state the proportionality of Newton’s second law as the equation: a

Fnet . m

where a is the acceleration, Fnet is the total or net force acting on the object, and m is the mass of the object. Since the acceleration is directly proportional to the imposed force, if we double the force acting on the object, we double the acceleration of the object. The same force acting on an object with a larger mass, however, will produce a smaller acceleration (fig. 4.5). Note that the acceleration is directly related to the imposed force, not the velocity. Aristotle did not make a clear distinction between acceleration and velocity. Many of us also fail to make the distinction when we think informally about motion. In Newton’s theory, this distinction is critical. Newton’s second law is the central idea of his theory of motion. According to this law, the acceleration of an object is determined by two quantities: the net force acting on the object and the mass of the object. In fact, the concepts of force and mass are, in part, defined by the second law. The net force acting on the object is the cause of its acceleration, and the magnitude of the force is defined by the size of the acceleration that it produces. Newton’s third law, discussed in section 4.4, completes the definition of force by

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m1 f=2N F

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F = 10 N

5 kg

figure 4.6

A block being pulled across a table. Two horizontal forces are involved.

a

example box 4.1 m2

F

Sample Exercise: Finding the Net Force a

figure 4.5

The smaller-mass object experiences a larger acceleration than the larger-mass object when identical forces are applied to the two objects.

noting that forces result from interaction of the object with other objects. The mass of an object is a quantity that tells us how much resistance an object has to a change in its motion, as indicated by the second law. We call this resistance to a change in motion inertia, following Galileo. (See everyday phenomenon box 4.1.) We can define mass as Mass is a measure of an object’s inertia, the property that causes it to resist a change in its motion.

The standard metric unit for mass is the kilogram (kg). We will say more about the determination of mass and its relationship to the weight of an object shortly (section 4.3). Units of force can also be derived from Newton’s second law. If we solve for Fnet by multiplying both sides of the second-law equation by the mass, it can be expressed as Fnet ma. The appropriate unit for force must therefore be the product of a unit of mass and a unit of acceleration, or in the metric system, kilograms times meters per second squared. This frequently used unit is called the newton (N). Accordingly, 1 newton 1 N 1 kg m/s2.

How do forces add? Our version of the second law implies that the imposed force is the total or net force acting on the object. Force is a vector quantity whose direction is clearly important. If there is more than one force acting on an object, as there often is, we must then add these forces as vectors, taking into account their directions. This process is illustrated in figure 4.6 and the sample exercise in example box 4.1. A block is being pulled across a table by a force of 10 N applied through a string attached to the block. A frictional force of 2 N acts on the block, a result

A block with a mass of 5 kg is being pulled across a tabletop by a force of 10 N applied by a string tied to the front end of the block (fig 4.6). The table exerts a 2-N frictional force on the block. What is the acceleration of the block? Fstring 10 N (to the right)

Fnet Fstring ftable

ftable 2 N (to the left)

10 N 2 N 8 N Fnet 8 N (to the right)

m 5 kg a? a

Fnet m 8N 5 kg

1.6 m/s2 (a 1.6 m/s2 to the right)

of contact with the table. What is the total force acting on the block? Is the net force the numerical sum of the two forces, 10 N plus 2 N or 12 N? Looking at the diagram in figure 4.6 should convince you that this cannot be true. The two forces oppose one another. Because the forces are in opposite directions, the net force is found by subtracting the frictional force from the force applied by the string, resulting in a net force of 8 N. We cannot ignore the directions of the forces involved. That forces are vectors whose directions must be taken into account when finding the net force is an important aspect of the second law. For forces restricted to one dimension, as in example box 4.1, finding the net force is not difficult. In problems involving forces in two or three dimensions, addition is more complex but can be accomplished using techniques described in appendix C. In this chapter we will only consider one-dimensional cases. A final point about Newton’s first and second laws bears repeating: the first law is contained within the second law, but it was very important for Newton to state the first law as a separate law to counter long-standing beliefs about motion. The relationship between the two laws can be demonstrated by asking what happens, according to the second law, when the net force acting on an object is zero. In this case, the acceleration

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Fnet must also be zero. If the acceleration is zero, the m velocity must be constant. The first law tells us that if the net force is zero, the object moves with constant velocity (or remains at rest). Newton’s first law addresses the special case of the second law in which the net force acting on an object is zero.

a

The central principle in Newton’s theory of motion is his second law of motion. This law states that the acceleration of an object is proportional to the net force applied to the object and inversely proportional to the mass of the object. The mass of an object is its inertia or resistance to change in motion. Newton’s first law follows from the second law when the net force acting on the object is zero. To find the net force acting on the object, we take into account the directions of the individual forces and add them as vectors.

everyday phenomenon

4.3 Mass and Weight What exactly is weight? Is your weight the same as your mass, or is there a difference in the meaning of these two terms? Clearly, mass plays an important role in Newton’s second law. Weight is a familiar term often used interchangeably with mass in everyday language. Here again, physicists make a distinction between mass and weight that is important to Newton’s theory.

How can masses be compared? From the role that mass plays in Newton’s second law, we can devise experimental methods of comparing masses. Mass is defined as the property of matter that determines how much an object resists a change in its motion. The greater the mass, the greater the inertia or resistance to change, and the smaller the acceleration provided by a given force. Imagine, for example, trying to decelerate a bowling

box 4.1

The Tablecloth Trick The Situation. When he was a child, Ricky Mendez saw a magician do the tablecloth trick. A full dinner place setting including a filled wineglass sat on a tablecloth covering a small table. The magician, with appropriate fanfare, pulled the tablecloth from the table without disturbing the dinnerware. Ricky ended up in the doghouse, however, when he tried this at home with disastrous results. More recently Ricky saw his physics instructor do a similar trick with a simpler place setting. The students were told that the demonstration had something to do with inertia. Why does the trick work, and how is inertia involved? Why did the trick not work when Ricky tried it at home as a child? The Analysis. The magician’s trick, which is frequently used as a physics demonstration, is indeed an illustration of the effects of inertia. Since the nature of frictional forces also plays a role, the choice of a smooth material for the tablecloth is important. (Butcher paper is sometimes substituted in physics demonstrations.) Some practice is usually essential to the successful execution of the trick. The performer, be it a magician, instructor, or student, must pull the cloth or paper very quickly, giving it a large initial acceleration. Pulling slightly downward across the

edge of the table helps to assure that there is no upward component to the acceleration and that the acceleration is reasonably uniform across the width of the tablecloth. As the tablecloth accelerates, it exerts a frictional force upon the tableware. If we pulled slowly, this frictional force would pull the dishes and glasses along with the tablecloth. Inertia is the tendency of an object (related to its mass) to resist a change in its motion. When an object is at rest, it remains at rest unless a force is applied. There is a force acting on the plates and glasses, however—the frictional force exerted by the tablecloth. If the tablecloth is pulled quickly enough, the frictional force is in effect for only a very short time so the acceleration of the objects is very brief. The objects will accelerate slightly, but not nearly as much as the tablecloth. There are two aspects of the frictional force that are important to our understanding of what happens. One is that the force of static friction (in effect when the surfaces are not sliding relative to one another) has a maximum value that is determined by the nature of the contacting surfaces and by the force pushing the surfaces together. The second is that once the objects start to slide, kinetic or sliding friction comes into play. The force of kinetic friction is usually smaller than that of static friction.

(continued)

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ball and a ping-pong ball that are moving initially with equal velocities (fig. 4.7). A much greater force is required to decelerate the bowling ball than the ping-pong ball because of the difference in mass. According to the second law, the force required is proportional to the mass. In effect, we are using Newton’s second law to define mass. If we used the same force to accelerate different masses, the different accelerations could be used to compare the masses involved. If we choose one mass as a standard, any other mass can be measured against the standard mass by comparing the accelerations produced by equal forces. We could, in principle, determine the mass of any object this way.

How do we define weight? In practice, the method just described is not convenient for comparing masses because of the difficulty of measuring acceleration. The more common method of comparing masses

When the tablecloth is given a large lateral acceleration, the force needed to also accelerate the tableware ( Fnet ma) exceeds the maximum force of static friction between the dish or glass and the tablecloth. The tablecloth then begins to slide underneath the dish, reducing the size of the frictional force. If the surfaces are smooth, the frictional force is never large enough to produce an acceleration of the dish or glass that is anywhere near the size of the acceleration of the tablecloth. In the fraction of a second that this force acts, it does not have a chance to increase the velocity very much or to move the object very far. (See synthesis problem 3.) Once the tablecloth is no longer in contact with the object, the frictional force exerted by the table quickly decelerates the object. You can test these ideas yourself with a pencil, cup, or similar object (preferably nonbreakable) and a sheet of smooth tablet paper. Place the paper on a smooth desk or table surface with the end of the paper extending over the edge. Grasping the paper with both hands near the corners, as shown in the drawing, pull it downward across the edge of the desk or table. Notice that a slow pull brings the object along with the paper, but a very rapid pull leaves the object essentially in place. (The objects will usually move slightly in the direction of the pull.) Before you graduate to tablecloths and full dinner place settings, a few cautions are in order. Objects that can tip, like filled wineglasses, are more difficult to work with. The bottom

F1

m1

v

a

m2 F2

v

a

figure 4.7

Stopping a bowling ball and a ping-pong ball. A much larger force is required to produce the same rate of change in velocity for the larger mass.

Grasp the paper near the corners and pull slightly downward across the edge of the table. A quick pull will leave the pencil near its initial position.

may start to move while the top portion (with its greater inertia) remains in place causing the glass to tip and spill the wine or water. Also, the larger the tablecloth, the more difficult it is to pull it clear of the table—your hands must move very rapidly through a large distance in the pull. Practice is essential, which is the case for most of the tricks that magicians (and physics instructors) perform.

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is to “weigh” the objects on a balance or scale (fig. 4.8). What we actually do in weighing is to compare the gravitational force acting on the mass we wish to measure with that acting on some standard mass. The gravitational force acting on an object is the weight of the object. As a force, weight has different units (newtons) than mass (kilograms). How is weight related to mass? From our discussion of gravitational acceleration in chapter 3, we know that objects of different mass experience the same gravitational acceleration near the Earth’s surface (g 9.8 m/s2). This acceleration is caused by the gravitational force exerted by the Earth on the object, which is the weight of the object. By Newton’s second law, the force (the weight) is equal to the mass times the acceleration or W mg .

example box 4.2 Sample Exercise: Computing Weights Suppose that a woman has a mass of 50 kg. What is her weight in a. newtons? b. pounds? a. m 50 kg W? b. W ? in pounds

figure 4.8

Wunknown

Comparing an unknown mass to standard masses on a balance.

W

490 N 4.45 N/lb

110 lb

The gravitational acceleration on the moon is approximately one-sixth that on the Earth’s surface. If we transported the woman whose weight we have just determined to the moon, her weight would decrease to about 18 lb (or 82 N), one-sixth her weight on Earth. The woman’s mass would still be 50 kg, provided that the trip did not take too much out of her. The mass of an object changes only if we add or subtract matter from it.

Why is the gravitational acceleration independent of mass? The distinction between weight and mass can provide insight into why the gravitational acceleration is independent of mass. Let’s turn to the case of a falling object and consider its motion using Newton’s second law. Reversing the argument that we used in defining weight, we use the gravitational force (the weight) to determine the acceleration. By Newton’s second law, the acceleration can be found by dividing the force (W mg) by the mass: a

Wstandards

(50 kg)(9.8 m/s2) 490 N

1 lb 4.45 N

The symbol W represents the weight. It is a vector whose direction is straight down toward the center of the Earth. If we know the mass of an object, we can then compute its weight. An example is provided in example box 4.2, where we show that a woman with a mass of 50 kg has a weight of 490 N. Since we are more used to expressing weights in the English system, we also convert her weight in newtons to pounds (lb), which yields a weight of 110 lb. The pound is most commonly used as a unit of force, not mass, in the English system. A mass of 1 kg weighs approximately 2.2 lb near the Earth’s surface. Although weight is proportional to mass, it also depends on the gravitational acceleration g. Since g varies slightly from place to place on the surface of the Earth—and has a much smaller value on the moon or the smaller planets— the weight of an object clearly depends on where that object is. On the other hand, the mass of an object is a property of the object related to the quantity of matter making up that object and does not depend on the location of the object.

W mg

mg g. m

Mass cancels out of the equation when we compute the acceleration for a falling object. The gravitational force is proportional to the mass, but by Newton’s second law, the acceleration is inversely proportional to the mass: these two effects cancel one another. This only holds true for falling objects. In most other cases, the net force does not depend directly on the mass. Force and acceleration are not the same, although they are closely related by Newton’s second law. A heavy object experiences a larger gravitational force (its weight) than a

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m2 m1

F=W a=g a=g F=W

figure 4.9

Different gravitational forces (weights) act on falling objects of different masses, but because acceleration is inversely proportional to mass, the objects have the same acceleration.

lighter object, but the two objects will have the same gravitational acceleration (fig. 4.9). Because the gravitational force is proportional to mass, we find the same acceleration for different masses. The gravitational force will be discussed further in chapter 5 when we take up Newton’s law of gravitation, a critical piece of his overall theory of motion.

Weight and mass are not the same. Weight is the gravitational force acting on an object, and mass is an inherent property related to the amount of matter in the object. Near the Earth’s surface, weight is equal to the mass multiplied by the gravitational acceleration (W mg), but the weight would change if we took the object to another planet where g has a different value. The reason that all objects experience the same gravitational acceleration near the Earth’s surface is that the gravitational force is proportional to the mass of the object, but acceleration is equal to the force divided by the mass.

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How does the third law help us to define force? If you push with your hand against a large chair or any large object, such as the wall of your room, you will feel the object push back against your hand. A force is acting on your hand that you can sense as it compresses your hand. Your hand is interacting with the chair or wall, and that object pushes back against your hand as you push against the object. Newton’s third law contains the idea that forces are caused by such interactions of two objects, each exerting a force on the other. It can be stated as If object A exerts a force on object B, object B exerts a force on object A that is equal in magnitude but opposite in direction to the force exerted on B.

The third law is sometimes referred to as the action/ reaction principle—for every action there is an equal but opposite reaction. Note that the two forces always act on two different objects, never on the same object. Newton’s definition of force includes the idea of an interaction between objects. The forces represent that interaction. If you exert a force F1 on the chair with your hand, the chair pushes back on your hand with a force F2 that is equal in size, but opposite in direction (fig. 4.10). Using this notation, Newton’s third law can be stated in symbolic form as F2 F1. The minus sign indicates that the two forces have opposite directions. The force F2 acts on your hand and partly determines your own motion, but it has nothing to do with the motion of the chair. Of this pair of forces, the only one that

F2 = – F1 F1

4.4 Newton’s Third Law Where do forces come from? If you push on a chair to move it across the floor, does the chair also push back on you? If so, how does that push affect your own motion? Questions like these are important to what we mean by force. Newton’s third law provides some answers. Newton’s third law of motion is an important part of his definition of force. It is an essential tool for analyzing the motion or lack of motion of real objects, but it is often misunderstood. For this reason, it is good to take a careful look at the statement and use of the third law.

F2

figure

4.10 The chair pushes back on the hand with a force F2 that is equal in size but opposite in direction to the force F1 exerted by the hand on the chair.

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affects the motion of the chair is the one acting on the chair, F1. Our definition of force is now complete. Newton’s second law tells us how the motion of an object is affected by a force, and his third law tells where forces come from. They come from interactions with other objects. With a suitable definition of mass, which also depends upon the second law, we know how to measure the size of forces by determining the acceleration that they produce (F ma). Both the second and third laws are necessary to define what we mean by force.

How can we use the third law to identify forces? How do we identify the forces that act on an object to analyze how that object will move? First, we identify other objects that interact with the object of interest. Consider a book lying on a table (fig. 4.11). What objects are interacting with the book? Since it is in direct contact with the table, the book must be interacting with the table, but it also interacts with the Earth through the gravitational attraction. The downward pull of gravity that the Earth exerts on the book is the book’s weight W. The object interacting with the book to produce this force is the Earth itself. The book and the Earth are attracted to one another (through gravity) with equal and opposite forces that form a third-law pair. The Earth pulls down on the book with the force W, and the book pulls upward on the Earth with the force W. Because of the Earth’s enormous mass, the effect of this upward force on the Earth is extremely small. The second force acting on the book is an upward force exerted on the book by the table. This force is often called the normal force, where the word normal means “perpendicular”

Can a mule accelerate a cart? Consider the story of the stubborn mule who, having had a brief exposure to physics, argued to his handler that there was no point in pulling on the cart to which he was connected. According to Newton’s third law, the mule argued, the harder he pulls on the cart, the harder the cart pulls back on him (fig. 4.13). The net result is, therefore, nothing. Is he right, or is there a fallacy in his argument?

–N

N

rather than “ordinary” or “usual.” The normal force N is always perpendicular to the surfaces of contact. The book, in turn, exerts an equal but oppositely directed downward force N on the table. These two forces, N and N, constitute another third-law pair. They result from the mutual compression of the book and table as they come into contact with one another. You could think of the table as a large and very stiff spring that compresses ever so slightly when the book is placed on it (fig. 4.12). The two forces acting on the book, the force of gravity and the force exerted by the table, also happen to be equal in size and opposite to one another, but this is not due to the third law. How do we know that they must be equal? Since the book’s velocity is not changing, its acceleration must be zero. According to Newton’s second law, the net force Fnet acting on the book must then be zero, since Fnet ma and the acceleration a is zero. The only way that the net force can be zero is for the two contributing forces, W and N, to cancel one another. They must be equal in magnitude and opposite in direction for their sum to be zero. Even though equal in size and opposite in direction, these two forces do not constitute a third-law action/reaction pair. They both act on the same object, the book, and the third law always deals with interactions between different objects. So, W and N are equal in size and opposite in direction in this case as a consequence of the second law rather than the third law. If they did not cancel one another, the book would accelerate away from the tabletop. (Both the second and third laws are critical to the analysis of the elevator example in everyday phenomenon box 4.2.)

W

–W Earth

figure 4.11

Two forces, N and W, act on a book resting on a table. The third-law reaction forces N and W act on different objects, the table and the Earth.

figure

4.12 An uncompressed spring and the same spring supporting a book. The compressed spring exerts an upward force on the book.

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everyday phenomenon

box 4.2

Riding an Elevator The Situation. We have all had the experience of riding an elevator and feeling sensations of heaviness or lightness as the elevator accelerates up or down. The feeling of lightness as the elevator accelerates downward is generally more striking, particularly if the acceleration is not smooth. Do we really weigh more or less than usual in these situations? If you took a bathroom scale into the elevator, would it read your true weight when the elevator is accelerating? How can we apply Newton’s laws of motion to explore these questions?

a

W

N

a

A free-body diagram of the woman in the elevator when accelerating upward. Why is the normal force N larger than the weight W?

A woman standing on a bathroom scale inside an accelerating elevator. Will she read her true weight on the scale?

The Analysis. The first step in analyzing any situation using Newton’s laws is to isolate the body of interest and carefully identify the forces that act on just that body. Different choices are possible for which objects to isolate, but some choices will be more productive than others. In this case, it makes sense to isolate the person standing on the scale, since her weight is the focus of our questions. The second drawing shows a free-body diagram of the woman indicating just those forces that act on her. In this case, just two other objects interact with the woman, resulting in two forces. The Earth pulls downward on the woman through the force of gravity W. The scale pushes upward on her feet with a force N, the normal force. The vector sum of these two forces determines her acceleration. If the elevator is accelerating upward with an acceleration a, the woman must also be accelerating upward at that rate. The net force must also be upward, which implies that the normal force N is larger than the gravitational force W. Using signs to indicate direction, and letting the positive direction be upward, Newton’s second law requires that Fnet N W ma.

What about the scale reading? By Newton’s third law, the woman exerts a downward force on the scale equal in size to the normal force N, but opposite in direction. Since this is the force pushing down on the scale, the scale should read the value N, the magnitude of the normal force. The woman’s true weight has not changed, but her apparent weight as measured by the scale has increased by an amount equal to ma. (Rearranging the second-law equation yields N W ma.) What happens when the elevator is accelerating downward? In that case, the net force acting upon the woman must be downward, and the normal force must be less than her weight. The scale reading N will then be less than the woman’s true weight by the amount ma, perhaps producing a smile rather than a scowl. If the elevator cable breaks, we have a particularly interesting special case. Both the woman and the elevator will accelerate downward with the gravitational acceleration g. Since the woman’s weight is all that is required to give her that acceleration, the normal force acting on her feet must then be zero. The scale reading will likewise be zero, and the woman is apparently weightless! The sensation of our own weight is produced in part by the pressure on our feet and forces in our leg muscles needed to maintain our posture. The woman will feel weightless in this situation even though her true weight (the gravitational force acting on her) has not changed. In fact, she would be able to float around in the elevator as the astronauts do in the orbiting space shuttle. (The space shuttle is also falling toward the Earth as it moves laterally in its orbit.) This happy scenario will come to a crashing halt for the woman, however, when the elevator reaches the bottom of the shaft.

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–F1

F1

f –f

figure 4.13

A mule and a cart. Does Newton’s third law prevent the mule from moving the cart?

4.14 The car pushes against the road, and the road, in turn, pushes against the car.

The fallacy is simple but perhaps not obvious. The motion of the cart is affected by only one of the two forces that the mule is talking about, namely, the force that acts on the cart. The other force in this third-law pair acts on the mule and must be considered in conjunction with other forces that act on the mule to determine how he will move. The cart will accelerate if the force exerted by the mule on the cart is larger than the frictional forces acting on the cart. Try placing yourself in the role of the handler and explain the fallacy to the mule.

recognizable with a little thought. The third law is the principle we use to identify any of these forces.

What force causes a car to accelerate?

4.5 Applications of Newton’s Laws

As with the mule, the reaction force to a push or pull exerted by an object is often extremely important in describing the motion of the object itself. Consider the acceleration of a car. The engine cannot push the car because it is part of the car. The engine drives either the rear or front axle of the car, which causes the tires to rotate. The tires in turn push against the road surface through the force of friction f between the tires and the road (fig. 4.14). According to Newton’s third law, the road must then push against the tires with an equal but oppositely directed force f. This external force causes the car to accelerate. Obviously, friction is desirable in this case. Without friction, the tires would spin, and the car would go nowhere. The case of the mule is similar. The frictional force exerted by the ground on his hooves causes him to accelerate forward. This frictional force is the reaction to his pushing against the ground. Think about this next time you find yourself walking. What external force causes you to accelerate as you start out? What is your role and that of friction in producing this force? How would you walk on an icy or slippery surface? To figure out what forces are acting on any object, we need first to identify the other objects with which it is interacting. Some of these will be obvious. Any object in direct contact with the object of interest will presumably contribute a force. Interactions producing other forces, such as air resistance or gravity, may be less obvious but still

figure

Newton’s third law of motion completes his definition of force. The third law notes that forces arise from interactions between different objects. If object A exerts a force on object B, object B exerts an equal-size but oppositely directed force on A. We use the third law to identify the external forces that act on an object in order to apply the second law of motion.

We have now introduced Newton’s laws of motion and discussed the definitions of force and mass within these laws. To appreciate their usefulness, however, we must be able to apply them to some familiar examples such as pushing a chair or throwing a ball. How do Newton’s laws help us make sense of these motions? Do they provide a satisfactory picture of what is going on?

What forces are involved in moving a chair? We have returned from time to time to the example of a chair being pushed but have not yet analyzed how and why it moves. As we indicated in section 4.4, the first step in any analysis is to identify the forces that act on the chair. As shown in figure 4.15, four forces act on the chair from four separate interactions:* 1. The force of gravity (the weight) W due to interaction with the Earth. 2. The upward (normal) force N exerted by the floor due to compression of the floor. 3. The force exerted by the hand of the person pushing, P. 4. The frictional force f exerted by the floor. Two of these forces, the normal force N and the frictional force f, are actually due to interactions with a single object, the *A figure such as figure 4.15, showing all the forces acting on an object, is often called a free body diagram. See also everyday phenomenon box 4.2 on page 69.

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The size of the force needed to keep the chair moving with constant velocity is determined by the strength of the frictional force, which, in turn, is influenced by the weight of the chair and the condition of the floor surface. If you fail to recognize the importance of the frictional force, you may be led, like Aristotle, to think that a force is always needed to keep an object moving. Frictional forces are almost always present, but they are not as obvious as the forces applied directly.

P

f

Does a sky diver continue to accelerate?

W N

figure

4.15 Four forces act on a chair being pushed across the floor, the weight W, the normal force N, the force P exerted by the person pushing, and the frictional force f. floor. Since they are due to different effects and are perpendicular to one another, they are usually treated separately. The effects of the two vertical forces acting on the chair, the weight W and the normal force N, cancel one another. Like the book on the table in section 4.4, this results because there is no acceleration of the chair in the vertical direction. By Newton’s second law, the sum of the vertical forces must then be zero, which implies that the weight W and the normal force N are equal in size but opposite in direction. They play no direct role in the horizontal motion of the chair. The other two forces, the push of the hand P and the frictional force f, do not necessarily cancel. These two forces together determine the horizontal acceleration of the chair. The push P must be larger than the frictional force f for the chair to accelerate. In the most likely scenario for moving the chair, you first give a push with your hand that is larger than the frictional force. This produces a total force, with magnitude P f, in the forward direction, causing the chair to accelerate. Once you have accelerated the chair to a reasonable velocity, you reduce the strength of your push P so that it is equal in size to the frictional force. The net horizontal force becomes equal to zero, and the horizontal acceleration is also zero by Newton’s second law. If you sustain the push at this level, the chair moves across the floor with constant velocity. Finally, you remove your hand and its push P, and the chair quickly decelerates to zero velocity under the influence of the frictional force f. If you happen to have a chair and a smooth floor handy, try to produce the motion that we have just been describing. See if you can feel differences in the force that you are exerting with your hand at various points in the motion. The force should be largest at the beginning of the motion.

In chapter 3, we considered the fact that an object falls with constant acceleration g if air resistance is not a significant factor. What about objects such as sky divers who fall for large distances? Do they continue to accelerate at this rate gaining larger and larger downward velocities? Any person with experience in sky diving knows that this does not happen. Why not? If air resistance were not a factor, a falling object would experience only the gravitational force (its weight) and would indeed continue to accelerate. In sky diving, air resistance is an important factor, and its effects get larger as the velocity of the sky diver (or any object) increases. The sky diver has an initial acceleration of g, but as her velocity increases, the force of air resistance becomes significant. Her acceleration decreases (fig. 4.16). For small velocities, the air-resistive force R is small, and the weight is the dominant force. As the velocity increases, the air-resistive force gets larger, causing the total magnitude of the downward force, W R, to decrease. Since the net force is responsible for the acceleration, the acceleration will also decrease. Ultimately, as the velocity continues to increase, the air-resistive force reaches a value equal in size to the gravitational force. The net force is then zero, and the sky diver stops accelerating. We say that she has reached terminal velocity, and from there on, she moves downward with constant velocity. This terminal velocity is usually between 100 and 120 MPH. Frictional or resistive forces play a critical role in analyzing the motion. Aristotle did not have the opportunity to try sky diving (nor have many of us), so this example was not a part of his experience. He did observe the terminal velocity, however, of very light objects such as feathers or leaves. The weight of such objects is small and the surface area is large relative to the weight, so the air-resistive force R becomes equal in size to the weight much sooner than for a heavier object. Try tearing a small corner from a piece of paper and watching it fall. Does it appear to reach a constant (terminal) velocity? It will flutter as it falls, but it does not seem to accelerate much for most of its downward motion. You can see why Aristotle concluded that heavier objects fall faster than lighter objects. Dropping heavier objects through water can also show the terminal velocity. Water exerts a larger resistive force at lower velocities than air.

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figure

the ball to a velocity that we often refer to as the initial velocity. The magnitude and direction of the initial velocity are determined by the strength and direction of the force P and the length of time that it acts on the ball. Since this force usually varies with time, a full analysis of the process of throwing gets quite complex. Once the ball leaves the hand, however, we are in the second time period, where P is no longer a consideration. During this interval, the gravitational force W and the airresistive force R produce changes in the ball’s velocity. From this point on, the problem becomes one of projectile motion (section 3.4). The gravitational force accelerates the ball downward, and the air-resistive force acts in a direction opposite to the velocity, gradually reducing the ball’s velocity. Contrary to Aristotle’s view, no forces are needed to keep the ball moving once it has been thrown. In fact, if an object is thrown in deep space, where air resistance is nonexistent and gravitational forces are very weak, it would keep moving with constant velocity, as stated in Newton’s first law. So, be careful with your tools when you are working in space outside of your spacecraft. Because the air-resistive force or the push exerted by a person throwing a ball varies with time, we have avoided working out numerical examples for these situations. Just identifying the forces involved and their causes due to third-law interactions with other objects provides a useful description of what is happening.

What happens when a ball is thrown?

How do we analyze the motion of connected objects?

Aristotle had trouble explaining the motion of a thrown object such as a ball, once it had left the thrower’s hand. Let’s reconsider this example from a Newtonian perspective. Do we need a force to keep the ball moving? Not according to Newton’s first law. Three forces, however, are involved in the flight of the ball: the initial push by the thrower, the downward pull of gravity, and (once again) air resistance (fig 4.17). To highlight Newton’s approach, it is best to break the motion down into two different spans of time. The first is the process of throwing, when the hand is in contact with the ball. During this interval, the force P exerted by the hand dominates the motion. The combined effects of the other forces (gravity and air resistance) must be smaller than the force P if the ball is to accelerate. Thus P accelerates

Verification of Newton’s laws of motion came initially from simpler examples that can be easily set up in the laboratory. One example not difficult to picture and set up in a physics laboratory (or even at home if suitable toys are available) is two connected carts accelerated by the pull of a string (fig. 4.18). To keep things simple, we will assume that the carts have excellent wheel bearings, so that they roll with very little friction. We will also assume that a scale is available to determine the masses of the carts and their contents. To measure the magnitude of the force applied by the string, we would have to insert a small spring balance somewhere between the hand and the carts. The trickiest part of the entire experiment is applying a steady force with this arrangement while the carts are accelerating.

R

W

R

W

R

W

4.16 The force of air resistance R acting on a sky diver increases as the velocity increases.

R W R P

W W

figure 4.17

Three forces act on a thrown ball, the initial push P, the weight W, and air resistance R.

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example box 4.3 m2

m1

F = 36 N

figure

4.18 Two connected carts being accelerated by a force F applied by a string. If we know the masses of the carts and their contents, and the magnitude of the force applied by the string, we should be able to predict the value of the acceleration of the system from Newton’s second law. (See example box 4.3.) For the masses given in the example, and an applied force of 36 N, we find an acceleration of 2.0 m/s2 for the two carts. The acceleration could be verified experimentally by measuring the time required for the carts to travel a fixed distance and using the equations developed for constant acceleration in chapter 2 to calculate an experimentally determined value. In example box 4.3, we first treated the two carts as a single system to find the acceleration. Suppose, however, that we wanted to know the magnitude of the force exerted by the hooks connecting the two carts. In this case, it makes sense to treat the motion of the individual carts separately. Once we know the acceleration, we again apply Newton’s second law to find the net force acting on each cart. This computation is done in the second part of example box 4.3 and is illustrated in figure 4.19. For the second cart, a force of 16 N is required to produce the acceleration of 2 m/s2. By Newton’s third law, there should then be a force of 16 N pulling back on the first cart. Combined with the forward force of 36 N applied by the string, this results in a net force of 20 N acting on the first cart (36 N 16 N). This is exactly the value required to give the first cart an acceleration of 2 m/s2. From this example, we see that Newton’s laws provide a completely consistent picture of the forces and accelerations of the different parts of the connected-cart system. This is a necessary condition for us to accept the laws as valid. Obviously, another condition is that any predictions be confirmed by experimental measurements. This has been done many times over by experiments similar to the one we have dealt with here. We could try many variations on this experiment in the laboratory to see if the results agree with predictions derived from Newton’s laws. Even with careful experimental technique using accurate stopwatches and balances, however, our results are unlikely to agree exactly with our predictions. It is impossible to eliminate the effects of friction completely, and none of our measurements can be made with infinite precision. The art of the experimentalist is to reduce these inaccuracies to a minimum as well as to predict how they affect our results.

Sample Exercise: Connected Objects Two connected carts are pulled across the floor under the influence of a force of 36 N applied by a string (fig. 4.18). The forward cart and its contents have a mass of 10 kg, and the second cart and contents have a mass of 8 kg. Assuming that frictional forces are negligible: a. What is the acceleration of the two carts? b. What is the net force acting on each cart? a. Defining the system as both carts, as discussed in text: m1 10 kg Fnet ma m2 8 kg Fnet 36 N F 36 N or: a m 10 kg 8 kg a ? 36 N 2.0 m/s2 18 kg a 2.0 m/s2 in the forward direction b. Treating each cart separately: first cart Fnet ? (for each cart) Fnet ma (10 kg)(2 m/s2) 20 N second cart Fnet ma (8 kg)(2 m/s2) 16 N

m2

16 N

16 N

m1

36 N

figure

4.19 The interaction between the two carts illustrates Newton’s third law. Newton’s laws of motion provide both qualitative and quantitative explanations of any familiar motion. First, we identify the forces acting on the object by examining interactions with other objects. The relative sizes of these forces, when added together, give the acceleration of the object. The acceleration may change as the forces change with time, as in the case of a sky diver. Newton’s laws have been verified many times over by experimental tests of their quantitative predictions. They are a much more consistent theory of the causes of motion than the older Aristotelian view.

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summary In 1685, Newton published his Principia, in which he introduced three laws of motion as the foundation of his theory of mechanics. These laws continue to serve as an extremely useful model for explaining the causes of motion and for predicting how objects will move in many familiar situations.

1

4

Newton’s third law. Newton’s third law completes the definition of force by showing that forces result from interactions between objects. If object A exerts a force on object B, then object B exerts an equal-size but oppositely directed force on object A.

A brief history. Newton’s theory was constructed on ground-work laid by Galileo and replaced a much earlier and less quantitative model developed by Aristotle to explain motion. Newton’s theory had much greater predictive power than Aristotle’s ideas. Although we now recognize its limitations, Newton’s theory is still used extensively to explain the motion of ordinary objects.

A

FA

FB

2

Newton’s first and second laws. Newton’s second law states that the acceleration of an object is proportional to the net external force acting on that object and inversely proportional to the mass of the object. The first law, a special case of the second law, describes what happens when the net force is zero. The acceleration must then be zero, and the object moves with constant velocity.

B

FA = – FB

5

Applications of Newton’s laws. In analyzing the motion of an object using Newton’s laws, the first step is to identify the forces that act on the object due to interactions with other objects. The strength and direction of the net force then determine how the object’s motion will change.

Fnet

m

P

a

a=

Fnet m

3 Mass and weight. Newton’s second law defines the inertial mass of an object as the property that causes the object to resist a change in its motion. The weight of an object is the gravitational force acting on the object and is equal to the mass multiplied by the gravitational acceleration g. The weight of an object may vary as g varies, but mass is an inherent property of the object related to its quantity of matter.

f W N

m

W

W = mg

key terms Force, 61 Newton’s first law of motion, 62 Frictional force, 62 Newton’s second law of motion, 62 Net force, 62

Mass, 63 Inertia, 63 Newton (unit of force), 63 Weight, 66 Newton’s third law of motion, 67

Action/reaction principle, 67 Normal force, 68 Free-body diagram, 69 Reaction force, 70 Terminal velocity, 71

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questions * more open-ended questions, requiring lengthier responses, suitable for group discussion Q sample responses are available in appendix D Q sample responses are available on the website

Q1. Did Galileo’s work on motion precede in time that of Aristotle or Newton? Explain. Q2. Why did Aristotle believe that heavier objects fall faster than lighter objects? Explain. Q3. Aristotle believed that a force was necessary to keep an object moving. Where, in his view, did this force come from in the case of a ball moving through the air? Explain. *Q4. How did Aristotle explain the continued motion of a thrown object. Does this explanation seem reasonable to you? Explain. Q5. Did Galileo develop a more complete theory of motion than that of Newton? Explain. Q6. Two equal forces act on two different objects, one of which has a mass ten times as large as the other. Will the more massive object have a larger acceleration, an equal acceleration, or a smaller acceleration than the less massive object? Explain. Q7. A 3-kg block is observed to accelerate at a rate twice that of a 6-kg block. Is the net force acting on the 3-kg block therefore twice as large as that acting on the 6-kg block? Explain. Q8. Two equal-magnitude horizontal forces act on a box as shown in the diagram. Is the object accelerated horizontally? Explain. –F

F

Q8 Diagram Q9. Is it possible that the object pictured in question 8 is moving, given the fact that the two forces acting on it are equal in size but opposite in direction? Explain. Q10. Suppose that a bullet is fired from a rifle in outer space where there are no appreciable forces due to gravity or air resistance acting on the bullet. Will the bullet slow down as it travels away from the rifle? Explain. Q11. Two equal forces act on an object in the directions pictured in the diagram below. If these are the only forces involved, will the object be accelerated? Explain, using a diagram.

F1

Q13. A car goes around a curve traveling at constant speed. a. Is the acceleration of the car zero in this process? Explain. b. Is there a non-zero net force acting on the car? Explain. *Q14. Is Newton’s first law of motion explained by the second law? Explain. Why did Newton state the first law as a separate law of motion? Q15. Is the mass of an object the same thing as its weight? Explain. Q16. The gravitational force acting on a lead ball is much larger than that acting on a wooden ball of the same size. When both are dropped, does the lead ball accelerate at the same rate as the wooden ball? Explain, using Newton’s second law of motion. Q17. The acceleration due to gravity on the moon is approximately one-sixth the gravitational acceleration near the Earth’s surface. If a rock is transported from Earth to the moon, will either its mass or its weight change in the process? Explain. Q18. Is mass a force? Explain. Q19. Two identical cans, one filled with lead shot and the other with feathers, are dropped from the same height by a student standing on a chair. a. Which can, if either, experiences the greater force due to the gravitational attraction of the Earth? Explain. b. Which can, if either, experiences the greater acceleration due to gravity? Explain. Q20. A boy sits at rest on the floor. What two vertical forces act upon the boy? Do these two forces constitute an action/ reaction pair as defined by Newton’s third law of motion? Explain. Q21. The engine of a car is part of the car and cannot push directly on the car in order to accelerate it. What external force acting on the car is responsible for the acceleration of the car on a level road surface? Explain. Q22. It is difficult to stop a car on an icy road surface. Is it also difficult to accelerate a car on this same icy road? Explain. Q23. A ball hangs from a string attached to the ceiling, as shown in the diagram. a. What forces act on the ball? How many are there? b. What is the net force acting on the ball? Explain. c. For each force identified in part (a), what is the reaction force described by Newton’s third law of motion?

F2

Q11 Diagram Q12. An object moving horizontally across a table is observed to slow down. Is there a non-zero net force acting on the object? Explain.

m

Q23 Diagram

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*Q24. When a magician performs the tablecloth trick, the objects on the table do not move very far. Is there a horizontal force acting on these objects while the tablecloth is being pulled off the table? Why do the objects not move very far? Explain. Q25. A sprinter accelerates at the beginning of a 100-meter race and then tries to maintain maximum speed throughout the rest of the race. a. What external force is responsible for accelerating the runner at the beginning of the race? Explain carefully how this force is produced. b. Once the runner reaches her maximum velocity, is it necessary to continue pushing against the track in order to maintain that velocity? Explain. Q26. A mule is attempting to move a cart loaded with rock. Since the cart pulls back on the mule with a force equal in size to the force that the mule exerts on the cart (according to Newton’s third law), is it possible for the mule to accelerate the cart? Explain.

Q30. Two blocks with the same mass are connected by a string and are pulled across a frictionless surface by a constant force, F, exerted by a string (see diagram). a. Will the two blocks move with constant velocity? Explain. b. Will the tension in the connecting string be greater than, less than, or equal to the force F? Explain.

F

Q30 Diagram *Q31. Suppose that a sky diver wears a specially lubricated suit that reduces air resistance to a small constant force that does not increase as the diver’s velocity increases. Will the sky diver ever reach a terminal velocity before opening her parachute? Explain.

Q27. The upward normal force exerted by the floor on a chair is equal in size but opposite in direction to the weight of the chair. Is this equality an illustration of Newton’s third law of motion? Explain.

Q32. If you get into an elevator on the top floor of a large building and the elevator begins to accelerate downward, will the normal force pushing up on your feet be greater than, equal to, or less than the force of gravity pulling downward on you? Explain.

Q28. A toy battery-powered tractor pushes a book across a table. Draw separate diagrams of the book and the tractor identifying all of the forces that act upon each object. What is the reaction force described by Newton’s third law of motion for each of the forces that you have drawn?

Q33. If the elevator cable breaks and you find yourself in a condition of apparent weightlessness as the elevator falls, is the gravitational force acting upon you equal to zero? Explain.

Q29. Two masses, m1 and m2, connected by a string, are placed upon a fixed frictionless pulley as shown in the diagram. If m2 is larger than m1, will the two masses accelerate? Explain.

m2 m1

Q29 Diagram

exercises E1. A single force of 40 N acts upon a 5-kg block. What is the magnitude of the acceleration of the block? E2. A ball with a mass of 2.5 kg is observed to accelerate at a rate of 6.0 m/s2. What is the size of the net force acting on this ball? E3. A net force of 20 N acting on a wooden block produces an acceleration of 4.0 m/s2 for the block. What is the mass of the block?

E4. A 3.0-kg block being pulled across a table by a horizontal force of 80 N also experiences a frictional force of 5 N. What is the acceleration of the block? E5. A pulled tablecloth exerts a frictional force of 0.6 N on a plate with a mass of 0.4 kg. What is the acceleration of the plate?

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Synthesis Problems

E6. A 6-kg block being pushed across a table by a force P has an acceleration of 3.0 m/s2. a. What is the net force acting upon the block? b. If the magnitude of P is 20 N, what is the magnitude of the frictional force acting upon the block? E7. Two forces, one of 50 N and the other of 30 N, act in opposite directions on a box as shown in the diagram. What is the mass of the box if its acceleration is 4.0 m/s2? 50 N 30 N

E7 Diagram E8. A 4-kg block is acted upon by three horizontal forces as shown in the diagram. a. What is the net horizontal force acting on the block? b. What is the horizontal acceleration of the block? 5N

10 N 2 kg

25 N

E8 Diagram E9. A 4-kg sled sliding freely on an icy surface experiences a 2-N frictional force exerted by the ice and an air-resistive force of 0.5 N. a. What is the net force acting on the sled? b. What is the acceleration of the sled? E10. What is the weight of a 40-kg mass? E11. What is the mass of a 196-N weight? E12. Jennifer has a weight of 110 lb. a. What is her weight in newtons? (1 lb 4.45 N) b. What is her mass in kilograms?

77

E13. The author of this text has a weight of 600 N. a. What is his mass in kilograms? b. What is his weight in pounds? (1 lb 4.45 N) E14. Who has the larger mass, a man weighing 145 lb or one weighing 735 N? E15. At a given instant in time, a 4-kg rock that has been dropped from a high cliff experiences a force of air resistance of 15 N. What are the magnitude and direction of the acceleration of the rock? (Do not forget the gravitational force!) E16. At a given instant in time, a 5-kg rock is observed to be falling with an acceleration of 7.0 m/s2. What is the magnitude of the force of air resistance acting upon the rock at this instant? E17. A 0.5-kg book rests on a table. A downward force of 6 N is exerted on the top of the book by a hand pushing down on the book. a. What is the magnitude of the gravitational force acting upon the book? b. What is the magnitude of the upward (normal) force exerted by the table on the book? (Is the book accelerated?) E18. An upward force of 18 N is applied via a string to lift a ball with a mass of 1.5 kg. a. What is the net force acting upon the ball? b. What is the acceleration of the ball? E19. A 60-kg woman in an elevator is accelerating upward at a rate of 1.2 m/s2. a. What is the net force acting upon the woman? b. What is the gravitational force acting upon the woman? c. What is the normal force pushing upward on the woman’s feet?

synthesis problems SP1. A constant horizontal force of 30 N is exerted by a string attached to a 5-kg block being pulled across a tabletop. The block also experiences a frictional force of 5 N due to contact with the table. a. What is the horizontal acceleration of the block? b. If the block starts from rest, what will its velocity be after 3 seconds? c. How far will it travel in these 3 seconds? SP2. A rope exerts a constant horizontal force of 250 N to pull a 60-kg crate across the floor. The velocity of the crate is observed to increase from 1 m/s to 3 m/s in a time of 2 seconds under the influence of this force and the frictional force exerted by the floor on the crate. a. What is the acceleration of the crate? b. What is the net force acting upon the crate? c. What is the magnitude of the frictional force acting on the crate? d. What force would have to be applied to the crate by the rope in order for the crate to move with constant velocity? Explain.

SP3. A dish with a mass 0.4 kg has a force of kinetic friction of 0.15 N exerted on it by a moving tablecloth for a time of 0.2 s. a. What is the acceleration of the dish? b. What velocity does it reach in this time, starting from rest? c. How far (in cm) does the dish move in this time? SP4. A 60-kg crate is lowered from a loading dock to the floor using a rope passing over a fixed support. The rope exerts a constant upward force on the crate of 500 N. a. Will the crate accelerate? Explain. b. What are the magnitude and direction of the acceleration of the crate? c. How long will it take for the crate to reach the floor if the height of the loading dock is 1.4 m above the floor? d. How fast is the crate traveling when it hits the floor?

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SP5. Two blocks tied together by a horizontal string are being pulled across the table by a horizontal force of 30 N as shown. The 2-kg block has a 6-N frictional force exerted on it by the table, and the 4-kg block has an 8-N frictional force acting on it. a. What is the net force acting on the entire two-block system? b. What is the acceleration of this system? c. What force is exerted on the 2-kg block by the connecting string? (Consider only the forces acting on this block. Its acceleration is the same as that of the entire system.) d. Find the net force acting on the 4-kg block and calculate its acceleration. How does this value compare to that found in part b? 2 kg

4 kg 30 N

6N

8N

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SP5 Diagram

SP6. A 60-kg man is in an elevator that is accelerating downward at the rate of 1.4 m/s2. a. What is the true weight of the man in newtons? b. What is the net force acting on the man required to produce the acceleration?

c. What is the force exerted on the man’s feet by the floor of the elevator? d. What is the apparent weight of the man in newtons? (This is the weight that would be read on the scale dial if the man were standing on a bathroom scale in the accelerating elevator.) e. How would your answers to parts b through d change if the elevator were accelerating upward with an acceleration of 1.4 m/s2? SP7. A sky diver has a weight of 750 N. Suppose that the airresistive force acting on the diver increases in direct proportion to his velocity such that for every 10 m/s that the diver’s velocity increases, the force of air resistance increases by 100 N. a. What is the net force acting on the sky diver when his velocity is 40 m/s? b. What is the acceleration of the diver at this velocity? c. What is the terminal velocity of the sky diver? d. What would happen to the velocity of the sky diver if for some reason (perhaps a brief down draft) his velocity exceeded the terminal velocity? Explain.

home experiments and observations HE1. Collect a variety of small objects such as coins, pencils, keys, and bottle caps. Ice cubes, if they are available, also make excellent test objects. Try sliding these objects across a smooth surface such as a tabletop or floor, being as consistent as possible in the initial velocity that you give to them. a. Do the objects slide the same distance after they leave your hand? What differences are apparent, and how are they related to the nature of the surface and size of the objects? Which objects come closest to demonstrating Newton’s first law of motion? b. What factors seem to be important in reducing the frictional force between the objects and the surface upon which they are sliding? If you see some general principle at work, test this idea by finding other objects that would support your hypothesis. HE2. Place a sheet of paper under a medium-sized book lying on a smooth tabletop or desktop. a. Try to accelerate the book smoothly by exerting a constant pull on the sheet of paper. What happens if you try to accelerate the book too rapidly? Can you pull the paper cleanly from underneath the book without moving the book? Explain your observations in terms of Newton’s laws of motion. b. Repeat these observations with a few books in a stack. How does increasing the mass of the books affect the results? c. Try other objects. Which objects move the least when the paper is pulled rapidly?

HE3. Falling objects whose surface area is large relative to their weight will reach terminal velocity more readily than a ball or a rock. Test several objects, such as a balloon, small pieces of paper, plant parts (leaves, flowers, or seeds), or whatever you think might work. Do these objects reach a terminal velocity? How far does each object fall before reaching constant velocity? How does the rate of fall differ for different objects when dropped at the same time? Which of the objects tested produces the clearest demonstration of terminal velocity, showing first a brief acceleration followed by a constant velocity? HE4. Using elevators in your dormitory or other campus buildings, observe the effects of the elevator’s acceleration. Most elevators accelerate briefly as they start and again as they stop (deceleration). Express elevators in high-rise buildings are best for observing the effects of acceleration. a. If you have a bathroom scale, see how much your apparent weight differs from your true weight when the elevator is stopping or starting. Can you estimate the rate of acceleration from this information? (See everyday phenomenon box 4.2 and synthesis problem 6.) b. Try holding your arm away from your body and maintaining it in this position as the elevator accelerates. How difficult is this to do for different conditions during the motion of the elevator? Explain your observations.

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Circular Motion, the Planets, and Gravity

5

chapter

chapter overview Using the example of a ball on a string, we first examine the acceleration involved in changing the direction of the velocity in circular motion (centripetal acceleration). Then we consider the forces involved in producing a centripetal acceleration in different cases, including that of a car rounding a curve. Kepler’s laws of planetary motion will then be examined and Newton’s law of universal gravitation will be introduced to explain the motion of the planets. We will also show how this gravitational force relates to the weight of an object and the gravitational acceleration near the Earth’s surface.

chapter outline

2 3 4 5

Centripetal acceleration. How can we describe the acceleration involved in changing the direction of an object’s velocity? How does this acceleration depend on the object’s speed? Centripetal forces. What types of forces are involved in producing centripetal accelerations in different situations? What forces are involved for a car rounding a curve? Planetary motion. How do the planets move around the sun? How has our understanding of planetary motion changed historically? What are Kepler’s laws of planetary motion? Newton’s law of universal gravitation. What is the fundamental nature of the gravitational force, according to Newton? How does this force help to explain planetary motion? The moon and other satellites. How does the moon orbit the Earth? How do the orbits of artificial satellites differ from the moon and from each other?

79

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1

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“T

he car failed to negotiate the curve.” How many times have you seen a phrase like that in an accident report in the newspapers? Either the road surface was slippery or the driver was driving too fast for the sharpness of the curve. In either case, poor judgment and probably a poor sense of the physics of the situation were at work (fig. 5.1). When a car goes around a curve, the direction of its velocity changes. A change in velocity means acceleration, and by Newton’s second law, an acceleration requires a force. The situation has much in common with a ball being twirled in a circle at the end of a string and other examples of circular motion. What forces keep a car moving around a curve? How does the force required depend on the speed of the car and the sharpness of the curve? What other factors are involved? Finally, what does the car rounding a curve have in common with the ball on a string and the motions of the planets around the sun? The motions of the planets around the sun and the moon around the Earth played important roles in the development of Newton’s theory of mechanics. Newton’s law of universal gravitation was a crucial part of that theory. The gravitational force explains the behavior of objects falling near the Earth’s surface, but it also explains

figure 5.1

The car failed to negotiate the curve. Newton’s first law at work.

why the planets move in curved paths about the sun. Circular motion is a very important special case of motion in two dimensions, both in the history of physics and in our everyday experience.

5.1 Centripetal Acceleration Suppose that we attach a ball to a string and twirl the ball in a horizontal circle (fig. 5.2). With a little practice it is not hard to keep the ball moving with a constant speed, but the direction of its velocity changes continually. A change in velocity implies an acceleration, but what is the nature of this acceleration? The key to this situation involves taking a careful look at what happens to the velocity vector as the ball moves in a circle. How does this vector change as the path of the ball changes direction? Can we evaluate the size of this change and how it is related to the speed of the ball or the radius of the curve? To define the concept of centripetal acceleration, we need to answer these questions.

figure 5.2

A ball being twirled in a horizontal circle. Is the ball accelerated?

What is a centripetal acceleration? What do we have to do to get the ball on the string to change its direction? If you try twirling a ball as pictured in figure 5.2, you will feel a tension in the string. In other words, you have to apply a force by pulling on the string to cause the change in direction of the ball’s velocity. What would happen if this force were not present? According to Newton’s first law of motion, an object will continue moving in a straight line with constant speed if there is no net force acting on the object. If the string breaks, or if we let go of the string, this is exactly what

will happen. The ball will fly off in the direction that it was traveling when the string broke (fig. 5.3). Without the pull of the string, the ball will move in a straight line. It will also fall, of course, as it is pulled down by the gravitational force. According to Newton’s second law of motion, if there is a net force, there must be an acceleration (Fnet ma). This acceleration is associated with the change in the direction of the velocity vector. In the case of the ball on the string, the string pulls the ball toward the center of the circle causing

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5.1 Centripetal Acceleration

v2

81

∆v

v1

v1 v2

v

v

figure 5.3

If the string breaks, the ball flies off in a straightline path in the direction it was traveling at the instant the string broke.

the direction of the velocity vector to change continually. The direction of the force, and of the acceleration that it produces, is toward the center of the circle. We call this acceleration the centripetal acceleration:

Centripetal acceleration is the rate of change in velocity of an object that is associated with the change in direction of the velocity. Centripetal acceleration is always perpendicular to the velocity vector itself and toward the center of the curve.

To find the size of the centripetal acceleration, we need to determine how fast the velocity is changing. You might guess that this depends on how rapidly you are twirling the ball, but it also depends on the radius of the curve—the size of the circle.

How do we find the change in velocity v? Figure 5.4 shows the ball and string as seen from above. The ball is moving in a horizontal circle. Velocity vectors are drawn on the circle at two positions separated by a short time interval. The velocity v2 occurs a short time after the velocity v1, as the ball moves counterclockwise around the circle. These two vectors are drawn with the same length, indicating that the speed of the ball is unchanged. The change in velocity, v, is the difference between the initial velocity and the final velocity for a given time interval. In other words, the change in velocity is a vector that is added to the initial velocity to produce the final velocity. Adding v to v1 produces v2. This vector addition is shown in the vector triangle to the right of the circle in figure 5.4. (See appendix C for a discussion of vector addition by graphical methods.)

figure 5.4

The velocity vectors for two positions of a ball moving in a horizontal circle. The change in velocity, v, adds to v1 to yield v2.

Note that the vector v has a direction different from either of the velocity vectors. If we choose a short enough time interval between the two positions, the direction of the change in velocity points toward the center of the circle, the direction of the instantaneous acceleration of the ball. (Acceleration always has the same direction as the change in velocity.) The ball is being accelerated toward the center of the circle, the direction of the tension in the string.

What is the size of the centripetal acceleration? But how large is this centripetal acceleration, and how does it depend on the speed of the ball and the radius of the curve? The triangle illustrating the vector addition in figure 5.4 can be used to explore these questions. There are three effects to consider: 1. As the speed of the ball increases, the velocity vectors become longer, which makes v longer. The triangle in figure 5.4 becomes larger. 2. The greater the speed of the ball, the more rapidly the direction of the velocity vector changes, because the ball reaches the second position in figure 5.4 more quickly. 3. As the radius of the curve decreases, the rate of change in velocity increases because the direction of the ball changes more rapidly. A tight curve (small radius) produces a large change, but a gentle curve (large radius) produces a small change. The first two effects both indicate that the rate of change in velocity will increase with an increase in the speed of the ball. Combining these two effects suggests that the centripetal acceleration should be proportional to the square of the speed. We need to multiply by the speed twice. The

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third effect suggests that the rate of change of velocity is inversely proportional to the radius of the curve. The larger the radius, the smaller the rate of change. Taken together, these effects produce the expression ac

v2 r

for the size of the centripetal acceleration, ac. It is proportional to the square of the speed and inversely proportional to the radius, r, of the curve. The direction of the centripetal-acceleration vector ac is always toward the center of the curve, the direction of the change in velocity v. The ball moving in a circle is accelerated, even though its speed remains constant. To change the direction of the velocity vector is to change the velocity, and an acceleration is involved. People often resist this idea: we use the term acceleration in everyday language to describe increases in speed without taking into account changes in direction.

What force produces the centripetal acceleration? Since an object moving in a circle is accelerated, a force must be acting to produce that acceleration, according to Newton’s second law. For the ball on the string, the tension in the string pulling on the ball provides the centripetal acceleration. A closer look shows that this tension has both horizontal and vertical components, since the string is not completely within the horizontal plane. As shown in figure 5.5, the horizontal component of the tension pulls the ball toward the center of the horizontal circle and produces the centripetal acceleration. The total tension in the string is determined by both the horizontal and the vertical components of the tension. The vertical component is equal to the weight of the ball, since the net force in the vertical direction should be zero. The ball stays in the horizontal plane of the circle and is not accelerated in the vertical direction. In example box 5.1,

Tv

Rev. confirming Pages

T

Th W

figure 5.5

The horizontal component of the tension is the force that produces the centripetal acceleration. The vertical component of the tension is equal to the weight of the ball.

example box 5.1 Sample Exercise: Circular Motion of a Ball on a String A ball has a mass of 50 g (0.050 kg) and is revolving at the end of a string in a circle with a radius of 40 cm (0.40 m). The ball moves with a speed of 2.5 m/s, or one revolution per second. (see Fig. 5.5) a. What is the centripetal acceleration? b. What is the horizontal component of the tension needed to produce this acceleration? a. v 2.5 m/s r 0.40 m ac ?

ac

v2 r

2 (2.5 m/s) (0.4 m)

15.6 m/s2 b. m 0.05 kg Th ?

Fnet Th ma (0.05 kg)(15.6 m/s2) 0.78 N

The horizontal component of the tension must equal 0.78 N in magnitude. The vertical component of the tension must equal the weight of the ball (0.50N) as discussed in the text.

the weight of the ball is approximately 0.50 N (W mg), so that becomes the value of the vertical component of the tension. The ball in example box 5.1 has a slow speed. Even at this low speed, the horizontal component of the tension is larger than the vertical component. As the ball twirls at a faster rate, the centripetal acceleration increases even more rapidly, since it is proportional to the square of the speed of the ball. The horizontal component of the tension then becomes much larger than the vertical component, which remains equal to the weight of the ball (fig. 5.6). These effects can be readily observed with your own ball and string. Give it a try. You will feel the tension increase with increasing speed. Centripetal acceleration involves the rate of change in the direction of the velocity vector. Its size is equal to the square of the speed of the object divided by the radius of the curve (ac v 2/r). Its direction is toward the center of the curve. Just as with any acceleration, there must be a force acting on the object to produce the centripetal acceleration. For a ball on a string, that force is the horizontal component of the tension in the string.

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5.2 Centripetal Forces

Low speed

Tv

String

T

Th W

High speed Tv

W

String T

Th

figure 5.6

At higher speeds, the string comes closer to lying in the horizontal plane because a large horizontal component of the tension is needed to provide the required centripetal force.

5.2 Centripetal Forces For a ball twirled at the end of a string, the string pulls inward on the ball, providing the force that causes the centripetal acceleration. For a car rounding a curve, however, there is no string attached. Different forces must be at work to provide the centripetal acceleration. A person riding on a Ferris wheel also experiences circular motion. What forces produce centripetal acceleration in these situations? The net force that produces a centripetal acceleration is often referred to as the centripetal force. This term is sometimes a source of confusion, because it implies that a special force is somehow involved. In fact, centripetal forces are any force, or combination of forces, that acts on an object in certain situations to produce the centripetal acceleration. Almost any force can play this role: pulls from strings, pushes from contact with other objects, friction, gravity, and so on. We need to analyze each situation separately to identify the forces and determine their effects.

For a flat road surface, friction alone produces the necessary centripetal acceleration. The tendency of the car to move in a straight line causes the tires to pull against the pavement as the car turns. By Newton’s third law, the pavement then pulls in the opposite direction on the tires (fig. 5.7). The frictional force acting on the tires points toward the center of the curve. If this force were not present, the car could not turn. The size of a frictional force depends on whether or not there is motion along the surfaces of contact producing the friction. If there is no motion in the direction of the force, we call it the static force of friction. If the object is sliding, as it might on a wet or icy surface, the kinetic force of friction is involved. Usually, the kinetic force of friction is smaller than the maximum possible static force of friction, so whether or not the car is skidding becomes an important factor. Unless the car has already begun to skid, the static force of friction produces the centripetal acceleration for the car rounding the curve. The part of the tire in contact with the road is momentarily at rest on the road; it does not slide along the road. If the tires do not move in the direction of the frictional force, the static force is in effect. How large is the required frictional force? It depends on the speed of the car and the radius of the curve. From Newton’s second law, we know that the magnitude of the required force is Fnet mac, where the centripetal acceleration ac is equal to v2/r. Putting these two ideas together, we see that the frictional force f must be equal to mv2/r, since it is the only force operating to produce the centripetal acceleration. The speed of the car is a critical factor in determining how large a force is needed, which is why we often slow down in approaching a curve.

What force helps a car negotiate a flat curve? What forces are involved in producing the centripetal acceleration for a car rounding a curve? It depends on whether or not the curve is banked. The easiest situation to analyze is when a curve is not banked, so that we deal with a flat road surface.

83

f

figure 5.7

f

The centripetal acceleration of a car rounding a level curve is produced by frictional forces exerted on the tires by the road surface.

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If the mass times the centripetal acceleration is greater than the maximum possible frictional force, we are in trouble. Because the square of the speed is involved in this relationship, doubling the speed would require a frictional force four times as large as that for the lower speed. Also, since we cannot control the frictional force, a sharper curve with a smaller radius r requires a lower speed. Both the speed and the radius must be considered in making driving judgments. What happens if the required centripetal force is larger than the maximum possible frictional force? The frictional force cannot produce the necessary centripetal acceleration, and the car begins to skid. Once it is skidding, kinetic friction comes into play rather than static friction. Since the force of kinetic friction is generally smaller than that of static friction, the frictional force decreases and the skid gets worse. The car, like the ball on the broken string, follows its natural tendency to move in a straight line.

everyday phenomenon

The maximum possible value of the frictional force is dictated by the road and tire conditions. Any factor that reduces the force of static friction will cause problems. Wet or icy road surfaces are the usual culprits. In the case of ice, the force of friction may diminish almost to zero, and an extremely slow speed will be necessary to negotiate a curve. There is nothing like driving on an icy road to give you an appreciation of the value of friction. Newton’s first law is illustrated vividly. (See also everyday phenomenon box 5.1.)

What happens if the curve is banked? If the road surface is properly banked, we are no longer totally dependent on friction to produce the centripetal acceleration. For the banked curve, the normal force between the car’s tires and the road surface can also be helpful (fig. 5.8).

box 5.1

Seat Belts, Air Bags, and Accident Dynamics The Situation. In automobile accidents, serious or fatal injuries are often the result of riders being thrown from the vehicle. Since the 1960s, federal regulations have required that cars be equipped with seat belts. More recently, frontseat air bags have also been required in an effort to reduce the carnage. Still, we often read of people being thrown from their vehicle in accident reports. How do air bags and seat belts help? If your car is equipped with air bags, as most now are, is it still necessary to wear your seat belt? In what situations are air bags most effective and when are seat belts essential?

In a head-on collision, the air bag inflates rapidly to prevent the rider from moving forward and colliding with the windshield or steering column.

The Analysis. Except in high-speed collisions where the passenger compartment of the vehicle is crushed, most injuries and fatalities are caused by motion of the rider within, and outside of, the vehicle. The vehicle stops or turns suddenly due to the collision and the rider continues to move in a straight line, following Newton’s first law of motion. In a head-on collision, the car stops while the rider continues to move forward unless constrained. In the absence of either seat belts or air bags, front-seat riders hit the windshield or the steering column, resulting in serious head or chest injuries. Seat belts can prevent this when used properly, but air bags are also designed to protect against these injuries. As the rider begins to move forward relative to the vehicle, the air bag inflates rapidly, providing a cushion between the rider and other objects in the car. The rider decelerates more gradually involving a smaller force and less trauma. (This idea is best understood in terms of the concept of impulse discussed in chapter 7.) Air bag usage has resulted in a significant reduction in serious head and chest injuries in head-on collisions with other vehicles or with fixed objects. Head-on collisions are not the most frequent type of serious accident, however. Rollover accidents involving single vehicles are common, and vehicles can also collide in intersections, providing impacts to the side of the car. In the latter case, the struck vehicle will often go into a spin. In both of these cases, the vehicle undergoes rotational motion while the rider moves forward in a straight line.

(continued)

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Nv N

θ

Nh

figure 5.8

The horizontal component of the normal force Nh exerted by the road on the car can help to produce the centripetal acceleration when the curve is banked.

