The Energy of Physics, Part I: Classical Mechanics and Thermodynamics [2nd ed.] 1516542452

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THE ENERGY OF PHYSICS

THE ENERGY OF PHYSICS

Part I: Classical Mechanics and Thermodynamics

By Christopher J. Fischer University of Kansas

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ACADEMIC PUBLISHING

Bassim Hamadeh, CEO and Publisher Carrie Montoya, Manager, Revisions and Author Care Kaela Martin, Project Editor Berenice Quirino, Associate Production Editor Jess Estrella, Senior Graphic Designer Alexa Lucido, Licensing Manager Natalie Piccotti, Director of Marketing Kassie Graves, Vice President of Editorial Jamie Giganti, Director of Academic Publishing Copyright © 2019 by Cognella, Inc. All rights reserved. No part of this publication may be reprinted, reproduced, transmitted, or utilized in any form or by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information retrieval system without the written permission of Cognella, Inc. For inquiries regarding permissions, translations, foreign rights, audio rights, and any other forms of reproduction, please contact the Cognella Licensing Department at [email protected]. Trademark Notice: Product or corporate names may be trademarks or registered trademarks, and are used only for identification and explanation without intent to infringe. Cover image copyright© 2012 Depositphotos/eskaylim. Printed in the United States of America. ISBN: 978-1-5165-4245-l(pbk)/ 978-l-5165-4246-8 (br) / 978-1-5165-9265-4 (al)

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ACADEMIC PUBLISHING

CONTENTS PREFACE

XIX

CHAPTER 1: WHY THINGS MOVE

l

1-1 Introduction

1

1-2 Energy

1

1-3 Energy Conservation

2

1-4 Why Do Things Move?

3

1-5 Potential Energy Curves

4

1-6 System Energy

5

1-7 Units

6

1-8 Looking Ahead

6

Summary

6

Problems

7

CHAPTER 2: KINEMATICS

9

2-1 Introduction

9

2-2 Position and Displacement

9

2-3 Velocity and Acceleration

12

VII

VIII

I

THE ENERGY OF PHYSICS, PART i: CLASSICAL MECHANICS AND THERMODYNAMICS

2-4 Kinematics 2-5 Constant Acceleration Kinematics

13 18

2-6 Relative Motion in 1-Dimension

20

2-7 Position, Velocity, and Acceleration for 2-Dimensional Motion

24

Summary

27

Problems

28

CHAPTER 3: ENERGY AND ENERGY CONSERVATION

33

3-1 Introduction

33

3-2 Translational Kinetic Energy

33

3-3 Gravitational Potential Energy

34

3-4 Spring Potential Energy

38

3-5 Energy Conservation in Isolated Systems

39

3-6 Energy Conservation in Non-Isolated Systems

45

3-7 Graphical Representations of Energy Conservation

49

Summary

51

Problems

52

CHAPTER 4: ENERGY-BASED MECHANICS

57

4-1 Introduction

57

4-2 Kinematics with Translational Kinetic Energy

57

4-3 1-Dimensional Kinematics with Constant Energy

62

CONTENTS

I

IX

4-4 Projectile Motion

69

4-5 Multiple Objects Moving Together

74

4-6 Energy Dissipation by Friction

78

4-7 Power

84

4-8 General Problem Solving Strategy

88

Summary

89

Problems

90

CHAPTER 5: CIRCULAR AND ROTATIONAL MOTION

97

5-1 Introduction

97

5-2 Variable Definitions

97

5-3 Circular Motion Kinematics

98

5-4 Constant Angular Acceleration Kinematics

101

5-5 Rotational Motion, Rotational Kinetic Energy, and Moment of Inertia

103

5-6 Systems with Both Rotational Kinetic and Translational Kinetic Energ y

111

5-7 Energy Conservation with Rotational Motion

114

5-8 Effective Mass

118

Summary

120

Problems

121

CHAPTER 6: OSCILLATORY MOTION

129

6-1 Introduction

129

X

J

THE ENERGY OF PHYSICS, PART i: CLASSiCA� MECHANICS M,jD THERMODYNAMICS

6-2 A Simple Oscillating System 6-3 Simple Harmonic Motion 6-4 Kinematics with Simple Harmonic Motion 6-5 Vertically-Mounted Springs

129 134 138 140 142

6-6 Simple Pendula 6-7 Effective Mass and Effective Spring Constant for Oscillating Systems

145

6-8 General Simple Harmonic Oscillator Approximation

151

6-9 Oscillations of Non-Isolated Systems

153

Summary

157

Problems

158

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7-1 Introduction

165

7-2 Equilibrium Conditions

165

7-3 What is a Force?

168

7-4 Conservative and Non-Conservative Forces

168

7-5 Forces Do Work

170

7-6 Work and Kinetic Energy

173

7-7 Simple Machines

180

7-8 Work, Kinetic Energy, Force, and Acceleration

185

Summary

186

Problems

187

CONTENTS I XI

CHAPTER 8: NEWTONIAN MECHANICS I

191

8-1 Introduction

191

8-2 Reference Frames

191

8-3 Newton's 1st Law

192

8-4 Newton's 2nd Law

192

8-5 Drawing Free Body Diagrams

195

8-6 Using Free Body Diagrams

197

8-7 Kinetic and Static Friction

198

8-8 Tension

203

8-9 Weight, Apparent Weight, and Fictitious Forces

205

Summary

209

Problems

210

CHAPTER 9: NEWTONIAN MECHANICS II

213

9-1 Introduction

213

9-2 Newton's 2nd Law

213

9-3 Redrawing Free Body Diagrams

217

9-4 Work

224

9-5 Power

228

9-6 General Problem-Solving Strategy

231

Summary

232

Problems

233

XII

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

CHAPTER 10: TORQUE AND ROTATIONAL MOTION

239

10-1 Introduction

239

10-2 Center of Mass

239

10-3 Torque

243

10-4 Vector Description of Torque and Rotational Motion

245

10-5 Newton's 2nd Law for Rotational Motion

246

10-6 Physical Pendula

250

10-7 Rolling Down a Ramp

256

10-8 Work Done by Torque

258

10-9 General Problem-Solving Strategy

266

Summary

267

Problems

269

CHAPTER 11 : FURTHER APPLICATIONS OF NEWTONIAN MECHANICS

275

11-1 Introduction

275

11-2 Static Equilibrium

275

11-3 Levers

280

11-4 Centripetal Acceleration

282

11-5 The Fictitious Centrifugal Force

290

Summary

295

Problems

296

CONTENTS

I

XIII

CHAPTER l 2: NEWTONIAN MECHANICS FOR SYSTEMS OF MOVING OBJECTS

303

12-1 Introduction

303

12-2 Newton's 3rd Law

303

12-3 Newton's Laws Applied to Systems with Multiple Moving Objects

306

12-4 Newton's Laws Applied to Systems with Multiple Moving Objects Connected by Ropes and Pulleys 311 12-5 External and Internal Forces

320

12-6 A System Approach to Newton's 2nd Law

322

Summary

328

Problems

329

CHAPTER 13: LINEAR MOMENTUM

335

13-1 Introduction

335

13-2 Linear Momentum and Newton's 2nd Law

335

13-3 Galilean Transformations

342

13-4 Linear Momentum Conservation

343

13-5 Collisions and Explosions

354

13-6 Conservation of Linear Momentum in Multiple Dimensions

362

13-7 Work, Kinetic Energy, and Linear Momentum

365

Summary

368

Problems

369

ANICS AND THERMODYNAMICS X!V I THE ENERGY OF PHYS!CS. PART i: CLASSICAL MECH

CHAPTER 14: ANGULAR MOMENTUM

377

14-1 Introduction

377

14-2 Angular Momentum

377

14-3 Newton's 2nd Law with Angular Momentum

381

14-4 Angular Momentum Conservation

383

14-5 Work, Rotational Kinetic Energy, and Angular Momentum

389

14-6 The Fictitious Coriolis Force

396

14-7 Hamiltonian Mechanics

399

Summary

402

Problems

403

CHAPTER ! 5:

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LAW OF THERf/\ODYN,6-lv\lCS

41 l

15-1 Introduction

411

15-2 Microscopic vs. Macroscopic

411

15-3 Statistical Descriptions of Systems

412

15-4 Heat and Temperature

414

15-5 The 1st Law of Thermodynamics

415

15-6 Heat Capacity and Equipartition Theorem

417

15-7 The Equation of State of an Ideal Gas

421

15-8 Work Done by an Ideal Gas

425

15-9 Heat Capacity Redux

428

15-10 Cyclic Processes with an Ideal Gas

431

15-11 Free Expansion

437

CONTENTS

I

XV

Summary

438

Problems

439

CHAPTER 16: ENTROPY AND THE 2ND LAW OF THERMODYNAMICS

443

16-1 Introduction

443

16-2 Entropy

443

16-3 The 2nd Law of Thermodynamics

446

16-4 Heat Engines

450

16-5 The Entropy of an Ideal Gas

454

16-6 Equilibrium Conditions

458

16-7 General Thermodynamic Relationships

460

16-8 Diffusion and Boltzmann Factors

462

16-9 Chemical Potential and Latent Heat

466

16-10 Looking Ahead

470

Summary

472

Problems

474

CHAPTER 17: FLUIDS

481

17-1 Introduction

481

17-2 Definitions

481

17-3 Surface Tension

487

17-4 Buoyancy

490

-

XVI

I

THE ENERGY OF PHYSICS PART I: CLASSICAL MECHANICS AND THERMODYNAMICS 1

17-5 The Continuity Equation

498

17-6 Bernoulli's Equation

501

17-7 Real Gases

509

Summary

513

Problems

515

APPENDIX A

519

A-1 SI System of Units

519

A-2 Physical Constants

520

A-3 Useful Mathematics

520

A-4 Greek Alphabet

522

APPENDIX B

523

B-1 Definition of Vectors

523

B-2 1-Dimensional Vectors

523

B-3 General Descriptions of Vectors

524

B-4 Components and Decomposition

524

B-5 Simple Vector Mathematics

525

B-6 The Dot Product

525

B-7 The Cross Product

526

CONTENTS

I

XVII

APPENDIX C

528

C-1 Expressions for Instantaneous Values

528

C-2 Expressions for Average Values

529

C-3 Constant Acceleration Equations

530

C-4 Translational and Rotational Kinetic Energy

532

APPENDIX D

534

D-1 Definition of the Moment of Inertia

534

D-2 Sample Calculation of Moment of Inertia

535

D-3 Common Moments of Inertia

536

D-4 Definition of the Center of Mass

536

D-5 Applications of the Center of Mass

537

D-6 Parallel Axis Theorem

538

D-7 Connection to Statistics

539

APPENDIX E

540

E-1 Work and Kinetic Energy

540

E-2 Newton's 2nd Law for Rotational Motion

542

APPENDIX F

544

F-1 Exact and Inexact Differentials

544

F-2 Principle of Minimum Potential Energy

545

ANSWERS

549

PREFACE

I was motivated to write this book because I felt that the standard presentation of introductory clas­ sical mechanics and thermodynamics was disjointed, and, furthermore, did not effectively emphasize the generally applicable concepts of the material. In contrast, energy conservation is the common theme applied to all the material in this book Having this unifying theme enables students to de­ velop a more universally applicable strategy for solving physics problems, to better understand the relationship between classical mechanics, thermodynamics, and statistical mechanics, and, finally, to be well prepared for advanced STEM courses. This presentation also allows for this book to be shorter in length since space need not be devoted to the discussion of one special case after another. Instead, this book demonstrates how basic principles can be generalized to solve different specific problems and, thereby, helps students to develop their critical thinking skills. This book is intended for the first semester of a two semester calculus-based physics curriculum. As such, I assume that students using this book are comfortable with simple differentiation and integra­ tion. Furthermore, since I anticipate that students using this book will likely also be simultaneously enrolled in one or more additional calculus courses, applications of more advanced calculus, such as partial differentiation, are introduced in the last chapters of the book Indeed, another motivation for writing this book was to develop an introductory physics curriculum that would give students a better opportunity to practice and further develop their skill with calculus. I would like to thank all of the people who contributed to this book and/or supported me as I wrote it. First, I wish to thank Professor Matthew Antonik for originally proposing that students in intro­ ductory physics be taught about energy and energy conservation before being taught about forces. Thanks, as well, to Professor Michael Murray and Professor Phil Baringer for putting up with me while I tried out this approach in our general physics class. I appreciate the fact that neither of you ever tried to "pull rank" and insist that I teach that course the standard way. I would also like to thank my department chair, Professor Hume Feldman, for fostering such a creative and supportive environment in the department.

XIX

XX

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THE ENERGY OF PHYSICS, PART I: ClASS!CAL MECHANICS AND THERMODYNAMICS

I would especially like to thank graduate students Allen Eastlund and Koan Briggs for taking time away from their research to help me with the writing and editing of this book. I've appreciated your counsel throughout the development of this book and the associated in-class material. Finally, I would like to thank my family for their support, especially my wife and children. I love all of you very much.

CHAPTER ONE Why Things Move The function of education is to teach one to think intensively and to think critically. -Dr. Martin Luther King, Jr.

1-1 Introduction

T

he focus of this book is classical mechanics, which involves both the description and causes of motion. This is a broad area of study encompassing everything from the motion of a baseball to the orbits of planets. The beauty and elegance of physics is that the movements of baseballs and planets are both governed by the same set of fundamental physical laws. Indeed, because of these fundamental laws we are able to form general conclusions about the motion of objects and, more importantly, predict the future motion of these same objects. One of these fundamental physical laws is the law of conservation of energy. The concept of energy conservation is central to all of physics and, therefore, will be used as the basis for all discussion and problem solving throughout this book Thus, the first step to each problem is to consider the following question: What happens to the energy?

1-2 Energy Because we will employ an energy-based approach in this book, we must begin with a definition of energy. Ironically, despite its fundamental usefulness, energy is a concept that is difficult to define. The most practical, though not very informative or insightful, definition of energy is that it is a scalar quantity associated with the state of an object or collection of objects. Energy: A scalar quantity associated with the state of an object or collection of objects. Scalar: A mathematical quantity specified by a magnitude [or size) only. The definition of the state (or condition) of an object or collection of objects naturally depends upon the scale of the investigation. For example, when describing the energy of a pencil lying on a desk, we might be interested in the energy associated with the position of the pencil relative to

2

I

AND THERMODYNAMICS THE ENERGY OF PHYSICS, PART i: ClASSICAl MECHANICS

ual atoms that constitute the surface of the Earth, with the energies associated with the individ ly restrict ourselves to a the pencil, or, perhaps, all of these energies. In this book, we will initial refer to as a macroscopic scale and set of objects corresponding to everyday life, which we will scopic motion and scale. Accordingly, we will define the state of objects in terms of their macro s to objects position only. Later in this book, we will apply a smaller scale, or microscopic, analysi atoms. and les molecu their and consider also the energies of their constitutive parts, such as Throughout this book, we will consider two main forms of energy: energy that is associated with the motion of an object, referred to as kinetic energy, and energy associated with the position of an object, referred to as potential energy. Kinetic energy: Energy associated with motion. Kinetic energies are denoted by the variable K Potential energy: Energy associated with position. Potential energies are denoted by the variable U. We will use subscripts to label the specific type of kinetic or potential energy as well as the object with which the energy is associated. Since the motion of an object is associated with a change in the object's position, it is not surprising that the kinetic and potential energies of an object are related to each other. Furthermore, it is this connection between an object's kinetic and potential energies that allows us to describe and predict how the object will move and, thus, will be the starting point for our discussion of classical mechanics. Throughout this book, we will use the word system to denote the object or collection of objects that is of interest to a particular problem or discussion. System: An object or a collection of objects. We w�ll then define the energy of the system to be the sum of all the kinetic and potential energies _ associated with all of the objects in the system1. Energy of a system: The sum of all kinetic and potential energies of all objects in a system. The total energy of a system will be denoted by the letter E.

E ergy has a very �nteresting fundamental property: it can be neither created nor destroyed. This � operty of energy is referre d to as the law of conser vation of energy or the 1st law of thermodynam­ � Energy can be conv rted from one type to another or transferred betwe en objects, however. : ;;� example, the P tential energy of a skydiver is converted into kinetic energy as she falls toward � the ground. Or durmg a collision, the k'met1c energy of one car may be transferred to the kinetic energy of another car. 1 we will interchangeably refer to this quantity as the total energy of the system. We will also sometimes refer to it as _ the internal energy of the system.

CHAPTER ONE

I

3

We can divide these changes in energy or transfers of energy into those that occur within a system and those that result from interactions between the system and something outside the system (more generally, with the environment outside the system). When a ball is thrown up into the air and then falls back down to the ground again, the initial kinetic energy of the ball is converted into potential energy, which is then converted back into kinetic energy. These changes all occur internal to the system of the balF. In contrast, a person changed the energy of this system through the external interaction of throwing the ball. If all external interactions with the outside environment are prevented, the system is referred to as an isolated system. Isolated system: A system for which interactions with the outside environment are prevented. The total energy of an isolated system is constant.

Conservation principles, such as the conservation of energy, are very powerful and can be used to generate other physical laws. Over the next several chapters we will also see how the conservation of energy can lead to equations describing the motions of systems. But where do conservation principles come from? Well, it turns out that the conservation of energy is the result of time being homogeneous, meaning that time is smoothly and evenly flowing, and that there is no absolute origin for time3 • Later on, we will learn about momentum, which is another intrinsic property of systems that is governed by a conservational principle. The conservation of momentum results from the homogeneity of space, which is also smooth and has no absolute origin3 • '

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1-4 Wliy [)o Things Move?'

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The law of conservation of energy explains how energy can be converted or transferred but doesn't explain why these processes occur. It is the 2nd law of thermodynamics that explains why. We will discuss the 2nd law of thermodynamics more formally in Chapter 16; for now we will express it as: , Systems will always arrange themselves in order to minimize their total potential energy.

This is often referred to as the principle of minimum potential energy. A ball thrown up into the air will fall back down because by doing so the potential energy of the ball is minimized. In Chapter 3, we will learn our first definitions of the different kinds of kinetic and potential energy and apply them to understanding why objects move. Before that, however, we must learn how to describe the position and motion of objects mathematically since this is required for quantitative calculations involving kinetic and potential energies. This will be the subject of Chapter 2.

2 3

We will discuss this example more carefully in Section 3-5 and Section 13-4. Please be inspired to take additional physics classes to learn why this is true.

4

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

1-5 Potential Energy Curves Even without a quantitative description of position and motion, we can nevertheless make qualitative predic­ tions about the behavior of a system from plots of its potential energy. Consider a system that consists of an object whose potential energy depends on its height above the ground, as shown in Figure 1.1. Let's imagine that we release the object at an initial position above the ground (see Figure 1.1). According to the principle of minimum potential energy, this system will always arrange itself in order to minimize its poten­ tial energy. In this case, the object will move closer to the ground (i.e., it will decrease its height above the ground). Metaphorically, we can think of the object sliding down the potential energy curve to the minimum accessible potential energy. It follows that an object would not change its position if it were released at rest at a point where its potential energy curve had no slope (e.g., if placed on a flat surface, a ball will not move away from where it was released). We refer to these locations as equilibrium points. Equilibrium point: A location where the potential energy curve has no slope. As shown in Figure 1.2, there are three varieties of equilibrium points, which are differentiated based upon the curvature of the potential energy curve at that location. At neutral equilibrium points the potential energy curve is flat. This is an example of a ball on a flat surface: it will not move when released at rest. At unstable equilibrium points the poten­ tial energy curve is convex; if you place a ball on top of a pole, it can and will fall off with only the slightest push or bump. Finally, at stable equilibrium points the potential energy curve is concave. If you

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Stable Equilibrium Local Minimum Stable Equilibrium Global Minimum

Increasing Separation Figure 1.2: Identification of equilibrium points on a potential e nergy curve. All points correspond to locations where the 1 e o the potential energy curve is zero. They ore further· � f�j° IS!hingul tshe? by the curvature of the potential energy curve at t at ocat1on. ----------- ------ - - - �-

CHAPTER ONE

I

5

place a ball at the bottom of a bowl, it will always return to that same position if moved slightly away from it. Neutral equilibrium point: An equilibrium location of the system where the system's potential energy curve is flat. Unstable equilibrium point: An equilibrium location of the system where the system's potential energy curve is convex. Stable equilibrium point: An equilibrium location of the system where the system's potential energy curve is concave. Stable equilibrium points can be further classified as local or global minima depending upon the associated magnitude of the potential energy at that point; the global minimum corresponds to the lowest possible potential energy, whereas local minima are other stable equilibrium points that, while stable, have higher potential energy than the global minimum.