These are the accidents in which the rider is likely to be thrown from the vehicle. Will air bags help in these situations? Air bags are most effective in head-on collisions and do not provide much protection against sideways motion of the rider. (Some newer vehicles do come equipped with air bags in the front-seat doors, which can protect against sideways movement, but air bags are not usually provided for the rear seats.) In a rollover accident, the vehicle goes into a spin about an axis through its long dimension. The doors will sometimes open or the windows will shatter during the first roll, providing openings for the rider to fly through as he or she continues to move forward while the vehicle turns. In some cases, the rider is thrown from the vehicle and the vehicle then rolls over the victim. Seat belts can make a big difference. Because the vehicle is turning rapidly in a rollover accident, a centripetal force acting on the rider is necessary to hold the rider against the seat rather than moving forward in a straight line. In the absence of such a force, the rider is thrown outward against the sides of the vehicle. Attempts by riders to brace themselves are usually totally inadequate to provide the required centripetal force. The seat belt and shoulder harness, on the other hand, can provide the force necessary to hold the rider in place. Statistics on accident fatalities are compelling. In rollover accidents, riders who are wearing their seat belts generally survive, while those who are not using their belts and shoulder harnesses are frequently killed or seriously injured. Often those killed are thrown from the vehicle, but even when they

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The normal force N is always perpendicular to the surfaces involved, so it points in the direction shown in the diagram. The total normal force acting on the car (indicated in the diagram) is the sum of those for each of the four tires. Since the car is not accelerated vertically, the net force in the vertical direction must be zero. The vertical component of the normal force Nv must be equal in magnitude to the weight of the car to yield a net vertical force of zero. This fact determines how large the normal force will be. Only the horizontal component of the normal force Nh is in the appropriate direction to produce the centripetal acceleration. The angle of the banking and the weight of the car determine the size of the normal force. They also determine the size of its horizontal component. At the appropriate speed, this horizontal component pushing on the tires of the car is all that is needed to provide the centripetal acceleration.

As the vehicle rolls, a rear-seat passenger is thrown against the side of the vehicle (viewed from the back). A properly adjusted seat belt and shoulder harness can prevent this.

remain inside the vehicle, trauma from being thrown around inside the vehicle can be fatal. Statistics indicate that a high percentage of the deaths in rollover accidents involve riders ejected from the vehicle. Newton’s first law of motion is vividly illustrated in automobile accidents. An object keeps moving in a straight line with constant speed unless acted upon by an external force. Air bags and seat belts can provide that force, but seat belts provide better protection for all passengers in rollover accidents.

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The higher the speed, the steeper the required banking angle because a steeper angle produces a larger horizontal component for the normal force. Fortunately, since both the normal force and the required centripetal force are proportional to the mass of the car, the same banking angle will work for vehicles of different mass. A banked curve is designed for a particular speed. Since friction is also usually present, the curve can be negotiated at a range of speeds above and below the intended speed. Friction and the normal force combine to produce the required centripetal acceleration. If the road is icy and there is no friction, the curve can still be negotiated at the intended speed. Speeds higher than that speed will cause the car to fly off the road, just as on a flat road surface. Speeds too low, on the other hand, will cause the car to slide down the icy banked incline toward the center of the curve.

What forces are involved in riding a Ferris wheel? Riding a Ferris wheel is another example of circular motion that many of us have experienced. On a Ferris wheel, the circular motion is vertical, unlike the horizontal circles of our previous examples. Figure 5.9 shows the forces exerted on the rider at the bottom of the circle as the Ferris wheel turns. At this point in the ride, the normal force acts upward and the weight downward. Since the centripetal acceleration of the rider is directed upward, toward the center of the circle, the net force acting on the rider must also be upward. In other

words, the normal force of the seat pushing on the rider must be larger than the weight of the rider. By Newton’s second law, the net force must be equal to the mass times the centripetal acceleration. In this case, the centripetal force is the difference of two forces, the upward normal force and the downward weight of the rider, so Fnet N W mac. Since the normal force is larger than her weight, she feels heavy in this position (N W mac). The situation is similar to that in an upward accelerating elevator (see everyday phenomenon box 4.2). As the rider moves up or down along the sides of the circle, a horizontal component of the normal force is needed to provide the centripetal acceleration. This horizontal component may be provided by the frictional force exerted by the seat on the rider, by the seat back pushing on the rider on the left side of the cycle, or by a seat belt or hand bar on the right side of the cycle. The latter case is more exciting. At the top of the cycle, the weight of the rider is the only force (other than a possible seat-belt force) in the appropriate direction to produce the centripetal acceleration. Again, from Newton’s second law, the net force must equal the mass of the rider times the centripetal acceleration, which is now directed downward. This yields the relationship Fnet W N mac. As the speed gets larger and the centripetal acceleration, ac v2/r, increases, the normal force must get smaller to increase the total force. Usually, the top speed of the Ferris wheel is adjusted so that the normal force is small when the rider is at the top of the cycle. Since the force exerted by the seat on the rider is small, the rider feels light, part of the thrill of the ride. If there is one nearby, take a break and go ride a Ferris wheel. There is nothing like direct experience to bring home the ideas we have just described. As you ride, try to sense the direction and magnitude of the normal force. The light feeling at the top and the sense of plunging outward in the downward portion of the cycle are what the price of the ride is all about.

N

W

figure 5.9

At the bottom of the cycle, the weight of the rider and the normal force exerted by the seat combine to produce the centripetal acceleration for a rider on a Ferris wheel.

A centripetal force is any force or combination of forces that produces the centripetal acceleration for an object moving around a curve. In the case of a car moving on a flat road surface, the centripetal force is provided by friction. If the road surface is banked, the normal force of the road pushing on the tires of the car also helps. In the case of a Ferris wheel, the weight of the rider and the normal force exerted by the seat on the rider combine to provide the centripetal force. We use Newton’s laws of motion to identify the forces and analyze each situation.

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5.3 Planetary Motion Have you ever watched Venus or Mars in the night sky and wondered how and why their positions change from night to night? From the standpoint of the history of science, the motions of the planets are the most important examples of centripetal acceleration. These objects are a part of our everyday experience, yet many of us are surprisingly unaware of how they move. How do the sun, the stars, and the planets move? How can we make sense of the motions?

Early models of the heavens Observing the heavens was probably a more popular pastime when there were fewer roofs over our heads. If you have ever spent a night in a sleeping bag under the stars, you probably experienced a sense of wonder and amazement at all of those bright objects out there. If you spent night after night observing the stars, you might notice, as the ancients did, that some of the brightest objects move relative to the other stars. These wanderers are the planets. The so-called fixed stars always maintain the same relative position to one another as they move across the sky (fig. 5.10). The Big Dipper never seems to change its shape, but the planets roam about with respect to the fixed stars in a regular but curious fashion. Their motions excited the curiosity of ancient observers of the heavens. They were carefully tracked and often incorporated into religious and cultural beliefs. Suppose you were an early philosopher-scientist trying to make sense of these motions. What kind of model might you develop? Some features seem simple and regular. The sun, for example, moves across the sky each day, from east to west, as if it were at the end of an enormously long and invisible rope tethered at the center of the Earth. The stars follow a similar pattern. Their apparent motion as seen from Earth could be explained by picturing them as lying on a giant sphere that revolves around the Earth. This Earth-centered or geocentric view of the universe seemed natural and reasonable. The moon also moves across the sky in an apparently circular orbit around the Earth. Unlike the stars, the moon does not reappear in the same position each night. Instead, it goes through a series of regular changes in position and phase in a cycle of approximately 30 days. How many of us can provide a clear explanation of the phases of the moon? The motion of the moon will be considered more fully in the final section of this chapter. Early models of the motions of the heavenly bodies developed by Greek philosophers involved a series of concentric spheres centered on the Earth. Plato and others of his time viewed spheres and circles as ideal shapes that would reflect the beauty of the heavens. The sun, the moon, and the five planets known then each had its own sphere. The fixed stars were on the outermost sphere. These spheres were thought to revolve around the Earth in ways that explained the positions of the heavenly bodies.

figure

5.10 A time-lapse photograph showing the apparent motion of stars in the northern sky. Polaris (the “North star”) lies near the center of the pattern and does not appear to move very much. The entire pattern appears to rotate during the night about a point near Polaris.

figure

5.11 An example of the retrograde motion of Mars relative to the background of fixed stars. These changes take place over a period of several months.

Unfortunately, the planets do not behave as though they are on a continuously revolving sphere. The planets sometimes appear to move backward relative to their normal direction of motion against the background of the fixed stars. We call this retrograde motion. It takes a few months for Mars to trace one of these retrograde patterns (fig. 5.11). To explain the apparent retrograde motion of these planets, Ptolemy (Claudius Ptolemaeus), working in the second century A.D., devised a more sophisticated model than the one used by earlier Greek philosophers. Ptolemy’s model used circular orbits rather than spheres but was still geocentric. He invented the idea of epicycles, circles that rolled along the larger basic orbit of the planet around the

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Epicycle Planet

Earth

Planet

Earth

figure

5.12 Ptolemy’s epicycles were circles rolling along the circular orbits of the planets. This model explained the retrograde motion observed for the outer planets. Earth (fig. 5.12). The epicycles accounted for the retrograde motion and could also be used to explain other irregularities in planetary orbits. Ptolemy’s model accurately predicted where to find the planets at any given time of any year. As more accurate observations became available, refinements were needed, however, to improve the predictions. In some cases, this meant adding epicycles to epicycles, but the basic scheme of circles was retained. Ptolemy’s system became part of the accepted knowledge during the Middle Ages and was incorporated, along with many of Aristotle’s works, into the teachings of the Roman Catholic Church and the emerging European universities.

How did the Copernican model differ from Ptolemy’s conception? Ptolemy’s model is not the one that you were introduced to in elementary school. It has been superseded. During the sixteenth century, a Polish astronomer, Nicolaus Copernicus (1473–1543) put forth a sun-centered or heliocentric view, later championed by Galileo. Copernicus was not the first to suggest such a model, but earlier heliocentric versions had not taken hold. Copernicus spent many years working out the details of his model, but he did not publish it until within a year of his death.

Galileo was an early advocate of the Copernican model and promoted it more vigorously than Copernicus himself. In 1610, hearing of the invention of the telescope, Galileo built his own improved version and turned it to the heavens. He discovered that the moon has mountains, that Jupiter has moons, and that Venus goes through phases like our moon. He showed that the phases of Venus could be explained better by the Copernican model than by a geocentric model. Galileo became famous throughout Europe for his discoveries and ended up in trouble with church authorities, a problem not to be taken lightly in his day. People had been burned at the stake for similar offenses. Copernicus placed the sun at the center of the circular orbits of the planets and demoted the Earth to the status of just another planet. Also, the Copernican model requires that the Earth rotate on an axis through its center—thus explaining the daily motions of the sun and the other heavenly bodies (including the fixed stars). This idea was revolutionary at the time. Why are we not blown away by the enormous winds that rotation might produce? Perhaps the air near the Earth’s surface is dragged along with the Earth. The advantage of the Copernican view is that it does not require complicated epicycles to explain retrograde motion, although epicycles were still used to make other adjustments to planetary orbits. Retrograde motion comes about because the Earth is orbiting the sun along with the other planets. The position of Mars appears to change as both Mars and Earth move in the same direction against the background of the fixed stars (fig. 5.13). As the more rapidly moving Earth passes Mars, Mars slips behind and briefly appears to move backward. Accepting the Copernican model meant giving up the Earth-centered view of the universe to endorse what seemed to some to be an absurd proposition: that the Earth rotates, with a frequency of one cycle per day. Since an approximation of the radius of Earth was known (6400 km), rotation implied that we must be moving at roughly 1680 km/h (or just over 1000 MPH) if we are standing near the equator on the Earth! We certainly do not feel that motion. Because Copernicus assumed the planets’ orbits to be circular, the accuracy of his model for predicting was no better than Ptolemy’s model. In fact, it required some adjustments (for which Copernicus used epicycles) just to make it agree with already known astronomical data. Settling the controversy generated by the competition between the two models called for more accurate observations, a project undertaken by a Danish astronomer, Tycho Brahe (1546–1601). Tycho was the last great naked-eye astronomer. He developed a large quadrant (fig. 5.14) that he used to make very accurate sightings of the positions of the planets and stars. It was capable of measuring these positions to an accuracy of 1⁄ 60 of a degree, considerably better than previously available data. Tycho spent several years painstakingly collecting data on the precise positions of the planets and other bodies—all without the benefit of a telescope.

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Mars

Sun Earth

Lines of sight

figure

5.13 As the Earth passes the more slowly moving Mars, Mars appears to move backward as seen against the background of the much more distant fixed stars. (Not drawn to scale.)

Focus

Focus Semimajor axis

figure

5.15 An ellipse can be drawn by fixing a string at two points (foci) and moving a pencil around the path permitted by the string.

figure

5.14 Tycho Brahe’s large quadrant permitted accurate measurement of the positions of the planets and other heavenly bodies. Kepler’s laws of planetary motion Analyzing the data collected by Tycho fell to his assistant, Johannes Kepler (1571–1630), after Tycho’s death. It was an enormous task requiring the transformation of the data to coordinates around the sun and then numerical trial and error to find regular planetary orbits. It was already known that these orbits were not perfect circles. Kepler was able to show that the orbits of the planets around the sun were ellipses, with the sun at one focus. An ellipse can be drawn by attaching a string between two fixed foci and then moving a pencil around the perimeter

of the path allowed by the string (fig. 5.15). A circle is a special case of an ellipse in which the two foci coincide. The orbits of most of the planets are very close to being circles, but Tycho’s data were so precise that they showed a difference between a perfect circle, on one hand, and an ellipse with two closely spaced foci. Kepler’s first law of planetary motion states that the orbits of the planets are ellipses. Kepler’s other two laws of planetary motion came after even more laborious numerical trial and error with Tycho’s data. Kepler’s second law describes how the planets move faster when they are nearer to the sun, so that an imaginary line drawn from the sun to the planet moves through equal areas in equal times regardless of where it is in its orbit (fig. 5.16)*. The first two laws were published in 1609. The third law (published in 1619) states a relationship between the average radius of the orbit and the time taken *The second law turns out to be a consequence of conservation of angular momentum, which is discussed in chapter 8.

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example box 5.2 Sample Exercise: Using Kepler’s Third Law 25 days

}

25 days

How long does it takes Mars to complete one orbit around the sun? The distance of Mars from the sun is approximately 1.5 astronomical units (AU). (An AU is the average distance from the Earth to the sun; thus the radius of the earth’s orbit, REarth, is just 1 AU.) REARTH 1AU the distance from Earth to the sun

figure 5.16

Because planets move faster when nearer to the sun, the radius line for each planet sweeps out equal areas in equal times (Kepler’s second law). In other words, the two blue sections each cover the same span of time and have the same area. (Not drawn to scale.)

for one complete cycle around the sun (the period of the orbit). Kepler found his third law after trying many other possible relationships between the periods, T, and the average radii of the planetary orbits, r. To a high degree of accuracy, he found that the ratio of the square of the period to the cube of the radius (T 2/r3) was the same for all of the known planets (See example box 5.2). The behavior of the planets is surprisingly regular. Kepler published his findings in papers that also contained elaborate speculations on numerical mysticism and musical harmonies associated with the planets. Some of these ideas must have seemed strange to Galileo and others who admired Kepler’s work. Kepler’s laws added to the accuracy with which we can predict the positions of the planets as they appear to wander among the fixed stars. Like the Copernican model, Kepler’s model was heliocentric (sun-centered), so it supported Galileo’s efforts to overthrow the geocentric (Earth-centered) model of Ptolemy. More importantly, however, Kepler’s laws described a new set of precisely stated relationships that called for explanation. The stage was set for Isaac Newton to incorporate these relationships into a grand theory that explains both celestial mechanics (the motion of the heavenly

Kepler’s Laws of Planetary Motion 1. The planets all move in elliptical orbits about the sun, with the sun located at one focus of the ellipse. 2. An imaginary line drawn from the sun to any planet moves through equal areas in equal intervals of time. 3. If T is the amount of time taken for the planet to complete one full orbit around the sun (period) and if r is the average radius of the distance of the orbit around the sun for each planet, then the ratio of the square of the period to the cube of the radius (T2/r3) is the same for all of the known planets.

RMars 1.5 AU the distance from Mars to the sun TEarth 1 Earth year (yr) TMars ? (in Earth years) Kepler’s third law for this case can be stated as: R3Mars R3Earth 2 . 2 T Mars T Earth Cross multiplying, we find R3 # T 2 R3 # T 2 Earth

Therefore, T2Mars

R3Mars # T2Earth R3Earth

Mars

Mars

Earth.

.

Inserting the given values: (1.5AU) 3# (1yr) 2 T2Mars , (1AU) 3 T2Mars 3.4 yr2 (notice that the AU units cancel) TMars 23.4 yr2 L 1.8 Earth years.

bodies) and the more mundane motion of everyday objects near the Earth’s surface. Many of the early models for describing the motion of the planets were geocentric (Earth-centered). Ptolemy’s model included epicycles to explain the apparent retrograde motion of the planets. Copernicus introduced a heliocentric (sun-centered) model, which explained retrograde motion more simply. This model was championed by Galileo. Galileo was one of the first scientists to use a telescope systematically, and he made significant discoveries supporting the heliocentric view. Kepler refined the heliocentric model by showing that planetary orbits are ellipses with some surprising regularities.

5.4 Newton’s Law of Universal Gravitation Planetary motion and centripetal acceleration lead us to the next question. If the planets are moving in curved paths around the sun, what force must be present to produce the centripetal acceleration? You are probably aware that gravity

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is involved, but that involvement was not at all obvious when Newton began his work. How did Newton put it all together?

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land. At very large launch velocities, the curvature of the Earth becomes a significant factor. In fact, if the launch velocity is large enough, the projectile would never reach the Earth’s surface. It keeps falling, but the curvature of the Earth falls away, too. The projectile goes into a circular orbit around the Earth. Newton’s insight was that the moon, under the influence of gravity, is actually falling, just as a projectile does. The moon, of course, is at a distance from the Earth much greater than the height of any mountain. The same force that accounts for the acceleration of objects near the Earth’s surface, as described by Galileo, explains the orbit of the moon.

What was Newton’s breakthrough? Newton realized that there is a similarity between the motion of a projectile launched near the Earth’s surface and the orbit of the moon. To illustrate this point, Newton produced a famous drawing similar to that shown in figure 5.17. The idea is simple but earthshaking. Imagine, as Newton did, a projectile being launched horizontally from an incredibly high mountain. The larger the launch velocity, the farther away from the base of the mountain the projectile will

Newton’s law of universal gravitation From Galileo’s work, Newton knew that near the Earth’s surface the gravitational force is proportional to the mass of the object, F mg. Mass, then, should be involved in any more general expression for the gravitational force. Does the gravitational force vary with distance, though, and, if so, how? The idea that a force could influence two masses separated by a large distance was hard to accept in Newton’s day (and, in some ways, even now). If such a force exists, we would expect that this force “acting at a distance” would decrease in strength as the distance increases. Using geometrical reasoning (fig. 5.18), other scientists had speculated that the force might be inversely proportional to the square of the distance r between the masses, but they could not prove it. At this point, Kepler’s laws of planetary motion and the concept of centripetal acceleration came into play. Newton was able to prove mathematically that Kepler’s first and third laws of planetary motion could be derived from the assumption that the gravitational force between the planets and the sun falls off with the inverse square of the distance. The proof involved setting the assumed 1/r 2 force equal to the required centripetal force in Newton’s second law of motion. All of Kepler’s laws are consistent with this assumption.

figure 5.17

In a diagram similar to this, Newton imagined a projectile fired from an incredibly high mountain. If fired with a large enough horizontal velocity, the projectile falls toward the Earth but never gets there.

4A1

9A1

A1

m

r1 2r1 3r1

figure 5.18

If lines are drawn radiating outward from a point mass, the areas intersected by these lines increase in proportion to r 2. Does this suggest that the force exerted by the mass on a second mass might become weaker in proportion to 1/r 2?

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m1

m2 F1

F2

Earth (me)

r

figure

5.19 The gravitational force is attractive and acts along the line joining the center of the two masses. It obeys Newton’s third law of motion (F2 F1).

F

F

Gm1m2 r2

,

where G is a constant. The direction of the force is attractive and lies along the line joining the centers of the two masses (fig. 5.19).

For this statement to be completely valid, the masses in question must be either point masses or perfect spheres. In Newton’s law of gravitation, G is the universal gravitational constant. It has the same value for any two objects. Newton did not actually know the value of this constant, because he did not know the masses of the Earth, the sun, and the other planets. Its value was determined more than a hundred years later in an experiment done by Henry Cavendish (1731–1810) in England. Cavendish measured the very weak gravitational force between two massive lead balls for different distances of separation. In metric units, the value of G is G 6.67 1011 N·m2/kg2. The power-of-10 notation (see appendix B) is useful here because G is a very small number. The power 11 means that the decimal point is located eleven places to the left of where it is shown. If we did not use power-of-10 notation, the number would appear as G 0.000 000 000 066 7 N·m2/kg2. Because of the small size of this constant, the gravitational force between two ordinary-sized objects, such as people, is extremely small and not usually noticeable. Cavendish’s experiment required real ingenuity to measure such a weak force.

Object (m)

re

The proof that Kepler’s laws could be explained by a gravitational force proportional to the masses of two interacting objects, and inversely proportional to the square of the distance between the objects, led to Newton’s law of universal gravitation. This law and Newton’s three laws of motion are the fundamental postulates of his theory of mechanics. The law of gravitation can be stated as The gravitational force between two objects is proportional to the mass of each object and inversely proportional to the square of the distance between the centers of the masses:

F

figure

5.20 For the Earth and an object near the Earth’s surface, the distance between the centers of the two objects is equal to the radius of the Earth. How is weight related to the law of gravitation? Suppose that one of the masses is a planet or other very large object. The force of gravity then can be quite large because one of the masses is very large. Consider the force exerted on a person standing on the surface of the Earth. As figure 5.20 illustrates, the distance between the centers of the two objects, the person and the Earth, is essentially the radius of the Earth, re. From Newton’s law of gravitation, the force on the person must be F Gmme /re2, where m is the mass of the person and me is the mass of Earth. Since this gravitational force is the weight of the person, we can also express the force as F W mg. For these two expressions for F to be the same, g, the gravitational acceleration, must be related to the universal gravitational constant G by g Gme /re2. The gravitational acceleration near the Earth’s surface g is therefore not a universal constant. It will be different on different planets and even slightly different at different points on the Earth because of variations in the radius of the Earth and other factors. The constant G is a universal constant of nature that can be used to find the gravitational acceleration for any planet if we know the radius and mass of the planet. If we know the gravitational acceleration near the Earth’s surface, it is easier to use the expression F mg to compute a weight than to use the law of universal gravitation. This computation is done both ways in the sample exercise in example box 5.3. Either way, we get the same result. The weight of the 50-kg person is approximately 490 N. The mass of Earth, 5.98 1024 kg, is a very large number that was first determined by Cavendish when he

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example box 5.3 Sample Exercise: Gravity, Your Weight, and the Weight of the Earth The mass of the Earth is 5.98 1024 kg, and its average radius is 6370 km. Find the gravitational force (the weight) of a 50-kg person standing on Earth’s surface a. by using the gravitational acceleration. b. by using Newton’s law of gravitation. a. m 50 kg

F W mg

g 9.8 m/s2

(50 kg)(9.8 m/s2)

F ?

490 N

b. me 5.98 1024 kg re 6.37 106 m F W Gmme /re2 (6.67 10 11 N#m2/kg2)(50 kg)(5.98 1024 kg) (6.37 106 m)2 490 N Most scientific calculators will handle the scientific notation directly. The powers add for multiplication and subtract in division.

measured the universal constant G. In a sense, Cavendish weighed the Earth by making that measurement. If we wanted to know the gravitational force exerted on a 50-kg person in a space capsule several hundred kilometers above the Earth, we would have to use the more general expression in Newton’s law of gravitation. Likewise, if we wanted to know the weight of this person when standing on the moon, we would need to use the mass and radius of the moon in place of those of the Earth in our calculation. The weight of a 50-kg person on the moon is only about 1⁄ 6 the value of 490 N that we computed for the same person standing on Earth. The expression F mg is valid only near the surface of the Earth. The weaker gravitational force and acceleration of the moon are explained by the moon’s smaller mass. Since our muscles are adapted to conditions on Earth, we would find that our smaller weight on the moon makes some amazing leaps and bounds possible. The smaller gravitational force on objects near the moon’s surface also explains why the moon has essentially no atmosphere. Gas molecules escape the gravitational pull of the moon much more readily than they can from the Earth. Newton recognized that the moon is falling toward the Earth much like projectiles moving near the Earth’s surface. He proposed that the gravitational force that explains projectile motion is also involved in the motions of the planets around the sun and of the moon around the Earth.

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Newton’s law of universal gravitation states that the gravitational force between two masses is proportional to the product of the masses and inversely proportional to the square of the distance between them. Using this law and his laws of motion, Newton was able to explain Kepler’s laws of planetary motion as well as the motion of ordinary objects near the Earth’s surface.

5.5 The Moon and Other Satellites The moon has fascinated people as long as humanity has existed and wondered about nature. In the twentieth century, we have actually visited the moon for the first time and brought back samples from its surface. That visit has not dulled the romance that the moon holds for us, but it may have reduced its mystery. How are the phases of the moon associated with changes in its position? Are Kepler’s laws of planetary motion valid for the moon? How are the orbits of other satellites of Earth similar to the moon’s?

How do we explain the phases of the moon? The moon was the only Earth satellite available to Newton and his predecessors to study. The moon played a pivotal role in Newton’s thinking and in the development of his law of gravitation. Observations of the moon and its phases, however, go back much farther than Newton’s day. The moon figures in many early religions and rituals. Its course must have been carefully followed even in prehistoric times. How do we explain the phases of the moon? Is the time that the moon rises in the evening related to whether it will be a full moon or not? Moonlight is reflected sunlight. So, to understand the moon’s phases, we have to take into account the positions of the sun, the moon, and the observer (fig. 5.21). When the moon is full, it is on the opposite side of the Earth from the sun, and we see the side that is fully illuminated by the sun. The full moon rises in the east about the same time that the sun sets in the west. These events are determined by the Earth’s rotation. Because Earth and the moon are both small compared to the distances between Earth, the moon, and the sun, they do not usually get in the way of light coming from the sun. When they do, however, there is an eclipse. During a lunar eclipse, Earth casts a full or partial shadow on the moon. From figure 5.21, we can see that a lunar eclipse can only occur during a full moon. A solar eclipse happens when the moon is in the right position to cast a shadow on the Earth. During what phase of the moon will this occur? At other times during the moon’s 27.3-day revolution around the Earth, we do not see all of the illuminated side of the moon; we see a crescent or a half-moon or some shape in between (fig. 5.22). The new moon occurs when the moon is on the same side of Earth as the sun and is more or less invisible. When we are a few days on either side of the new moon, we see the familiar crescent.

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Half-moon

Sun

Earth

New moon

Full moon

figure

5.22 Photographs of different phases of the moon. When during the day will each rise and set? Half-moon

figure

5.21 The phases of the moon depend on the positions of the sun, the moon, and the Earth. (Not drawn to scale.)

When the moon is between full moon and new moon, it can often be seen during daylight. In particular, when it is near half-moon, it rises around noon and sets around midnight (or vice versa, depending on where it is in its cycle). Under just the right conditions near sunset or sunrise, we can sometimes see the dark portion of the crescent moon illuminated by earthshine.

everyday phenomenon

The next time you see the moon, think about where it is in the sky, when it will rise and set, and how this is related to its phase. Better yet, try explaining this to a friend. You too can be the wizard who predicts the motions of the heavens.

Does the moon obey Kepler’s laws? The moon’s orbit around Earth is more complicated than those of the planets because two bodies, Earth and the sun, exert strong forces on the moon, rather than just one (fig. 5.23). Earth is much closer to the moon than the sun is, but the sun has a much larger mass than Earth, so the sun’s effect is still appreciable. First, let’s consider just the effects of Earth on the moon’s motion.

box 5.2

Explaining the Tides The Situation. Anyone who has lived near the ocean is familiar with the regular variation of the tides. Roughly twice a day the tides go in and go out again. The actual cycle of two high tides and two low tides is closer to 25 hours. Sometimes high tide is higher and low tide is lower than at other times— these times correspond to the full moon or the new moon. The times when high tides and low tides happen shift from day to day because of their 25-hour cycle, but the pattern repeats monthly. How do we explain this behavior?

High tide

The Analysis. The monthly cycle and the correlations of the highest tides with the phase of the moon suggest a lunar influence. Both the moon and the sun exert gravitational forces on the Earth. The sun exerts the stronger force because of its much larger mass, but the moon is much closer, and variations in its distance from Earth may be significant. The gravitational force depends on 1/r 2, so its strength will vary as the distance r varies, as indicated in the drawing on page 95.

Low tide

High tide and low tide produce different water levels at the dock.

(continued)

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5.5 The Moon and Other Satellites

The physics of the situation is the same as for the orbits of the planets around the sun. The gravitational attraction between the moon and Earth provides the centripetal acceleration to keep the moon moving in its roughly circular orbit. By Newton’s law of gravitation, the gravitational force acting on the moon is proportional to 1/r2, where r is the distance between the center of the moon and the center of Earth. The tides can be explained by this dependence of the gravitational force on distance. (See everyday phenomenon box 5.2.) Like the planets, the moon’s orbit is an ellipse but with the Earth at one focus of the ellipse rather than the sun. The sun also exerts a force on the moon that distorts the ellipse, causing the moon’s orbit around Earth to oscillate about a true elliptical path as the moon and Earth orbit together around the sun. Calculating these oscillations was a problem that kept mathematical physicists busy for many years. Kepler’s first and second laws of planetary motion are approximately true for the moon, provided that we substitute the Earth for the sun in the statement of these laws.

Moon

r2 r r1 Because it depends on distance, the gravitational force per unit mass exerted by the moon on different parts of the Earth (and water in the oceans) gets weaker as we move from the side nearer the moon to the far side. (Not drawn to scale; the bulges here are greatly exaggerated.)

Since water is a fluid (except when frozen), the water that makes up the oceans moves over the more rigid crust of Earth. The primary force acting on the water is the gravitational attraction of Earth that holds the water to the Earth’s surface. The gravitational force exerted by the moon on the water is also significant, however, and its strength per unit mass is greatest on the side of Earth closest to the moon and weakest on the opposite side of Earth because of the difference in distance. This difference in strength of the moon’s pull produces a bulge in the water surface on both sides of the Earth. The bulge on the side nearest the moon results from the water being pulled toward the moon by a stronger force per unit

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Kepler’s third law shows some differences between the moon and the planets. When Newton derived the expression for the ratio in Kepler’s third law, he arrived at the expression T2 4 2 r3 Gms where ms is the mass of the sun. For the moon, we would replace the mass of the sun with the mass of Earth. We get a different ratio for the moon’s orbit around Earth than for the orbits of the planets around the sun.

Orbits of artificial satellites Any satellite orbiting the Earth must have the same value for the ratio T 2/r 3 as the moon. Kepler’s third law holds for any satellite of Earth, then, as long as we keep in mind that the ratio will not have the same value as it does for the orbits of the planets. The value of this ratio for Earth satellites is calculated either from the Earth’s mass or from the

mass than the force per unit mass exerted on the rest of the Earth. This produces a high tide. The water will rise nearer to the top of the dock. On the opposite side of the Earth, it is the Earth that is being pulled by the moon with a stronger force per unit mass than the water. Since the Earth is pulled away (slightly) from the water, this also produces a high tide. The forces exerted by the moon are small compared to the force that the water and the Earth exert on each other but are still large enough to produce the tides. When the sun and the moon both line up with the Earth during the new moon or full moon, the sun also contributes to this difference in forces and produces bulges on either side of the Earth, adding to those produced by the moon. The highest tides occur during a full moon or a new moon because of this combination of the moon and sun. Why is the cycle 25 hours rather than 24 hours? The hightide bulges occur on either side of Earth along the line joining the moon and the Earth. The Earth rotates underneath these bulges with a period of 24 hours, but in this time, the moon also moves, since it orbits Earth with a period of 27.3 days. In one day, therefore, the moon has moved through roughly 1/ 27 of its orbital cycle, causing the time when the moon again lines up with a given point on Earth to be a little longer than 1 day. This additional time is approximately 1/ 27 of 24 hours, or a little less than an hour. This model was conceived by Newton and accounts neatly for the major features of the tides. The variation of the gravitational force with distance is the key to the explanation.

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Sun

Moon

Earth 150 million km

figure

5.23 The moon is influenced by gravitational attraction to both the Earth and the sun. (Distances and sizes not drawn to scale.)

Polar orbit

values of the period and average distance of the moon’s orbit. Any artificial satellite of Earth must have the same value for this ratio. If its distance from the center of the Earth r is smaller than the moon’s distance, its orbital period T must also be smaller to keep the ratio T 2/r 3 the same. Using this ratio, we can calculate the appropriate distance from Earth for any satellite if we know its orbital period. For example, a satellite with a synchronous orbit has a period of 24 hours, which keeps it above the same point on the Earth as Earth rotates. From the third-law ratio, we find a distance r of 42 000 km for such a satellite (measured from the center of the Earth). Since Earth’s radius is 6370 km, this is roughly seven times the radius of the Earth. Quite a ways up, but not nearly as high as the moon. Most artificial satellites are even closer to the Earth. The original Russian satellite, Sputnik, for example, had a period of about 90 minutes or 1.5 h. Using the third-law ratio, this yields an average distance from the center of the Earth of 6640 km. Subtracting Earth’s radius, 6370 km, indicates that this distance is only 270 km above the Earth’s surface. The shorter the period, the closer the satellite is to the Earth. The orbital period cannot be much shorter than Sputnik’s before atmospheric drag becomes too large for motion to be sustained. Obviously, the orbit cannot have a radius smaller than Earth’s radius. The orbits of different satellites are planned to meet different objectives. Some are close to circular, others much more elongated ellipses (fig. 5.24). The plane of the orbit can pass through the poles of the Earth (polar orbit) or take any orientation between the poles and the equator. It all depends on the mission of the satellite. Artificial satellites have become a routine feature of today’s world that did not exist before 1958 when Sputnik was launched. Their uses are many, including communications, surveillance, weather observations, and various military

figure

5.24 The orbits of different artificial satellites can have different orientations and elliptical shapes.

applications. The basic physics of their behavior is accounted for by Newton’s theory. If Newton could return, he might be amazed at the developments, but for him, the analysis would be routine. The motion of the moon around Earth is governed by the same principles as that of the planets around the sun. The gravitational force provides the centripetal acceleration that keeps the moon in an approximately elliptical orbit. The moon is illuminated primarily by the sun, and the phases of the moon can be explained by the moon’s position with regard to the sun and Earth. The full moon occurs when the sun and moon are on opposite sides of the Earth. Other satellites of Earth are governed by these principles, but Kepler’s third-law ratio has a different value for satellites of Earth (including the moon) than it does for the planets. The moon is no longer alone; it has been joined by many much smaller objects buzzing around the Earth in lower orbits.

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summary Objects moving in circular paths are accelerated because the direction of the velocity vector continually changes. The forces involved in producing this centripetal acceleration were examined for the motion of a ball on a string, cars rounding curves, a rider on a Ferris wheel, and finally the planets moving around the sun. The force providing the centripetal acceleration for planetary motion is described by Newton’s law of gravitation.

motion. Kepler’s three laws of planetary 3motionPlanetary describe the orbits of the planets around the sun. The orbits are ellipses that sweep out equal areas in equal times (the first and second laws). The third law states a relationship between the period of the orbit and the distance of the planet from the sun.

1 Centripetal acceleration. Centripetal acceleration is the acceleration involved in changing the direction of the velocity

r

vector. It is proportional to the square of the speed of the object and inversely proportional to the radius of the curve.

r r

T2 —— = constant r3

∆v

v2 v1

v2

v1

law of universal gravitation. Newton’s 4law ofNewton’s universal gravitation states that the gravitational force between two masses is proportional to each of the masses and inversely proportional to the square of the distance between the masses. Using this law with his laws of motion, Newton could derive Kepler’s laws of planetary motion.

2

v ac = –– r

m1

m2 F2

F1

2

Centripetal forces. A centripetal force is any force or combination of forces that acts on a body to produce the centripetal acceleration, including friction, normal forces, tension in a string, or gravity. The net force is related to the centripetal acceleration by Newton’s second law.

Fc

v

r m1m 2 F = G _____ r2

moon and other satellites. The moon’s orbit 5aroundThe Earth can also be described by Kepler’s laws, provided we substitute the mass of Earth for that of the sun in the expression for the period. Artificial satellites have the same ratio T 2/r3 as that for the moon. Polar orbit

Fc = mac

key terms Centripetal acceleration, 81 Centripetal force, 83 Static force of friction, 83 Kinetic force of friction, 83 Retrograde motion, 87

Epicycle, 87 Heliocentric, 88 Ellipse, 89 Period, 90 Newton’s law of universal gravitation, 92

Universal gravitational constant, 92 Phases of the moon, 93 Eclipse, 93 Synchronous orbit, 96

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questions * more open-ended questions, requiring lengthier responses, suitable for group discussion Q sample responses are available in appendix D Q sample responses are available on the website

Q1. Suppose that the speed of a ball moving in a horizontal circle is increasing at a steady rate. Is this increase in speed produced by the centripetal acceleration? Explain. Q2. A car travels around a curve with constant speed. a. Does the velocity of the car change in this process? Explain. b. Is the car accelerated? Explain. Q3. Two cars travel around the same curve, one at twice the speed of the other. After traveling the same distance, which car, if either, has experienced the larger change in velocity? Explain. Q4. A car travels the same distance at constant speed around two curves, one with twice the radius of curvature of the other. For which of these curves is the change in velocity of the car greater? Explain. *Q5. The centripetal acceleration depends upon the square of the speed rather than just being proportional to the speed. Why does the speed enter twice? Explain. Q6. A ball on the end of a string is whirled with constant speed in a counterclockwise horizontal circle. At point A in the circle, the string breaks. Which of the curves sketched below most accurately represents the path that the ball will take after the string breaks (as seen from above)? Explain. 4

3

2 A 1

Q6 Diagram Q7. Before the string breaks in question 6, is there a net force acting upon the ball? If so, what is its direction? Explain. Q8. For a ball being twirled in a horizontal circle at the end of a string, does the vertical component of the force exerted by the string produce the centripetal acceleration of the ball? Explain. Q9. A car travels around a flat (nonbanked) curve with constant speed. a. Sketch a diagram showing all of the forces acting on the car. b. What is the direction of the net force acting on the car? Explain.

Q10. Is there a maximum speed at which the car in question 9 will be able to negotiate the curve? If so, what factors determine this maximum speed? Explain. Q11. If a curve is banked, is it possible for a car to negotiate the curve even when the frictional force is zero due to very slick ice? Explain. *Q12. If a ball is whirled in a vertical circle with constant speed, at what point in the circle, if any, is the tension in the string the greatest? Explain. (Hint: Compare this situation to the Ferris wheel described in section 5.2.) Q13. Sketch the forces acting upon a rider on a Ferris wheel when the rider is at the top of the cycle, labeling each force clearly. Which force is largest at this point, and what is the direction of the net force? Explain. *Q14. In what way did the heliocentric view of the solar system proposed by Copernicus provide a simpler explanation of planetary motion than the geocentric view of Ptolemy? Explain. Q15. Did Ptolemy’s view of the solar system require motion of the Earth, rotational or otherwise? Explain. *Q16. Heliocentric models of the solar system (Copernican or Keplerian) require that the Earth rotate on its axis producing surface speeds of roughly 1000 MPH. If this is the case, why do we not feel this tremendous speed? Explain. Q17. How did Kepler’s view of the solar system differ from that of Copernicus? Explain. Q18. Consider the method of drawing an ellipse pictured in figure 5.15. How would we modify this process to make the ellipse into a circle, which is a special case of an ellipse? Explain. Q19. Does a planet moving in an elliptical orbit about the sun move fastest when it is farthest from the sun or when it is nearest to the sun? Explain by referring to one of Kepler’s laws. Q20. Does the sun exert a larger force on the Earth than that exerted on the sun by the Earth? Explain. Q21. Is there a net force acting on the planet Earth? Explain. Q22. Three equal masses are located as shown in the diagram. What is the direction of the net force acting upon m2? Explain. m1

m2

m3

Q22 Diagram Q23. Two masses are separated by a distance r. If this distance is doubled, is the force of interaction between the two masses doubled, halved, or changed by some other amount? Explain.

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Q24. A painter depicts a portion of the night sky as shown in the diagram below, showing the stars and a crescent moon. Is this view possible? Explain.

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Q27. Are we normally able to see the new moon? Explain. Q28. During what phase of the moon can a solar eclipse occur? Explain. *Q29. A synchronous satellite is one that does not move relative to the surface of the Earth; it is always above the same location. Why does such a satellite not just fall straight down to the Earth? Explain. Q30. Is Kepler’s third law valid for artificial satellites orbiting about the Earth? Explain. Q31. Since the Earth rotates on its axis once every 24 hours, why don’t high tides occur exactly twice every 24 hours? Explain.

Q25. At what times during the day or night would you expect the new moon to rise and set? Explain.

Q32. Why is there a high tide rather than a low tide when the moon is on the opposite side of the Earth from the ocean and the gravitational pull of the moon on the water is the weakest? Explain.

Q26. At what times of the day or night does the half-moon rise or set? Explain.

Q33. Would tides exist if the gravitational force did not depend upon the distance between objects? Explain.

Q24 Diagram

exercises E1. A ball is traveling at a constant speed of 5 m/s in a circle with a radius of 0.8 m. What is the centripetal acceleration of the ball?

b. What is the magnitude of the net force required to produce this centripetal acceleration for a rider with a mass of 70 kg?

E2. A car rounds a curve with a radius of 25 m at a speed of 20 m/s. What is the centripetal acceleration of the car?

E9. What is the ratio of the Earth’s orbital period about the sun to the Earth’s period of rotation about its own axis?

E3. A ball traveling in a circle with a constant speed of 3 m/s has a centripetal acceleration of 9 m/s2. What is the radius of the circle?

E10. Joe has a weight of 720 N (about 162 lb) when he is standing on the surface of the Earth. What would his weight (the gravitational force due to the Earth) be if he doubled his distance from the center of the Earth by flying in a spacecraft?

E4. How much larger is the required centripetal acceleration for a car rounding a curve at 60 MPH than for one rounding the same curve at 30 MPH? E5. A 0.25-kg ball moving in a circle at the end of a string has a centripetal acceleration of 4 m/s2. What is the magnitude of the centripetal force exerted by the string on the ball to produce this acceleration? E6. A car with a mass of 1200 kg is moving around a curve with a radius of 40 m at a constant speed of 20 m/s (about 45 MPH). a. What is the centripetal acceleration of the car? b. What is the magnitude of the force required to produce this centripetal acceleration? E7. A car with a mass of 1000 kg travels around a banked curve with a constant speed of 27 m/s (about 60 MPH). The radius of curvature of the curve is 40 m. a. What is the centripetal acceleration of the car? b. What is the magnitude of the horizontal component of the normal force that would be required to produce this centripetal acceleration in the absence of any friction? E8. A Ferris wheel at a carnival has a radius of 12 m and turns so that the speed of the riders is 8 m/s. a. What is the magnitude of the centripetal acceleration of the riders?

E11. Two masses are attracted by a gravitational force of 0.36 N. What will the force of attraction be if the distance between the two masses is tripled? E12. Two 200-kg masses (440 lb) are separated by a distance of 1 m. Using Newton’s law of gravitation, find the magnitude of the gravitational force exerted by one mass on the other. E13. Two masses are attracted by a gravitational force of 0.14 N. What will the force of attraction be if the distance between these two masses is halved? E14. The acceleration of gravity at the surface of the moon is approximately 1⁄ 6 that at the surface of the Earth (9.8 m/s2). What is the weight of an astronaut standing on the moon whose weight on Earth is 180 lb? E15. The acceleration of gravity on the surface of Jupiter is 26.7 m/s2. What is the weight on Jupiter of a woman whose weight on Earth is 110 lb? E16. The time separating high tides is 12 hours and 25 minutes. If high tide occurs at 2:10 P.M. one afternoon: a. At what time will high tide occur the next afternoon? b. When would you expect low tides to occur the next day?

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synthesis problems SP1. A 0.20-kg ball is twirled at the end of a string in a horizontal circle with a radius of 0.60 m. The ball travels with a constant speed of 4.0 m/s. a. What is the centripetal acceleration of the ball? b. What is the magnitude of the horizontal component of the tension in the string required to produce this centripetal acceleration? c. What is the magnitude of the vertical component of the tension required to support the weight of the ball? d. Draw to scale a vector diagram showing these two components of the tension and estimate the magnitude of the total tension from your diagram. (See appendix C.) SP2. A Ferris wheel with a radius of 12 m makes one complete rotation every 8 seconds. a. Using the fact that the distance traveled by a rider in one rotation is 2r, the circumference of the wheel, find the speed with which the riders are moving. b. What is the magnitude of their centripetal acceleration? c. For a rider with a mass of 40 kg, what is the magnitude of the centripetal force required to keep that rider moving in a circle? Is the weight of the rider large enough to provide this centripetal force at the top of the cycle? d. What is the magnitude of the normal force exerted by the seat on the rider at the top of the cycle? e. What would happen if the Ferris wheel is going so fast that the weight of the rider is not sufficient to provide the centripetal force at the top of the cycle? SP3. A car with a mass of 900 kg is traveling around a curve with a radius of 60 m at a constant speed of 25 m/s (56 MPH). The curve is banked at an angle of 15 degrees. a. What is the magnitude of the centripetal acceleration of the car? b. What is the magnitude of the centripetal force required to produce this acceleration? c. What is the magnitude of the vertical component of the normal force acting upon the car to counter the weight of the car? d. Draw a diagram of the car (as in fig. 5.8) on the banked curve. Draw to scale the vertical component of the normal force. Using this diagram, find the magnitude of the total normal force, which is perpendicular to the surface of the road. e. Using your diagram, estimate the magnitude of the horizontal component of the normal force. Is this component sufficient to provide the centripetal force?

SP4. Assume that a passenger in a rollover accident must turn through a radius of 3.0 m to remain in the seat of the vehicle. Assume also that the vehicle makes a complete turn in 1 second. a. Using the fact that the circumference of a circle is 2r, what is the speed of the passenger? b. What is the centripetal acceleration? How does it compare to the acceleration due to gravity? c. If the passenger has a mass of 60 kg, what is the centripetal force required to produce this acceleration? How does it compare to the passenger’s weight? SP5. The sun’s mass is 1.99 1030 kg, the Earth’s mass is 5.98 1024 kg, and the moon’s mass is 7.36 1022 kg. The average distance between the moon and the Earth is 3.82 108 m, and the average distance between the Earth and the sun is 1.50 1011 m. a. Using Newton’s law of gravitation, find the average force exerted on the Earth by the sun. b. Find the average force exerted on the Earth by the moon. c. What is the ratio of the force exerted on the Earth by the sun to that exerted by the moon? Will the moon have much of an impact on the Earth’s orbit about the sun? d. Using the distance between the Earth and the sun as the average distance between the moon and the sun, find the average force exerted on the moon by the sun. Will the sun have much impact on the orbit of the moon about the Earth? SP6. The period of the moon’s orbit about the Earth is 27.3 days, but the average time between full moons is approximately 29.3 days. The difference is due to the motion of the Earth about the sun. a. Through what fraction of its total orbital period does the Earth move in one period of the moon’s orbit? b. Draw a sketch of the sun, the Earth, and the moon with the moon in the full moon condition. Then, 27.3 days later sketch the moon’s position again for the new position of the Earth. If the moon is in the same position relative to Earth as it was 27.3 days earlier, is this a full moon? c. How much farther would the moon have to go to reach the full moon condition? Show that this represents approximately an extra two days.

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home experiments and observations HE1. Tape a string half a meter or so in length securely to a small rubber ball. Practice whirling the ball in both horizontal and vertical circles and make these observations: a. For horizontal motion of the ball, how does the angle that the string makes with the horizontal vary with the speed of the ball? b. If you let go of the string at a certain point in the circle, what path does the ball follow after release? c. Can you feel differences in tension in the string for different speeds of the ball? How does the tension vary with speed? d. For a vertical circle, how does the tension in the string vary for different points in the circle? Is it greater at the bottom than at the top when the ball moves with constant speed?

HE3. Observe the position and phase of the moon on several days in succession and at regularly chosen times during the day and evening. (It is probably best to choose a point near the first quarter of the moon’s cycle when the moon is visible in the afternoon and evening.) a. Sketch the shape of the moon on each successive day. Does this shape change for different times in the same day? b. Can you devise a method for accurately noting changes in the position of the moon at a set time, say, 10 P.M., on successive days? A fixed sighting point, a meter stick, and a protractor may be useful. Describe your technique. c. By how much does the position of the moon change from one day to the next at your regular chosen time?

HE2. Tie a small paper cup to a string, attaching it at two points near the rim as shown in the diagram. Take a marble or other small object and place it in the cup. a. Whirl the cup in a horizontal circle. Does the marble stay in the cup? (Be careful! A flying marble can be dangerous.) b. Whirl the cup in a vertical circle. Does the marble stay in the cup? What keeps the marble in the cup at the top of the circle? c. Try slowing the cup down. Does the marble stay in the cup? d. If you are brave, try replacing the marble with water. Under what conditions does the water stay in the cup?

HE4. Consult your instructor or other sources to find out what planets are observable in the evening during the current month. Venus, Jupiter, or Mars are usually the best candidates. a. Locate the planet visually and observe it with binoculars if possible. How does the planet differ in appearance from that of nearby stars? b. Sketch the position of the planet relative to nearby stars for several nights. How does this position change?

HE2 Diagram

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6

Energy and Oscillations

chapter overview We usually approach energy by first considering how it is added to a system. This involves the concept of work, which has a specialized meaning in physics. If a force does work on a system, the energy of the system increases. Work is a means of transferring energy. We begin by defining work and showing how to find it in simple cases. In different circumstances, work done on a system increases either the kinetic energy or the potential energy of the system. Finally, we will tie these ideas together by introducing the principle of conservation of energy and applying it to practical situations, including oscillations.

chapter outline

1

2 3 4 unit one

5

Simple machines, work, and power. What is a simple machine? How does the idea of work help us to understand the operation of simple machines? How do physicists define work, and how is work related to power? Kinetic energy. What is kinetic energy? When and how does work change the kinetic energy of an object? Potential energy. What is potential energy? When and how does work change the potential energy of an object? Conservation of energy. What is the total energy of a system, and when is it conserved? How can we use these ideas to explain the motion of a pendulum and other phenomena? Springs and simple harmonic motion. How is the motion of a mass on a spring like a pendulum? What is simple harmonic motion?

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H

ave you ever watched a ball on the end of a string swing back and forth? A pendant on the end of a chain (fig. 6.1), a swing in the park, and the pendulum on a grandfather clock all display the same hypnotic motion. Galileo (it is said) amused himself during boring sermons in church by watching the chandeliers sway slowly back and forth at the end of their chains. What intrigued Galileo is the way a pendulum always seems to return to the same position at the end of each swing. It may fall a little short of the earlier position in successive swings, but the motion goes on for a long time before coming to a complete stop. On the other hand, the velocity is continually changing, from zero at the end points of the swing to a maximum at the low point in the path. How can the pendulum go through such changes in velocity and yet always return to its starting point? Evidently, something is being saved or conserved. The quantity that remains constant (and is conserved) turns out to be what we now call energy. Energy did not play a role in Newton’s theory of mechanics. It was not until the nineteenth century that energy and energy transformations were elevated to the central position that they now hold in our understanding of the physical world. The motion of a pendulum and other types of oscillation can be understood using the principle of conservation of mechanical energy. The potential energy that the pendulum has at its end points is converted to kinetic energy at the low point—and then back to potential energy. What is energy, though, and how does it get

figure 6.1

A pendant swinging at the end of a chain. Why does it return to approximately the same point after each swing? into the system in the first place? Why does energy now play a central role in physics and all of science?

Energy is the basic currency of the physical world. To spend energy wisely, we must understand it. That understanding begins with the concept of work.

6.1 Simple Machines, Work, and Power If you make a pendulum by fastening a ball to the end of a string (fig. 6.2), what do you do to start it swinging? In other words, how do you get energy into the system? Usually, you would start by pulling the ball away from the center position directly below the point from which the string is suspended. To do so, you must apply a force to the ball with your hand and move the ball some distance. To a physicist, applying a force to move an object some distance involves doing work, even though the actual exertion may be slight. Doing work on a system increases the energy of the system, and this energy can then be used in the motion of the pendulum. How do we define work, though, and how can simple machines demonstrate the usefulness of the idea?

F

figure 6.2

The force applied does work to move the ball from its original position directly below the point of suspension.

What are simple machines? An early application of work was the analysis of the devices such as levers, pulley systems, or inclined planes that we call simple machines. A simple machine is any mechanical device that multiplies the effect of an applied force. A lever is one example of a simple machine. By applying a small

force at one end of a lever, a larger force can be exerted on the rock at the opposite end (fig. 6.3). What price do you pay for this multiplying effect of the applied force? To move the rock a small distance, the other end of the lever must move through a larger distance.

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F1

F2

T

d1

T

T

d2

figure 6.3

A lever is used to lift a rock. A small force F1 generates a larger force F2 to lift the rock, but F1 acts through a larger distance d1 than does F2. W

Generally, with simple machines, we get by with a small force if we are willing to apply that force over a large distance. The output force at the other end may be large, but it acts only over a short distance. The pulley system shown in figure 6.4 is another simple machine that achieves a similar result. In this system, the tension in the rope pulls up on either side of the pulley supporting the weight. If the system is in equilibrium, the tension in the rope is only half the weight being lifted, since there are, in effect, two ropes pulling up on the pulley. But to lift the pulley and its load a certain height, the person must move the rope twice the distance that the load moves. (Both rope segments on either side of the pulley must decrease in length by an amount equal to the increase in the height of the load.) The net result of using the pulley system illustrated in figure 6.4 is that you can lift a weight a certain height by applying a force equal to only half the weight being lifted. However, we must pull the rope twice the distance the weight is lifted. This way the product of the force and the distance moved will be the same for the input force applied by the person to the rope as for the output force exerted on the load. The quantity force times distance is thus conserved (if frictional losses are small). We call this product work, and the result for an ideal simple machine is work output work input. The ratio of the output force to the input force is called the mechanical advantage of the simple machine. For our pulley system, the mechanical advantage is 2. The output force that lifts the load is twice the input force exerted by the person pulling on the rope.

How is work defined? Our discussion of simple machines shows that the quantity force times distance has a special significance. Suppose

figure 6.4

A simple pulley system is used to lift a weight. The tension in the rope pulls up on either side of the lower pulley, so the tension is only half the size of the supported weight.

that you apply a constant horizontal force to a heavy crate to move it across a concrete floor, as illustrated in figure 6.5. You would agree that you have done work to move the crate and that the farther you move it, the more work you will do. The amount of work that you do also depends on how hard you have to push to keep the crate moving. These are the basic ideas that we use in defining work: work depends both on the strength of the applied force and the distance that the crate is moved. If the force and the distance moved are in the same direction, then work is the applied force multiplied by the distance that the crate moves under the influence of this force, or work force distance W Fd, where W is the work and d is the distance moved. The units of work will be units of force multiplied by units of distance, or newton-meters (Nm) in the metric system. We call this unit a joule (J). The joule is the basic metric unit of energy. (1 J 1 Nm) The first part of the sample exercise in example box 6.1 shows how we find the work done in a simple case. A horizontal force of 50 N is used to pull a crate a distance of 4 m, resulting in 200 J of work done on the crate by the applied force. In doing this work, we transfer 200 J of energy to the crate and its surroundings from the person applying the force. The person loses energy; the crate and its surroundings gain energy.

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Initial position

105

Final position F

d

figure 6.5

A crate is moved a distance d across a concrete floor under the influence of a constant horizontal force F.

example box 6.1

a. F 50 N

(50 N)(4 m)

W ?

200 J

d 4m W ?

F

30 N

W Fd

d 4m b. Fh 30 N

40 N

50

A crate is pulled a distance of 4 m across the floor under the influence of a 50-N force applied by a rope to the crate. What is the work done on the crate by the 50-N force if a. the rope is horizontal, parallel to the floor? b. the rope pulls at an angle to the floor, so that the horizontal component of the 50-N force is 30 N (fig. 6.6)?

N

Sample Exercise: How Much Work?

W Fh d (30 N)(4 m)

d

figure 6.6

A rope is used to pull a box across the floor. Only the portion of the force that is parallel to the floor is used in computing the work.

120 J

Does any force do work? In our initial example, the force acting on the crate was in the same direction as the motion produced. What about other forces acting on the crate—do they do work? The normal force of the floor pushes upward on the crate, for example, but the normal force has no direct effect in producing the motion because it is perpendicular to the direction of the motion. Forces perpendicular to the motion, such as the normal force or the gravitational force acting on the crate, do no work when the crate moves horizontally. What if the force acting on an object is neither perpendicular nor parallel to the direction of the object’s motion? In this case, we do not use the total force in computing work. Instead, we use only that portion or component of the force in the direction of the motion. This idea is illustrated in figure 6.6 and in the second part of example box 6.1. In figure 6.6, the rope used to pull the crate is at an angle to the floor, so that part of the applied force is directed

upward, rather than parallel to the floor. The box does not move in the direction of the force. Picture the force as having two components, one parallel to the floor and the other perpendicular to the floor. Only the component of the force in the direction of motion is used in computing the work. The component perpendicular to the motion does no work. By taking direction into account, we can complete the definition of work: The work done by a given force is the product of the component of the force along the line of motion of the object multiplied by the distance that the object moves under the influence of the force.

How is power related to work? When a car accelerates, energy is transferred from the fuel in the engine to the motion of the car. Work is done to move the car, but often we are more concerned with how

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fast this work is accomplished. The rate at which this work can be done depends on the power of the engine. The shorter the time, the greater the power. Power can be defined as

so that this is indeed the case? Work serves as the starting point.

Power is the rate of doing work; it is found by dividing the amount of work done by the time required.

Imagine that you are pushing a crate across the floor (fig. 6.5). If you place the crate on rollers with good bearings, the frictional forces may be small enough to be ignored. The force that you apply will then accelerate the crate. If you knew the mass of the crate, you could find its acceleration from Newton’s second law of motion. As the crate gains speed, you will have to move faster to keep applying a constant force. For equal time intervals, the crate would move larger distances as its speed increases, and you would find yourself doing work more rapidly. For constant acceleration, the distance traveled is proportional to the square of the final speed. The work done is therefore also proportional to the square of the speed. Since the work done should equal the increase in kinetic energy, the kinetic energy must increase with the square of the speed. If the crate begins from rest, the exact relationship is

power P

work time W t

In the first part of the example in example box 6.1, we computed a work value of 200 J for moving a crate 4 m across the floor using a force of 50 N. If the crate is in motion for 10 seconds, the power is found by dividing 200 J by 10 seconds, yielding a power of 20 J/s. A joule per second (J/s) is called a watt (W), the metric unit of power. We use watts commonly in discussing electric power, but watts are also used more generally for any situation involving the rate of transfer of energy. Another unit of power still used to describe the power of automobile engines is horsepower (hp). One horsepower is equal to 746 watts or 0.746 kilowatt (kW). The day may come when we routinely compare the power of different engines in kilowatts rather than in horsepower, but we are not there yet. The relationship of horsepower to the typical horse is dubious, but comparing the iron horse to the fleshand-blood kind still has a certain appeal. Work is the applied force times the distance moved, provided that the force acts along the line of motion of the object. In simple machines, work output can be no greater than work input, even though the output force is larger than the input force. Power is the rate of doing work: the faster the work is done, the greater the power. Doing work on an object increases the energy of the object or system, as in our initial example of pulling the pendulum bob away from its equilibrium position.

6.2 Kinetic Energy Suppose that the force applied to move a crate is the only force acting on the crate in the direction of motion. What happens to the crate then? According to Newton’s second law, the crate will accelerate, and its velocity will increase. Doing work on an object increases its energy. We call the energy associated with the motion of the object kinetic energy. Since work involves the transfer of energy, the amount of kinetic energy gained by the crate should be equal to the amount of work done. How can we define kinetic energy

How do we define kinetic energy?

1

work done change in kinetic energy 2 mv2. We often use the abbreviation KE to represent kinetic energy. Kinetic energy is the energy of an object associated with its motion and is equal to one-half the mass of the object times the square of its speed. 1

KE 2 mv2

Figure 6.7 illustrates the process. If the crate is initially at rest, its kinetic energy is equal to zero. After being accelerated over a distance d, it has a final kinetic energy of 1 2 2 mv , which is equal to the work done on the crate. The work done is actually equal to the change in kinetic energy. If the crate was already moving when you began pushing, its increase in kinetic energy would equal the work done. In example box 6.2, we highlight these ideas by calculating the energy gained by the crate in two different ways.

KE = 0

KE =

1 – mv 2 2

v=0 F

m

m

v

d

figure 6.7

The work done on an object by the net force acting on the object results in an increase in the object’s kinetic energy.

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Note that the frictional force exerted on the car acts in the opposite direction to the motion of the car shown in figure 6.8. When this is so, we say that the work done on the car by the force is negative work, removing energy from the system (the car) rather than increasing its energy. For a frictional force of magnitude f, the work done is W fd, if the car moves a distance d while decelerating.

example box 6.2 Sample Exercise: Work and Kinetic Energy Starting from rest on a frictionless floor, you move a 100-kg crate by applying a net force of 50 N for a time of 4 s. This results in a final speed of 2 m/s after the crate moves a distance of 4 m (see synthesis problem 2). Find a. the work done on the crate. b. the final kinetic energy of the crate. a. F 50 N d 4m

b.

Stopping distance for a moving car

W Fd (50 N)(4 m)

W ?

200 J

m 100 kg

KE

1 2

mv2

1 2

(100 kg)(2 m/s)2

v 2.0 m/s KE ?

200 J

In the first method, we use the definition of work. In the second, we use the definition of kinetic energy. We find that 200 J of work done on the crate results in an increase in kinetic energy of 200 J. It is no accident that these values are equal. Our definition of kinetic energy guarantees this to be true.

What is negative work? If work done on an object increases its kinetic energy, can work also decrease the energy of an object? Forces can decelerate objects as well as accelerate them. Suppose, for example, that we apply the brakes to a rapidly moving car, and the car skids to a stop. Does the frictional force exerted by the road surface on the tires of the car do work? When the car skids to a stop, it loses kinetic energy. A decrease in kinetic energy can be thought of as a negative change in kinetic energy. If the change in kinetic energy is negative, the work done on the car should also be negative.

The kinetic energy of the car is not proportional to the speed but rather to the square of the speed. If we double the speed, the kinetic energy quadruples. Four times as much work must be done to reach the doubled speed as was done to reach the original speed. Likewise, if we stop the car, four times as much energy must be removed. A practical application is the stopping distances of cars traveling at different speeds. The amount of negative work required to stop the car is equal to the kinetic energy of the car before the brakes are applied. This amount of energy must be removed from the system. Since kinetic energy is proportional to the square of the speed, the work required (and the stopping distance) increases rapidly with the speed of the car. For example, the kinetic energy is four times as large for a car traveling at 60 MPH as for one traveling at 30 MPH. Doubling the speed requires four times as much negative work to remove the kinetic energy. The stopping distance at 60 MPH will be four times that required at 30 MPH, since the work done is proportional to the distance (assuming the frictional force is constant). In fact, the frictional force varies with the speed of the car. If you look at the stopping distances in driver-training manuals, you will see that they do indeed increase rapidly with speed, although not exactly in proportion to the square of the speed. The more kinetic energy present initially, the more negative work is required to reduce this energy to zero, and the greater the stopping distance. Kinetic energy is the energy associated with an object’s motion, and it is equal to one-half the mass of the object times the square of its speed. The kinetic energy gained or lost by an object is equal to the work done by the net force accelerating or decelerating the object.

v=0 v

f

d W = –fd = ∆KE

figure 6.8

107

Frictional forces exerted on the car’s tires by the road surface do negative work in stopping the car, resulting in a decrease in kinetic energy.

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6.3 Potential Energy Suppose that we lift a crate to a higher position on a loading dock, as in figure 6.9. Work is done in this process, but no kinetic energy is gained if the crate ends up just sitting on the dock. Has the energy of the crate increased? What happens to the work done by the lifting force? Drawing back a bowstring or compressing a spring are similar. Work is done, but no kinetic energy is gained: instead the potential energy of the system increases. How does potential energy differ from kinetic energy?

Gravitational potential energy To lift the crate in figure 6.9, we need to apply a force that pulls or pushes upward on the crate. The applied force will not be the only force acting on the crate. The gravitational attraction of the Earth (the weight of the crate) pulls down on the crate. If we lift the crate with a force exactly equal to the force of gravity but opposite in direction, the net force acting on the crate will be zero, and the crate will not accelerate. We actually accelerate the crate a little bit at the start of the motion and decelerate it at the end of the motion, moving it with constant velocity during most of the motion. The work done by the lifting force increases the gravitational potential energy of the crate. The lifting force and the gravitational force are equal in magnitude and opposite in direction, so the net force is zero and there is no acceleration. The lifting force does work by moving the

object against the gravitational pull. If we let go of the rope, the crate will accelerate downward, gaining kinetic energy. How much gravitational potential energy is gained? The work done by the lifting force is equal to the size of the force times the distance moved. The applied force is equal to the weight of the crate mg. If the crate is moved a height h, the work done is mg times h or mgh. The gravitational potential energy is equal to the work done, PE mgh, where we use the abbreviation PE to represent potential energy. The height h is the distance that the crate moves above some reference level or position. In example box 6.3, we have chosen the original position of the crate on the ground to be our reference level. We usually choose the lowest point in the probable motion of the object as the reference level to avoid negative values of potential energy. The changes in potential energy are what is important, however, so the choice of reference level does not affect the physics of the situation.

The essence of potential energy The term potential energy implies storing energy to use later on for other purposes. Certainly, this feature is present in the situation just described. The crate could be left indefinitely higher up on the loading dock. If we push it off the dock, though, it would rapidly gain kinetic energy as it fell. The kinetic energy, in turn, could be used to compress objects lying underneath, drive pilings into the ground, or for other useful mayhem (fig. 6.10). Kinetic energy also has this feature, however, so storing energy is not what distinguishes potential energy. Potential energy involves changing the position of the object that is being acted on by a specific force. In the case of gravitational potential energy, that force is the gravitational attraction of the Earth. The farther we move the object away from the Earth, the greater the gravitational potential energy. Other kinds of potential energy involve different forces.

example box 6.3

F

Sample Exercise: Potential Energy h = 2.0 m W

A crate with a mass of 100 kg is lifted onto a loading dock 2 m above ground level. How much potential energy has been gained? m 100 kg h 2m

figure 6.9

A rope and pulley are used to lift a crate to a higher position on the loading dock, resulting in an increase in potential energy.

PE mgh (100 kg)(9.8 m/s2)(2 m) (980 N)(2 m) 1960 J

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by the spring constant multiplied by the distance stretched or F kx,

Coconut

figure

6.10 The potential energy of the raised crate can be converted to kinetic energy and used for other purposes.

What is elastic potential energy?

PE

1 2

kx2.

The potential energy of the stretched-spring system is onehalf the spring constant times the square of the distance stretched. The same expression is valid when the spring is compressed. The distance x is then the distance that the spring is compressed from its original relaxed position.

kx

Force

What happens if we pull on a bowstring or stretch a spring? In these examples, work is done by an applied force against an opposing elastic force, a force that results from stretching or compressing an object. Imagine a spring attached to a post, as in figure 6.11, with a wooden block or similar object attached to the other end of the spring. If we pull the block from the original position where the spring was unstretched, the system gains elastic potential energy. If we let go, the block would fly back. Since a force must be applied over some distance to move the block, work is done in pulling against the force exerted by the spring. Most springs exert a force proportional to the distance the spring is stretched. The more the spring is stretched, the greater the force. This can be stated in an equation by defining the spring constant k that describes the stiffness of the spring. A stiff spring has a large spring constant. The force exerted by the spring is given

where x is the distance that the spring is stretched, measured from its original unstretched position. This is often called Hooke’s Law, named after Robert Hooke (1635–1703). The minus sign indicates that the force exerted by the spring pulls back on the object as the object moves away from its equilibrium position. Thus, if the mass is moved to the right, the spring pulls back to the left. If the spring is compressed, it pushes back to the right. How do we find the increase in potential energy of such a system? As before, we need to find the work done by the force involved in changing the position of the object. We want the block to move without acceleration so the net force acting on the block is zero. The applied force must be adjusted so that it is always equal in magnitude but opposite in direction to the force exerted by the spring. This means that the applied force must increase as the distance x increases (fig. 6.12). The increase in elastic potential energy is equal to the work done by the average force needed to stretch the spring. Figure 6.12 suggests that the average force is one-half the magnitude of the final force kx. The work done is the aver1 age force 2 kx times the distance x, so

x

0 Distance stretched

figure 6.11

A wooden block is attached to a spring tied to a fixed support at the opposite end. Stretching the spring increases the elastic potential energy of the system.

figure

6.12 The applied force used to stretch the spring varies with the distance stretched, going from an initial value of zero to a final value of kx. Work done is equal to the shaded area under the Force versus Distance curve.

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The potential energy stored in the spring can be converted to other forms and put to various uses. If we let go of the block when the spring is either stretched or compressed, the block will gain kinetic energy. Cocking a bow and arrow, squeezing a rubber ball, or stretching a rubber band are all familiar examples in which we generate elastic potential energy similar to the spring.

What are conservative forces? Potential energy can result from work done against a variety of different forces besides gravity and springs. Work done against frictional forces, however, does not result in an increase in the potential energy of the system. Instead, heat is generated, which either transfers energy out of the system or increases the internal energy of the system at the atomic level. As discussed in chapter 11, this internal energy cannot be completely recovered to do useful work. Forces such as gravity or elastic forces that lead to potential energy relationships are referred to as conservative forces. When work is done against conservative forces, the energy gained by the system is completely recoverable for use in other forms. Potential energy is an object’s energy by virtue of its position along the line of action of some conservative force (such as gravity or the spring force). Potential energy is stored energy associated with the position of the object rather than the object’s motion. We find the potential energy by computing the work done to move the object against the conservative force. The system is poised to release that energy, converting it to kinetic energy or work done on some other system.

energy of the ball, since the height of the ball above the ground increases as the ball is pulled to the side. The work done transfers energy from the person doing the pulling to the system consisting of the pendulum and the Earth. It becomes gravitational potential energy, PE mgh, where h is the height of the ball above its initial position (fig. 6.13). When you release the ball, this potential energy begins to change to kinetic energy as the ball begins its swing. At the bottom of the swing (the initial position of the ball when it was just hanging), the potential energy is zero, and the kinetic energy reaches its maximum value. The ball does not stop at the low point; its motion continues to a point opposite the release point. During this part of the swing, the kinetic energy decreases, and the potential energy increases until it reaches the point where the kinetic energy is zero and the potential energy is equal to its initial value before release. The ball then swings back, repeating the transformation of potential energy to kinetic energy and back to potential energy (fig. 6.13).

What does it mean to say that energy is conserved? As the pendulum swings, there is a continuing change of potential energy to kinetic energy and back again. The total mechanical energy of the system (the sum of the potential and kinetic energies) remains constant because there is no work being done on the system to increase or decrease its energy. The swing of the pendulum demonstrates the principle of conservation of energy: If there are no nonconservative forces doing work on a system, the total mechanical energy of the system (the sum of its kinetic energy and potential energy) remains constant.

6.4 Conservation of Energy The concepts of work, kinetic energy, and potential energy are now available to us. How can they help explain what is happening in systems like a pendulum? Conservation of energy is the key. The total energy, the sum of the kinetic and potential energies, is a quantity that remains constant (is conserved) in many situations. We can describe the motion of a pendulum by tracking the energy transformations. What can this tell us about the system?

Energy changes in the swing of a pendulum Imagine a pendulum consisting of a ball initially hanging motionless at the end of a string attached to a rigid support. You pull the ball to the side and release it to start it swinging. What happens to the energy of the system? In the first step, work is done on the ball by your hand. The net effect of this work is to increase the potential

E = PE = mgh h

E = PE = mgh v

h

E = KE = 1_ mv 2 2

figure

6.13 Potential energy is converted to kinetic energy and then back to potential energy as the pendulum swings back and forth.

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6.4 Conservation of Energy

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Work is pivotal. If no energy is added or removed by forces doing work, the total energy should not change. In symbols, this statement takes the form: If W 0,

E PE KE constant,

where E is the symbol commonly used to represent the total energy. A broader picture of the meaning of this very important principle is provided in everyday phenomenon box 6.1 We applied conservation of energy in describing the motion of the pendulum. Some points deserve close attention: for example, why do we not consider the work done by gravity on the pendulum? The answer is that the gravitational force becomes part of the system by including the gravitational potential energy of the ball in our description. Gravity is a conservative force already accounted for by potential energy. What other forces act on the ball? The tension of the string acts in a direction perpendicular to the motion of the ball (fig. 6.14). This force does no work, because it has no component in the direction of the motion. The only other force that need concern us is air resistance. This force does negative work on the ball, slowly decreasing the total mechanical energy of the system. The total energy of the system is not completely constant in this situation. It would be constant only if air resistance were negligible. The air-resistive effects are often small, however, and can be ignored.

everyday phenomenon

T v R W

figure

6.14 Of the three forces acting on the ball, only the force of air resistance does work on the system to change its total energy. The tension does no work, and the work done by gravity is already included in the potential energy. Why do we use the concept of energy? What are the advantages of using the principle of conservation of energy? Imagine trying to describe the motion of the pendulum by direct application of Newton’s laws of motion. You would have to deal with forces that vary

box 6.1

Conservation of Energy The Situation. Mark Shoemaker had just come out of a physics lecture on the conservation of energy, and he was confused. His instructor had noted that the principle of conservation of energy in its most general form implies that energy can be neither created nor destroyed - it is always conserved. On the other hand, Mark had been following news items regarding the need to conserve energy. If energy is always conserved, what is the problem that news commentators and environmentalists are harping on? Don’t they understand physics? The Analysis. When people talk about conserving energy in the context of everyday energy use, the term conservation of energy has a different meaning than that involved in the physics principle. Understanding these distinctions is critical to understanding issues involving energy and the environment that have been hot topics in recent years. The principle of conservation of energy stated in this section is limited to mechanical energy, the sum of the potential

and kinetic energies of a mechanical system. In chapters 10 and 11, we will see that heat is also a form of energy and must be included in a more general statement of the conservation principle. Later, in chapter 20, we will find that even mass must be considered a form of energy. In this most general context, energy is always conserved. However, not all forms of energy have equal value in our daily lives. To take an example, oil is an important energy resource. The energy stored in oil is a form of potential energy involving the electric forces that bind atoms together in molecules. When we use oil, we are converting this potential energy to other forms of energy depending upon the application. To release this potential energy, we usually burn the oil. Burning generates heat, which is a form of energy. At high temperatures, this heat can be very useful for running heat engines of various kinds (discussed in chapter 11). The common gasoline engine in our cars is a heat engine, as are the

(continued)

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diesel engines that power trucks and trains and the jet turbines that power airplanes. These engines transform some energy in the form of heat to kinetic energy for cars, trains, boats, or airplanes. But what happens, ultimately, to the kinetic energy associated with the moving vehicles? Eventually, it is transformed to lower-temperature heat due to frictional effects in the engine and the tires and air resistance. The energy has not disappeared—it warms up the surroundings. However, it is now in a form that is much less useful than the original potential energy stored in the fuel. Heat at temperatures near those of the surroundings is sometimes called low-grade heat—its uses are limited to heating our homes or similar applications. So what are we conserving when we talk about energy conservation in our daily lives? We are conserving high-value forms of energy by using them more wisely and, as much as possible, preventing them from being converted to less useful forms of energy. We are also limiting the environmental effects associated with burning oil and other fuels such as natural gas or coal. From the standpoint of physics, though, the energy itself is conserved in all situations. If you have a choice of commuting to work or school by walking, riding a bicycle, or driving a fuel-efficient car as opposed to driving a large car or sports utility vehicle (SUV), your choice should be clear. By walking or riding a bicycle, as pictured in the photograph, you convert some energy obtained from the food you eat to low-grade heat, but much less high-value energy is converted than if you are driving an SUV or car. And, of course, your impact on the environment is much smaller when you are walking or riding a bicycle. Later chapters in this book deal with many aspects of energy use. The laws of thermodynamics, discussed in chapters 10 and 11, are particularly important to understanding energy issues. Solar energy, geothermal power, and other methods of generating electricity are discussed in chapter 11. Chapter 13 addresses household electrical energy uses, chapter 14 deals with electrical power generation, and chapter 19 discusses nuclear power.

continually in direction and magnitude as the pendulum moves. A full description using Newton’s laws is quite complex. Using energy considerations, however, we can make predictions about the behavior of a system much more easily than by applying Newton’s laws. To the extent that we can ignore frictional effects, for example, we can predict that the ball will reach the same height at either end of its swing. The kinetic energy is zero at the end points of the swing where the ball momentarily stops, and at these points, the total energy equals potential energy. If no energy has been lost, the potential energy has the same value that it had at the point of release, which implies that the same height is reached (PE mgh).

The physics and economics of wise energy use are critical issues. To be involved in these debates, you should understand what energy conservation is all about. Energy can be neither created nor destroyed, but the ways in which it is transformed from one form to another are extremely important to the use of energy resources and to the environment.

Is energy conserved for all these commuters?

A demonstration sometimes performed in physics lecture rooms illustrates this idea dramatically by using a bowling ball as the pendulum bob. The bowling ball is suspended from a support near the ceiling so that, when pulled to one side, the ball is near the chin of the physics instructor. The instructor pulls the ball to this position, releases it to allow it to swing across the room, and stands without flinching as the ball returns and stops just a few inches from his or her chin (fig. 6.15). Not flinching requires some faith in the principle of conservation of energy! The success of this demonstration depends on the ball not be given any initial velocity when released—what happens if it is pushed?

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example box 6.4 Sample Exercise: The Swing of a Pendulum A pendulum bob with a mass of 0.50 kg is released from a position in which the bob is 12 cm above the low point in its swing. What is the speed of the bob as it passes through the low point in its swing? m 0.5 kg

The initial energy is

h 12 cm

E PE mgh

v ? (at the low point)

(0.5 kg)(9.8 m/s2)(0.12 m) 0.588 J

At the low point, the potential energy is zero, so E KE 0.588 J 1 2

figure 6.15

A bowling ball at the end of a cable suspended from the ceiling is released and allowed to swing across the room and back, stopping just in time.

We can also use the principle of conservation of energy to predict what the speed will be at any point in the swing. The speed is zero at the end points and has its maximum value at the low point of the swing. If we place our reference level for measuring potential energy at this low point, the potential energy will be zero there because the height is zero. All of the initial potential energy has been converted to kinetic energy. Knowing the kinetic energy at the low point allows us to compute the speed, as shown in example box 6.4. We could find the speed at any other point in the swing by setting the total energy at any point equal to the initial energy. Different values of the height h above the low point yield different values of the potential energy. The remaining energy must be kinetic energy. The system has only so much energy, either potential or kinetic energy or some of both, but it cannot exceed the initial value.

How is energy analysis like accounting? A sled on a hill and a roller coaster illustrate the principle of conservation of energy. Conservation of energy can be used to make predictions about the speed of the sled or roller coaster that would be hard to make by direct application of Newton’s laws. An energy accounting provides a better overview. The pole-vaulting example in everyday phenomenon box 6.2 can also be analyzed in this way. Consider the sled on the hill pictured in figure 6.16. A parent pulls the sled to the top of a hill, doing work on the sled and rider that increases their potential energy. At the top of the hill, the parent may do more work by giving

mv2 0.588 J

Dividing both sides by 12 m: v2

KE 1 2m

(0.588 J) 1 2

(0.5 kg)

2.35 m2/s2 Taking the square root of both sides: v 1.53 m/s

the sled a push, providing it with some initial kinetic energy. The total work done by the parent is the energy input to the system and equals the sum of the potential and kinetic energies shown in table 6.1. Where does this initial energy come from? It came from the body of the parent doing the pulling and pushing. Muscle groups were activated, releasing chemical potential energy stored in the body. That energy came from food, which in turn involved solar energy stored by plants. A parent who does not eat a good breakfast, or attempts too many trips up the hill, may not have enough energy to get to the top. If the sled and rider slide down the hill with negligible friction and air resistance, energy is conserved, and the total energy at any point during the motion should equal the initial energy. It is more realistic to assume that there is some friction as the sled slides down the hill (fig. 6.17). Although it is difficult to predict the amount of work done against friction precisely, we can make an estimate if we know the total distance traveled and make some assumptions about the size of the average frictional force. In the energy accounting done in table 6.1, we assume that

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everyday phenomenon

box 6.2

Energy and the Pole Vault The Situation. Ben Lopez goes out for track. He specializes in the pole vault and helps out sometimes with the sprint relays where his speed can be used to good advantage. His coach, aware that Ben is also taking an introductory physics course, suggests that Ben try to understand the physics of the pole vault. What factors determine the height reached? How can he optimize these factors? The coach knows that energy considerations are important in the pole vault. What type of energy transformations are involved? Could understanding these effects help Ben’s performance?

The flexibility of the pole and the point at which the vaulter grasps the pole are important to the success of the vault.

A pole-vaulter on the way up. What energy transformations are taking place?

The Analysis. It was not difficult for Ben to describe the energy transformations that take place in the pole vault: the vaulter begins by running down a path to the vaulting standard and pit. During this phase, he is accelerating and increasing his kinetic energy at the expense of chemical energy stored in his muscles. When he reaches the standard, he plants the end of the pole in a notch in the ground. At this point, some of his kinetic energy is stored in the elastic potential energy of the bent pole, which acts like a spring. The rest is converted to gravitational potential energy as he begins to rise over the standard. Near the top of the vault, the elastic potential energy in the bent pole converts to gravitational potential energy as the pole straightens out. The vaulter does some additional work with his arm and upper-body muscles to provide an extra boost. At the very top of his flight, his kinetic energy should be zero, with only a minimal horizontal velocity left to carry him over the standard. Too large a kinetic energy at this

point would indicate that he had not optimized his jump by converting as much energy as possible into gravitational potential energy. What can Ben learn from his analysis? First, the importance of speed. The more kinetic energy he generates during his approach, the more energy is available for conversion to gravitational potential energy (mgh), which will largely determine the height of his vault. Successful pole-vaulters are usually good sprinters. The characteristics of the pole and Ben’s grip on it are also important factors. If the pole is too stiff, or if he has gripped it too close to the bottom, he will experience a jarring impact in which little useful potential energy is stored in the pole, and some of his initial kinetic energy will be lost in the collision. If the pole is too limber, or if Ben’s grip is too far from the bottom, it will not spring back soon enough to provide useful energy at the top of his vault. Finally, upper-body strength is important in clearing the standard. Good upper-body conditioning should improve Ben’s pole-vaulting. Timing and technique are also critical and can be improved only through practice. As far as his coach is concerned, that may be the most important message.

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6.5 Springs and Simple Harmonic Motion

E = PE = mgh 0

E = PE + KE v

F

W ⇒ ∆PE h 0 = 20 m 15 m

E = KE = _1 mv 2 2

v

figure

6.16 Work done in pulling the sled up the hill produces an increase in potential energy of the sled and rider. This initial energy is then converted to kinetic energy as they slide down the hill. 2000 J of work has been done against friction by the time the sled reaches the bottom of the hill. The work done against friction removes energy from the system and shows up as an expenditure on the account sheet. The energy balance at the bottom of the hill is 8200 J, rather than 10 200 J. This will lead to a smaller, more realistic value for the speed of the sled and rider at the bottom of the hill than if we ignored friction. Although precise calculations are not always possible, energy accounting sets limits on what is likely and helps us understand the behavior of systems such as the sled on the hill.

table 6.1 Energy Balance Sheet for the Sled A parent pulls a sled and rider with a combined weight of 50 kg to the top of a hill 20 m high and then gives the sled a push, providing an initial velocity of 4 m/s. Frictional forces acting on the sled do 2000 J of negative work as the sled moves down the hill. Energy input Potential energy gained by work done in pulling sled up the hill:

PE mgh (50 kg)(9.8 m/s2)(20 m) Energy is the currency of the physical world; an understanding of energy accounting is relevant to both science and economics. Doing work on a system puts energy in the bank. Total energy is then conserved, provided that only conservative forces are at work. Many aspects of the motion of the system can be predicted from a careful energy accounting.

Kinetic energy gained by work done in pushing the sled at the top: KE 21 mv 2 12 (50 kg)(4 m/s)2 Total initial energy:

400 J 10 200 J

Energy expenditures Work done against friction as the sled slides down the hill: W fd

W = –fd

9 800 J

Energy balance:

2 000 J 8 200 J

f

v

figure

6.17 The work done by frictional forces is negative, and it removes mechanical energy from the system.

6.5 Springs and Simple Harmonic Motion If conservation of energy explains the motion of a pendulum, what about other systems that oscillate? Many systems involve springs or elastic bands that move back and forth, with potential energy being converted to kinetic

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energy and then back to potential energy repeatedly. What do such systems have in common? What makes them tick? A mass on the end of a spring is one of the simplest oscillating systems. This system, and the simple pendulum described in section 6.4, are examples of simple harmonic motion.

Oscillation of a mass attached to a spring If we attach a block to the end of a spring, as in figure 6.18, what happens when we pull it to one side of its equilibrium position? The equilibrium position is where the spring is neither stretched nor compressed. Doing work to pull the mass against the opposing force of the spring increases the potential energy of the spring-mass system. The poten1 tial energy in this case is elastic potential energy, 2 kx2, rather than the gravitational potential energy associated with the pendulum. Increasing the potential energy of the mass on the spring is similar to cocking a bow and arrow or slingshot. Once the mass is released, potential energy is converted to kinetic energy. Like the pendulum, the motion of the mass carries it beyond the equilibrium position, and the spring is compressed, gaining potential energy again. When the kinetic energy is completely reconverted to potential energy, the mass stops and reverses, and the whole process repeats (fig. 6.18). The energy of the system changes continuously from potential energy to kinetic energy and back again. If frictional effects can be ignored, the total energy of the system remains constant while the mass oscillates back and forth. Using a video camera or other tracking techniques, it is possible to measure and plot the position of a pendulum Potential energy

x=0 (Equilibrium point) v Kinetic energy 0

Potential energy

T Position (x)

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amplitude 0

Time (t)

figure 6.19

The horizontal position x of the mass on the spring is plotted against time as the mass moves back and forth. The resulting curve is a harmonic function.

bob or mass on a spring as it varies with time. If we plot the position of the mass against time, the resulting curve takes the form shown in figure 6.19. The mathematical functions that describe such curves are called “harmonic” functions, and the motion is called simple harmonic motion,* a term probably borrowed from musical descriptions of sounds produced by vibrating strings, reeds, and air columns. (See chapter 15.) The line at zero on the graph in figure 6.19 is the equilibrium position for the mass on a spring. Points above this line represent positions on one side of the equilibrium point, and those below the line represent positions on the other side. The motion starts at the point of release, where the distance of the mass from equilibrium is a maximum. As the mass moves toward the equilibrium position (x 0 on the graph) it gains speed, indicated by the increasing slope of the curve. (See section 2.4.) The object’s position changes most rapidly when it is near the equilibrium point, where the kinetic energy and speed are the greatest. As the mass passes through the equilibrium position, it starts to move away from equilibrium in the direction opposite to its initial position. The force exerted by the spring is now in the direction opposite to the velocity and is decelerating the mass. When the mass reaches the point farthest from its release point, the speed and kinetic energy are again zero, and the potential energy has returned to its maximum value. (See example box 6.5.) The slope of the curve is zero at this point, indicating that the mass is momentarily stopped (its velocity is zero). The mass continually gains or loses speed as it moves back and forth.

What are the period and the frequency?

If you look at the graph in figure 6.19, you will notice that the curve repeats itself regularly. The period T is the repeat time, or the time taken for one complete cycle. It is

6.18 Energy added by doing work to stretch the spring is then transformed back and forth between the potential energy of the spring and kinetic energy of the mass.

*If you have studied trigonometry, you may know that the curve plotted in figure 6.19 is a cosine function. Sines and cosines are collectively referred to as harmonic functions.

figure

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6.5 Springs and Simple Harmonic Motion

example box 6.5 Sample Exercise: Motion of a Mass on a Spring A 500-g mass (0.50 kg) is undergoing simple harmonic motion at the end of a spring with a spring constant of 800 N/m. The motion takes place on a frictionless horizontal surface as pictured in figure 6.18. The speed of the mass is 12 m/s when it passes through the equilibrium point. a. What is the kinetic energy at the equilibrium point? b. How far does the mass travel from the equilibrium point before it turns around? a.

m 0.50 kg

KE

v 12 m/s

KE

KE ? b.

x ? (when v 0)

1 2 1 2

mv2 (0.50 kg) (12 m/s)2

KE 36 J E KE PE 36 J but KE 0 at the turn-around point so PE

1 2

kx2 36 J

x2

2(36 J) k

x2

72 J 800 N/m

x2 0.09 m2 x 0.30 m 30 cm

usually measured in seconds. You can think of the period as the time between adjacent peaks or valleys on the curve. A slowly oscillating system has a long period, and a rapidly oscillating system has a short period. Suppose that the period of oscillation for a certain spring and mass is half a second. There are then two oscillations each second, which is the frequency of oscillation. The frequency f is the number of cycles per unit time, and it is found by taking the reciprocal of the period, f 1/T. A rapidly oscillating system has a very short period and thus a high frequency. The unit commonly used for frequency is the hertz, which is defined as one cycle per second. What determines the frequency of the spring-mass system? Intuitively, we expect a loose spring to have a low frequency of oscillation and a stiff spring to have a high frequency. This is indeed the case. The mass attached to the spring also has an effect. Larger masses offer greater resistance to a change in motion, producing lower frequencies. The period and frequency of oscillation of a pendulum depend primarily on its length, measured from the pivot point to the center of the bob. To measure the period, you usually measure the time required for several complete

117

swings and then divide by the number of swings to get the time for one swing. Simple experiments with a ball on a string will give you an idea of how the period and frequency change with length. Try it and see if you can find a trend (see home experiment 1). The motion is regular—you can keep time by the swing of a pendulum or the motion of a mass on a spring.

Will any restoring force produce simple harmonic motion? When a mass attached to a spring is moved to either side of equilibrium, the spring exerts a force that pulls or pushes the mass back toward the center. We call such a force the restoring force. In this case, it is the elastic force exerted by the spring. In any oscillation, there must be some such restoring force. As discussed in section 6.3, the spring force is directly proportional to the distance x of the mass from its equilibrium position (F kx). The spring constant k has units of newtons per meter (N/m). Simple harmonic motion results whenever the restoring force has this simple dependence on distance. If the force varies in a more complicated way with distance, we may get an oscillation but not simple harmonic motion, and it will not produce a simple harmonic curve (fig. 6.19). It is generally easiest to set up a spring-mass system by suspending the spring from a vertical support and hanging a mass on the end of the spring, as in figure 6.20. This arrangement avoids the frictional forces of the tabletop in a

x=0

figure

6.20 A mass hanging from a spring will oscillate up and down with the same period as a horizontal mass-spring system using the same spring and mass.

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horizontal arrangement. In the vertical setup, when the mass is pulled down and released, the system oscillates up and down rather than horizontally. Two forces then act on the mass, the spring force pulling upward and the gravitational force pulling downward. Since the gravitational force in the vertical setup is constant, it simply moves the equilibrium point lower. The equilibrium point is where the net force is zero—the downward pull of gravity is balanced by the upward pull of the spring. The variations in the restoring force are still provided by the spring. These variations are proportional to the distance from equilibrium just as they are in the horizontal case. This system also meets the condition for simple harmonic motion. The potential energy involved, however, is the sum of the gravitational and elastic potential energies. Gravity is the restoring force for the simple pendulum. When the pendulum bob is pulled to one side of its equilibrium position, the gravitational force acting on the bob pulls it back toward the center. The part of the gravitational force in the direction of motion is proportional to the displacement, if the displacement from equilibrium is not too

large. Thus, for small amplitudes of swing, the simple pendulum also displays simple harmonic motion. Amplitude is the maximum distance from the equilibrium point. Look around for systems that oscillate. There are many examples, ranging from a springy piece of metal to a ball rolling in a depression of some kind. What force pulls back toward the equilibrium position in each case? Is the motion likely to be simple harmonic motion, or a more complicated oscillation? What kind of potential energy is involved? The analysis of vibrations such as these forms an important subfield of physics that plays a role in music, communications, analysis of structures, and other areas. Any oscillation involves a continuing interchange of potential and kinetic energies. If there are no frictional forces removing energy from the system, the oscillation will go on indefinitely. A restoring force that increases in direct proportion to the distance from the equilibrium position results in simple harmonic motion, with simple curves (harmonic functions) describing the position, velocity, and acceleration of the object over time.

summary The concept of work is central to this chapter. Energy is transferred into a system by doing work on the system, which can result in an increase in either the kinetic energy or the potential energy of the system. If no additional work is done on the system, the total energy of the system remains constant. This principle of conservation of energy allows us to explain many features of the behavior of the system.

3

Potential energy. If work done on an object moves the object against an opposing conservative force, the potential energy of the object is increased. Two types of potential energy were considered, gravitational potential energy and elastic potential energy.

machines, work, and power. Work is de1fined asSimple force times the distance involved in moving an object.

m

Only the portion of the force in the direction of the motion is used. In simple machines, work output cannot exceed work input. Power is the rate of doing work.

x

F PE = mgh (gravitational)

d W P=— t

W = Fd,

2

Kinetic energy. The work done by the net force acting on an object is used to accelerate the object, and the object gains kinetic energy. Kinetic energy is equal to one-half the mass of the object times the square of its speed. Negative work removes kinetic energy. m

k

h

PE =

(elastic)

4

Conservation of energy. If no work is done on a system, the total mechanical energy (kinetic plus potential) remains constant. This principle of conservation of energy explains the behavior of many systems that involve exchanges of kinetic and potential energy. The system can be analyzed by energy accounting.

v

KE = 1_ mv 2 2

1 –kx 2 2

E = KE + PE = constant

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5 Springs and simple harmonic motion. The motions of a simple pendulum and of a mass on a spring both illustrate the

F

principle of conservation of energy, but they involve different kinds of potential energy. They are also examples of simple harmonic motion, which results whenever the restoring force is proportional to the distance of the object from its equilibrium position.

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F

x

x F = –kx

key terms Potential energy, 108 Gravitational potential energy, 108 Elastic force, 109 Spring constant, 109 Elastic potential energy, 109 Conservative forces, 110

Simple machine, 103 Mechanical advantage, 104 Work, 104 Power, 106 Kinetic energy, 106 Negative work, 107

Conservation of energy, 110 Simple harmonic motion, 116 Period, 116 Frequency, 117 Restoring force, 117 Amplitude, 118

Q1. Equal forces are used to move blocks A and B across the floor. Block A has twice the mass of block B, but block B moves twice the distance moved by block A. Which block, if either, has the greater amount of work done on it? Explain. Q2. A man pushes very hard for several seconds upon a heavy rock, but the rock does not budge. Has the man done any work on the rock? Explain. Q3. A string is used to pull a wooden block across the floor without accelerating the block. The string makes an angle to the horizontal as shown in the diagram. a. Does the force applied via the string do work on the block? Explain. b. Is the total force involved in doing work or just a portion of the force? Explain.

the ball moving in the circle at constant speed. Does the force exerted by the string on the ball do work on the ball in this situation? Explain. *Q7. A man walks across the room. What forces act on the man during this process? Which, if any, of these forces do work on the man? Explain. Q8. A woman uses a pulley arrangement to lift a heavy crate. She applies a force that is one-fourth the weight of the crate, but moves the rope a distance four times the height that the crate is lifted. Is the work done by the woman greater than, equal to, or less than the work done by the rope on the crate? Explain. Q9. A lever is used to lift a rock as shown in the diagram. Will the work done by the person on the lever be greater than, equal to, or less than the work done by the lever on the rock? Explain. F

F

Q3 Diagram d

Q3 Diagram Q4. In the situation pictured in question 3, if there is a frictional force opposing the motion of the block, does this frictional force do work on the block? Explain. Q5. In the situation pictured in question 3, does the normal force of the floor pushing upward on the block do any work? Explain. Q6. A ball is being twirled in a circle at the end of a string. The string provides the centripetal force needed to keep

*Q10. A crate on rollers is pushed up an inclined plane into a truck. The pushing force required is less than half the force that would be needed to lift the crate straight up into the truck. Does the inclined plane serve as a simple machine in this situation? Explain. Q11. A boy pushes his friend across a skating rink. Since the frictional forces are very small, the force exerted by the boy on his friend’s back is the only significant force acting on the friend in the horizontal direction. Is the change in kinetic energy of the friend greater than, equal to, or less than the work done by the force exerted by the boy? Explain.

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Q12. A child pulls a block across the floor with force applied by a horizontally held string. A smaller frictional force also acts upon the block, yielding a net force on the block that is smaller than the force applied by the string. Does the work done by the force applied by the string equal the change in kinetic energy in this situation? Explain.

a. What form of energy is added to the system prior to its release? Explain. b. At what points in the motion of the pendulum after release is its kinetic energy the greatest? Explain. c. At what points is the potential energy the greatest? Explain.

Q13. If there is just one force acting on an object, does its work necessarily result in an increase in kinetic energy? Explain.

Q23. For the pendulum in question 22––when the pendulum bob is halfway between the high point and the low point in its swing––is the total energy kinetic energy, potential energy, on both? Explain.

Q14. Two balls of the same mass are accelerated by different net forces such that one ball gains a velocity twice that of the other ball in the process. Is the work done by the net force acting on the faster-moving ball twice that done on the slower-moving ball? Explain. Q15. A box is moved from the floor up to a tabletop but gains no speed in the process. Is there work done on the box, and if so, what has happened to the energy added to the system? Q16. When work is done to increase the potential energy of an object without increasing its kinetic energy, is the net force acting on the object greater than zero? Explain. Q17. Is it possible for a system to have energy if nothing is moving in the system? Explain. Q18. Suppose that work is done on a large crate to tilt the crate so that it is balanced on one edge, as shown in the diagram, rather than sitting squarely on the floor as it was at first. Has the potential energy of the crate increased in this process? Explain.

Q24. Is the total mechanical energy conserved in the motion of a pendulum? Will it keep swinging forever? Explain. Q25. A sports car accelerates rapidly from a stop and “burns rubber.’’ a. What energy transformations occur in this situation? b. Is energy conserved in this process? Explain. Q26. A man commutes to work in a large sport utility vehicle (SUV). a. What energy transformations occur in this situation? b. Is mechanical energy conserved in this situation? Explain. c. Is energy of all forms conserved in this situation? Explain. Q27. Suppose that we burn a barrel of oil just to warm our hands on a cold day. a. From the standpoint of physics, is energy conserved in this process? Explain. b. Why is this a bad idea from an economic or environmental standpoint? Explain. *Q28. A bird grabs a clam, carries it in its beak to a considerable height, and then drops it on a rock below, breaking the clam shell. Describe the energy conversions that take place in this process. Q29. A mass attached to a spring, which in turn is attached to a wall, is free to move on a friction-free horizontal surface. The mass is pulled back and then released. a. What form of energy is added to the system prior to the release of the mass? Explain. b. At what points in the motion of the mass after its release is its potential energy the greatest? Explain. c. At what points is the kinetic energy the greatest? Explain.

Q18 Diagram *Q19. Which has the greater potential energy: a ball that is 10 feet above the ground, or one with the same mass that is 20 feet above the bottom of a nearby 50-foot-deep well? Explain.

Q30. Suppose that the mass in question 26 is halfway between one of the extreme points of its motion and the center point. In this position, is the energy of the system kinetic energy, potential energy, or a combination of these forms? Explain. *Q31. A spring gun is loaded with a rubber dart, the gun is cocked, and then fired at a target on the ceiling. Describe the energy transformations that take place in this process.

Q20. When a bow and arrow are cocked, a force is applied to the string in order to pull it back. Is the energy of the system increased? Explain.

Q32. Suppose that a mass is hanging vertically at the end of a spring. The mass is pulled downward and released to set it into oscillation. Is the potential energy of the system increased or decreased when the mass is lowered? Explain.

Q21. Suppose that the physics instructor pictured in figure 6.15 gives the bowling ball a push as she releases it. Will the ball return to the same point or will her chin be in danger? Explain.

*Q33. A sled is given a push at the top of a hill. Is it possible for the sled to cross a hump in the hill that is higher than its starting point under these circumstances? Explain.

Q22. A pendulum is pulled back from its equilibrium (center) position and then released.

*Q34. Can work done by a frictional force ever increase the total mechanical energy of a system? (Hint: Consider the acceleration of an automobile.) Explain.

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Q35. Suppose that a pulley system is used to lift a heavy crate, but the pulleys have rusted and there are frictional forces acting on the pulleys. Will the useful work output of this system be greater than, equal to, or less than the work input? Explain.

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Q36. Is the elastic potential energy stored in the pole the only type of potential energy involved in pole-vaulting? Explain. Q37. If one pole-vaulter can run faster than another, will the faster runner have an advantage in the pole vault? Explain.

exercises E1. A horizontally directed force of 40 N is used to pull a box a distance of 2.5 m across a tabletop. How much work is done by the 40-N force? E2. A woman does 160 J of work to move a table 4 m across the floor. What is the magnitude of the force that the woman applied to the table if this force is applied in the horizontal direction? E3. A force of 60 N used to push a chair across a room does 300 J of work. How far does the chair move in this process? E4. A rope applies a horizontal force of 180 N to pull a crate a distance of 2 m across the floor. A frictional force of 60 N opposes this motion. a. What is the work done by the force applied by the rope? b. What is the work done by the frictional force? c. What is the total work done on the crate? E5. A force of 50 N is used to drag a crate 4 m across a floor. The force is directed at an angle upward from the crate so that the vertical component of the force is 30 N and the horizontal component is 40 N as shown in the diagram. a. What is the work done by the horizontal component of the force? b. What is the work done by the vertical component of the force? c. What is the total work done by the 50-N force?

30 N

50

N

40 N 4m

E5 Diagram E6. A net force of 60 N accelerates a 4-kg mass over a distance of 10 m. a. What is the work done by this net force? b. What is the increase in kinetic energy of the mass? E7. A 0.4-kg ball has a velocity of 20 m/s. a. What is the kinetic energy of the ball? b. How much work would be required to stop the ball? E8. A box with a mass of 5.0 kg is lifted (without acceleration) through a height of 2.0 m, in order to place it upon the shelf of a closet. a. What is the increase in potential energy of the box?

b. How much work was required to lift the box to this position? E9. A spring with a spring constant k of 40 N/m is stretched a distance of 20 cm (0.20 m) from its original unstretched position. What is the increase in potential energy of the spring? E10. To stretch a spring a distance of 0.20 m, 40 J of work is done. a. What is the increase in potential energy of the spring? b. What is the value of the spring constant k of the spring? E11. Which requires more work: lifting a 2-kg rock to a height of 4 m without acceleration, or accelerating the same rock horizontally from rest to a speed of 10 m/s? E12. At the low point in its swing, a pendulum bob with a mass of 0.2 kg has a velocity of 4 m/s. a. What is its kinetic energy at the low point? b. Ignoring air resistance, how high will the bob swing above the low point before reversing direction? E13. A 0.20-kg mass attached to a spring is pulled back horizontally across a table so that the potential energy of the system is increased from zero to 120 J. Ignoring friction, what is the kinetic energy of the system after the mass is released and has moved to a point where the potential energy has decreased to 80 J? E14. A sled and rider with a combined mass of 50 kg are at the top of a hill a height of 15 m above the level ground below. The sled is given a push providing an initial kinetic energy at the top of the hill of 1600 J. a. Choosing a reference level at the bottom of the hill, what is the potential energy of the sled and rider at the top of the hill? b. After the push, what is the total mechanical energy of the sled and rider at the top of the hill? c. If friction can be ignored, what will be the kinetic energy of the sled and rider at the bottom of the hill? E15. A roller-coaster car has a potential energy of 450 000 J and a kinetic energy of 120 000 J at point A in its travel. At the low point of the ride, the potential energy is zero, and 50 000 J of work have been done against friction since it left point A. What is the kinetic energy of the roller coaster at this low point? E16. A roller-coaster car with a mass of 1200 kg starts at rest from a point 20 m above the ground. At point B, it is 12 m above the ground. a. What is the initial potential energy of the car? b. What is the potential energy at point B? c. If the initial kinetic energy was zero and the work done against friction between the starting point and point B is 30 000 J, what is the kinetic energy of the car at point B?

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E17. A 200-g mass lying on a frictionless table is attached to a horizontal spring with a spring constant of 400 N/m. The spring is stretched a distance of 40 cm (0.40 m). a. What is the initial potential energy of the system? b. What is the kinetic energy of the system when the mass returns to the equilibrium position after being released?

E18. The time required for one complete cycle of a mass oscillating at the end of a spring is 0.25 s. What is the frequency of oscillation? E19. The frequency of oscillation of a pendulum is 8 cycles/s. What is the period of oscillation?

synthesis problems SP1. Suppose that two horizontal forces are acting upon a 0.25-kg wooden block as it moves across a laboratory table: a 5-N force pulling the block and a 2-N frictional force opposing the motion. The block moves a distance of 1.5 m across the table. a. What is the work done by the 5-N force? b. What is the work done by the net force acting upon the block? c. Which of these two values should you use to find the increase in kinetic energy of the block? Explain. d. What happens to the energy added to the system via the work done by the 5-N force? Can it all be accounted for? Explain. e. If the block started from rest, what are its kinetic energy and velocity at the end of the 1.5-m motion? SP2. As described in example box 6.2, a 100-kg crate is accelerated by a net force of 50 N applied for 4 s. a. What is the acceleration of the crate from Newton’s second law? b. If it starts from rest, how far does it travel in the time of 4 s? (See section 2.5 in chapter 2.) c. How much work is done by the 50-N net force? d. What is the velocity of the crate at the end of 4 s? e. What is the kinetic energy of the crate at this time? How does this value compare to the work computed in part c? SP3. A slingshot consists of a rubber strap attached to a Y-shaped frame, with a small pouch at the center of the strap to hold a small rock or other projectile. The rubber strap behaves much like a spring. Suppose that for a particular slingshot a spring constant of 600 N/m is measured for the rubber strap. The strap is pulled back approximately 40 cm (0.4 m) prior to being released. a. What is the potential energy of the system prior to release? b. What is the maximum possible kinetic energy that can be gained by the rock after release? c. If the rock has a mass of 50 g (0.05 kg), what is its maximum possible velocity after release? d. Will the rock actually reach these maximum values of kinetic energy and velocity? Does the rubber strap gain kinetic energy? Explain. SP4. Suppose that a 200-g mass (0.20 kg) is oscillating at the end of a spring upon a horizontal surface that is essentially friction-free. The spring can be both stretched and compressed

and has a spring constant of 240 N/m. It was originally stretched a distance of 12 cm (0.12 m) from its equilibrium (unstretched) position prior to release. a. What is its initial potential energy? b. What is the maximum velocity that the mass will reach in its oscillation? Where in the motion is this maximum reached? c. Ignoring friction, what are the values of the potential energy, kinetic energy, and velocity of the mass when the mass is 6 cm from the equilibrium position? d. How does the value of velocity computed in part c compare to that computed in part b? (What is the ratio of the values?) SP5. A sled and rider with a total mass of 40 kg are perched at the top of a hill as pictured in the diagram. The top of this hill is 40 m above the low point in the path of the sled. A second hump in the hill is 30 m above this low point. Suppose that we also know that approximately 2000 J of work is done against friction as the sled travels between these two points. a. Will the sled make it to the top of the second hump if no kinetic energy is given to the sled at the start of its motion? Explain. b. What is the maximum height that the second hump could be in order for the sled to reach the top, assuming that the same work against friction will be involved and that no initial push is provided? Explain.

40 m 30 m

SP5 Diagram

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SP6. Suppose you wish to compare the work done by pushing a box on rollers up a ramp to the work done if you lift the box straight up to the same final height. a. What work is required to lift a 100N box (about 22 lbs) up to a table which is 1 m off the floor? b. Let’s assume you also have a ramp available that makes an angle of 30° with the horizontal, as shown in the figure to the right. The ramp is 2 m long. The weight of the box (100 N) is due to the Earth pulling on the box. This 100 N is a force directed straight down. If you push it up a ramp, you are doing work against only the component of this weight along the ramp, which is 50 N, as shown in the diagram. How much work does it require to push the box up the ramp, assuming no friction? c. Which situation (pushing up the ramp or lifting straight up) requires more work? d. Which situation requires more force? e. For which situation is the distance moved greater? f. What is the change in the gravitational potential energy of the box for each situation? g. What advantage, if any, is there to using the ramp? Explain.

123

2m 50

N

86.6 N

30°

1m

100 N

SP6 Diagram

home experiments and observations HE1. You can construct a simple pendulum easily by attaching a ball to a string (with tape or a staple) and fixing the other end of the string to a rigid support. (A pencil taped firmly to the end of a desk or table will do nicely.) a. The frequency of oscillation can be measured by timing the swings. The usual method is to use a watch to measure the time required for ten or more complete swings. The period T (the time required for one swing) is then the total time divided by the number of swings counted and the frequency f is just 1/T. b. How does the frequency change if you vary the length of your pendulum? (Try at least three different lengths.) HE2. A ramp for a marble or small steel ball can be made by bending a long strip of cardboard into a V-shaped groove. Two such ramps can be placed end to end, as pictured in the diagram, to produce a track in which the marble will oscillate. a. Can you measure a frequency of oscillation for this system? Does this frequency depend upon how high up the ramp the marble starts? b. How high up the second ramp does the marble go? Is more energy lost per cycle in this system than for a pendulum?

HE2 Diagram HE3. The height to which a ball bounces after being dropped provides a measure of how much energy is lost in the collision with the floor or other surface. A small portion of the energy is lost to air resistance as the ball is moving, but most is lost in the collision. a. Trying a number of different balls that you may have available, test the height of the bounce using the same height of release for all of the balls tested. Which ball loses the most energy and which the least? b. Can you explain why many balls return to a higher height than a marble will? What characteristics of the balls tested give the best bounce? c. For a ball that bounces several times, does the period (time between bounces) change with each bounce? Does the bouncing ball undergo simple harmonic motion?

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Momentum and Impulse

chapter overview In this chapter, we explore momentum and impulse and examine the use of these concepts in analyzing events such as a collision between a fullback and defensive back. The principle of conservation of momentum is introduced and its limits explained. A number of examples will shed light on how these ideas are used, particularly conservation of momentum. Momentum is central to all of these topics—it is a powerful tool for understanding a lot of life’s sudden changes.

chapter outline

1

2 3 4

Conservation of momentum. What is the principle of conservation of momentum, and when is it valid? How does this principle follow from Newton’s laws of motion? Recoil. How can we explain the recoil of a rifle or shotgun using momentum? How is this similar to what happens in firing a rocket? Elastic and inelastic collisions. How can collisions be analyzed using conservation of momentum? What is the difference between an elastic and an inelastic collision? Collisions at an angle. How can we extend momentum ideas to two dimensions? How does the game of pool resemble automobile collisions?

unit one

5

Momentum and impulse. How can rapid changes in motion be described using the ideas of momentum and impulse? How do these ideas relate to Newton’s second law of motion?

124

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125

T

he word momentum is overused by sports announcers to mean changes in the flow of a game. The “old mo” that announcers talk about bears only a metaphorical relationship to the physical concept of momentum. There are plenty of real examples of changes in momentum for us to consider in both the world of sports and the world more generally. Take the collision between a hard-charging fullback and a defensive back on the football field (fig. 7.1). If they meet head-on, the velocity of the fullback is sharply reduced, although the two players might continue moving briefly in the original direction of the fullback’s velocity. If the defensive back is moving before the collision, his velocity also changes abruptly. There must be strong forces at work to produce these accelerations, but these forces act for only an instant. How do we use Newton’s laws to analyze this event? Momentum, impulse, and conservation of momentum figure in any discussion of collisions. The total momentum of the fullback and defensive back is involved in predicting what will happen after the collision. How is momentum defined, and what does conservation of momentum have to do with Newton’s laws? How is conservation of momentum useful in predicting what

figure 7.1

A collision between a running back and a defensive back (in red, center of the photograph). How will the two players move after the collision? happens in collisions? These questions will be addressed as we examine a variety of collisions and other highimpact events.

7.1 Momentum and Impulse Imagine a baseball heading toward the catcher’s mitt when its flight is rudely interrupted by the impact of a bat. In a very short time, the velocity of the ball changes direction and is accelerated in the direction opposite its original flight. Similar changes happen when a tennis racket hits a ball or when a ball bounces off a wall or the floor. In many everyday situations, a brief impact causes a rapid change in an object’s velocity. The forces responsible for such rapid changes in motion can be large, but they act for very short times and are difficult to measure. Not only are they brief, but they may change rapidly during the collision.

v2

v1

What happens when a ball bounces? Consider the seemingly simple example of dropping a tennis ball. The ball is initially accelerated downward by the gravitational force. When it reaches the floor, its velocity quickly changes in direction, and the ball heads back up toward you (fig. 7.2). There must be a strong force exerted on the ball by the floor during the short time that they are in contact. This force provides the upward acceleration necessary to change the direction of the ball’s velocity. If we used a high-speed camera to catch the action during the time the ball is in contact with the floor, we would see that the ball’s shape is distorted (fig. 7.3). The ball behaves

figure

7.2 A tennis ball bouncing off the floor. There is a rapid change in the direction of the velocity when the ball hits the floor. like a spring, first compressing as it moves downward, then expanding (springing back) as it begins to move upward. A quick test (squeezing the ball with your hands) will persuade you that a strong force is required to distort the ball. What we have, then, is a strong force acting for a very brief time producing a rapid acceleration that quickly changes the ball’s velocity from a downward direction to an upward

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does, we must use the average value of the force over this time interval. Impulse is the average force multiplied by its time interval of action: impulse Ft.

figure 7.3

A high-speed photograph of a ball hitting the floor. The ball is compressed like a spring.

one. The magnitude of the ball’s velocity decreases rapidly to zero and then increases equally rapidly in the opposite direction. This all happens in a time so short that we would miss it if we blinked.

How can we analyze such rapid changes? We have described the collision of the ball with the floor using force and acceleration, and we could also use Newton’s second law to predict how the velocity actually changes. The problem with this approach is that the time of the interaction is very short, and the force itself varies during this short time, so it is hard to describe the collision accurately. It is more productive to look at the total change in motion in this brief interaction. We introduced Newton’s second law in chapter 4 using the expression Fnet ma. The acceleration a is the rate of change in velocity, which can be expressed as the change in velocity v divided by the time interval t required to produce that change. The time interval is important: the shorter the time for a given change in velocity, the larger the acceleration and the force needed to produce this change. We can restate Newton’s second law as Fnet m a

¢v b, ¢t

expressing the acceleration in terms of the change in velocity. Multiplying both sides of this equation by the time interval t recasts the second law as Fnett mv. While this is still Newton’s second law, rewriting it offers us a different way of looking at events. This new view is more convenient for describing the overall change in motion.

Since force is a vector quantity, impulse is also a vector in the direction of the average force. How a force changes the motion of an object depends on both the size of the force and how long the force acts. The stronger the force, the larger the effect, and the longer the force acts, the greater its effect. Multiplying the two factors together to get the impulse shows the overall effect of the force. On the right side of our recast second law, mv is the mass of the object multiplied by the change in velocity produced by the impulse. This product is the change in the quantity of motion, to use Newton’s own term. We now call this product the change in momentum of the object, where momentum is defined as Momentum is the product of the mass of an object and its velocity, or p mv.

The symbol p is often used for momentum. If the mass of the object is constant, the change in momentum is the mass times the change in velocity or p mv. Like velocity, momentum is a vector and has the same direction as the velocity vector. Two different objects traveling in the same direction can have different masses and velocities but still have the same momentum. For example, a 7-kg bowling ball moving with the relatively slow speed of 2 m/s would have a momentum of 14 kg·m/s. On the other hand, a tennis ball with a mass of just 0.07 kg, moving with the much larger velocity of 200 m/s, has the same momentum as the bowling ball, 14 kg·m/s (fig. 7.4). Using these definitions of impulse and momentum, we can state our recast form of Newton’s second law as impulse change in momentum p. This statement of the second law is sometimes called the impulse-momentum principle:

What are impulse and momentum? Impulse shows up as the quantity on the left side of the recast second law, Fnett. Impulse is the force acting on an object multiplied by the time interval over which the force acts. If the force varies during this time interval, and it often

The impulse acting on an object produces a change in momentum of the object that is equal in both magnitude and direction to the impulse.

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127

m = 7 kg v = 2 m /s

p = 14 kg•m /s

m = 0.07 kg v = 200 m /s p = 14 kg•m /s

figure 7.4

A bowling ball and a tennis ball with the same momentum. The tennis ball with its smaller mass must have a much

larger velocity.

This principle is not a new law but another way of expressing Newton’s second law of motion. It is particularly useful for looking at collisions.

How do we apply the impulse-momentum principle? The impulse-momentum principle applies to almost any collision. Whacking a golf ball with a golf club is a good example (fig. 7.5). The impulse delivered by the golf club produces a change in the golf ball’s momentum, also described in example box 7.1. Note that the units of impulse (force multiplied by time, or N·s) must equal those of momentum (mass times velocity, or kg·m/s). Does the momentum of the bouncing tennis ball we discussed earlier change when it hits the floor? Even if the ball loses no energy in its collision with the floor and bounces back with the same speed and kinetic energy it had just before hitting the floor, the momentum changes because its direction changes. The momentum decreases to

example box 7.1 Sample Exercise: The Momentum and Impulse of Golf A golf club exerts an average force of 500 N on a 0.1-kg golf ball, but the club is in contact with the ball for only a hundredth of a second. a. What is the magnitude of the impulse delivered by the club? b. What is the change in velocity of the golf ball? a. F 500 N

impulse Ft

t 0.01 s

(500 N)(0.01 s)

impulse ?

5 N·s

b. m 0.1 kg v ?

impulse p mv v impulse m 5 N#s 0.1 kg 50 m/s

p = pf

impulse

figure 7.5

An impulse is delivered to the golf ball by the head of the club. If the initial momentum of the ball is zero, the final momentum is equal to the impulse delivered.

Since the golf ball started at rest, this change in velocity equals the velocity of the ball as it leaves the face of the club. The direction of this velocity is the same as the impulse of the force exerted by the club.

zero as the ball comes to a momentary halt, and it changes again as the ball gains momentum in the opposite direction (fig. 7.6). The total change in momentum is larger than the change that would happen if the tennis ball stopped and did not bounce. When the tennis ball bounces back with the same speed, the total change in momentum is twice the value of the momentum just before the ball hits the floor. Its final momentum is mv, where the direction of v is upward, but

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pf ∆p pi

Impulse Before

During

After

figure 7.6

The impulse exerted by the floor on the tennis ball produces a change in its momentum.

its initial momentum was mv because the initial velocity was directed downward. We find the change in momentum by subtracting the initial value from the final value: mv (mv) 2mv. The impulse required to produce this change in momentum is twice as large as what is needed simply to stop the ball. There are many practical lessons involving impulse and change in momentum. Why does it help to pull your hand back as you catch a hard-thrown ball? When you pull your hand back, you lengthen the time interval t. This reduces the average force that your hand must exert on the ball, since impulse is the product of the force and the time interval (Ft). If the time interval is longer, the force can be smaller yet still produce the same impulse and change in momentum. It hurts less this way! A padded dashboard or an air bag similarly lessens injury to passengers by increasing the time interval required to bring them to a halt in a collision. Another practical lesson involving impulse and change in momentum is described in everyday phenomenon box 7.1. Everything that we have done in this section is just another way of working with Newton’s second law of motion. In fact, by dividing both sides of the impulse-momentum principle by the time interval t, Newton’s second law can be expressed in the form that most nearly captures the meaning of Newton’s original statement of the second law, Fnet p/t. In words, this form of the second law says that the net force acting on an object is equal to the rate of change in momentum of the object. This form covers a wider range of situations than the more familiar Fnet ma. Momentum and impulse are most useful for evaluating events such as collisions, where powerful forces act briefly to produce striking changes in the motion of objects. The impulse-momentum principle states that the change in momentum is equal to the impulse. This is a different way of stating Newton’s second law. The impulse, the product of the average force and the time interval that it is applied, allows us to predict the change in momentum of the object. Large impulses yield large changes in momentum.

figure 7.7

Two football players colliding. The impulses acting on the two players are equal in magnitude but opposite in direction.

7.2 Conservation of Momentum How do impulse and momentum help to explain the collision between the defensive back and fullback mentioned in the chapter introduction? The conditions described in section 7.1 are certainly present. The defensive back exerts a sizable but brief force on the fullback (fig. 7.7), and the momentum of both players changes rapidly in the collision. The principle of conservation of momentum provides the key to understanding such a collision. This principle arises when we apply Newton’s third law to impulse and changes in momentum. Conservation of momentum allows us to predict many features of collisions without requiring detailed knowledge of the forces of impact.

Why and when is momentum conserved? Let’s take a more detailed look at the head-on collision between the hard-charging fullback and the defensive back. To simplify the situation, we assume that the two players meet in midair and that after the collision they move together, with the fullback held in the tackle of the defensive back (fig. 7.7). What happens when they collide? During the collision, the defensive back exerts a strong force on the fullback, and by Newton’s third law, the fullback exerts a force equal in magnitude but opposite in direction on the defensive back. Since the time interval of action t is the same for both forces, the impulses Ft must also be equal in magnitude but opposite in direction. From the impulse-momentum principle (Newton’s second law), changes in the momentum p for each player are also equal in magnitude but opposite in direction. If the two players experience equal but oppositely directed momentum changes, the total change in momentum of the two players together is zero. We look at the overall system and define the total momentum of that system as the sum of the momentum values of the two players. There

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everyday phenomenon

129

box 7.1

The Egg Toss The Situation. Have you ever competed to see how far you and a partner can successfully throw and catch a raw egg? The most successful technique involves moving your hand back as the egg lands in it. Why does this technique reduce the likelihood of the egg breaking? What does this have to do with momentum and impulse? How can your physics knowledge reduce the chances of a raw egg bath? These same principles apply when catching water balloons. The Analysis. When you throw an egg or a balloon toward your partner, you apply a force and give it momentum. When your partner catches it and brings it to a stop, the object experiences a change in momentum. That change in momentum is equal to the impulse, as discussed on page 126. The impulse is also equal to the product of the force of the hand on the object multiplied by the time that force is applied. Therefore, by the impulse-momentum principle, the change in momentum is equal to the force of the hand on the egg multiplied by the time that force is applied: p = Ft. Unlike a golf swing or a bat hitting a baseball, the idea here is not to increase the change in momentum; the idea is to minimize the force applied to the egg so that it won’t break. It is the strength of the force that will break the egg. The larger the force applied by your hand in catching the egg, the more likely it is that you will be splattered. Once the egg leaves the thrower’s hand, it has a certain momentum that stays the same as it travels from the thrower to the catcher. We say this momentum is fixed. The impulse required to reduce this momentum to zero is therefore also fixed, it does not change. Since Ft is fixed, if you wish to decrease the force applied to the egg, the time interval t involved in the catch must be increased. We say that force and time are inversely proportional to one another. As one increases, the other decreases in proportion. F is proportional to 1/t. (F is proportional to the inverse of t.) The idea of inverse proportionality comes up frequently in everyday life but is not always recognized or understood. To take a simple example, let us say that you have $2.00 to spend on candy. If you buy candy pieces that cost 10¢ each, you will get 20 pieces of candy. If, instead, you buy candy pieces that cost 25¢ each, you can only buy eight pieces of candy. The more expensive the candy, the fewer pieces can be purchased, a fact that most of us can easily recognize. In this example, the amount of money you have to spend is fixed, and it must equal the total cost of the candy, which is the product of the price per piece times the number of pieces. The number of pieces of candy you can buy and the price of each piece of candy are therefore inversely proportional to one another. As one increases, the other must decrease in proportion.

As we have already indicated, inverse proportionality is involved in catching the egg. The impulse is fixed, so force and time are inversely proportional to one another. As you increase one (say the time to stop it), the other (the force applied) decreases. Your objective is then to make the time involved in stopping the egg as large as possible. How can this be done? You can increase the time involved in the catch substantially by moving your hand back as the egg reaches it. This should be done as smoothly as possible so that the velocity (and momentum) of the egg decreases to zero much more gradually than if your hand were stationary. Using this technique, the average force applied to the egg can be made much smaller than that involved in a sudden stop. With luck, this force will be small enough so the egg will not break! This same principle has many applications. Air bags in cars reduce the force applied to your head in a collision by increasing the time it takes your head to come to a stop. Gym floors have much more ‘give’ to them to reduce the force on players’ knees when jumping and landing on the floor. Dropping a wine glass on a carpeted floor will make a mess by staining the carpet, but it is far less likely to break the glass than if it is dropped onto a concrete floor. In all of these examples, the force has decreased because the time it takes to stop the object increases. You may not be able to measure the decrease in time with a stopwatch because the times involved are very short, but you can see the evidence of the increased time in the result. The next time you have a picnic, bring some raw eggs or water balloons and apply the impulse-momentum principle with your friends and family. With water balloons, good technique will keep you dry, but sometimes it is more fun to get wet.

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is no change in the momentum of this system because the changes in the momentums of the parts cancel one another. The total momentum of the system is conserved. To reach this conclusion, we ignored external forces (produced by other objects) acting on the two players and assumed that the only significant forces were their own forces of interaction. The forces that they exert on one another are internal to the system consisting of both players. The principle of conservation of momentum can therefore be stated as If the net external force acting on a system of objects is zero, the total momentum of the system is conserved.

example box 7.2 Sample Exercise: A Head-on Collision A 100-kg fullback moving straight downfield with a velocity of 5 m/s collides head-on with a 75-kg defensive back moving in the opposite direction with a velocity of 4 m/s. The defensive back hangs on to the fullback, and the two players move together after the collision. a. What is the initial momentum of each player? b. What is the total momentum of the system? c. What is the velocity of the two players immediately after the collision? p mv

a. fullback:

The forces of interaction between the objects in a system are internal forces whose effects on the total momentum cancel one another, because of Newton’s third law of motion. Different portions of the system can exchange momentum without affecting the total momentum of the system. If there is a net external force acting on the system because of interaction with some object that is not part of the system, the entire system will be accelerated—and the momentum of the system will change.

Conservation of momentum and collisions Using the principle of conservation of momentum, what information can we obtain about the results of a collision (like the one between two football players)? If we know the masses of the players and their initial velocities, we can find how fast and in what direction the players will move after they collide. We do not need to know anything at all about the details of the strong forces involved in the collision itself. The sample exercise in example box 7.2 treats a headon collision between a fullback and a defensive back using realistic numerical values. The fullback has a mass of 100 kg (equivalent to a weight of about 220 lb) and is moving straight downfield with a velocity of 5 m/s through the hole created by his linemen. The somewhat smaller defensive back charges up to meet him with a velocity in the opposite direction of 4 m/s (fig. 7.8). The minus sign indicates direction: we have chosen the fullback’s direction of motion to be positive. The total momentum of the system before the collision in example box 7.2 is found by adding the initial momentum of the fullback to the momentum of the defensive back, taking into account the difference in sign. If we assume that both players’ feet leave the ground just before the collision (so that there are no frictional forces between their feet and the ground), momentum should be conserved in the collision. The total momentum of the two players moving together after the collision has the same value as it did immediately before the collision (fig. 7.9).

m 100 kg

(100 kg)(5 m/s)

v 5 m/s

500 kg·m/s

p ? defensive back:

p mv

m 75 kg

(75 kg)(4 m/s)

v 4 m/s

300 kg·m/s

b. ptotal ?

ptotal pfullback pdefensive back 500 kg·m/s (300 kg·m/s) 200 kg·m/s

c. v ? (for both players after the collision) m 100 kg 75 kg 175 kg

p mv v

ptotal m

200 kg # m/s 175 kg

1.14 m/s

The positive value of the momentum after the collision means that the two players are traveling in the direction of the fullback’s initial motion. The fullback had a larger initial momentum than the defensive back, so his direction of motion prevails when the two values are added. The defensive back will be carried backward briefly before the two players hit the turf. Conservation of momentum results when the changes in momentum of different parts of a system cancel each other by Newton’s third law. If there are no external forces acting on the system, its total momentum is conserved. The principle applies to all sorts of situations involving collisions and explosions or other forms of brief but forceful interaction between objects.

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figure 7.8

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The two players before the collision, with velocity and momentum vectors for each.

figure 7.9

The two players after the collision, with velocity and momentum vectors indicated.

7.3 Recoil Why does a shotgun slam against your shoulder when fired, sometimes with painful consequences? How can a rocket accelerate in empty space when there is nothing there to push against but itself? These are examples of the phenomenon of recoil, a common part of everyday experience. Conservation of momentum is the key to understanding recoil.

What is recoil? Imagine two ice skaters facing one another and pushing against each other with their hands (fig. 7.10). The frictional forces between their skates and the ice are presumably very small, so we can neglect them. The upward normal force and the downward force of gravity cancel one another, too, since we know that there is no acceleration in the vertical direction. The net external force acting on the system of the two skaters is effectively zero, and conservation of momentum should apply. How do we apply conservation of momentum in this situation? Since neither skater is moving before the push-off,

figure

7.10 Two skaters of different masses prepare to push off against one another. Which one will gain the larger velocity?

the initial total momentum of the system is zero. If momentum is conserved, the total momentum of the system after the push-off will also be zero. How can the total momentum be zero when at least one of the skaters is moving? Both skaters must move with momentum values equal in magnitude but opposite in direction p2 p1. The momentum of the second skater p2 must be opposite that of the first skater p1. When added together to find the total momentum of the system, these individual values will cancel each other to produce a total momentum of zero. After the push-off, the two skaters move in opposite directions with momentum vectors equal in magnitude (fig. 7.11), but their velocities are not of equal magnitude. Since momentum is mass times velocity (p mv), the skater with the smaller mass must have the larger velocity

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Recoil 100 kg

50 kg –p

–p

p

p

figure

v1

v2

7.12 The shot and the shotgun have equal but oppositely directed momentums after the gun is fired.

example box 7.3 figure

7.11 The two skaters after pushing off, with the velocity and momentum vectors indicated.

Sample Question: Is Momentum Conserved When Shooting a Shotgun? Question: When a shotgun is held firmly against your shoulder, is the momentum of the system conserved?

to yield the same magnitude of momentum as the larger skater. Suppose that the smaller skater’s mass is just half the mass of the larger skater. The smaller skater’s velocity will then be twice as large as the larger skater after pushing off. The ice skaters illustrate the basics of recoil. A brief force between two objects causes the objects to move in opposite directions. The lighter object attains the larger velocity to equalize the magnitudes of the momentums of the two objects. The total momentum for the system after the push-off equals zero, the value of the momentum of the system before the push-off if the objects were initially at rest. The total momentum of the system is conserved and does not change.