1-6 System Ene.rgy It is always important to remember that energy is an extensive parameter. This means that the energy of a system is the sum of all the energies of the individual objects that constitute the system. Extensive parameter: A parameter of a system that is the sum ofproperties of the components of the system. The motions and/or positions of objects in a system can often be related to each other. As we will see, this allows us to use relationships between the variables denoting the motions and positions of ob­ jects in a system to simplify the equation for the energy of that system. It is always beneficial to take advantage of these relationships when they exist. Consider, for example, the system shown in Figure 1.3 that consists of two blocks, A and 8, which are connected to each other by a rope that passes over a pulley. Because the rope connects the two blocks, we know that the two blocks cannot move independently of one another. For example, block A will move to the right if block 8 moves down. Similarly, block 8 would move upwards if block A moves to the left. In either case, it is important to note that the distance moved by block A must be equal to the distance moved by block 8 (even though these two motions are occurring along different directions) since the two blocks are connected by a rope4• It is this kind of correlation between the distances travelled by objects in systems that will be the basis for our future discussion of the coupled motions of these objects. Figure 1.3: A system con­ Furthermore, since the positions and motions of the two blocks in the sisting of two blocks, a system shown in Figure 1.3 are coupled to each other, so, too, are the po­ rope, and a pulley. tential and kinetics energies of these blocks. In other words, it is possible 4

We will assume that the ropes behave ideally and, thus, do not stretch.

6

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

to write expressions for the kinetic and potential energies of the two blocks in Figure 1.3 using a common set of variables. Of course, this, in turn, then also allows for a more simplified mathematical description of the energy conservation occurring within the system. Furthermore, the energy of the system is also a state function5 • This means that the energy depends upon the arrangement of the system (i.e., the positions and motions of the objects in the system) but does not depend upon the process by which the system arrived at its arrangement (i.e., how each object obtained its position and motion). State function: A property ofa system that depends upon only the arrangement of the system. The energy of the system shown in Figure 1.3 depends upon only the positions and motions of the two blocks, the pulley, and the rope. Knowledge of these quantities for these objects is sufficient to determine the energy of the system regardless of how these objects came to have them.

1-7 Units Physics is an experimental science, and experimental measurements require units. The system of units used in science is the International System of Units (known by its French acronym, SI). This system is commonly referred to as the metric system. We are already familiar with some metric units in our daily lives (e.g., seconds, liters), whereas others may not be so familiar (e.g., candela). A list of the seven basic SI units and several additional derived units is shown in Appendix A. Also in Appendix A, is a list of the prefixes that are used to describe very large or very small quantities with SI units and a summary of the Greek alphabet, as we will be using those letters occasionally in the course.

1-8 Looking Ahead The energy-based approach for understanding classical mechanics presented in this book will provide a common framework for the treatment of a wide variety of problems. In this sense, it is a generally applicable way to understand physics. As with any physics course, it is always important to pay close attention to terminology and to be careful with your mathematics! And remember, the first step to each problem you encounter is to consider the following question: What happens to the energy?

Summary • • • •

5

System: A single object or a collection of objects. E?er�: An extensive scalar quantity associated with the location and motion of an object. Kinet1� energy: Energy associated with the motion of an object. Potential energy: Energy associated with the position of an objec t.

We will discuss state functions in more detail in Chapter 15 and Chapter 16.

CHAPTER ONE

I

7

• Energy of a system: The sum of all the kinetic and potential energies associated with all of the objects in the system. The energy of a system is a state function, which means that it depends upon only the current positions and motions of the objects in the system and not how these objects came to possess these quantities. • Isolated system: A system that cannot exchange energy with the outside environment. The total energy of an isolated system is, therefore, constant. • The law of conservation of energy: Energy can be neither created nor destroyed but can be converted from one type to another or transferred between objects. This is also referred to as the 1st law of thermodynamics. • The principle of minimum potential energy: All systems will arrange themselves in order to minimize their total potential energy. This follows from the 2nd law of thermodynamics. • Equilibrium point: A location where the potential energy curve for a system has no slope. A neutral equilibrium point occurs when the potential energy curve is flat; an unstable equi­ librium point occurs when the potential energy curve is convex; and a stable equilibrium point occurs when the potential energy curve is concave.

Problems 1. Do you get better gas mileage with your car when the air conditioning is turned on or when it is turned off? 2. A system consists of two long parallel wires carrying electricity in opposite directions. The presence of this electricity in the two wires results in a potential energy being created for the system. This potential energy has a dependence on the separation dis­ tance between the two wires that can be described using Figure 1.4. The two wires are initially held in place close together and then released. After they are released, what will happen? Specifically, will the two wires move closer together, move farther apart, or stay in the same position?

) Increasing Separation Figure 1.4

8

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNA/v'dCS

3. The potential energy existing between two atoms in a diatomic molecule has a dependence upon the separation of the two atoms shown in Figure 1.5. Three points are labeled on this potential energy curve. Describe the behavior of the system at each point. In other words, how will the atoms move if released from rest at that point? Are any of the points equilibrium points? If so, what kind of equilibrium point?

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Increasing Separation Figure i

4. A system consists of two blocks, A and B, connected by a single rope that passes over two pulleys as shown in Figure 1.6. How far does block A move to the right if block B moves down 2m? Figure i .6

5

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CHAPTER TWO Kinematics Learning is not attained by chance, it must be sought for with ardor and diligence. -Abigail Adams

A�

2- 1 Introduction

discussed in t�e previ�us chapter, potential ener�ies are fu�ctions of the positions of ob­ _ _ jects, and kinetic energies are functions of the motions of objects. Therefore, using energy as a basis for a mathematical description of the world around us requires mathematical definitions of both position and motion. We first recognize that defining where an object is located or how an object is moving requires two pieces of information. For example, we could describe a chair as being "1 m from the northeast corner of the room." This description includes a specification of both magnitude (1 m) and direction (from the northeast corner). Similarly, if we state that a car is moving "west at 20 mph," the magni­ tude is 20 mph and the direction is west. Descriptions such as these that include both a magnitude and a direction are referred to as vector quantities. Vector: A mathematical quantity specified by both a magnitude and a direction. The direction of a vector can be described in many ways and with respect to many different reference points. For example, while someone in St. Louis would describe the position of Chicago as being 300 miles to the north-northeast, someone in Detroit would describe the position of Chicago as being 250 miles to the west. Because of this potential subjectivity, we must always designate the reference point, or origin, for our vectors as the first step in using them. We will limit our initial discussion to !-dimensional vectors and 1-dimensional motion but will return to multi-dimensional vectors in Chapter 9. A summary of vectors and vector mathematics can be found in Appendix B.

2-2 Position and Displacement Since position is a vector, identifying the location of an object requires specifying the origin (or refer­ ence point) for the measurement. For !-dimensional motion, we will denote the origin of a !-dimensional coordinate axis as the zero point for referencing the magnitude of the position 9

lO

I

THE E�-.iERGY OF PHYSICS, PART i: CI.ASSiCAL MECHANICS Ai'JD THERMC)DYNAMICS

(i.e., the distance from the origin) and a positive (+) or Positive Direction negative ( - ) sign to denote the direction. For example, I )I El in Figure 2.1 we denote moving to the right along the 3 1 2 0 -1 -2 -3 coordinate axis to be the positive direction and moving to the left to be the negative direction. The increments Negative Direction along this axis ( -1, 0, 1, etc.) indicate measurements of distance from the origin. Although there are many units coo; of distance that would be applicable for describing posi­ tion (miles, feet, etc.), most scientific fields measure dis­ tances using the SI unit meter, which is denoted by the symbol m. We will follow this convention, too. Let's now label the 1-dimensional axis the Xgrno•1 = +1 m xblue = -2 m "x-axis." As shown in Figure 2.2, we can then denote the position of the blue circle that is 2 meters t,) for cJiffe,ent coordinole oxes. !_n 1he fixed coc�dinote _axis (Panel/-> leh) rhe position of the police cm and speeder both change w:th time, however. 1n t_!,e r,10•11r;g 8 1is.yhtL the

coordinote ox.is

of the

cu 1 is

ot rrv�

111

coordinate axis that has its origin at a fixed position on the ground ( e.g., at the initial position of the police car), we can instead choose a coordinate axis whose origin moves with the position of one of the objects in our system. For example, we could choose the origin of the x-axis to always be at the position of the police car. Let's refer to the x-axis in which the origin moves with the police car as coordinate system 1. The x-axis in which the origin is fixed at the initial position of the police car will be referred to as coordi­ nate system 2. We can then use subscripts to denote the coordinate system in which the position measurement is being made, as shown in Figure 2.10. For example, x52 would be the position of the speeder relative to coordinate system 2, and x.s: 1 would be the position of the police relative to coor­ dinate system 1. As shown in Figure 2.10, we can then relate the positions of the speeder relative to the two different coordinate systems with the -i---t-�i---+I -+---+I --,•x1 following equation: ----o+---+i

®

Xs,2

®

Xs, 1

i

0

--tl--+---+--+l--1--+1-,!IIJ,,x2

o----

Figure 2 ·10: The relotionships be­ tween the positions of the speede, measured in tvvo different coordi­ note systems.

In this equation, x1 _2 is the position of coordinate system 1 relative to coordinate system 2, which is the same as the position of the police car relative to coordinate system 2. x1,2

i =xS,l Figure 2. l l: Mnemonic for writing relative position equations.

= xP,2

In addition to using graphical representations of the position vectors to determine the equation relating the position of an object in different coordinate systems, as in Figure 2.10, we can also rely upon a simple mnemonic to remember how to construct these equations. As shown in Figure 2.11, the order of the subscripts for the different positions dictates their location in the relative position equation. The equation that relates the velocities of the speeder relative to the two different

CHAPTER TWO

I

23

coordinate systems is found through differentiation of the equation relating the positions of the speeder relative to the two different coordinate systems.

d d d -xS, =-x S, +-x , dt dt dt 2

1

1 2

In this last equation, v1,2 is the velocity of coordinate system 1 relative to coordinate system 2. Let's now use these equations for relative position and velocity to solve our problem with the speeder and the police car using coordinate system 1. The first step is to solve for the velocity of the speeder relative to this coordinate system.

vS, l =(4s

m

S

)-(so

m

S

)=-S

m

S

As expected, the direction of the velocity of the speeder relative to the police car is negative (i.e., the velocity is directed toward the police car), as shown in Figure 2.98. The equations for the position of the police car and speeder in this coordinate system are x1 ,, =x;, =Om and x1 ,, =500m-(5:

)M,

We can now use these equations to determine the time interval until the speeder is caught.

As expected this is the sam e solution as we obtained using coordinate system 2.

Example 2-7: Problem: A train is m oving in 1-di m ension with a speed of a 30 m /s. A person on the train is walking in the same direction as that of the train's velocity with a speed of 2 m /s relative to the train. What is the velocity of this person relative to the ground? Solution: The velocity of the person relative to the ground ( vP,c) is related to the velocity of the person relative to the train ( vP,T ) through the following equation:

24

I

THE ENERGY OF PHYSICS PART I: CLASSICAL MECHJi.NICS AND THERMODYNAMICS 1

In this equation, v is the velocity ofthe train relative to the ground. Let's define the p�sitive direction for our ;�tion to be the direction ofthe train's velocity. Substitution then yields m

m

m

=2-+30- � V PP =32S S S PP

V

Using relative motion to describe the relationship between the position and velocity of objects in different coordinate systems is also our first example of what is referred to as a coordinate transformation. Coordinate transformation: A change in the coordinate system used to describe a problem. This can include a change in the origin of the coordinate system and/or a change in the definition of the positive directions for the coordinate axes. We will use additional simple coordinate transformations in Chapter 4 and Chapter 5 for other 1-dimensional motion calculations and then return to a discussion ofrelative motion in Chapter 13 and Chapter 14 when we learn about momentum.

-

·

2-1 Position, Velocity, and Acceleration for 2-Dimensional Motion

Throughout this course we will rely upon our ability to describe multi-dimensional motion as a collection of Final separate 1-dimensional motions. Consider, for example, C Y2 __________.. the 2-dimensional motion depicted in Figure 2.12. If the x-axis and y-axis are perpendicular to each other (i.e., if 0 the angle between the two axes is 90 °), then an object C. � Y1 +----·tJ can move along the x-axis without changing its position Initial along the y-axis and move along the y-axis without changing its position along the x-axis. Because perpenX1 X2 dicular axes allow for this straightforward description of x position multi-dimensional motions as a combination ofseparate and independent 1-dimensional motions, we will always Figure 2.12: 2-dimensional motion from x 1 to x2 and from y 1 to y2 . use them in our coordinate systems. With the appropriate perpendicular axes for our coordinate system� we can· n���d���;ib-;th; 2-dimensional motion in Figure 2.12 as a combination oftwo independent 1-dimensional motions: 1-dimensional n_iotion along the x-axis from x 1 to x2 , and 1-dimensional motion along the y-axis from Y1 toy2. Ifthe obJect moves from its initial position to its final position during a time interval !it then the average velocity for these 1-dimensional motions would be

CHAPTER TWO

I

25

Example 2-8: Problem: An object m oves from x.I = 0 m andyi = 0 to x = 4 m andy = 6 m in 2 s What are f f • the average x-axis and y-axis velocities for this motion? Solution: Following the derivation above we have 4m-Om � 2s 6 m-O m Vy,avg = � 2s Vx,avg

=

Vx,avg

=2

m

Vy,avg

=3

m

s s

Although we can describe the 2-dimensional motion in Figure 2.12 as a combination of inde­ pendent 1-dimensional motions, there are, of course, several ways in which we can combine those 1-dimensional motions (i.e., there are several paths by which the object can m ove between its initial and final position). Three possible paths are shown in Figure 2.13. Path 2 Final In path 1, the x-axis and y-axis motion occur simultane­ C y2 0 ously; in path 2, the y-axis motion occurs first and then the x-axis motion occurs; and in path 3, the x-axis motion 'iii 0 � occurs first and then the y-axis motion occurs. There are, C. Path 3 � y1 Initial of course, countless m ore paths that connect the initial and final position of the object. As we will discover throughout the course of this book, there are so m e param eters used to describe a x position system that depend upon the path taken by the system as it changes in configuration (i.e., as the positions and Figure 2. 13: 2-dimensional motion from x 1 to x 2 and from y 1 to y2 can occur over velocities of the objects in the system change), and there different paths. are some parameters that are independent of the path --------·-------- ---�-·-·--------··--·taken by the system (Section 9-4 and Section 15-10). For this reason, we must always be aware of the initial state of our syste m, the final state of our system, and the path by which the system transitions between those two states.

I

26

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

Figure 2. 14: A system consi5t­ ina of h,10 blocks, a rope, ond a pulley -· �-- ------ .

I

x=O -----'--'---y=O Figure 2. l 5: Description of the posi­ tions of the in the system shown in F1gu1e 2 i 4 -----------------------

Finally, let's consider the system shown in Figure 2.14 that consists of two blocks, A and B, which are connected to each other by a rope that passes over a pulley 1 . We can describe the position of the blocks using two separate 1-dimen­ sional coordinate axes. For block A: the 1-dimensional axis, which we will denote as the x-axis, will be parallel to the horizontal surface on which block A sits and will have a positive direction pointing from block A to the pulley (i.e., to the right in Figure 2.14). We can define an origin for the x-axis (x = 0) on this surface, and then denote the position of block A as xA with respect to this origin (Figure 2.15). For block B: the 1-dimensional axis will be the vertical direction in Figure 2.14, which we refer to as the y-axis. Furthermore, we can specify the floor or the ground to be the origin of they-axis (y = 0) and denote the position of block B asy8 with respect to this origin(Figure 2.15). As mentioned in Section 1-6, it is clear that the presence of the rope in this system provides a constraint on the motion of the two blocks. Specifically, if block A moves 1 m to the right (positive x-axis direction) then block B moves 1 m down(nega­ tivey-axis direction). We can express this relationship between the displacements of the two blocks mathematically as

Since the rope maintains a fixed length (e.g., the rope does not stretch) we can express this relationship equivalently as

This is analogous to the average and instantaneous velocities being equal when the velocity is constant. Of course, had we defined different directions to be positive for the x-axis and y-axis in this problem, this derivative could have had a different(±) sign.

1

We have seen this system before (in Section 1-6) and will see it many times again.

CHAPTER TWO

I

27

Example 2-9: Problem: A system consists of two blocks, A and B, con­ nected by a single rope that passes over two pulleys, as shown in Figure 2.16. The x andy axes for describing the positions of the two blocks are also shown in the figure. For these axes, what is the value of dys? c1x.

Solution: It is clear from Figure 2.16 that as block A moves in the positive x-axis direction, block B will move in the negative y-axis direction. From the orientation of the rope we also see that if block A moves 1 m, block B will move only 0.5 m. Thus, 1 dy �Y ------8

9

_

LlXA

x=O

f

Ys

-----------------y- 0 Figure 2.16: A system consisting of two blocks, a rope, and a pulley.

_

dXA

2

Summary • Vector: A mathematical quantity specified by both a magnitude and a direction. • Position: A vector quantity specifying the location of an object. • Displacement: The change in the position of an object. It is a vector quantity that is indepen­ dent of the origin of the coordinate system in which it is measured. Ax=xjinal-xinitia/

�

llx=xf-xi

• Velocity: The rate of change of position. Specifically, it is the first derivative of position with respect to time. Velocity is a vector quantity. The equation for the instantaneous velocity is dx dt

v=-

The equation for the average velocity is Ax =­ �t avg

V

The average and instantaneous velocities are equal when the velocity is constant.

28

I

CS

HANICS AND THERMODYNAMI THE ENERGY OF PHYSICS, PART 1: CLASSICAL MEC

e o f elo � ity � ec i fica lly, it i s the first �er ivativ • Acceleration: The rate of change of veloc ity. Sp rati o n 1s a e l Acce . e tim o t ct e o s it io n with res� _ with respect to time or the second der ivative of p elerati o n 1s vector quant ity. The equatio n for the instantaneo us acc

dv d 2 x a=-=-2 dt dt The equation for the average acceleration is tJ.v aavg =tJ.t The average and instantaneous accelerations are equal when the acceleratio n is constant. • Equations for constant acceleration kinematics in one dimension. The follow ing equa­ tions can be used to relate the posit ion, velo city, and accelerat io n of an o bject mov ing in o ne dimension with an accelerat ion that is constant in t i me. tJ.v = atJ.t

vf2 = vi2 + 2atJ.x These equati ons are also presented in Appendix C in their vector form. • The motion of objects in a system can be described using a coordinate axis with an origin external to the system or with an origin internal to the system (e.g., an origin at the loca­ tion of one of the objects in the system). • Multi-dimensional motion of an object can be described in terms of the independent motion of the object along corresponding perpendicular 1-dimensional axes. • An object's speed will increase if its velocity and acceleration vectors are in the same direction. Similarly, an object's speed will decrease if its velocity and acceleration vec­ tors are In opposite directions.

Problems Conceptual 1. What does the speedometer in your car measure? a. Average Speed b. Instantaneous Speed

CHAPTER TWO

I

29

c. Average Velocity d. Instantaneous Velocity 2. Figure 2.17 shows the positions of two different moving objects as a function of time. Do objects A and B ever have the same instantaneous velocity? If so, during which time interval do the objects have the same instantaneous velocity? X (m)

A

t(s)

Figure 2. 17

3. Figure 2.18 shows the positions of two different moving objects as a function of time. Do objects A and B ever have the same instantaneous velocity? If so, during which time interval do the objects have the same instantaneous velocity? X

(m)

1

2

3

4

t (s)

Figure 2.18

4. You jump off the diving board into the pool. As you sink into the water your speed decreases because of interactions between you and the water. What is the direction of your acceleration while descending into the depths of the pool? 5. A space probe is landing on the moon. As it approaches the surface of the moon it slows its descent (i.e., decreases its speed) by using its rocket motor. What is the direction of the probe's acceleration? 6. In a movie a car is shown accelerating to the right. If the film is run backwards, will it show the car accelerating to the right or to the left?

Section 2-3

Quantitative

7. An object moves along a straight line at 5 m/s for 4 s. It then continues along in the same direc­ tion at 4 m/s for 6 s. What is the average velocity of the object for the 10 s it traveled? What is the average acceleration of the object during the 10 s it traveled? 8. An object moves along a straight line for 10 m at 5 m/s. It then continues along in the same direction for 21 m at 7 m/s. What is the average velocity of the object during the total 31 m it traveled? What is the average acceleration during the 31 m it traveled?

30

I

THE ENERGY OF PHYSICS, PART i 0 CL;\SSICAL N1ECHANiCS AND THERMODYNAMICS

Section 2-4

ment 9. The velocity of an object as a function of time is shown in Figure 2.19. What is the displace of the object between t = Os and t = 3 s? What is the acceleration of the object between t = 0 s and t = 3 s? v(m/s) 3 2

-1-----+----+----"I- t (s) 3

2

2 19

10. The velocity of an object as a function of time is shown in Figure 2.20. What is the displacement of the object between t = 0 s and t = 3 s? v(m/s) 2

-2

2 20

11. The acceleration of an object as a function of time is shown in Figure 2.21. The initial velocity of the particle is 2 m/s at t = 0 s. What is the particle's velocity at t = 3 s? a(m/s2) 6

2

3

12. The position of an object as a function of time is given by the following equation:

What is the velocity of the object at t = 4 s? What is the acceleration of the object at t = 5 s?

CHAPTER TWO

I

31

13. The velocity of an object as a function of time is given by the following equation:

What is the displacement of the object between t = 0 s and t = 2 s? What is the acceleration of the object at t = 2 s? 14. The acceleration of an object as a function of time is given by the following equation:

What is the change in the velocity of the object between t = 1 s tot= 3 s? What is the displace­ ment of the object between t = 1 s tot= 3 s if the velocity of the object is zero at t = 0 s? 15. The speed of an object as a function of time is given by the following equation:

What is the magnitude of the object's acceleration att = S s? 16. The speed of an object as a function of position is given by the following equation: V

2

= (7

1)

-;z

X

2

What is the magnitude of the object's acceleration atx = 10 m? 17. The speed of an object as a function of position is given by the following equation:

v'=(a;} What is the magnitude of the object's acceleration atx = 10 m? 18. The speed of an object as a function of time is given by the following equation:

What is the magnitude of the object's acceleration at t = 3 s?

Section 2-5

2 • How much time is 19. A car that is moving at 32 m/s starts to accelerate at a constant rate of-4 m/s to a stop? required for the car to come to a stop? What distance will the car travel while coming 2 • How much time is required for the 20. A car initially at rest accelerates at a constant rate of 3 m/s car to reach a speed of 30 m/s? What distance will the car have travelled when it has reached

this speed? 2 21. A car accelerates from 10 m/s to 30 m/s at a constant rate of 4 m/s • How much time is required for the car to accelerate from 10 m/s to 30 m/s? What distance does the car travel while accelerating? 22. A car starts from rest and travels 25 m in 5 s. Assuming that the car's acceleration was constant while it traveled the 25 m, what is the speed of the car after it has driven the 25 m? 23. A motorcycle that is initially at rest accelerates in a straight line at a constant rate of 2 m/s2 until it obtains a speed of 20 m/s. At that point, the motorcycle accelerates at a constant rate of -4 m/s2 until it stops. What distance does the car travel between when it starts and when it stops? 24. A car starts from rest and accelerates at a constant rate travelling a distance of 100 m in 5 s. The car continues to accelerate at the same rate for an additional 300 m. What is the speed of the car after it has travelled the full 400 m? Section 2-6

25. You are walking on a moving walkway in an airport. The length of the walkway is 58 m. If your speed relative to the walkway is 2.3 m/s and the walkway moves with a speed of 1.7 m/s, how long will it take you to reach the other end of the walkway? Assume that you are walking in the same direction as the walkway is moving. 26. A football player running at 4 m/s is chasing another football player running at 2 m/s. At one instant they are 12 m apart. How much time will pass from this instant until they collide? 27. A protoceratops is running at +4 m/s. The velociraptor chasing the protoceratops is running at +12 m/s. What is the relative velocity of the velociraptor in protoceratops's reference frame? 28. A protoceratops is running at +4 m/s. The velociraptor chasing protoceratops is running at +12 m/s. At one instant, the protoceratops is 160 m ahead of the velociraptor. How much time will pass from this instant until the protoceratops is caught? 29. Police driving with a velocity of 50 m/s decide to chase a speeder who is 3 km ahead and moving at 55 m/s. The police car accelerates at 2 m/s2 • Instantly the speeder becomes aware that he is being chased and starts to accelerate at 1 m/s2 • How much time passes until the police catch the speeder? 30. While you are driving, a pedestrian steps out onto the road 100 m ahead of you. Because you were texting while driving, you are not immediately aware that the pedestrian is in your path and don't step on the brakes until 1.5 s later. While skidding toward the pedestrian, the magnitude of your car's acceleration is 5 m/s2 • What is the maximum initial speed you could have had to avoid hitting the pedestrian?

CHAPTER THREE Energy and Energy Conservation The foundation of every state is the education of its youth. -Diogenes

3-1 Introduction

A

s we have discussed in the two previous chapters, each object in a system may have kinetic energy and/or potential energy depending upon its location and motion. Furthermore, since energy is an extensive parameter, the total energy in a system of objects is the sum of all the separate energies of the constituent objects. In this chapter, we will introduce three of these energies that are related to position and motion.

3-2 Translational Kinetic Energy Kinetic energy is an energy related to the motion of an object (Section 1-2). If an object is moving in any direction at any speed, it has kinetic energy. Because there are several ways in which objects can move (in a straight line, in a circle, etc.), there is, naturally, more than one type of kinetic energy. We define translational kinetic energy as the energy associated with the linear movement of an object (i.e., the motion of an object along a 1-dimensional axis). Translational kinetic energy: Energy associated with the linear motion of an object. The equation for translational kinetic energy for 1-dimensional motion is 1 2 K=-mv

(3-1)

In Equation 3-1, mis the mass ofthe object measured in kg, and v2 is the square ofthe speed with units of (m/s)2. Notice that since the translational kinetic energy depends upon v2, the direction of the ve­ locity, which is indicated by a (:±:) sign for 1-dimensional motion (Section 2-2), is not relevant. Indeed, a velocity of-5 m/s would have the same value ofv2 as a velocity of +5 m/s, specifically 25 m2 /s2 • All that matters is the magnitude of the velocity (i.e., the speed), which means that translational 33

34

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

kinetic energy is a scalar quantity. The SI unit of translational kinetic energy is the joule, named after James Joule, and is denoted by J. kgm 2 1J=1-2 s

Example 3- l : Problem: A 2 kg block is moving with a velocity of 2 m/s. The translational kinetic energy of the block then increases by 21 J. What is the new speed of the block? Solution:

The ( ±) sign in our solution reminds us that translational kinetic energy is independent of the direction of the velocity1 • Furthermore, since this question asked for the speed of the object, we need only report the magnitude of our solution. V

m =Sf s

3-3 Gravitational Potential Energy The gravitational interaction between objects gives rise to a potential energy that depends upon the masses of the objects and the distance between them. Gravitational potential energy is, thus, different from translational kinetic energy since it exists because of the interaction between 1 Another way of interpreting this result is that the assignment of the positive and negative directions for the coordinate axis for measurements of position is arbitrary.

I

CHAPTER THREE

35

two objects. The equation for the gravitational potential energy of two objects (with masses m 1 and m 2 ) separated by a distance r is mm U =-G-1 _2 u r

(3-2)

In this equation, G is the gravitational constant of the universe. m G = 6.67x10-11 J 2 kg It follows that the SI unit of gravitational potential energy is also the joule. Furthermore, since the magnitude of the gravitational constant is so small, large masses and/or small distances are required for objects to have significant values for gravitational potential energy. For example, in this course, we will often be concerned with the gravitational potential energy of macroscopic objects (blocks, balls, etc.) interacting with the Earth, which has a very large mass. Strictly speaking, Equation 3-2 is valid only when the distance r is much larger than the physical dimension (length, width, etc.) of the objects. In other words, Equation 3-2 is valid for calculating the gravitational potential energy of interacting point particles only. However, throughout most of this book, we will use Equation 3-2 even when this requirement is not satisfied. This is the first instance in which we will use an approximation. An approximation such as this is a helpful tool to simplify systems and/or the mathematics associated with their descriptions. In this case, we will be invoking what is commonly referred to as the point particle approximation in which we treat objects as though they were points without any physical size. When using the point particle approximation, we will usually define the location of the "point" corresponding to an object to be at the geometric center of the object2 •

Example 3-2: Problem: Consider a 2 kg block lying on the ground. What is the gravitational potential energy of this block and (i) a 4 kg block lying on the ground 3 m away; (ii) the Earth? Solution: We answer both questions using Equation 3-2. (i) For the 2 kg block and the 4 kg block, we have U =-(6.67x10-11�2 )( u kg

2

We will discuss this further in Section 10-5.

2kg 4kg ) � U =-1.78x10-10 J )( u 3m

36

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS (ii) To determine the gravitational potential energy of the 2 kg block and the Earth, we define the point corresponding to the Earth to be located at the center of the Earth. Thus, the distance between the object and the Earth that we will substitute into Equation 3-2 is the radius of the Earth.

U =-(6.67x10-11 Jm)( 9 kg 2

2kg . 7x1024 kg ) � U =-1.25x10sJ )(S 9 6.37x106 m 9

Note that gravitational potential energy of the 2 kg block and the Earth is many orders of magnitude larger than the gravitational potential energy of the 2 kg block and the 4 kg block.

When calculating the gravitational potential energy between the Earth and objects near the surface of the Earth, it is common to use a simplified form of Equation 3-2. In order to derive this expression, let's consider a small object of mass m that is a heighty above the surface of the Earth; we will again invoke a point particle approximation for both the Earth and the object. The gravitational potential between this object and the Earth is U9 = -c

Mm � U = -c : 9 RE Y

(

M

E

m

RE 1+L R

)

E

In this equation ME and RE denote the mass and radius of the Earth, respectively. We can expand the term in parenthesis in the denominator using the binomial expansion

(

l+x

)n

= l+nx+

n(n-1) 21.

n(n-l)(n-2)

x 2 +�---'--'---'-X 3 + ...

3!

Thus

Now l�t's restrict our interest to those instances in which the values ofy are much smaller than _ R E; this will be tru� for �ost problems we'll encounter in this course. This assumption allows us to _ approximate the bmomial expansion with only the first two terms. Under these conditions we have

y)

ME m U ,:::-G-( 1-9 RE RE

I

CHAPTER THREE

37

We, therefore, have two terms for our potential energy equation. ME m ME m G--y --t U = (U ) +m gy UB=-G-+ RE g 0 g R2 E

(3-3)

In Equation 3-3, we have defined the scalar g as

ME g= G-2 RE

m

4 g=9.82

s

The first term in Equation 3-3 is the gravitational potential energy between the object and the Earth when the object is at the surface of the Earth.

(u)

9 0

I kg

(5.97x1024 kg)m

=- (6.67x10-11� )�---�--t 6 2

6.37x10 m

(u)

9 0

J kg

=- (6.3x10 7 - ) m

The second term in Equation 3-3 is the additional gravitational potential energy between the object and the Earth when the object is not on the surface of the Earth. As we shall see, this second term is more important than the first for most of our calculations.

Exaf!1ple 3-3: Problem: A system consists of the Earth and a 2 kg block The 2 kg block is initially 5 m above the ground and when released from rest falls down to the ground. What is the change in the gravitational potential energy of this system when the block reaches the ground? Solution: 11uB = (u ) -(u ) g f g i !1U9 =((uu)o +mgy1)-((uut +mgy;) !1U9=((u0 t- (uu)J+(mgy1-mgy;)

au, =(21

au, =-98J

The negative sign in our solution informs us that the gravitational potential energy of the system has decreased.

38

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

t

It is important to note that although the ( U9 term in Equation 3-3 is much larger than the mgy term, it does not contribute to calculations of !).U since it is a constant and, thus, always cancels out by subtraction, as shown in Example 3-3, �r through differentiation, as we shall see in Chapter 4. Indeed, the change in gravitational potential energy of the system in Example 3-3 was dependent upon only the vertical displacement of the block (!).y). In general, it is changes in the energy of a system rather than absolute values of the energy that are important for mechanics. Because of this, we can effectively ignore the contributions of ( U 9 to the gravitational potential energy of systems in solutions to problems in classical mechanics. We can also consider ignoring the contributions of ( U9 as "redefining" the zero point of the gravi­ tational potential energy equation (Equation 3-3) to be located at the surface of the Earth, rather than at r = oo (as Equation 3-2 indicates). Indeed, since the change in the gravitational potential energy of the object depends upon only the vertical displacement of the object (Example 3-3), it must be independent of the origin of the coordinate system used to define the vertical position of the object 3 •

t

t

3-4 Spring Potential Energy The potential energy of a spring is a function of both the "stiffness" of the spring and the extent to which the spring has been extended or compressed. The equation for this potential energy is: = !k(l-10 ) s 2

u

2 (3-4)

In this equation, k is the spring constant, and 10 is the normal length of the spring (i.e., 10 corresponds to the length of the spring when it is neither compressed nor extended). The normal length of a spring is often also referred to as its unstretched length. The spring constant is a measure of the "stiffness" of the spring and has units4 of J/m2 ; the larger the value of k, the more difficult it is to compress or extend the spring. Finally, it follows from Equation 3-4 that the SI unit of spring potential energy is also the joule.

Example

3-4:

Problem: A spring with a sp:ing constant of SO J/m2 and a normal length of SO cm is initially compressed to 40 cm. What 1s the change in the potential energy of the spring if it is further compressed to 30 cm?

3 Recall from Section 2-2 that displacements are independent of the origin of the coor ct·mat e system used to measure position. 4 We will introduce alternative units for the spring constant in Section 7-2.

CHAPTER THREE

I

39

Solution: !J.Us =(U) sf -(U) s ;

�

!J.Us = _!2 k(lf-IO ) 2

!J.Us = _! ) 2 k((1f -IO ) -(I-/ i 0

2

2

_

_!k(l -1 ) Z ; o

2

)

&U, =½(SO� )((o.3m-O.Sm )' -(0.4m-O.Sm )') '

&U, =( 25 � )(o.04m'-O.Olm') -, &U, =0.75] '

The potential energy stored in the spring has increased since the length of the spring has been moved further away from its unstretched length.

Throughout this book we will assume that springs behave ideally. That is, their potential energy is described by Equation 3-4 regardless of the magnitude of the compression or extension. In real life, however, very large compressions or extensions of springs can result in behavior that cannot be described by this equation alone. In other words, Equation 3-4 is an approximation that is valid for small compressions and extension of the spring. 5 Furthermore, we will assume that all springs are massless so that we do not need to consider their kinetic and gravitational potential energies.

3-5 Energy Conservation in Isolated Systems We recall from Section 1-3 that the total energy of an isolated system will be constant. Therefore, any energy conversion that occurs for an isolated system must occur within the system and not involve interactions between the system and the outside environment. This fact forms the basis of a powerful problem-solving strategy that we will employ in this book. However, the successful application of this strategy requires us to define the system correctly and usefully. Let's consider an isolated system that consists of the Earth and a small block that is initially held in place a short distance above the ground (e.g., the system discussed in Example 3-3). When the block is released from rest, this system will react in order to minimize its total potential energy; this is a consequence of the principle of minimum potential energy (Section 1-4). Interestingly, this reaction will consist of both the blockfalling down toward the Earth and the Earth moving up toward the block. Because this system is isolated, as the block and the Earth move toward each other, their 5 For the sake of completeness, we note that Equation 3-1 is also an approximation that is valid for only small values ofv. You should feel inspired to take additional physics courses to learn more about this.

40

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

gravitational potential energy is converted into the separate and distinct translational kinetic ener­ gies of the block and the Earth (Section 1-3). However, because the Earth has a much larger mass than the block, nearly all of this converted energy will be transferred to the translational kinetic energy of the block; we will discuss this further in Chapter 13. Because of this large asymmetry in the energy conversion process, we can ignore the motion of the Earth and approximate this system to consist of only the block. In this definition of the system, the gravitational potential energy that exists between the Earth and the block is defined to be "possessed" by the block only since it will be converted almost exclusively into the translational kinetic energy of the block. Because so many of our examples will involve a gravitational interaction with the Earth, we will continue to use this assumption and limit our definition of systems to include only the objects that are undergoing significant changes in energy as well as anything that connects these objects (ropes, springs, pulleys, etc.).

Example 3-5: Problem: A 4 kg block is released from rest 3.6 m above the ground. What is the speed of the block when it hits the ground? You may treat the block as an isolated system. Solution: In this problem, the system consists of only the block. Therefore, the energy of this system is the sum of the kinetic and potential energies of the block alone. Specifically, it is the sum of the block's translational kinetic energy and gravitational potential energy. E=K+U

g

Because this system is isolated, its energy will be constant. In other words, the energy of the system will not change as the block falls to the ground.

Thus, the translational kinetic energy of the block increases as the gravitational potential energy of the block decreases. Another way of saying this is that as the block falls, the gravitational potent al �nergy of the block is converted into the translational kinetic energy � of the block. Subst1tut1on of the equations for the translational kinetic and gravitational potential energy gives us M

mz v z =70.561 S2

Since the question asked for the speed of the block when it hit the ground, we need report only the magnitude of our result. m V =8.4J S

Example 3-6: Problem: A block with a mass of 4 kg is moving across a horizontal surface with a speed of 5 m/s when it collides with a horizontally mounted massless spring with a spring constant of 400 J/m2 and a normal length of 75 cm, as shown in Figure 3.1. What is the minimum length of the spring following the collision of the block with the spring? You may treat the block and spring together as an isolated system. Sm/s --+

Figure 3. i: The system in Example 3-6. Solution: For this problem the system consists of the block and the spring. The energy of the system is the sum of the kinetic and potential energies of these two objects. Specifically, is the sum of the block's translational kinetic energy, the block's gravitational potential energy, and the spring's potential energy.

41

42

I

· CS ' AM1 ·� DYN · '"· M n · ' THi=R · AND . -,-·HAI'JI··cs - NII:�.. THE ENERGY OF PHYSICS, PART I: CLASSICAL

E=K+UB +Us Because this system is isolated, its energy will be constant.

Since the spring is massless, the only contribution to the gravitational potential energy .and the translational kinetic energy of the system comes from the block. Furthermore, smce the block is moving across a horizontal surface, there will be no change in its gravitational potential energy as it moves.

Thus, our equation for energy conservation becomes M +l!.U =0 In other words, the potential energy of the spring increases as the translational kinetic energy of the block decreases. To put it another way, the translational kinetic energy of the block is converted into the potential energy of the spring as the block compresses the spring. Substitution of Equation 3-1 and Equation 3-4 gives us the following expression: 1 ) )2 ( 1 1 1 l -1 )2 --k -k ( ( l. -1 + -mv2 --mv.2 = 0 2 / 0 2 1 0 2 t z I The maximum compression of the spring occurs when the potential energy of the spring is the largest. Therefore, the maximum compression of the spring will occur when the trans­ lational kinetic energy of the block has decreased to its smallest value. The smallest trans­ lational kinetic energy of the block is zero, corresponding to when the block has stopped moving6 • Let's take this to be the final point in our energy conservation equation and use a time before the collision for the initial point. Before the collision has occurred, the length of the spring will be equal to its normal (i.e., unstretched) length, and the potential energy associated with the spring will be zero.

6 This makes sense. If the block were moving then the length of the spring would be changing. Only when the block has stopped moving is the spring at its minimum length.

CHAPTER THREE I 43 Substitution yields

The (±) sign is there since the spring can store potential energy by being compressed Ur < 10) or by being extended Ur > 10). In this case, the spring must be compressed, and, there­ fore, the negative sign is appropriate.

Example 3-7: Problem: A 1.5 kg block is released from rest in con­

tact with a spring with a spring constant of 200 J/m2 and an initial compression of 20 cm, as shown in the Figure 3.2. What is the maximum distance travelled by the block up the nearby ramp? You may treat this as an isolated system.

.......

(

....

....

\\,.

\;d,' '

° .______....,._.,j,___\,.r.. 60 ___

Figure 3.2: The system in Example 3-7.

Solution: Let's choose as the initial and final points for our energy conservation equation

to be the instant the block is released and when the block reaches its highest point on the ramp, respectively. When the block has travelled its maximum distance up the ramp, its speed will be zero and thus its translational kinetic energy will be zero. Before the block is released, it is also at rest and therefore has no translational kinetic energy. Thus, both the initial and final translational kinetic energies of the system are zero, and we can write our energy conservation equation as

When the block is at its highest point on the ramp, the spring is no longer in contact with the block and, thus, must be at its normal length. Therefore,

44

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

( 200 �' )(o.2m)

'

�Y = �-�.,...-----, � �y = 27.2cm

l(

2( 1.5kg 9.8;) We can then use trigonometry to relate the distance up the ramp to the change in the height of the block. As shown in Figure 3.3, we can designate a new 1-dimensional coordinate axis, which we will call the s-axis, to be aligned parallel to the surface of the ramp with a positive direction pointing up the ramp. This change in the coordinate axis used to describe the position of the block is another example of coordinate transformation, as discussed in Section 2-6. We can relate changes in position along the s-axis to changes along they-axis using trigonometry.

, ' · r, ' , · i , p oeci Fi(JUrf-; ";.J· i\e1ot;c,n::>11 tv;ePn the soxis and y-axis for the systern in Exon-1p!e 3-7. 1

sin ( 60 ° ) = Y � sin ( 60 ° ) s = y � sin (60 ° ) as- = �y s Thus,

2 M=31.4cm

Finally, let's consider what happens to a rocket launched from the surface of the Earth. In order for the rocket to "break free" from the Earth's gravity (and not eventually fall back to the Earth), the ini­ tial kinetic energy of the rocket must be larger than the change in the gravitational potential energy that the Earth-rocket system will experience as the rocket moves from the surface of the Earth to an infinite distance away from the Earth. The minimum speed required for this process is referred to as the escape speed for the Earth; i.e., if the initial speed of the rocket equals the escape speed, then the speed of the rocket is zero when the rocket is infinitely far away from the Earth. If we assume that the Earth-rocket system is isolated (i.e., if we ignore all other gravitational interactions for the rocket and the Earth) then the equation for energy conservation for this system can be written as �E=0 � M+�U =0

CHAPTER THREE

I

45

Let's choose the initial point of our calculation to be when the rocket is on the surface of the Earth and the final point to be when the rocket is infinitely far away from the Earth. The final speed of the rocket, final kinetic energy of the rocket, and final gravitational potential energy of the system will all be zero.