Recoil of a shotgun If you have ever fired a shotgun without holding it firmly against your shoulder, you have probably had a painful experience of recoil. What happened? The explosion of the powder in the shotgun causes the shot to move very rapidly in the direction of the gun’s aim. If the gun is free to move, it will recoil in the opposite direction with a momentum equal in magnitude to the momentum of the shot (fig. 7.12). Even though the mass of the shot is considerably less than the shotgun, the momentum of the shot is quite large as a result of its large velocity. If the external forces acting on the system can be ignored, the shotgun recoils with a momentum equal in magnitude to the momentum of the shot. The recoil velocity of the shotgun will be smaller than the shot’s velocity because of the larger mass of the gun, but it is still sizable. As the gun slams back against your shoulder, you will know that it has recoiled. How can you avoid a bruised shoulder? The trick is to hold the gun firmly against your shoulder. (See example

Answer: It depends on how you define the system. If the system is defined as just the shotgun and the pellets, there is then a strong external force exerted on the system by the shoulder of the shooter. Since the condition for conservation of momentum is that the net external force acting on the system be zero, the momentum of this system is not conserved. If we included the shooter and the Earth in our system, then momentum would be conserved because all of the forces would be internal to this system. The change in the momentum of the Earth would be imperceptible, however.

box 7.3.) Your own mass then becomes part of the system. This increased mass will produce a smaller recoil velocity, even if you happen to be standing on ice with no frictional forces between your feet and the Earth. More important, the shotgun will not move against your shoulder.

How does a rocket work? The firing of a rocket is another example of recoil. The exhaust gases rushing out of the tail of the rocket have both mass and velocity and, therefore, momentum. If we ignore external forces, the momentum gained by the rocket in the forward direction will equal in magnitude the momentum of the exhaust gases in the opposite direction (fig. 7.13). Momentum is conserved, just as in our other examples of recoil. The rocket and the exhaust gases push against one another, and Newton’s third law applies. The difference between a rocket and our earlier examples of the skaters and the shotgun is that firing a rocket is usually a continuous process. The rocket gains momentum gradually rather than in a single short blast. The mass

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–p

p

figure

7.13 If a short blast is fired, the rocket gains momentum equal in magnitude but opposite in direction to the momentum of the exhaust gases.

of the rocket also changes as fuel is consumed and gases exhaust from the rocket engines. Computation of the final velocity becomes more difficult than for the skaters. For a brief blast of the rocket, though, the same analysis can be used. Recoil works in empty outer space: the two objects need only push against each other, just as with the skaters and the shotgun. Rocket engines can be used for space travel, unlike the propeller engines or jet engines used on airplanes. Airplane engines depend on the presence of the atmosphere, both as a source of oxygen used in burning fuel and as something to push against. An airplane propeller pushes against the air, and the air, by Newton’s third law, pushes against the propeller. This interaction accelerates the airplane. A rocket, on the other hand, is self-contained. It exerts a force on its own exhaust gases, and by the third law, the exhaust gases exert a force on the rocket. During recoil, objects push against one another, moving in opposite directions. If external forces can be neglected, momentum is conserved. The total momentum before and after the interaction equals zero. After the interaction, the two objects move away with equal but oppositely directed momentum vectors that cancel one another. Recoil is one of many kinds of brief interaction to which conservation of momentum applies.

7.4 Elastic and Inelastic Collisions As the example involving football players showed, collisions are one of the most fruitful areas for applying conservation of momentum. Collisions involve large forces of interaction acting for very brief times, and they produce dramatic changes in the motion of the colliding objects. Because the forces of interaction are so large, any external forces acting on the system usually are unimportant by comparison: momentum is conserved. Different kinds of collisions produce different results. Sometimes the objects stick together and sometimes they bounce apart. What distinguishes these different cases, and what do the terms elastic, inelastic, and perfectly inelastic mean when applied to collisions? Is energy conserved as well as momentum? Railroad cars, bouncing balls, and billiard balls help illustrate the differences.

133

What is a perfectly inelastic collision? The easiest type of collision to analyze is one where two objects collide head-on and stick together after the collision, like the two football players discussed earlier. Because they stuck together and moved as one object after the collision, we had just one final velocity to contend with. A collision in which the objects stick together after collision is called a perfectly inelastic collision. The objects do not bounce at all. If we know the total momentum of the system before the collision (and external forces are ignored), we can readily compute the final momentum and velocity of the now-joined objects. Coupling railroad cars are another example of this type of collision. Example box 7.4 uses conservation of momentum to predict the final momentum and velocity of coupled railroad cars from knowledge of the momentum of the system before the collision. The process is much the same as the one used to predict the final velocity of the football players in section 7.2. In both cases, the separate objects move as one following the collision. In example box 7.4, the total mass of the coupled cars after the collision is five times that of car 5, so the final velocity of the coupled cars must be one-fifth that of car 5 to conserve momentum. The momentum of the system immediately after the collision is equal to that just before the collision, but the velocities have changed. The “final” velocity that we calculated is valid immediately after the collision. As the cars continue to move following the collision, frictional forces will gradually decelerate them until they come to rest.

Is energy conserved in collisions? Is the kinetic energy after the railroad cars collide equal to the original kinetic energy of car 5 in the example in 1 example box 7.4? Using the relationship KE 2 mv2 introduced in chapter 6, we can compute the kinetic energy before and after the collision. The original kinetic energy of car 5 is 2250 kJ. (A kilojoule, kJ, is a thousand joules.) Immediately after the collision, the kinetic energy of the five cars moving together is just 450 kJ. (You can check these values.) A portion of the original kinetic energy is lost in any perfectly inelastic collision. If we put a large spring on the front of the moving railroad car and allowed it to bounce off the other four cars rather than coupling, we will find that a greater portion of the kinetic energy is retained in the collision. When the objects bounce, the collision is either elastic or only partially inelastic rather than perfectly inelastic. The distinction is based on energy. An elastic collision is one in which no energy is lost. A partially inelastic collision is one in which some energy is lost, but the objects do not stick together. The greatest portion of energy is lost in the perfectly inelastic collision, when the objects stick.

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example box 7.4 a. m5 20 000 kg

Sample Exercise: When Railroad Cars Couple Four railroad cars, all with the same mass of 20 000 kg, sit on a track, as shown in the drawing. A fifth car of identical mass approaches them with a velocity of 15 m/s (to the right). This car collides and couples with the other four cars. a. What is the initial momentum of the system? b. What is the velocity of the five coupled cars after the collision?

pinitial m5v5

v5 15 m/s

(20 000 kg)(15 m/s)

pinitial ?

300 000 kg·m/s (before the collision)

b. mtotal 100 000 kg vfinal pfinal pinitial vfinal ?

pfinal m total 300 000 kg # m/s 100 000 kg

3 m/s (for the five cars after the collision)

car 5 AN

pinitial

TN

CN

DN

EN

A railroad car approaches four others at rest on the track. What is the velocity of the cars after they couple?

In most collisions, some kinetic energy is lost because the collisions are not perfectly elastic. Heat is generated, the objects may be deformed, and sound waves are created, all of which involve conversions of the kinetic energy of the objects to other forms of energy. Even if the objects bounce, we cannot assume that the collision is elastic. More likely, the collision will be partially inelastic, implying that some of the initial kinetic energy has been lost. A ball bouncing off a floor or wall with no decrease in the magnitude of its velocity is an example of an elastic collision. Since the magnitude of the velocity does not change (only the direction changes), the kinetic energy does not decrease. No energy has been lost. More likely, of course, some energy will be lost in such a collision, and the magnitude of the ball’s velocity after the collision will be a little smaller than before. The opposite extreme to an elastic collision of a ball with the wall would be a perfectly inelastic collision in which the ball sticks to the wall. In this case, the velocity of the ball after the collision is zero. So is its kinetic energy. All of the kinetic energy is lost (fig. 7.14).

What happens when billiard balls bounce? Very little energy is lost when billiard balls collide with one another. (Time spent playing pool can be justified as a form of experimental physics. Your intuition about elastic

vi Elastic vf = vi vf

Perfectly inelastic

figure

vi vf = 0

7.14 An elastic collision and a perfectly inelastic collision of a ball with a wall. The ball sticks to the wall in the perfectly inelastic collision.

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collisions can be improved in the process!) The collisions are basically elastic. Both momentum and kinetic energy are conserved in most collisions of billiard balls. When colliding objects such as billiard balls bounce off one another, we must deal with two final velocities rather than one. We can readily compute the total momentum of the system before and after the collision from our knowledge of the initial momentum values of the objects. More information is needed to determine the individual velocities of the objects after the collision, however, since one value is not enough to determine two unknown velocities. (This is why the case of the perfectly inelastic collision, where objects stick together, is particularly easy to analyze.) In an elastic collision, conservation of energy provides the additional information. For billiard balls, the simplest case is the white cue ball colliding head-on with a second ball that is not moving before it is hit (the eleven ball in fig. 7.15). What happens? If spin is a minor factor in the collision, the cue ball stops dead on impact, and the eleven ball moves forward with a velocity equal to that of the cue ball before the collision. If the eleven ball acquires the same velocity that the cue ball had before the collision, it also has the same momentum mv as the initial momentum of the cue ball because both balls have the same mass. Momentum is conserved. Kinetic energy is also conserved. The cue ball had a 1 kinetic energy of 2 mv2 before the collision. After the collision, the velocity and kinetic energy of the cue ball are 1 zero, but the eleven ball now has a kinetic energy of 2 mv2, since its mass and speed are the same as the cue ball before the collision. Given the equal masses of the two balls, the only way that both momentum and kinetic energy can be conserved is for the cue ball to stop and the eleven ball to move forward with the same momentum and kinetic energy that the cue ball had before the collision. This effect is familiar to any pool player. The same phenomenon is involved in the familiar swinging-ball demonstration often seen as a decorative toy on mantels or desktops (fig. 7.16). A row of steel balls hangs

135

figure

7.16 The swinging-ball apparatus provides an example of collisions that are approximately elastic. by threads from a metal or wooden frame. If one ball is pulled back and released, the collision with the other balls results in a single ball from the other end of the chain flying off with the same velocity as the first ball just before the collision. Both momentum and kinetic energy are conserved. If two balls on one side are pulled back and released, two balls fly off from the opposite side of the row of balls after the collision. Again, both momentum and kinetic energy are conserved by this result. You can explore a variety of other combinations. It can be entertaining as well as addictive. Collisions between hard spheres, such as billiard balls or the steel balls in the swinging-ball apparatus, will generally be more or less elastic. Most collisions involving everyday objects, though, are inelastic to some degree. Some kinetic energy is lost. Momentum will be conserved, however, as long as our concern is with the values of momentum and velocity immediately before and after the collision. Conservation of momentum is the primary tool used in understanding collisions. External forces can be ignored for the brief time of the collision, when the collision forces are dominant, and the law of conservation of momentum applies. Kinetic energy can also be conserved if the collision is elastic, as it is approximately for billiard balls or other hard spheres. Most collisions involving familiar objects are partially inelastic and involve some loss of energy. The greatest proportion of energy is lost in perfectly inelastic collisions where objects stick together.

p1

p2

7.5 Collisions at an Angle

figure

7.15 A head-on collision between the white cue ball and the eleven ball initially at rest. The cue ball stops, and the eleven ball moves forward.

What happens when objects such as billiard balls or automobiles collide at an angle, rather than head-on? Some interesting applications of conservation of momentum arise

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when motion is not confined to a straight line. It becomes more apparent that momentum is a vector when objects are free to move in two dimensions. Balls on a pool table, cars colliding at an intersection, or football players tackling all provide interesting examples.

/s

•m

g 3k

58

p2 = 300 kg•m/s

31° p1 = 500 kg•m/s

figure

An inelastic two-dimensional collision Two football players, originally traveling at right angles to one another, collide and stick together, as in figure 7.17. What will be their direction of motion after the collision? How do we apply conservation of momentum in this two-dimensional case? In figure 7.17, we assume that the two players have the same masses and initial speeds as in our earlier example (section 7.2), but we no longer have a head-on collision. The momentum of the defensive back is now directed across the field as the fullback heads downfield. Because momentum is a vector, we need to add the individual momentum vectors of the fullback and the defensive back to get the total momentum of the system before the collision. This can be done most readily by using a vector diagram with the vectors drawn to scale. The vectors can then be added graphically as we have done before. As shown in figure 7.18, the total momentum of the system before the collision is the hypotenuse of the right triangle formed by adding the other two momentum vectors. If momentum is conserved in the collision, the total momentum of the two players after the collision will equal the total momentum before the collision. Since the two players move together after the collision, they will travel in the

p1 = 500 kg•m/s

p2 = 300 kg•m/s

figure

7.17 The fullback and the defensive back approaching each other at a right angle.

7.18 The total momentum of the two football players prior to the collision is the vector sum of their individual momentums.

direction of the total momentum vector shown in figure 7.18. (See appendix C for a review of vector addition.) The direction of motion of both players changes as a result of the collision. The larger momentum of the fullback before the collision dictates that the final direction of motion is more downfield than across the field, but it is some of both. This result makes intuitive sense if you imagine yourself as one of the players. The final direction of motion of the two football players after the collision depends on their momentum values before the collision. If the defensive back were bigger or moving faster than we assumed initially, he would have a larger momentum, and his tackle would cause a more impressive change in the direction of the fullback’s motion. On the other hand, if the defensive back is small and moving slowly, his effect on the fullback’s direction will be small. Adjusting the length of the momentum arrow p2, the momentum of the defensive back in figure 7.18, will illustrate these changes. Everyday phenomenon box 7.2 describes a similar situation. Two cars approaching an intersection at right angles to one another collide and stick together. Working backward from information about the final direction of travel, the investigator can draw conclusions about the initial velocities of the two cars. Conservation of momentum is extremely important in accident analysis.

What happens in elastic two-dimensional collisions? When billiard balls collide, they do not stick together after the collision: when objects bounce, we have to contend with two final velocities with different directions. Although many real collisions are like this, analysis is more complicated for them than for the perfectly inelastic examples we have discussed. More information is needed to predict the final velocities. If we know that the collision is elastic, however, conservation of kinetic energy can provide that additional information. Experimental physics on the pool table comes through again with an interesting example (and one of practical value to any pool player). Suppose that the cue ball strikes the

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everyday phenomenon

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box 7.2

An Automobile Collision

Main St.

The Situation. Officer Jones is investigating an automobile collision at the intersection of Main Street and 19th Avenue. Driver A was traveling east on 19th Avenue when she was struck from the side by driver B who was traveling north on Main Street. The two cars stuck together after the collision and ended up against the lamppost on the northeast corner of the intersection. Both drivers claim they started up just as the light in their direction changed to green and then collided with the other driver, who was running a red light and speeding. There are no other witnesses. Which driver is telling the truth?

3. Both cars have about the same mass (both are compacts of roughly the same vintage and size). 4. Conservation of momentum should determine the direction of the momentum vector after the collision. After sketching the diagram and noting the direction of the final momentum vector, Officer Jones concludes that B is lying. Why? The final momentum vector must be equal to the sum of the initial momentum vectors of the two cars before the collision. Since the cars were traveling at right angles to one another, the two initial momentum vectors form the sides of a right triangle whose hypotenuse is the total momentum of the system. The diagram clearly shows that the momentum of B’s car must have been considerably larger than the momentum of A’s car.

Pole Main St.

19th Ave.

vA

Pole

19th Ave.

vf

pA

pB vB

pT

N E

N

The collision at Main Street and 19th Avenue.

The Analysis. Officer Jones, having taken a physics course during college and being trained in the art of accident investigation, makes these observations: 1. The point of impact is well marked by the shards of glass from the headlights of B’s car and other debris. Officer Jones indicates this point on the diagram in her accident report form. 2. The direction the two cars are traveling after the impact is also obvious. (She indicated this by a line drawn from the point of impact to the cars’ final resting spot.)

E

Officer Jones’s accident report contains a vector diagram derived from conservation of momentum.

Since both drivers claimed to have just started from a complete stop after the traffic signal changed, the driver with the larger velocity before the collision is not telling it like it was. Driver B is the one who had the larger velocity, and so was presumably speeding through the red light. Driver B is thus cited by Officer Jones.

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stationary eleven ball at an angle (off-center), as in figure 7.19. What happens to the two balls after the collision? The combined effects of conservation of momentum and conservation of kinetic energy lead to a unique result wellknown to serious pool players. The initial momentum of the system is simply that of the cue ball, the only one moving. Its direction is indicated by the arrow labeled pi in figure 7.19 and in the drawings in figure 7.20. The force of interaction (and the impulse) between the two balls is along a line joining the centers of the balls at the point of impact. The eleven ball moves off along this line because the force of contact pushes it in that direction. The total momentum of the system after the collision must still be in the direction of the initial momentum because momentum is conserved in the collision. Conservation of momentum also restricts the possible momentum and direction of the cue ball’s motion after the collision (fig. 7.20). The momentum vectors of the two balls after the collision are added here to give the total momentum of the system, ptotal, which must be equal in both magnitude and direction to the initial momentum of the system. Since the collision is elastic, the initial kinetic energy of the cue ball, 12 mv2, must also equal the sum of the kinetic energies of the two balls after the collision. Since the masses of the two balls are equal, conservation of kinetic energy in the collision requires that*

pi

figure

7.19 The cue ball is aimed at a point off-center on the second ball to produce an angular collision.

p1

pi

v2 (v1)2 (v2)2 where v is the speed of the cue ball before the collision, and v1 and v2 are the speeds of the two balls afterward. The velocity vectors form a triangle like the one formed by the momentum vectors in figure 7.20. If the sum of the squares of the two sides equals the square of the third side of the triangle, this triangle must be a right triangle according to the Pythagorean theorem from plane geometry. If the velocity vectors form a right triangle, so do the momentum vectors, which have the same directions as the corresponding velocity vectors. Conservation of momentum requires that the momentum vectors add to form a triangle, but conservation of kinetic energy dictates that it be a right triangle. The cue ball will move off at a right angle (90°) to the direction of motion of the eleven ball after the collision. In playing pool, this is an important piece of intelligence if you are planning your next shot. Conservation of momentum and conservation of kinetic energy determine the shot. If you have a pool table handy, test these conclusions using a variety of impact angles. (Marbles or steel balls are

1

1

*Conservation of kinetic energy requires that 2 mv2 2 m(v1)2 1 1 2 2 m(v2) , but the mass and the factor of 2 can be divided out of the equation.

p2 Before

During

After

p1 p2

pi = pt

ota

l

Vector addition

figure

7.20 The momentum vectors of the two balls after the collision add to give the total (initial) momentum of the system. The paths of the two balls are approximately at right angles after the collision.

a suitable substitute.) You may not get perfect 90° angles after the collision: the collision is not perfectly elastic, and spin can sometimes be a factor. The angle between the two final velocities, though, will usually be within a few degrees of a right angle. Seeing is believing—give it a try.

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summary

139

imagine the direction and magnitude of the original momentum vector, you will have some sense of the outcome. These conservation laws are powerful predictors of what happens when people, billiard balls, cars, and even subatomic particles or stars collide.

Conservation of momentum requires that the direction of the momentum vector be conserved as well as its size. When collisions occur at an angle, this requirement restricts the directions and velocities of the resulting motions. If the collision is elastic, as with billiard balls, conservation of energy adds another restriction. If you can

summary In this chapter, we recast Newton’s second law in terms of impulse and momentum to describe interactions between objects, such as collisions, that involve strong interaction forces acting over brief time intervals. The principle of conservation of momentum, which follows from Newton’s second and third laws, plays a central role.

Recoil. If an explosion or push occurs between two ob3jects initially at rest, conservation of momentum dictates that the total momentum after the event must still be zero if there is no net external force. The final momentum vectors of the two objects are equal in size but opposite in direction.

1

Momentum and impulse. Newton’s second law can be recast in terms of momentum and impulse, yielding the statement that the net impulse acting on an object equals the change in momentum of the object. Impulse is defined as the average force acting on an object multiplied by the time interval during which the force acts. Momentum is defined as the mass of an object times its velocity.

p2

p1

p2 = –p1 ∆p

Impulse

Fnet∆t = ∆p,

4

p = mv

2

Conservation of momentum. Newton’s second and third laws combine to yield the principle of conservation of momentum: if the net external force acting on a system is zero, the total momentum of the system is a constant. Before

Elastic and inelastic collisions. A perfectly inelastic collision is one in which the objects stick together after the collision. If external forces can be ignored, the total momentum is conserved. An elastic collision is one in which the total kinetic energy is also conserved.

vi

vi

After

P

vf

P

Elastic If Fexternal = 0 Ptotal = constant

Perfectly inelastic

5

Collisions at an angle. Conservation of momentum is not restricted to one-dimensional motion. When objects collide at an angle, the total momentum of the system before and after the collision is found by adding the momentum vectors of the individual objects.

pf

Before

After pf = pi

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key terms Impulse, 126 Momentum, 126 Impulse-momentum principle, 126

Conservation of momentum, 128 Recoil, 132 Perfectly inelastic collision, 133

Elastic collision, 133 Partially inelastic collision, 133

study hint Except for the examples involving impulse, most of the situations described in this chapter highlight the principle of conservation of momentum. The basic ideas used in applying conservation of momentum are: 1. External forces are assumed to be much smaller than the very strong forces of interaction in a collision or other brief event. If external forces acting on the system can be ignored, momentum is conserved. 2. The total momentum of the system before the collision or other brief interaction pinitial is equal to the momentum after the event pfinal. Momentum is conserved and does not change.

3. Equality of momentum before and after the event can be used to obtain other information about the motion of the objects. For review, look back at how these three points are used in each of the examples in this chapter. The total momentum of the system before and after the event is always found by adding the momentum values of the individual objects as vectors. You should be able to describe the magnitude and direction of this total momentum for each of the examples.

Q1. Does the length of time that a force acts on an object have any effect on the strength of the impulse produced? Explain. Q2. Two forces produce equal impulses, but the second force acts for a time twice that of the first force. Which force, if either, is larger? Explain. Q3. Is it possible for a baseball to have as large a momentum as a much more massive bowling ball? Explain. Q4. Are impulse and force the same thing? Explain. Q5. Are impulse and momentum the same thing? Explain. Q6. If a ball bounces off a wall so that its velocity coming back has the same magnitude that it had prior to bouncing: a. Is there a change in the momentum of the ball? Explain. b. Is there an impulse acting on the ball during its collision with the wall? Explain. Q7. Is there an advantage to following through when hitting a baseball with a bat, thereby maintaining a longer contact between the bat and the ball? Explain. Q8. What is the advantage of a padded dashboard compared to a rigid dashboard in reducing injuries during collisions? Explain using momentum and impulse ideas.

Q9. What is the advantage of an air bag in reducing injuries during collisions? Explain using impulse and momentum ideas. *Q10. If an air bag inflates too rapidly and firmly during a collision, it can sometimes do more harm than good in lowvelocity collisions. Explain using impulse and momentum ideas. Q11. If you catch a baseball or softball with your bare hand, will the force exerted on your hand by the ball be reduced if you pull your arm back during the catch? Explain. Q12. A truck and a bicycle are moving side by side with the same velocity. Which, if either, will require the larger impulse to bring it to a halt? Explain. Q13. Is the principle of conservation of momentum always valid, or are there special conditions necessary for it to be valid? Explain. Q14. A ball is accelerated down a fixed inclined plane under the influence of the force of gravity. Is the momentum of the ball conserved in this process? Explain. Q15. Two objects collide under conditions where momentum is conserved. Is the momentum of each object conserved in the collision? Explain. Q16. Which of Newton’s laws of motion are involved in justifying the principle of conservation of momentum? Explain.

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*Q17. A compact car and a large truck have a head-on collision. During the collision, which vehicle, if either, experiences: a. the greater force of impact? Explain. b. the greater impulse? Explain. c. the greater change in momentum? Explain. d. the greater acceleration? Explain. Q18. A fullback collides midair and head-on with a lighter defensive back. If the two players move together following the collision, is it possible that the fullback will be carried backward? Explain. Q19. Two ice skaters, initially at rest, push off one another. What is the total momentum of the system after they push off? Explain. Q20. Two shotguns are identical in every respect (including the size of shell fired) except that one has twice the mass of the other. Which gun will tend to recoil with greater velocity when fired? Explain. *Q21. When a cannon rigidly mounted on a large boat is fired, is momentum conserved? Explain, being careful to clearly define the system being considered. Q22. Is it possible for a rocket to function in empty space (in a vacuum) where there is nothing to push against except itself? Explain. Q23. Suppose that you are standing on a surface that is so slick that you can get no traction at all in order to begin moving across this surface. Fortunately, you are carrying a bag of oranges. Explain how you can get yourself moving. Q24. Suppose an astronaut in outer space suddenly discovers that the tether connecting her to the space shuttle is cut and she is slowly drifting away from the shuttle. Assuming that she is wearing a tool belt holding several wrenches, how can she move herself back toward the shuttle? Explain. *Q25. Suppose that on a perfectly still day, a sailboat enthusiast decides to bring along a powerful battery-operated fan in order to provide an air current for his sail, as shown in the diagram. a. What are the directions of the change in momentum of the air at the fan and at the sail? b. What are the directions of the forces acting on the fan and on the sail due to these changes in momentum? c. Would the sailor in this picture be better off with the sail furled (down) or unfurled (up)? Explain.

Q26. A skateboarder jumps on a moving skateboard from the side. Does the skateboard slow down or speed up in this process? Eaplain, using conservation of momentum. Q27. A railroad car collides and couples with a second railroad car that is standing still. If external forces acting on the system are ignored, is the velocity of the system after the collision equal to, greater than, or less than that of the first car before the collision? Explain. Q28. Is the collision in question 24 elastic, partially inelastic, or perfectly inelastic? Explain. Q29. If momentum is conserved in a collision, does this indicate conclusively that the collision is elastic? Explain. Q30. A ball bounces off a wall with a velocity whose magnitude is less than it was before hitting the wall. Is the collision elastic? Explain. *Q31. A ball bounces off a wall that is rigidly attached to the Earth. a. Is the momentum of the ball conserved in this process? Explain. b. Is the momentum of the entire system conserved? Explain, being careful to clarify how you are defining the system. Q32. A cue ball strikes an eight ball of equal mass, initially at rest. The cue ball stops and the eight ball moves forward with a velocity equal to the initial velocity of the cue ball. Is the collision elastic? Explain. Q33. Two lumps of clay traveling through the air in opposite directions collide and stick together. Their momentum vectors prior to the collision are shown in the diagram. Sketch the momentum vector of the combined lump of clay after the collision, making the length and direction appropriate to the situation. Explain your result.

1

p1

p2

2

Q33 Diagram Q34. Two lumps of clay, of equal mass, are traveling through the air at right angles to each other with velocities of equal magnitude. They collide and stick together. Is it possible that their velocity vector after the collision is in the direction shown in the diagram? Explain. Before v3

v1

v2

Q25 Diagram

141

Q34 Diagram

After

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Q35. Two cars of equal mass collide at right angles to one another in an intersection. Their direction of motion after the collision is as shown in the diagram. Which car had the greater velocity before the collision? Explain.

A v

Q36. A car and a small truck traveling at right angles to one another with the same speed collide and stick together. The truck’s mass is roughly twice the car’s mass. Sketch the direction of their momentum vector immediately after the collision. Explain your result. *Q37. A cue ball strikes a glancing blow against a second billiard ball initially at rest. Sketch the situation indicating the magnitudes and directions of the momentum vectors of each ball before and after the collision.

B

Q35 Diagram

exercises E1. An average force of 300 N acts for a time interval of 0.04 s on a golf ball. a. What is the magnitude of the impulse acting on the golf ball? b. What is the change in the golf ball’s momentum? E2. What is the momentum of a 1200-kg car traveling with a speed of 27 m/s (60 MPH)? E3. A bowling ball has a mass of 6 kg and a speed of 1.5 m/s. A baseball has a mass of 0.12 kg and a speed of 40 m/s. Which ball has the larger momentum? E4. A force of 45 N acts on a ball for 0.2 s. If the ball is initially at rest: a. What is the impulse on the ball? b. What is the final momentum of the ball? E5. A 0.12-kg ball traveling with a speed of 40 m/s is brought to rest in a catcher’s mitt. What is the size of the impulse exerted by the mitt on the ball? E6. A ball experiences a change in momentum of 24 kg·m/s. a. What is the impulse acting on the ball? b. If the time of interaction is 0.15 s, what is the magnitude of the average force acting on the ball? E7. A 60-kg front-seat passenger in a car moving initially with a speed of 18 m/s (40 MPH) is brought to rest by an air bag in a time of 0.4 s. a. What is the impulse acting on the passenger? b. What is the average force acting on the passenger in this process? E8. A ball traveling with an initial momentum of 2.5 kg·m/s bounces off a wall and comes back in the opposite direction with a momentum of 2.5 kg·m/s. a. What is the change in momentum of the ball? b. What impulse would be required to produce this change? E9. A ball traveling with an initial momentum of 4.0 kg·m/s bounces off a wall and comes back in the opposite direction with a momentum of 3.5 kg·m/s. a. What is the change in momentum of the ball? b. What impulse is required to produce this change?

E10. A fullback with a mass of 100 kg and a velocity of 3.5 m/s due west collides head-on with a defensive back with a mass of 80 kg and a velocity of 6 m/s due east. a. What is the initial momentum of each player? b. What is the total momentum of the system before the collision? c. If they stick together and external forces can be ignored, what direction will they be traveling immediately after they collide? E11. An ice skater with a mass of 80 kg pushes off against a second skater with a mass of 32 kg. Both skaters are initially at rest. a. What is the total momentum of the system after they push off? b. If the larger skater moves off with a speed of 3 m/s, what is the corresponding speed of the smaller skater? E12. A rifle with a mass of 1.2 kg fires a bullet with a mass of 6.0 g (0.006 kg). The bullet moves with a muzzle velocity of 600 m/s after the rifle is fired. a. What is the momentum of the bullet after the rifle is fired? b. If external forces acting on the rifle can be ignored, what is the recoil velocity of the rifle? E13. A rocket ship at rest in space gives a short blast of its engine, firing 50 kg of exhaust gas out the back end with an average velocity of 400 m/s. What is the change in momentum of the rocket during this blast? E14. A railroad car with a mass of 12 000 kg collides and couples with a second car of mass 18 000 kg that is initially at rest. The first car is moving with a speed of 12 m/s prior to the collision. a. What is the initial momentum of the first car? b. If external forces can be ignored, what is the final velocity of the two railroad cars after they couple? E15. For the railroad cars in example box 7.4: a. What is the kinetic energy of car 5 before the collision? b. What is the kinetic energy of all five cars just after the collision? c. Is energy conserved in this collision?

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E16. A 4000-kg truck traveling with a velocity of 10 m/s due north collides head-on with a 1200-kg car traveling with a velocity of 20 m/s due south. The two vehicles stick together after the collision. a. What is the momentum of each vehicle prior to the collision? b. What are the size and direction of the total momentum of the two vehicles after they collide?

143

E18. A truck with a mass of 4000 kg traveling with a speed of 10 m/s collides at right angles with a car with a mass of 1500 kg traveling with a speed of 20 m/s. a. Sketch to proper scale and direction the momentum vectors of each vehicle prior to the collision. b. Using the graphical method of vector addition, add the momentum vectors to get the total momentum of the system prior to the collision.

E17. For the two vehicles in exercise 16: a. Sketch to scale the momentum vectors of the two vehicles prior to the collision. b. Add the two vectors on your sketch graphically.

synthesis problems SP1. A fast ball thrown with a velocity of 40 m/s (approximately 90 MPH) is struck by a baseball bat, and a line drive comes back toward the pitcher with a velocity of 60 m/s. The ball is in contact with the bat for a time of just 0.04 s. The baseball has a mass of 120 g (0.120 kg). a. What is the change in momentum of the baseball during this process? b. Is the change in momentum greater than the final momentum? Explain. c. What is the magnitude of the impulse required to produce this change in momentum? d. What is the magnitude of the average force that acts on the baseball to produce this impulse?

SP4. A car traveling at a speed of 18 m/s (approximately 40 MPH) crashes into a solid concrete wall. The driver has a mass of 90 kg. a. What is the change in momentum of the driver as he comes to a stop? b. What impulse is required in order to produce this change in momentum? c. How does the application and magnitude of this force differ in two cases: the first, in which the driver is wearing a seat belt, and the second, in which he is not wearing a seat belt and is stopped instead by contact with the windshield and steering column? Will the time of action of the stopping force change? Explain.

SP2. A bullet is fired into a block of wood sitting on a block of ice. The bullet has an initial velocity of 500 m/s and a mass of 0.005 kg. The wooden block has a mass of 1.2 kg and is initially at rest. The bullet remains embedded in the block of wood afterward. a. Assuming that momentum is conserved, find the velocity of the block of wood and bullet after the collision. b. What is the magnitude of the impulse that acts on the block of wood in this process? c. Does the change in momentum of the bullet equal that of the block of wood? Explain.

SP5. A 1500-kg car traveling due north with a speed of 25 m/s collides head-on with a 4500-kg truck traveling due south with a speed of 15 m/s. The two vehicles stick together after the collision. a. What is the total momentum of the system prior to the collision? b. What is the velocity of the two vehicles just after the collision? c. What is the total kinetic energy of the system before the collision? d. What is the total kinetic energy just after the collision? e. Is the collision elastic? Explain.

SP3. Consider two cases in which the same ball is thrown against a wall with the same initial velocity. In case A, the ball sticks to the wall and does not bounce. In case B, the ball bounces back with the same speed that it came in with. a. In which of these two cases is the change in momentum the largest? b. Assuming that the time during which the momentum change takes place is approximately the same for these two cases, in which case is the larger average force involved? c. Is momentum conserved in this collision? Explain.

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home experiments and observations HE1. Take two marbles or steel balls of the same size and practice shooting one into the other. Make these observations: a. If you produce a head-on collision with the second marble initially at rest, does the first marble come to a complete stop after the collision? b. If the collision with a second marble occurs at an angle, is the angle between the paths of the two marbles after the collision a right angle (90°)? c. If marbles of different sizes and masses are used, how do the results of parts a and b differ from those obtained with marbles of the same mass? HE2. If you have access to a pool table, try parts a and b of the observations in home experiment 1 on the pool table. What effect does putting spin on the first ball have on the collisions? HE3. If you have both a basketball and a tennis ball, try dropping the two of them onto a floor with a hard surface, first individually and then with the tennis ball placed on top of the basketball before the two are dropped together. a. Compare the height of the bounce of each ball in these different cases. The case where the two are dropped together may surprise you.

b. Can you devise an explanation for these results using impulse and Newton’s third law? (Consider the force between the basketball and the floor as well as that between the tennis ball and the basketball for the case where they are dropped together.) HE4. Place a cardboard box on a smooth tile or wood floor. Practice rolling a basketball or soccer ball at different speeds and allowing the ball to collide with the box. Observe the motion of both the box and the ball just after the collision. a. How do the results of the collision vary for different speeds of the ball (slow, medium, fast)? b. If we increase the weight of the box by placing books inside, how do the results of the collision change for the cases in part a? c. Can you explain your results using conservation of momentum?

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Rotational Motion of Solid Objects

8

chapter overview Starting with a merry-go-round—and making use of the analogy between linear and rotational motion—we first consider what concepts are needed to describe rotational motion. We then turn to the causes of rotational motion, which involve a modified form of Newton’s second law. Torque, rotational inertia, and angular momentum will be introduced as we proceed. Our goal is to develop a clear picture of both the description and causes of rotational motion. After studying this chapter, you should be able to predict what will happen in many common examples of spinning or rotating objects, such as ice skaters and divers. The world of sports is rich in examples of rotational motion.

chapter outline

2 3 4 5

What is rotational motion? How can we describe rotational motion? What are rotational velocity and acceleration, and how are they related to similar concepts used to describe linear motion? Torque and balance. What determines whether a simple object such as a balance beam will rotate? What is torque, and how is it involved in causing an object to rotate? Rotational inertia and Newton’s second law. How can Newton’s second law be adapted to explain the motion of rotating objects? How do we describe rotational inertia, an object’s resistance to changes in rotational motion? Conservation of angular momentum. What is angular momentum, and when is it conserved? How do spinning skaters or divers change their rotational velocities? Riding a bicycle and other amazing feats. Why does a bicycle remain upright when it is moving but not when it is stationary? Can we treat rotational velocity and angular momentum as vectors?

145

unit one

1

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I

n the park next door to the author’s house, there is a child-propelled merry-go-round (fig. 8.1). It consists of a circular steel platform mounted on an excellent bearing so that it rotates without much frictional resistance. Once set in rotation, it will continue to rotate. Even a child can start it moving and then jump on (and sometimes fall off). Along with the swings, the slide, and the little animals mounted on heavy-duty springs, the merry-go-round is a popular center of activity in the park. The motion of this merry-go-round bears both similarities and differences to motions we have already considered. A child sitting on the merry-go-round experiences circular motion, so some of the ideas discussed in chapter 5 will come into play. What about the merry-go-round itself, though? It certainly moves, but it goes nowhere. How do we describe its motion? Rotational motion of solid objects such as the merrygo-round is common: the rotating Earth, a spinning skater, a top, and a turning wheel all exhibit this type of motion. For Newton’s theory of motion to be broadly useful, it should explain what is happening in rotational motion as well as in linear motion (where an object moves from one point to another in a straight line). What causes rotational motion? Can Newton’s second law be used to explain such motion?

8.1 What Is Rotational Motion? A child begins to rotate the merry-go-round described in the introduction. She does so by holding on to one of the bars on the edge of the merry-go-round (fig. 8.1) as she stands beside it. She begins to push the merry-go-round, accelerating as she goes, until eventually she is running, and the merry-go-round is rotating quite rapidly. How do we describe the rotational motion of the merrygo-round or that of a spinning ice skater? What quantities would we use to describe how fast they are rotating or how far they have rotated?

Rotational displacement and rotational velocity How would you measure how fast the merry-go-round is rotating? If you stood to one side and watched the child pass your position, you could count the number of revolutions that the child makes in a given time, measured with your watch. Dividing the number of revolutions by the time in minutes yields the average rotational speed in revolutions per minute (rpm), a commonly used unit for describing the rate of rotation of motors, Ferris wheels, and other rotating objects. If you say that the merry-go-round rotates at a rate of 15 rpm, you have described how fast an object is turning. The rate is analogous to speed or velocity, quantities used

figure 8.1

The merry-go-round in the park is an example of rotational motion. How do we describe and explain this motion?

We will find that there is a useful analogy between the linear motion of objects and rotational motion. The questions just posed can be answered best by making full use of this analogy. Taking advantage of the similarities between rotational motion and linear motion saves space in our mental computers, thus making the learning process more efficient.

to describe how fast an object is moving in the case of linear motion. We usually use the term rotational velocity to describe this rate of rotation. Revolutions per minute is just one of several units used to measure this quantity. In measuring the rotational velocity of the merry-goround, we describe how far it rotates in revolutions or complete cycles. Suppose that an object rotates less than one complete revolution. We could then use a fraction of a revolution to describe how far it has turned, but we might also use an angle measured in degrees. Since there are 360° in one complete revolution or circle, revolutions can be converted to degrees by multiplying by 360°/rev. The quantity measuring how far an object has turned or rotated is an angle, often called the rotational displacement. It can be measured in revolutions, degrees, or a simple but less familiar unit used in mathematics and physics called the radian.* The three units commonly used to describe rotational displacement are summarized in figure 8.2.

*The radian is defined by dividing the arc length through which the point travels by the radius of the circle on which it is moving. Thus, in figure 8.2, if the point on the merry-go-round moves along the arc length a distance s, the number of radians involved is s/r where r is the radius. Since we are dividing one distance by another, the radian itself has no dimensions. Also, since the arc length s is proportional to the radius r, it does not matter how large a radius we choose. The ratio of s to r will be the same for a given angle. By definition of the radian, 1 revolution (rev) 360° 2 radians, and 1 radian (rad) 57.3°.

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What is rotational acceleration?

r 1 revolution = 360° = 2π radians

s 1 radian = 57.3°

In our original description of the child pushing the merrygo-round, the rate of rotation increased as she ran alongside. This involves a change in the rotational velocity, which suggests the concept of rotational acceleration. The Greek letter alpha () is the symbol used for rotational acceleration. It is the first letter in the Greek alphabet and corresponds to the letter a used to represent linear acceleration. Rotational acceleration can be defined similarly to linear acceleration (see chapter 2):

1°

Rotational acceleration is the rate of change in rotational velocity. It is found by dividing the change in rotational velocity by the time required for this change to occur, a

figure 8.2

Revolutions, degrees, or radians are different units for describing the rotational displacement of the merrygo-round.

Rotational displacement is analogous to the distance traveled by an object in linear motion. If we include the direction of travel, this distance is sometimes called the linear displacement. The symbols used to describe rotational quantities mainly come from the Greek alphabet. Greek letters are used to avoid confusion with other quantities represented by letters of our ordinary Roman alphabet. The Greek letter theta (u) is commonly used to represent angles (rotational displacements), and the Greek letter omega () is used to represent rotational velocities. The quantities that we have just introduced for describing the motion of an object such as the merry-go-round can be summarized as

¢ . t

The units of rotational acceleration are rev/s2 or rad/s2. These definitions for both rotational velocity and rotational acceleration actually yield the average values of these quantities. To get instantaneous values, the time interval t must be made very small, as in the linear-motion definitions of instantaneous velocity and instantaneous acceleration (see sections 2.2 and 2.3). This then yields the rate of change of either displacement or velocity at a given instant in time. You will remember these definitions of rotational displacement, velocity, and acceleration better if you keep in mind the complete analogy that exists between linear and rotational motion. This analogy is summarized in figure 8.3. In one dimension, distance d represents the change in position or linear displacement, which corresponds to rotational displacement u. Average velocity and acceleration for linear motion are defined as before, with the corresponding definitions of rotational velocity and acceleration shown on the right side of the diagram in figure 8.3.

Constant rotational acceleration Rotational displacement u is an angle showing how far an object has rotated.

and Rotational velocity is the rate of change of rotational displacement. It is found by dividing the rotational displacement by the time taken for this displacement to happen

u . t

In describing rotational velocity, we usually use either revolutions or radians as the measure of rotational displacement. Degrees are less commonly used.

In chapter 2, we introduced equations for the special case of constant linear acceleration because of its many important applications. By comparing linear and rotational quantities, we can write similar equations for constant rotational acceleration by substituting the rotational quantities for the corresponding linear quantities in the equations developed for linear motion. Table 8.1 shows the results beside corresponding equations for linear motion. Example box 8.1 is an application of the equations for constant rotational acceleration. The merry-go-round in example box 8.1 starts from rest and rotates through nine complete revolutions in 1 minute, a good effort by the person pushing. It is unlikely that this rate of acceleration could be sustained much longer than 1 minute. The rotational velocity reached in this time is a

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Chapter 8 Rotational Motion of Solid Objects

Rotational motion

Linear motion

d v

θ

ω

Displacement = d Velocity = d –=v t __ = a Acceleration = ∆v t

Displacement = θ θ Velocity = – = ω t ∆ω = α Acceleration = __ t

figure 8.3

There is a close resemblance between quantities used to describe linear motion and those used to describe rotational motion.

little less than a third of a revolution per second, a fast rotational velocity for such a merry-go-round.

How are linear and rotational velocity related? How fast is the rider going when the merry-go-round in example box 8.1 is rotating with a velocity of 0.30 rev/s? The answer to this question depends on where the rider is sitting. He or she will move faster when seated near the edge of the merry-go-round than near the center. The question involves a relationship between the linear speed of the rider and the rotational velocity of the merry-go-round. Figure 8.4 shows two circles on the merry-go-round with different radii representing different positions of riders. The rider seated at the greater distance from the center travels a larger distance in 1 revolution than the rider near the center because the circumference of his circle is greater. The outside rider is therefore moving with a greater linear speed than the rider near the center.

The farther the rider is from the center, the farther he travels in 1 revolution, and the faster he is moving. The circumference of the circle on which the rider is traveling increases in proportion to the radius of the circle r, the distance of the rider from the center. If we express the rotational velocity in radians per second (rad/s), the relationship for the linear speed of the rider takes the form v r.

example box 8.1 Sample Exercise: Rotating a Merry-Go-Round Suppose that a merry-go-round is accelerated at a constant rate of 0.005 rev/s2, starting from rest. a. What is its rotational velocity at the end of 1 min? b. How many revolutions does the merry-go-round make in this time? a. 0.005 rev/s2 0 0

table 8.1

0 (0.005 rev/s2)(60 s)

t 60 s

Constant Acceleration Equations for Linear and Rotational Motion Linear motion

Rotational motion

v v0 at

0 t

d v0t

0 t

1 2

at2

u 0t

1 2

t2

b. u ?

0.30 rev/s Since 0 is equal to zero, u

1 2

t 2

1 2

(0.005 rev/s2)(60 s)2

9 rev

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pushes straight in toward the center, nothing happens. How do we apply a force to produce the best effect? Unbalanced torques cause objects to rotate. What are torques, though, and how are they related to forces? A look at a simple scale or balance can help us get at the idea.

When is a balance balanced?

figure 8.4

The rider near the edge travels a greater distance in 1 revolution than one near the center.

The linear speed v of a rider seated a distance r from the center of a merry-go-round is equal to r times the rotational velocity of the merry-go-round. (For this simple relationship to be valid, however, the rotational velocity must be expressed in radians per second rather than revolutions or degrees per second.) The rate at which the merry-go-round or other object turns will affect how fast a point on the rotating object will move—in other words, its linear speed. Linear speed will depend on the distance from the axis of rotation. A child out at the edge of the merry-go-round will get a bigger thrill from the ride than one more timidly parked near the middle. Rotational displacement, rotational velocity, and rotational acceleration are the quantities that we need to fully describe the motion of a rotating object. They describe how far the object has rotated (rotational displacement), how fast it is rotating (rotational velocity), and the rate at which the rotation may be changing (rotational acceleration). These definitions are analogous to similar quantities used to describe linear motion. They tell us how the object is rotating, but not why. Causes of rotation are considered next.

8.2 Torque and Balance What causes the merry-go-round to rotate in the first place? To get it started, a child has to push it, which involves applying a force. The direction and point of application of force are important to the success of the effort. If the child

Consider a balance made of a thin but rigid beam supported by a fulcrum or pivot point, as in figure 8.5. If the beam is balanced before we place weights on it, and if we put equal weights at equal distances from the fulcrum, we expect that the beam will still be balanced. By balanced, we mean that it will not tend to rotate about the fulcrum. Suppose that we wish to balance unequal weights on the beam. To balance a weight twice as large as a smaller weight, would we place the two weights at equal distances from the fulcrum? Intuition suggests that the smaller weight needs to be placed farther from the fulcrum than the larger weight for the system to be balanced, but it may not tell you how much farther (fig. 8.6). Trial and error with a simple balance will show that the smaller weight must be placed twice as far from the fulcrum as the larger (twice as large) weight. Try it yourself using a ruler for the beam and a pencil for the fulcrum. Coins can be used as the weights. Experiments will show that both the weight and the distance from the fulcrum are important. The farther a weight sits from the fulcrum, the more effective it will be in balancing larger weights on the other side of the fulcrum. Weight times distance from the fulcrum determines the effect. If this product is the same for weights placed on either side of the fulcrum, the balance will not rotate.

F1

F2

figure 8.5

A simple balance with equal weights placed at equal distances from the fulcrum.

F1 F2

figure 8.6

A simple balance with unequal weights placed at different distances from the fulcrum. What determines whether the system will be balanced?

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Chapter 8 Rotational Motion of Solid Objects

What is a torque? This product of the force and the distance from the fulcrum—which describes the tendency of a weight to produce a rotation, is called the torque. More generally: The torque, t, about a given axis or fulcrum is equal to the product of the applied force and the lever arm, l; t Fl. The lever arm is the perpendicular distance from the axis of rotation to the line of action of the applied force.

The symbol t is the Greek letter tau and is commonly used for torque. The length l is the distance from the fulcrum to the point of application of the force and must be measured in a direction perpendicular to the line of action of the force. This distance is called the lever arm or moment arm of the force in question. The strength of the torque depends directly on both the size of the force and the length of its lever arm. If the torques produced by weights on either side of the fulcrum of our balance are equal in magnitude, the scale is balanced. It will not rotate. Most of us have tried to turn a nut with a wrench at some time. We exert the force at the end of the wrench, in a direction perpendicular to the handle (fig. 8.7). The handle is the lever, and its length determines the lever arm. A longer handle is more effective than a shorter one because the resulting torque is greater. As the term suggests, lever arm comes from our use of levers to move objects. Moving a large rock with a crowbar,

for example, involves leverage. The applied force is most effective if it is applied at the end of the bar and perpendicular to the bar. The lever arm l is then just the distance from the fulcrum to the end of the bar. If the force is applied in some other direction, as in figure 8.8, the lever arm is shorter than it would be if the force is applied perpendicular to the bar. The lever arm is found by drawing the perpendicular line from the fulcrum to the line of action of the force, as indicated in figure 8.8.

How do torques add? The direction of rotation associated with a torque is also important. Some torques tend to produce clockwise rotations and others counterclockwise rotations about a particular axis. For example, the torque due to the heavier weight on the right side of the fulcrum in figure 8.6 will produce a clockwise rotation about the fulcrum if it acts by itself. This is opposed by the equal-magnitude torque of the weight on the left side of the fulcrum, which would produce a counterclockwise rotation. The two torques cancel one another when the system is balanced. Since torques can have opposing effects, we assign opposite signs to torques that produce rotations in opposite directions. If, for example, we chose to call torques that produce a counterclockwise rotation positive, torques producing clockwise rotations would be negative. (This is the conventional choice—it is unimportant which direction is chosen as positive as long as you are consistent in a given situation.) Identifying the sign of the torque indicates whether it will add or subtract from other torques. In the case of the simple balance, the net torque will be zero when the beam is balanced, because the two torques are equal but have opposite signs. The condition for balance or equilibrium is that the net torque acting on the system be zero. Either no torques act or the sum of the positive

F F l

F l

l

figure 8.7

A wrench with a long handle is more effective than one with a short handle because of the longer lever arm for the longer wrench.

figure 8.8

When the applied force is not perpendicular to the crowbar, the lever arm is found by drawing a perpendicular line from the fulcrum to the line of action of the force.

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8.2 Torque and Balance

torques equals the sum of the negative torques, canceling one another by adding up to zero. In example box 8.2, we find the distance that a 3-N weight must be placed from a fulcrum to balance a 5-N weight producing a net torque of zero. (Since W mg, a 5-N weight has a mass of approximately 0.5 kg or 500 g.) The units of torque are those of force times distance, newtonmeters (N·m) in the metric system.*

151

Center of gravity of the plank Wc Wp

What is the center of gravity of an object? Often, the weight of an object is itself an important factor in whether the object will rotate. How far, for example, could the child in figure 8.9 walk out on the plank without

example box 8.2 Sample Exercise: Balancing a System Suppose we have a 3-N weight that we want to balance against a 5-N weight on a beam, which is balanced when no masses are in place. The 5-N weight is placed 20 cm to the right of the fulcrum. a. What is the torque produced by the 5-N weight? b. How far would we have to place the 3-N weight from the fulcrum to balance the system?

3N

20 cm

?

5N F Where should the 3-N weight be placed on the beam to balance the system?

a. F 5 N

t Fl

l 20 cm 0.2 m

(5 N)(0.2 m)

t ?

1 N·m

The minus sign indicates that this torque would produce a clockwise rotation. b. F 3 N l ?

t Fl l

t F 1 N#m 3N

0.33 m (33 cm) *Although the product of a Newton-meter (N.m) equals a Joule (J) when we are using it as an energy unit, when a N.m is used as a torque unit we state it as N.m, not Joules.

figure 8.9

How far can the child walk without tipping the plank? The entire weight of the plank can be treated as though it is located at the center of gravity.

the plank tipping? The weight of the plank is important in this case, and the concept of center of gravity is useful. The center of gravity is the point about which the weight of the object itself exerts no net torque. If we suspend the object from its center of gravity, there would be no net torque at the suspension or support point. The object would be balanced. We can locate the center of gravity of a rodlike object by finding the point where it balances on your finger or other suitable fulcrum. For a more complex two-dimensional (planar) object, you can locate the center of gravity by suspending the object from two different points, drawing a line straight down from the point of suspension in each case, and locating the point of intersection of the two lines, as figure 8.10 illustrates. In the case of the plank (fig. 8.9), the center of gravity is at the geometric center of the plank, provided that the plank is uniform in density and cut. The pivot point will be the edge of the supporting platform, the point to consider when computing torques. The plank will not tip as long as the counterclockwise torque produced by the weight of the plank about the pivot point is larger than the clockwise torque produced by the weight of the child. The weight of the plank is treated as though it is concentrated at the center of gravity of the plank. The plank will verge on tipping when the torque of the child about the edge equals the torque of the plank in magnitude. This determines how far the child can walk on the plank before it tips. As long as the torque of the plank about the edge of the platform is larger than the torque of the child, the child is safe. The platform keeps the plank from rotating counterclockwise. The location of the center of gravity is important in any effort at balancing. If the center of gravity lies below the pivot point, as in the balancing toy in figure 8.11, the toy will automatically regain its balance when disturbed. The center of gravity returns to the position directly below the pivot point, where the weight of the toy produces no torque.

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Chapter 8 Rotational Motion of Solid Objects

1

2 1

Center of gravity

figure 8.10

Locating the center of gravity of a planar object. The center of gravity does not necessarily lie within the object.

Torques determine whether or not something will rotate. A torque is found by multiplying a force by its lever arm (the perpendicular distance from the axis of rotation to the line of action of the force). If the torque tending to produce a clockwise rotation equals the torque tending to produce a counterclockwise rotation, there is no rotation. If one of these torques is larger than the other, the torque will be unbalanced and the system will rotate.

8.3 Rotational Inertia and Newton’s Second Law When a child runs beside a merry-go-round, starting it to rotate, the force exerted by the child produces a torque about the axle. From our discussion in section 8.2, we know that the net torque acting on an object determines whether or not it will begin to rotate. Can we predict the rate of rotation by knowing the torque? In linear motion, net force and mass determine the acceleration of an object, according to Newton’s second law of motion. How do we adapt Newton’s second law to cases of rotational motion? In this case, torque determines the rotational acceleration. A new quantity, the rotational inertia, takes the place of mass.

What is rotational inertia?

Pivot point Center of gravity W

figure

8.11 The clown automatically returns to an upright position because the center of gravity is below the pivot point.

In this position, the lever arm for the weight of the clown and bar is zero. Similarly, the location of your center of gravity is important in performing various maneuvers, athletic or otherwise. Try, for example, touching your toes with your back and heels against a wall. Why is this apparently simple trick impossible for most people to do? Where is your center of gravity relative to the pivot point determined by your feet? Center of gravity and torque are at work here.

Let’s return to the merry-go-round. The propulsion system (one energetic child or tired parent) applies a force at the edge of the merry-go-round. The torque about the axle is found by multiplying this force by the lever arm, in this case the radius of the merry-go-round (fig 8.12). If the frictional torque at the axle is small enough to be ignored, the torque produced by the child is the only one acting on the system. This torque produces the rotational acceleration of the merry-go-round. How would we find this rotational acceleration? To find the linear acceleration produced by a force acting on an object, we use Newton’s second law, Fnet ma. By analogy, we can develop a similar expression for rotational motion, where the torque t replaces the force and the rotational acceleration replaces the linear acceleration. But what quantity should we use in place of the mass of the merry-go-round? In linear motion, mass represents the inertia or resistance to a change in motion. For rotational motion, a new concept is needed, rotational inertia, also referred to as the moment of inertia. The rotational inertia is the resistance of an object to change in its rotational motion. Rotational inertia is related to the mass of the object but also depends on how that mass is distributed about the axis of rotation.

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8.3 Rotational Inertia and Newton’s Second Law

F r m

figure

Top view

8.13 A single concentrated mass at the end of a very light rod is set into rotation by the applied force F. Use Newton’s second law to find the acceleration.

= Fl l=r

the size of the mass, the rotational inertia of a concentrated mass is F

rotational inertia mass square of distance from axis I mr2,

figure

8.12 The child exerts a force at the rim of the merry-go-round that produces a torque about the axle.

To get a feeling for a concept, physicists often use the trick of considering the simplest possible situation. For rotational motion, the simplest case is a single concentrated mass at the end of a very light rod, as in figure 8.13. If a force is applied to this mass in a direction perpendicular to the rod, the rod and mass will begin to rotate about the fixed axis at the other end of the rod. For the rod and mass to undergo a rotational acceleration, the mass itself must have a linear acceleration. Like riders on a merry-go-round, however, the farther the mass is from the axis, the faster it moves for a given rotational velocity (v r). To produce the same rotational acceleration, a mass at the end of the rod must receive a larger linear acceleration than one nearer the axis. It is harder to get the system rotating when the mass is at the end of the rod than when it is nearer to the axis. Applying Newton’s second law to this situation, we find that the resistance to a change in rotational motion depends on the square of the distance of this mass from the axis of rotation. Since the resistance to change also depends on

where I is the symbol commonly used for rotational inertia, and r is the distance of the mass m from the axis of rotation. The total rotational inertia of an object like the merry-go-round can be found by adding the contributions of different parts of the object lying at different distances from the axis.

Newton’s second law modified for rotational motion By analogy to Newton’s second law, Fnet ma, we can state the second law for rotational motion as The net torque acting on an object about a given axis is equal to the rotational inertia of the object about that axis times the rotational acceleration of the object, or tnet I.

To put it differently, the rotational acceleration produced is equal to the torque divided by the rotational inertia, tnet /I. The larger the torque, the larger the rotational acceleration, but the larger the rotational inertia, the smaller the rotational acceleration. Rotational inertia dictates how hard it is to change the rotational velocity of the object.

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Chapter 8 Rotational Motion of Solid Objects

To get a feel for these ideas, consider a simple object such as a twirler’s baton. A baton consists of two masses at the end of a rod (fig. 8.14). If the rod itself is light, most of the baton’s rotational inertia comes from the masses at either end. If you hold the baton at the center, a torque can be applied with your hand, producing a rotational acceleration and starting the baton to rotate. Suppose that we could move these masses along the rod. If we moved the masses toward the center of the rod so that the distance from the center is half the original distance, what happens to the rotational inertia? The rotational inertia decreases to one-fourth of its initial value, ignoring the contribution of the rod. Rotational inertia depends on the square of the distance of the mass from the axis. Doubling the distance quadruples the rotational inertia. Halving the distance divides the rotational inertia by four. The baton will be four times as hard to get to rotate when the masses are at the ends as when they are halfway from the ends. In other words, the torque needed to produce a rotational acceleration will be four times as large when the masses are at the ends as when they are at the intermediate positions. If you had a rod with adjustable masses, you could feel the difference in the amount of torque needed to start it rotating. Try a pencil with lumps of clay for the masses as a substitute.

Finding the rotational inertia of the merry-go-round Finding the rotational inertia of an object like a merry-goround is more difficult than just multiplying the mass by the square of the radius. Not all of the mass of the merrygo-round is at the outer edge—some of it is closer to the

axis and will make a smaller contribution to the rotational inertia. Imagine breaking the merry-go-round down into several pieces, finding the rotational inertia of each piece, and adding the rotational inertias for each piece together to get the total. Results of this process for a few simple shapes are shown in figure 8.15. The equations illustrate the ideas we have discussed. For example, a solid disk has a smaller rotational inertia than a ring of the same mass and radius, because the mass of the disk is, on average, closer to the axis. The location of the axis is also important. A rod has a larger rotational inertia about an axis through one end than about an axis through the middle. When the axis of rotation is at the end of the rod, there is more mass at greater distances from the axis. Depending on how it is constructed, the merry-go-round might be like a solid disk. A child sitting on the merrygo-round will also affect the rotational inertia. If several children all sit near the edge of the merry-go-round, their rotational inertia makes it more difficult to get the merrygo-round moving. If the children cluster near the center, they provide less additional rotational inertia. If you are feeling tired, have the children sit near the middle. You will save some effort.

Rod with axis through end

Rod with axis through center

Ι = 1_ ml 2

1 Ι=_ ml 2

3

12

l

l

(a)

m

m

(b)

Uniform disk or cylinder

Ring

Uniform solid sphere

Ι = 1_ mr 2

Ι = mr 2

Ι = 2_ mr 2

2

r

(c)

figure 8.14

The rotational inertia of a baton is determined largely by the masses at either end.

figure

5

r

(d)

r

(e)

8.15 Expressions for the rotational inertia of several objects, each with a uniform distribution of mass over its volume. The letter m is used to represent the total mass of the object.

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8.4 Conservation of Angular Moment

8.4 Conservation of Angular Momentum

example box 8.3 Sample Exercise: Turning a Merry-Go-Round and a Rider A simple merry-go-round has a rotational inertia of 800 kg·m2 and a radius of 2 m. A child with a mass of 40 kg sits near the edge of the merry-go-round. a. What is the total rotational inertia of the merrygo-round and the child about the axis of the merrygo-round? b. What torque is required to give the merry-go-round a rotational acceleration of 0.05 rad/s2? a. Imerry-go-round 800 kg·m2

Ichild mr2

mchild 40 kg

(40 kg)(2 m)2

r 2m

160 kg·m2

The total rotational inertia is Itotal Imerry-go-round Ichild 800 kg·m2 160 kg·m2 960 kg·m2 b. 0.05 rad/s2 tnet ?

155

tnet I (960 kg·m2)(0.05 rad/s2) 48 N·m

Example box 8.3 attaches some numbers to these quantities. A child sitting on a merry-go-round is being accelerated by a push at a rate of 0.05 rad/s2.* A torque of 48 N·m is needed to produce this rotational acceleration. A force of 24 N applied at the edge would have a lever arm of 2 m and produce the necessary torque of 48 N·m, a reasonable force for a child to generate if the child is not too small.

Have you ever watched an ice skater go into a spin? She starts the spin with her arms and one leg extended, then brings them in toward her body. As she brings her arms in, the rate of the spin increases; as she extends her arms again, her rotational velocity decreases (fig. 8.16). The concept of angular or rotational momentum is useful in situations like this. The principle of conservation of angular momentum explains a variety of phenomena like the ice skater, including tumbling divers or gymnasts as well as the motion of planets around the sun. How can we use the analogy between linear and rotational motion to understand these ideas?

What is angular momentum? If you were asked to invent the idea of angular (rotational) momentum, how might you go about it? Linear momentum is the mass (the inertia) times the linear velocity of an object (p mv). An increase in either the mass or the velocity increases the momentum. Since it is a measure both of how much is moving and how fast it is moving, Newton called momentum the quantity of motion. What is momentum’s rotational equivalent? In comparing rotational and linear motion, rotational inertia plays the role of mass and rotational velocity replaces linear velocity. By analogy, we can define angular momentum as Angular momentum is the product of the rotational inertia and the rotational velocity, or L I ,

where L is the symbol used for angular momentum. The term angular momentum is more common than rotational momentum, but either can be used.

Rotational inertia is the resistance to change in rotational motion. It depends on both the mass of the object and the distribution of that mass about the axis of rotation. The rotational form of Newton’s second law, tnet I , shows the quantitative relationship between torque, rotational inertia, and rotational acceleration. Torque takes the place of force, rotational inertia replaces mass, and rotational acceleration replaces linear acceleration.

*To use Newton’s second law for rotational motion, the rotational acceleration must be stated in radians per second squared. If the rotational acceleration is provided in rev/s2 or some other angular unit, we convert it to rad/s2 before proceeding.

ω1

figure

ω2

8.16 The rotational velocity of the skater increases as she pulls her arms and leg in toward her body.

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Chapter 8 Rotational Motion of Solid Objects

Like linear momentum, angular momentum is the product of two quantities, an inertia and a velocity. A bowling ball spinning slowly might have the same angular momentum as a baseball spinning much more rapidly, because of the larger rotational inertia I of the bowling ball. With its enormous rotational inertia, the Earth has a huge angular momentum associated with its daily turn about its axis, even though the rotational velocity is small.

When is angular momentum conserved? We have used the analogy between linear and rotational motion to introduce angular momentum. Can we also use it to state the principle of conservation of angular momentum? In chapter 7, we found that linear momentum is conserved when there is no net external force acting on a system. When would angular momentum be conserved? Since torque takes the role of force for rotational motion, we can state the principle of conservation of angular momentum as If the net torque acting on a system is zero, the total angular momentum of the system is conserved.

Torque replaces force, and angular momentum replaces ordinary or linear momentum. Table 8.2 lists some important parallels between linear and rotational motion.