(o-!m rock 2

et

)+(o-(-G v� 1 , rocket

m

M E rocket

R E

JJ=O

In this equation ME and R denote the mass and radius of the Earth, respectively, as in Section 3-3. We will ignore the change in the kinetic energy of the Earth since it will be much smaller than the change in the kinetic energy of the rocket. Simplifying this equation gives us E

m M 1 -mrocke vi2, ocket = G ro ket 2

v.,,

rocket

= �2G

t

,

M

RE

c

R E

r

-> v.,,

rocket

E

�

M E =2Gv� ,,rocket R E

.98x10 24 kg m J = 2(6.67x10-11 J 2 )(5 6 .37x106m kg

V.1,rocket

km =11.2S

The initial speed of the rocket must be greater than or equal to 11.2 km/s for the rocket to escape Earth's gravity. Notice that this answer is independent of the mass of the rocket. This occurs because the energy of this system is converted between translational kinetic energy and gravitational poten­ tial energy, both of which have the same linear dependence upon the mass of the rocket.

3-6 Energy Conservation in Non-Isolated Systems In non-isolated systems, energy is exchanged with the outside environment (Section 1-3). A common example is the energy exchange resulting from friction, which is an interaction occurring when two surfaces move past each other. Friction opposes the motion of these surfaces and, thus, decreases the kinetic energy of the associated moving objects. We refer to this decrease in energy as a dissipation ofenergy from the system and can model it as a rate of energy decrease per distance traveled by the surfaces (i.e., by the corresponding objects). When objects move through fluids like air or water, they interact directly with the molecules of the fluid (e.g., through collisions). This interaction, which also results in energy dissipation, is referred to as drag. Drag is distinguished from friction in that the magnitude of the drag interaction depends upon the velocity of the object moving in the fluid, whereas the magnitude of the friction interaction is independent of the velocity of the moving surfaces.

46

I

THE ENERGY OF PHYS/CS, PART I· CLASSICAL tv\ECH,-:\NICS AND THERN\ODYNA/v\lCS

Friction and drag are both examples of mechanical interactions (i.e., pushes or pulls). A change in energy of a system that results from a mechanical is referred to as work. Work: A change in energy associated with a mechanical interaction. Work is denoted by the variable W. This is another example of words having specific definitions in physics that may differ from the definitions in everyday language. Specifically, work is not a measure of energy but rather a measure of a change in energy.

Example 3-8: Problem: A block with a mass of 4 kg is sliding across a horizontal surface with an initial speed of 5 m/s. Because of friction, the energy of the block will decrease linearly with the distance travelled by the block at a constant rate of 2.5 J/m. How far will the block slide before coming to a rest? Solution: The system in this problem consists of only the block. The energy of this system is thus the sum of the kinetic and potential energies of the block. Specifically, it is the sum of the block's translational kinetic energy and gravitational potential energy. E=K+Ug

The change in the energy of this system resulting from the friction interaction is equal to the work done on the system by friction. M =W

._,on •••

fri

�

M + t1..Ug = wfriction

Because the surface along which the block slides is horizontal, there is no change in the gravitational potential energy of the block as it slides t1..y=O � t1..U =0 g

When the block comes to a rest, its speed and translational kinetic energy are zero. Let's define this to be the final point for the energy conservation equation. The initial point will correspond to when the block had a speed of 5 m/s. Thus,

CHAPTER THREE

Wl" i.-;ction

=M �

.. Wfriction

=!mvt -!mv; 2 2 2

m)

1 Wfriction . . =--(4kg) (52

2

2

S

�

I

47

wfriction __ ,! mvi2 2

� Wfriction .. =-50J

We can determine the distance the object slides by dividing this work by the rate at which friction dissipates energy from the system. �=

-50J

-2.s1-

� �=20m

m

This solution to Example 3-8 demonstrates a useful general statement. The work done on a system byfriction is always negative. We will return to this statement in Chapter 7 and Chapter 9 to understand why it is true. For now, it is sufficient for us to recognize that the work done by friction or drag is negative since these interactions "pull back" on moving objects and, thus, reduce both the speed and the translational kinetic energy of these objects. Similarly; positive work would be done on an object if an object was pushed forward (in the same direction as its velocity), and the object's speed and translational kinetic energy increased. Furthermore, we recall from Section 2-3 and Section 2-4 that an object's speed will increase if its accelera­ tion and velocity vectors are in the same direction, and an object's speed will decrease if its acceleration and velocity vectors are in opposite directions. Since an object's kinetic energy is determined from its speed (Equation 3-1), we conclude that there must be a relationship between mechanical interactions (pushes or pulls, e.g.) and acceleration. We also conclude that work will be positive if the direction of this associated acceleration is the same as the direction of the velocity; and that work will be negative if the direction of this associated acceleration is opposite the direction of the velocity.

Example 3-9: Problem: A pitcher is able to accelerate a 14 0 g baseball from rest to a speed of 40 m/s over a distance of 1.5 m. What is the work done on the ball by the pitcher? Solution: The system in this problem consists of only the baseball. The energy of this system is thus the sum of the kinetic and potential energies of the ball. Specifically; it is the sum of the ball's translational kinetic energy and gravitational potential energy.

E=K+Ug

48

ICS AND THERMODYNAMICS THE ENERGY OF PHYSICS, PART !: CLASS!CA.L MECHAN The pitcher is able to change the energy of the system by doing work on th� b�ll; specifically, by pushing or throwing the ball. we will assume that the system is otherwise isolate�, so ":"'e _ can ignore any work done by drag (i.e., air resistance). We will also use the approx1mat10n _ . that there is no net change in the vertical position of the ball durmg the p1t�h, and, therefore, that there is no change in the gravitational potential energy of the ball. With these assump­ tions the equation for energy conservation for this system is

� AfO W,

�

ei GI

ij

il GI

2

u

·1 u

.5

Increasing Height

Initial

.5

Figure.. 3. 13

- · · --------- --·-· --------�-------·-·---�------------

Increasing Height ----------

Initial

---·---

- --------·--

Quantitative

Section 3-2, Section 3-3, and Section 3-4

8. A 5000 kg asteroid is hurtling toward the Earth at 100 km/s. What is the asteroid's translational kinetic energy in terms of the number of tons of TNT producing the same energy? The energy released in the explosion ofone ton ofTNT is 4 x 109}. 9. The masses of the Sun and the Earth are 2 X 1030 kg and 6 X 10 24 kg, respectively. What is the gravitational potential energy oftheir interaction at their average separation of1.5 X 1011 m? 10. A spring with a spring constant of 250 J/m2 and a normal length of40 cm is initially compressed to 20 cm. What is the change in the potential energy ofthe spring ifit is then extended to 60 cm? Section 3-5

11. A block with a mass of 2 g is launched straight up into the air with a speed of5 m/s. What is the maximum height obtained by the block? You may treat this as an isolated system. 12. A ball is thrown straight up with an initial speed of20 m/s. The ball was released 2 m above the ground, but when it returns back down, it falls into a hole 11.5 m deep. What is the ball's speed at the bottom of the hole? You may treat this as an isolated system. 13. A 10 kg object is moving in a straight line with an initial speed of2 m/s. What distance will it take for the speed ofthe object to increase to 15 m/s ifits kinetic energy increases at a rate of5 J/m? You may treat this as an isolated system. 14. A 10 kg object is moving in a straight line with an initial speed of 2 m/s. How much time is required for the speed ofthe object to increase to 10 m/s ifits kinetic energy increases at a rate of20 J/s? You may treat this as an isolated system.

54

I

S AND THER/V\ODYNAMiCS THE ENERGY OF PHYSICS, PART I: CLASSiCAL MECHANIC

· . . of 50 cm is initially extended to a 15 A sprmg with a sprmg constant of 250 J/m2 and a normal length . . . spring 1f 1t 1s then compressed · length of 70 cm. What is the change in the potential energy of this to a length of 20 cm? You may treat this as an isolated system. . . e as shown 1� Figure 3.� 4. surfac d incline an of top the at rest from ed releas is block kg 5 _ 16 A e 1s 6? If the spn�g _ · The initial height of the block is 1.5 m and the angle of the incline� surfac sprmg (1.e., what 1s the of ss10n compre um maxim the is what J/m2 700 is spring the of t constan L).1t)? You may treat this as an isolated system.

.

1 1. s

m

so• '----------�-------------------Figure 3. 14 ----- ·---··· --·---·--

17. A 2 x 106 kg rocket is launched from the surface of the Earth. What is the escape speed of the rocket with respect to its gravitational interaction with the Sun? The initial distance of the rocket from the Sun is 1.50 x 1011 m and the mass of the Sun is 1.99 X 1030 kg. You may ignore all other gravitational interactions and assume that the system is isolated. 18. What would the radius of the Earth have to be in order for the escape speed of the Earth to equal the speed of light (3 x 108 m/s)? You may ignore all other gravitational interactions and assume that the Earth-rocket system is isolated.

Section 3-6 19. A 3 kg block is sliding across a horizontal surface. The initial speed of the block is 4 m/s, but because of friction the block's speed will decrease at a constant rate (i.e., constant acceleration) until the block finally comes to a stop after sliding 8 m. What is the magnitude of the acceleration of the block? 20. You lift a 2 kg block from the ground to a height of 2.2 m. How much work did you do in raising the block? 21. A 10 kg object is moving in a straight line with an initial speed of 10 m/s. What work must be done on the object for the object's speed to decrease to 4 m/s? 22. A block with a mass of 6 kg is sliding across a horizontal surface with an initial speed of 7 m/s. Because of friction, the energy of the block will decrease linearly with the distance travelled by the block at a constant rate of 4.2 Jim. What distance will the block slide before coming to a rest? 23. A 5 kg block is released from rest at the top of an inclined surface as shown in Figure 3.14 . The initial height of the block is 1.5 m and the angle of the inclined surface is 60°. As the block slides down the ramp, friction causes energy to dissipate at a rate of 10 J/m. The horizontal surface is frictionless. If the spring constant of the spring is 700 J/m2 , what is the maximum compression of the spring (i.e., what is 1).9?

CHAPTER THREE

I

55

24. A 2 kg block is launched up a 30 ° ramp with an initial speed of 10 m/s. The block will lose energy to friction at a constant rate of 2 J/m as it slides on the ramp (regardless of the direction). What is the speed of the block when it has slid back down to its starting point? 25. A spring is used to propel a 1.5 kg block up a ramp as shown in Figure 3.15. The horizontal surface along which the block initially moves is frictionless, but there is friction on the ramp which causes energy to be dissipated at a rate of 2 Jim. The block is initially held in place with the spring compressed 20 cm. When released, the block will slide up the ramp and then back down again, eventually recompressing the spring. What is the magnitude of the final compression of the spring (i.e., what is 111t)?

t�kg(

\ ·-ifi�0�ti;t

Figure 3. 15

CHAPTER FOUR Energy-based Mechanics I'm not afraid of storms, for I'm learning how to sail my ship. -Louisa May Alcott

4- 1 Introduction

I

n Chapter 2, we learned the terms and equations required for a quantitative description of 1-dimensional motion. In this chapter, we will combine this knowledge of kinematics together with the definitions of energy from Chapter 3 to begin our discussion of classical mechanics. Specifically, we will develop methods for using the total energy of a system to make predictions about how the positions and velocities of objects in the system will change over time.

4-2 Kinematics with Translational Kinetic Energy When discussing and calculating equations of motion (i.e., kinematic equations), it is natural to start with the energy of motion, kinetic energy. This will form the basis of our exploration of the physics of kinematics. We have already seen, for example, that the speed of an object can be determined from the object's translational kinetic energy.

Understanding how translational kinetic energy varies with time and/or position can, therefore, allow us to calculate directly the relationship between position and time.

dx dt

)½

.!

dx v=� -= ( - K 2 dt

2

m

Alternatively, an equation for the translational kinetic energy can be used to determine directly the corresponding acceleration. 57

58

I

THE ENERGY OF PHYSICS, PART I: CLASSiCf.,J MECHA!"iiCS Ai',JD THcFi/v\CDY,"lNv\:C

Let's consider three simple cases: and time. (i) Translational kinetic energy is constant in both position nt, then the speed ofthe object consta both Ifan object's mass and translational kinetic energy are the object's acceleration must must also be constant. It is tempting to conclude that this implies that not speed. Changes in only also be zero; however, acceleration is defined as a change in velocity, tional kinetic energy. We the direction of the velocity of an object will not affect the object's transla discuss this further in Chapter 5. (ii) Translational kinetic energy is a function of position. If an object's mass is constant, we can take the derivative with respect to position (with respect to the x-axis, e.g.) ofthe object's translational kinetic energy to determine the object's acceleration. 1 K=-m v2 2

�

dK dv dK dv dt -=mv- � -=m v-dx dx dx dt dx

dK 1 dK -=m va- � -=ma dx v dx 1 dK a=-­ m dx

This leads to two interesting conclusions: If an object's mass is constant, but its translational kinetic energy is linearly dependent upon position, the object's acceleration is constant. If an object's mass is constant, then the first derivati ve of the object's translational kinetic energy with respect to position is the product ofthe object's mass and the object's acceleration. (iii) Translational kinetic energy is a function of time. Simi�arly, we start with the derivative ofan object's translational kinetic energy with respect to time and agam assume that the object's mass is constant. K dv dK K=.!.m vz � d =m v � =m va 2 dt dt dt 1 dK a=-m v dt

CHAPTER FOUR

I

59

Thus, if an object's mass is constant, but its translational kinetic energy is linearly dependent upon time, the object's acceleration is not constant. This leaves us with an interesting tool at our disposal: we can visually examine any equation for translational kinetic energy to determine if the corresponding object is experiencing constant acceleration. Similarly, if given a plot of an object's translational kinetic energy with respect to time or position, a simple glance will tell us if the relationship is linear or not and, thus, allow us to draw conclusions about the object's acceleration.

Example 4-1: Problem: The translational kinetic energy of an 8 kg object moving in 1-dimension along the x-axis is given by the equation:

In this equation, x denotes the position of the object, and the velocity of the object is in the positive x-axis direction. How much time is required for the object to move fromx = 0 m to X = 3.5 m?

Solution: By looking at this equation, we can see that the translational kinetic energy is linearly dependent upon position. Therefore, since the object's mass is constant, we know that the acceleration of the object is constant. We can determine the magnitude of this ac­ celeration by differentiating the equation for kinetic energy with respect to x. 1

m

a= _!_ dK � a=(- )(16 .l ) � a=2 2 8kg m m dx s

From the constant acceleration kinematics equations in Section 2-5, we know that

We need only determine the initial velocity of the object (atx = 0 m) to determine the time interval. From the equation for translational kinetic energy, we have

60

ODYNAMICS THE ENERGY OF PHYSICS, PART !: CLASSICAL MECHANICS AND THERM

positive and one This solution from the quadratic equation has left us with two roots, one we will use the n, directio negative. Since the velocity of the object is in the positive x-axis positive root. Substitution then yields,

This equation has two solutions. M=ls Lit=-3.Ss However, only the positive root is physically real. Therefore, Lit=ls Alternative Solution: We could have also solved this problem using a more general approach (i.e., one that is applicable even when the acceleration is not constant). We begin with the relationship between speed and translational kinetic energy.

Substitution of the equation for translational kinetic energy in this problem then yields

w_e can now integrate both sides of this equation to relate a change in position (i.e., a displacement) to a change in time.

1( ( X;

dx

J 25J+ 16)X m

)½ =

I( t,

1 4 kg

)½ dt

CHAPTER FOUR

I

In this integration, we define the object to be at the position x. at time t. and at the position x1 at time t1 Performing the integration yields I

I

Substitution of the values of x1 and x; then gives us

�t=1s

Example 4-2: Problem: The translational kinetic energy of a 3 kg object moving across a horizontal fric­ tionless surface increases linearly with time, according to the equation:

What is the displacement of the object between t = 0 s and t = 9 s? Solution: Since the mass of the object is constant and the translational kinetic energy is a linear function of time and not a linear function of position, we know that the acceleration will not be constant. So, instead of using the constant acceleration kinematics equations, we will begin with the general relationship between speed and kinetic energy. v=

dx

dt

(

2 � v= m

)½ K

.!2

61

62

THERMODYNAMICS THE ENERGY OF PHYSICS, PART I: CIASSICAL MECHANICS AND Substitution of the equation for translational kinetic energy in this problem then yields

we can now integrate both sides of this equation to relate a change in position (i.e., a _ displacement) to a change in time. In this integration, we define the object to be at the posi­ tion x, at time t1 and at the position x1 at time t1

Performing the integration gives us

Substitution of the values of t1 and t1 then gives us

�=36m

4-3 1-Dimensional Kinematics with Constant Energy The examples of the previous section demonstrate that the kinematics of a system can be readily calculated if an equation for the translational kinetic energy of a system as a function of position or time is known. An equation for translational kinetic energy of this form can often be obtained from the potential energy associated with the system; recall that potential energy is a function of the position of an object or a collection of objects. Let's consider a block that is falling down toward the surface of the Earth with no energy dissi­ pated because of air resistance; we previously considered such a situation in Example 3-5. We refer

CHAPTER FOUR

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63

to this motion as free-fall. The block is moving in 1-dimension with the axis of motion corresponding to the vertical direction, as shown in Figure 4.1. As discussed in Section 3-5, we can ignore the motion of C: the Earth in this problem and define the system to consist 0 of the block only. The energy of this system is, consequently, ; (.) the sum of the kinetic and potential energies of the block. Specifically; it is the sum of the block's translational kinetic C energy and gravitational potential energy. u, 0 a.

E=K+U

g

y

Figure 4. 1: An block in free--fa!I.

E=.!_mv 2 +mgy+(u) 2 g 0

The equation for the energy of this system is a function of the speed and the vertical position of the block. We also see from this equation that when released from rest, the block will fall in order to minimize its gravitational potential energy in accordance with the principle of minimum potential energy (Section 1-4). In other words, a decrease in the value of the variable y, which is a measure of the position of the object above the surface of the Earth, will result in a decrease in the potential energy of the system. This is, of course, also consistent with our many observations of objects falling down and not up. Let's include an additional subscript ''.Y" for the speed in the translational kinetic energy term in our expression for the energy of the system; it will remind us that the translational kinetic energy of the block results from its movement along the y-axis. E= .!_ mv 2 +mgy+(u) g 0 2 y Since we are ignoring energy dissipation from friction, we can treat the block as an isolated system. Hence, during free-fall, the energy ofthe block is always the same (i.e., is constant) regardless ofits po­ sition, its speed, or how much time has passed since it was released. We can express these conditions mathematically in several different ways. For example, dE dE dE =0 =0 or =O or dt dv dy y

Let's first apply the condition that the energy is constant regardless ofthe vertical position ofthe block

2

d 1 dE 0 � -mv +mgy+ U = g dy ( 2 y dy

( ))

=0

0

!!_(.!.mv 2 )+ !!_ (mgy)+ !!_ (u ) = O dy g 0 dy dy 2 y

64 I THE ENERGY OF PHYSICS, PART !: CLASSICAL MECHN'-IICS AND THERMODYNAMICS The first term in this equation is the differentiation of the translational kinetic energy of the block with respect to the position of the block. We know from Section 4-2 that this derivative will be equal to the product of the mass of the block and the acceleration of the block.

.!!___(.!.mv 2 ) = maY dy 2 Y We have, again, included the subscript 'Y' to denote that the acceleration is directed along the y-axis. The remaining two derivatives are

.!!__(u ) =O BO d y

Putting it all together, we have dB =0 � may +mg=0 dy

This solution tells us that the magnitude of the acceleration is equal to the constant g. The direction of this acceleration is indicated by the negative sign. This occurred because there is an implicit vector definition present in our equation for the gravitational potential energy. Namely, that moving up from the surface of the Earth corresponds to the positive vertical direction (an increase in the gravi­ tational potential energy), as shown in Figure 4.1. We will further discuss the relationship between potential energy and the direction of acceleration in Chapter 7. The same value for the acceleration is obtained using the other expressions for energy conserva­ tion. Next, let's consider that the energy of the system is constant in time. d d dB =0 � .!!_(.!.mv 2 +mgy+(u) )=o � mv vY +mg y +0=0 dt dt 2 y B O y dt dt

This equation has the following two solutions: V

y

=0

CHAPTER FOUR

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65

Because we know that the block will be moving after we release it, only the second solution is physi­ cally real. Finally, let's use the constraint that the energy of the system is constant regardless of the speed of the block. d E =0 � ..!!.._(.!.mv 2 +mgy+(u) J=o � mv +mg dy +0=0 9 o dvy dvy 2 Y Y dv y

dy dt 1 mv +mg--=0 � mv +mgv -=0 � Y Y Ya dt dvy

a

y

Y

=-g

Of course, we could have guessed that the acceleration would be constant from our original equation for the energy of the system. Since energy is conserved, we have M'=O � M+dU =0 � M=-dU g

g

We know that the acceleration of the system will be con­ stant because the translational kinetic energy changes linearly with the position of the block (Section 4-2). It is important to note that the ( U ) term in the 9 equation for gravitational potential en°ergy did not contribute to the solution since it is a constant and, thus, has a derivative of zero regardless of whether we are dif­ ferentiating with respect to position, time, or speed. This is consistent with ( U9 ) not contributing to calculations of t!.U (Section 3-3). jndeed, we would have obtained the sa�e solution for our problem had we redefined the zero point of the gravitational potential energy to be at .As the surface of the Earth and completely ignored ( U0 discussed in Section 3-3, such a redefinition of the origin of the coordinate system is another example of how the use of a coordinate transformation (Section 2-6) can simplify calculations. We can use a graphical representation of the energy of this system, as shown in Figure 4.2, to complement the previous mathematical description of the associated energy conservation process. It is clear from Figure 4.2 that the block will fall down to the ground (i.e., decrease

t

-JldCS 1

system. In this case, the energy of the system is a sum of the translational kinetic energy and gravita­ tional energy of the block, and the elastic potential energy of the spring. E= .!. mv 2 +mg�+ .!. k(l-l° ) 2 2 /

2

We have defined the zero point of the gravitational potential energy in this equation to be the horizontal surface across which the block oscillates. We have included an additional subscript "/" for the speed in the translational kinetic energy term to remind us that the translational kinetic energy of the E block results from its movement along the /-axis. A plot of the potential energy for this system as a function of Cl) C the position of the block is shown in Figure 6.2. Because w the system is isolated, its energy will be constant as a function of the position of the block, as shown in Figure 6.2. Accordingly, the block will oscillate back and forth between the two turning points, / 1 and /2 • energy of the The next step in our general problem solving strategy Fi�Jure 6.2: tuncrion of in F is to determine the acceleration of the system through r;osdicn of .._,, differentiation of the equation for the energy of the is constant since the svstem is :so!ote,j The system with respect to the position of the block Since turnino 001n1·s, t CFld i' 1 a:. vve!i 05) the stable {:;qu:!ibriun-i pc:,fnr, :'_., c;: else :r:dicated. energy is constant for isolated systems, we have ,J

'

Since the block is oscillating on a horizontal surface

We can see that the acceleration of this system is not constant but, instead, depends upon the posi­ tion of the block This is expected because the kinetic energy of the system is not linearly dependent upon the position of the block but, instead, depends upon the square of the position 2. A non-constant acceleration is also consistent with the back and forth oscillation of the block, as indicated by Figure 6.2. Let's now define a new variable, co, called the angular frequency of the oscillation.

2

See Section 4-2.

I

CHAPTER SIX

l 31

The angular frequency is a measure of how quickly the block moves back and forth across the horizontal surface; angular frequency is always a positive scalar. Using this definition for angular frequency, we can write the kinematic equation for the motion of our block as d2l 2 -=-a> (1-l) 0 dt 2 You can verify that a solution to this equation is3

The variable A in this equation denotes the amplitude of the oscillation, and 0 )

(6-3)

Example 6-2: Problem: An objec t oscillates in one dimension with an amplitude of 10 cm. At t = 0 s, its position is x = 5 cm, and it is moving with a velocity of -2 cm/s. What is the angular frequency of the oscillation? Solution: We begin with Equation 6-3. x(t) = A sin( cot+t/)0 )

�

v(t) = Acocos( cot +¢0 )

Att= 0,we have

There are two solutions for this equation: !

0

c:;'j. .. •>

... >I

·················�

,-4IN

I

141

I

142

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

2 1 1 E=-mv 2 +mgy+-k ( y-10 ) 2 2 Y

Because the system is isolated, we know that energy will be conserved.

k d 2y -(y-10 )-g a =-=2 m dt Y You can verify that a solution to this equation is y(t)=Asin(wt+�0 )+(1,� '": J

as before. Through comparison with Equation 6-1 we see that the equilibrium length where o/ = !!_, m of the spring in this oscillating system is given by m 1e =lO - g k As expected, this system will not oscillate around the normal length of the spring but rather, around a compressed length. Nevertheless, the angular frequency of the oscillation is the same regardless of the orientation of the system.

,

, , ,

Figure 6,8: A simple pendulum,

,

1-i Sim�e Pencwla We can now discuss other oscillating systems that are not based on the motion on a spring but still exhibit harmonic motion. An example of such a system is the simple pendulum, which consists of a small object of mass m (called the bob) suspended from a horizontal surface by a massless rope of length L (Figure 6.8). The width of the bob is much smaller than the length of the string so that the bob may be treated as a point particle (Equation 5-12). As the bob swings back and forth, the energy of the system will be continuously converted back and forth between rotational kinetic energy and gravitational potential energy.

CHAPTER SIX

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143

To determine the equation of motion that describes the oscillation of the bob, we start with an equa­ tion for the energy of this system. 1 2

E=-Ioi+TnlJY When written in this form, the equation for the energy y=O _________ of the system is a combination of angular descriptions U g =O of motion (i.e., the angular velocity co) and transla­ tional descriptions of motion (i.e., the position along LcosO they-axis). T he subsequent differentiation and algebra required to determine the angular frequency of the os­ cillations of the pendulum can be simplified if instead, we express the energy in terms of only one description of motion. For pendula, it is simplest if we choose an Figure 6. 9: Parameter definitions for O simpie pendulum. angular description of the motion. Let's define the ori- --------- ·-------------.---gin of the y-axis and the zero point of the gravitational potential energy of the system to both be at the horizontal surface to which the string is attached (Figure 6.9). As shown in Figure 6.9, the equation that relates the vertical position of the bob to its angular position is y=-Lcos0

Hence,

1

E =-lco 2 -mglcos0 2

We can determine the acceleration of the bob by differentiating this equation for the energy of the system with respect to the position of the bob (i.e., with respect to 0). Since the system is isolated, we have mgL . dE 0 -=0 � Ia+mgLsin 0 = 0 � a=---sm

I

d0

Because the bob is a point particle, its moment of inertia can be determined using Equation 5-8. l=ml2

Thus, g . mgl . 0 0 � a=--sm a=---sm 2 L ml

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THE ENERGY OF PHYSICS, PART I: CLASSiCAL MECHANICS AND THERMODYNAMICS

This is an equation for oscillatory motion, but it is not simple harmonic motion because it does not meet the definition of simple harmonic motion (Section 6-3). Specifically, the acceleration of this system does not depend linearly on its displacement from equilibrium. However, if the amplitude of the oscillation is small, simple pendula do behave as simple harmonic oscillators. To understand why this is true, let's expand the trigonometric function in the equation for the angular acceleration of the pendulum using a Taylor series. 3

a=-B(e-0 + 9 L

31

s

1

-9 S! 7!

+ ...)

Ifwe limit the oscillation to only small angles, we can ignore all but the leading term in this expan­ sion. Thus, the equation for the angular acceleration of the system can be approximated as d 20 - 90 a=-= 2 L dt

(6-6)

This is commonly referred to as the small angle approximation7. We see that Equation 6-6 is an equation for simple harmonic motion and, furthermore (upon comparison to Equation 6-2)8, that the angular frequency of this oscillation is

Example 6-6: Problem: A simple pendulum consists of a 0.2 kg bob and a 0.5 m massless rope. What is the period of small amplitude oscillations of this system about its equilibrium point? Solution: We start with calculating the angular frequency of the oscillations. Since we are able to use the small angle approximation, we have m 9.82 2 i B � o l = __s_ � l = 1 6 rad 9 o = . o L 0.Sm s2

7 The small a�gle approximation is frequently written as sin8 - 8 and cos8 - 1. Bo� �quat1on 6-� and Equation 6-6 relate the second derivative of an object's position ( defined in terms of x or co) to th8e ob1ect s accelerat10n (either translational or angular).

CHAPTER SIX I 145

Therefore, the period of the oscillations (Equation 6-4) is 2nrad

2 T= 1t' -+ T=

ra 2 19.6 � s

(J)

-+ T=1.42s

6-7 Effective Mass and Effective Spring Constant for Oscillating Systems Because of the generality of our energy-based approach to kine­ matics, we can readily expand our previous derivations to include more complicated systems that undergo simple harmonic oscilla­ tion. Consider the isolated system shown in Figure 6.10 that con­ sists of two blocks connected by a massless rope that passes over a massless pulley. One of the blocks is connected to a horizontally­ mounted spring. Because of the existance of a spring in the system, when the blocks are displaced slightly from their equilbirium posi­ tions, this system will also undergo simple harmonic motion. To determine the angular frequency of this oscillation, we begin, as always, by writing down an equation for the energy of the system. The system in this problem consists of the two blocks, the rope, the pul­ ley, and the spring. Following the solution presented in Section 4-5 9, we can describe the positions of the blocks and the spring using two separate !-dimensional coordinate axes, as shown in Figure 6.11. With these definitions of the coordinate axes, the length of the spring is equal to the position of block A, and, thus, our expression for the energy of this system is

Figure 6. 10: A system consisting of two blocks, A and B, connect­ ed by a massless rope that passes over a massless pulley. Block A is also attached to a horizontally­ mounted spring.

Figure 6. 11 : The definitions of for the

the x-axis and y-axis In this equation, we have defined the zero point of the gravitational system in Figure 6.10. potential energy to be on the ground (i.e., aty8 = OJ. Because the rope connects the two blocks, the speeds of the two blocks must be equal. 2

2

VA,x =VB, y

9

This approach is also consistent with the general problem-solving strategy outlined in Section 4-8.

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THE ENERGY OF PHYSICS, PART i: CLASSiCL'..L MECHAi'-,i!CS Ai'lD THERMODYNAMICS

Hence,

we can now determine the acceleration of block A by differentiating this equation for the energy of the system with respect to the position of block A. Since the system is isolated, dE -=0 -"7 dXA

(m

+mB A

)a

dy

dy

(

B A +mag-+k x A -lo +mgA A ,x dX A dX A

) =0

The following two relationships we know from inspection of Figure 6.11:

Putting it all together we have

You can verify that a solution for this equation is

Therefore, block A will undergo simple harmonic motion with an angular frequency and equilibrium spring length of

Ie

mag

= IO + k

As expected, the spring is stretched initially. We could have guessed that this would be the angular frequency for this oscillation by compar­ ing this system to the system consisting of a single block attached to a spring (Section 6-2 and Section 6-5). The only difference here is that the total mass of the oscillating system in this example has increased through the addition of the second block.

CHAPTER SIX

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l 47

This result can also be readily understood in terms of the effective mass of the oscillating system (Section 5-8). Indeed, a general expression for the angular frequency of oscillation for a system undergoing simple harmonic motion can be written as oi =

k

ejfective

(6-7)

mejfective

In Equation 6-7, keffective denotes the effective spring constant of the system. As we shall see, the number of springs (or effective springs like pendula) in a system, their associated spring constants, and their location to the axis of rotation all contribute to the value of keffective . . We can also consider the effective spring constant to be a measure of the steepness of the total potential energy of the system as a function of the position of the oscillating object; the steeper the curve, the larger the value of keffective ..

Example 6-7: Problem: A spring with a spring constant of 288 J/m2 is attached to a uniformly dense solid cylinder with a mass of 3 kg that rolls without slipping across a horizontal surface around an axis through its center, as shown in Figure 6.12. What is the period of simple harmonic oscillations of this system about its equilib­ rium position?

Figure 6.12: The oscillating system in Example 6-7.

Solution: We begin with an equation for the energy of the system. 1 1 1 E =-lro 2 +-mv 2 +mgy+-kx 2 2 2 X 2 In this equation, we have defined the zero point of the gravitational potential energy to be the horizontal surface. We define the x-axis for describing the position of the cylinder to be parallel to the horizontal surface with a positive direction pointing to the right in Figure 6.12 and an origin where the spring is at its normal length. We have included an additional subscript "x" for the speed in the translational kinetic energy term to remind us that the translational kinetic energy of the block results from its movement along the x-axis (i.e., parallel to the horizontal surface). Due to the constraint that the cylinder rolls without slip­ ping, we know that the angular and translational velocities of the cylinder are related to each other by the radius of the cylinder (Equation 5-13). 2 2 vX =±mR � vX =ro R

2

�

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THE ENERGY OF PHYSICS, PART I: ClASSICAL MECHANICS AND THERMODYNAMICS

Substitution yields 1

v

2

)

1

1

E = -/ ( _.3._ +-mv 2 +mgy+-kx2 2 2 X 2 R2

1 /

1

� E=-(-+m ) v X2 +mgy+-kx 2 2 R2

z

Because the system is isolated, we know its energy will be constant dy / dE -=0 � (-+m ) ax +mg-+kx=O 2 dx dxA R Since the cylinder is rolling across a horizontal surface we have dy =0 dx

�

2 k ]x (_i_+m)a +kx=O � a = d x =-[ X dt2 X R2 I -+m Rz

Upon comparison to Equation 6-2, we see that the angular frequency of the oscillation is

As expected, the angular frequency of this oscillation is less than that of a block attached to a spring (Section 6-2) since the effective mass of this oscillator is larger. The period of oscillations of this system is

The equation for the moment of inertia for the cylinder can be found :!.mR'+mR' lm T=2rc 2 � T=2rc L R2k k Substitution of the values for the mass and the spring constant yields �(3kg) T=2rc -=---- � T=0.79s 288-z1 m

in Table D-2.

CHAPTER SIX

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149

Example 6-8: Problem: A 5 kg bob is connected to a thin massless but rigid rod of length L = 1.3 m to form a simple pendulum, as shown in Figure 6.13. The bob is also connected to a nearby vertical wall by a spring with spring constant k = 75 J/m2 • What is the angular frequency of small amplitude oscillations of this isolated system? When the spring is at its equilibrium position (i.e., at its unstretched length), 0 = 0.

m

Solution: We begin by writing down an equation for the energy of the system.

Figure 6 I 3 The oscilloting system in Exornple 6-8.

1 2

1 2

·-�------· - . ·-·- ------- ··-- -� -- -- ---------------

2 E=-Jo/ +-kx +mgy

As discussed earlier, it is simplest if the equation for the energy of the pendulum is written in terms of angular descriptions of motion only. If we define the zero point of the y-axis to be at the horizontal surface of the system as before (Figure 6.9), we have 1 2

y

=

o--------

1 2

E =-Io> 2 +-kx 2 -mglcos0

As shown in Figure 6.14, if the amplitude of the oscillation is small, the displacement of the spring can be related to the angle of the oscillation by x = Lsin0

x=O

Figure 6. 14 The reloiionship be­ tween x, y and 11 for the system in Figure 6.13. ---

Substitution of this expression gives us 2 E = .!.10/ +.!.kl2 sin 0-mglcos0

2

2

Since the system is isolated, this energy must be constant. dE d0

=0 � /a+kL2sin0cos0+mglsin0=0

2 2 kL sin0cos0+mglsin0 � a= (kL cos0+mgl) sm . 0 -__::::.a=_.:.:=.�-=-=-=--I

I

150

I THE ENERGY OF PHYSICS, PART I: CLASSiCA.L N\ECHANICS .AND THERMODYN/l-MICS Because we are interested in small amplitude oscillations, we can apply the small angle approximation to this equation.

a

co= kL2 +mgL =d 2 0=- kL2+mgL � J ( 0 I I dt 2

Substitution of the equation for the moment of inertia for a point particle gives us

Hence, for our system

CO=

co= kL2+mgL � co=JkL+mg ml mL2

(1s;;(,}.3m)+(Skg)( 9.8�) (skg)(1.3m)

__,

rad co=4.75s

The solution to Example 6-8 identifies another conclusion about simple harmonic oscillators. The angular frequency for small amplitude oscillations of the system in Figure 6.13 can be written as

CO2 =CO2 _ +CO2 pendulum spnng In other words, the value for ai for a system is the sum of the individual values of ai correspond­ ing to the different oscillating elements in the system. This is true since oi is determined from the acceleration of the system, which is determined from the equation for the energy of the system (Section 4-8). Since we can linearly sum the energies of the objects in a system to determine the equation for the energy of the system, it follows that we can linearly sum the values of ai for the individual oscillating elements in a system to determine the value of oi for the entire system 10.

10 Although this approach will work for the simple oscillating systems in this book, it is not universally valid and cannot be applied to more complex oscillating systems (such as coupled oscillators). Please be inspired to take additional physics classes to learn why this is true.

CHAPTER. SIX

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151

6-8 General Simple Harmonic Oscillator Approximation As discussed in Section 1-5, stable equilibrium points of a system occur where the system's potential energy curve is concave. Because of this curvature, systems can undergo oscillations around stable equilibrium points. Depending upon the specific shape of the potential energy curve, these oscilla­ tions may or may not be simple harmonic oscillations. However, as with the case of simple pendula, if the amplitude of the oscillation is small, then the oscillation can be approximated as simple harmonic motion. To understand why this occurs, let's consider a system whose potential energy is a function of the position of an object along the x-axis. The minimum of this potential energy occurs when the object is at x = x0 • We can write an equation for this potential energy as a function of position relative to x0 using a Taylor series expansion.

u(x)-u(x,)+[ it}-x,)+ :,( �� }-x,)' + ... ,., The potential energy is a minimum when x = x0 •

Hence,

If we limit ourselves to only small amplitude oscillations (i.e., oscillations in which x is always very close to x0), we can ignore all higher order terms in this expansion, thus, leaving us with

Therefore, under these conditions, the potential energy will depend upon the square of the displace­ ment of the system from its equilibrium position. Because of this, the associated oscillations will be simple harmonic (Section 6-3). This leads to a very powerful conclusion. Small amplitude oscillations of any system near stable equilibrium points can be modeled as simple harmonic oscillations. Furthermore, the effective spring constant for these oscillations is dependent upon the curvature (i.e., the steepness) of the potential energy function near the equilibrium point.

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THE ENERGY OF PHYSICS, PART I: CtASSICAL. MECHANiCS AND THERMODYNAMICS

d 2U keffective =dx 2

x=x0

This is consistent with the discussion in Section 6-6. It's also worth mentioning here that since th effective spring constant for a system is determined from the second derivative of the potentic energy of that system, potential energy terms with linear dependencies on position, such as th gravitational potential energy in Equation 3-3, do not affect the effective spring constant and thus di not affect the angular frequency ofoscillation. This is, indeed, what we have observed (Section 6-5;

Example 6-9: Problem: An isolated system consists ofa 1.5 kg mass moving in the presence ofthe follow­ ing potential energy function.

What is the period ofsmall amplitude oscillations ofthis system about its stable equilibrium point? Solution: We begin by determining the location of the equilibrium points of the potential energy function.

We ignore the additional complex roots since they are not physically relevant. We can deter­ m n ifthe remaining real roots correspond to stable or unstable equilibrium points by deter­ � � mmmg the value ofthe second derivative of the potential energy function at those locations. = (20 �s :� ,.,. (

}•+ �, JL. )x 3 -(2-l ))x= �l ( m 1

=-2 �, unstable

, =( 2 a7L(�)3m ( 0 ms

�

m2

=6-J2 stable m

Th� e�ective spring constant for small amplitude oscillatio ns around the stable equilibrium pomt 1s, therefore, 1 keffective =6m2

CHAPTER SIX

I 153

The period of oscillations is, thus, T = 2,r m

� T = --;::::=2,r== -> . k,,.. e ecti.ve JJ

r�

+

21C

6 J _nL 1.5kg

->

T=1 w' = ':__3'_ ---> w' = wspnTlfJ ' . m 4m 2 m 4m 2

-(..!]_) 2m

2

2J

Following the discussion at the end of Section 6-6, we can identify the term ( : in this equation as the value of corresponding to the oscillating energy loss in the system. Recall that since the rate of energy loss in this system is proportional to the speed of the block, the rate of energy loss will oscillate just as the position and speed of the block oscillate. Furthermore, since the ( ::n r term is associated with dissipation of energy of the system, it should decrease the angular frequency of oscillation for the system (i.e., it should appear with a negative sign in the equation for m2). This is analogous to how the presence of friction decreased the magnitude of the acceleration of a block sliding down a ramp (Example 4-8).

w

Example 6-10: Problem: A block with a mass of 2 kg is attached to a horizontally-mounted spring with a spring constant of 200 J/m2 , as shown in Figure 6.6. The surface along which the block oscillates is perfectly horizontal. As the block oscillates, it dissipates energy at a rate of dE = -17vX dx What value of 1J will result in a 50% reduction in the angular frequency of oscillation of the system? Solution: According to our previous derivation, we have

2

2

2

( !)

k -11 k q' � q' 3k co=-� --=-----=2 m 4m 4 m 4m2 4m 4m

CHAPTER SIX

I

g 11 =34.6 k s

We refer to an oscillating system in which energy is continually added as a driven oscillating system. Let's, again, consider as a model the oscillating system shown in Figure 6.6. The equation for the energy of this system is 1 1 E =-mv 2 +-kx2 +mg y 2 2 X

In this equation, we have defined the zero point of the gravitational potential energy to be the horizontal surface on which the block oscillates. We will describe the addition of energy to this system using the time dependent sinusoidal function

In this equation, the variable E (with units of J/m) is the small rate of energy addition that has an angular frequency of ro0• Differentiating the equation for the energy of the system with respect to the position of the block gives us dE = ma +kx +mg dy � ma + kx +mg dy = £ cos ( ro l) dx x dx x dx d2x d -1[=0 � maX +kx=Ecos(rol) � m-2 +kx=Ecos(rol) dx dt

You can verify that a solution to this equation is x(t) = Asin

(( f/k);;; ] t+¢

0

+

£cos( rol)

m(;-w;)

The second term in this equation results from the periodic addition of energy to the system. As expected, the magnitude of this term depends upon the difference between the angular frequency of the mass-spring oscillator ( rosprino ) and the angular frequency of the energy ad­ dition(%); it will be largest when rosprino = ro0 • If these two angular frequencies are identical,

l 55

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHAN!CS AND THERMODYNAMICS

we say that the frequency of energy addition is in resonance with the natural frequency of the system. As an example of a driven oscillator in resonance, consider a parent pushing her child on a swing. The parent must synchronize her pushing to the swinging motion of her child in order to increase the amplitude of each swing. For our model system, we see that during resonance (i.e., when wspring = m0), the amplitude of the oscillation becomes infinite. There is, of course, always some energy dissipation pres­ ent in all oscillating systems, and, thus, it is not be possible to achieve an infinite amplitude oscillation. Indeed, a more realistic equation to describe a driven oscillating system would be dx d2 x m-2 +17-+kx = ecos ( OJi ) dt dt A solution to this equation is

X

()

t =Ae

-(i} . ( Zm

)

e( k-mm�)cos(mi)+11w0 sin(mi))

( sm rot+0. We see immediately from Figure 7.11 that 1 Av A:,vblock -2 :, F

Hence, ( 6kg)( 9.8

;)

F=---"---� 2 F=29.4N

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS J..ND THERMODYNAMICS

We can define the mechanical advantage of a simple machine as the "amplification" of the force obtained by increasing the distance over which the force acts.