When the skater’s arms and one leg are extended, they contribute a relatively large portion to her total rotational inertia—their average distance from her axis of rotation is much larger than for other portions of her body. Rotational inertia depends on the square of the distance of various portions of her mass from the axis (I mr2). The effect of this distance is substantial, even though her arms and one leg are only a small part of the total mass of the skater. When the skater pulls her arms and leg in toward her body, their contribution to her rotational inertia decreases, and therefore, her total rotational inertia decreases. Conservation of angular momentum requires that her angular momentum remain constant. Since angular momentum is the product of the rotational inertia and rotational velocity, L I, if I decreases, must increase for angular momentum to stay constant. She can slow her rate of spin by extending her arms and one leg again, which she does at the end of the spin. This increases her rotational inertia and decreases her rotational velocity: angular momentum is conserved. These ideas are illustrated in example box 8.4. This phenomenon can be explored using a rotating platform or stool with good bearings to keep the frictional torques small (fig. 8.17). In these demonstrations, we often have the students hold masses in their hands, which increase the changes in rotational inertia that happen as the arms are drawn in toward the body. A striking increase in rotational velocity can be achieved!

Changes in the ice skater’s rate of spin Conservation of angular momentum is the key to understanding what happens when the spinning ice skater increases her rotational velocity by pulling in her arms. The external torque acting on the skater about her axis of rotation is very small, so the condition for conservation of angular momentum exists. Why does her rotational velocity increase?

table 8.2 Corresponding Concepts of Linear and Rotational Motion Concept Inertia Newton’s second law Momentum Conservation of momentum Kinetic energy

Linear motion

Rotational motion

m

I

Fnet ma

tnet I

p mv

L I

If Fnet 0, p constant 1

KE 2 mv 2

If tnet 0, L constant 1

KE 2 I2

example box 8.4 Sample Exercise: Some Physics of Figure Skating An ice skater has a rotational inertia of 1.2 kg·m2 when her arms are extended and a rotational inertia of 0.5 kg·m2 when her arms are pulled in close to her body. If she goes into a spin with her arms extended and has an initial rotational velocity of 1 rev/s, what is her rotational velocity when she pulls her arms in close to her body? I1 1.2 kg·m2 I2 0.5 kg·m2

Since angular momentum is conserved:

1 1 rev/s

Lfinal Linitial

2 ?

I22 I11

Dividing both sides by I2, 2 (I1 /I2)1 (1.2 kg·m2/0.5 kg·m2)(1 rev/s) (2.4)(1 rev/s) 2.4 rev/s

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8.4 Conservation of Angular Moment

157

ω

figure

8.17 A student holding masses in each hand while sitting on a rotating stool can achieve a large increase in rotational velocity by bringing his arms in toward his body.

A similar effect is at work when a diver pulls into a tuck position to produce a spin. In this case, the diver starts with her body extended and a slow rate of rotation about an axis through her body’s center of gravity (fig. 8.18). As she goes into a tuck, the rotational inertia about this axis is reduced, and rotational velocity increases. As her dive nears completion, she comes out of the tuck, increasing the rotational inertia and decreasing the rotational velocity. (The torque about the center of gravity due to the gravitational force acting on the diver is zero.) There are many examples of varying the rotational velocity by changing the rotational inertia. It is much easier to produce a change in the rotational inertia of a body than to change the mass of the body. We simply change the distance of various portions of the mass from the axis of rotation. Conservation of angular momentum provides a quick explanation for these phenomena.

Kepler’s second law Conservation of angular momentum also plays a role in the orbit of a planet about the sun, and in fact, it can be used to explain Kepler’s second law of planetary motion (see section 5.3). Kepler’s second law says that the radius line from the sun to the planet sweeps out equal areas in equal times. The planet moves faster in its elliptical orbit

figure

8.18 The diver increases her rotational velocity by pulling into a tuck position, thus reducing her rotational inertia about her center of gravity.

when it is nearer to the sun than when it is farther from the sun. The gravitational force acting on the planet produces no torque about the sun, because its line of action passes directly through the sun (fig. 8.19). The lever arm for this force is zero, and the resulting torque must also be zero. Angular momentum, therefore, is conserved. When the planet moves nearer to the sun, its rotational inertia I about the sun decreases. To conserve angular momentum, the rotational velocity of the planet about the sun (and thus its linear velocity*) must increase to keep the product L I constant. This requirement results in equal areas being swept out by the radius line in equal times. The

*For a compact mass rotating about some axis, the definition of angular momentum reduces to L mvr, where mv is the linear momentum and r is the perpendicular distance from the axis of rotation to the line along which the object is moving at that instant. If r decreases, v must increase to conserve angular momentum.

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Chapter 8 Rotational Motion of Solid Objects

everyday phenomenon

box 8.1

Achieving the State of Yo The Situation. A physics professor noticed that one of his students often carried a yo-yo to class and was proficient at putting the yo-yo through its paces. The professor challenged the student to explain the behavior of the yo-yo using the principles of torque and angular momentum. In particular, the professor asked the student to explain why the yo-yo sometimes comes back but sometimes can be made to “sleep,” or continue to rotate, at the end of the string. What are the differences in these two situations?

T

W A cut-away diagram showing the forces acting on the yo-yo when it is falling. Its weight and the tension in the string are the only significant forces. A yo-yo will come back to your hand, or with sufficient skill, you can make it “sleep” at the end of its string.

The Analysis. The student carefully examined the yo-yo’s construction and how the string is attached. He noticed that the string is not tied tightly to the axle of the yo-yo, but ends in a loose loop around the axle instead. When the yo-yo is at the end of its string, the string can slip on the axle. When wound around the axle, on the other hand, the string is less likely to slip. Usually, the yo-yo is started with the string wound around the axle and looped around the middle finger. When the yo-yo is released from the hand, the string unwinds, and the yo-yo gains rotational velocity and angular momentum. The student reasoned that a torque must be at work, and he drew a force diagram for the yo-yo that looked like the one shown here. Two forces act on the yo-yo, its weight acting downward and the tension in the string acting upward. Since the yo-yo is accelerated downward, the weight must be greater than the tension to produce a downward net force. The weight does not produce a torque about the center of gravity of the yo-yo, though, because its line of action passes through the center of gravity, and the lever arm is zero. The tension acts along a line that is off-center and produces a torque that will cause a counterclockwise rotation about the center of gravity, as in the drawing. The torque due to the tension in the string produces a rotational acceleration, and the yo-yo gains rotational velocity

and angular momentum as it falls. The yo-yo has a sizable angular momentum when it reaches the bottom of the string, and in the absence of external torques to change this angular momentum, it will be conserved. This is what happens when the yo-yo “sleeps” at the bottom of the string: the only torque acting is the frictional torque of the string slipping on the axle, and this will be small if the axle is smooth. What happens, however, when the yo-yo returns to the student’s hand? The yo-yo artist (yo-yoist?) jerks lightly on the string at the instant that the yo-yo reaches the bottom of the string. This jerk provides a brief impulse and upward acceleration of the yo-yo. Since it is already spinning, the yo-yo continues spinning in the same direction and the string rewinds itself around the axle of the yo-yo. The line of action of the tension in the string is now on the opposite side of the axle, though, and its torque causes the rotational velocity and angular momentum to decrease. The rotation should stop when the yo-yo slips back into the student’s hand. When the yo-yo is rising, the net force acting on the yo-yo is still downward, and the linear velocity of the yo-yo decreases along with its rotational velocity. The only time that a net force acts upward is when the upward impulse is delivered by jerking on the string. The situation is similar to a ball bouncing on the floor—the net force is downward except during the very brief time of contact with the floor. Our ability to affect the nature and timing of the impulse through the string causes the yo-yo either to sleep or return. This is what the “art of yo” is all about.

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8.5 Riding a Bicycle and Other Amazing Feats

ω

159

ω

F V

L = mvr

figure

8.20 The direction of the rotational-velocity vector for the counterclockwise rotation is defined to be upward along the axis of rotation, as indicated by the thumb on the right hand with the fingers curled in the direction of rotation.

figure 8.19

The gravitational force acting on the planet produces no torque about an axis through the sun because the lever arm is zero for this force.

can angular momentum have direction, and how is this direction involved in explaining the behavior of a bicycle, a spinning top, or other phenomena? velocity of the planet must be larger when the radius gets smaller to keep the area being swept out the same. You can observe a related effect in a simple experiment with a ball on a string. If you let the string wrap around your finger as it rotates, which produces a smaller radius of rotation, the ball will increase its rotational velocity about your finger. The rotational velocity increases as the rotational inertia I decreases because of the decreased radius. Angular momentum is conserved. Try it! Everyday phenomenon box 8.1 provides an example in which angular momentum is conserved at some points in the motion of a yo-yo. At other points the angular momentum changes under the influence of torques. By analogy to linear momentum, angular momentum is the product of the rotational inertia and the rotational velocity. Angular momentum is conserved when the net external torque acting on a system is zero. Decreases in rotational inertia lead to increases in rotational velocity, as demonstrated by the spinning ice skater. A spinning diver, a ball rotating at the end of a string, and a planet orbiting the sun are other examples of this effect.

8.5 Riding a Bicycle and Other Amazing Feats Have you ever wondered why a bicycle remains upright when it is moving but promptly falls over when not moving? Angular momentum is involved, but some additional wrinkles are needed in the explanation. The direction of angular momentum is an important consideration. How

Is angular momentum a vector? Linear momentum is a vector, and the direction of the momentum p is the same as for the velocity v of the object. Since angular momentum is associated with a rotational velocity, the question comes down to whether rotational velocities have direction. How would we define the direction of a rotational velocity? If a merry-go-round (or just a disk) is rotating in a counterclockwise direction, as in figure 8.20, how might we indicate that direction with an arrow? The term counterclockwise indicates the direction of rotation as seen from a certain perspective, but it does not define a unique direction. To complete our description, we would also have to specify the axis of rotation and our perspective or viewpoint. An object seen rotating counterclockwise when viewed from above is seen rotating clockwise when viewed from below. We could draw an axis of rotation and a curved arrow around it, as we often do, but it would be more desirable to specify direction with a simple straight arrow. The usual solution to this problem is to define the direction of the rotational-velocity vector as being along the axis of rotation and in the upward direction for the counterclockwise rotation in figure 8.20. A rule for whether the vector should point up or down along the axis can be defined with the help of your right hand. If you hold your right hand with the fingers curling around the axis of rotation in the direction of the rotation, your thumb points in the direction of the rotational-velocity vector. If the merrygo-round were rotating clockwise (instead of counterclockwise), your thumb would point down, the direction of the rotational-velocity vector.

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axis of rotation of the wheel

L1 ∆L

L2 axis of rotation of the tilt

figure 8.22

The change in angular momentum ( L) associated with a leftward tilt points straight back, parallel to the line of contact of the tires with the road. This change causes the angular momentum vector (and the wheel) to turn to the left.

L

L

figure

8.21 The angular-momentum vector for each wheel is horizontal when the bicycle is upright.

The direction of the angular-momentum vector is the same as the rotational velocity, since L I. Conservation of angular momentum requires that the direction of the angular-momentum vector remain constant, as well as its magnitude.

Angular momentum and bicycles Most of us have had some experience with riding a bicycle. The wheels of a bicycle acquire angular momentum when the bicycle is moving. Torque is applied to the rear wheel by the pedals and chain to produce a rotational acceleration. If the bicycle is moving in a straight line, the direction of the angular-momentum vector is the same for both wheels and is horizontal (fig. 8.21). To tip the bike over, the direction of the angularmomentum vector must change, and that requires a torque. This torque would normally come from the gravitational force acting on the rider and the bicycle through their center of gravity. When the bicycle is exactly upright, this force acts straight downward and passes through the axis of rotation for the falling bike. This axis of rotation is the line along which the tires contact the road. The torque about this axis will be zero, because the line of action of the force passes through the axis of rotation and the lever

arm is zero. The direction and magnitude of the original angular momentum are conserved. If the bike is not perfectly upright, a gravitational torque acts about the line of contact of the tires with the road. As the bike begins to fall, it acquires a rotational velocity and angular momentum about this axis. By our “right-hand rule,” the direction of that angular-momentum vector is along the axis and points forward or backward depending on the direction of tilt. If the bike tilts to the left as seen from behind, the change in angular momentum associated with this torque points straight back, as in figure 8.22. If the bike is standing still, that is all there is to it—the gravitational torque causes the bike to fall. When the bike is moving, however, the change in angular momentum L produced by the gravitational torque adds to the angular momentum already present (L1) from the rotating tires. As shown in figure 8.22, this causes a change in the direction of the total angular-momentum vector (L 2). This change in direction can be accommodated simply by turning the wheel

study hint Visualizing these angular momentum vectors and their changes can be an abstract and difficult task. The effect will seem much more real if you can directly experience it. If a bicycle wheel mounted on a hand-held axle (such as that pictured in figure 8.23) is available, try the tilt effect yourself. Grasp the wheel with both hands by the handles on each side and have someone give it a good spin with the wheel in a vertical plane. Then try tilting the wheel downward to the left to simulate a fall. The wheel will seem to have a mind of its own and will turn to the left as suggested by figure 8.22.

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of the bicycle rather than letting the bike fall. We compensate for the effects of the gravitational torque by turning the bicycle towards the direction of the impending fall. The larger the initial angular momentum, the smaller the turn required. The angular momentum of the wheels is a major factor in stabilizing the bicycle. This result may be surprising—yet all of us who have ridden bicycles take advantage of it routinely. When the bike is moving slowly, sharp turns of the wheel can keep it from falling while you shift your weight. Smaller adjustments suffice when the bike is moving more rapidly. By leaning into a curve, you use the gravitational torque to change the direction of angular momentum, helping to round the curve. Likewise, if you roll a coin along a tabletop, you will see it curve as it begins to fall. The path curves in the direction that the coin is tilting. You can also observe this effect of torque in changing the direction of an angular-momentum vector by holding a bicycle upright on its rear wheel and having a friend spin the front wheel. It is harder to change the direction of this wheel when it is spinning rapidly than when it is spinning slowly or not at all. You will also get the feeling that the wheel has a mind of its own. As you try to tilt the wheel, it will tend to turn in a direction perpendicular to the tilt. A bicycle tire mounted on a hand-held axle is even more effective for sensing the effects of torques applied to the axle. This is a common demonstration apparatus, but usually the tire is filled with steel cable rather than air. The steel cable gives the wheel a larger rotational inertia and a larger angular momentum for a given rate of spin. If you hold the axle on either side while the wheel is spinning in a vertical plane and then try to tip the wheel, you get a sense of what happens when you are riding a bicycle. It also demonstrates how hard it is to change the direction of the angular momentum of a rapidly spinning wheel. Everyday phenomenon box 8.2 discusses how torques are involved in the gear system of a bicycle.

figure

8.23 A student holds a spinning bicycle wheel while sitting on a stool that is free to rotate. What happens if the wheel is turned upside down?

rotating initially. The sum of the angular-momentum vector of the wheel and the vector of the student and stool add to yield the original angular momentum (fig. 8.24). This will be true if the angular momentum gained by the student and stool is exactly twice the original angular momentum of the wheel. The student can stop the rotation of the stool by flipping the wheel axis back to its original direction. The direction of angular momentum and its conservation are important in many other situations. The angular

Before wheel is flipped

Rotating stools and tops The hand-held bicycle wheel is good for other demonstrations that highlight angular momentum as a vector. If a student holds the wheel with its axle in the vertical direction while sitting on a rotating stool, conservation of angular momentum produces striking results. It is best to start the wheel spinning while holding the stool so that the stool does not rotate initially. We then have the student turn the wheel over, as in figure 8.23, reversing the direction of the angularmomentum vector of the wheel. Can you imagine what happens then? To conserve angular momentum, the original direction of the angularmomentum vector must be maintained. The only way this can happen is for the stool with the student volunteer to begin to rotate in the same direction that the wheel was

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After wheel is flipped

Lw

–Lw

Ls ⇒ Lw

figure

8.24 The angular momentum of the student and stool, Ls , adds to that of the wheel, Lw , to yield the direction and magnitude of the original angular momentum, Lw .

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everyday phenomenon Bicycle Gears The Situation. Most modern bicycles come equipped with the ability to change gears. When we are climbing a hill, we shift into a low gear making pedaling easier. When we are on level ground or a downgrade, we shift into higher gears, allowing us to cover more ground per turn of the pedal crank. How do these gears work? How does the torque exerted on the rear wheel change when we change gears? What is the advantage of having many different gears? Adding a rotational twist to the concept of simple machines can help in understanding how gears work. Similar ideas apply to an automobile transmission.

There are seven different sprockets on the rear-wheel gear assembly of a 21-speed bicycle. The two smaller sprockets on the pedal wheel lie behind the largest sprocket.

box 8.2 The Analysis. The photograph shows the pedal wheel and the rear-wheel gear assembly for a 21-speed bicycle. There are seven sprockets (toothed wheels) of different sizes on the rearwheel hub. There are also three different sprockets on the pedal wheel, only one of which is fully visible in the photograph. A pulley and lever mechanism (called a derailleur) allows us to move the chain from one sprocket to another. This is controlled by levers mounted on the handlebars that are linked to the derailleurs by cables. When we pedal the bicycle we apply a torque to the pedal sprocket by pushing on the pedals. If our feet push perpendicularly to the pedal shaft, then the lever arm is just the length of the shaft. That will be the case when the pedal is in the forward position where we get maximum torque as shown in the drawing. This maximum-torque position alternates from the left foot to the right foot as the crank turns. The torque exerted on the pedal chain ring causes the sprocket to accelerate rotationally provided that this torque is larger than the opposing torque exerted by the tension in the chain pulling back on the sprocket. This tension, in turn, produces a torque on the rear wheel via the rearwheel sprocket. The size of the torque transmitted to the rear wheel depends on which of the several sprockets is engaged with the chain. A larger sprocket radius yields a larger torque because of the greater lever arm. (The lever

(continued)

momentum of the helicopter’s rotors, for example, is an extremely important factor in helicopter design. The motion of a top also shows fascinating effects. If you have a top, observe what happens to the direction of the angularmomentum vector as the top slows down. As the top begins to totter, the change in direction of the angular-momentum vector causes the rotation axis of the top to rotate (precess) about a vertical line. Does this remind you of what happens with a bicycle wheel? Angular momentum and its direction are also central to atomic and nuclear physics. The particles that make up atoms have spins, and these spins imply angular momentum. The ways these angular-momentum vectors add are used to explain a variety of atomic phenomena. While the size scales differ enormously, it is useful to recognize the common

ground that atoms and nuclei share with bicycle wheels and the solar system. Like linear momentum, angular momentum is a vector. Its direction is the same as that of the rotational-velocity vector, which is along the axis of rotation, with the “righthand rule” specifying which way it points along that axis. Conservation of angular momentum requires that both the magnitude and direction of the angular-momentum vector be constant (if there are no external torques). Many interesting phenomena can be explained using these ideas, including the stability of a moving bicycle, the motion of a spinning top, and the behavior of atoms and galaxies.

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Summary

T T r1

F

r2

A force applied to the pedal produces a torque on the pedal wheel. This torque produces a tension in the chain that exerts a smaller torque on the rear-wheel sprocket due to its smaller radius.

arm is equal to the radius of the sprocket engaged.) As with an automobile, the rear wheel pushes against the road surface via friction, and by Newton’s third law, the frictional force pushes forward on the bicycle. How are simple-machine ideas involved? Suppose that the chain is engaged with the smallest sprocket on the rear-wheel assembly. The torque exerted on the rear wheel by the chain is then relatively small because of the small lever arm. The wheel turns several times, however, for each turn of the pedal sprocket. If, for example, the radius of the pedal sprocket is five times that of the rear-wheel sprocket, then the rear wheel turns five times for each turn of the pedal crank. (The circumference of each sprocket (2r) is proportional to the radius,

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and the circumference determines how far the chain must travel for each turn.) In this case, then, a small torque turns the rear wheel through a large angle while a larger input torque turns the pedal sprocket through a smaller angle. By analogy to linear work (force times distance moved), rotational work can be defined as the torque times the angle through which the sprocket turns (t u). As for any simple machine, the work output equals the work input, ignoring frictional torques at the wheel axles. Would the situation we have just described represent a high gear or a low gear? Since we are getting several turns of the rear wheel for each turn of the pedal crank, this is a high gear. For a lower gear, we need to move the chain to a larger sprocket on the rear wheel (or a smaller sprocket on the pedal wheel). This will transmit a larger torque to the rear wheel at the expense of turning the wheel through a smaller angle and moving the bicycle through a smaller distance for each turn of the crank. When we are going uphill, we need this larger torque to overcome the pull of gravity. For a 21-speed bike, there are three sprocket sizes on the pedal wheel and seven on the rear wheel allowing twentyone (3 7) different ratios between the two sprockets. The advantage of having all these choices is that we can adjust the mechanical advantage of our gear system to the conditions we encounter, thus adjusting the force that we need to apply to the pedals to achieve the desired torque. If this force is too large, we will quickly tire. When it is small, however, we may not be taking maximum advantage of the easier riding conditions. We can go faster in a higher gear.

summary We have considered the rotational motion of a solid object and what causes changes in rotational motion. We have used an analogy between linear motion and rotational motion to develop many of the concepts. The key ideas are summarized here:

1

What is rotational motion? Rotational displacement is described by an angle. Rotational velocity is the rate of change of that angle with time. Rotational acceleration is the rate of change of rotational velocity with time.

and balance. A torque is what causes an 2object Torque to rotate. It is defined as a force times the lever arm of the force, which is the perpendicular distance from the line of action of the force to the axis of rotation. If the net torque acting on an object is zero, the object will not change its state of rotation. F

θ ω

t ω = –θ , t

l

∆ω α = ––– t = Fl

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3

Rotational inertia and Newton’s second law. In the form of Newton’s second law that relates to rotation, torque takes the place of force, rotational acceleration replaces ordinary linear acceleration, and rotational inertia replaces mass. Rotational inertia depends on the distribution of mass about the axis of rotation.

Riding a bicycle and other amazing feats (angu5lar momentum as a vector). The direction of the rotationalvelocity and angular-momentum vectors are defined by the right-hand rule. These vectors explain the stability of a moving bicycle and other phenomena. If there are no external torques, the direction of the angular momentum is conserved, as well as its magnitude.

F

r

m ω

α net = Ι α,

Ι = mr 2

of angular momentum. By analogy 4to linearConservation momentum, angular momentum is defined as the rotational inertia times the rotational velocity. It is conserved when no net external torque acts on the system.

L = Iω If ext = 0, L = constant

key terms Linear motion, 146 Rotational velocity, 146 Rotational displacement, 146 Radian, 146 Linear displacement, 147

Rotational acceleration, 147 Fulcrum, 149 Torque, 150 Lever arm, 150 Center of gravity, 151

Rotational inertia, 152 Moment of inertia, 152 Angular momentum, 155 Conservation of angular momentum, 156

Q5. Is the rotational velocity of a child sitting near the center of a rotating merry-go-round the same as that of another child sitting near the edge of the same merry-go-round? Explain.

Q1. Which units would not be appropriate for describing a rotational velocity: rad/min2, rev/s, rev/h, m/s? Explain.

Q6. Is the linear speed of a child sitting near the center of a rotating merry-go-round the same as that of another child sitting near the edge of the same merry-go-round? Explain.

Q2. Which units would not be appropriate for describing a rotational acceleration: rad/s, rev/s2, rev/m2, degrees/s2? Explain. Q3. A coin rolls down an inclined plane gaining speed as it rolls. Does the coin have a rotational acceleration? Explain. Q4. The rate of rotation of an object is gradually slowing down. Does this object have a rotational acceleration? Explain.

Q7. If an object has a constant rotational acceleration, is its rotational velocity also constant? Explain. *Q8. A ball rolls down an inclined plane gaining speed as it goes. Does the ball experience both linear and rotational acceleration? How far does the ball travel in one revolution?

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How is the linear velocity of the ball related to its rotational velocity? Explain. Q9. Which, if either, will produce the greater torque: a force applied at the end of a wrench handle (perpendicular to the handle) or an equal force applied in the same direction near the middle of the handle? Explain. Q10. Which of the forces pictured as acting upon the rod in the diagram will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod? Explain. Axis F1

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*Q19. A tall crate has a higher center of gravity than a shorter crate. Which will have the greater tendency to tip over if we push near the top of the crate? Explain with a force diagram. Where is the probable axis of rotation? Q20. Two objects have the same total mass, but object A has its mass concentrated closer to the axis of rotation than object B. Which object will be easier to set into rotational motion? Explain. Q21. Is it possible for two objects with the same mass to have different rotational inertias? Explain. Q22. Can you change your rotational inertia about a vertical axis through the center of your body without changing your total weight? Explain. Q23. A solid sphere and a hollow sphere made from different materials have the same mass and the same radius. Which of these two objects, if either, will have the greater rotational inertia about an axis through its center? Explain.

F2

Q24. Is angular momentum always conserved? Explain.

Q10 Diagram Q11. The two forces in the diagram have the same magnitude. Which orientation will produce the greater torque on the wheel? Explain. F1 F2

Q11 Diagram Q12. Is it possible to balance two objects of different weights on the beam of a simple balance resting upon a fulcrum? Explain. *Q13. Is it possible for the net force acting on an object to be zero, but the net torque to be greater than zero? Explain. (Hint: The forces contributing to the net force may not lie along the same line.) Q14. You are trying to move a large rock using a steel rod as a lever. Will it be more effective to place the fulcrum nearer to your hands or nearer to the rock? Explain. Q15. A pencil is balanced on a fulcrum located two-thirds of the distance from one end. Is the center of gravity of this pencil located at its center point? Explain. Q16. A solid plank with a uniform distribution of mass along its length rests on a platform with one end of the plank protruding over the edge. How far out can we push the plank before it tips? Explain. *Q17. A uniform metal wire is bent into the shape of an L. Will the center of gravity for the wire lie on the wire itself? Explain. Q18. An object is rotating with a constant rotational velocity. Can there be a net torque acting on the object? Explain.

Q25. A metal rod is rotated first about an axis through its center and then about an axis passing through one end. If the rotational velocity is the same in both cases, is the angular momentum also the same? Explain. Q26. A child on a freely rotating merry-go-round moves from near the center to the edge. Will the rotational velocity of the merry-go-round increase, decrease, or not change at all? Explain. *Q27. Moving straight inward, a large child jumps onto a freely rotating merry-go-round. What effect will this have on the rotational velocity of the merry-go-round? Explain. Q28. Is it possible for an ice skater to change his rotational velocity without involving any external torque? Explain. Q29. Suppose you are rotating a ball attached to a string in a circle. If you allow the string to wrap around your finger, does the rotational velocity of the ball change as the string shortens? Explain. Q30. Does the direction of the angular-momentum vector of the wheels change when a bicycle goes around a corner? Explain. Q31. An ice skater is spinning counterclockwise about a vertical axis when viewed from above. What is the direction of her angular-momentum vector? Explain. Q32. A pencil, balanced vertically on its eraser, falls to the right. a. What is the direction of its angular-momentum vector as it falls? b. Is its angular momentum conserved during the fall? Explain. *Q33. A top falls over quickly if it is not spinning, but will stay approximately upright for some time when it is spinning. Explain why this is so. Q34. Can a yo-yo be made to “sleep” if the string is tied tightly to the axle? Explain.

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exercises E1. Suppose that a merry-go-round is rotating at the rate of 10 rev/min. a. Express this rotational velocity in rev/s. b. Express this rotational velocity in rad/s.

E10. A weight of 3 N is located 10 cm from the fulcrum on the beam of a simple balance. What weight should be placed at a point 5 cm from the fulcrum on the opposite side in order to balance the system?

E2. When the author was a teenager, the rate of rotation for popular music records on a record player was 45 RPM. a. Express this rotational velocity in rev/s. b. Through how many revolutions does the record turn in a time of 5 s?

E11. Two forces are applied to a merry-go-round with a radius of 1.2 m as shown in the diagram below. One force has a magnitude of 80 N and the other a magnitude of 50 N. a. What is the torque about the axle of the merry-go-round due to the 80-N force? b. What is the torque about the axle due to the 50-N force? c. What is the net torque acting on the merry-go-round?

E3. Suppose that a disk rotates through three revolutions in 4 seconds. a. What is its displacement in radians in this time? b. What is its average rotational velocity in rad/s? E4. The rotational velocity of a merry-go-round increases at a constant rate from 0.6 rad/s to 1.8 rad/s in a time of 4 s. What is the rotational acceleration of the merry-go-round?

1.2 m

E5. A bicycle wheel is rotationally accelerated at the constant rate of 1.2 rev/s2. a. If it starts from rest, what is its rotational velocity after 4 s? b. Through how many revolutions does it turn in this time? E6. The rotational velocity of a spinning disk decreases from 6 rev/s to 3 rev/s in a time of 12 s. What is the rotational acceleration of the disk?

80 N

50 N

E11 Diagram E12. A net torque of 60 N·m is applied to a disk with a rotational inertia of 12.0 kg·m2. What is the rotational acceleration of the disk?

E7. Starting from rest a merry-go-round accelerates at a constant rate of 0.2 rev/s2. a. What is its rotational velocity after 5 s? b. How many revolutions occur during this time?

E13. A wheel with a rotational inertia of 4.5 kgm2 accelerates at a rate of 3.0 rad/s2. What net torque is needed to produce this acceleration?

E8. A force of 50 N is applied at the end of a wrench handle that is 24 cm long. The force is applied in a direction perpendicular to the handle as in the diagram. a. What is the torque applied to the nut by the wrench? b. What would the torque be if the force were applied half way up the handle instead of at the end?

E14. A torque of 60 N·m producing a counterclockwise rotation is applied to a wheel about its axle. A frictional torque of 10 N·m acts at the axle. a. What is the net torque about the axle of the wheel? b. If the wheel is observed to accelerate at the rate of 2 rad/s2 under the influence of these torques, what is the rotational inertia of the wheel?

24

cm

50 N

E15. Two 0.2-kg masses are located at either end of a 1-m long, very light and rigid rod as in the diagram. What is the rotational inertia of this system about an axis through the center of the rod? Axis

0.2 kg

E8 Diagram E9. A weight of 30 N is located a distance of 10 cm from the fulcrum of a simple balance beam. At what distance from the fulcrum should a weight of 20 N be placed on the opposite side in order to balance the system?

0.2 kg 1m

E15 Diagram

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E16. A mass of 0.8 kg is located at the end of a very light and rigid rod 50 cm in length. The rod is rotating about an axis at its opposite end with a rotational velocity of 3 rad/s. a. What is the rotational inertia of the system? b. What is the angular momentum of the system? E17. A uniform disk with a mass of 4 kg and a radius of 0.2 m is rotating with a rotational velocity of 20 rad/s. a. What is the rotational inertia of the disk? (See fig. 8.15.) b. What is the angular momentum of the disk?

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E18. A student, sitting on a stool rotating at a rate of 20 RPM, holds masses in each hand. When his arms are extended, the total rotational inertia of the system is 4.5 kg·m2. He pulls his arms in close to his body, reducing the total rotational inertia to 1.5 kg·m2. If there are no external torques, what is the new rotational velocity of the system?

synthesis problems SP1. A merry-go-round in the park has a radius of 1.8 m and a rotational inertia of 900 kg·m2. A child pushes the merrygo-round with a constant force of 80 N applied at the edge and parallel to the edge. A frictional torque of 12 N·m acts at the axle of the merry-go-round. a. What is the net torque acting on the merry-go-round about its axle? b. What is the rotational acceleration of the merry-goround? c. At this rate, what will the rotational velocity of the merry-go-round be after 15 s if it starts from rest? d. What is the rotational acceleration of the merry-goround if the child stops pushing after 15 s? How long will it take for the merry-go-round to stop turning? SP2. A 4-m long plank with a weight of 80 N is placed on a dock with 1 m of its length extended over the water, as in the diagram. The plank is uniform in density so that the center of gravity of the plank is located at the center of the plank. A boy with a weight of 150 N is standing on the plank and moving out slowly from the edge of the dock. a. What is the torque exerted by the weight of the plank about the pivot point at the edge of the dock? (Treat all the weight as acting through the center of gravity of the plank.) b. How far from the edge of the dock can the boy move until the plank is just on the verge of tipping? c. How can the boy test this conclusion without falling in the water? Explain.

a. What is the rotational inertia of the children about the axle of the merry-go-round? What is the total rotational inertia of the children and the merry-go-round? b. The children now move inward toward the center of the merry-go-round so that their average distance from the axle is 0.5 m. What is the new rotational inertia for the system? c. If the initial rotational velocity of the merry-go-round was 1.2 rad/s, what is the rotational velocity after the children move in toward the center, assuming that the frictional torque can be ignored? (Use conservation of angular momentum.) d. Is the merry-go-round rotationally accelerated during this process? If so, where does the accelerating torque come from? SP4. A student sitting on a stool that is free to rotate but is initially at rest, holds a bicycle wheel. The wheel has a rotational velocity of 5 rev/s about a vertical axis, as shown in the diagram. The rotational inertia of the wheel is 2 kg·m2 about its center and the rotational inertia of the student and wheel and platform about the rotational axis of the platform is 6 kg·m2. a. What are the magnitude and direction of the initial angular momentum of the system? b. If the student flips the axis of the wheel, reversing the direction of its angular-momentum vector, what is the rotational velocity (magnitude and direction) of the student and the stool about their axis after the wheel is flipped? (Hint: See fig. 8.23.) c. Where does the torque come from that accelerates the student and the stool? Explain.

3m 1m

SP2 Diagram SP3. In the park, several children with a total mass of 240 kg are riding on a merry-go-round that has a rotational inertia of 1500 kg·m2 and a radius of 2.2 m. The average distance of the children from the axle of the merry-go-round is 2.0 m initially, since they are all riding near the edge.

SP4 Diagram

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home experiments and observations HE1. If there is a park nearby containing a freely rotating child’s merry-go-round, take some time with a friend to observe some of the phenomena discussed in this chapter. In particular, make these observations: a. What is a typical rotational velocity that can be achieved with the merry-go-round? How would you go about measuring this? b. How long does it take for the merry-go-round to come to rest after you stop pushing? Could you estimate the frictional torque from this information? What other information would you need? c. If you or your friend are riding on the merry-go-round, what happens to the rotational velocity when you move inward or outward from the axis of the merry-go-round? How do you explain this? HE2. Create a simple balance using a ruler as the balance beam and a pencil as the fulcrum. (A pencil or pen with a hexagonal cross section is easier to use than one with a round cross section.) a. Does the ruler balance exactly at its midpoint? What does this imply about the ruler? b. Using a nickel as your standard, what are the ratios of the weights of pennies, dimes, and quarters to that of the nickel? Describe the process used to find these ratios. c. Is the distance from the fulcrum necessary to balance two nickels on one side with a single nickel on the opposite side exactly half the distance for the single nickel? How would you account for any discrepancy? HE3. You can make a simple top by cutting a circular piece of cardboard, poking a hole through the center, and using a short dull pencil for the post. A short wooden dowel with a rounded end works even better than a pencil.

a. Try building such a top and testing it. How far up the pencil should the cardboard disk sit for best stability? b. Observe what happens to the axis of rotation of the top as it slows down. What is the direction of the angularmomentum vector, and how does it change? c. What happens to the stability of your top if you tape two pennies near the edge on opposite sides of the cardboard disk? HE4. Try spinning a quarter or other large coin about its edge on a smooth tabletop or other similar surface. Describe the motion that follows, paying particular attention to the direction of the angular-momentum vector. HE5. As described in the study hint on page 160, use a bicycle wheel mounted on a hand-held axle (probably available from your friendly physics department) to study the angular momentum vectors and their changes. In particular, make these observations: a. Grasp the wheel with both hands by the handles on each side and have someone give it a good spin with the wheel in a vertical plane. Then try tilting the wheel downward to the left to simulate a fall. What direction does the wheel turn? Describe what happens. b. Grasp the wheel with both hands by the handles on each side and hold it in the vertical orientation. Without spinning the wheel, rotate it into the horizontal plane. Now repeat this, but first have someone give it a good spin with the wheel in the vertical plane. Is it harder or easier to rotate it into the horizontal plane when the wheel is spinning? Explain how this can be related to riding a bicycle.

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Two Fluids and Heat

E

he lacked the crucial insights of the first law. The first law involves conservation of energy, but that idea came into its own only around 1850, about thirty years later. A comprehensive theory had to wait until then for the insights that resulted from combining the first and second laws of thermodynamics. The first and second laws of thermodynamics emerged during the 1850s through the work of many scientists, most prominently Rudolph Clausius (1822–1888) and William Thompson (later Lord Kelvin) (1824–1907). Another important contributor was James Prescott Joule (1818–1889), who measured the heating effect of mechanical work, a key idea in the first law of thermodynamics. Together, the first and second laws of thermodynamics place limits on what can be accomplished with energy in the form of heat. The laws of thermodynamics play an enormous role in any discussion of the use of energy resources. Energy is still a critical issue—our heavy dependence on fossil fuels cannot go on indefinitely. Concerns about global warming (associated with the greenhouse effect) and other environmental issues related to energy use have become hot political issues. Understanding the science underlying these issues is important for economists, politicians, environmentalists, and citizens in general. Chapters 9, 10, and 11 address these ideas.

nergy issues have become increasingly important as the worldwide oil supply tightens. Although Newton’s theory of mechanics was introduced in the seventeenth century, the physics underlying our understanding of energy utilization did not make great strides until the nineteenth century. Some of these developments, particularly in the fields of fluid mechanics and thermodynamics, were stimulated by the industrial revolution. This revolution could not have happened without the invention of steam engines, which were used to run factories, trains, and ships. Understanding the behavior of fluids, particularly gases, is critical to our understanding of many kinds of engines including steam engines. Thermodynamics is also crucial because it describes the energy conversions that take place in heat engines and other systems. The study of fluid mechanics and thermodynamics is a central part of the training of the mechanical engineers who design these systems. The history of the field of thermodynamics has taken some strange twists and turns. A French scientist and engineer, Sadi Carnot (1796–1832), developed a theory of heat engines around 1820, but Carnot’s theory raised more questions than it answered. Carnot had some understanding of what we now call the second law of thermodynamics, but

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9

The Behavior of Fluids

chapter overview Our first objective in this chapter is to explore the meaning of pressure. We will then investigate atmospheric pressure and how pressure varies with depth in a fluid. Those ideas will prepare us to explore the behavior of floating objects as well as what happens when fluids are in motion. Moving fluids are described by Bernoulli’s principle, which helps us to explain why a curveball curves and many other phenomena.

chapter outline

1

2 3 4

Atmospheric pressure and the behavior of gases. How do we measure atmospheric pressure, and why does it vary? Why can we compress gases more readily than liquids? What is Boyle’s law? Archimedes’ principle. What is Archimedes’ principle? How is it related to differences in pressure? Why does a steel boat float but a lump of steel sink? Fluids in motion. What special characteristics can we observe in moving fluids? What is viscosity? How does the velocity of a moving fluid vary if we change the width of its pipe or stream? Bernoulli’s principle. What is Bernoulli’s principle, and how is it related to conservation of energy? How can Bernoulli’s principle be used to explain the motion of a curveball and other phenomena?

unit two

5

Pressure and Pascal’s principle. What is pressure? How is it transmitted from one part of a system to another? How does a hydraulic jack or press work?

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oats hold a special fascination for many of us. As a child, you probably floated twigs or sticks in streams. A stick or toy boat will follow the current of a stream, sometimes moving swiftly and other times getting caught in an eddy or stranded near the bank. You were observing some characteristics of fluid flow. You probably also noticed that some things float and some do not. Stones sink quickly to the bottom of the stream. As you grew older, you may have wondered why a steel boat floats, while a piece of metal dropped in the water quickly sinks (fig. 9.1). Does the shape of the material have something to do with whether it floats or not? Can you make a boat of concrete? Things can float in air as well as in water. A balloon filled with helium pulls up on the string, but a balloon filled with air drifts down to the floor. What makes the difference? The behavior of things that float (or sink) in water or air is one aspect of the behavior of fluids. Water and air are both examples of fluids, although one is a liquid and the other a gas. They both flow readily and conform to the shapes of their containers, unlike solids, which have shapes of their own. Although liquids are

9.1 Pressure and Pascal’s Principle A small woman wearing high-heel shoes sinks into soft ground, but a large man wearing size-13 shoes may walk across the same ground without difficulty (fig. 9.2). Why is this so? The man weighs much more than the woman, so he must exert a larger force on the ground. But the woman’s high heels leave much deeper indentations in the ground. Clearly weight alone is not the determining factor. How the force is distributed across the area of contact between the shoes and the Earth is more important. The woman’s shoes have a small area of contact, while the man’s shoes have a much larger area of contact. The force exerted on the ground by the man’s feet due to his weight is distributed over a larger area.

How is pressure defined? What is happening when you stand or walk on soft ground? If you are not accelerating in the vertical direction, your weight

figure 9.2

A woman’s high heels sink into the soft ground, but the larger shoes of the much bigger man do not.

figure 9.1

A steel boat floats, but a piece of metal dropped in the water quickly sinks. How do we explain this?

usually much denser than gases, many of the principles that apply to liquids also apply to gases, so it makes sense to consider them together under the common heading of fluids. Pressure plays a central role in describing the behavior of fluids. We will explore pressure thoroughly in this chapter. Pressure is involved in Archimedes’ principle, which explains how things float, but pressure is also important in other phenomena we will consider, including the flow of fluids.

must be balanced by the normal force exerted upward on your feet by the ground. By Newton’s third law, you exert a downward normal force on the ground equal to your weight. The quantity that determines whether the soil will yield, letting your shoes sink into the ground, is the pressure exerted on the soil by your shoes. The total normal force does not matter as much as the force per unit area: Pressure is the ratio of the force to the area over which it is applied: P

F A

Pressure is measured in units of newtons per meter squared (N/m2)—the metric unit of force divided by the metric unit of area. This unit is also called a pascal (1 Pa 1 N/m2). The heel area of a woman’s high-heel shoe can be as small as just 1 or 2 square centimeters (cm2). When you walk, there are times when almost all of your weight is supported by your heel. The weight shifts from heel to toe as you move forward with your other foot off the ground. Take a few steps to test this statement. The woman’s weight divided by the small area of her heel produces a large pressure on the ground. The man’s shoe, on the other hand, may have a heel area of as much as 100 cm2 (for a size-13 shoe). Since his heel area may be 100 times larger than the woman’s heel, he can weigh two or three times as much as the woman yet exert a pressure on the ground (P F/A) that is a small fraction of the pressure exerted by the woman. His smaller pressure leads to a smaller indentation in the ground.

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2 cm 1 cm

figure 9.3

A square with sides 2 cm long has an area that is four times as large as a square with sides 1 cm long.

The area over which a force is distributed is the critical factor in pressure. The area of a surface increases more rapidly than the linear dimensions of the surface. A square that is 1 cm on each side has an area of 1 cm2, for example, but a square that is 2 cm on a side has an area of 4 cm2 (2 cm 2 cm), four times as large as the smaller square (fig. 9.3). For a circle, the area is equal to times the square of the radius of the circle (A r2). A circle with a radius of 10 cm has an area 100 times larger (102) than a circle with a radius of 1 cm.

Pascal’s principle What happens inside a fluid when pressure is exerted on it? Does pressure have direction? Does it transmit a force to the walls or bottom of a container? These questions point to another important feature of pressure in fluids. When we apply a force by pushing down on the piston in a cylinder, as in figure 9.4, the piston exerts a force on the fluid. By Newton’s third law, the fluid also exerts a force (in the opposite direction) on the piston. The fluid inside the cylinder will be squeezed and may decrease in volume somewhat. It has been compressed. Like a compressed spring, the compressed fluid will push outward on the walls and bottom of the cylinder as well as on the piston. Although the fluid behaves like a spring, it is an unusual spring. It pushes outward uniformly in all directions when compressed. Any increase in pressure is transmitted

uniformly throughout the fluid, as figure 9.4 indicates. If we ignore variations in pressure due to the weight of the fluid itself, the pressure that pushes upward on the piston is equal to the pressure that pushes outward on the walls and downward on the bottom of the cylinder. The ability of a fluid to transmit the effects of pressure uniformly is the core of Pascal’s principle and the basis of the operation of a hydraulic jack and other hydraulic devices. Blaise Pascal (1623–1662) was a French scientist and philosopher whose primary contributions were in the areas of fluid statics and probability theory. Pascal’s principle is usually stated as Any change in the pressure of a fluid is transmitted uniformly in all directions throughout the fluid.

How does a hydraulic jack work? Hydraulic systems are the most common applications of Pascal’s principle. They depend on the uniform transmission of pressure, as well as on the relationship between pressure, force, and area stemming from the definition of pressure. The basic idea is illustrated in figure 9.5. A force applied to a piston with a small area can produce a large increase in pressure in the fluid because of the small area of the piston. This increase in pressure is transmitted through the fluid to the piston on the right in figure 9.5, which has a much larger area. Since pressure is force per unit area, the force exerted on the larger piston by this pressure is proportional to the area of the piston (F PA). Applying the same pressure to the much larger area of the second piston results in a much larger force on the second piston. Since it is feasible to build a system in which the area of the second piston is more than 100 times larger than the first piston, we can produce a force on the second piston more than 100 times larger than the input force. The mechanical advantage, the ratio of the output force to the input force, of a hydraulic system will usually be large compared to a lever or other simple machine. (See section 6.1.)

50 kg F1

P

A2

A1

P

P P

F2

F2 A2 = F1 A1

figure 9.4

The pressure exerted on the piston extends uniformly throughout the fluid, causing it to push outward with equal force per unit area on the walls and bottom of the cylinder.

figure

9.5 A small force F1 applied to a piston with a small area produces a much larger force F2 on the larger piston. This allows a hydraulic jack to lift heavy objects.

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A small force applied to the input piston of a hydraulic jack can be multiplied to produce a force large enough to lift a car (fig. 9.6). These ideas are illustrated in example box 9.1. As discussed in chapter 6, we pay a price for getting a larger output force. The work done by the larger piston in lifting the car cannot be any greater than the work input to the handle of the jack. Since work is equal to force times distance (W Fd), a large output force means that the larger piston moves just a small distance. Conservation of energy requires that if the output force is 50 times larger than the input force, the input piston must move 50 times farther than the output piston. With a hand-pumped hydraulic jack, you have to move the smaller piston several times, letting the jack’s chamber refill with fluid after each stroke. The total distance moved by the smaller piston is the sum of the distances moved on each stroke. The larger piston inches upward as you pump. A similar process occurs in larger hydraulic jacks. Hydraulic systems and fluid are also used in the brake systems of cars and in many other applications. Oil is more effective than water as a hydraulic fluid because it is not corrosive and can also lubricate, ensuring smooth operation of the system. Hydraulic systems take good advantage of the ability of fluids to transmit changes in pressure, as described by Pascal’s principle. They also use the multiplying effect produced by pistons of different areas. A hydraulic system is a good example of the concept of pressure in action. Pressure is the ratio of the force to the area over which it is applied. A force applied to a small area exerts a much larger pressure than the same force applied to a larger area. Changes in pressure are transmitted uniformly through a fluid, and the pressure pushes outward in all directions, according to Pascal’s principle. These ideas explain the operation of a hydraulic jack and other hydraulic systems.

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example box 9.1 Sample Exercise: Some Basics of Jacks A force of 10 N is applied to a circular piston with an area of 2 cm2 in a hydraulic jack. The output piston for the jack has an area of 100 cm2. a. What is the pressure in the fluid? b. What is the force exerted on the output piston by the fluid? F1 a. F1 10 N P A1 A1 2 cm2 10 N 0.0002 m2 0.0002 m 2 P ? 50 000 N/m2 50 kPa A kilopascal (kPa) is 1000 Pa, or 1000 N/m2. b. A2 100 cm2 0.01

m2

P 50 kPa

F2 PA2 (50 000 N/m2)(0.01 m2) 500 N

F2 ? The mechanical advantage of this jack is 500 N divided by 10 N, or 50: the output force is 50 times larger than the input force.

9.2 Atmospheric Pressure and the Behavior of Gases Living on the surface of the Earth, we are at the bottom of a sea of air. Except for smog or haze, air is usually invisible. We seldom give it a second thought. How do we know it is there? What measurable effects does it have? We feel the presence of air, of course, when riding a bike or walking on a gusty day. Skiers, bicycle racers, and car designers are all conscious of the need to reduce resistance to the flow of air past themselves and their vehicles. The labored breathing of a mountain climber is partly due to the thinning of the atmosphere near the top of a high mountain. How do we measure atmospheric pressure and its variations with weather or altitude?

How is atmospheric pressure measured?

figure 9.6

A hydraulic jack can easily lift a car.

Atmospheric pressure was first measured during the seventeenth century. Galileo noticed that the water pumps he designed were capable of pumping water to a height of only 32 feet, but he never adequately explained why. His disciple, Evangelista Torricelli (1608–1647), invented the barometer as he attempted to answer this question. Torricelli was interested in vacuums. He reasoned that Galileo’s pumps created a partial vacuum and that the pressure of the air pushing down on the water at the pump intake

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was responsible for lifting the water. In thinking about how to test this hypothesis, he was struck by the idea of using a much denser fluid than water. Mercury or quicksilver was the logical choice, because it is a fluid at room temperature and has a density approximately 13 times that of water. A given volume of mercury therefore has a mass (and weight) that is 13 times an equal volume of water. Densities of a few common substances are listed in table 9.1 Density is the mass of an object divided by its volume. The metric units of density are kg/m3 or g/cm3.

In his early experiments, Torricelli used a glass tube about 1 meter in length, sealed at one end and open at the other. He filled the tube with mercury and then, holding his finger over the open end, inverted the tube and placed this end in the open container of mercury (fig. 9.7). The mercury flowed from the tube into the container until an equilibrium was reached, leaving a column of mercury in the tube approximately 760 mm (76 cm or 30 in.) high. The pressure of air pushing down on the surface of the mercury in the open container apparently was strong enough to support a column of mercury 760 mm high. Torricelli was careful to demonstrate experimentally that there was a vacuum in the space at the top of the tube above the column of mercury. The reason that the mercury column does not fall is that the pressure at the top of the column is effectively zero, while the pressure at the bottom of the column is equal to atmospheric pressure. We still often quote atmospheric pressure in either millimeters of mercury or

table 9.1 Densities of Some Common Substances Material

Density (g/cm3)

Water

1.00

Ice

0.92

Aluminum

2.7

Iron, steel

7.8

Mercury

13.6

Gold

19.3

inches of mercury (the commonly used unit in the United States). How are these units related to the pascal? We can establish the relationship between these units if we know the density of mercury. From this, we can find the weight of the column of mercury being supported by the atmosphere. Dividing this weight by the cross-sectional area of the tube gives us force per unit area, or pressure. Using this reasoning, we find that a column of mercury 760 mm in height produces a pressure of 1.01 105 Pa, known as standard atmospheric pressure. Atmospheric pressure is approximately 100 kilopascals (kPa) at sea level, or 14.7 pounds per square inch (psi). (Pounds per square inch is the British unit of pressure still used in the United States for measuring tire pressure and many other pressures of practical interest 1 psi 6.9 kPa.) Living at the bottom of this sea of air, you have 14.7 pounds pushing on you for every square inch of your body. Why do you not notice this? Fluids permeate your body and push back out—the interior and exterior pressures are essentially equal. (See everyday phenomenon box 9.1.) A famous experiment designed to demonstrate the effects of air pressure was performed by Otto von Guericke (1602–1686). Von Guericke designed two bronze hemispheres that could be smoothly joined together at their rims. He then pumped the air out of the sphere formed from the two hemispheres, using a crude vacuum pump that he had invented. As shown in figure 9.8, two eight-horse teams

760 mm Mercury

figure 9.7

Torricelli filled a tube with mercury and inverted it into an open container of mercury. Air pressure acting on the mercury in the dish can support a column of mercury 760 mm in height.

figure 9.8

Two teams of eight horses were unable to separate von Guericke’s evacuated metal sphere. What force pushes the two hemispheres together?

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everyday phenomenon

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box 9.1

Measuring Blood Pressure The Situation. When you visit your doctor’s office, the nurse will almost always take your blood pressure before the doctor spends time with you. A cuff is placed around your upper arm (as shown in the photograph) and air is pumped into the cuff, producing a feeling of tightness in your arm. Then the air is slowly released while the nurse listens to something with a stethoscope and records some numbers, such as 125 over 80.

through the compressed artery at the peak of the heart’s cycle. The lower reading, the diastolic pressure, is taken when blood flow occurs even at the low point in the cycle. There are distinctive sounds picked up by the stethoscope at these two points. The pressure recorded is actually the pressure in the air cuff for these two conditions. It is a gauge pressure, meaning that it is the pressure difference between the pressure being measured and atmospheric pressure. It is recorded in the units mm of mercury, which is the common way of recording atmospheric pressure. Thus a reading of 125 means that the pressure in the cuff is 125 mm of mercury above atmospheric pressure. A mercury manometer that is open to the air on one side (see the drawing) will measure gauge pressure directly.

Open end

Cuff Release valve

Having your blood pressure measured is a standard procedure for most visits to a doctor’s office. How does this process work?

What is the significance of these two numbers? What is blood pressure and how is it measured? Why are these readings an important factor, along with your weight, temperature, and medical history, in assessing your health? Stethoscope

The Analysis. Your blood flows through an elaborate system of arteries and veins in your body. As we all know, this flow is driven by your heart, which is basically a pump. More accurately, the heart is a double pump. One-half pumps blood through your lungs, where the blood cells pick up oxygen and discard carbon dioxide. The other half of the heart pumps blood through the rest of your body to deliver oxygen and nutrients. Arteries carry blood away from the heart into small capillaries that interface with other cells in muscles and organs. The veins collect blood from the capillaries and carry it back to the heart. We measure the blood pressure in a major artery in your upper arm at about the same height as your heart. When air is pumped into the cuff around your upper arm, it compresses this artery so that the blood flow stops. The nurse places the stethoscope, a listening device, near this same artery at a lower point in the arm and listens for the blood flow to restart as the air in the cuff is released. The heart is a pulsating pump that pumps blood most strongly when the heart muscle is most fully compressed. The pressure therefore fluctuates between high and low values. The higher reading in the blood pressure measurement, the systolic pressure, is taken when the blood just begins to spurt

An open-ended manometer can be used to measure the gauge pressure of the cuff. The stethoscope is used to listen for sounds indicating the restart of blood flow.

High blood pressure can be a symptom of many health problems, but most specifically, it is a warning sign for heart attacks and strokes. When arteries become constricted from the buildup of plaque deposits inside, the heart must work harder to pump blood through the body. Over time this can weaken the heart muscle. The other danger is that blood vessels might burst in the brain, causing a stroke, or blood clots might break loose and block smaller arteries in the heart or brain. In any case, high blood pressure is an important indicator of a potential problem. Low blood pressure can also be a sign of problems. It can cause dizziness when not enough blood is reaching the brain. When you stand up quickly, you sometimes experience a feeling of “light-headedness” because it takes a brief time for the heart to adjust to the new condition where your head is higher. Giraffes have a blood pressure about three times higher than humans (in gauge pressure terms). Why do you suppose this is so?

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were unable to pull the hemispheres apart. When the stopcock was opened to let air back into the evacuated sphere, the two hemispheres could easily be separated.

Variations in atmospheric pressure If we live at the bottom of a sea of air, we might expect pressure to decrease as we go up in altitude. The pressure that we experience results from the weight of the air above us. As we go up from the Earth’s surface, there is less atmosphere above us, so the pressure should decrease. Similar reasoning led Blaise Pascal to try to measure atmospheric pressure at different altitudes shortly after Torricelli’s invention of the mercury barometer. Because Pascal was in poor health through most of his adult life, and not up to climbing mountains, he sent his brother-in-law to the top of the Puy-de-Dome mountain in central France with a barometer similar to Torricelli’s. Pascal’s brother-in-law found that the height of the mercury column supported by the atmosphere was about 7 cm lower at the top of the 1460-m (4800-ft) mountain than at the bottom. Pascal also had his brother-in-law take a partially inflated balloon to the top of the mountain. As Pascal predicted, the balloon expanded as the climbers gained elevation, indicating a decrease in the external atmospheric pressure (fig. 9.9). Pascal was even able to show a decrease in pressure within the city of Clermont between the low point in town and the top of the cathedral tower. The decrease was small but measurable. Using the newly invented barometer, Pascal also observed variations in pressure related to changes in the

figure 9.9

A balloon that was partially inflated near sea level expanded as the experimenters climbed the mountain.

weather. Column heights were lower on stormy days than on clear days. This pressure variation has been used ever since to indicate changes in weather. Falling atmospheric pressure points to stormy weather ahead. Readings are usually corrected to sea level so that variations in altitude will not mask changes related to weather.

The weight of a column of air Can we calculate the variation in atmospheric pressure with altitude the same way that we compute the pressure at the bottom of a column of mercury? We would need to know the weight of the column of air above us. Even though mercury and air are both fluids, there is a significant difference (besides the difference in density) between the behavior of a column of mercury and a column of air. Mercury, like most liquids, is not readily compressible. In other words, increasing the pressure on a given amount of mercury does not change the volume of the mercury much. The density of the mercury (mass per unit volume) is the same near the bottom of the column of mercury as near the top. A gas like air, on the other hand, is easy to compress. As the pressure changes, the volume changes, and so does the density. Therefore, we cannot use a single value for the density to compute the weight of a column of air. Its density decreases as we rise in the atmosphere (fig. 9.10). This major difference between gases and liquids comes from differences in their atomic or molecular “packing.” Except at very high pressures, the atoms or molecules of a gas are separated by large distances compared to the size of the atoms themselves, as in figure 9.11. The atoms in a liquid, on the other hand, are closely packed, much like those in a solid. They cannot be easily squeezed. Gases are springy. They can readily be compressed to a small fraction of their initial volume. They can also expand if pressure is reduced, like the gas in Pascal’s balloon. Changes in temperature are likely to affect the volume or

figure

9.10 The density of a column of air decreases as altitude increases because air expands as pressure decreases.

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Liquid

177

Gas

V1

V2

figure

9.11 The atoms in a liquid are closely packed while those in a gas are separated by much larger distances. (a)

pressure of a gas much more than they affect a liquid. If the temperature is held constant, however, the volume changes with changes in pressure.

How does the volume of a gas change with pressure? Variations in the volume and density of a gas that accompany changes in pressure were studied by Robert Boyle (1627–1691) in England, as well as by Edme Mariotte (1620–1684) in France. Boyle’s results were first reported in 1660 but went unnoticed on the European continent where Mariotte published his similar conclusions in 1676. Both were interested in the springiness, or compressibility, of air. Both experimenters used a bent glass tube sealed on one end and open on the other (fig. 9.12). In Boyle’s experiment, the tube was partially filled with mercury, so that air was trapped in the closed portion of the tube. He allowed air to pass back and forth initially so that the pressure in the closed side of the tube was equal to atmospheric pressure and the column of mercury on either side was at the same height. As Boyle added mercury to the open end of the tube, the volume of the air trapped on the closed side decreased. When he added enough mercury to increase the pressure to twice atmospheric pressure, the height of the air column on the closed side decreased by one-half. In other words, doubling the pressure caused the volume of air to decrease by half. Boyle discovered that the volume of a gas is inversely proportional to the pressure. We can express Boyle’s law in symbols: PV constant, where P is the pressure in the gas and V is the volume of the gas. If the pressure increases, the volume decreases in inverse proportion to keep the product of pressure and volume constant. We often write Boyle’s law (also known as Mariotte’s law in continental Europe) as P1V1 P2V2 , where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume (see example

(b)

figure

9.12 In Boyle’s experiment, adding mercury to the open side of the bent tube caused a decrease in the volume of the trapped air in the closed side. box 9.2). For Boyle’s law to be valid, a fixed mass or quantity of gas must be kept at a constant temperature while the pressure and volume change. As we gain altitude, atmospheric pressure decreases, and the volume of a given mass of air increases. Since density is the ratio of the mass to the volume, the density of the air must be decreasing as the volume increases. In computing the weight of a column of air, we have to take the change in density into account, and the computation becomes more complex than for the column of mercury. The density of a gas also depends on temperature, which usually decreases as we gain altitude, a further complication.

example box 9.2 Sample Exercise: How Does a Gas Change with Pressure? A fixed quantity of gas is held in a cylinder capped at one end by a movable piston. The pressure of the gas is initially 1 atmosphere (101 kPa) and the volume is initially 0.3 m3. What is the final volume of the gas if the pressure is increased to 3 atmospheres at constant temperature? P1 1 atm V1 0.3

m3

P2 3 atm V2 ?

P1V1 P2V2 constant V2

P1V1 P2

3 (1 atm)(0.3 m ) 3 atm

13 (0.3 m3) 0.1 m3

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We live at the bottom of a sea of air, whose pressure we can measure with barometers. The earliest barometers were a column of mercury in a closed glass tube. The height of the column supported by atmospheric pressure is a measure of the pressure. Atmospheric pressure decreases with increasing altitude, because it is determined by the weight of the column of air above. Determining the weight of a column of air is harder than determining the weight of a column of mercury, however, because air is compressible, and its density varies with altitude. Boyle’s law describes how the volume of a gas changes with pressure: increasing the pressure decreases the volume in inverse proportion.

9.3 Archimedes’ Principle Why do some things float while others do not? Is floating determined by the weight of the object? A large ocean liner floats, but a small pebble quickly sinks. Clearly, it is not a matter of the total weight of the object. The density of the object is the key. Objects that are denser than the fluid they are immersed in will sink—those less dense will float. The complete answer to the question of why things sink or float is found in Archimedes’ principle, which describes the buoyant force exerted on any object fully or partly immersed in a fluid.

What is Archimedes’ principle? A block of wood floats, but a metal block of the same shape and size sinks. The metal block weighs more than the block of wood, even though it is the same size, which means that the metal is denser than wood. Density is the ratio of the mass to the volume (or mass per unit volume). The metal has a greater mass than wood for the same volume. Since weight is found by multiplying the mass by the gravitational acceleration g, the metal also has a greater weight for the same volume. If we compared the densities of the metal and wood blocks to water, we would find that the metal block has a density greater than water and that the density of the wood block is less than water. The average density of an object compared to a fluid is what determines whether the object will sink or float in that fluid. If we push down on a block of wood floating in a pool of water, we can clearly feel the water pushing back up on the block. In fact, it is hard to submerge a large block of wood or a rubber inner tube filled with air. They keep popping back up to the surface. The upward force that pushes such objects back toward the surface is called the buoyant force. If the block of wood is partially submerged at first and you push down to submerge it farther, the buoyant force gets larger as more of the block is underwater. Legend has it that Archimedes was sitting in the public baths observing floating objects when he realized what determines the strength of the buoyant force. When an object

is submerged, its volume takes up space occupied by water: it displaces the water, in other words. The more water displaced by the object as you push downward, the greater the upward buoyant force. Archimedes’ principle can be stated as The buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object.

If the fluid has a greater density than the object, the weight of the fluid displaced when the object is fully submerged will be greater than the weight of the object. By Archimedes’ principle, the buoyant force will be greater than the weight of the object, and there will be a net upward force on the object that pushes it toward the surface.

What is the source of the buoyant force? The source of the buoyant force described by Archimedes’ principle is the increase in pressure that occurs with increasing depth in a fluid. When we swim to the bottom of the deep end of a swimming pool, we can feel pressure building on our ears. The pressure near the bottom of the pool is larger than near the surface for the same reason that the pressure of the atmosphere is greater near the surface of the Earth than at higher altitudes. The weight of the fluid above us contributes to the pressure that we experience. By Pascal’s principle, the atmospheric pressure pushing on the surface of the pool extends uniformly throughout the fluid. To get the total pressure at some depth, we add the excess pressure resulting from the weight of the water to the atmospheric pressure. For many purposes, the excess pressure above atmospheric pressure is of most interest. Since our bodies have internal pressures equal to atmospheric pressure, our eardrums are sensitive to the increase in pressure beyond atmospheric pressure rather than to the total pressure. To find the excess pressure at a certain depth in a liquid, we need to determine the weight of the liquid above that depth. The problem is similar to finding the pressure at the bottom of a column of mercury discussed in section 9.2. If we imagine a column of water in a swimming pool (fig. 9.13), the weight of the column depends on the volume of the column and the density of the water. The volume of the column is directly proportional to its height h and therefore so is its weight. The excess pressure increases in direct proportion to the depth h below the surface.* A large can filled with water demonstrates this variation of pressure with depth. If we punch holes in the can at different depths, the water shoots from a hole near the bottom *Since the weight W mg Vrg, and V Ah, the excess pressure, rgAh W ¢P rgh. The symbol for density, r, is the Greek letter rho. A A

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V = Ah h W T

A

figure

9.13 The weight of a column of water is proportional to the volume of the column. The volume V is equal to the area A times the height h. of the can with a much greater horizontal velocity than from a hole punched near the top (fig. 9.14). Since the can is submerged in the atmosphere, the excess pressure above atmospheric pressure is most significant. A larger excess pressure provides a larger accelerating force for the emerging water. How does the fact that pressure increases with depth explain the buoyant force? Imagine a rectangular-shaped object, such as a steel block, submerged in water. If we suspend the block from a string, as in figure 9.15, the pressure of the water will push on the block from all sides. Because the pressure increases with depth, however, the pressure at the bottom of the block is greater than at the top. This greater pressure produces a larger force (F PA) pushing up on the bottom of the block than that pushing down on the top. The difference in these two forces is the buoyant force. The buoyant force is proportional to both the height and the cross-sectional area of the block, and thus to its volume, Ah. The volume of the fluid displaced by the object is directly related to the weight of the fluid displaced, which leads to the statement of Archimedes’ principle.

figure

9.14 Water emerging from a hole near the bottom of a can filled with water has a larger horizontal velocity than water emerging from a hole near the top.

W

figure

9.15 The pressure acting on the bottom of the suspended metal block is greater than that acting on the top due to the increase of pressure with depth. What forces act on a floating object? If there are no strings attached, or other forces pushing or pulling on an object partially or fully submerged in a fluid, only the weight of the object and the buoyant force determine what happens. The weight is proportional to the density and volume of the object, and the buoyant force depends on the density of the fluid and the volume of fluid displaced by the object. The motion of the object is governed by the sizes of the buoyant force pushing upward and the weight pulling downward. There are just three possibilities: 1. The density of the object is greater than the density of the fluid. If the object has an average density greater than the fluid it is submerged in, the weight of the object will be greater than the weight of the fluid displaced by the fully submerged object, because the same volume is involved. Since the weight, which acts downward, is greater than the buoyant force, which acts upward, the net force acting on the object will be downward and the object will sink (unless it is supported by another force such as one exerted by a string tied to the block). 2. The density of the object is less than the density of the fluid. If the object’s density is less than the fluid’s, the buoyant force will be larger than the weight of the object when the object is fully submerged. The net force acting on the object will be upward, and the object will float to the top. When it reaches the surface of the fluid, just enough of the object remains submerged so that the weight of the fluid displaced by the submerged portion of the object (the buoyant force) will equal the weight of the object. The net force is zero, and the object is in equilibrium. It has no acceleration. 3. The density of the object equals the density of the fluid. The weight of the object then equals the weight of the fluid displaced when the object is submerged. The object floats when fully submerged, rising or sinking in the fluid by

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changing its average density slightly, which is what a fish or a submarine does. The average density of a submarine can be increased or decreased by taking in or releasing water.

Why does a boat made of steel float? Steel is a lot denser than water, and a large steel boat is a very heavy object. Why does it float? The answer is that the boat is not made of solid steel all the way through. There are open spaces filled with air and other materials in the boat. A solid piece of steel will quickly sink, but if the average density of the boat is less than the water it is displacing, a steel boat will certainly float. Because of the air spaces and other materials, the average density of the boat will be much less than that of steel. According to Archimedes’ principle, the buoyant force that acts on the boat must equal the weight of the water displaced by the hull of the boat. For the boat to be in equilibrium (with a net force of zero), the buoyant force must equal the weight of the boat. As we load the boat with cargo, the total weight of the boat increases. So must the buoyant force. The amount of water displaced by the hull must increase, so the boat sinks lower in the water. There is a limit to how much weight can be added to a boat (often expressed as tons of displacement). A fully loaded oil tanker will ride much lower in the water than an unloaded tanker (fig. 9.16). Other important considerations in designing a boat are the shape of the hull and how the boat will be loaded. If the

figure

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9.16 A fully loaded tanker rides much lower in the water than an empty tanker.

center of gravity of the boat is too high or if the boat is unevenly loaded, there is some danger that the boat will tip over. Wave action and winds add to this danger, so a margin of safety must be included in the design. Once water enters the boat, the overall weight of the boat and its average density increase. When the boat’s average density becomes larger than the average density of water, down she goes.

When will a balloon float? Buoyant forces also act on objects submerged in a gas such as air. If a balloon is filled with a gas whose density is less than air, the average density of the balloon is less than air, and the balloon will rise. Helium and hydrogen are the two common gases with densities less than air, but helium is more commonly used in balloons even though its density is somewhat greater than hydrogen. Hydrogen can combine explosively with the oxygen in air, making its use dangerous. The average density of the balloon is determined by the material the balloon is made of as well as by the density of the gas with which it is filled. Ideal materials for making balloons stretch very thin without losing strength. They should also remain impermeable to the flow of gas, so that the helium or other gas will not be lost rapidly through the skin of the balloon. Balloons made of Mylar (which is often coated with aluminum) are much less permeable than ordinary latex balloons. Hot-air balloons take advantage of the fact that any gas will expand when it is heated. If the volume of the gas increases, its density decreases. As long as the air inside the balloon is much hotter than the air surrounding the balloon, there will be an upward buoyant force. The beauty of a hot-air balloon is that we can readily adjust the density of the air within the balloon by turning the gas-powered heater on or off. This gives us some control over whether the balloon will ascend or descend. Buoyant forces and Archimedes’ principle are useful in applications besides boats and balloons. We can use Archimedes’ principle to determine the density of objects or the fluid in which they are submerged. This, in fact, was Archimedes’ original application. Archimedes is said to have used his idea to determine the density of the king’s crown to ascertain whether it was truly pure gold. (The king suspected a goldsmith of fraud.) Gold is denser than cheaper metals that might have been substituted for gold. Archimedes’ principle states that the buoyant force acting on an object is equal to the weight of the fluid displaced by the object. If the average density of the object is greater than the density of the fluid being displaced, the weight of the object will exceed the buoyant force and the object will sink. The buoyant force results because the pressure on the bottom of the object is greater than on top since pressure increases with depth. Archimedes’ principle can be used to understand the behavior of boats, balloons, or objects floating in a bathtub or stream.

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9.4 Fluids in Motion

t, which gives us LA/t. Since L/t is the speed v of the water, we get

9.4 Fluids in Motion If we return to the bank of the stream mentioned in the introduction to this chapter, what else can we see? When a stick or toy boat floats down the stream, we might notice that the speed of the current varies from point to point in the stream. Where the stream is wide, the flow is slow. Where the stream narrows, the speed increases. Also, the speed is usually greater near the middle of the stream than close to the banks. Eddies and other features of turbulent flow may be observed. These are all characteristics of the flow of fluids. The speed of flow is affected by the width of the stream and by the viscosity of the fluid, a measure of the frictional effects within the fluid. Some of these features are easy to understand, while others, particularly the behavior of turbulent flow, are still areas of active research.

Why does the water speed change? One of the most obvious features of the stream’s current is that the water speed increases where the stream narrows. Our stick or toy boat moves slowly through the wider portions of the stream but gains speed when it passes through a narrow spot or rapids. As long as no tributaries add water to the stream and there are no significant losses through evaporation or seepage, the flow of the stream is continuous. In a given time, the same amount of water that enters the stream at some upper point leaves the stream at some lower point. We call this continuity of flow. If flow were not continuous, water would collect at some point or, perhaps, be lost somewhere within that segment of the stream, something that does not usually happen. How would we describe the rate of flow of water through a stream or pipe? A volume flow rate is a volume divided by time, so many gallons per minute or, in the metric system, liters per second or cubic meters per second. As figure 9.17 shows, the volume of a portion of water of length L flowing past some point in a pipe is the product of the length times the cross-sectional area A, or LA. The speed with which this volume moves determines the rate of flow. What is the rate of flow through the pipe? To find the rate, we divide the water’s volume LA by the time interval L

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rate of flow vA. This expression is valid for any fluid and makes intuitive sense: the greater the speed, the greater the rate of flow, and the larger the cross section of the pipe or stream, the greater the rate of flow. How does the rate of flow explain changes in the speed of the water? If the flow through the pipe is continuous, the rate of flow must be the same at any point along the pipe. The same number of gallons per minute flow past each point. If the cross-sectional area A decreases, the speed v increases to keep the rate vA constant. If the cross-sectional area increases, the speed decreases to maintain the same rate of flow. The same principle applies to a stream. Where the stream is narrower, the cross-sectional area of the stream will generally be smaller than at a wider point in the stream. The stream may be deeper at the narrow places, but usually not enough to make the cross-sectional area as large as at the wider places. If the cross-sectional area decreases, the fluid speed must increase to maintain the rate of flow.

How does viscosity affect the flow? Up to this point, we have ignored any variation in the fluid speed across its cross-sectional area. We mentioned that the water speed will usually be greatest near the middle of the stream. The reason is the frictional or viscous effects between layers of the fluid itself and between the fluid and the walls of the pipe or the banks of the stream. Imagine the fluid as made up of layers, and you can see why the speed will be greatest near the center. Figure 9.18 shows different layers of a fluid moving through a trough. Since the bottom of the trough is not moving, it exerts a frictional force on the bottom layer of fluid, which moves more slowly than the layer immediately above it. This layer exerts a frictional drag, in turn, on the layer above it, which flows more slowly than the next one above it, and so on. The viscosity of the fluid is the property of fluid that determines the strength of the frictional forces between the layers of the fluid—the larger the viscosity, the larger the frictional force. The magnitude of the frictional force also

A

v v Rate of flow = vA

figure

9.17 The rate at which water moves through a pipe is defined by the volume divided by time. This is equal to the speed of the water times the cross-sectional area.

figure

9.18 Because of the frictional or viscous forces between layers, each layer of fluid flowing in the trough moves more slowly than the layer immediately above.

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depends on the area of contact between layers and the rate at which the speed is changing across the layers. If these other factors are the same, a fluid with a high viscosity like molasses experiences a larger frictional force between layers than a fluid with a low viscosity such as water. A thin layer of fluid that does not move at all is usually found next to the walls of the pipe or trough. The fluid speed increases as the distance from the wall increases. The exact variation with distance depends on the viscosity of the fluid and the overall rate of flow of the fluid through the pipe. For a fluid with a low viscosity, the transition to the maximum speed occurs over a short distance from the wall. For a fluid with a high viscosity, the transition takes place over a larger distance, and the speed may vary throughout the pipe or trough (fig. 9.19). The viscosity of different fluids varies enormously. Honey, thick oils, and syrup all have much larger viscosities than water or alcohol. Most liquids have much higher viscosities than gases. The viscosity of a given fluid also changes substantially if its temperature changes. An increase in temperature usually produces a decrease in viscosity. Heating a bottle of syrup, for example, makes it less viscous, causing it to flow more readily.

Laminar and turbulent flow One of the most fascinating questions about the flow of fluids is why the flow is smooth or laminar under some conditions but turbulent in others. Both kinds of flow can be observed in rivers or creeks. How do they differ—and what determines which type of flow prevails? In sections of a stream where flow is smooth or laminar, there are no eddies or other similar disturbances. The flow of the stream can be described by streamlines that indicate the direction of flow at any point. The streamlines for laminar

Low viscosity

v

High viscosity

v

figure

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9.19 The velocity increases rapidly from the wall inward for a low-viscosity fluid but more gradually for a high-viscosity fluid.

Laminar flow

Turbulent flow

figure

9.20 In laminar flow, the streamlines are roughly parallel to one another. In turbulent flow, the flow patterns are much more complicated. flow are roughly parallel, as in figure 9.20. The speeds of different layers may vary, but one layer moves smoothly past another. As the stream narrows and the fluid speed increases, the simple laminar-flow pattern disappears. Ropelike twists in the streamlines appear, then whorls and eddies: the flow becomes turbulent. In most applications, turbulent flow is undesirable, because it greatly increases the fluid’s resistance to flow through a pipe or past other surfaces. It does make river rafting much more exciting, though. If the density of a fluid and the width of the pipe or stream do not vary, the transition from laminar to turbulent flow is predicted by two quantities, the average fluid speed and the viscosity. Higher speeds are more likely to produce turbulent flow, as we would expect. On the other hand, higher viscosities inhibit turbulent flow. Larger fluid densities and pipe widths are also more conducive to turbulent flow. From experiments, scientists have been able to use these quantities to predict with some accuracy the speed at which the transition to turbulent flow will begin. You can observe the transition from laminar to turbulent flow in many common phenomena. The higher water speeds at the narrowing of a stream often produce turbulent flow. The transition can also be seen in the flow of water from a spigot. A small flow rate will usually produce laminar flow, but as the flow rate increases, the flow becomes turbulent. Flow may be smooth near the top of the water column but turbulent lower down, as the water is accelerated by gravity. Try it next time you are near a sink. You can also see this phenomenon in the smoke rising from a candle or incense stick. Near the source, the upward flow of the smoke is usually laminar. As the smoke accelerates upward (due to the buoyant force), the column widens, and the flow becomes turbulent (fig. 9.21). The whorls and eddies are much like those you see in a stream.

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figure

9.22 Whorls and eddies, including the giant red spot (lower left), can be seen in the atmosphere of the planet Jupiter.

figure

9.21 Smoke rising from an incense stick first exhibits laminar flow, and then as the speed increases and the column widens, turbulent flow appears. The conditions that produce turbulence are well understood, but until recently, scientists could not explain why the flow patterns develop as they do. Some surprising patterns can be discerned in the seemingly chaotic behavior of turbulent flow in different situations. Recent theoretical advances in the study of chaos have produced a much better understanding of the reasons for these patterns. The study of chaos and the regular behaviors that appear in turbulent flow have provided new insights into global weather patterns including hurricanes and other phenomena. Perhaps the most striking examples of atmospheric flow patterns are the photographs sent back from the Voyager flyby of the planet Jupiter. Whorls and eddies can be seen in the flow of the atmospheric gases on Jupiter. These include the famous red spot, which is now thought to be a giant, and very stable, atmospheric eddy (fig. 9.22). The rate of flow of a stream is equal to the fluid speed multiplied by the cross-sectional area through which the stream is flowing. For the flow to be continuous, the speed must increase where the stream narrows to pass through a smaller cross-sectional area. The fluid speed of the stream also varies across its area because of viscosity: the fluid speed is largest at the center and smallest near the banks or pipe walls. Smooth or laminar flow gives way to turbulent flow with its eddies and whorls as the fluid speed increases or the viscosity decreases.

9.5 Bernoulli's Principle Have you ever wondered why a spinning ball curves? This and many other interesting phenomena can be explained by a principle regarding the flow of fluids that was published in 1738 by Daniel Bernoulli (1700–1782) in a treatise on hydrodynamics. Although energy ideas had not been fully developed at that time, Bernoulli's principle is really a result of conservation of energy.

What is Bernoulli’s principle? What happens if we do work on a fluid, increasing its energy? This increase may show up as an increase in kinetic energy of the fluid, leading to an increase in the fluid’s speed. It could also appear as an increase in potential energy if the fluid is squeezed (elastic potential energy) or if the fluid is raised in height (gravitational potential energy). Bernoulli considered all of these possibilities. Bernoulli’s principle is a direct result of applying conservation of energy to the flow of fluids. The most interesting examples of Bernoulli’s principle involve changes in kinetic energy. If a fluid that is not compressible (not squeezable) is flowing in a level pipe or stream, any work done on this fluid will increase its kinetic energy. (If a fluid is compressible, some of the work is used to squeeze the fluid.) To accelerate the fluid and increase its kinetic energy, there must be a net force doing work on the fluid. This force is associated with a difference in pressure from one point to another within the fluid. If there is a difference in pressure, the fluid accelerates from the region of higher pressure toward the region of lower

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pressure, because that is the direction of the force acting on the fluid. We expect to find higher fluid speeds in the regions of lower pressure. In the simple case of an incompressible fluid flowing in a level pipe or stream,* work and energy considerations lead to a statement of Bernoulli’s principle:

h1 h2

The sum of the pressure plus the kinetic energy per unit volume of a flowing fluid must remain constant.

P

1 2

v1

v2

rv2 constant

Here, P is the pressure, r is the density of the fluid, and v is the fluid’s speed. The second term in this sum is the kinetic energy per unit volume (kinetic energy divided by volume) of the fluid, since density is mass divided by volume. A fuller statement of Bernoulli’s principle would include the effects of gravitational potential energy, allowing for changes in the height of the fluid. However, many of the most interesting effects can be investigated using the form just stated. In applying Bernoulli’s principle, the association of lower pressures with higher fluid speeds is often the key point.

How does the pressure vary in pipes and hoses? Consider a pipe with a constriction in its center section, as in figure 9.23. Would you expect the pressure of water flowing in the pipe to be greatest at the constriction or in the wider sections of the pipe? Intuition leads you to suspect that the pressure is greater in the constricted section, but this is not the case. We know that the speed of the water will be greater in the constricted section (where the cross-sectional area is smaller) than in the wider portions of the pipe, because of the concept of continuous flow. What does Bernoulli’s 1 principle tell us? To keep the sum P 2 rv2 constant, the pressure must be larger where the fluid speed is smaller. In other words, if the speed increases, the pressure must decrease. Open tubes at different places in the pipe (fig. 9.23) can serve as simple pressure gauges. For the water in these open tubes to rise, fluid pressure must be greater than atmospheric pressure. The height that the fluid rises will depend on how much greater its pressure is. The water level in these tubes will reach a greater height above the main pipe where the pipe is wider rather than at the constriction, indicating that the pressure is higher in the wider portion of the pipe. This result goes against our intuition because we tend, incorrectly, to associate higher pressure with higher speed.

*If the stream is not level, an additional term, rgh, must be included to account for changes in potential energy.

figure

9.23 Vertical open pipes can serve as pressure gauges. The height of the column of water is proportional to the pressure. The pressure of a moving fluid is greater where the fluid velocity is smaller.

Another confirming example is the nozzle on a hose. The nozzle constricts the area of flow and increases the fluid’s speed. By Bernoulli’s principle, the pressure of the water is smaller at the narrow end of the nozzle than farther back in the hose, contrary to what we expect. If you place your hand in front of the nozzle, you will feel a force on your hand as the water strikes it. This force results from the change in the velocity and momentum of the water as it strikes your hand. By Newton’s second law, a force is required to produce this change in momentum, and by Newton’s third law, the force exerted on the water by your hand is equal in magnitude to the force exerted on your hand by the water. That force is not directly associated with the fluid pressure in the hose—the pressure is actually greater farther back in the hose where the water is not moving very fast.

Air flow and Bernoulli’s principle Bernoulli’s principle, as we have stated it, is only valid for fluids whose density does not change (noncompressible fluids), but we often extend it to investigate effects of the motion of air and other compressible gases. Even for compressible fluids, a larger fluid speed is usually associated with a smaller fluid pressure. A simple demonstration will make a believer of you. Take half a sheet of tablet paper (or even a facial tissue) and hold it in front of your mouth, as in figure 9.24. The paper should hang down limply in front of your chin. If you blow across the top of it, the paper rises, and if you blow hard enough, the paper may even stand straight out horizontally. What is happening? Blowing across the top of the paper makes the air flow across the top with a greater speed than underneath the paper. The air underneath presumably is not moving much at all. This greater speed leads to a reduction in pressure.

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The pressure increases away from the center of the air column toward regions where the air is moving less rapidly. If the ball moves out of the center of the air column, a larger pressure and a larger force act on the outer side of the ball than on the side nearer the center of the column. The ball moves back into the center. The upward force of air hitting the bottom of the ball holds it up, while the low pressure in the center of the column keeps the ball near the center. You can produce the same effect with a small ball and a hair dryer (fig. 9.25). The motion of a curveball discussed in everyday phenomenon box 9.2 provides another example of Bernoulli’s principle at work. In all of these phenomena, we see the effects of a reduction in fluid pressure associated with an increase in fluid speed, as predicted by Bernoulli’s principle. We do not always expect this. We are tempted to think that blowing between two pieces of paper will push them apart, for example, but a simple experiment shows otherwise. Understanding such surprises is part of the fun of physics.

figure

9.24 Blowing across the top of a limp piece of paper causes the paper to rise, demonstrating Bernoulli’s principle.

Since the air pressure is then greater on the bottom of the paper than on the top, the upward force on the bottom of the paper is larger than the downward force on the top, so the paper rises. Contrary to intuition, blowing between two strips of paper makes them move closer together rather than farther apart, for the same reason. Try it. Bernoulli's principle is frequently used to explain the lift force on an airplane wing. The wing shape and tilt can produce a greater air speed across the top of the wing than across the bottom, leading to greater pressure on the bottom of the wing than on the top. Although the simplicity of this explanation can be appealing, it is also misleading and cannot yield accurate predictions. The effects involved in producing the lift forces on a wing are actually quite complex. They have been thoroughly explored and analyzed with the help of wind tunnels.

Bernoulli’s principle is derived from energy considerations and says that the sum of the pressure plus the kinetic energy per unit volume remains constant from one point in a fluid to another. This holds true if the fluid is not compressible and changes in height are not involved. Pressure is then lower where fluid speed is higher. This effect explains many surprising phenomena involving pipes, hoses, airplane wings, and the fact that blowing across the top of a piece of paper causes the paper to rise rather than drop.

What keeps the department-store ball in the air? Another demonstration of Bernoulli’s principle that uses airflow is often seen in department stores when vacuum cleaners are being advertised. A ball can be suspended in an upward-moving column of air produced by a vacuum cleaner. As the air moves upward, the speed of airflow is greatest in the center of the stream and falls off to zero away from the center. Once again, Bernoulli’s principle requires that the pressure will be smallest in the center, where the speed is greatest.

figure

9.25 A ball is suspended in an upward-moving column of air produced by a hair dryer. The air pressure is smallest in the center of the column where the air is moving with the greatest speed.

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everyday phenomenon

box 9.2

Throwing a Curveball The Situation. Baseball players know that they can be badly fooled by a well-thrown curveball. The idea that a fastmoving curveball can be deflected by as much as a foot on its way to the plate is hard for some people to accept. Many people have insisted, over the years, that the curve is just an illusion. Does the path of a curveball really curve? If so, how can we explain it?

Top view F

v

The whirlpool of air created by the spin of the ball causes the air to move more rapidly on one side of the ball than on the other. This produces a deflecting force, as predicted by Bernoulli’s principle.

A batter is badly fooled by a curveball. Does the path of the ball really curve?

The Analysis. There is no secret to throwing a curveball. A right-handed pitcher throws a curveball with a counterclockwise spin (as viewed from above), so that it curves away from a right-handed batter. The pitch is most effective when it starts out looking as if it is headed toward the inside of the plate and then curves over the plate and down and away from the batter. Bernoulli’s principle can explain the deflection of the ball’s path. Because the surface of the spinning ball is rough, it drags a layer of air around with it, creating a whirlpool of air near the ball. The ball is also moving toward the plate, which produces an additional flow of air past the ball in the direction opposite to the velocity of the ball. The whirlpool created by the spin of the ball causes the air to move more swiftly on the side opposite the right-handed batter, where the two effects add, than on the side nearer the batter, as shown in the drawing. By Bernoulli’s principle, a greater speed of airflow is associated with a lower pressure: the air pressure is lower on the side of the ball opposite the batter than on the near side.

This difference in pressure produces a deflecting force on the ball, pushing it away from the right-handed batter. Although Bernoulli’s principle gives a good explanation of the direction of the deflecting force and the direction of the curve, it cannot be used to make accurate quantitative predictions. Air is a compressible fluid, and the usual form of Bernoulli’s principle is valid only for noncompressible fluids such as water or other liquids. More accurate methods of treating the effects of airflow past the ball must be used for predicting the degree of curvature. Both theoretical computations and experimental measurements have confirmed that there is a deflecting force on the ball and that the path of the ball does indeed curve. The degree of curvature depends on the rate of spin on the ball and the roughness of the surface of the ball, as you might expect from Bernoulli’s principle. Some pitchers have been known to cheat by roughening the ball’s surface with sandpaper hidden in their gloves. Debate continues about whether the orientation of the seams of the baseball also has an effect. The pitcher’s grip on the ball is an important factor in how much spin he can put on the ball. Once the ball is released, however, experimental evidence suggests that the orientation of the seams is not important in the strength of the deflecting force. A good discussion of the theory and the experimental evidence is found in an article by Robert Watts and Ricardo Ferrer in the January 1987 issue of the American Journal of Physics, pages 40–44. The curved motion of spinning balls is important in other sports, such as golf and soccer. A good athlete needs to be able to recognize and take advantage of the effects of these curves.

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summary Fluid pressure is central to understanding the behavior of fluids, which include both liquids and gases. This chapter has focused on the effects of pressure in determining the behavior of both stationary fluids and fluids in motion.

1

Pressure and Pascal’s principle. Pressure is defined as the ratio of force to area exerted on or by a fluid. According to Pascal’s principle, pressure extends uniformly in all directions through a fluid, which explains the operation of hydraulic systems.

4

Fluids in motion. The rate of flow of a fluid is equal to the speed times the cross-sectional area, vA. In continuous flow, the speed increases if the area decreases. Fluids with high viscosity have a greater resistance to flow than those with low viscosity. As the flow speed increases, the flow may change from laminar (smooth) flow to turbulent flow.

F

v1

v2

A

A2

A1

v1A1 = v2A2

principle. Energy considerations require 5that theBernoulli’s kinetic energy per unit mass of a fluid increase as the _ P=F

A

pressure and the behavior of gases. 2We canAtmospheric measure atmospheric pressure by determining the height

pressure decreases. This is Bernoulli’s principle for cases in which changes in height are not important. Higher fluid speeds are associated with lower pressures. These ideas can be used to explain the curve of a curveball and other phenomena.

of a column of mercury supported by the atmosphere. Atmospheric pressure decreases with increasing altitude. The density of air also changes with altitude, because a lower pressure leads to a larger volume, according to Boyle’s law.

V1

V2 P ρv 2 constant 2 1

P1V1

= P2V2

Archimedes’ principle. In a fluid, pressure increases 3with depth, producing a buoyant force on objects submerged in the fluid. Archimedes’ principle states that this force is equal to the weight of the fluid displaced by the object. If the buoyant force is less than the weight of the object, it will sink in the fluid. Otherwise, it will float.

W FB

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key terms Pressure, 171 Pascal’s principle, 172 Atmospheric pressure, 173 Density, 174

Boyle’s law, 177 Buoyant force, 178 Archimedes’ principle, 178 Viscosity, 181

Turbulent flow, 182 Bernoulli’s principle, 184

Q1. Is it possible for a 100-lb woman to exert a greater pressure on the ground than a 250-lb man? Explain. Q2. If we measure force in pounds (lb) and distance in feet (ft), what are the units of pressure in this system? Explain. Q3. The same force is applied to two cylinders that contain air. One has a piston with a large area, and the other has a piston with a small area. In which cylinder will the pressure be greater? Explain. Q4. A penny and a quarter are embedded in the concrete bottom of a swimming pool filled with water. Which of these coins experiences the greater downward force due to water pressure acting on it? Explain. *Q5. Why are bicycle tires often inflated to a higher pressure than automobile tires, even though the automobile tires must support a much larger weight? Explain. Q6. The fluid in a hydraulic system pushes against two pistons, one with a large area and the other with a small area. a. Which piston experiences the greater force due to fluid pressure acting on it? Explain. b. When the smaller piston moves, does the larger piston move through the same distance, a greater distance, or a smaller distance than the smaller piston? Explain. Q7. If the output piston in a hydraulic pump exerts a greater force than one applied to the input piston, is the pressure at the output piston also larger than at the input piston? Explain.

Q12. If you filled an airtight balloon at the top of a mountain, would the balloon expand or contract as you descend the mountain? Explain. *Q13. When you go over a mountain pass in an automobile, your ears often “pop” both on the way up and on the way down. How can you explain this effect? Q14. The plunger of a sealed hypodermic syringe containing air is slowly pulled out. Does the air pressure inside the syringe increase or decrease when this happens? Explain. Q15. Helium is sealed inside a balloon impermeable to the flow of gas. If a storm suddenly comes up, would you expect the balloon to expand or contract? Explain. (Assume that there is no change in temperature.) Q16. Is it possible for a solid metal ball to float in mercury? Explain. Q17. A rectangular metal block is suspended by a string in a beaker of water so that the block is completely surrounded by water. Is the water pressure at the bottom of the block equal to, greater than, or less than the water pressure at the top of the block? Explain. Q18. Is it possible for a boat made of concrete to float? Explain. Q19. A block of wood is floating in a pool of water. a. Is the buoyant force acting on the block greater than, less than, or equal to the weight of the block? Explain. b. Is the volume of the fluid displaced by the block greater than, less than, or equal to the volume of the block? Explain.

Q8. When a mercury barometer is used to measure atmospheric pressure, does the closed end of the tube above the mercury column usually contain air? Explain.

Q20. A large bird lands on a rowboat that is floating in a swimming pool. Will the water level in the pool increase, decrease, or remain the same when the bird lands on the boat? Explain.

*Q9. Could we use water instead of mercury to make a barometer? What advantages and disadvantages would be associated with the use of water? Explain.

Q21. Is it possible that some objects might float in salt water but sink in fresh water? Explain.

Q10. A blood pressure reading is given as 130/75. a. What units are implied by these numbers? b. Are these numbers total pressures or gauge pressures? Explain. Q11. If you climbed a mountain carrying a mercury barometer, would the level of the mercury column in the glass tube of the barometer increase or decrease (compared to the mercury reservoir) as you climb the mountain? Explain.

*Q22. A rowboat is floating in a swimming pool when the anchor is dropped over the side. When the anchor is dropped, will the water level in the swimming pool increase, decrease, or remain the same? Explain. Q23. If an object has a smaller density than water, will the object stay fully submerged, partly submerged, or rise completely out of the water when it is released underwater? Explain.

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Q24. A steady stream of water flowing in a narrow pipe reaches a point where the pipe widens. Does the speed of the water increase, decrease, or remain the same when the pipe widens? Explain. Q25. Why does the stream of water flowing from a faucet often get more narrow as the water falls? Explain. Q26. Does a stream of liquid with a high viscosity flow more rapidly under the same conditions as a stream with low viscosity? Explain. Q27. If the speed of flow in a stream decreases, is the flow likely to change from laminar to turbulent flow? Explain. Q28. Why is the flow of smoke from a cigarette often laminar near the source but turbulent farther from the source? Explain.

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Q29. If you blow between two limp pieces of paper held hanging down a few inches apart, will the pieces of paper come closer together or move farther apart? Explain. *Q30. A wind gust blows sideways across an outward-swinging door that is slightly ajar. Will this cause the door to slam shut or swing open? Explain. Q31. A hair dryer can be used to create a stream of air. Is the air pressure in the center of the stream greater than, less than, or equal to the air pressure at some distance from the center of the stream? Explain. Q32. From the perspective of a right-handed batter, does a rising fastball spin clockwise or counterclockwise? Explain.

exercises E1. A force of 40 N pushes down on the movable piston of a closed cylinder containing a gas. The piston’s area is 0.5 m2. What pressure does this produce in the gas? E2. A 110-lb woman puts all of her weight on one heel of her high-heel shoes. The heel has an area of 0.4 in2. What is the pressure that her heel exerts on the ground in pounds per square inch (psi)? E3. A 250-lb man supports all of his weight on a showshoe with an area of 200 in2. What pressure is exerted on the snow (in pounds per square inch)? E4. The pressure of a gas contained in a cylinder with a movable piston is 450 Pa (450 N/m2). The area of the piston is 0.2 m2. What is the magnitude of the force exerted on the piston by the gas? E5. In a hydraulic system, a force of 400 N is exerted on a piston with an area of 0.001 m2. The load-bearing piston in the system has an area of 0.2 m2. a. What is the pressure in the hydraulic fluid? b. What is the magnitude of the force exerted on the loadbearing piston by the hydraulic fluid? E6. The load-bearing piston in a certain hydraulic system has an area 50 times as large as the input piston. If the larger piston supports a load of 6000 N, how large a force must be applied to the input piston? E7. A column of water in a vertical pipe has a cross-sectional area of 0.2 m2 and a weight of 450 N. What is the increase in pressure (in Pa) from the top to the bottom of the pipe? E8. With the temperature held constant, the pressure of a gas in a cylinder with a movable piston is increased from 10 kPa to 90 kPa. The initial volume of the gas in the cylinder is 0.36 m3. What is the final volume of the gas after the pressure is increased?

E9. With the temperature held constant, the piston of a cylinder containing a gas is pulled out so that the volume increases from 0.1 m3 to 0.3 m3. If the initial pressure of the gas was 80 kPa, what is the final pressure? E10. A 0.5-kg block of wood is floating in water. What is the magnitude of the buoyant force acting on the block? E11. A block of wood of uniform density floats so that exactly half of its volume is underwater. The density of water is 1000 kg/m3. What is the density of the block? E12. A certain boat displaces a volume of 4.5 m3 of water. (The density of water is 1000 kg/m3.) a. What is the mass of the water displaced by the boat? b. What is the buoyant force acting on the boat? E13. A rock with a volume of 0.2 m3 is fully submerged in water having a density of 1000 kg/m3. What is the buoyant force acting on the rock? E14. A stream moving with a speed of 0.5 m/s reaches a point where the cross-sectional area of the stream decreases to one-fourth of the original area. What is the water speed in this narrowed portion of the stream? E15. Water emerges from a faucet at a speed of 1.5 m/s. After falling a short distance, its speed increases to 3 m/s as a result of the gravitational acceleration. By what number would you multiply the original cross-sectional area of the stream to find the area at the lower position? E16. An airplane wing with an average cross-sectional area 10 m2 experiences a lift force of 60 000 N. What is the difference in air pressure, on the average, between the bottom and top of the wing?

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synthesis problems SP1. Suppose that the input piston of a hydraulic jack has a diameter of 2 cm and the load piston a diameter of 25 cm. The jack is being used to lift a car with a mass of 1400 kg. a. What are the areas of the input and load pistons in square centimeters? (A r 2) b. What is the ratio of the area of the load piston to the area of the input piston? c. What is the weight of the car in newtons? (W mg) d. What force must be applied to the input piston to support the car? SP2. Water has a density of 1000 kg/m3. The depth of a swimming pool at the deep end is about 3 m. a. What is the volume of a column of water 3 m deep and 0.5 m2 in cross-sectional area? b. What is the mass of this column of water? c. What is the weight of this column of water in newtons? d. What is the excess pressure (above atmospheric pressure) exerted by this column of water on the bottom of the pool? e. How does this value compare to atmospheric pressure? SP3. A steel block with a density of 7800 kg/m3 is suspended from a string in a beaker of water so that the block is completely submerged but not resting on the bottom. The block is a cube with sides of 3 cm (0.03 m). a. What is the volume of the block in cubic meters? b. What is the mass of the block? c. What is the weight of the block? d. What is the buoyant force acting on the block? e. What tension in the string is needed to hold the block in place?

SP4. A flat-bottomed wooden box is 3 m long and 1.5 m wide, with sides 1 m high. The box serves as a boat carrying five people as it floats on a pond. The total mass of the boat and the people is 1200 kg. a. What is the total weight of the boat and the people in newtons? b. What is the buoyant force required to keep the boat and its load afloat? c. What volume of water must be displaced to support the boat and its load? (The density of water is 1000 kg/m3.) d. How much of the boat is underwater? (What height of the sides must be submerged to produce a volume equal to that of part c?) SP5. A pipe with a circular cross section has a diameter of 8 cm. It narrows at one point to a diameter of 5 cm. The pipe is carrying a steady stream of water that completely fills the pipe and is moving with a speed of 1.5 m/s in the wider portion. a. What are the cross-sectional areas of the wide and narrow portions of the pipe? (A r2, and the radius is half the diameter.) b. What is the speed of the water in the narrow portion of the pipe? c. Is the pressure in the narrow portion of the pipe greater than, less than, or equal to the pressure in the wider portion? Explain.

home experiments and observations HE1. Using an awl or drill, punch three similar holes in an empty plastic milk jug at three different heights. Plug the holes and fill the jug with water. a. Placing the jug at the edge of the sink, unplug the holes and let the water flow out. Which hole produces the greatest initial speed for the emerging water? b. Which stream travels the greatest horizontal distance? What factors determine how far the water will go horizontally? c. Observe the shape of the water streams as they fall into the sink. Do the streams narrow as the water falls? If so, how can this be explained? HE2. If you have some modeling clay available, try building a boat from the clay. a. Does the clay sink when it is rolled into a ball? What does this indicate about the density of the clay? b. Is a flat boat more effective than a canoe in carrying the maximum load of steel washers or other weights? Try them both. What are the problems with each?

HE3. If you have a hair dryer handy, try supporting a ping-pong ball in a vertical column of air coming from the dryer. a. Can you get the ping-pong ball to stay in place? How far from the center can the ball wander and still return? b. Will a tennis ball or a small balloon work? What differences do you note in each of these cases? HE4. Using paper clips and a pen or pencil as a support rod, suspend two full-sized pieces of paper a few inches apart so that they hang vertically. a. Blow downward between the two pieces and observe the effect. How does increasing the spacing affect the results? b. Try blowing at different rates. How does this affect the results?

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chapter

Temperature and Heat

10

chapter overview The first two sections of this chapter explore the ideas of temperature, heat, and the relationship between these two distinct concepts. We then introduce the first law of thermodynamics, which helps us explain why a drill bit gets hot as it works, as well as many aspects of gas behavior. Finally, we discuss the ways heat is transferred from one object to another.

chapter outline

2 3 4 5

Temperature and its measurement. What is temperature? How do we go about measuring it? Where should we place the zero of a temperature scale? Heat and specific heat capacity. What is heat? How does it differ from temperature? Does adding heat always change the temperature of a substance? How is heat involved in changes of phase? Joule’s experiment and the first law of thermodynamics. Are there other ways of changing the temperature of an object besides adding heat? What does the first law of thermodynamics say and mean? Gas behavior and the first law. How can the behavior of gases be explained using the first law of thermodynamics? What is an ideal gas? The flow of heat. What are the different ways that heat can be transferred from one object to another? How do these ideas apply to heating or cooling a house?

unit two

1

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H

ave you ever touched a drill bit right after drilling a hole in a piece of wood or metal (fig. 10.1)? Chances are that you removed your hand quickly, since the bit was probably hot, particularly if you had been drilling in metal. The same phenomenon is at work when brakes get hot on your bicycle or car, or in any process where one surface is rubbing on another. We could also make the drill bit hot by placing it in a pot of boiling water or the flame of a blowtorch. In either case, when the bit gets hotter, we say that its temperature has increased. What is temperature, though, and how do we go about comparing one temperature to another? Is the final condition of the drill bit any different if it has been warmed by drilling rather than by being placed in contact with hot water? Questions such as these lie in the realm of thermodynamics, the study of heat and its effects on matter. Thermodynamics is ultimately about energy but in a broader context than mechanical energy, which we discussed in chapter 6. The first law of thermodynamics, introduced in this chapter, extends the principle of conservation of energy to include the effects of heat. What is the difference between heat and temperature? Our everyday language often blends these two ideas. A true appreciation for this distinction did not emerge until about the middle of the nineteenth century when the laws of thermodynamics were developed. Heat, temperature, and the distinction between them

figure

10.1 The drill bit feels hot after drilling a hole in a piece of metal. What causes the increase in its temperature? are critical, however, to understanding why things get hot or cold and why some remain that way longer than others. The same ideas involved in cooling a drink are important to understanding global weather patterns.

10.1 Temperature and Its Measurement Suppose that, as child, you were not feeling well and were looking for an excuse to skip school. You thought you might have had a fever, and your mother, by placing a hand on your forehead, agreed that you felt a little hot. To confirm this hunch, however, she took your temperature (fig. 10.2). She told you that you had a temperature of 101.3°. What exactly did this tell you? Since units are important, but often unstated, the measurement indicated that your body temperature is 101.3° Fahrenheit. This is above 98.6° F, which is generally considered normal, so you had a fever. (In most other parts of the world, your temperature would have been read as 38.5° Celsius, above the normal temperature of 37.0° C.) You were probably justified in spending the rest of the day in bed. The thermometer provides a quantitative measure of how hot or cold things are and a basis for comparing your current temperature to your normal temperature. This quantitative measure usually carries more credibility with the boss or a professor than the vaguer statement that you felt like you had a fever.

How do we measure temperature? Taking a temperature or reading a thermometer is a common experience. Besides the obvious function of comparing the “hotness” or “coldness” of different objects, however,

figure

10.2 Taking a temperature. What does the thermometer tell us?

do the numbers have any more fundamental meaning? To ask a basic question, how do hot objects differ from cold objects? What is temperature? Although we all have an intuitive feel for what hot and cold mean, putting that impression into words can be hard. Try it before you read on: how would you define the term hot? Even our senses can mislead us. The pain that we feel when we touch something very hot can be hard to distinguish

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from the pain that we feel when we touch a very cold object. A metal block feels colder to the touch than a wooden block, even though the two blocks may have the same temperature. In the end, temperature measurement is a comparison, and the comparative terms hotter and colder have more meaning than hot or cold themselves. A closer look at what happens when we use a thermometer is helpful. The traditional clinical thermometer was a sealed glass tube partially filled with some fluid, usually mercury. (These have now been largely replaced by digital thermometers.) The inside diameter of the tube was very small, but it widened at the bottom into a reservoir containing most of the mercury (fig. 10.3). Usually you placed the thermometer in your mouth under your tongue and waited a few minutes until the thermometer reached the same temperature as the inside of your mouth. At first, the mercury rose in the thin tube. When the mercury level in the tube was no longer changing, you assumed that it had reached the same temperature as your mouth. Why does the mercury rise? Most materials expand as they get warmer, and mercury and many other liquids expand at a greater rate than glass. As it expands, the mercury in the reservoir must go somewhere, so it rises in the narrow tube. We use the physical property of the thermal expansion of mercury to give an indication of temperature; by placing marks along the tube, we create a temperature scale. Any other physical property that changes with temperature could, in principle, be used. Such properties include changes in electrical resistance, thermal expansion of metals, and even changes in color. In this whole process, we assume that if two objects are in contact with one another long enough so that the physical properties (such as volume) of the objects are no longer changing, the two objects have the same temperature. When we take a temperature with the clinical thermometer, we wait until the mercury stops rising before reading the

figure

10.3 The traditional clinical thermometer contained mercury in a thin glass tube with a wider reservoir of mercury at the bottom.

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thermometer. This procedure gives us part of the definition of temperature by defining when two or more objects have the same temperature. When the physical properties are no longer changing, the objects are said to be in thermal equilibrium. Two or more objects in thermal equilibrium have the same temperature. This assumption is sometimes referred to as the zeroth law of thermodynamics, because it underlies the definition of temperature and the process of temperature measurement.

How were temperature scales developed? What do the numbers on a thermometer mean? When the first crude thermometer was made, the numbers marking the divisions on the scale were arbitrary. They were useful for comparing temperatures only if the same thermometer was used. To compare a temperature measured in Germany with one thermometer to another measured in England with a different thermometer, a standard temperature scale was needed. The first widely used temperature scale was devised by Gabriel Fahrenheit (1686–1736) in the early 1700s. Anders Celsius (1701–1744) invented another widely used scale somewhat later, in 1743. Both of these scales use the freezing point and boiling point of water to anchor the scales. The modern Celsius scale uses the triple point of water, where ice, water, and water vapor are all in equilibrium. This point varies only slightly from the freezing point at normal atmospheric pressure. Fahrenheit set the freezing point of water at 32° and the boiling point at 212° on his scale. These two points are set at 0° and 100° on the Celsius scale (fig. 10.4). As figure 10.4 shows, the Celsius degree is larger than the Fahrenheit degree. Only 100 Celsius degrees span the temperature range between the freezing point and boiling point of water, while 180 Fahrenheit degrees (212° 32°) are needed to span the same range. The ratio of Fahrenheit degrees to Celsius degrees is therefore 180 ⁄ 100, or 9⁄ 5. The Fahrenheit degree is 5⁄ 9 the size of the Celsius degree, so more Fahrenheit degrees are needed to cover the same range in temperature. The Celsius scale is used in science and in most of the world. Because Fahrenheit temperatures are still more familiar in the United States, we commonly have to convert from one scale to the other. Since the zero points and the size of the degree differ, we take both factors into account in the conversion. For example, an ordinary room temperature of 72° Fahrenheit is 40 degrees above the 32°F freezing point of water. Because it takes only 5⁄ 9 as many Celsius degrees to span the same range, this temperature would be (5⁄ 9)(40°) or 22 Celsius degrees above the freezing point of water. The Celsius scale has its zero at the freezing point of water, so 72°F is equal to 22°C. Let’s recap. First, we found how many Fahrenheit degrees this temperature is above the freezing point of water by subtracting 32. Then we multiplied by the factor 5⁄ 9,

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Celsius 100°

Fahrenheit 212° Boiling point of water 200°

80° 60° 40°

150°

100°

20° 50° 0°

32° Freezing point of water

–20°

0°

–40°

–40°

figure

10.4 The Fahrenheit and Celsius scales use different numerical values for the freezing and boiling points of water. The Celsius degree is larger than the Fahrenheit degree.

Is there an absolute zero? Do the zero points on either scale have any special significance? Although the zero point on the Celsius scale is the freezing point of water, there is no special significance to these points. In fact, the temperature goes below zero on either scale, as happens frequently in places like Minnesota and Alaska in the winter. Fahrenheit’s zero point was based on the temperature of a mixture of salt and ice in a saturated salt solution. Although the zero points of both the Fahrenheit and Celsius scales were selected arbitrarily, there is an absolute zero with more fundamental significance. Absolute zero was not proposed until 100 years or so after these scales were originally developed. The first indication of an absolute zero for temperature measurements came from studying changes in pressure and volume that take place in gases when the temperature is changed. If we hold the volume of a gas constant while increasing the temperature, the pressure of the gas will increase. Pressure is another physical property that changes with temperature, just like the volume of mercury in the case of a mercury thermometer. (See chapter 9 for a discussion of pressure.) Because the pressure of a gas of constant volume increases with temperature, we can use this property as a means of measuring temperature (fig. 10.5). If we plot the pressure of a gas as a function of the temperature measured on the Celsius scale, we get a graph like figure 10.6. If the temperatures and pressures are not too high, a striking feature emerges from such graphs. The

which is the ratio of the size of the Fahrenheit degree to the Celsius degree. We know it takes fewer Celsius degrees to span this range, since the Celsius degree is larger than the Fahrenheit degree. We can express this conversion in an equation: TC

5 9

(TF 32).

Using the same logic to perform the reverse conversion from Celsius to Fahrenheit, or simply rearranging the equation, it becomes TF

9 5

Mercury

Gas

TC 32.

Multiplying the temperature in Celsius by 9⁄ 5 tells us how many Fahrenheit degrees above the freezing point it is. This value then is added to 32°F, the freezing point on the Fahrenheit scale. Using this relationship, you can confirm that the boiling point of water (TC 100°C) equals 212°F. Normal body temperature (98.6°F) is equivalent to 36°C. A body temperature of 38°C would be enough to keep you home from school or work.

Flexible tubing

figure

10.5 A constant-volume gas thermometer allows the pressure to change with temperature while the volume is held constant. The difference in height of the two mercury columns is proportional to the pressure.

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10.2 Heat and Specific Heat Capacity

Pressure

A

x x x as B x G x x x x x Gas

–273° –300°

Gas C –200°

–100°

100°

195

200° T (°C)

figure

10.6 The pressure plotted as a function of temperature for different amounts of different gases. When extended backward, the lines all intersect the temperature axis at the same point regardless of gas type or quantity.

curves for different gases or amounts all are straight lines, and when these lines are extended backward to zero pressure, they all intersect the temperature axis at the same point. This is true regardless of the gas used (oxygen, nitrogen, helium, and so on) or the quantity of gas. The curves all intersect the axis at 273.2°C. Since a negative pressure has no meaning, this intersection suggests that the temperature cannot get any lower than 273.2°C. It is important to keep in mind that most gases will condense to liquids and then solidify well before this point is reached. We refer to this temperature, 273.2°C, as absolute zero. The Kelvin, or absolute, temperature scale has its zero at this point and uses intervals the same size as the Celsius degree. To convert Celsius temperatures to kelvins, we simply add 273.2 to the Celsius temperature, or

You are already late for class or work, but your morning cup of coffee is too hot to drink. What can you do to cool it to the point where you won’t scald your tongue? You can blow on it, but blowing is not a very efficient way of cooling coffee quickly. If you take milk with your coffee, you can pour a little cold milk into the cup, and the temperature of the coffee will be lowered (fig. 10.7). What happens when objects or fluids at different temperatures come in contact with one another? The temperature of the colder object increases, and the temperature of the warmer one decreases, with both objects eventually coming to the same intermediate temperature. Something is flowing from the hotter to the colder body (or vice versa). But what is flowing?

What is specific heat capacity? Early attempts at explaining phenomena like our coffeecooling example involved the idea that an invisible fluid called caloric flowed from the hotter to the cooler object. The amount of caloric transferred dictated the extent of the temperature change. Different materials were thought to store different amounts of caloric in the same amount of mass, which helped to explain some observations. Although the caloric model successfully explained many simple phenomena involving temperature changes, there were also problems with it, which we will discuss in section 10.3.

TK TC 273.2. Room temperature of 22°C is an absolute temperature of approximately 295 K. (The term degree and the degree symbol are not used in expressing absolute temperatures; they are simply called kelvins.) The absolute temperature scale currently in use is basically the same as the one originally suggested by observing the behavior of gases. As we cool a substance to near absolute zero, all molecular motion is ceasing. Absolute zero is a temperature that we can approach but never really reach. It represents a limiting value. Speaking of anything getting any colder than absolute zero is not meaningful. Temperature measurements are based on physical properties that change with temperature, such as the volume of a liquid or the pressure of a gas. The Fahrenheit and Celsius scales have different zero points and differentsized degrees. The Kelvin or absolute scale starts at absolute zero and uses intervals the same size as the Celsius degree.

figure 10.7

Pouring cold milk into a hot cup of coffee lowers the temperature of the coffee.

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We now use the term heat for the quantity of energy that flows from one object to another when two objects of different temperatures come in contact. As we will see in section 10.3, heat flow is a form of energy transfer between objects. The idea of exchange of heat sheds light on a large variety of phenomena. If you drop a 100-gram steel ball, initially at room temperature, into your cup of coffee, you will find that it is less effective in cooling your coffee than 100 grams of room-temperature milk or water. Steel has a lower specific heat capacity than milk or water. (Milk is mostly water, as is coffee.) Less heat is needed to change the temperature of 100 grams of steel than to change the temperature of an equal mass of water by the same amount. The specific heat capacity of a material is the quantity of heat needed to change a unit mass of the material by a unit amount in temperature (for example, to change 1 gram by 1 Celsius degree). It is a property of the material.

The specific heat capacity of a material is the relative amount of heat needed to raise its temperature. This number is determined experimentally for each substance. For example, the specific heat capacity of water is 1 cal/g·C°— that is, it takes 1 calorie of heat to raise the temperature of 1 gram of water 1°C. The calorie is a commonly used unit of heat. It is defined as the quantity of heat required to raise the temperature of 1 gram of water 1 Celsius degree. Likewise, if 1 calorie is removed from 1 gram of water, its temperature decreases 1°C. Water happens to have an unusually large specific heat capacity. Therefore, we use water as a reference material for measuring the specific heat capacities of other materials, a few of which are listed in table 10.1.

The specific heat capacity of steel is approximately 0.11 cal/g·C°, much less than water. Dropping 100 grams of room-temperature steel shot into our cup of coffee is much less effective in cooling the coffee than pouring 100 grams of room-temperature water into the cup. The steel absorbs less heat from the coffee for each degree of temperature that it changes than the water does (fig. 10.8). When the cold steel is dropped into the hot water, heat flows from the water into the steel shot. From the definition of specific heat capacity, we can find how much heat must be absorbed by the steel to change its temperature by a given amount. Since specific heat capacity is the amount of heat per unit mass per unit temperature change, the total heat required is Q mcT, where Q is the standard symbol for a quantity of heat, m is the mass, c is the symbol for specific heat capacity, and T is the change in temperature. Because the specific heat capacity c is much larger for water than for steel, a larger quantity of heat is needed to warm 100 grams of water to a given temperature than 100 grams of steel. Since this heat comes from the hot coffee, water is more effective than steel in cooling the coffee. The large specific heat capacity of water is partly responsible for its moderating effect on temperatures near the coast of an ocean or large lake. A large body of water, with its large specific heat capacity, requires the addition or removal of a large quantity of heat to change its temperature. The body of water will have a moderating influence on the air temperature in its vicinity. Nights will be warmer and days cooler than at some distance inland, because it is difficult to change the temperature of the water.

What is the distinction between heat and temperature? When two objects at different temperatures are placed in contact, heat will flow from the object with the higher temperature to the one with the lower temperature (fig 10.9).

table 10.1 Specific Heat Capacities of Some Common Substances Substance

Specific heat capacity in cal/g·C°

water

1.0

ice

0.49

steam

0.48

ethyl alcohol

0.58

glass

0.20

granite

0.19

steel

0.11

aluminum

0.215

lead

0.0305

Water

figure

Steel shot

10.8 One-hundred grams of room-temperature water (20°C) is more effective than 100 grams of roomtemperature steel shot in cooling a hot beaker of water.

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T1 > T2

T1

Q T2

m1, c1

m2, c2

figure

10.9 Heat flows from the hotter object to the colder one when two objects at different temperature are placed in contact. The changes in temperature that result depend on the quantities of material and the specific heat capacity of each object.

Heat added increases the temperature, and heat removed decreases the temperature—heat and temperature are not the same. The quantity of heat added or removed to produce a given change in temperature depends on the amount of material involved (its mass) and the specific heat capacity of the material. Temperature is a quantity that tells us which direction the heat will flow. If two objects are at the same temperature, no heat will flow. If they are at different temperatures, the direction of heat flow is from the higher to the lower temperature. The quantity of heat transferred depends on the temperature difference between the two materials as well as on their masses and specific heat capacities. Heat and temperature are closely related concepts but play different roles in our explanation of heating and cooling processes:

drink using ice or boil water for tea or coffee we produce changes of phase. What happens? Ice, liquid water, and water vapor are different phases of the same substance, water. If the temperature of water is cooled to 0°C, and we continue to remove heat, the water will freeze into ice. Likewise, if we warm ice to 0°C and continue to add heat, the ice will melt (fig. 10.10). As we add heat, the temperature of the ice and water remains at 0°C while the melting is taking place. Heat is being added, but no change in temperature occurs. Apparently, adding or removing heat produces changes other than just temperature changes. Careful measurements have shown that it takes approximately 80 calories of heat to melt 1 gram of ice. This ratio, 80 cal/g, is referred to as the latent heat of fusion of water and is often denoted by the symbol Lf . Latent heat changes the phase of water without changing its temperature. Likewise, approximately 540 calories of heat is required to turn 1 gram of water at 100°C into steam. This ratio, 540 cal/g, is called the latent heat of vaporization, Lv . These values are valid only for water. Other substances have their own particular latent heats of fusion and vaporization. What happens when we cool a glass of water by adding ice? At first, the ice and water are at different temperatures, the ice somewhere below 0°C and the water somewhere

Heat is energy that flows from one object to another when there is a difference in temperature between the objects.

Temperature is the quantity that indicates whether or not, and in which direction, heat will flow. Objects at the same temperature are in thermal equilibrium, and no heat will flow from one to the other.

How is heat involved in melting or freezing? Can heat be added to or removed from a substance without changing the temperature at all? This does indeed occur when substances go through a change of phase or state. The melting of ice and the boiling of water are the most familiar examples of changes in phase. Whenever we cool a

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figure 10.10

Adding heat to a mixture of water and ice at 0°C melts the ice without changing the temperature of the mixture.

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above this temperature. Heat flows from the water to the ice until the ice reaches a temperature of 0°C. At this point, the ice begins to melt as heat continues to flow from the water to the ice. If enough ice is present, this flow of heat will proceed until the water also reaches a temperature of 0°C. Example box 10.1 illustrates these ideas. If the glass were well insulated so that heat could not flow into the glass from the warmer surroundings, the ice and water mixture would remain at 0°C after both the water and the ice have reached this temperature, and no more ice would melt. Usually, though, heat flows into the system from the surroundings, and the ice continues to melt more slowly. Once the ice is gone, the drink will begin to warm and will eventually reach room temperature as heat flows in from the surrounding air. Once it reaches thermal equilibrium, an ice-water mixture, well stirred, remains at 0°C. Small quantities of heat may flow into or out of the mixture from the surroundings, and small quantities of ice may melt or freeze, but the temperature will not change. This is one reason why the temperature 0°C is useful as a reference point for our temperature scales. It is a stable and reproducible temperature. When we cook food by boiling, we take advantage of the fact that water boils at 100°C—and that we can continue to add heat without changing the temperature. Adding heat from the burner causes the liquid water to change to water vapor while the temperature remains at 100°C.

example box 10.1 Sample Exercise: Changing Ice to Water If the specific heat capacity of ice is 0.5 cal/g·C°, how much heat would have to be added to 200 g of ice, initially at a temperature of 10°C, to: a. raise the ice to the melting point? b. completely melt the ice? a. Heat required to raise the temperature: m 200 g 0.5 cal c g. C T 10°C

Q mcT 0.5 cal (200 g)a g. b(10°C) C 1000 cal

Qraise ? b. Heat required to melt the ice: Lf

80 cal g

Qmelt ?

Q mLf (200 g)a

80 cal b g

16 000 cal The total heat required to raise the ice to 0°C and to melt it is 1000 cal 16 000 cal 17 000 cal or 17 kcal.

Since the temperature is constant, the cooking time required to boil an egg or a potato is also fairly constant, depending on the size of the egg or potato. Cooks should be aware, however, of the effects of altitude on the boiling temperature. In cities like Denver or Albuquerque, both located approximately a mile above sea level, water boils at a temperature of about 96°C because of the lower atmospheric pressure at this altitude. It takes longer to boil an egg or potato in these places than near sea level, where the boiling point is 100°C. If you do not make the adjustment in cooking time, your potatoes will be crunchy in the center. Another familiar phenomenon involving the latent heat of vaporization is that of cooling by perspiration (sweating). The cooling is caused by evaporation, the fact that the sweat on your skin changes from a liquid to a vapor. Heat is drawn from your body to provide the heat of vaporization needed to evaporate the sweat, thus cooling your body. One of the reasons we find climates of high humidity so uncomfortable is that the greater amount of water vapor in the air inhibits the evaporation of sweat from our bodies. Temperature changes can be caused by the flow of energy as heat from one object to another. The amount of heat required to produce a given change in temperature depends on the mass and specific heat capacity of the object as well as the temperature change. Temperature is a quantity that tells us when and in which direction heat will flow: heat flows from the hotter to the cooler body. When a substance changes phase, as from ice to water or water to steam, heat is added or removed without changing the temperature. The amount of heat needed per unit mass to produce a phase change is called the latent heat.

10.3 Joule’s Experiment and the First Law of Thermodynamics Can the temperature of an object be increased without placing the object in contact with a warmer object? What is going on when things get warmer from rubbing? These questions were a subject of scientific debate during the first half of the nineteenth century. Their resolution came about through the experimental work of James Prescott Joule (1818–1889), which was followed by the statement of the first law of thermodynamics around mid-century. One of the first persons to raise these questions forcefully was the colorful American-born scientist and adventurer, Benjamin Thompson (1753–1814), also known as Count Rumford. Thompson, finding himself on the losing side of the American Revolutionary War, migrated to Europe, where he served the king of Bavaria as a consultant on weaponry. In Bavaria, he received the title Count Rumford. From his experience in supervising the boring of cannon barrels, Rumford knew firsthand that the barrels and drill bits became quite hot during drilling (fig. 10.11). He once

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Insulation

m

figure 10.11

Rumford’s cannon-drilling apparatus. What is the source of heat that raises the temperature of the drill and the cannon barrel?

demonstrated that water could be boiled by placing it in contact with the cannon barrel as it was being drilled. What was the source of the heat that caused the increase in temperature? No hotter body was present that heat might flow from. Horses were used to turn the drill, but their temperature, although warm, was certainly below the boiling point of water. The horses performed mechanical work on the drilling engine. Was this work capable of generating heat?

What did Joule’s experiments show? Rumford’s demonstration took place in 1798 and was discussed by many scientists during the early part of the nineteenth century. Quantitative research on Rumford’s observations was not done until the 1840s, when Joule performed a famous series of experiments. Joule showed that various ways of performing mechanical work on a system had a consistent and predictable effect in raising the temperature of the system. In the most dramatic of Joule’s experiments, he turned a simple paddle wheel in an insulated beaker of water and measured the increase in temperature (fig. 10.12). A system of weights and pulleys turned the paddle wheel, transferring energy from the weights to the water. As the weights fell, losing gravitational potential energy, the paddle wheel did work against the resisting viscous forces of the water. This work was equal to the loss in potential energy of the weights. A thermometer placed in the water measured the increase in temperature. Joule found that 4.19 J of work was required to raise the temperature of 1 gram of water by 1 C°. (The use of the joule (J) as an energy unit did not come about until well after Joule’s work. He stated his results in more archaic energy units. See chapter 6 for the definition of the joule.) Since we can also raise the temperature of 1 gram of water 1 C° by adding 1 calorie of heat,

figure 10.12

A falling mass turns a paddle in an insulated beaker of water in this schematic representation of Joule’s apparatus for measuring the temperature increase produced by doing mechanical work on a system.

Joule’s experiment implies that 4.19 J of work is equivalent to 1 calorie of heat. Joule used several ways of performing work in his experiments, as well as several kinds of systems. The results were always the same: 4.19 J of work produced the same temperature increase as 1 calorie of heat. The final state of the system provided no clue as to whether the temperature was increased by adding heat or by doing mechanical work.

The first law of thermodynamics At the time of Joule’s work, several people had already suggested that the transfer of heat was a transfer of kinetic energy between the atoms and molecules of the systems involved. Joule’s experiments supported this view and led directly to the statement of the first law of thermodynamics. The idea underlying the first law is that both work and heat represent transfers of energy into or out of a system. If energy is added to a system either as work or as heat, the internal energy of the system increases accordingly. The change in internal energy is equal to the net amount of heat and work transferred into the system. An increase in temperature is one way an increase in internal energy might show up. The first law of thermodynamics summarizes these ideas: The increase in the internal energy of a system is equal to the amount of heat added to a system minus the amount of work done by the system.

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In symbols, we often write the first law of thermodynamics as U Q W, where U represents the internal energy of the system, Q represents the amount of heat added to the system, and W represents the amount of work done by the system (fig. 10.13). The minus sign in this statement is a direct result of the convention of choosing work done by the system to be positive rather than work done on the system. Work done on the system adds energy to the system, but work done by the system on its surroundings removes energy from the system and reduces the internal energy. This sign convention is convenient for discussing heat engines, the main focus of chapter 11. On its surface, the first law seems like a statement of conservation of energy, and indeed it is. Its apparent simplicity obscures the important insights that it contains. First and foremost among these insights is that heat flow is a transfer of energy, an idea strongly reinforced by Joule’s experiments. Before 1850, the concept of energy had been restricted to mechanics. In fact, it was only then coming into its own in mechanics and did not play an important role in Newton’s original theory. The first law of thermodynamics extended the concept of energy into new areas.

What is internal energy? The first law also introduced the concept of the internal energy of a system. An increase in internal energy can show up in a variety of ways: one is an increase in temperature, as in Joule’s experiment with the paddle wheel. A temperature increase is related to an increase in the average kinetic energy of the atoms or molecules making up the system. A change in phase, such as melting or vaporization, is another way that an increase in the internal energy of a system affects the system. In a phase change, the average potential energy of the atoms and molecules is increased as the atoms and molecules are pulled farther away from each other. No temperature change occurs, but there is still an increase in internal energy.

An increase in internal energy, then, can show up as either an increase in the kinetic energy or potential energy (or both) of the atoms or molecules making up the system: The internal energy of the system is the sum of the kinetic and potential energies of the atoms and molecules making up the system.

Internal energy is a property of the system uniquely determined by the state of the system. If we know that the system is in a certain phase, and the temperature, pressure, and quantity of material are specified, there is only one possible value for the internal energy. The amount of heat or work that have been transferred to the system, on the other hand, are not uniquely determined by its state. As Joule’s experiment showed, either heat or work could be transferred to the beaker of water to produce the same increase in temperature. The system under study can be anything we wish it to be—a beaker of water, a steam engine, or an elephant. It is best, however, to consider a system that has distinct boundaries, so that we can easily define where it begins and ends. Choosing the system is similar to the process of isolating an object in mechanics to apply Newton’s second law. In mechanics, forces define the interaction of the chosen object with other objects. In thermodynamics, the transfer of heat or work defines the interaction of a system with other systems or the surroundings. Example box 10.2 applies the first law of thermodynamics. The system is a beaker containing water and ice (fig. 10.14). The internal energy of the system is increased by both heating on a hot plate and doing work by stirring. Notice that the work done by stirring is a negative quantity, since it represents work done on the system. The increase in internal energy results in the melting of the ice.

Counting food calories A final note on energy units is worth mentioning. When we eat food, we add energy to our system by taking in material that can release potential energy in chemical reactions. We often measure that energy in a unit called a Calorie. This unit is actually a kilocalorie, or 1000 calories. The calorie counting that is done for food uses the unit: 1 Cal 1 kilocalorie 1000 cal

Q

figure 10.13

U

–W

The internal energy U of a system is increased by the transfer of either heat or work into the system.

The capital C used in the abbreviation indicates that this is a larger unit than the ordinary calorie. Caloric values given for different types of food are the amount of energy these foods release when they are digested and metabolized. This energy may be stored in various ways within the body, including fat cells. Our bodies are constantly converting energy stored in muscles to other forms of energy. When we do physical work, for example, we transfer mechanical energy to other systems and warm the surroundings by heat flow from our bodies. When you

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count calories, you are counting energy intake. If your output is less than your intake, you should expect to get larger.

example box 10.2 Sample Exercise: Applying the First Law of Thermodynamics A hot plate is used to transfer 400 cal of heat to a beaker containing ice and water. 500 J of work are also done on the contents of the beaker by stirring. a. What is the increase in internal energy of the ice-water mixture? b. How much ice melts in this process? a. We first convert the heat units to joules: Q 400 cal W 500 J U ?