Mechanical advantage: A measure of the force amplification obtained by using a tool, mechanical device, or machine. For the ramp in Example 7-7 and the system ofpulleys in Example 7-8, the mechanical advantage is 2. In general, the mechanical advantage ofa ramp is equal to the length ofthe ramp divided by the height ofthe ramp, and the mechanical advantage ofa system offixed and moving pulleys is equal to twice the number ofmoving pulleys9 • Furthermore, the total mechanical advantage of a system is the product of the mechanical advantages of each component of the system.

Example 7-9: Problem: What is the mechanical advantage of the system of fixed and moving pulleys shown in Figure 7.12? Solution: Since the system consists of three moving pulleys, the mechanical advantage of this simple machine will be 6. Thus, the magnitude of the external force required to the rope to lift the block in Figure 7.12 would be equal to 1/6 ofthe magnitude ofthe force of Fit1u1e 7.12: The simple gravity acting on the block. mcchine in Exarnpie 7-9. A screw is a simple machine that works as a modified ramp. Indeed, the thread of the screw can be considered to be a ramp that wraps around the shaft of the screw. It follows from the derivation ofthe mechanical advantage oframp that the mechanical advantage ofa screw would be equal to the ratio ofthe circumference ofthe screw to the pitch of the screw (see Figure 7.13).

Figure 7. 1 3: A screw with o diameter d and pitch h. The mechuni­ cal odvcmtoge of the screw is equal to Tf ���. --�------�·-·-·--·---�----U_ 1

9

_

This simple calculation assumes that the pulleys are mass Jess and fnct1onless · · , and that any ropes are

also massless.

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A crank is an arm attached at a right angle to a rotating disk or shaft, as shown in Figure 7.14. Cran_ks are _a compone�t of devices ranging from mechanical pencil sharpen to bicycles to auto­ ers mobile engmes. What 1s the mechanical advantage of turning the disk by applying constant force a to the arm (F2 in Figure 7 .14) rather than applying a constant force directly to the disk (F in Figure 7.14)? In other words, what is the mechanical advantage ofusing the crank's arm to turn {he disk?

F,

FiS:;1u1e / 14 A sirnpie machine consisting of a crank arm attached to a disk.

Since the arm is attached to the disk, the angular displacement of the arm and the disk must be equal.

The work done by F1 or F2 to change the rotational kinetic energy of the disk can be determined using Equation 7-9. The work done by these forces must be equal to generate the same change in rotational kinetic energy of the disk and thus to cause the same angular displacement of the disk. MF. =MF. 1

2

� WF. =WF. 1

2

� F/1.s 1 =F./ls2

The variable & in this expression denotes a tangential distance through which the force has acted. These tangential distances can be related to angular distances using Equation 5-1.

And since the angular displacements are the same we have

!J.01 =!J.02

�

F1(r1 1102 )=F2(r211 02)

�

r/'1

=

r2F2

�

r

� = /F; 1

The mechanical advantage of the crank is thus equal to the ratio of the radius of the disk and the length of the arm. We can build upon this solution to determine the mechanical advantage of a bicycle. As shown in Figure 7.15, we can model the bicycle as_ a crank (�n �rm oflength r2 attached to a disk of radius rJ attached by a belt to a disk of radius r3 • T�1s latter _disk 1s then �ttached to a l�rger _ disk of radius r4• What is the mechanical advantage of turnmg the disk with radms r4 by applymg a

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directly to the disk constant force to the arm (F2 in Figure 7.15), rather than applying a constant force (F4 in Figure 7.15)?

Figure 7.15: Modeling a bicycle as a combination of two simple machines.

The work done by F2 or F4 to change the rotational kinetic energy of the disk with radius r4 can be determined using Equation 7-9. The work done by these forces must be equal to generate the same change in rotational kinetic energy of the disk and thus to cause the same angular displacement of the disk.

The angular displacement of the disk with radius r4 must be the same as the angular displacement of the disk with radius r3 •

Since a belt connects the disk with radius r3 to the disk with radius r1, the tangential speed of the belt on the surface ofthese two disks must be equal. Ifwe assume that the belt passes over both disks without slipping, the tangential speeds of the disks at their surfaces must be equal.

Substitution ofthis result yields

ll.8

3

= '2 /l.8 r_

3

2

�

F

1

) ( )

(r ll.8 = F 1

1

4

r r

4 2 ll.8 r_

3

2

Finally, we recognize the angular displacement of the disk with radius r must be the same as the 1 angular displacement of the arm with length r2 •

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As expected, the mechanical advantage of the bicycle is thus the product of the mechanical advantage of the crank (r/r2 ) and the mechanical of the other two attached disks (r/r

J.

7-8 Work, Kinetic Energy, Force, and Acceleration Using Equation 7-10, we can now define an isolated system as a system upon which no non-con­ servative forces are acting. Since energy is conserved for isolated systems, for an object moving in 1-dimension along the x-axis we have

max - LF;,x = O � LF;.x = max � Fnet ,x = max i

The sum of all of the conservative forces acting on an isolated system is equal to the product of the mass of the system and the system's acceleration.

Example 7-10: Problem: An isolated system consists of a block with a mass of 4 kg that is released from rest a height of 5 m above the ground. What is the acceleration of the block as it falls to the ground? Solution: Since this is an isolated system, the acceleration can be determined from the sum of the conservative forces acting on the system. In this case, the only force is the force of gravity. Let's denote the vertical direction for the acceleration and the force to bey-axis. Fnet,y =maY � -mo=maY � aY =-g This result is consistent with what we obtained in Section 4-3.

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This relationship between the sum of the forces acting on a system and the acceleration of the system is one of Newton's laws of motion that we will discuss in Chapter 8.

Summary • A force is: (i) a push or a pull; (ii) has an agent that does the pushing or pulling; (iii) acts on an object; (iv) is a vector; and (v) results either from the contact between objects or from a long-range interaction. • Net force: The sum of all the forces acting on an object. • The work done by a force acting on an object is equal to the integral of the force with respect to the displacement of the object.

• Conservative forces are associated with potential energy functions. For example, dif­ ferentiating a potential energy function with respect to x will give the component of the associated force along the x-axis. dU F' =--I dx The work done by a conservative force acting on an object is equal to the negative of the change in the associated potential energy of the object.

Wconservative =-ll.U Thus, the work done by a conservative force acting on an object is independent of the path of the object's displacement. • Non-conservative forces are not associated with potential energy functions. The work done by a non-conservative force acting on an object depends upon the path of the object's displacement. The most common non-conservative forces are: external pushes and pulls; ten­ sion; the normal force; and friction. • The total potential energy of a system is a minimum when the system is at equilibrium. Thus, no net force acts on as system when the system is at equilibrium. • The net work done on a system is equal to the change in the kinetic energy of the system.

M w, the woman will feel heavier as she accelerates upwards.

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Example 8-10: Problem: A 60 kg woman is n·d·mg m · an elevator that is accelerating downwards at , 2 . the woman s apparent weight? 9. 8 m/s . What is Solution: We can use the same f ree bodY d"1agram as m Example 8-9 (Figure 8.13). Hence, . the apparent weight of the woman is, again, wapp =m(a +g) Y Since the wom an is accelerating downwards, the com ponent of her acceleration along the y-axis is negative. Thus, m aY =-9.8- � wapp =(60kg)(-9s m +98 m ) S2

s2

•

. s2

wapp =ON Since wapp < w, the woman will feel lighter as the she accelerates downwards. In fact, since her apparent weight is zero, she is weightless.

Now, let's imagine that the woman riding in the accelerating elevator in Example 8-9 drops a block above the floor of the elevator and m easures the motion of the block as it falls. Since the woman is inside the elevator, she would use a reference frame moving with the elevator for her m easurements of the position, velocity, and acceleration of the block What would she perceive for the acceleration of the block within this reference frame? The acceleration of the block with respect to the elevator and the Earth can be written in terms of a coordinate transformation (Section 2-6). ablock ,Earth =ablock ,elevator +aelev tor , a

Earth

The woman is measuring the motion of the block relative to the elevator. In acceleration of the block is

ablock ,elevator -a - block ,

Earth

-aelev tor , a

Earth

this reference frame, the

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AMICS THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYN

5 law to express Now:, since the Earth is an inertial reference frame , we can use Newton's 2nd a in terms of the mass of the block and the magnitude of the net force acting on the block r:i;�;;; to the reference frame of the Earth. This net force is simply the magnitude of the force of gravity of the Earth on the block, which we will denote as F9 •

F ablock ,elev tor =-9---aelev tor , mblock a

a

The equation for the net force acting on the block ( mb/ock her reference frame inside the elevator) is, therefore,

a

Earth

block,eleva tor

mblock ablock,elev tor =Fg -mblock aelev tor , a

a

) according to the woman (and

Earth

The second term on the right side of this expression ( mb/ock elev tor, arth ) is referred to as a fictitious force; although it has the units of a force, we cannot attribute any agent to it (Section 7-3). a

a

E

Fictitious force: An apparent force acting on an object in a non-inertial reference frame. Fictitious forces arise whenever we apply Newton's 2nd law within non-inertial reference frames, such as the accelerating elevator6 • The acceleration of the elevator gives rise to an apparent force that both affects the acceleration of the block and makes the woman feel heavier or lighter depending upon the direction of the elevator's acceleration. In Example 8-10, the fictitious force acting on the woman has the same magnitude but opposite direction of the force of gravity acting on her. Thus, with respect to the non-intertial reference frame of the accelerating elevator, the net force acting on he woman is zero and thereforev the acceleration of the woman in that reference frame is also zero. Similarly, let's consider what happens to you as you drive your car around a turn. You feel yourself being pushed toward the outside of the turn and interpret this sensation as resulting from the action f a force. However, no such force exists since we cannot attribute any agent to it (Section 7-3). nstead, what is happening is that you are continuing to move in a straight line in accordance with ewton's 1st law but as the car turns you "collide" with it. There is no force pushing you around nside the car, but rather the apparent force you feel is just a result of the fact that you are trying to pply Newton's laws within a non-inertial reference frame (i.e., a reference frame fixed within the interior of the car, which is accelerating since the direction of its velocity is changing). Indeed, someone standing by the side of the road watching you drive your car around the turn would perceive your motion differently; she would describe everything from her point of view with respect to her inertial reference frame7 • As you drive your car around the turn, she would see you continue to move in a straight line (in agreement with Newton's 1st law) only to hit the car turning "beneath" you. She does not need any fictitious force to explain why you end up pressed against the side of your car as you drive through the turn. Rather, she interprets your motion relative to the car as resulting from the fact that the car is accelerating and you are not. 5 6 7

A reasonable approximation for this situation (see Section 8-2). Indeed, if aelevator,Earth = 0, there is no fictitious force. Again, we will assume that the surface of the Earth can be treated as an inertial reference frame for this situation.

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Lastly, let's consider the case of a weightless astronaut in orbit around the Earth. Since the as­ tronaut is under the influence of a gravitational force from the Earth, there must be an additional fictitious force acting on her to make her weightless. Just as in the case of the weightless woman in the elevator (Example 8-10), the fictitious force acting on the astronaut arises because she is accelerating. Indeed, since the astronaut is moving around the Earth rather than in a straight line, we know that the direction of the astronaut's velocity is changing and, thus, that she is accelerating. Furthermore, based upon the solution to Example 8-10, we can conclude that astronauts in orbit around the Earth are weightless because they are accelerating downwards towards the Earth with an acceleration whose magnitude is exactly equal to the magnitude of the gravitational acceleration at their location. It is interesting to note, however, that since the astronauts stay in orbit, this "downward" acceleration is not causing a change in their position. We will resolve this conundrum in Chapter 11 when we discuss an additional acceleration that is associated with circular motion and the associated fictitious force.

Summary • Reference frame: A system of coordinate axes used to describe the position, velocity, and acceleration of a system. • Inertial reference frame: A reference frame that is not accelerating. Newton's laws are valid in inertial reference frames only. • Non-inertial reference frame: A reference frame that is accelerating. Fictitious forces appear to act in non-inertial reference frames. • Newton's 1st law: An object that is at rest will remain at rest, or an object that is moving will continue to move in a straight line with constant speed if and only if the net force acting on the object is zero. • Newton's 2nd law: The net force acting on an object is equal to the time rate of change of that object's linear momentum. dp dt For systems of constant mass, we can express Newton's 2nd law as Fnet =

Fnet =ma • Free body diagram: A pictorial representation of the forces that act on a_ system. • Kl netic Frtcti. on.. The force of friction that opposes the continued motion of an obJect. The · of k"met1c · · t·10n · fnc magmtude of the force of k"met'1c friction is determined from the coe ff"1C1ent and the magnitude of the normal force fk =µkn e the direction of the velocity. . The direction of the force of kinet IC friction is always opposit . · de of the · t·wn that opposes the start of mot10n. The magmtu • Static Friction·· The force O f f nc · de of . . · ed from the coefficient of static friction and the magmtu force of static friction .1s determm the normal force

J: ::;;; µsn

· tion is as required to prevent the start of motion. · f nc The direction of the force of static

210 • • • •

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THE ENERG'{ OF PHYSiCS f1ART ! 1

when they are taut. Tension: The force that exists within ropes (and cables, strings, etc.) Weight: The magnitude of the force of gravity acting on an object. s to support an ob1ect. Apparent weight: The magnitude of the normal force that a _sur�ace exer_t _ rt1al reference frame. non-me ma Fictitious force: An apparent force that appears to act on an obJect

Conceptual 1. Two constant forces are applied to a 2 kg block that is at rest on a frictionless and horizontal surface as shown in Figure 8.14. What is the direction of the resulting acceleration of the block? 2. The work done by static friction a.

is always positive

b.

is always negative

c.

can be positive or negative

d.

is always zero

+x

�,2kgl�

3. The work done by kinetic friction a.

is always positive

b.

is always negative

c.

can be positive or negative

d.

is always zero

4. An astronaut is in an elevator that is accelerating upward (i.e., away from the ground). Is her apparent weight greater than, less than, or equal to her weight?

Section 8-4

Quantitative

5. A net force of 46 N acts on a 2 kg object. What is the magnitude of the acceleration of the object? 6. A 25 kg object is accelerating at a rate of 4 m/s2• What is the magnitude of the net force that acts on the object? 7. An object has an acceleration of 7 m/s2 when pushed with a constant 35 N net force. What is the mass of the object? 8. After landing with an initial speed of 80 m/s an airplane comes to a stop after moving 500 m down the runway under constant acceleration. What is the magnitude of the average net force that a 75 kg person riding on the airplane would experience as the airplane slows to a stop? 9. A block with a mass of 6 kg is at rest on a horizontal and frictionless surface. A constant 12 N force is then applied to the block as shown in Figure 8.15. What is the displacement of the block 4 s after the force is applied? What is the work done by the 12 N force during the first 4 s the force

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I

211

is applied to the block? What is the change in the kinetic energy of the _ 12 N block durmg the first 4 s the force is applied to the block? 10. Two c�n�tant forces are applied to a 5 kg block that is initially at rest on a frictionless and horizontal surface as shown in Figure 8.16. What u is the displacement of the object 2 s after the forces are applied? What Fig !.�-��L�------ ______ _ is the work done by the 2 N force during the first 2 s after the forces are a�plied? What is the work done by the 12 N force � v,yt '. ,Jt � _ durmg the first 2 s after the forces are applied? What is the change in the kinetic energy of the block during the first 2 s �!��-=--�---1___�--after the forces are applied? 11. The velocity of a 2 kg object is given by the following equation:

)�.:�,A

What is the net force acting on the object at t = S s? 12. The following time-dependent force is applied to a 2 kg block that is initially at rest.

What is the work done by this force during the first 3 s it is applied? What is the displacement of the block after 3 s if this is the net force acting on the block? 13. The following time-dependent net force acts on a 2 kg object that is initially at rest.

What is the work done by this net force during the first 4 s it acts on the object?

Section 8-7 14. A 2 kg block moving along a horizontal surface at 4 m/s is subject to kinetic friction with a coef­ ficient of µk = 0.16. What distance will the block travel before coming to a stop? What is the work done by the force of kinetic friction as the block slows to a stop? 15. A block with a mass of 4 kg is initially at rest on a horizontal surface. The coefficient of kinetic friction between the block and the surface is µ k = 0.2. What is the magnitude of the acceleration of the block when a horizontal force with a magnitude of 10 N is applied to it? What net work has been done on the block after the block has been sliding for 4 s? What work has been done on the block by the 10 N force after the block has been sliding for 4 s? What work has been done on the block by the force of kinetic friction after the block has been sliding for 4 s? 16. A block with a mass of 4 kg is pushed across a horizontal surface by a constant force; the block started from rest. The coefficient of kinetic friction between the block and the horizontal surface is µk = 0.2. After having been pushed 7 m along the horizontal surface, the speed of the block is 4 m/s. What is the magnitude of the pushing force?

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THE ENERGY OF PHYS!CS 1 PJ\RT I:

AND THERtv'\!'.JDYNJ\MlCS

Section 8-8 17. A block with a mass of 4 kg is pulled across a horizontal surface at constant speed by a horizon­ tally directed rope as shown in Figure 8.11. The coefficient of kinetic friction between the block and the surface is µk = 0.2. What is the tension in the rope? What work has been done on the block by the tension in the rope after the block has slid 5 m? What net work has been done on the block after the block has slid 5 m? 18. A block with a mass of 4 kg is pulled across a horizontal surface by a horizontally directed rope as shown in Figure 8.11. The coefficient of kinetic friction between the block and the surface is µk = 0.2. What is the acceleration of the block if the tension in the rope is 12 N? What work has been done on the block by the tension after the block has been sliding for 4 s?

Section 8-9 19. A 78 kg woman is riding in an elevator that is accelerating upwards at 1.8 m/s2• What is the woman's apparent weight? 20. A 60 kg woman is riding in an elevator that is accelerating downwards at 3.8 m/s2• What is the magnitude of the ficticious force acting on the woman?

CHAPTER NINE Newtonian Mechanics II The roots of education are bitter, but the fruit is sweet. -Aristotle

9- 1 Introduction

N

ewton's laws provide the bridge between a description of the forces acting on a system and the associated kinematic description of the motion of the system. To take full advantage of the usefulness of this relationship, we must include a description of force, position, velocity, and acceleration as vector quantities. Specifically, we will employ the vector notation presented in Appendix B to perform the vector mathematics inherent in Newton's 2nd law and in the associated kinematic calculations. For this reason, it is advisable to review the material presented in Appendix B and Appendix C before moving ahead with the material in this chapter.

9-2 Newton's 2nd Law As discussed in Section 8-3, Newton's 2nd law defines the relationship between the net force acting on a system and that system's linear momentum. In Equation 9-1, we express that relationship for systems of constant mass using the appropriate vector notation.

F =ma net

(9-1)

Example 9-1: Problem: The position of a 2 kg object is given by the following equation.

213

D.

What is the net force that acts on this object? Solution: we can determine the acceleration ofthe object using

Equation C-1 and Equation C-3.