201

Q (400 cal)(4.19 J/cal) 1680 J

Joule discovered that doing 4.19 joules of work on a system increased the temperature of the system by the same amount as adding 1 calorie of heat. This discovery convinced scientists that the flow of heat was a form of energy transfer and led to the statement of the first law of thermodynamics. The first law says that the change in the internal energy of a system is equal to the net amount of heat and work transferred into the system. The internal energy is the sum of the kinetic and potential energies of the atoms making up the system. These ideas extend the principle of conservation of energy.

U Q W 1680 J (500 J) 2180 J

b. Lf 80 cal/g

U mLf

335 J/g m ?

m

¢U Lf 2180 J 335 J/g

6.5 g (amount of ice melted) Since the internal energy was expressed in joules, we converted the latent heat to units of J/g by multiplying by 4.19 J/cal.

figure 10.14

The ice in the beaker can be melted either by adding heat from a hot plate or by doing work stirring the ice and water mixture.

10.4 Gas Behavior and the First Law What happens when we compress air in a cylinder? How does a hot-air balloon work? The behavior of gases in these and other situations can be investigated using the first law of thermodynamics, with the help of some additional facts about the nature of gases. Our atmosphere is another interesting system for exploring the laws of thermodynamics.

What happens when we compress a gas? Suppose that a gas is contained in a cylinder with a movable piston, as in figure 10.15. If the piston is pushed inward by an external force, work is done by the piston on the gas in the cylinder. What effect does this have on the pressure or temperature of the gas? From the first law of thermodynamics, we know that doing work on a system adds energy to the system. By itself, work would increase the internal energy of the gas, but the change in internal energy also depends on the heat flow into or out of the gas. Knowing how much work is done on the gas provides only part of the picture. From the definition of work (section 6.1), we know that the work done on the gas equals the force exerted by the piston times the distance that the piston moves. Since pressure is the force per unit area, the force exerted on the piston by the gas equals the pressure of the gas times the area of the piston (F PA). If we move the piston without accelerating it, the net force acting on the piston is zero, and the external force applied to the piston equals the force exerted by the gas on the piston. Putting these ideas together, the magnitude of the work done on the gas is the force times the distance d that the piston moves, or W Fd (PA)d. The motion of the piston produces a change in the volume of the gas, and this change in volume equals the area of the piston times the distance that it moves, V Ad (see fig. 10.15). The work done by the gas in an expansion is therefore W P V.

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Internal energy (U )

before

d

after

Absolute temperature (T )

F P ∆V

figure 10.15

A movable piston compresses a gas in a cylinder. The magnitude of the work done is W Fd PV.

If the gas is being compressed, the volume of the gas is decreasing, and the change in volume V will be negative, indicating that the work done is also negative. Work is negative when it is done on the system. From the first law, U Q W, negative work increases the energy of the system, since the minus signs cancel. We do work on the gas in compressing it, and that work increases the internal energy of the gas. If the gas is expanding, positive work is done by the gas on its surroundings, and the internal energy of the gas decreases. Compressing a gas causes its internal energy to increase in proportion to the change in volume of the gas, provided that there is no heat flow across the boundaries of the container. What effect does this have on the gas? How does a change in internal energy affect the gas?

How is internal energy related to temperature? Many natural processes involving gases are adiabatic, meaning that no heat flows into or out of the gas during the process. If we compress a gas in a cylinder, for example, the compression can occur quickly enough that there is no time for a significant amount of heat to be transferred. The increase in internal energy is equal to the work done on the gas. What happens to the gas? In general, internal energy is made up of both the kinetic and potential energies of the atoms and molecules in the system. In a gas, the internal energy is almost exclusively kinetic energy, because the molecules are so far apart (on average) that the potential energy of interaction between them is negligible. An ideal gas is one for which the forces between atoms (and the associated potential energy) are small enough to be completely ignored. This is a good approximation for gases in many situations. For an ideal gas, the internal energy of the gas is exclusively kinetic energy: increasing the internal energy increases

figure 10.16

The internal energy is plotted as a function of absolute temperature for an ideal gas. The internal energy increases in direct proportion to the temperature.

the kinetic energy of the gas molecules. Absolute temperature is directly related to the average kinetic energy of the molecules of a system. If the internal energy of an ideal gas increases, the temperature also increases in direct proportion. A plot of internal energy as a function of absolute temperature for an ideal gas looks like the graph in figure 10.16. The temperature of a gas will therefore increase in an adiabatic compression. According to the first law of thermodynamics, the internal energy of the gas increases by an amount equal to the work done on the gas. If we know the specific heat capacity and mass of the gas, we can calculate the increase in temperature from the energy increase. The reverse is also true. If we allow the gas to expand against the piston, doing work on its surroundings, the internal energy of the gas decreases. The temperature of the gas decreases in an adiabatic expansion—this is how a refrigerator works. A pressurized gas is allowed to expand, lowering its temperature. The cool gas then circulates through coils inside the refrigerator, removing heat from the contents of the refrigerator.

How can we keep the temperature of a gas from changing? Is it possible for the temperature of a gas to remain constant during a compression or expansion? In an isothermal process, the temperature does not change. (Iso means equal, and thermal refers to warmth or temperature.) Because of the direct relationship between temperature and internal energy in an ideal gas, the internal energy must be constant if there is no change in temperature. The change in internal energy, U, is zero for an isothermal process in an ideal gas. If U is zero, the first law of thermodynamics (U Q W) requires that W Q. In other words, if an amount of heat Q is added to the gas, an equal amount of work W will be done by the gas on its surroundings if the temperature and the internal energy are to remain constant. It is possible to add heat to a gas without changing its temperature, as long as the gas does an equal amount of work on

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its surroundings by expanding. (This is not an adiabatic process since heat is flowing into the system!) Here again, the reverse is also true. If we compress a gas while holding the temperature constant, heat is removed from the gas, according to the first law of thermodynamics. For the internal energy and the temperature to remain constant, energy added to the gas in the form of work must result in an equal amount of energy in the form of heat being removed. Isothermal compressions and expansions are important processes in heat engines, which we will consider in chapter 11.

What happens to the gas in a hot-air balloon? When the gas is heated in a hot-air balloon (fig. 10.17), the pressure, not the temperature, remains constant. The pressure of the gas inside the balloon cannot be significantly larger than the surrounding atmosphere. When pressure remains constant in a process, it is called isobaric (baric refers to pressure). The internal energy increases as the gas is heated, and so does the temperature. The gas also expands in this process, however, which removes some internal energy. The gas expands in the isobaric heating process because of another property of ideal gases discovered near the beginning of the nineteenth century. A series of experiments showed that the pressure, volume, and absolute temperature of

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example box 10.3 Sample Exercise: Using the Ideal Gas Law Suppose that the pressure of an ideal gas mixture remains constant, and the volume of the mixture is decreased from 4 liters to 1 liter. If the initial temperature is 600K, what is the final temperature? This question can be answered using the ideal gas law: PV = NkT. Since we assume no gas escapes the container, we assume N, the number of molecules of gas, remains constant. Boltzmann’s constant also does not change. The ideal gas law, PV NkT, can be rewritten as V/T = Nk/P = a constant. Vinitial = 4 liters Vfinal = 1 liter Tinitial = 600K Tfinal = ?

Vinitial Tinitial

V final T final

Cross multiplying: T final V final Tinitial Vinitial T final Tinitial

1 1l 4l 4

Therefore: 1 T 4 initial 1 T final (600K) 150K 4

T final

Notice that even though you may not know how many molecules of gas there are (N) or the value of Boltzmann’s constant, k (which you could look up), you can still use the ideal gas law to discover how the pressure, temperature, and volume change relative to one another.

figure 10.17

Heat is added to the air in a hot-air balloon by the propane burner above the gondola. The heated air expands making it less dense than the surrounding air.

an ideal gas are related by the equation PV NkT where N is the number of molecules in the gas and k is a constant of nature called Boltzmann’s constant. (Ludwig Boltzmann (1844–1906) was an Austrian physicist who developed a statistical theory of gas behavior.) This equation is called the equation of state of an ideal gas. Example box 10.3 uses this equation of state. The temperature T used in the equation of state must be the absolute temperature. Recall that the concept of absolute temperature originally arose in the study of the behavior of gases (see section 10.1). The equation of state combines Boyle’s law (section 9.2) with the effect of temperature on pressure discussed in section 10.1. The equation of state shows that if the temperature increases while the pressure and number of molecules are held constant, the volume of the gas also must increase. In other words, the gas expands. Likewise, if the gas expands at constant pressure, the temperature must increase. For an isobaric expansion, the first law of thermodynamics indicates that the heat added to the gas must be

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greater than the amount of work done by the gas in expanding. The internal energy, U, increases when the temperature increases. Since U Q W, Q must be larger than W (which is positive, since the gas is expanding) for the internal energy to increase. The first law of thermodynamics and the equation of state for an ideal gas can then be used to compute how much heat must be added to the system to produce a given amount of expansion. Since the gas inside the balloon has expanded, the same number of molecules now occupy a larger volume, and the density of the gas has decreased. In other words, the balloon rises because the density of the gas inside the balloon is less than the density of the air outside the balloon. The buoyant force acting on the balloon is described by Archimedes’ principle (section 9.3). The same phenomenon occurs on a much larger scale in the atmosphere. A warm air mass will rise compared to cooler air masses for the same reason that a hot-air balloon rises. Air heated near the ground on a warm summer day expands, becomes less dense, and rises in the atmosphere. Rising air carries water vapor that provides the moisture for the formation of clouds at higher elevations. The rising columns of warm air also create air turbulence, which can become a problem for aircraft. These thermals are a real boon, however, to hang-glider enthusiasts and soaring eagles. The behavior of gases and the implications of the first law of thermodynamics play important roles in our understanding of weather phenomena. Latent heat is involved in the evaporation and condensation of water as well as in the formation of ice crystals in snow or sleet. The source of energy for all of these processes is the sun: surface water and the atmosphere of the Earth are a huge engine driven by the sun. The first law of thermodynamics explains many processes involving gases. The absolute temperature of a gas is directly related to the internal energy of the gas: in the case of an ideal gas, the internal energy is the kinetic energy of the gas molecules. The amount of heat added and the work done by the gas determine its temperature, yielding different results for adiabatic, isothermal, and isobaric processes. The first law, together with the equation of state, can be used to predict the behavior of gases in heat engines, hot-air balloons, and air masses in the atmosphere.

T1 > T2 Metal bar T1

Q

T2

figure 10.18

In thermal conduction, energy flows through a material when there is a temperature difference across the material.

figure 10.19

When a metal block and a wooden block, both at room temperature, are picked up, the metal block feels cooler.

Conduction, convection, and radiation are the three basic means of heat flow. All three are important in heat loss (or gain) from a house. An understanding of these mechanisms can help you achieve comfort while saving fuel in your house, residence-hall room, or workplace.

10.5 The Flow of Heat

Heat flow by conduction

In most areas of the world we need to heat our homes and buildings in the winter. We warm our houses by burning wood or fossil fuels such as coal, oil, or natural gas in a furnace, or by electric heating. We pay for the fuel that we use and thus have an incentive to keep the heat it generates from escaping. How does heat flow away from a house or any other warm object? What mechanisms are involved, and how can we lessen the loss of heat by our awareness of these mechanisms?

In conduction, heat flows through a material when objects at different temperatures are placed in contact with one another (fig 10.18). Heat flows directly from the hotter body to the cooler body at a rate that depends on the temperature difference between the bodies and a property of the materials called the thermal conductivity. Some materials have much larger thermal conductivities than others. Metals, for example, are much better heat conductors than wood or plastic.

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10.5 The Flow of Heat

Imagine that you pick up a block of metal in one hand and a block of wood in the other (fig. 10.19). The two blocks are at room temperature, but your body is typically 10 to 15 Celsius degrees warmer than room temperature, so heat will flow from your hands to the blocks. The metal block, however, will feel colder than the wood block. Since the two blocks are initially at the same temperature, the difference lies in the thermal conductivity, not in the temperature, of the two materials. Metal conducts heat more readily than wood does, so the heat flows more rapidly from your hand into the metal than into the wood. Since contact with the metal cools your hand more rapidly than contact with the wood, the metal block feels colder. Heat flow is a flow of energy, the actual transfer of kinetic energy of atoms and electrons. Because of the higher temperature of your hands, the average kinetic energy of the atoms in your hands is larger than the average kinetic energy of the atoms in the blocks. As the atoms bump into one another, kinetic energy is transferred, and the average kinetic energy of the atoms in the blocks increases at the expense of the atoms in your hands. Free electrons in metals play an important role in this process: metals are good conductors of both heat and electrical charge. In discussing home insulation, we often use the concept of thermal resistance or R values to compare the effectiveness of different materials as insulators. The greater the conductivity, the lower the R value, since good thermal conductors are poor thermal insulators. The R value, in fact, relates both to the thickness of the material and to its thermal conductivity. The insulating effect of a material increases as its thickness increases, so R values increase with the thickness of the material. Still air is a good thermal insulator. Porous materials with trapped pockets of air like rock wool or fiberglass insulation make excellent insulators and have high R values. Their low thermal conductivities result from the air pockets trapped in these materials rather than from the material itself. When we use such materials to insulate the walls, ceilings, and floors of a dwelling, we reduce their ability to conduct heat and lessen heat loss from the house.

What is convection? If we heat a volume of air, then move the air by blowers or natural flow, we transfer heat by convection. In heating a house, we often move air, water, or steam through pipes or ducts to carry heat from a central furnace to different rooms. Convection transfers heat by the motion of a fluid containing thermal energy. Convection, therefore, is the main way of heating a house. Warm air is less dense than cooler air, so it tends to rise from radiators or baseboard heaters toward the ceiling. As it does, the warm air sets up air currents within the room that distribute energy around the room. Warm air rises along the wall containing the heat source, and cooler air falls along the opposite wall (fig. 10.20). This is called natural convection; The use of fans would be forced convection.

205

figure 10.20

In convection, thermal energy is transferred by the motion of a heated fluid. In a room, this occurs when heated air flows from a radiator or heater duct.

Convection is also involved in heat loss from buildings. When warm air leaks out of a building or cooler air leaks in from outside, we say that we are losing heat to infiltration, a form of convection. Infiltration can be prevented, to some extent, by weatherstripping doors and windows and by sealing other cracks or holes. It is not desirable, however, to eliminate infiltration completely. A turnover of the air in a house is needed to keep the air fresh and free of odors.

What is radiation, and how does it transfer heat? Radiation involves the flow of energy by electromagnetic waves. Wave motion and electromagnetic waves are discussed in chapters 15 and 16. The electromagnetic waves involved in the transfer of heat lie primarily in the infrared portion of the electromagnetic wave spectrum. Infrared wavelengths are shorter than those of radio waves but longer than the wavelengths of visible light. (See fig. 16.5.) While conduction requires a medium to travel through, and convection a medium to be carried along with, radiation can take place across a vacuum, such as the evacuated barrier in a thermos bottle (fig. 10.21). The radiation is reduced to a minimum by silvering the facing walls of the evacuated space. Silvering causes the electromagnetic waves to be reflected rather than absorbed and reduces the energy flow into or out of the container. The same principle is involved in the use of foil-backed insulation in housing construction. Some of the heat flow across the dead-air space within a wall is due to radiation and some to conduction through the air. Using foil-backed insulation reduces heat flow from radiation. Thin sheets of foilbacked insulation may have R values equal to that of thicker sheets of insulation without the foil backing.

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Chapter 10 Temperature and Heat

everyday phenomenon

box 10.1

Solar Collectors and the Greenhouse Effect The Situation. Flat-plate solar collectors are often used for collecting solar energy to heat water or houses. The flat-plate collector consists of a metal plate with water-carrying tubes bonded to its surface. The plate and tubes are painted black and are held in a frame insulated below the plate, and with a glass or transparent plastic cover on top. How does the flatplate collector work? What heat-flow processes are involved, and what relationship, if any, does the solar collector have to the much-discussed greenhouse effect? The Analysis. The flat-plate collector receives energy from the sun as electromagnetic radiation, the only form of heat flow that can take place across the vacuum of space. Electromagnetic waves emitted by the sun are mainly in the visible part of the electromagnetic spectrum, the wavelengths to which our eyes are sensitive. These waves pass easily through the transparent cover plate of the collector. The black surface of the collector plate absorbs most of the visible light that falls on it, reflecting very little. This absorbed energy raises the internal energy and temperature of the

Glass cover plate Visible light

Metal collector plate

Insulation

The black-surfaced metal collecting plate of a flat-plate solar collector is insulated below and covered with a glass or transparent plastic plate above.

plate. Water traveling through the tubes bonded to the plate is heated by conduction. The plate must be at a higher temperature than the water for heat flow to take place. The

(continued)

Infrared radiation

Insulation

Vacuum

figure 10.21

Radiation is the only mechanism that can carry thermal energy across the evacuated space in a thermos bottle. Silvering the walls reduces the flow of radiation.

Radiation is also involved in the loss of energy from roofs or from road surfaces to the dark (and cold) night sky. Blacktopped road surfaces in particular, and black surfaces in general, are effective radiators of electromagnetic waves. The road surface radiates energy to the sky, cooling the surface more rapidly than the surrounding air. Even when the air temperature is still above freezing, the surface of a blacktopped road can drop below the freezing point, creating icy conditions. Good absorbers of electromagnetic radiation are also good emitters: black materials both absorb more solar energy in the summer and lose more heat in the winter. A black roof may look nice, but it is not the best color for energy conservation. A lighter color is more effective both in keeping your home cool in the summer and warm in the winter.

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10.5 The Flow of Heat

insulation below the plate reduces heat flow to the surroundings. Since it is warm, the metal plate will radiate heat in the form of infrared electromagnetic waves. Glass (or plastic) is opaque to infrared radiation, so these waves cannot pass through the glass cover. The glass cover also reduces heat loss from convection, since it prevents air currents from passing directly over the metal plate. The glass enclosure of a greenhouse serves the same purpose. Besides reducing convection losses, the glass allows visible sunlight in while preventing heat loss from radiation at longer wavelengths. It is an energy trap. This one-way transmission of radiation at visible wavelengths, but not at longer infrared wavelengths, is the greenhouse effect. It is what causes your car to heat up on a sunny day when the windows are closed. The Earth also resembles a large greenhouse. Carbon dioxide gas, as well as certain other gases present in the atmosphere in smaller quantities, is transparent to visible radiation but not to infrared radiation. Carbon dioxide is a natural by-product of the burning of any carbon-based fuel such as oil, natural gas, coal, or wood. Carbon dioxide is taken up by plant life and converted to other carbon compounds, so that the global plant cover is also a factor in determining the amount of carbon dioxide in the atmosphere. Our heavy use of fossil fuels such as oil, coal, and natural gas has increased the amount of carbon dioxide in the upper atmosphere. The greenhouse effect associated with this increase in carbon dioxide and other greenhouse gases is slowly increasing the Earth’s average temperature (global warming). The loss of forests and other plant life that absorb carbon dioxide may also contribute to the carbon dioxide buildup. Although modeling of the earth’s atmosphere is a complex problem, there is ample evidence to support the conclusion that global warming is occuring.

Knowledge of the basic mechanisms of heat transfer is useful in designing a house. Conduction, convection, and radiation are all important in understanding how heat flows into and out of a house, as well as how it circulates inside the house. Large windows produce high heat loss, even when double-pane windows are used. If the window is on the south side of the house, however, this heat loss may be partially offset by the heat gain by radiation from the sun. Heat-flow mechanisms are also important in understanding how to make the best use of solar energy, discussed in everyday phenomenon box 10.1.

Carbon dioxide

Visible light from sun

Infrared radiation Earth Carbon dioxide and other gases in the atmosphere of the Earth play the same role as the cover plate on a solar collector. Visible light passes through, but infrared radiation does not.

With a slow increase in the Earth’s temperature, we can expect that the polar ice caps will begin to melt, and this appears to be happening. Thawing of the ice caps will cause the oceans to rise, and low-lying coastal regions may be flooded. Changes in global weather patterns may also affect crop production. The precise effects are difficult to predict because other factors are also at work that can either accelerate or moderate these effects. The solution to the global warming problem lies primarily in reducing worldwide use of fossil fuels. This can be achieved through energy conservation (see everyday phenomenon box 6.1) and through development of other energy resources that do not depend upon fossil fuels. To develop rational enviromental policies, we need to carefully monitor the greenhouse effect and other factors that can increase or decrease the Earth’s temperature.

The flow of heat can occur by three distinct means— conduction, convection, and radiation. In conduction, energy is transferred through a material: some materials are better thermal conductors than others. Convection involves transfer of heat by the flow of a fluid containing thermal energy. Radiation is the flow of energy through electromagnetic waves. All three of these processes are important in any application of the first law of thermodynamics, since the flow of heat is one way the internal energy of a system can change.

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Chapter 10 Temperature and Heat

summary Heat, which plays an important role in the first law of thermodynamics, is the focus of this chapter. We used the first law of thermodynamics to explain changes in temperature or phase of a substance as well as the behavior of an ideal gas. The distinction between the concepts of heat and temperature is critical.

by adding heat or by doing work on the system. Increasing the internal energy increases the potential and kinetic energies of the atoms and molecules making up the system, and it may show up as an increase in temperature, a change of phase, or as changes in other properties of the system.

Temperature and its measurement. Temperature 1scales have been devised to provide a consistent means of describing how hot or cold an object is. When objects are at the same temperature, no heat flows between them, and the physical properties remain constant. The idea of an absolute zero of temperature emerged from studies of gas behavior. 100°C

Q

U

212°F

W

∆U = Q – W

4

0°C

Gas behavior and the first law. The internal energy of an ideal gas equals the kinetic energy of the gas molecules. It varies directly with the absolute temperature of the gas. Work done by the gas is proportional to its change in volume, so the first law can be used to predict how the temperature or pressure will change when work is done or heat is added.

32°F

∆V Q

T

Heat and specific heat capacity. The energy that 2flows between objects because they are at different temperatures is called heat. The amount of heat required to cause a one-degree change in temperature to one unit of mass is called the specific heat capacity. Latent heat is the amount of heat per unit mass required to change the phase of a substance without changing the temperature. T1 > T2

W = P ∆V PV = NkT

5

The flow of heat. There are three basic mechanisms of heat flows from one system to another. Conduction is the direct flow of heat through materials. Convection is the flow of heat carried by a moving fluid. Radiation is the flow of heat by electromagnetic waves. T1 > T2

T1 Q

T2

T1

T2 Conduction

Q = mc∆T Convection

3

Joule’s experiment and the first law of thermodynamics. The first law of thermodynamics states that the

internal energy of a substance or system can be increased either

Radiation

key terms Thermodynamics, 192 Thermal equilibrium, 193 Zeroth law of thermodynamics, 193 Absolute zero, 195 Absolute temperature, 195 Heat, 196 Specific heat capacity, 196 Temperature, 197

Change of phase, 197 Latent heat, 197 First law of thermodynamics, 199 Internal energy, 200 Adiabatic, 202 Ideal gas, 202 Isothermal, 202 Isobaric, 203

Equation of state, 203 Conduction, 204 Thermal conductivity, 204 Convection, 205 Radiation, 205 Greenhouse effect, 207

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Questions

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Q17. Which represents the greater amount of energy, 1 J or 1 cal? Explain.

Q1. Is an object that has a temperature of 0°C hotter than, colder than, or at the same temperature as one that has a temperature of 0°F? Explain.

*Q18. Suppose that the internal energy of a system has been increased, resulting in an increase in the temperature of the system. Is it possible to tell from the final state of the system whether the change in internal energy was due to the addition of heat or of work to the system? Explain.

Q2. Which spans a greater range in temperature, a change in temperature of 10 Fahrenheit degrees or a change of 10 Celsius degrees? Explain.

Q19. Is it possible for the internal energy of a system to be greater than the kinetic energy of the molecules and atoms making up the system? Explain.

Q3. The volume of a gas held at constant pressure increases in a predictable way when its temperature increases. Could such a system be used as a thermometer? Explain.

*Q20. Based upon his experiments, Joule proposed that the water in a pool at the bottom of a waterfall should have a higher temperature than that at the top. Why might this be so? Explain.

*Q4. We sometimes attempt to determine whether another person has a fever by placing a hand on their forehead. Is this a reliable procedure? What assumptions do we make in this process? Q5. Is it possible for a temperature to be lower than 0°C? Explain. Q6. Is it possible for a temperature to be lower than 0 K on the Kelvin temperature scale? Explain. Q7. Is an object with a temperature of 273.2 K hotter than, colder than, or at the same temperature as an object with a temperature of 0°C? Explain. Q8. Two objects at different temperatures are placed in contact with one another but are insulated from the surroundings. Will the temperature of either object change? Explain. Q9. Is it possible for the final temperature of the objects discussed in question 8 to be greater than the initial temperatures of both objects? Explain. Q10. Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object. Will the final temperature of the two objects necessarily be the same? Explain. Q11. Two cities, one near a large lake and the other in the desert, both reach the same high temperature during the day. Which city, if either, would you expect to cool down more rapidly once the sun has set? Explain. Q12. Is it possible to add heat to a substance without changing its temperature? Explain. Q13. What happens if we add heat to water that is at the temperature of 100°C? Does the temperature change? Explain. Q14. What happens if we remove heat from water at 0°C? Does the temperature change? Explain. Q15. Is it possible to change the temperature of a glass of water by stirring the water, even though the glass is insulated from its surroundings? Explain. Q16. A hammer is used to pound a piece of soft metal into a new shape. If the metal is thermally insulated from its surroundings, will its temperature change due to the pounding? Explain.

Q21. An ideal gas is compressed without allowing any heat to flow into or out of the gas. Will the temperature of the gas increase, decrease, or remain the same in this process? Explain. Q22. Is it possible to decrease the temperature of a gas without removing any heat from the gas? Explain. Q23. Heat is added to an ideal gas, and the gas expands in the process. Is it possible for the temperature to remain constant in this situation? Explain. Q24. Heat is added to an ideal gas maintained at constant volume. Is it possible for the temperature of the gas to remain constant in this process? Explain. Q25. Heat is added to a hot-air balloon causing the air to expand. Will this increased volume of air cause the balloon to fall? Explain. *Q26. Heat is added to ice causing the ice to melt but producing no change in temperature. Because water expands when it freezes into ice, the volume of the water obtained from melting the ice is less than the initial volume of ice. Does the internal energy of the ice-water system change in this process? Explain. Q27. A block of wood and a block of metal have been sitting on a table for a long time. The block of metal feels colder to the touch than the block of wood. Does this mean that the metal is actually at a lower temperature than the wood? Explain. Q28. Heat is sometimes lost from a house through cracks around windows and doors. What mechanism of heat transfer (conduction, convection, or radiation) is involved? Explain. Q29. Is it possible for water on the surface of a road to freeze even though the temperature of the air just above the road is above 0°C? Explain. Q30. What heat transfer mechanisms (conduction, convection, or radiation) are involved when heat flows through a glass windowpane? Explain. Q31. Is it possible for heat to flow across a vacuum? Explain.

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Q32. What property does glass share with carbon dioxide gas that makes them both effective in producing the greenhouse effect? Explain. Q33. What mechanism of heat flow (conduction, convection, or radiation) is involved in transporting useful thermal energy away from a flat-plate solar collector? Explain.

Q34. Will a solar power plant (one that generates electricity from solar energy) have the same tendency to increase the greenhouse effect in the atmosphere as a coal-fired power plant? Explain.

exercises E1. An object has a temperature of 45°C. What is its temperature in degrees Fahrenheit?

E11. If 200 cal of heat are added to a system, how much energy has been added in joules?

E2. The temperature on a winter day is 14°F. What is the temperature in degrees Celsius?

E12. If 600 J of heat are added to 50 g of water initially at 20°C, a. how much energy is this in calories? b. what is the final temperature of the water?

E3. The temperature in a residence-hall room is 24°C. What is the temperature of the room on the absolute (Kelvin) temperature scale? E4. The temperature on a warm summer day is 95°F. What is this temperature a. in degrees Celsius? b. on the absolute (Kelvin) scale? E5. The temperature of a beaker of water is 318.2 K. What is this temperature a. in degrees Celsius? b. in degrees Fahrenheit?

E13. While a gas does 300 J of work on its surroundings, 800 J of heat is added to the gas. What is the change in the internal energy of the gas? E14. The volume of an ideal gas is increased from 0.8 m3 to 2.4 m3 while maintaining a constant pressure of 1000 Pa (1 Pa 1 N/m2). How much work is done by the gas in this expansion? E15. If the initial temperature in exercise E14 is 700K, what is the final temperature?

E6. How much heat is required to raise the temperature of 70 g of water from 20°C to 80°C?

E16. Work of 1500 J is done on an ideal gas, but the internal energy increases by only 800 J. What is the amount and direction of heat flow into or out of the system?

E7. How much heat must be removed from a 200-g block of copper to lower its temperature from 150°C to 30°C? The specific heat capacity of copper is 0.093 cal/g·C°.

E17. If 500 cal of heat are added to a gas, and the gas expands doing 500 J of work on its surroundings, what is the change in the internal energy of the gas?

E8. How much heat must be added to 60 g of ice at 0°C to melt the ice completely?

E18. Work of 600 J is done by stirring an insulated beaker containing 100 g of water. a. What is the change in the internal energy of the system? b. What is the change in the temperature of the water?

E9. If 600 cal of heat are added to 50 g of water initially at a temperature of 10°C, what is the final temperature of the water? E10. How much heat must be added to 120 g of water at an initial temperature of 60°C to a. heat it to the boiling point? b. completely convert the 100°C water to steam?

synthesis problems SP1. Heat is added to an object initially at 30°C, increasing its temperature to 80°C. a. What is the temperature change of the object in Fahrenheit degrees? b. What is the temperature change of the object in kelvins? c. Is there any difference in the numerical value of a heat capacity expressed in cal/g·C° from one expressed in cal/g·K? Explain. SP2. A student constructs a thermometer and invents her own temperature scale with the ice point of water at 0°S (S for

student) and the boiling point of water at 50°S. She measures the temperature of a beaker of water with her thermometer and finds it to be 15°S. a. What is the temperature of the water in degrees Celsius? b. What is the temperature of the water in degrees Fahrenheit? c. What is the temperature of the water in kelvins? d. Is the temperature range spanned by 1 Student degree larger or smaller than the range spanned by 1 Celsius degree? Explain.

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Home Experiments and Observations

SP3. The initial temperature of 150 g of ice is 20°C. The specific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·C°. The latent heat of fusion of water is 80 cal/g. a. How much heat is required to raise the ice to 0°C and completely melt the ice? b. How much additional heat is required to heat the water (obtained by melting the ice) to 25°C? c. What is the total heat that must be added to convert the 80 g of ice at 20°C to water at 25°C? d. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C? Explain. SP4. A 150-g quantity of a certain metal, initially at 120°C, is dropped into an insulated beaker containing 100 g of water at 20°C. The final temperature of the metal and water in the beaker is measured as 35°C. Assume that the heat capacity of the beaker can be ignored. a. How much heat has been transferred from the metal to the water? b. Given the temperature change and mass of the metal, what is the specific heat capacity of the metal? c. If the final temperature of the water and this metal is 70°C instead of 35°C, what quantity of this metal (initially at 120°C) was dropped into the insulated beaker?

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SP5. A beaker containing 400 g of water has 1200 J of work done on it by stirring and 200 cal of heat added to it from a hot plate. a. What is the change in the internal energy of the water in joules? b. What is the change in the internal energy of the water in calories? c. What is the temperature change of the water? d. Would your answers to the first three parts differ if there had been 200 J of work done and 1200 cal of heat added? Explain. SP6. Suppose that the pressure of an ideal gas mixture remains constant at 1800 Pa (1 Pa = 1N/m2) and the temperature is increased from 250K to 750K. a. If the original volume of the gas was 0.15 m3, what is the final volume? (See example box 10.3.) b. What is the change in volume V for this process? c. How much work does the gas do on the surroundings during the expansion? d. If the initial volume was 0.24m3 and the same temperature change occurred, would the work done be the same as in the first case? Show by repeating the steps of the first three parts. e. Is the same amount of gas involved in these two situations? Explain.

home experiments and observations HE1. Look around your home to see what kinds of thermometers you can find. Indoor, outdoor, clinical, or cooking thermometers can be found in most homes. For each thermometer, note: a. What temperature scales are used? b. What is the temperature range of the thermometer? c. What is the smallest temperature change that can be read on each thermometer? d. What changing physical property is used to indicate change in temperature for each thermometer? HE2. Take two Styrofoam cups, partially fill them with equal amounts of cold water, and drop equal amounts of ice into each cup to obtain a mixture of ice and water. a. Stir the water and ice mixture in one of the cups vigorously with a nonmetallic stirrer until all of the ice has been melted. Note the time that it takes to melt the ice. b. Set the second cup aside and observe it every 10 minutes or so until all the ice has melted. Note the time that it takes for the ice to melt. c. How do the times compare? Where is the energy coming from to melt the ice in each cup?

HE3. Fill a Styrofoam cup with very hot water (or coffee). Take objects made of different materials such as a metal spoon, a wooden pencil, a plastic pen, or a glass rod. Put one end of each object into the water and make these observations: a. How close to the surface of the water do you have to hold the object for it to feel noticeably warm to the touch? b. From your observations, which material would you judge to be the best conductor of heat? Which is the worst? Can you rank the materials according to their ability to conduct heat? HE4. Conduct an energy survey of your residence-hall room or other similar living space. Note: a. What is the heat source (or sources) for the room and how does this heat flow into the room? What heat-flow mechanisms are involved? b. How is heat lost from the room? What heat-flow mechanisms are involved? c. What could be done to reduce heat loss or to otherwise improve the energy efficiency of the room?

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chapter

11

Heat Engines and the Second Law of Thermodynamics

chapter overview After a discussion of heat engines, we will explore the second law of thermodynamics. The second law is central to understanding the efficiency of heat engines. Using both the first and second laws of thermodynamics, we then consider the use of energy resources in today’s global economy. These issues have fundamental significance to the quality of our environment as well as to the health of our economy. The laws of thermodynamics are critical to making intelligent choices about energy policy.

chapter outline

1 2 3

unit two

4 5

Heat engines. What is a heat engine? What does the first law of thermodynamics tell us about how heat engines work? The second law of thermodynamics. How would an ideal heat engine operate if we could build one? How is the concept of an ideal engine related to the second law of thermodynamics—what does the second law say and mean? Refrigerators, heat pumps, and entropy. What does a refrigerator or heat pump do? What is entropy? How is it related to limitations on the use of heat energy? Thermal power plants and energy resources. How can we generate power efficiently? What are the implications of the second law of thermodynamics in our use of energy resources such as fossil fuels and solar energy? Perpetual motion and energy frauds. Is perpetual motion possible according to the laws of thermodynamics? How can we judge the claims of inventors?

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11.1 Heat Engines

213

M

any of us spend a good portion of our lives driving or riding in automobiles. We talk about how many miles per gallon we get or discuss the merits of different kinds of engines. We regularly drive into gas stations to buy unleaded gasoline, and we generally have a good idea of what this fuel costs at different places (fig. 11.1). How many of us really understand what is going on inside that engine, though? As we turn the ignition switch and depress the gas pedal, the engine roars into action, consumes fuel, and powers the car at our command, without requiring the driver to have a detailed knowledge of its principles of operation. Ignorance may be bliss—until something goes wrong—and then some understanding of the engine may be useful. What does an automobile engine do, and how does it work? The internal combustion engine used in most modern automobiles is a heat engine. The science of thermodynamics arose in an attempt to better understand the principles of operation of steam engines, the first practical versions of heat engines. Building more efficient engines was the primary objective. Although steam engines are usually external-combustion engines, whose fuel is burned outside of the engine rather than inside, the basic principles of operation of steam engines and automobile engines are the same.

figure

11.1 Filling up the family chariot. How does the engine use this fuel to produce motion? How do heat engines work? What factors determine the efficiency of a heat engine, and how can efficiency be maximized? The second law of thermodynamics plays a central role. Heat engines will lead us to an exploration of the second law and the related concept of entropy.

11.1 Heat Engines Think for a moment about what the internal-combustion engine in your car does. We have said that both steam engines and gasoline engines are heat engines. What does this mean? We know that fuel is burned in the engine and that work is done to move the car. Somehow, work is generated from heat released in burning the fuel. Can we develop a model that describes the basic features of all heat engines? v

What does a heat engine do? Let’s sketch a description of what the gasoline engine in your car does: Fuel in the form of gasoline is mixed with air and introduced into a can-shaped chamber in the engine called a cylinder. A spark produced by a spark plug ignites the gas-air mixture, which burns rapidly (fig. 11.2). Heat is released from the fuel as it burns, which causes the gases in the cylinder to expand, doing work on the piston. Through mechanical connections, work done on the pistons is transferred to the drive shaft and to the wheels of the car. The wheels push against the road surface, and by Newton’s third law, the road exerts a force on the tires that does the work to move the car. Not all of the heat obtained from burning fuel is converted to work done in moving the car. The exhaust gases

Spark plug

figure

11.2 Heat released by burning gasoline in the cylinder of an automobile engine causes the piston to move, converting some of the heat to work.

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emerging from the tailpipe of your car are hot, so some heat is released into the environment. Unused heat is a general feature of the operation of a heat engine. This general description shows you the features common to all heat engines: Thermal energy (heat) is introduced into the engine. Some of this energy is converted to mechanical work. Some heat is released into the environment at a temperature lower than the input temperature. Figure 11.3 presents these ideas schematically. The circle represents the heat engine. The box at the top is the hightemperature source of heat, and the box at the bottom is the lower-temperature environment into which the waste heat is released.

Efficiency of a heat engine How much useful mechanical work can an engine produce for a given energy input of heat? It is important to know how productive or efficient an engine is. Efficiency is defined as the ratio of the net work done by the engine to the amount of heat that must be supplied to accomplish this work. In symbols, W , e QH

example box 11.1 Sample Exercise: How Efficient Is This Engine? A heat engine takes in 1200 J of heat from the hightemperature heat source in each cycle and does 400 J of work in each cycle. a. What is the efficiency of this engine? b. How much heat is released into the environment in each cycle? W a. QH 1200 J e QH W 400 J 400 J e ? 1200 J

1 0.33 3

33% b. QC ?

W QH QC so QC QH W 1200 J 400 J 800 J

where e is the efficiency, W is the net work done by the engine, and QH is the quantity of heat taken in by the engine

TH QH

W

TC QC

figure

11.3 A schematic representation of a heat engine. Heat is taken in at high temperatures, TH . Some heat is converted to work, and the remainder is released at a lower temperature, TC .

from the high-temperature source or heat reservoir. The work W is positive, since it represents work done by the engine on the surroundings. (The subscripts H and C used in this and subsequent sections stand for hot and cold.) In example box 11.1, a heat engine takes in 1200 J of heat from a high-temperature source and does 400 J of work in each cycle, resulting in an efficiency of 1⁄ 3. We usually state efficiency as a decimal fraction, 0.33 or 33% in this case. An efficiency of 33% is greater than that of most automobile engines but less than that of the steam turbines powered by coal or oil used in many electric power plants. In computing this efficiency, we used heat and work values for one complete cycle. An engine usually functions in cycles where the engine repeats the same process over and over. It is necessary to use a complete cycle or an average of several complete cycles to compute efficiency, because heat and work exchanges occur at different points within the cycle.

What does the first law of thermodynamics tell us about heat engines? The first law of thermodynamics places some limits on what a heat engine can do. Since the engine returns to its initial state at the end of each cycle, its internal energy at the end of the cycle has the same value as at the beginning.

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TH QH = 1200 J

W = 400 J

TC QC = 800 J

figure

11.4 The arrow widths depict the quantities of energy in the sample exercise in example box 11.1.

reservoir. The arrow representing the released heat is twice as wide as the arrow depicting the work, and the combined width of these two arrows equals the width of the arrow of the original heat input. Automobile engines, diesel engines, jet engines, and the steam turbines used in power plants all are heat engines. You can create a simple steam turbine by placing a pinwheel in front of the spout of a tea kettle, as in figure 11.5. The input heat is supplied to the tea kettle by the stove or hot plate. Some of this heat is converted to work in turning the pinwheel. The rest of the heat is released into the room and does no work on the pinwheel, although it does warm the room slightly. If a string or thread is attached to the shaft of the pinwheel, work done on the pinwheel could be used to lift a small weight. This pinwheel steam turbine would not be powerful or efficient. Designing better engines with both high power (the rate of doing work) and high efficiency (the fraction of input heat converted to work) has been the goal of scientists and mechanical engineers for the last 200 years. Everyday phenomenon box 11.1 discusses a recent development in this effort. Discovering the factors that have an impact on efficiency is an important part of this quest.

The change in the internal energy of the engine for one complete cycle is zero. The first law states that the change in internal energy is the difference between the net heat added and the net work done by the engine (U Q W). Since the change in internal energy U is zero for one complete cycle, the net heat Q flowing into or out of the engine per cycle must equal the work W done by the engine during the cycle. Energy is conserved. Net heat is the difference between the heat input from the higher-temperature source QH and heat released to the lower-temperature (or colder) environment QC . The first law of thermodynamics shows that the net work done by the heat engine is:

W QH 0 QC 0 ,

since the work done equals the net heat.* In the sample exercise in example box 11.1, 800 J of heat are released to the environment. The width of the arrows in figure 11.4 corresponds to the quantities of heat and work involved in this example. Start with the wide arrow at the top: in each cycle, 1200 J of heat are taken in, 400 J (1⁄ 3 of the total) are converted to work, and 800 J of heat (2⁄ 3 of the total) are released to the lower-temperature

figure

*We are using the absolute value of QC , as indicated by the vertical bars. Since QC flows out of the engine, it is often considered a negative quantity. The relationship is clearer if we make the minus sign explicit.

215

11.5 Steam issuing from the kettle makes the pinwheel turn in this simple steam turbine. Work could be done to lift a small weight with such an engine.

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everyday phenomenon

box 11.1

Hybrid Automobile Engines The Situation. Automobile companies have been working for years to develop better engines and drive systems. The goals are to achieve greater efficiency and lower exhaust emissions without excessive increases in cost. Federal and state environmental regulations provide incentive. Electric cars have been one solution. Although electric motors produce essentially no exhaust emissions, electric cars suffer from limited range, weight issues associated with the storage batteries, and the need to recharge for several hours with relatively high-cost electric power. The introduction of hybrid systems that involve both an electric motor and a gasoline engine is a more recent development. How do these hybrid systems work? What are their advantages and disadvantages?

Gasoline engine

The Analysis. Although different arrangements are possible, most hybrid designs allow both the electric motor and the gasoline engine to turn the transmission, which transmits power to the drive wheels of the car. The gasoline engine is a heat engine, but the electric motor is not. The electric motor converts electrical energy stored in batteries to mechanical energy to drive the car. Either or both engines can power the car depending upon conditions. A sophisticated powersplitting transmission is needed to direct the energy flows. In city-driving conditions, the electric motor is used to power the car when starting from a complete stop or accelerating at low speeds. This avoids the exhaust emissions associated with using a gasoline engine at speeds at which it is not very efficient. It also avoids the need for a large gasoline engine that

Gasoline tank

Generator Electric motor

Power-splitting transmission

Battery pack

The gasoline engine and the electric motor can both drive the wheels of a hybrid car. The gasoline tank and battery pack represent different ways of storing energy.

(continued)

A heat engine takes in heat from a high-temperature source, converts part of this heat into work, and releases the remaining heat to the environment at a lower temperature. Gasoline engines in cars, jet engines, rocket engines, and steam turbines are all heat engines. The efficiency of an engine is the ratio of work done by the engine to the amount of heat taken in from the high-temperature source. Since the change in internal energy is zero for a complete cycle, the first law of thermodynamics requires that the work done in one cycle equal the net heat flow into and out of the engine.

11.2 The Second Law of Thermodynamics If the efficiency of a typical automobile engine is less than 30%, we seem to be wasting a lot of energy. What is the best efficiency that we could achieve? What factors determine efficiency? These questions are as important today for automotive engineers or designers of modern power plants as they were for the early designers of steam engines.

What is a Carnot engine? One of the earliest scientists to be intrigued by these questions was a young French engineer named Sadi Carnot

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would otherwise be required for good acceleration. The gasoline engine can even be turned off in these situations. In highway driving, the gasoline engine provides the primary power with the electric motor kicking in when extra acceleration is needed. At times when the full power of the gasoline engine is not needed to drive the car, it is also used to drive a generator to produce electricity to charge the batteries. (See chapter 14 for a discussion of electric motors and generators.) Electric power can also be generated by running the electric motor in reverse when the car is going downhill or braking for a stop. In this way we recapture and store energy in the batteries that would otherwise be lost as low-grade heat. The flow chart shows the different possible directions of energy flow in a hybrid vehicle.

Battery pack

Generator

Transmission Gasoline engine

Electric motor

Power-splitting transmission

Flow chart showing the possible directions of energy flow in a hybrid vehicle. During braking, energy flows from the wheels to the battery pack via the electric motor.

(1796–1832). Carnot was inspired by the work of his father, also an engineer, who had studied and written about the design of water wheels, an important source of mechanical power at the time. The workings of a water wheel provided Carnot with the basis for modeling an ideal heat engine. Carnot’s father realized that bringing in all of the water at the highest point and releasing it at the lowest point maximized the efficiency of the water wheel. Carnot reasoned that the greatest efficiency of a heat engine would be obtained by taking in all of the input heat at a single high temperature and releasing all of the unused heat at a single low temperature.

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The advantages of the hybrid system are: 1. Because we are not using the gasoline engine for the peak power needed for acceleration, we can get by with a small gasoline engine. The smaller the engine, the greater the fuel economy. 2. The batteries of the hybrid car are charged by the gasoline engine and by recapturing energy in the braking process. They do not need to be recharged overnight with high-cost electric power as the batteries for an all-electric car must be. Also, we do not need as large a battery pack for the hybrid vehicle as that required for decent range in an all-electric car. 3. Because it is used mainly for highway driving and running the generator, the gasoline engine can operate at its most efficient speed. This leads to lower exhaust emissions, which are greatest at low speeds and high accelerations. The disadvantages of hybrid systems lie primarily in the extra cost of having both a gasoline engine and a relatively large electric motor, as well as a sophisticated transmission for directing the power from these two sources. (Standard gasoline-engine automobiles require an electric motor as the starter, but the electric motor used in hybrid cars must have higher power and greater size.) The hybrid car also has an expensive battery pack that ultimately will need to be replaced. These disadvantages are reflected in higher cost for the hybrid vehicle than that of a standard gasoline vehicle of similar size and features. Despite this, hybrid cars have gained quick acceptance from people who are aware of the environmental benefits of better fuel economy and lower emissions. More stringent environmental regulations will make hybrid vehicles an even more attractive option.

This would maximize the effective temperature difference within the limits of the temperatures of the heat source and the environment. Carnot imposed another requirement on his ideal engine: all of the processes had to occur without undue turbulence or departure from equilibrium. This requirement also paralleled his father’s ideas on water wheels. In an ideal heat engine, the working fluid of the engine (steam, or whatever else might be used) should be roughly in equilibrium at all points in the cycle. This condition means that the engine is completely reversible—it can be turned around and run the other way at any point in the cycle, because it

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is always near equilibrium. This was Carnot’s ideal—a real engine might depart considerably from these conditions. When Carnot published his paper on the ideal heat engine in 1824, the energy aspects of heat and the first law of thermodynamics were not yet understood. Carnot pictured caloric flowing through his engine just as water flowed through a water wheel. It was not until after the development of the first law around 1850 that the full impact of Carnot’s ideas became apparent. It then became clear that heat does not simply flow through the engine. Some of the heat is converted to mechanical work done by the engine.

What are the steps in a Carnot cycle? The cycle devised by Carnot for an ideal heat engine is illustrated in figure 11.6. Imagine a gas or some other fluid contained in a cylinder with a movable piston. In step 1 of the cycle (the energy input), heat flows into the cylinder at a single high temperature TH . The gas or fluid expands isothermally (at constant temperature) during this process and does work on the piston. In step 2, the fluid continues to expand, but no heat is allowed to flow between the cylinder and its surroundings. This expansion is adiabatic (with no heat flow). Step 3 is an isothermal compression—work is done by the piston on the fluid to compress the fluid. This is the exhaust step, since heat QC flows out of the fluid at a single low temperature TC during this compression. The final step, step 4, returns the fluid to its initial condition by an additional compression done adiabatically. All four steps must be done slowly so that the fluid is in approximate equilibrium at all times. The complete process is then reversible, a crucial feature of the Carnot engine. When the fluid is expanding in steps 1 and 2, it is doing positive work on the piston that can be transmitted by mechanical links for another use. In steps 3 and 4, the fluid is being compressed, which requires that work be done on the engine by external forces. The work added in steps 3

and 4 is less than the amount done by the engine in steps 1 and 2, however, so that the engine does a net amount of work on the surroundings.

What is the efficiency of a Carnot engine? The first law of thermodynamics made it possible to compute the efficiency of the Carnot cycle by assuming that the working fluid was an ideal gas. The process involves computing the work done on or by the gas in each step and the quantities of heat that must be added and removed in steps 1 and 3. Using the definition of efficiency from section 11.1, we obtain TH TC

ec

TH

,

where ec is the Carnot efficiency and TH and TC are the absolute temperatures (for example, temperatures in Kelvin) at which heat is taken in and released. Example box 11.2 illustrates these ideas. For the temperatures given, a Carnot efficiency of approximately 42% is obtained. According to Carnot’s ideas, this would be the maximum efficiency possible for any engine operating between these two temperatures. Any real engine operating between these same two temperatures has a somewhat lower efficiency because it is impossible to run a real engine in the completely reversible manner required for a Carnot engine.

The second law of thermodynamics The absolute temperature scale was developed in the 1850s by Lord Kelvin in England. Kelvin was aware of Joule’s experiments and had a hand in the statement of the first law of thermodynamics. He was in an excellent position to take a fresh look at Carnot’s ideas on heat engines. By combining Carnot’s ideas with the new recognition that heat flow is a transfer of energy, Kelvin put forth a

1

2

3

4

Isothermal expansion

Adiabatic expansion

Isothermal compression

Adiabatic compression

v v v TC

TH

QH

figure

v

QC

11.6 The Carnot cycle. Step 1: isothermal expansion; step 2: adiabatic expansion; step 3: isothermal compression; step 4: adiabatic compression.

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11.2 The Second Law of Thermodynamics

example box 11.2

TH QH

Sample Exercise: Carnot Efficiency A steam turbine takes in steam at a temperature of 400°C and releases steam to the condenser at a temperature of 120°C. a. What is the Carnot efficiency for this engine? b. If the turbine takes in 500 kJ of heat in each cycle, what is the maximum amount of work that could be generated by the turbine in each cycle? a. TH 400°C

ec

673 K TC 120°C e ?

TH TC 673 K 393 K 673 K

280 K 673 K

0.416 (41.6%) b. QH 500 kJ W ?

e

W , so W eQH QH (0.416)(500 kJ) 208 kJ

general principle or law of nature. This principle, which we now call the second law of thermodynamics, is usually stated as No engine, working in a continuous cycle, can take heat from a reservoir at a single temperature and convert that heat completely to work.

In other words, it is not possible for any heat engine to have an efficiency of 1 or 100%. Using the second law of thermodynamics, we can also show that no engine operating between two given temperatures can have a greater efficiency than a Carnot engine operating between the same two temperatures. This proof, which involves the reversible feature of the Carnot cycle, justifies Carnot’s contention that the Carnot engine is the best that can be achieved. In fact, if some engine did have an efficiency greater than the Carnot efficiency, the second law of thermodynamics would be violated. This argument is illustrated in figure 11.7. Here is how the argument goes. An engine with an efficiency greater than the Carnot engine would produce a greater amount of work than the Carnot engine for the same amount of heat input QH. Some of this work could be used to run the Carnot engine in reverse, returning the heat released by the first engine to the higher-temperature reservoir. To run the engine in reverse, we reverse the directions

Wexcess

Carnot engine

TH

393 K

WCarnot

QC

QC

More efficient engine

TC

figure

11.7 A diagram showing that if some engine had a greater efficiency than a Carnot engine operating between the same two temperatures, the second law of thermodynamics would be violated. of the arrows without changing the quantities of heat and work. The remaining work (Wexcess in figure 11.7) would be available for external use, and no heat would end up in the lower-temperature reservoir. The two engines operating in tandem would take a small quantity of heat from the higher-temperature reservoir and convert it completely to work. This violates the second law of thermodynamics. Therefore, no engine can have a greater efficiency than a Carnot engine operating between these two reservoirs. The Carnot efficiency is thus the maximum possible efficiency for any heat engine operating between these two temperatures. Since the proof follows from the second law of thermodynamics, the Carnot efficiency is sometimes referred to as the second-law efficiency. The second law of thermodynamics cannot be proved. It is a law of nature, which, as near as we know, cannot be violated. It sets a limit on what can be achieved with heat energy. Time has shown it to be an accurate statement. The second law is consistent with what we know about heat transfer, heat engines, refrigerators, and many other phenomena. Sadi Carnot developed the concept of a completely reversible, ideal heat engine. The Carnot engine takes in all of its heat at a single high temperature and releases the unused heat at a single low temperature; its efficiency depends on this temperature difference. The second law of thermodynamics as stated by Lord Kelvin says that no continuously operating engine can take in heat at a single

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temperature and convert that heat completely to work. Relying on the second law, we can show that no engine can have a greater efficiency than a Carnot engine operating between the same two temperatures.

11.3 Refrigerators, Heat Pumps, and Entropy

QC

To a teenager, the automobile and the refrigerator just may be the two most important inventions. The gasoline engine in an automobile is a heat engine. What is a refrigerator? In section 11.2, we talked about running a heat engine in reverse, with the result that work was used to pump heat from a colder to a hotter reservoir. Is this what a refrigerator does? Is there a relationship between heat engines and refrigerators?

QH

What do refrigerators and heat pumps do? The term refrigerator requires little explanation. We are all familiar with refrigerators, even if we do not understand how they function. A refrigerator keeps food cold by pumping heat out of the cooler interior of the refrigerator into the warmer room (fig. 11.8). An electric motor or gas-powered engine does the necessary work. A refrigerator also warms the room, as you can tell by holding your hand near the coils on the back of the refrigerator when it is running. Figure 11.9 shows a diagram of a heat engine run in reverse: It is the same diagram used for heat engines, but the directions of the arrows showing the flow of energy have been reversed. Work W is done on the engine, heat QC is removed from the lower-temperature reservoir, and a greater quantity of heat QH is released to the higher-temperature reservoir. A device that moves heat from a cooler reservoir to a warmer reservoir by means of work supplied from some external source is called a heat pump or a refrigerator. The first law of thermodynamics requires that, for a complete cycle, the heat released at the higher temperature must equal the energy put into the engine in the form of both heat and work. As before, the engine returns to its initial condition at the end of each cycle—the internal energy of the engine does not change. More heat is released at the higher temperature than was taken in at the lower temperature (fig. 11.9). For example, if 200 J of work are used to move 300 J of heat from the lower-temperature reservoir, then 500 J of heat would be delivered to the highertemperature reservoir. While a refrigerator is a heat pump, the term heat pump usually refers to a device that heats a building by pumping heat from the colder outdoors to the warmer interior (fig. 11.10). An electric motor does the work needed to run the pump. The amount of heat energy available to heat the house is greater than the work supplied because QH is equal in magnitude to the sum of the work W and the heat QC removed from the outside air. We can get a larger amount of

figure

11.8 A refrigerator pumps heat from the cooler interior of the refrigerator to the warmer room. The heat exchange coils that release heat to the room are usually on the back side of the refrigerator.

TH QH

W

TC QC

figure

11.9 A diagram of a heat engine run in reverse. In reverse, the heat engine becomes a heat pump or refrigerator.

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A heat pump can deliver a quantity of heat to the inside of the building that is often two to three times the amount of electric energy supplied as work. Again, the first law of thermodynamics is not violated, since the extra energy is being supplied from the outside air. The work applied to the heat pump allows us to move thermal energy in the direction opposite to its natural tendency, much as a water pump moves water uphill.

The Clausius statement of the second law of thermodynamics QC

QH

Heat normally flows from hotter objects to colder objects. This natural tendency is the basis for another statement of the second law of thermodynamics. Often called the Clausius statement after its originator, Rudolf Clausius (1822–1888), it takes the form:

figure 11.10

A heat pump removes heat from the outside air and pumps it into the warmer house.

figure 11.11

The exterior heat-exchange coils for an air-toair heat pump are usually located behind the house or building.

heat from the heat pump than by converting the electrical energy directly to heat, as is done in an electric furnace. Two sets of coils are used for heat exchange in the heat pump—one sits outside (fig. 11.11) and removes heat from the outside air, while the other releases heat to the air inside the building. Many heat pumps are designed to pump heat in either direction and can be used as air conditioners in the summer. (Most air conditioners are heat pumps that pump heat from inside a building to the warmer exterior.) Heat pumps are most effective for heating houses in climates where the winters are mild and the difference between the outside temperature and inside temperature is not too large.

Heat will not flow from a colder body to a hotter body unless some other process is also involved.

In the case of a heat pump, the other process is the work used to pump the heat against its usual direction of flow. Although this statement of the second law sounds very different from the Kelvin statement, they both express the same fundamental law of nature. They both place limits on what can be done with heat, and the limits in each statement are equivalent. This can be shown by an argument similar to the one that confirmed that no engine can have a greater efficiency than a Carnot engine operating between the same two temperatures. Figure 11.12 illustrates the argument. If it were possible for heat to flow from the colder to the hotter reservoir without any work, heat released by the heat engine on the right side of the diagram could flow back to the hotter reservoir. Some heat could then be removed from the hotter reservoir and converted completely to work. Heat would not be added to the cooler reservoir. This result violates the Kelvin statement of the second law of thermodynamics, which says that taking heat from a reservoir at a single temperature and converting it completely to work is impossible. A violation of the Clausius statement of the second law is therefore a violation of the Kelvin statement. A similar argument will show that a violation of the Kelvin statement is a violation of the Clausius statement. (You may want to try your hand at developing this argument.) These two statements express the same fundamental law of nature in two different ways.

What is entropy? Is there something inherent in heat that leads to the limitations described in the two statements of the second law of thermodynamics? Both statements of the second law describe processes that do not violate the first law of thermodynamics

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TH

TH

TH

Q

QH

QH

W W

QC

QC

TC TC

figure 11.12

If we assume that heat can flow spontaneously from a colder to a hotter reservoir, the Kelvin statement of the second law of thermodynamics is violated.

(conservation of energy)—but apparently are not possible. Certain things cannot be done with heat energy. Suppose that we had a quantity of heat in a hightemperature reservoir (perhaps just a container of hot water). We could imagine two different ways to remove heat from the source. First, we could simply let the heat flow through a good heat conductor into the cooler environment, as shown in figure 11.13a. Second, we could use this heat to run a heat engine and do some useful work (fig. 11.13b). If we were to use a Carnot engine, the process is completely reversible and we have obtained the maximum possible work from the available heat. The first process, heat simply flowing through a conductor to the lower-temperature reservoir, is irreversible. The system is not in equilibrium while this heat flow takes place, and the energy is not converted to useful work. In the irreversible process, we lose some ability to do useful work. Entropy is the quantity that describes the extent of this loss. As entropy increases, we lose the ability to do work. Entropy is sometimes defined as a measure of the disorder of the system. The entropy of a system increases any time the disorder or randomness of the system increases. A system organized into two reservoirs at two distinct temperatures is, in this sense, more organized than having the energy all at a single intermediate temperature. If we use the heat available in the hotter reservoir to run a completely reversible Carnot engine, there is no increase in the entropy of the system and its surroundings. We get the maximum useful work from the available energy. In isolated systems (and in the universe as well), entropy remains

TC a.

QC b.

figure 11.13

Heat can be removed from a hightemperature source either by direct flow through a conducting material or by being used to run a heat engine.

constant in reversible processes but increases in irreversible processes where conditions are not in equilibrium. The entropy of a system decreases only if it interacts with some other system whose entropy is increased in the process. (This happens, for example, in the growth and development of biological organisms.) The entropy of the universe never decreases. This statement is yet another version of the second law of thermodynamics: The entropy of the universe or of an isolated system can only increase or remain constant. Its entropy cannot decrease.

The randomness of heat energy is responsible for the limitations in the second law of thermodynamics. The entropy of the universe would decrease if heat could flow by itself from a colder to hotter body—but the Clausius statement of the second law says that this cannot happen. Likewise, the entropy of the universe would decrease if heat at a single temperature could be converted completely to work, violating the Kelvin statement of the second law. The thermal energy of a gas consists of the kinetic energy of the molecules. The velocities of these molecules are randomly directed, however, as in figure 11.14. Only some of them move in the proper direction to push the piston to produce work. If the molecules all moved in the same direction, we could convert their kinetic energy completely to work. This would be a lower-entropy (more organized) condition, but it does not represent the normal condition

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What bearing do the laws of thermodynamics and the efficiency of heat engines have on these questions? v

figure 11.14

The random directions of gas molecules prevent us from completely converting their kinetic energy to useful mechanical work.

for the thermal energy in a gas. The basic disorganization of heat energy is responsible for the limitations encompassed by the three statements of the second law. Like internal energy, entropy can be computed for any state of a given system. The notion of disorder or randomness is at the heart of the matter. The state of my office or of a student’s residence-hall room tends to become more disordered. This natural tendency of entropy to increase can only be countered by introducing energy in the form of work to straighten things up again. A heat pump or refrigerator is a heat engine run in reverse: work is supplied to pump heat from a colder to a hotter body. The Clausius statement of the second law of thermodynamics, which is equivalent to the Kelvin statement, says that heat normally flows from a hotter to a colder body and cannot flow in the opposite direction without some other process being involved. The limitations on the use of heat expressed in the second law come about because of the disorganized nature of heat energy. Entropy is a measure of the disorder of a system. The entropy of the universe can only increase.

11.4 Thermal Power Plants and Energy Resources We use electric power so routinely that most of us do not stop to think about where that energy comes from. Concerns about the greenhouse effect (see everyday phenomenon box 10.1) and other environmental issues have recently pushed questions about how we generate electric power back into public debate. How is electric power generated in your area? If hydroelectric power is not a major contributor, most of your power comes from thermal power sources that use heat engines. Thermodynamics plays an extremely important role in any discussion of the use of energy. What are the most efficient ways of using energy resources such as coal, oil, natural gas, nuclear energy, solar energy, or geothermal energy?

How does a thermal power plant work? The most common way of producing electric power in this country is a thermal power plant fueled by coal, oil, or natural gas (fossil fuels). The heart of such a plant is a heat engine. The fuel is burned to release heat that causes the temperature of the working fluid (usually water and steam) to increase. Hot steam is run through a turbine (fig. 11.15) that turns a shaft connected to an electric generator. Electricity is then transmitted through power lines to consumers (such as homes, offices, and factories). Because the steam turbine is a heat engine, its efficiency is inherently limited by the second law of thermodynamics to the maximum given by the Carnot efficiency. The efficiency of a real heat engine is always less than this limit because it falls short of the ideal conditions of a Carnot engine. In any real engine, there are always irreversible processes taking place. A steam turbine generally comes closer to the ideal, however, than the internal-combustion engines used in automobiles. Rapid burning of gasoline-and-air mixtures in an automobile engine are highly turbulent and irreversible processes. Since the maximum possible efficiency is dictated by the temperature difference between the hot and cold reservoirs, heating the steam to as high a temperature as the materials will permit is advantageous. The materials the boiler and turbine are made from set an upper limit on the temperatures that can be tolerated. If these materials begin to soften or melt, the equipment will obviously deteriorate. For most steam turbines, the upper temperature limit is around 600°C, well below the melting point of steel. In practice, most turbines operate at temperatures below this limit—550°C is typical. If a steam turbine is operating between an input temperature of 600°C (873 K) and an exhaust temperature near the boiling point of water (100°C or 373 K) where the steam condenses to water, we can compute the maximum possible efficiency for this turbine: ec

TH TC TH

.

The difference between these two temperatures is 500 K. Dividing this by the input temperature of 873 K yields a Carnot efficiency of 0.57 or 57%. This is the ideal efficiency. The actual efficiency will be somewhat lower and usually runs between 40% and 50% for modern coal- or oil-fired power plants. At best, we can convert about half of the thermal energy released in burning coal or oil to mechanical work or electrical energy. The rest must be released into the environment at temperatures too low for running heat engines or most other functions except space heating. The exhaust side of the turbines must be cooled to achieve maximum

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Steam Boiler

Turbine

QH

Generator

Condenser

Fuel Water

QC

figure 11.15

A diagram showing the basic components of a thermal-electric power plant. Heat from the boiler generates steam, which turns the steam turbine. The turbine does work to generate electric power in the generator.

efficiency, and the cooling water is either returned to some body of water or run through cooling towers, where the heat is dissipated into the atmosphere (fig. 11.16). This is the waste heat often referred to in discussions of the environmental impact of power plants. If the waste heat is dumped into a river, the temperature of the river will rise, with possible effects on the fish and other wildlife.