The net force acting on the object is then found using Equation 9-1.

'

pie Problem: Two perpendicularly aligned forces are acting on a 5 kg block, as shown in Figure 9.1. What is the magnitude of the resulting acceleration of the block? Solution: Let's define a reference frame for this system consisting of perpendicular x and y axes that are parallel to 8 N and 6 N forces, respectively. Furthermore, let's define the positive direction for these axes to be the same as the direction of these forces, as shown in Figure 9.2. Using this reference frame, we can express the net force acting on this block as

The acceleration of the block can then be determined using Equation 9-1.

_ [8m],. [6m],.

a=

s;z x+ s;z y

!6N

r:::"J

8N

)�

8N

1

6N

f;::l

--"»�

+x

l

+y

CHAPTER NINE

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The magnitude 1 of the acceleration is, thus, a=

(� 5

2

m s2 )

+(� sm) 5

2

2

m � a=2-2 s

We also could have determined the magnitude of the acceleration from the magnitude of the net force.

Fnet

=ma � 10N=(Skg)a � a=2;

As discussed in Appendix 8, vector arithmetic is greatly simplified by vector decomposition (i.e., by expressing the vector in terms of its components along the coordinate axes of a reference frame) . We will, therefore, always decompose all forces into their components along the coordinate axes of the free body diagram before applying Newton's 2nd law to that free body diagram.

Example 9-3: Problem: A block is suspended from the ceiling by two ropes, as shown in Figure 9.3. What is the magnitude of the tension in each rope? Solution: Let's define a reference frame for this sys­ tem in which the y-axis denotes the vertical axis with a positive direction pointing up from the ground. The x-axis of this reference frame will be perpendicular to they-axis with a positive direction pointing to the right in Figure 9.3. In the free body diagram for the block, we will denote the magnitude of the tension in rope 1 as T1, the magnitude of the tension in rope 2 as T2 , and the magnitude of the force of gravity as F9 • With these definitions, the free body diagram for the block is shown in Figure 9.4. 1

See Section B-4.

---- Lllf-----

Rope ��--�A"ope 2

Figure 9.3: A block suspended from the ceiling by two ropes in Example 9-3. -------- ------�---" ------·--- ________________.,_ ,

.

y

Figure 9.4: The free body diagram for the block in Figure 9.3.________ ____

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THE ENERGY OF PHYSICS, P,t,,RT I: Ci.ASSiCAl MECHfaJ'·�iCS J\l'-ID THE:RMODYNAMICS

In order to simplify our application of Newton's 2nd law to this free body diagram, we next decompose the two ten­ sion vectors into their components along the x-axis and y-axis of the reference frame. The free body diagram for the block after this decomposition is shown in Figure 9.5. Applying Newton's 2nd law to each of the coordinate axes in this free body diagram yields the following equations 2 :

y

-E----e-----l�

T 1 cos0 1

X

T2 cos0 2

Fg

(Fnrt ) y = may � T1 sin01 + T2 sin02 - FB = maY

We can safely assume that the block will remain at rest and, thus, that the components of its acceleration along the x-axis and y-axis will both be zero. Substitution of aX = 0 and ay = O into the equations above yields

We can now algebraically solve for T1 and T2• cos01

T2 =T1 cos02

�

J.

. cos01 T1 sm01 + ( T-sm02 =mg 1 cos02

Similarly, T2 =

mg sin02 + cos02 tan01

R cal that we will denote the x-axis component of the net force as F ( ne,,) and the y-axis component of the net force as '. ; ( ) net y

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9-3 Redrawing Free Body Diagrams Although vector decomposition is an important step in the application of Newton's 2nd law to free body diagrams, it will be of limited effectiveness if the reference frames of these free body diagrams are not appropriate for the solution of the associated problems. The most useful reference frames for free body diagrams are those which have a coordinate axis aligned with the direction of motion of the system. Whenever possible, free body diagrams should be constructed such that at least one of their axes corresponds to an axis along which the system is moving. Let's consider a system that consists of a block being pulled across a horizontal and frictionless surface by a rope, as shown in Figure 9.6. The direction of the block's displacement is parallel to the horizontal surface. Because of this, we should choose one of the coordinate axes in the reference frame for the block to be aligned with this direction. Let's define this to be the x-axis and, further, specify that the positive direction for the x-axis points to the right in Figure 9.6. We can then define they-axis to be perpen­ dicular to the x-axis with a positive direction pointing up from the horizontal surface. In the free body diagram for the block, we will denote the magnitude of the tension in the rope as T, the magni­ tude of the normal force as n, and the magnitude of the force of gravity as F . With these definitions, the free body diagram for the block is shown in Figure 9. 7. The next step in our solution is to simplify this free body diagram by decomposing the vector for the tension. The free body diagram for the block after this decomposition is shown in Figure 9.8. Applying Newton's 2nd law to each of the coordinate axes in this free body diagram yields the following equations:

Figure 9 .6: A block being pulled across a horizontal surface by a rope. y

Figure 9.7: The free body dia­ gram for th_�_�lockin Fig��e 9.6.__

y

(Fnet ) X = maX � Tcos0 = max (Fnet )

y

= ma

y

� n+Tsin0-FB = may � n+Tsin0-mg = maY

Now, let's assume that as we pull the block along the horizontal surface the tension in the rope is not sufficient to lift the block up off �f the surface. In other words, let's assume that there is no displacement of the block along the y-axis in Figure 9.8 and, therefore, that the component of the block's acceleration alon� the y-axis is zero. Substitution of aY = 0 into the equation above yields ay =0 � n+Tsin8 = mg � n = mg-Tsin0

Tsin0

n

Tcos0

-----------,-x

Figure 9.8: The free body diagram for the block in Figure 9.6. It was obtained by decomposing the ten­ sion vector in Figure 9.7.

---- --------- - -----------·--------

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECH/ANICS AND THERMODYr,4AMICS

Since the magnitude ofthe normal force cannot be less than zero3, there is a limit for the magnitude ofthe tension in the rope such that aY = 0. In other words, there is an upper limit with how strongly we can pull on the rope without lifting the block off the surface. mg n�0 � mg-Tsin0�0 � mg�Tsin0 � T5, sin0 Let's examine the predictions of this equation under two extreme values of the angle fJ. First, let's consider the situation in which the rope is oriented parallel to the horizontal surface (i.e., 0 = 0°).

In other words, it is impossible to lift the block off ofthe horizontal surface in Figure 9.6 if the rope is aligned parallel to that surface. No matter how strongly we pull on the rope (up to infinite force), the block will never move in the vertical direction. This is consistent with our physical intuition. Next, let's consider the situation in which the rope is oriented perpendicular to the horizontal surface (i.e., 0 = 90 ° ). 0=90° � T5,

mg � T5,mg sin( 90° )

We now see that there is an upper limit to the magnitude ofthe tension. This indicates that the block will not lift off the surface unless the magnitude of the tension force is greater than the magnitude of the force of gravity acting on the block. Again, this is consistent with our physical intuition. It is possible, however, that our initial choice ofrefer­ ence frame may not be ideal for the final solution of the problem. In these situations, we must choose a new ref­ erence frame and redraw our free body diagram. Let's consider the system shown in Figure 9.9 that consists of a block sliding down a frictionless ramp. There are two forces acting on the block: the force of gravity and the normal force. In the reference frame for __ ?''.':'_:��?-���----�--�:ng �(J_,�'-"�� this system, let's denote the vertical axis as they-axis with a positive direction for they-axis pointing up from the surface of the Earth. The x-axis for the reference frame will be perpendicular to they-axis with a positive direction for the x-axis pointing to the right in Figure 9.9. In the free body diagram for the block, we will denote the magnitude of

3 4

The magnitude of a vector cannot be less than zero (see Section B-4). As mentioned earlier, whenever possible, always check your solution in a few limiting cases.

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the normal force as n and the magnitude of the force of gravity as F . y With these definitions, the free body diagram for the block is show� n in Figure 9.10. As the block slides down the ramp, it will be moving along both the x-axis and y-axis of this reference frame. As shown in the solution to Example 4-3, this motion can be described most easily with respect to a coordinate axis that is parallel to the surface of the ramp rather than with respect to the reference frame in Figure 9.10. Thus, it would be best if one of the axes of the reference frame and associated free body diagram for the block were directed parallel to the surface of the Figure 9. l 0: The free body ramp. To accomplish this, we can redraw our free body diagram with diagram for the block in Fiaure 9. 9. The surface of the a new x-axis defined to be parallel to the surface of the ramp with a ro'i'np is indicoted by the positive direction pointing up the surface of the ramp5 and a new y­ dashed line. axis defined to be perpendicular to the surface of the ramp (i.e., perpendicular to the new x-axis) with a y positive direction pointing up from y the surface of the ramp. This is the ) n n equivalent of rotating our original reference frame and associated free -�-----x body diagram (Figure 9.10) clockwise through the angle 0 (i.e., the angle of the ramp). The redrawn free body dia­ gram is shown in Figure 9.11. Figure 9.11: Free body diagram for the obiect in Figure 9.9. In Figure 9.11, the force of gravity The force of gravity is, subsequently, decomposed into its x-axis is further decomposed into its compo­ and y-axis components. ----nents along the x-axis and y-axis of the reference frame. Applying Newton's 2nd law to each of the coordinate axes in this free body diagram yields the following equations: (Fnet ) =maX � -Fg sin0=maX � -mgsin0=max X

a =-gsin0 X

(Fnet )y =maY � n-F9 cos0=maY � n-mgcos0=may Since the object doesn't spontaneously lift off the ramp as it is sliding down, the component of the block's acceleration along they-axis will be zero. Substitution of a1 = 0 into our equation above yields ay =0 � n-mgcos0=0 � n=mgcos0

5

This is the orientation of the s-axis in Example 4-3.

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

It is worth noting that the magnitude of the component of the acceleration along the x-axis is identical to that which we determined in Example 4-3 using an energy-based approach to this problem.

Example 9-4: Problem: A 4 kg block is sliding down a ramp, as shown in Figure 9.9. The angle of the ramp is 0 = 30°, and the coefficient of kinetic friction between the block and the surface is µk = 0.2. What is the magnitude of the acceleration of the block? Solution: In the reference frame for this system, let's denote the vertical direction as the y-axis with a positive direction for they-axis pointing up from the surface of the Earth. The x-axis will be perpendicular to the y-axis with a positive direction for the x-axis pointing to the right in Figure 9.9. In the free body diagram for the block, we will denote the magnitude of the normal force as n, the magnitude of the force of kinetic friction as fk' and the magnitude of the force of gravity as F9• y With these definitions, the free body diagram for the block is n shown in Figure 9.12. However, as discussed earlier, since the block is sliding down the ramp, it would be best to use a reference frame in which one of the coordinate axes was directed parallel to the direction of the block's velocity (i.e., parallel to the sur­ face of the ramp). As before, we can obtain such a reference frame and corresponding free body diagram by rotating Figure 9. 12 The free body the free body diagram in Figure 9.12 clockwise through an diagram for the block in angle 0. This new free body diagram is shown in Figure 9.13. Example 9-4. The surface of Applying Newton's 2nd law to each of the coordinate axes the ramp is indicated by the dashed line. ·---·--------- ----in this free body diagram yields the following equations: y

y

n

)

Figure 9. 13: The free body diagram for the block in Example 9-4. The force of 9rav1ty 1s, subsequently, decomposed into its x-axis and y-axis � 0 s 60.3 ° 0max =60.3 °

We can now modify the problem-solving strategy presented in Section 9-6 to include the application of the rotational form of Newton's 2nd law. Step 1: Specify the reference frame for the system. Step 2: Identify the forces that act on the system. Step 3: Draw the free body diagram for the system. All of the identified forces in Step 2 should be included with their correct orientation within the reference frame of the system. Step 4: If necessary, redraw the free body diagram. Whenever possible, free body diagrams should be constructed such that at least one of their coordinate axes corresponds to an axis along which the system is moving. If the original reference frame does not satisfy this goal, it should be redrawn with new coordinate axes that do satisfy this goal.

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Step 5: Decompose all vectors into their components along the coordinate axes of the free body diagram. Step 6: Apply Newton's 2nd law to each of the coordinate axes of the free body diagram. This will generate a single algebraic equation for each coordinate axis. A(::!:: ) sign can be used to denote the direction of the forces with respect to the coordinate axes. Step 7: Determine the torque associated with each of the forces acting on the object and then combine these torques to determine the net torque acting on the object. Step 8: Apply Newton's 2nd law to this net torque to determine the angular acceleration of the object. Step 9: Determine the relationship between the angular and translational accelerations of the object. a=±aR

Be careful with the(::!::) sign conventions for the accelerations! Step 10: Use algebra to relate the magnitudes of the forces to each other or to the ac­ celeration of the object.

Summar)"

, ,,

• Center of mass: The mass weighted center of a system.

The center of mass has the following important properties: . . , geo• The center of mass of any symmetrical object of uniform density is at the physical (1.e. metric) center of the object. on the center of • The force of gravity acting on a system can be modeled as though it is acting mass of the system. . d'irectly ab ave the . 1s • An object can balance on a pivot only if the center of mass of the obJect r ference frame for the sys­ • ��:!enter of mass is invariant with respect to the origin of the � rotat10n for the system. tem and with respect to the location of any potential axis of its center of mass. • An object that is freely rotating will always rotate around ion of a system around an axis, • Torque: A measure of the ability of a force to cause the rotat ent of force. fulcrum, or pivot. Torque is also frequently called mom

f=rxF

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

The magnitude of the torque can also be expressed as

-r=rFsin0 Positive torques cause counterclockwise rotation, and negative torques cause clockwise rota­ tions. • Newton's 2nd law: The net torque acting on an object with constant moment of inertia is equal to the product of the object's moment of inertia and angular acceleration. -rnet =la

• The work done by a torque acting on an object is the integral of the dot product of the torque with the object's angular displacement.

The work done by the net torque acting on a system is equal to the change in the rotational kinetic energy of the system. Mrot =Jfnet •d0

These equations for work and change in rotational kinetic energy can also be expressed in terms of the average torque and average net torque, respectively.

w =f

avg

-�o

Mrot =(fnet )

avg

·�0

Similarly, the instantaneous power supplied by a torque acting on an object is equal to the dot product of the torque and the angular velocity of the object.

pavg =favg •OJavg

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Problems Conceptual

1. A spaceship is in a perfectly circular orbit around the Earth. What is the magnitude of the torque caused by the gravitational force of the Earth on the spaceship if the radius of the spaceship's orbit is Rand the mass of the spaceship is M? 2. Is the torque associated with the force acting on the object in Figure 10.29 positive, negative, or zero? 3. Is the torque associated with the force acting on the object in Figure 10.30 positive, negative, or zero? 4. Is the power supplied by the force of gravity in Figure 10.23 positive, negative, or zero? Is the power supplied by the normal force in Figure 10.23 positive, negative, or zero? Is the power supplied by the force of friction in Figure 10.23 positive, negative, or zero? 5. Is the power supplied by the torque associated with the force of gravity in Figure 10.23 positive, negative or zero? Is the power supplied by the torque associated with the force of friction in Figure 10.23 positive, nega­ tive or zero?

F

Figure

l 0.29

Figure l 0.30

Quantitative Section 10-2 6. Two point particles are connected by a thin massless rod of length L = 2 m as shown in Figure 10.31. The mass of particle A is 4 kg and the mass of particle B is 2 kg. How far is the center of mass of this system from particle A? 7. For the system of three point particles shown in Figure 10.32, de­ termine the distance between the 5 kg point particle and the center of mass of the system. 8. The system in Figure 10.33 consists of 5 point particles. Four identi­ cal point particles, each with mass m, are positioned at the corners of a square with sides d. The fifth point particle has a mass 2m and is located a distance 2d from the center of the square formed by the other four point particles, along a direction perpendicular to the plane of the square. What is the distance from the point particle with mass 2m to the center of mass of this system if d = 12 cm and m = 0.01 kg?

1.5m

4kg

5kg Figure

l 0.32 m 2d

d

m Figure l 0.33

2m

270

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THE ENERGY OF PHYSICS, PART i: CLASSICAL MECHAN!CS AND THERMODYl'·�AM!CS

9. The object shown in Figure 10.34 consists of two uniformly dense thin disks, one ofradius Rand one ofradius 2R, which are attached together. The densities of the two disks are the same and R = 5 cm. What is the position ofthe center ofmass ofobject along the x-axis? 10. The object shown in Figure 10.35 consists ofa uniformly dense thin disk with radius 2R out of which a disk of radius R has been cut. What is the position of the center of mass of the object along the x-axis if R = 6 cm?

y

1034 y

Section 10-3

11. A force of F=(4N)x+(-3N)y

is exerted on a particle at

r =(2m )x +(-2m )S,. What is the torque on the particle about the origin?

12. A force of F=(-sN)x+(6N)z is exerted on a particle at What is the torque on the particle about the origin?

r=(2m)y.

Section 10-4

J N t 13. A uniformly dense rod with a length of3 m and a mass of 8 N t-----:--E_1_m_.... 1....1__s 1------+ 6 N • 4 kg is free to rotate around a frictionless axle through the center of the rod. Three forces are acting on the rod 3m as shown in Figure 10.36. What is the net torque acting on the rod? 14. A uniformly dense rod with a length of3 m and a mass of 10 N 4 kg is free to rotate around a frictionless axle through the center of mass of the rod. Four forces act on the rod 1-------'�•1....· __..;.....;.;...,r--�BN as shown in Figure 10.37. What is the net torque acting ----• 6 N i+----3m on the rod? 5. A uniformly dense rod with a length of 4 m and a mass of F,gure 1 (i 37 6 kg is free to rotate around a frictionless axle through 2m the center of the rod. Three forces are acting on the rod as shown in Figure 10.38. What is the net torque acting SN on the rod? 16. A uniformly dense rod with a length of4 m and a mass of -----1 10N 4N 6 kg is free to rotate around a frictionless axle through 2m leo· the center of the rod. Three forces are acting on the rod as shown in Figure 10.39. What is the net torque acting f-iqure 10.38 on the rod? 2m

1 0.8m

+-------1 4N

•

5 N

2m

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271

Section 10-5 17. A uniformly dense rod with a length of 3 m and a mass of 4 kg is free to rotate around one end. What is the angular acceleration of the rod if a net torque of 6 Nm is applied to the rod? 18. The following time-dependent net torque acts on a uniformly dense rigid rod:

The rod is free to rotate around a frictionless axle located at one end of the rod. The mass and length of the rod are 4 kg and 1.5 m, respectively. If the rod starts from rest, what is the angular velocity of the rod after the net torque has been applied for 3 s? 19. A uniformly dense solid sphere with a radius of 3 m and a mass of 4 kg is free to rotate around an axis through its center of mass. The sphere is initially at rest. If a constant net torque of 4 Nm is applied to the sphere, how much time is required for the sphere to complete 4 revolutions? 20. A uniformly dense solid rod with a length of 4 m and a mass of 5 kg is free to rotate around a fric­ tionless axle located at one end of the rod. The rod is initially oriented horizontally (i.e., parallel to the ground) and then released from rest. What is the magnitude of the angular acceleration of the rod resulting from the gravitational torque acting on the rod the instant the rod is released? 21. A uniformly dense solid cylinder with a radius of 3 m and a mass of 4 kg is free to rotate around an axis through its center of mass. The cylinder is initially at rest. If a constant net torque of 9 Nm is applied to the cylinder, how much time is required for the cylinder to complete 3 revolutions? 22. A uniformly dense rod with a length of 3 m and a mass of 4 kg is free to rotate around a friction­ less axle through the center of mass of the rod. Four forces act on the rod as shown in Figure 10.37. What is the angular acceleration of the rod? 23. A uniformly dense rod with a length of 4 m and a mass of 6 kg is free to rotate around a friction­ less axle through the center of the rod. Three forces are acting on the rod as shown in Figure 10.38. What is the magnitude of the angular acceleration of the rod? 24. A uniformly dense rod with a length of 4 m and a mass of 6 kg is free to rotate around a friction­ less axle through the center of the rod. Three forces are acting on the rod as shown in Figure 10.39. What is the angular acceleration of the rod? 25. A uniformly dense 50 cm rod with a total mass of 0.2 kg is standing perfectly vertically on a horizontal surface. When given a slight (i.e., negligible) push, the rod begins to topple over as shown in Figure 10.40. Let's assume that static friction is sufficient to allow the rod to fall over without slipping at the base. What is the magnitude of the angular acceleration of the rod when it is at 30 ° relative to the vertical axis?