Alternatives to fossil fuels Nuclear power plants, discussed in chapter 19, also generate heat to run steam turbines. Because of the effects of radiation on materials, however, it is not feasible to run the turbines in a nuclear plant at temperatures as high as those in fossil-fuel plants. The thermal efficiencies for nuclear plants are somewhat lower, typically between 30% and 40%. For equal amounts of energy generated, the amount of heat released into the environment is somewhat larger for a nuclear plant than for a fossil-fuel plant. On the other hand, nuclear power plants do not release carbon dioxide and other exhaust gases into the atmosphere and do not contribute to the greenhouse effect. Processing and disposal of nuclear wastes and concerns about nuclear accidents continue to be significant issues. Heat is also available from other sources such as geothermal energy, which is heat that comes from the interior of the Earth. Hot springs and geysers indicate the presence of hot water near the surface of the Earth that might be used for power production, but usually the temperature of the water is not much greater than 200°C. In places where steam is available from geysers, low-temperature steam turbines can be run, as in northern California at the Geysers power plant (fig. 11.17).

figure 11.16

The cooling towers that are a common feature of many thermal power plants transfer heat from the cooling water to the atmosphere.

If the water temperature is below 200°C, steam turbines are not effective, and some other fluid with a lower boiling temperature than water is preferable for running a heat engine. Isobutane has been studied as a possible working fluid for low-temperature heat engines. The efficiency of such an engine would be quite low, however. If water were available at a temperature of 150°C (423 K), for example, and cooling water was available at 20°C (293 K), the Carnot efficiency would be 31%. In practice, the efficiency will be

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figure 11.17

The Geysers power plant in California uses geothermal energy to run steam turbines.

even less, so it is common to find efficiencies of only 20% to 25% in proposed geothermal power plants. Warm ocean currents are yet another source of heat. Prototypes have been developed of ocean power plants that take advantage of the temperature difference between warm water at the surface of an ocean and cooler water drawn from greater depths. Example box 11.3 illustrates a possible scenario. Although the efficiency is low (6.7% in this case), so is the cost of water warmed by the sun. It still may be economically feasible to produce power in this manner. The sun is an energy source with enormous potential for development if the costs become competitive. The temperatures that can be achieved with solar power depend on the type of collection system used. The ordinary flat-plate collector only achieves relatively low temperatures of 50°C to

example box 11.3 Sample Exercise: What is the efficiency of an ocean power plant? As described in the text, a power plant can take advantage of the temperature difference between warm water on the surface of the ocean and the cooler water found at greater depths. A typical surface temperature in the tropics is 25°C, and deep water can be 5°C. What is the Cannot efficiency for a heat engine with this temperature difference? TH TC TH 25°C e TH 298 K 298 K 278 K TC 5°C 298 K 278 K 20 K e ? 298 K 0.067 (6.7%)

225

figure 11.18

At a solar thermal power plant near Seville, Spain, high temperatures are achieved by focusing sunlight from a large array of mirrors onto a boiler located at the top of the tower.

100°C, but concentrating collectors that use mirrors or lenses to focus the sunlight provide much higher temperatures. Figure 11.18 shows a solar power plant in Spain, where an array of mirrors focuses sunlight on a boiler on a central tower. The temperatures generated are comparable to those of fossil-fuel plants, so similar steam turbines can be used.

High-grade and low-grade heat The temperature makes a big difference in how much useful work can be extracted from heat. The second law of thermodynamics and the related Carnot efficiency define the limits. What effects do these factors have on our national energy policies and our day-to-day use of energy? Clearly, heat at temperatures around 500°C or higher is much more useful for running heat engines and producing mechanical work or electrical energy than heat at lower temperatures. Heat at these high temperatures is sometimes called high-grade heat because of its potential for producing work. Even then, only 50% or less of the heat can actually be converted to work. Heat at lower temperatures can produce work but with considerably lower efficiency. Heat at temperatures around 100°C or lower is generally called low-grade heat. Lowgrade heat is better used for purposes like heating homes or buildings (space heating). Space or water heating are the optimal uses for heat collected by flat-plate solar collectors, and even from geothermal sources, if they are near enough to buildings that need heat. Geothermal heat is used for space heating in Klamath Falls, Oregon, as well as in certain other parts of the world where conditions are favorable. Much low-grade heat, such as the low-temperature heat released from power plants, goes to waste simply because it is not economical to transport it to places where space heating may be needed. Nuclear power plants, for example, are not usually built in populated areas. There are other possible uses of low-grade heat, such as in agriculture or aquaculture, but we have not gone far in developing such uses.

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The main advantage of electrical energy is that it can be easily transported through power lines to users far from the point of generation. Electrical energy can also be readily converted to mechanical work by electric motors. Electric motors operate at efficiencies of 90% or greater because they are not heat engines. The efficiency involved in producing the electrical power in the first place, of course, may have been considerably lower. Electrical energy can also be converted back to heat if desired. Electricity is used for space heating in regions like the Pacific Northwest, where electric power has been relatively cheap because of the extensive hydroelectric resources on the Columbia River and its tributaries. The development of these resources has been subsidized by the federal government. Low prices encourage the use of a high-grade form of energy for purposes that might be just as well served by lower-grade sources. Energy has been relatively cheap in this country and in many other parts of the world. Continued economic development and depletion of fossil-fuel resources will gradually change this picture. As scarcity occurs, questions of the optimal uses of energy resources will become critical. Wise decisions will depend on the participation of an informed, scientifically literate citizenry.

Thermal power plants use heat engines—steam turbines— to generate electric power. These engines are limited by the Carnot efficiency, which depends on input and output temperatures. The best that can be achieved is the 50% or so efficiency of fossil-fuel plants. The temperatures that can be obtained with other energy resources dictate the best way of using these resources. Lower temperatures are more suitable for space heating and similar uses than for power generation. The laws of thermodynamics set limits on power generation.

Perpetual-motion machines of the first kind A proposed engine or machine that would violate the first law of thermodynamics is called a perpetual-motion machine of the first kind. Since the first law of thermodynamics involves conservation of energy, a perpetual-motion machine of the first kind is one that puts out more energy as work or heat than it takes in. If the machine or engine is operating in a continuous cycle, the internal energy must return to its initial value, and the energy output of the engine must equal its energy input, as we have already seen. In figure 11.19, the total magnitude of the work and heat output is greater than the magnitude of the heat input (as represented by the width of the arrows). This could only happen if there was some source of energy, such as a battery, within the engine itself. If this were so, the energy in the battery would gradually be depleted, and the internal energy of the engine would decrease. The engine could not run indefinitely. The fact that physicists reject such an engine as impossible does not keep inventors from proposing them. From time to time, we see claims reported in newspapers and other popular media of engines that can run indefinitely fueled only by a gallon of water or a minuscule quantity of gasoline. Given the sometimes high price of gasoline, such claims have an obvious appeal and often attract investors

TH QH

11.5 Perpetual Motion and Energy Frauds The idea of perpetual motion has long fascinated inventors. The lure of inventing an engine that could run without fuel, or from some plentiful source like water, is like finding gold. If such an engine could be developed and patented, the inventor would become even richer than if he or she had discovered gold. Is such an engine possible? The laws of thermodynamics impose some limits. Since these laws are consistent with everything that physicists know about energy and engines, any claim that violates the laws of thermodynamics should be suspect. We can analyze claims of perpetual motion or miracle engines by testing to see if they violate either the first or second laws of thermodynamics.

W

TC QC

figure 11.19

A perpetual-motion machine of the first kind. The energy output exceeds the input and therefore violates the first law of thermodynamics.

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TH QH

W

TC

figure 11.20

A perpetual-motion machine of the second kind. Heat is extracted from a reservoir at a single temperature and converted completely to work, thus violating the second law of thermodynamics.

and other interest. With your knowledge of physics, though, you could ask the inventor some simple questions: Where is the energy coming from? How can the machine put more energy out than went in? Hang on to your wallet—it looks like a poor investment.

What is a perpetual-motion machine of the second kind? Engines that would violate the second law of thermodynamics without violating the first law often are a little more subtle. Their inventors have learned how to answer the questions in the previous paragraph. They have a source of energy—perhaps they plan to extract heat from the atmosphere or the ocean. These claims must be evaluated in terms of the second law of thermodynamics. If the second law is violated, the inventor has proposed a perpetualmotion machine of the second kind (fig. 11.20). The second law states that it is not possible to take heat from a reservoir at a single temperature and convert it completely to work. We must have a lower-temperature reservoir available, and some heat must be released into that reservoir. In addition, even if we have a lower-temperature reservoir, the efficiency of any engine running between the two reservoirs will be quite low if the temperature difference between the reservoirs is not great. Any claim of

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efficiency greater than the Carnot efficiency (for the available temperature difference) also violates the second law of thermodynamics. The laws of thermodynamics are useful in evaluating an inventor’s claims, and they also guide our attempts to build better engines and to use energy more efficiently. Better engines can and will be developed, and their inventors may indeed be enriched in the process. A host of possibilities exist that involve relatively low-grade energy sources or other special circumstances. Engineers and scientists are also working on developing special materials and innovative designs for engines that might use even higher temperatures than are common in today’s fossil-fuel plants. These efforts do not violate the laws of thermodynamics. Most physics departments receive inquiries from local inventors seeking endorsement or help with their schemes for producing or using energy. The inventors are often sincere and sometimes quite well informed. Occasionally their ideas have merit, although they are often based on misunderstandings of the laws of thermodynamics (see everyday phenomenon box 11.2). Unfortunately, it is often hard to persuade inventors that their ideas will not work, even when they involve clear violations of the laws of thermodynamics. Other cases are more clearly the work of charlatans. Inventors who begin with good intentions sometimes discover that their ideas attract money from eager investors even when the invention fails to work. Some inventors have managed to raise several million dollars for the design and testing of prototype engines. Somehow, the engines are never quite finished, or tests are inconclusive and more money is needed to continue the work. The inventors, in the meantime, live quite well and find that promoting their inventions is more lucrative than actually building and testing them. Investors beware. We know of no circumstances that violate the laws of thermodynamics. The failure of repeated attempts to violate them reinforces our belief in their validity. The laws cannot be proved, but their proven ability to accurately describe experimental results gives physicists confidence that they will continue to be useful indicators of what is possible. A perpetual-motion machine of the first kind violates the first law of thermodynamics because more work is obtained than the energy input. A perpetual-motion machine of the second kind violates the second law of thermodynamics either by converting heat completely to work or by claiming an efficiency greater than the Carnot efficiency. Physicists do not believe that either of these options is possible, but inventors keep trying and investors keep wasting money on such schemes. The laws of thermodynamics place limits on what we can do and guide our attempts to build better engines.

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everyday phenomenon

box 11.2

A Productive Pond The Situation. A local farmer consulted with the author about an idea he had for generating electricity on his farm. The farmer had a pond that he thought could be used to run a water wheel, which could power an electric generator. He brought a sketch that looked something like the drawing here. The sketch showed a drain pipe at the bottom of the pond. The farmer was aware that water would flow down such a drain with considerable velocity. His plan was to direct water through a pipe to the side of the pond and then up above the pond level, where it would flow onto a water wheel, powering the generator. The water would return to the pond after leaving the water wheel, so there would be no need to replenish the water supply, except to replace water lost to evaporation or seepage. How would you advise the farmer? Will this plan work? Does it represent a perpetual-motion machine, and, if so, of what kind?

Water turbine

Generator

A sketch of the farmer’s plan for obtaining electrical power from his pond. Which law of thermodynamics does he propose to violate?

The Analysis. Just by looking at the overall result, we see that work is being obtained to turn the generator without the input of any energy in the form of heat or work. Because the pond returns to its initial state (with the same internal energy), the first law of thermodynamics (or the principle of conservation of energy) is violated by the proposal. The farmer’s plan is a perpetual-motion machine of the first kind. If we look at the mechanics in more detail, we see that the water will indeed gain kinetic energy as it flows down the drain. This gain in kinetic energy comes at the expense of a loss in potential energy as the pond level is lowered. Directing the water upward causes it to regain potential energy at the expense of kinetic energy: it will slow down. If there are no losses due to friction, its velocity should reach zero at the point where it reaches the original level of the pond. In an initial surge when the valve is opened, the water might overshoot the original pond level, but water cannot be raised above this level in a continuous process. Eventually the water in the vertical pipe will reach the same level as the pond, and no water will flow. The proposal will not work. Although the author explained these ideas to the farmer carefully, and the farmer was an educated and intelligent person, the farmer still was not persuaded that his idea would not work. Because the farmer was not convinced by the theoretical arguments, the author encouraged him to make a small scale model before investing any money in plumbing his pond. Models or prototypes are a good way to test ideas and can sometimes be more convincing than theoretical arguments. (Whether or not the farmer actually tried out his plan was never reported to the author.)

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summary Attempts to build better steam engines led to a more general analysis of heat engines and, ultimately, to a statement of the second law of thermodynamics. The second law explains the workings of heat engines and refrigerators, and the concept of entropy. An understanding of the laws of thermodynamics is crucial to making wise energy choices and policy decisions.

thought of as a measure of the disorder of a system. This disorder is responsible for the limitations expressed in the second law. Irreversible processes always increase the entropy of the universe. TH

QH

1

Heat engines. A heat engine is any device that takes in heat from a high-temperature source and converts some of this heat to useful mechanical work. In the process, some heat is always released at lower temperatures into the environment. The efficiency of a heat engine is defined as the work output divided by the heat input. TH

W

TC

QC

QH

power plants and energy resources. 4Power Thermal plants that use heat generated from coal, oil, natural gas, W

nuclear fuels, geothermal sources, or the sun to produce electrical energy are all examples of thermal power plants. Their efficiency cannot be greater than the Carnot efficiency, so high input temperatures are desirable.

TC QC e =W — QH QH

The second law of thermodynamics. The Kelvin 2statement of the second law says that it is not possible for an engine, working in a continuous cycle, to take in heat at a single temperature and convert that heat completely to work. The maximum possible efficiency for any engine operating between two given temperatures is that of a Carnot engine. It is always less than 1 or 100%. TH

QC

Perpetual motion and energy frauds. Any pro5posed device that would violate the first law of thermodynamics is called a perpetual-motion machine of the first kind. One that would violate the second law, but not the first, is called a perpetualmotion machine of the second kind. Neither will work.

QH

Second kind

First kind W

TH

TH

QH

TH – TC ec = ––––––– TH

heat pumps, and entropy. A heat 3pump orRefrigerators, refrigerator is a heat engine run in reverse—it uses work to move heat from a low-temperature body to a hotter one. The Clausius statement of the second law says that heat will not flow spontaneously from a colder to a hotter body. Entropy can be

W

TC

QC

QH

W

TC

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key terms Low-grade heat, 225 Perpetual-motion machine of the first kind, 226 Perpetual-motion machine of the second kind, 227

Second law of thermodynamics, 219 Heat pump, 220 Entropy, 222 Thermal power plant, 223 High-grade heat, 225

Heat engine, 213 Efficiency, 214 Reversible, 217 Carnot engine, 218 Carnot efficiency, 218

Q1. Which of these types of motors or engines are heat engines? a. an automobile engine b. an electric motor c. a steam turbine Explain your reasons for classifying or not classifying each of these as a heat engine. Q2. Could a simple machine such as a lever, a pulley system, or a hydraulic jack be considered a heat engine? Explain. Q3. In applying the first law of thermodynamics to a heat engine, why is the change in the internal energy of the engine assumed to be zero? Explain. Q4. Is the total amount of heat released by a heat engine to the low-temperature reservoir in one cycle ever greater than the amount of heat taken in from the high-temperature reservoir in one cycle? Explain. Q5. From the perspective of the first law of thermodynamics, is it possible for a heat engine to have an efficiency greater than 1? Explain.

Q8. Is it possible for a heat engine to operate as shown in the following diagram? Explain, using the laws of thermodynamics. TH

QH

W

TC

QC

Q8 Diagram Q9. Is it possible for a heat engine to operate as shown in the following diagram? Explain, using the laws of thermodynamics. QH

W

Q6. Which law of thermodynamics requires the work output of the engine to equal the difference in the quantities of heat taken in and released by the engine? Explain. Q7. Is it possible for a heat engine to operate as shown in the following diagram? Explain, using the laws of thermodynamics. TH

TH

TC

Q9 Diagram QH Q10. Is it possible for the efficiency of a heat engine to equal 1? Explain. W

Q11. Can a Carnot engine operate in an irreversible manner? Explain. Q12. Does a gasoline-burning automobile engine operate in a reversible manner? Explain.

TC

Q7 Diagram

Q13. Which would have the greater efficiency—a Carnot engine operating between the temperatures of 400°C and 300°C, or one operating between the temperatures of 400 K and 300 K? Explain.

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Questions

*Q14. If we want to increase the efficiency of a Carnot engine, would it be more effective to raise the temperature of the high-temperature reservoir by 50°C or lower the temperature of the low-temperature reservoir by 50°C? Explain. Q15. Is a heat pump the same thing as a heat engine? Explain. Q16. Is a heat pump essentially the same thing as a refrigerator? Explain. Q17. When a heat pump is used to heat a building, where does the heat come from? Explain. Q18. Is it possible to cool a closed room by leaving the door of a refrigerator open in the room? Explain. Q19. Is it ever possible to move heat from a cooler to a warmer temperature? Explain. Q20. Is it possible for a heat pump to operate as shown in the diagram? Explain, using the laws of thermodynamics. TH

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Q23. Which has the higher entropy, a deck of cards in which the cards are organized by suit or a shuffled deck of cards? Explain. Q24. A hot cup of coffee is allowed to cool down, thus warming its surroundings. Does the entropy of the universe increase in this process? Explain. *Q25. When a substance freezes, the molecules become more organized and the entropy decreases. Does this involve a violation of the entropy statement of the second law of thermodynamics? Explain. Q26. Which would normally have the greater thermal efficiency, a coal-fired power plant or a geothermal power plant? Explain. Q27. In what ways is a nuclear power plant similar to a coalfired plant? Explain. Q28. What is the distinction between high-grade heat and lowgrade heat? Explain.

QH

*Q29. Electric motors convert electric energy to mechanical work at a much higher efficiency than gasoline engines can convert heat to work. Why might it not make sense, then, to run all of our vehicles on electric power? Where does the electric power originate? Explain.

TC

Q30. Is heat obtained from a flat-plate solar collector best used for running a heat engine or for space heating? Explain. QC

Q31. Is an automobile engine a perpetual-motion machine? Explain.

Q20 Diagram Q21. Is it possible for a heat pump to operate as shown in the diagram? Explain, using the laws of thermodynamics. TH

QH

W

TC

QC

Q21 Diagram Q22. Is it possible for a heat pump to operate as shown in the diagram? Explain, using the laws of thermodynamics. TH

QH

W

TC

QC

Q22 Diagram

Q32. An engineer proposes a power plant that will extract heat from warm surface water in the ocean, convert some of it to work, and release the remaining heat into cooler water at greater depths. Is this a perpetual-motion machine and, if so, what kind? Explain. Q33. An engineer proposes a device that will extract heat from the atmosphere, convert some of it to work, and release the remaining heat back into the atmosphere at the same temperature as the input heat. Is this a perpetual-motion machine and, if so, what kind? Explain.

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exercises E1. In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-temperature reservoir, and does 400 J of work. What is its efficiency? E2. A heat engine with an efficiency of 25% does 400 J of work in each cycle. How much heat must be supplied from the high-temperature source in each cycle? E3. In one cycle, a heat engine takes in 900 J of heat from a high-temperature reservoir and releases 600 J of heat to a lower-temperature reservoir. a. How much work is done by the engine in each cycle? b. What is its efficiency? E4. A heat engine with an efficiency of 40% takes in 600 J of heat from the high-temperature reservoir in each cycle. a. How much work does the engine do in each cycle? b. How much heat is released to the low-temperature reservoir? E5. In one cycle, a heat engine does 400 J of work and releases 500 J of heat to a lower-temperature reservoir. a. How much heat does it take in from the highertemperature reservoir? b. What is the efficiency of the engine? E6. A Carnot engine takes in heat at a temperature of 650 K and releases heat to a reservoir at a temperature of 350 K. What is its efficiency? E7. A Carnot engine takes in heat from a reservoir at 400°C and releases heat to a lower-temperature reservoir at 150°C. What is its efficiency?

E9. A heat pump takes in 300 J of heat from a low-temperature reservoir in each cycle and uses 150 J of work per cycle to move the heat to a higher-temperature reservoir. How much heat is released to the higher-temperature reservoir in each cycle? E10. In each cycle of its operation, a refrigerator removes 18 J of heat from the inside of the refrigerator and releases 30 J of heat into the room. How much work per cycle is required to operate this refrigerator? E11. A typical electric refrigerator has a power rating of 400 W, which is the rate (in J/s) at which electrical energy is supplied to do the work needed to remove heat from the refrigerator. If the refrigerator releases heat to the room at a rate of 900 W, at what rate (in watts) does it remove heat from the inside of the refrigerator? E12. A typical nuclear power plant delivers heat from the reactor to the turbines at a temperature of 540°C. If the turbines release heat at a temperature of 200°C, what is the maximum possible efficiency of these turbines? E13. An ocean thermal-energy power plant takes in warm surface water at a temperature of 22°C and releases heat at 10°C to cooler water drawn from deeper in the ocean. Is it possible for this power plant to operate at an efficiency of 8%? Justify your answer. E14. An engineer designs a heat engine using flat-plate solar collectors. The collectors deliver heat at 70°C and the engine releases heat to the surroundings at 35°C. What is the maximum possible efficiency of this engine?

E8. A Carnot engine operates between temperatures of 600 K and 400 K and does 150 J of work in each cycle. a. What is its efficiency? b. How much heat does it take in from the highertemperature reservoir in each cycle?

synthesis problems SP1. Suppose that a typical automobile engine operates at an efficiency of 25%. One gallon of gasoline releases approximately 150 MJ of heat when it is burned. (A megajoule, MJ, is a million joules.) a. Of the energy available in a gallon of gas, how much energy can be used to do useful work in moving the automobile and running its accessories? b. How much heat per gallon is released to the environment in the exhaust gases and via the radiator? c. If the car is moving at constant speed, how is the work output of the engine used? d. Would you expect the efficiency of the engine to be greater on a very hot day or on a cold day? Explain. (There may be competing effects at work.)

SP2. Suppose that a certain Carnot engine operates between the temperatures of 500°C and 150°C and produces 30 J of work in each complete cycle. a. What is the efficiency of this engine? b. How much heat does it take in from the 450°C reservoir in each cycle? c. How much heat is released to the 150°C reservoir in each cycle? d. What is the change, if any, in the internal energy of the engine in each cycle?

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SP3. A Carnot engine operating in reverse as a heat pump moves heat from a cold reservoir at 5°C to a warmer one at 25°C. a. What is the efficiency of a Carnot engine operating between these two temperatures? b. If the Carnot heat pump releases 200 J of heat into the higher-temperature reservoir in each cycle, how much work must be provided in each cycle? c. How much heat is removed from the 5°C reservoir in each cycle? d. The performance of a refrigerator or heat pump is described by a “coefficient of performance” defined as K Qc /W. What is the coefficient of performance for our Carnot heat pump? e. Are the temperatures used in this example appropriate to the application of a heat pump for home heating? Explain.

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SP5. Suppose that an oil-fired power plant is designed to produce 100 MW (megawatts) of electrical power. The turbine operates between temperatures of 650°C and 240°C and has an efficiency that is 80% of the ideal Carnot efficiency for these temperatures. a. What is the Carnot efficiency for these temperatures? b. What is the efficiency of the actual oil-fired turbines? c. How many kilowatt-hours (kW·h) of electrical energy does the plant generate in 1 h? (The kilowatt-hour is an energy unit equal to 1 kW of power multiplied by 1 h.) d. How many kilowatt-hours of heat must be obtained from the oil in each hour? e. If one barrel of oil yields 1700 kW·h of heat, how much oil is used by the plant each hour?

SP4. In section 11.3, we showed that a violation of the Clausius statement of the second law of thermodynamics is a violation of the Kelvin statement. Develop an argument to show that the reverse is also true: a violation of the Kelvin statement is a violation of the Clausius statement.

home experiments and observations HE1. If there is a refrigerator handy in your residence hall or home, study the construction of the refrigerator to make these observations: a. What energy source (gas or electricity) provides the work necessary to remove the heat? Can you find a power rating somewhere on the refrigerator? b. Where are the heat-exchange coils that are used to release heat into the room? (They are usually on the back of the refrigerator, often near the bottom.) c. Use a thermometer to determine the temperatures inside the refrigerator, inside the freezer compartment, near the heat-exchange coils outside, and in the room at some distance from the refrigerator. You should take measurements at different times to see how much these temperatures vary. HE2. Find a local car dealer who sells hybrid automobiles. (Honda and Toyota are currently the best bets.) Collect literature describing the specifications for the hybrid vehicle as well as for other vehicles of similar size. (This information may also be obtained on the Internet.) a. Compare the gasoline engine size, horsepower, and estimated gas mileage for the hybrid vehicle and the standard vehicle of similar size.

b. What are the voltage and current ratings for the batteries in hybrid and standard vehicles? c. What tradeoffs do you see in purchasing the hybrid versus the standard vehicle of similar size? HE3. What types of power plants are used to generate electricity in your area? Do they employ heat engines, and if so, what is the source of heat? Are there cooling towers or other cooling structures used to release waste heat into the atmosphere? (Your local utility company should be able to provide such information.)

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Three Electricity and Magnetism

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chemical solution, were capable of producing sustained electric currents. As often happens, this new device made many new experiments and investigations possible. The invention of the battery led to the discovery of the magnetic effects of electric currents by Hans Christian Oersted (1777–1851) in 1820. Oersted’s discovery made a formal connection between electricity and magnetism, leading to the modern term electromagnetism. Electromagnetism was the hot area of research for physicists during the 1820s and 1830s, and key advances were made by Ampère, Faraday, Ohm, and Weber. In 1865, the Scottish physicist James Clerk Maxwell (1831–1879) published a comprehensive theory of electric and magnetic fields that brought together the insights of many of these other scientists. Maxwell invented the concepts of electric and magnetic fields, ideas that proved to be tremendously productive. Electromagnetism is still an active area of research. Its ramifications are important in radio and television, computers, communications, and other areas of technology. Despite this importance, the invisibility of the underlying phenomena makes the subject of electromagnetism seem abstract and mysterious to many people. The basic ideas are not difficult, though, and can be understood by carefully examining familiar phenomena.

o area of physics has had a greater impact on the way we live than the study of electricity and magnetism. Using electricity and electronic devices has become second nature to us now but would have been hard to imagine 200 years ago. The invention and design of television sets, microwave ovens, computers, cell phones and thousands of other familiar appliances and devices have all required an understanding of the basic principles of electricity and magnetism. Although the effects of magnets and static electricity had been known for a long time, our basic knowledge of electricity and magnetism is mainly a product of the nineteenth century. A key invention at the turn of the century opened the door to these developments. In 1800, the Italian scientist Alessandro Volta (1745–1827) invented the battery. Volta’s invention grew out of the work of an Italian physician, Luigi Galvani (1737–1798), who had discovered effects that he called animal electricity. Galvani found that he could produce electrical effects by probing a frog’s leg with metal scalpels. Volta discovered that the frog was not necessary. Two different kinds of metal separated by a suitable chemical solution were sufficient to produce many of the electrical effects observed by Galvani. Volta’s voltaic piles, which consisted of alternating plates of copper and zinc separated by paper impregnated with a

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12

Electrostatic Phenomena

chapter overview The main purpose of this chapter is to describe and explain the electrostatic force. We will also investigate concepts such as electric field and electric potential. These concepts will be made more vivid by describing and interpreting some simple experiments. Experimentation can lead to a better understanding of electric charge, the distinction between conductors and insulators, and many other ideas.

chapter outline

1 2 3

4 unit three

5

Effects of electric charge. What is electric charge? How is it involved in the electrostatic force? How do objects acquire charge? Conductors and insulators. How do insulating materials and conducting materials differ? How can we test this difference? How does charging by induction work? Why are bits of paper or Styrofoam attracted to charged objects? The electrostatic force: Coulomb’s law. What is the electrostatic force described by Coulomb’s law? How does it depend on charge and distance? How is it similar to (and different from) the gravitational force? The electric field. How is the concept of electric field defined? Why is it useful? How can we use electric field lines to help us visualize electrostatic effects? Electric potential. How is the concept of electrostatic potential defined? How is it related to potential energy? What is voltage? How are the concepts of electric field and electric potential related?

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ost of us have had the experience of running a comb through our hair on a dry winter day and hearing a crackling sound or, if the room is dark enough, even seeing sparks. Depending on the length and looseness of the hair, strands of it may have stood on end, as in figure 12.1. This phenomenon is intriguing but also annoying. You often have to wet the comb to get your hair to behave. What is happening? Some force seems to be at work, causing the individual hairs to repel one another. You would probably identify that force as an electrostatic force, or at least, you might say that static electricity is the cause of your unruly hair. To name a phenomenon is not the same as explaining it, though. Why does static electricity occur under these conditions, and what is the force in question? Hair-and-comb fireworks are just one example of electrostatic phenomena that are part of our everyday experience. Why does a piece of plastic refuse to leave your hand after you peeled it off a package? Why do you get a slight shock after you walk across a carpeted floor and touch a light switch? The phenomena range from static cling and similar annoyances to dramatic displays of lightning in a big thunderstorm. They are familiar and yet, to many of us, mysterious.

figure

12.1 Hair sometimes seems to have a mind of its own when combed on a dry winter day. What causes the hairs to repel one another? Despite its familiarity, many of us are surprisingly unaware of what static electricity is. People are often put off by a fear of electric shock as well as by the abstract nature of the subject. It need not be so. Many of the phenomena can be explained with ideas accessible to all of us.

12.1 Effects of Electric Charge What do the hair-combing example and the other phenomena mentioned in the chapter introduction have in common? The comb passing through your hair, your shoes passing over the rug, or plastic being peeled from a box all involve different materials rubbing against one another. Perhaps, this rubbing is the cause of the phenomena. How can we test such an idea? One approach would be to collect different materials, rub them together in various combinations, and see what we generate in the way of sparks or other observable effects. Experiments like these could help to establish which combinations of materials are most effective in producing electrostatic crackles and pops. Of course, we would also need some consistent way of gauging the strength of these effects.

What can we learn from experiments with pith balls? A common way of demonstrating electrostatic effects is to rub plastic or glass rods with different furs or fabrics. If we rub a plastic or hard rubber rod with a piece of cat fur, for example, we can see the hairs on the fur standing out and snapping at one another. Like the hair-combing fireworks, these effects are most striking on a dry winter day when little moisture is in the air. Another piece of equipment often used in these demonstrations is a small stand to which two pith balls are attached by threads (fig. 12.2).

figure

12.2 Pith balls suspended from a small stand are attracted to a charged plastic rod. Pith balls are small wads of dry, paperlike material light enough to be strongly influenced by electrostatic forces. An interesting sequence of events happens when a plastic rod, vigorously rubbed with cat fur on a dry day, is brought near the pith balls. At first, the pith balls are attracted to the rod like bits of iron to a magnet. After contacting the rod,

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and perhaps sticking to it for a few seconds, the pith balls dance away from the rod. They are repelled by the rod at this point and also by each other. The threads supporting the pith balls now form an angle to the vertical, as in figure 12.3, rather than hanging straight down. How can we explain what has happened? A repulsive force must be acting between the two pith balls after they have been in contact with the rod. We might imagine that the balls have received something (call it electric charge) from the rod that is responsible for the force we observe. This charge, whatever it is, was somehow generated by rubbing the rod with the cat fur. The force that is exerted by one stationary charge on another is called the electrostatic force. We can observe further changes if we rub a glass rod with a piece of synthetic fabric such as nylon. If the glass rod is brought near the pith balls that had been charged by the plastic rod, the pith balls are now attracted to the glass rod. They are still repelled by the plastic rod. If we allow the pith balls to touch the glass rod, they repeat the sequence of events that we observed earlier with the plastic rod. At first, they are attracted and stick briefly to the rod. Then, they dance away and are repelled by the rod and by each other. If we bring the plastic rod near, we now find that the balls are attracted to the plastic rod. These observations complicate the picture somewhat. Apparently, there are at least two types of charge: one generated by rubbing a plastic rod with fur and another generated by rubbing a glass rod with nylon. Could there be

more kinds of charge? Further experiments with different materials would indicate that these two types are all that we need to explain the effects. Other charged objects will cause either an attraction or repulsion with the two types of charge already identified and can be placed in one of these two categories.

What is an electroscope? A simple electroscope consists of two leaves of thin metallic foil suspended face-to-face from a metal hook (fig. 12.4). At one time, gold foil was used, but aluminized mylar is now preferred. The leaves are connected by the hook to a metal ball that protrudes from the top of the instrument. The foil leaves are protected from air currents and other disturbances by a glass-walled container. If the foil leaves are uncharged, they will hang straight down. If we bring a charged rod in contact with the metal ball on the top, however, the leaves immediately spread apart. They will remain spread apart even after the rod is taken away, presumably because the leaves are now charged. If any charged object is now brought near the metal ball, the electroscope shows what type of charge is on the object and gives a rough indication of how much charge is present. If an object with the same type of charge as the original rod is brought near the ball, the leaves will spread farther apart. An object with the opposite charge will make the leaves come closer together. A larger charge produces a larger effect. Although less dramatic than the pith balls, the electroscope offers several advantages for detecting the type and strength of charge. First, the foil leaves do not dance around

Metal ball and post F

F

Foil leaves

figure

12.3 Once the pith balls have moved away from the rod, they also repel each other, indicating the presence of a repulsive force.

figure

12.4 A simple electroscope consists of two metallicfoil leaves suspended from a metal post inside a glass-walled container.

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and become tangled together as pith balls often do. The metal ball on the electroscope offers a stationary point for testing other objects. Also, the distance the leaves move apart or come together when a charged object is brought near the ball gives a consistent indication of the strength of the charge. An electroscope gives further information about the charging process. When we charge the plastic rod by rubbing it with the cat fur, the fur also is charged but with a charge of the opposite type. We can verify this by bringing the fur close to the ball on the electroscope after the electroscope has been charged by the rod. When the fur approaches the metal ball, the leaves move closer together. They move farther apart when the rod approaches. Similar experiments show that the glass rod and the nylon fabric also acquire opposite types of charge when rubbed together. Example box 12.1 provides another example of charging by rubbing.

Benjamin Franklin’s single-fluid model By the middle of the eighteenth century, it was well known that there were two types of charge that produced electrostatic forces between charged objects. The forces were either attractive or repulsive according to the simple rule:

There was less agreement, however, on what to call these two types of charge. Charge-produced-on-a-rubberrod-when-rubbed-by-cat-fur and charge-produced-on-aglass-rod-when-rubbed-with-silk are unwieldy. Positive and negative, the names we now use, were introduced by the American statesman and scientist Benjamin Franklin (1706– 1790) around 1750. During the 1740s, Franklin had performed a series of experiments on static electricity like those we have described. Franklin proposed that the facts, as they were known, could be explained by the action of a single fluid that was transferred from one object to another during charging. The charge acquired when a surplus of this fluid was gained was positive () and the charge associated with a shortage of this fluid was negative () (fig. 12.5). Which of these was which was not clear, since the fluid itself was invisible and not otherwise detectable. Franklin arbitrarily proposed that the charge on a glass rod when rubbed with silk (there were no synthetic fabrics then) be called positive. Besides simplifying the names of the two types of charge, Franklin’s model offered a picture of what might be happening during charging. In his model, two objects become oppositely charged because some neutral or stable amount of the invisible fluid is present in all objects, and rubbing objects together transfers some of this fluid from one object to the other. During rubbing, one object gains

Like charges repel each other, and unlike charges attract each other. Like charges

Unlike charges

example box 12.1 Sample Question: Carpet Fireworks Question: Why do sparks fly when we shuffle across a carpet on a dry day and then touch a light switch? Answer: Shuffling our feet on a carpet represents a rubbing process between two different materials. The soles of our shoes are often made of a rubberlike material and the carpet fibers are usually a synthetic fabric. The rubbing process creates a separation of charge similar to those we have been describing using rods and fabric. The charge that our shoes acquire can flow to other parts of our bodies, which are good conductors. When we touch a light switch, the charge that we have acquired discharges producing the sparks. This is because, when properly wired, the switch is grounded. This means that the external parts of the switch are connected via wires and other conductors to the Earth, which serves as a large sink for charge. The excess charge on our bodies is discharged to ground via the switch.

F

+

+

F

+

F

–

figure

–

F

F

–

F

12.5 Like charges repel and unlike charges attract. The plus and minus signs were introduced in Franklin’s singlefluid model.

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an excess of the fluid while the other experiences a shortage. Franklin’s model was simpler than an earlier theory that had proposed two different substances for the two types of charge. Franklin’s model comes surprisingly close to our modern view of what takes place during charging. We now know that electrons are transferred between objects when they are rubbed together. Electrons are small, negatively charged particles present in all atoms and, therefore, in all materials. A negatively charged object has a surplus of electrons, and a positively charged object has a shortage of electrons. The atomic or chemical properties of materials dictate which way the electrons flow when objects are rubbed together. Rubbing different materials together sometimes causes electric charge to move from one material to the other. This charge then produces attractive or repulsive electrostatic forces between objects. Like charges repel one another, and unlike charges attract one another. An electroscope consistently gauges the sign and strength of charge present. Using Benjamin Franklin’s labels, there are two types of charge, positive and negative. The positive and negative labels originally meant a surplus or shortage of an invisible fluid in Franklin’s model, but we now know that negatively charged electrons are transferred during rubbing.

12.2 Conductors and Insulators The experiments described in section 12.1 gave us some basic information about the electrostatic force. Different properties of materials are also important, however, in understanding the range of electrostatic phenomena. Why were the pith balls initially attracted to the rod, for example,

Metal

figure

even when they were not charged? Why are the leaves in the electroscope made of metal? The distinction between insulators and conductors is a big piece of the puzzle.

How do insulators differ from conductors? Suppose that you touch the electroscope with a charged plastic or glass rod. The leaves repel one another. What happens if you then touch the metal ball on top of the electroscope with your finger? The leaves of the electroscope immediately droop straight down. You have discharged the electroscope by touching the ball with your finger (fig. 12.6). Suppose that you charge up the electroscope again. Now, touch the metal ball on top with an uncharged rod made of plastic or glass. There is no effect on the leaves of the electroscope. If, however, you touch the ball with a hand-held metal rod, the leaves immediately droop straight down. The electroscope is discharged. How can we explain these observations? Apparently, both the metal rod and your finger allow charge to flow from the leaves of the electroscope to your body. Your body is a large neutral sink for charge. You can easily absorb the charge on the electroscope without much change in your overall charge. The plastic or glass rods, on the other hand, do not seem to permit the flow of charge from the electroscope to your body. The metal rod and our bodies are examples of conductors, materials through which charge can flow readily. Plastic and glass are examples of insulators, materials that do not ordinarily permit the flow of charge. By performing similar experiments using the charged electroscope, we could test many other materials. We would discover that all metals are good conductors, while glass, plastic, and most nonmetallic materials are good insulators. Table 12.1 lists some examples of insulating and conducting materials.

Glass

12.6 Touching the ball on top of a charged electroscope with either your finger or a metal rod causes the electroscope to discharge. Touching a charged electroscope with an uncharged glass rod produces no effect.

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table 12.1 Some Common Conductors, Insulators, and Semiconductors Conductors

Insulators

Semiconductors

copper

glass

silicon

silver

plastic

germanium

iron

ceramics

gold

paper

salt solution

oil

acids

The difference in the ability of good conductors and good insulators to conduct electric charge is amazingly large. Charge flows much more readily through several miles of copper wire than it does through the few inches of ceramic material used as an insulator on electric transmission lines. Semiconductors, are materials that are intermediate between a good conductor and a good insulator. Silicon is probably the most familiar example. A wooden rod behaves like a semiconductor in discharging the electroscope. With the appropriate amount of moisture in the wood, the rod will cause a slow discharge of the electroscope. Although semiconductors are far less common than good conductors or good insulators, their importance to modern technology is enormous. The ability to control the level of conduction in these materials by mixing in small amounts of other substances has led to the development of miniaturized electronic devices like transistors and integrated circuits. The entire computer revolution has relied on the use of these materials, mostly silicon. Chapter 21 gives a closer look.

figure

12.7 The negatively charged rod is brought near a metal ball mounted on an insulating post, thus producing a separation of charge on the ball.

Charging a conductor by induction Can you charge an object without actually touching it with another charged object? It turns out that you can. The process is called charging by induction, and it involves the conducting property of metals. Suppose that you charge a plastic rod with cat fur and then bring the rod near a metal ball mounted on an insulating post, as in figure 12.7. The free electrons in the metal ball will be repelled by the negatively charged rod. Free to flow within the ball, they produce a negative charge (excess electrons) on the side opposite the rod and a positive charge (shortage of electrons) on the side near the rod. The overall charge of the metal ball is still zero. To charge the ball by induction, you now touch the ball with your finger on the side opposite the rod, still holding the rod near, but not touching, the ball. The negative charge flows from the ball to your body, since it is still being repelled

figure

12.8 Touching a finger to the opposite side of the metal ball draws off the negative charge, leaving the ball with a net positive charge.

by the negative charge on the rod. If you now remove your finger and then the rod (in that order), a net positive charge will be left on the ball (fig. 12.8). You can easily test this fact by bringing the metal ball near an electroscope that has been given a negative charge from the plastic rod. The leaves will come closer together when the ball charged by induction approaches, indicating the presence of positive charge on this ball.

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In this experiment, the proper sequence of events is important for the experiment to work. The charged rod must be held in place while the finger touches the ball on the opposite side and is removed. Only after the finger has been removed can the charged rod be moved away. The ball ends up with a charge opposite to the charge on the rod. This would also be true if we had used a positively charged glass rod. Then, the metal ball would end up with a negative charge. The process of charging by induction illustrates the mobility of charges on a conducting object such as the metal ball. The process will not work with a glass ball. Charging by induction is an important process in machines used for generating electrostatic charges, and in many other practical devices. It also explains some of the phenomena associated with lightning storms. (See page 252.)

Why are insulators attracted to charged objects? In our initial experiments with the pith balls, we noted that the pith balls were attracted to the charged rod before they had a chance to become charged themselves (fig. 12.2). How can we explain this phenomenon? What happens inside an insulating object when it is brought near a charged object? Unlike the charges in the metal ball, the electrons in the pith ball or other insulating material are not free to migrate through the material. Instead, they are tied to the atoms or molecules of the material. Within an atom or molecule, however, charges have some freedom of movement. The distribution of charge within the atom or molecule can change. Without delving into details of atomic structure, we can develop a rough picture of what happens to the charge in atoms when an insulating material is brought near a charged object. The basic idea is illustrated in figure 12.9, which magnifies the pith ball and greatly exaggerates the size of the atoms. Within each atom, a small distortion of the charge distribution takes place. The negative charge in the atom is attracted to the positively charged rod, and the positive charge is repelled. Each atom has now become an electric dipole, in which the center, or average location, of the negative charge is separated by a slight distance from the center of the positive charge. The atom now has positive and negative poles— hence the term dipole—and we say that the material has become polarized. Overall, these atomic dipoles within the insulating material produce a slight negative charge on the surface of the pith ball near the positively charged rod and a slight positive charge on the opposite surface. The adjacent positive and negative charges within the material cancel one another. The pith ball itself then becomes an electric dipole in the presence of the charged rod. Since the negatively charged surface is closer to the rod than the positively charged surface, it experiences a stronger electrostatic force, with the

+

+ + + +++ + +

–+ –+ –+ –+ –+ –+ –+ + + –+ –+ –+ – – –+ –+ + + – – –+ –+ –+ –+ –+ – + – + – + – + – + – +– + –+ –+ –+ –+ –+ –+ –+ –+ –+ + – –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ –+ – + –+ –+ –+ –+ –+ –+ –+ – – – –+ –+ –+ + + + –+ –+ –+ – – –+ –+ + + –+ –+– + –+ –+ –+ –+

figure

12.9 The negative charge in the atoms is attracted to the positively charged glass rod, while the positive charge is repelled. This produces a polarization of the charge in the atoms. The size of the atoms is grossly exaggerated. result that the pith ball is attracted to the charged rod. At this point, the overall charge on the pith ball is still zero. Once the ball comes in contact with the charged rod, however, some of the charge on the rod can be transferred to the pith ball, which becomes positively charged like the rod. The pith balls are then repelled by the rod, as we observed earlier. The ability to become polarized is an important property of insulating materials. Polarization explains why small bits of paper or Styrofoam will be attracted to a charged object such as a synthetic-fabric sweater rubbed against some other material (fig. 12.10). Induced polarization is also a factor in the operation of electrostatic precipitators used to remove particles from smoke in industrial smoke stacks. (See everyday phenomenon box 12.1.)

figure 12.10

Packing “peanuts” made of plastic foam are attracted to a charged rod.

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everyday phenomenon

box 12.1

Cleaning Up the Smoke The Situation. The effects of static electricity often seem to be just a nuisance. They cause our hair to misbehave and clothing to cling and sometimes produce annoying shocks. Do these effects have any useful applications? Smoke emissions from burning coal and from other industrial processes (as shown in the photograph) have been a problem ever since the industrial revolution began. These emissions usually contain many small particles or soot, which blackened the cities of many industrialized countries beginning in the early 1800s. The problem persists, although we have made great strides in finding ways of removing these small particles from smoke. How is static electricity helpful in addressing this problem?

The Analysis. Electrostatic precipitators are an important technology for removing small particles from smokestack emissions. The first electrostatic precipitator was patented in 1907 by Frederick Cottrell, who was then a chemistry professor at the University of California in Berkeley. They have been used ever since, with continuing improvements in design and variety. The design of early precipitators consisted of a row of highly charged wires (usually negatively charged) spaced between positively-charged parallel conducting plates (see diagram). The highly-charged wires cause the gas molecules flowing through the apparatus to become ionized by gaining electrons. The ions are attracted by induced polarization of the particles of ash or dust in the exhaust gas and attach to these particles, causing them to become negatively charged. The negatively-charged ash particles are then attracted to the positively-charged conducting plates and are removed from the stream of gas flowing through the precipitator. The actual design of modern precipitators is somewhat more complex, but the basic principle remains the same. The conducting plates are often corrugated or have protruding fins to give them a greater surface area. If the conducting plates were not cleaned somehow, they would quickly become coated with ash or dust, so “rappers” must be used to vibrate the plates at regular time intervals. This causes the ash to fall into hoppers, which then must be emptied from time to time.

243

Charged wires

+

+

– +– – + – + +– + – + – –

+

– – – – – – – –

– – – – – – – –

–

+ + +

+ + +

+ + +

Collector plate

Weights The particle-laden exhaust gases pass through an array of parallel collecting plates (positively charged) with the negatively-charged wires suspended between the plates.

The ionized gas that is produced in the vicinity of the highly-charged wires is called a corona. The geometry of the wires produces a very strong electric field (see section 12.4) in the immediate area around the wires. The situation is like that pictured in figures 12.14 or 12.15, where the field lines become very closely spaced near the charges. It is this very strong electric field in the vicinity of the wires that is responsible for the ionization of the gas molecules and the resulting corona. Electrostatic precipitators are effective in removing small particles from the smoke produced in metal smelters, cement plants, power plants, and many other industrial applications. Sometimes they are also used for removing suspended dust from the air in homes or other buildings. They are not, by themselves, effective for removing other pollutants such as sulfur, mercury, or organic molecules from exhaust gases. For these other purposes more sophisticated strategies must be used. Many of these applications employ devices called scrubbers. A scrubber will pass the exhaust gases through a mist of water or other liquid (in a wet scrubber) or fine particles in the case of a dry scrubber. Through chemical processes, pollutants such as sulfur will be attracted to the injected mist droplets or particles. (Sulfur is of particular concern in coal-burning power plants.) If a dry scrubber is used, an electrostatic precipitator might then be used to remove the particles holding the pollutant. Depending upon how much money we are willing to spend, most pollutants can be removed from exhaust gases by the use of electrostatic precipitators and scrubbers. However, these processes increase the cost of electric power and industrial products and often involve some loss in efficiency. One pollutant that cannot be readily removed is carbon dioxide because it is a basic combustion product and is in the form of a gas (see chapter 18). Carbon dioxide is a primary “greenhouse” gas associated with global warming discussed in everyday phenomenon box 10.1. Whenever we burn carbon-based fuels such as coal, oil, and natural gas, carbon dioxide is released into the atmosphere. The only recourse is to pump it underground or “sequester” it in other ways.

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Experiments in discharging an electroscope show that different materials vary widely in their ability to allow the flow of electric charge. Most metals are good conductors, but glass, plastic, and many other nonmetallic materials are poor conductors and, therefore, good insulators. A few materials, called semiconductors, have intermediate abilities to conduct charge. Conductors can be charged by induction without actually touching the conductor to another charged object. Insulators become polarized in the presence of charged objects, which explains why they are attracted to charged objects.

Thin wire

12.3 The Electrostatic Force: Coulomb’s Law Although we cannot see electric charge, we can see the effects of the forces exerted on charged objects. Pith balls and foil leaves refuse to hang straight down when they carry like charges. A repulsive force pushes them apart. Can we describe this force in a quantitative way? How does it vary with distance and quantity of charge? Is it similar in some way to the gravitational force? Questions such as these were actively explored by scientists in the latter part of the eighteenth century. Like the gravitational force, the electrostatic force was apparently present even when objects were not in contact—it acted at a distance. Although others had already speculated about the form of the force law, the experiments of the French scientist Charles Coulomb (1736–1806) during the 1780s settled the matter, establishing what we now know as Coulomb’s law.

How did Coulomb measure the electrostatic force? At first glance, measuring the strength of a force such as the electrostatic force might seem like a simple exercise. In reality, it is not an easy thing to do. Although much stronger than the gravitational force between ordinary-sized objects, the electrostatic force is still relatively weak, so Coulomb needed to develop techniques for measuring small forces. In addition, he was faced with the problem of defining how much charge was present, far from a trivial matter. Coulomb’s answer to the problem of measuring weak forces was to develop what we now call a torsion balance, shown in figure 12.11. Two small metal balls are balanced on an insulating rod suspended at the middle from a thin wire. The balls and wire are contained in a glass-walled enclosure to avoid disturbance from air currents. A force applied to either ball perpendicular to the rod produces a torque that causes the wire to twist. If we have previously determined how much torque is required to produce a given angle of twist, we have a method of measuring weak forces. To measure the electrostatic force, we must somehow charge one of the balls. A third ball, also charged, can then be inserted on the end of an insulating rod (fig. 12.11). If it

Glass enclosure

Charged balls + +

figure 12.11

A diagram of Coulomb’s torsion balance. The degree of twist of the wire provides a measure of the repulsive force between the two charges.

has the same type of charge as the ball on the end of the suspended rod, the two charges will repel one another, and the resulting torque will twist the wire. By adjusting the distance between the two charged balls, we can measure the strength of the repulsive force at different distances. The problem of determining the amount of charge placed on the balls required even more ingenuity. A simple electroscope gave only a rough indication of the quantity of charge present, and no consistent set of units or procedures for specifying quantity of charge were in use at the time of Coulomb’s work. His solution was to develop a system of charge division. He started with an unknown quantity of charge on a single metal ball mounted on an insulating stand, as in figure 12.12. He touched this ball to an identical ball mounted on a similar stand. He reasoned that the two balls would now contain equal quantities of charge (half of the original amount on the first ball), which was verified by bringing the balls near an electroscope. If a third identical metal ball touched one of the first two, the charge would again be divided equally, and these two balls would each have one-half the charge of the ball that did not participate in this second exchange. If several more identical balls are available, splitting the charge can be continued. Although this process does not provide an absolute measure of charge, we can say with confidence that one ball contains twice as much charge, or perhaps four times as much charge, as another. Coulomb used this process to test the effects of different amounts of charge on the electrostatic force. If the

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q1

–F

r

+

q2

+

245

F

figure

12.13 Two positive charges exert equal but oppositely directed forces upon one another, according to Coulomb’s law and Newton’s third law of motion. The force is inversely proportional to the square of the distance r between the two charges.

example box 12.2 Sample Exercise: Calculating Electrostatic Force Two positive charges, one 2 C and the other 7 C, are separated by a distance of 20 cm. What is the magnitude of the electrostatic force that each charge exerts upon the other?

figure 12.12

By bringing two identical metal balls into contact, one charged and the other one initially uncharged, equal quantities of charge are obtained on the two balls.

balls used in splitting the charge were identical in size to those in the torsion balance, touching one of the torsionbalance balls with one of the others was just one more division of charge. Using these procedures, Coulomb was able to determine how the strength of the electrostatic force varied with the quantity of charge on each object and the distance between the two charged objects.

What were the results of Coulomb’s measurements? The results of Coulomb’s work can be stated in a relationship usually referred to as Coulomb’s law: The electrostatic force between two charged objects is proportional to the quantity of each of the charges and inversely proportional to the square of the distance between the charges, or F

kq1q2 r2

.

Here the letter q represents the quantity of charge, k is a constant (called Coulomb’s constant) whose value depends on the units used, and r is the distance between the centers of the two charges. Figure 12.13 illustrates Coulomb’s law. The forces obey Newton’s third law—the two charges experience equal but oppositely directed forces. The two charges shown are both positive, so the force is repulsive, since like charges repel. If one charge were negative and the other positive, the directions of both forces would be reversed. Although Coulomb himself used different units, we now usually express charge in units called coulombs (C). If distance is measured in meters, the value of Coulomb’s

q1 2 C

(1 C 106 C 1 microcoulomb)

q2 7 C r 20 cm 0.2 m F ? F

kq1q2 r2

(9.0 109 Nm2/C2) (2 106 C) (7 106 C) (0.2 m)2 2 0.126 Nm 0.04 m2

3.15 N

constant turns out to be approximately k 9 10 9 N·m2/C2. The coulomb itself is determined by measurements involving electric current discussed in chapter 14. A force computation using Coulomb’s law is shown in example box 12.2. The coulomb (C) is a relatively large unit of charge. At the atomic level, the basic quantity of charge is 1.6 × 10-19C, the magnitude of the charge on an electron (which is negative). A pith ball might ordinarily hold a charge of about 10-10C, which represents the absence or presence of a billion or so (109) electrons. Had the two positive charges in example box 12.2 been coulomb-sized charges instead of microcoulombs, we would have found an enormous force.

Coulomb’s law compared to Newton’s law of gravity The electrostatic force has the same inverse-square dependence on distance as Newton’s law of gravitation (see chapter 5). If we double the distance between the two charges, the force between the two charges falls to only one-fourth what it was before the distance was doubled. Tripling the distance produces a force that is one-ninth the force at the original distance, and so forth. The strength of the interaction between the two charges falls off rapidly when the distance between them is increased.

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Since the gravitational force and the electrostatic force are two of the fundamental forces of nature, it is interesting to compare them. How do they differ? Placing their symbolic expressions side by side highlights both the similarities and the differences: Fg

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Gm1m2 kq1q2 and Fe . r2 r2

An obvious difference is that the gravitational force depends on the product of the masses of the two objects, and the electrostatic force depends on the product of the charges of the two objects. Otherwise, the forms of these two force laws are similar. A more subtle difference involves the direction. The gravitational force is always attractive. As far as we know, there is no such thing as negative mass. The electrostatic force, on the other hand, can be either attractive or repulsive depending on the signs of the two charges. The rule that like charges repel and unlike charges attract determines the direction. Another difference has to do with the strengths of these two forces. For objects of ordinary size, and for subatomic particles, the gravitational force is much weaker than the electrostatic force. At least one of the objects must have an enormous mass (like the Earth) to produce a significant gravitational force. For charged particles at the atomic or subatomic level, the electrostatic force is far more important than the much weaker gravitational force. It is the electrostatic force that holds atoms together and binds one atom to another in liquids and solids. Although the basic forms of these two force laws have been known for more than 200 years, physicists are still trying to understand the underlying reasons for the relative strengths of these and other fundamental forces of nature. The search for a unified field theory that would explain the relationships between all of the fundamental forces is a major area of research in modern theoretical physics (see chapter 21). Some of today’s students will undoubtedly play important roles in this search. Charles Coulomb designed a torsion balance to measure the strength of the electrostatic force between two charges. He found that the force was proportional to the size of each charge and inversely proportional to the square of the distance between the two charges. Coulomb’s force law has a form very similar to Newton’s law of gravitation describing the force between two masses. The gravitational force is always attractive, however, and generally much weaker than the electrostatic force.

12.4 The Electric Field Coulomb’s law tells us how to find the force between any two charged objects if the objects are small compared to the distance between them. These forces act at a distance:

the charges do not have to be in contact for the force to be exerted. Does the presence of electric charge somehow modify the space surrounding the charge? How can we describe the effects of a large distribution of charges on some other charge? The concept of electric field describes the effect of such a distribution of charges on another individual charge. The usefulness of the idea of a field has made it a central concept in modern theoretical physics. To most of us, the word field suggests a field of wheat or a meadow filled with wild flowers. The concept of electric field in physics is somewhat more abstract. An example involving just a few charges can help to introduce the ideas.

Finding the force exerted by several charges Using Coulomb’s law, we can find the magnitude of the electrostatic force between any two charged objects. If the charged objects are small compared to the distance between them, we often call them point charges. If there are more than two point charges, we can compute the net force on any one of them by adding (as vectors) the forces due to each of the other charges. In the first part of example box 12.3, we find the force on a charge q0 exerted by two other charges for the situation shown, where the charge q0 lies between the charges q1 and q2. The resulting forces are added as vectors to obtain the net force on q0. The forces due to each of the other charges must be computed separately using Coulomb’s law (see example box 12.2 for details of this process). Notice that the force F1 on the charge q0 exerted by the charge q1 is considerably larger than the force F2 exerted by q2, mainly because q2 is farther from q0 than q1 is. The electrostatic force described by Coulomb’s law is inversely proportional to the square of the distance between the two charges. The two forces acting on q0 are in opposite directions, yielding a net force of 9 N in the direction of the larger force.

What is an electric field? Suppose that we wanted to know the force on some other charge placed at the same location as q0. Would we have to use Coulomb’s law again, find the forces due to each of the other charges, and add them as we did in example box 12.3? There is an easier way that involves the concept of electric field. Think of the charge q0 as a test charge that has been inserted at this particular location to assess the strength of the electrostatic effect at that point. By the nature of Coulomb’s law, the force on this test charge will be proportional to the magnitude of the charge selected. If we divide the net force by the magnitude of the test charge, we find the force per unit charge at this location. (See the second part of the sample exercise in example box 12.3.) Knowing the

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12.4 The Electric Field

example box 12.3 Sample Exercise: An Electric Field Two point charges with charges q1 3 C and q 2 2 C are separated by a distance of 30 cm as shown in the drawing. A third charge q 0 4 C is placed between the initial two charges 10 cm from q1. From Coulomb’s law, the force exerted by q1 on q 0 is 10.8 N, and the force exerted by q 2 on q 0 is 1.8 N. q1

+

F1 = 10.8 N

q0

+

q2

F2 = 1.8 N

10 cm

+ 20 cm

a. What is the net electrostatic force acting on the charge q0? b. What is the electric field (force per unit charge) at the location of the charge q 0 due to the other two charges? a. F1 10.8 N to the right Fnet F1 F2 F2 1.8 N to the left Fnet ?

b. E ?

10.8 N 1.8 N 9N

Fnet 9 N to the right Fnet E q0

9N 4 106 C

2.25 106 N/C E 2.25 106 N/C to the right

force per unit charge permits us to compute the force on any other charge placed at that same point. Using the force per unit charge as a measure of the strength of the electrostatic effect at some point in space lies at the heart of the concept of electric field. In fact, we define electric field as The electric field at a given point in space is the electric force per unit positive charge that would be exerted on a charge if it were placed at that point. Fe q It is a vector having the same direction as the force on a positive charge placed at that point. E

The symbol E represents the electric field.

247

In other words, the ratio Fnet/q0 that we computed in example box 12.3 is the magnitude of the electric field at that point. We can then use the field to find the force on any other charge placed at that point by multiplying the charge by the electric field, Fe qE. If the charge happens to be negative, the minus sign indicates that the direction of the force on a negative charge is opposite to the direction of the field. The direction of the electric field at any point in space is the direction of the force that would be exerted on a positive charge placed at that point. Keep in mind that the electric field and the electrostatic force are not the same. We can talk about the field at a point in space even if there is no charge at that point. The field tells us the magnitude and direction of the force that would be exerted on any charge placed at that point. There must be a charge if there is to be a force, but the field exists regardless of whether there is a charge at that point or not. The electric field can exist even in a vacuum. When we use the field concept, we shift our focus from the interactions between particles or objects to the way a charged object affects the space surrounding it. The field concept is not restricted to electrostatics. We can also define a gravitational field or a magnetic field, as well as others.

How are electric field lines used? The concept of electric field was formally introduced by the Sottish physicist, James Clerk Maxwell (1831–1879), around 1865 as part of his highly successful theory of electromagnetism. The idea had already been used informally by Michael Faraday (1791–1867), who developed the concept of what we now call field lines as an aid in visualizing electric and magnetic effects. Faraday was not trained in mathematics but was a brilliant experimentalist who made good use of mental pictures. To illustrate field lines, we can use a positive test charge to assess the field direction and strength at various points around a single positive charge. We will find that the test charge is repelled by the original positive charge wherever we place it around that charge. If we draw lines to indicate the direction of the force on the test charge (which is also the direction of the field), we obtain a drawing like figure 12.14. Figure 12.14 is only a two-dimensional slice of a threedimensional phenomena. The electric field lines associated with a single positive charge radiate in all directions from the charge. If we also adopt the convention that we always start the lines on positive charges and end them on negative charges, the density of the field lines is proportional to the strength of the field. The closer together the field lines are, the stronger the field. Field lines are a means of visualizing both the direction and the strength of the field. Figure 12.15 shows a twodimensional slice of the electric field lines associated with

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E

F

+

q0

–

+

+

figure 12.16

The electric field lines associated with two equal but opposite-sign charges (an electric dipole).

figure 12.14

The direction of the electric field lines around a positive charge can be found by imagining a positive test charge q0 placed at various points around the source charge. The field has the same direction as the force on a positive test charge.

E

+

q0

F

–

a negative charge. Here the field lines end on the charge and must be directed inward to indicate the proper direction for the force on a positive test charge. As a final example, consider the field lines associated with the electric dipole in figure 12.16. An electric dipole is two charges of equal magnitude but opposite sign, separated by a small distance. The field lines originate on the positive charge and end on the negative charge. Imagine a positive test charge placed at various points around the dipole. Do the field lines agree with your expectations for the correct direction of the force on the test charge? This is the acid test for any field-line diagram. The concept of electric field focuses our attention on how the space surrounding an electric charge is affected by that charge. The field is defined as the force per unit charge that would be exerted on a test charge placed at some point in space in the vicinity of the source charges. The direction of the electric field is the same as the direction of the force on a positive test charge placed at that point in space. The field is a useful concept for treating the effects of many point charges on some other charge. Electric field lines can be used to visualize these effects.

12.5 Electric Potential

figure 12.15

The electric field lines associated with a negative charge are directed inward, as indicated by the force on a positive test charge, q0.

In chapter 6, we defined the potential energy associated with the gravitational force, as well as the potential energy of a spring. Can we define the potential energy of a charged particle being acted upon by an electrostatic force? The electrostatic force is a conservative force, which means that we can define an electrostatic potential energy. This potential energy leads to the related concept of electric potential, often referred to simply as voltage. Voltage

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is familiar because we use it to discuss batteries and household circuits. What is voltage, and how is it related to electrostatic potential energy?

Finding the change in potential energy of a charge To see how the potential energy of a charge changes with position, the easiest case to consider is a charged particle moving in a uniform electric field. In a uniform field, the electric field lines are parallel and evenly spaced. The field does not vary in direction or strength as we move from point to point within the region where the field exists: it is constant within this region. How do we go about producing a uniform electric field? Two parallel metal plates with opposite charges, arranged as in figure 12.17, will do the trick. If one plate is positively charged and the other has an equal negative charge, the field lines are straight lines originating at the positive charges and terminating on the negative charges, as shown. You can test this conclusion yourself by thinking about the force on a positive test charge placed between the two plates, as we did in section 12.4. Two conductors separated by an insulating material like air represent a useful means of storing charge. If the two conductors have charges of opposite sign, these charges are held in place by the attractive electrostatic force between the charges on the opposing plates. We call this arrangement a capacitor, which is basically a device to store charge. Capacitors have many applications, particularly in electric circuits. Suppose that we now place a positive charge in the uniform-field region between the two plates of the parallelplate capacitor. This charge will experience an electrostatic force in the direction of the electric field. It will be attracted toward the negative charges on the bottom plate and repelled by the positive charges on the top plate. If we release the charge, it accelerates toward the bottom plate. (This acceleration has nothing to do with gravity. We can assume that the gravitational force on the charged particle is very small compared to the electrostatic force.)

+

+

+ +

+ +

+ +

+

+ E

– –

figure 12.17

– –

– –

– –

– –

Two parallel metal plates containing equal but opposite-sign charges produce a uniform electric field in the region between the plates.

+

+

+

+

–

–

+ d

F q

+ qE

–

–

figure

12.18 An external force F, equal in magnitude to the electrostatic force qE, is used to move the charge q a distance d in a uniform field. If we apply an external force to move the charge in the opposite direction to the electric field, as in figure 12.18, this external force does work on the charge. This work increases the potential ener