Section 10-6 26. A uniformly dense rod with a mass of 0.3 kg and a length of 1.5 m is pivoted on a frictionless hinge at one end and held horizontally by a spring with spring

hinge

10.41

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANiCS AND THERMODYNAMICS

constant 10 N/m attached at the other end as shown in Figure 10.41. What is the frequency ofsmall up and down oscillations of this system? 27. A uniformly dense rod with a mass of 3 kg and a length of 2 m is free to rotate around a frictionless axis that passes through its Figure l 0.42 center of mass. A spring with spring constant k = 150 N/m is connected vertically between one end of the rod and the ceiling as shown in Figure 10.42. When the rod is in equilibrium it is parallel to the ceiling. What is the period of small amplitude os­ cillations ofthis system? 28. A uniformly dense rigid rod with mass of 6 kg and length of 1.5 m is free to rotate around a frictionless axis that passes through Figure ·10.43 its center ofmass. Identical springs (each with a spring constant of49 J/m2) are attached to each end ofthe rod as shown in Figure 10.43. What is the angular frequency of small amplitude oscilla­ tions ofthis system? 29. A 1.5 m long uniformly dense rigid rod with a mass of 3 kg is suspended from a horizontal surface by a hinge at one end and connected to a vertical surface by a spring with spring constant 10 N/m attached at the other end as shown in Figure 10.21. When the rod hangs straight down the spring is at its normal Figure 10.44 -----length. This pendulum is initially displaced by an angle of 10° relative to the vertical and then released from rest. What is the hinge maximum angular velocity of the rod when it is then vertically ·1 oriented again? 30. A 2.1 m long uniformly dense rigid rod with a mass of 3 kg is suspended from a horizontal surface by a hinge at one end and \{ attached to vertical surfaces by two identical springs with spring constant 21 N/m attached at the other end as shown in Figure 10.44. When the rod hangs straight down the springs are at their 1 .4S- ______ l C__ norma11engths. What 1s ' the per10 · d of smaII osciIIations of the Figure _v rod about its equilibrium position? 31. A 1.5 m long uniformly dense rigid rod with a mass of5 kg is sus­ pended from a horizontal surface by a hinge at one end and at­ tached to vertical surfaces by two springs with spring constants k1 = 10 N/m and k2 = 20 N/m attached at the other end as shown in Figure 10.45. When the rod hangs straight down the springs are at their normal lengths. What is the angular frequency of small amplitude oscillations ofthis system? --Figure - - 10.46 · - -- - --- ---- - - �- · -32. A thin uniformly dense rod oflength 2.8 m oscillates as a physical pendulum on a frictionless hinge as shown in Figure 10.46. What value ofthe distance x between the rod's center of mass and hinge corresponds to the smallest period of oscillation?

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273

Section 1 0-7 33. A uniformly dense solid cylinder with a mass of 2 kg and a radius of 5 cm is initially at rest at the top of a ramp. What is the maximum angle the ramp can have in order for the cylinder to roll without slipping down the ramp? In other words, if the ramp is set at a larger angle the cylinder will roll down the ramp while also slipping/sliding. The coefficient of static friction between the cylinder and the ramp is 0.4. 34. A 5 kg sphere with a radius of 10 cm is thrown along a level surface so that initially it slides with a linear speed of 2 m/s, but does not rotate. As it slides, it begins to spin, and eventually rolls without slipping. How much time is required for the sphere to begin rolling without slipping? The coefficients of static and kinetic friction between the sphere and the floor are 0.3 and 0.02, respectively.

Section 1 0-8 35. Two ropes, with tensions T1 = 5 N and T2 = 2.5 N, pull on a pulley as shown in Figure 10.12. The pulley is a uniformly dense cylinder with a radius of 5 cm and a mass of 2 kg that rotates around a frictionless axle that passes through its center. The pulley is initially at rest but begins to rotate after the tensions are applied. What is the net work done on the pulley by the tensions during the first 5 s that the tensions are applied? 36. A physical pendulum made using a uniformly dense rod with a mass of 1.5 kg and a length of 0. 75 m pivots around a frictionless hinge at one end of the rod as shown in Figure 10.28. What is the work done by the gravitational torque as the rod swings from 0 = 15 ° to 0 = 0° ? 37. The following angle-dependent net torque acts on a uniformly dense rigid rod with a length of 1.5 m and a mass of 4 kg that rotates around an axis through its center of mass: rnet (0) = (o.3

Nm 2 )0 rad 2

The rod is initially at rest. What is the change in the rotational kinetic energy of the rod during its first 2 revolutions? 38. Two point particles, A and B, with masses 4 kg and 6 kg respectively, are connected by a rigid, massless rod with a length of 0.6 m. The entire system is able to freely rotate, but is initially at rest. The following net torque acts on the system:

What is the work done by this net torque during the first 2 s it acts on this system?

27 4

I

THE ENERGY OF PHY.SICS, PART I: CLA.SS!(j\L

39. Two point particles, A and B, with masses 4 kg and 6 kg respectively, are connected by a uniformly dense rigid rod with a mass of 15 kg and a length of 0.4 m. The entire system is able to rotate around a frictionless axle through the center of the rod, but is initially at rest. The following net torque acts on the system: '� (t )=(0.15 N m), s What is the average power supplied by this net torque during the first 4 s it acts on this system? 40. The following ti me-dependent net torque acts on a uniformly dense solid sphere with a radius of 5 cm and a mass of 10 kg that rotates around an axis through its center of mass:

The sphere is initially at rest. What is the average power supplied by this net torque during the first 6 seconds it acts on the object?

CHAPTER ELEVEN Further Applications of Newtonian Mechanics Education is the passport to the future, for tomorrow belongs to those who prepare for it today. -MalcomX

11-1 Introduction

I

n the preceding four chapters, we developed a framework for solving mechanics problems using forces. We further showed that, often, the solution obtained by this force-based approach was identical to the solution previously obtained using an energy-based approach. In this chapter, we turn our attention to specific classes of problems for which an energy-based approach does not work at all or at least not nearly as well as a force-based approach. In each of these cases, the system is either not moving (and thus the energy of the system is not changing) or the motion of the system does not cause a change in the energy of the system. In either case, it is simplest to describe the mechanics of the system using forces.

11-2 Static Equilibrium A system that is at rest is said to be in static equilibrium if no net force and no net torque act upon it. A system is in static equilibrium if Fnet = O and fnet = 0 for the system.

This condition is commonly referred to as a balance of forces ( Fnet = O) and a balance of torques (inet =O).

Consider the system shown in Figure 11.1 that consists of a rigid and uniformly dense beam with a mass of 10 kg and a length of 3 m that rests on two supports. Let's define · a reference frame for this system in which the y-axis denotes the verti­ cal axis with a positive direction pointing up from the ground. The x-axis of this reference frame will be perpendicular to the y-axis with

1m

Support A

1.5m

Support B

Figure 11. 1: A beam resting on two supports.

275

276

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THE ENERGY OF PHYSICS 1 PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

a positive direction pointing to the right in Figure 11.1. In the free body diagram for the beam, we will denote the magnitude ofthe force exerted by support A on the beam as nA, the magnitude of the force exerted by support Bon the beam as n 8, and the magnitude ofthe force ofgravity as Fg. With these definitions, the-free body diagram for the beam is shown in Figure 11.2. Applying the Fnet = 0 condition for static equilibrium to the free body diagram in Figure 11.2 yields

(Fnet ) y = 0 �

y

Figure 1 l. 2: The free body dioorom for the beam in

To apply the fnet = 0 condition for static equilibrium, we need to specify Fig�1e i 1.1. the axis with respect to which the torques will be determined (Equation 10-3). It doesn't matter where we choose to locate this axis, however, since static equilibrium re­ quires that fnet = 0 for all potential axes ofrotation 1 . For example, we could place the axis ofrotation at the point where support A contacts the beam (Figure 11-3). Applying the fnet = 0 condition for static equilibrium with respect to this axis ofrotation yields2

Support A

Support B

Figu1e l I .3: The r;gid beorn from Figure l l. l with an axis of rotation located at the point 'Nhere su_pport A contacts the beam (red circle).

1 If a net torque ex!sted for an axis of rotation, the object would experience angular accelerat ion around that axis. Thus, to ensure that the 0�1ect has no angular acceleration, the net torque must be zero for all possible axes of rotation. 2 Recall from Section 10-5 that we can model the force of gravity to act at the center of mass of an object.

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277

Notice that the force nA produces no torque since it has no moment arm for rotations about this axis (Equation 10-5). Substitution of the values of our problem then gives us

We can then combine this result with the equation derived from the ft = O condition for static net equilibrium to determine the value for nA'

nA =58.8N

Alternatively, we could have solved this problem with the axis of rotation located at the center of mass of the beam (Figure 11.4). Applying the fnet =0 condition for static equilibrium with respect to this axis of rotation yields

Support A

Support B

Support A

Support B

F, ire l l. l with an axis of rotation localed at the · ·d 'b earn +.r om re '1,l . 4 . 7·1_r·,2_,191, ,··:F;q ' :r \, center ot moss ot the beam (rea c1rc,e1.

fnet

= O � ( Om )mg+( 1.0m )nA sin(270°)+(1.Sm )n8 sin( 90°) = 0

-(1.om)nA +(1.5m)n8 =0 � (1.om)nA =(1.Sm)n8

�

2 nB = nA 3

For this axis of rotation, there is no torque generated by the force of gravity since the_re is no �oment arm for this force for rotations around this axis (Equation 10-5). Substitution of this result mto the • • • & · m gives us i rm 1or static eqm·1·b equation derived from the Fnet =0 condit10n

278

I THE ENERGY OF PHYSICS, PART I: CLASSiCAL MECHANICS AND THf.RMODYt'-lAMICS

2

nA +-n 3 A =mg �

nA = 58.8N

r is of rotation. As anticipated, the results are identical to those obtained wit� �he_ oth� � condition for static eqmhbrmm 1s mdependent ofthe locat1on · the app 1·1cat·10n ofthe ;;• net -0 Smee · of rotat'10n, we should always place the axis of rotation where one_ or more. forces act on of the axis . the system. Placing the axis of rotation there will simplify our equation for r net = 0 smce there will be no torques from the forces acting there.

Example 1 ·1-1: Problem: A rigid and uniformly dense beam oflength L and mass mis supported by a hinge at one end and by a mass­ less rope on the other end, as shown in Figure 11.5. What is the tension in the rope? What is the force exerted by the hinge on the beam? Solution: Since the beam is in static equilibrium, we can apply the conditions of Fnet = 0 and f net = 0 to the beam to determine the magnitudes of these forces. Let's begin , · by defining a reference frame for this system in which the ed y-axis denotes the vertical axis with a positive direction pointing up from the ground. The x-axis of this reference frame will be perpendicular to the y-axis with a positive direction pointing to the right in Figure 11.5. In the free body diagram for the beam, we will denote the magnitude of the tension as T, the magnitude of the horizontal component of force exerted by the hinge on the beam as nx' the magnitude of the vertical component offorce exerted by the hinge on the beam as n 3, and the magnitude y of the force of gravity as F. With these definitions, the free body diagram for the beam is g shown in Figure 11.6. Applying the Fnet = 0 condition for static equilibrium to the free body diagram in Figure 11-6 yields C

I

3 It is worth noting that we do not know a priori that both n and n exist but it is safe for us to include them in our free body diagram. We can determine from the solutio� of ou? subs�quent calculations if these compo· nents are present. For example, ifwe determine that n = o, we will know that this compone nt does not exist. Y

CHAPTER ELEVEN

y

I

y

Tsin0 Tease

Figure 11.6: The free body diagram for the beam in Figure 11.5.

ny +Tsin0=mg The next step in our solution is to choose an axis of rotation for the fnet = 0 condition for static equilibrium. The best location for this axis is at the hinge since two of the forces in our free body diagram (nx and nY) act on the beam there; hence, there will be no torque from these forces for this axis of rotation. Furthermore, if the axis ofrotation is located at the hinge, then only the y-axis components of the forces in the free body diagram of the beam are capable of causing rotation (Figure 11-7). Tsin8

U2 270 °

cm

L

''

r

0 ... °('\9

.

--

°

mg

L

Tcos8 Figure 1 l .7: Only the components of forces oriented along the y-ax!s in the fr_ee body diag ram in Fi�ur: l !�c:i_� �?pable of musing to�c:i�e around the OXIS of rotation (red circle). _ _ _

279

2 so I THE ENERGY OF PHYSICS, PART 1: CLASSICAL MECHANICS AND THERMODYNAMICS As shown in Figure 11. 7, the net torque acting on the beam about this axis of rotation is fnet

½

=0 � (o)ny +(L}(Tsin0)sin(90 °)+( )mgsin(270 °)=0

moment arm The ny force does not generate any torque about this axis of rotation since its • • is zero. Simplifying this equation gives us

Tsin0 = mg � T=_!!!ff_ 2sin0 2 We can then substitute this result into our equations for

Fnet

=0 to determine nx and nY.

nx =(_!!!!L)cos0 � nx =(mg)cot0 2sin0 2 mg _ mg n Y =mg-Tsin0 � n Y =mg-- � n Y 2 2 The force exerted by the hinge on the beam has components along both the horizontal and vertical directions in Figure 11.5 (the x-axis and y-axis, respectively, in Figure 11.6). Interestingly, the magnitudes of nx' nY, and Tare all independent of the length of the beam.

11-3 Levers The conditions for static equilibrium also provide the basis for understanding the function of another simple machine: the lever. Recall from Section 7-7 that simple machines are devices that allow for the reduction in the magnitude of the force necessary to do a specific amount of work by increasing the displacement over which the force acts. Levers are simple machines that consist of a beam or rigid rod that is pivoted around a fixed hinge or fulcrum. The position of an object (referred to as the load) on the beam can be changed by applying an external force (referred to as the effort) to the beam. As shown in Figure 11.8, levers are divided into three classes based upon the relative position of the fulcrum, the effort, and the load.

CHAPTER ELEVEN

Class 1

I

281

�

Class 2

Class 3

Figure 11.8: The three classes of levers. The lood is denoted by the squor� with ihe label L and the effort as an the externally applied force �-

Common examples of the different classes of levers are:

Class 1 levers: scissors, pliers, and crowbar. Class 2 levers: wheelbarrow, nutcracker, and bottle opener. Class 3 levers: human arm, tweezers, and human jaw. The mechanical advantage of a lever depends upon the positions of the load and effort relative to the fulcrum, which defines the axis of rotation for the system. When the lever is in static equilibrium, the torque generated by the effort is equal in magnitude but opposite in direction to the torque gen­ erated by the load4 • Let's consider the situation shown in Figure 11.9 when the beam is horizontally oriented and in static equilibrium. I rF I

i

I

I

I

i�

!

Class 3

Class 2

Class 1

F 1 ou,e l l 9: The three classes of levers in static equiiibrium with the b:orn parallel to the ground ---

.

---

Applying the f = o condition for static equilibrium with respect to an axis of rotation located at the fulcrum for eaa; of the levers in Figure 11. 9 yields the following equation:

fnet = O � rF F9 = rF F � •

(l)F r Fg

= Fg

t by which the Hence, the mechanical advantage of the lever is the ratio ;� as this is the amoun r than 1 or le�s applied force (i.e., the effort) is amplified. For class 1 levers, this ratio can be greate and all class 3 levers will than 1. All class 2 levers will have a mechanical advantage greater than 1, have a mechanical advantage less than 1. on the beam since there is no moment arm for 4 There is no torque generated by the •,orce of th e fulcrum pushing up this force.

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.

'

l l-.4 ND THERMODYNAMICS

= A.0L � T

2-r net A.0 L

It is interesting to note that this equation is independent of the masses of the point particles. The maximum angular displacement can then be found from the maximum tension the rod can support/maintain.

T�Tmax

�

� (A0)

2-rnetA.0 �Tmax L

LT =� 2-rnet max

As expected, the maximum angular displacement is directly proportional to the maximum tension that the rod can support/maintain.

11-5 The Fictitious Centrifugal Force Imagine that you are swinging a bucket full of water in a vertical circle. Based upon your intuition (or perhaps personal experience), you suspect that if the bucket moves too slowly, the water will fall out when the bucket is inverted, but if you swing the bucket fast enough, the water will always stay in the bucket. What is the minimum angular velocity with which you can swing the bucket without the water falling out? Let's begin our solution to this problem by defining a reference frame for this system in which they­ axis denotes the vertical axis with a positive direction pointing up from the ground. The x-axis for this reference frame will be perpendicular to y the y-axis and have a positive direction pointing parallel to the direc­ tion of the bucket's velocity at the top of the vertical circle; of course, the velocity of the water is the same as the velocity of the bucket. In the free body diagram for the water; ·we will denote the magnitude of the normal force exerted by the bucket on the water as n and the ---..---x magnitude of the force of gravity as F9• With these definitions, the free n body diagram for the water when it is at the top of the vertical circle is shown in Figure 11.17. Applying Newton's 2nd law to each of the coordinate axes in this free body diagram yields the following equations:

(Fner)

y

=may �

-F g

-n=may � -mg-n=ma

y

Figure 11. i 7. The free body diagram for the wate;· when the bucket is ot the top of the vertical circle.

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Since the water is moving in a circle, centripetal acceleration must be acting on the water to change the direction of its velocity. The direction of this acceleration must be toward the center of the circle, which is the negativey-axis direction for the free body diagram in Figure 11.17. Hence, from Equation 11-3, we have

In this equation, the varaible R denotes the radius of the vertical circle around which you are swinging the bucket. Not surprisingly, the magnitude of the normal force exerted by the bucket on the water is a func­ tion of the angular velocity of the water. Indeed, it is this force that is required to change the direction of the tangential velocity of the water and, thereby, keep the water moving in a circle10 • An increase in the tangential velocity of the water (which, according to Equation 5.4, is equivalent to an increase in the angular velocity of the water) will, naturally, result in an increase in the magnitude of the cen­ tripetal acceleration required to change the direction of the water's velocity. This, in turn, requires an increase in the magnitude of the net force acting radially and, hence, an increase in the magnitude of the normal force exerted by the bucket on the water when the bucket is at the top of its vertical circle. As long as the normal force exists (i.e., as long as n � 0 ), the water is in contact with the bucket and will not fall out. Therefore, the minimum angular velocity for keeping the water in the bucket occurs when the magnitude of the normal force is zero. n=O � ro=±Ji

The ( ±) sign in this equation indicates that the result is valid for either direction of the angular velocity11 • Since the magnitude of the normal force acting on the water is also equal to the apparent weight12 of the water, we see that this minimum angular velocity corresponds to the condition where the water is weightless. 2 It is interesting to note that if the angular velocity is small enough such that a> R < g, then the magnitude of the normal force exerted by the bucket on the water is less than zero. Since the magnitude of a vector cannot be negative13 , we interpret this result as indicating the water has lost contact with the bucket. In other words, if we spin the bucket too slowly, the water will fall out of the bucket. At the top of the vertical circle the force of gravity pulls the water straight down toward the center of the circle. Therefore, if the bucket was not moving, the water would fall straight down to the ground. If the bucket is spinning around in the circle, the force of gravity acting on the water line unless a net force 10 Recall from Newton's 1st Jaw (Section 8-3) that the water will continue to move in a straight acts upon it. 11 See Section 5-2. 12 See Section 8-9. 13 See Section 8-4.

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will contribute the net force required for the water to move with the bucket (i.e., for the water's centripetal acceleration). However, since the magnitude of the force of gravity is constant regard­ less of the angular velocity of the bucket, there is a lower limit to the magnitude of the net force that can act on the water; this occurs when the magnitude of the normal force is zero. If the magnitude of this minimum net force (i.e., the magnitude of the force of gravity acting on the water) is larger than what is required for the centripetal acceleration of the water, the total radial acceleration of the water will be larger than the centripetal acceleration of the water alone. The remaining radial acceleration (the difference between the total radial acceleration and the centripetal acceleration) will cause the water to have a displacement toward the center of the circle and fall out of the bucket. You can also think of this as both the water and the bucket undergoing circular motion but around circles of different radii. This difference in radii results in a displacement of the water relative to the bucket. As discussed in Section 8-9, fictitious forces appear to act inside non-inertial (i.e., accelerating) reference frames. A common example of an accelerating reference frame is one that moves with a rotating object or an object moving in a circle. For example, a person moving with the block as it passes over the hill (Example 11-3), who is using a reference frame that is fixed with the block, would perceive the normal force and the force of gravity acting on the block but not the acceleration 14. Hence, this person would need to add a fictitious force pointing up (in the positive y-axis direction in Figure 11.13) to obtain a "net" force of zero and, thus, no acceleration for the block. This ficti­ tious force, associated with the centripetal acceleration of the system, is referred to as the fictitious centrifugal force since it is directed away from the center of the circle.

-

FcentrifuBal

( 2 )A

= mw R r

(11-4)

This fictitious centrifugal force can also be considered to give rise to the change in the apparent weight of the water, as shown in the vertically spinning bucket or the tension in the string of the pendulum in Example 11-4. An object on the surface of the Earth will experience a fictitious centrifugal force pushing it out from the surface of the Earth due to the Earth's rotation. As shown in Figure 11.18, we can describe the radius of the circular motion of the object associated with the Earth's rotation (i.e., the value of R in Equation 11-4) in terms of the radius of the Earth and the polar angle 0 describing the position of the object.

14 Since the reference frame is moving with the block, there is no movement of the block relative to the reference frame. Hence, in that reference frame, the block has no acceleration.

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North Pole

South Pole !--!(J·_, ...�

1 l

">-

on

/;_,;·:

tc;-; �;- -.../!!! exper"ier:i.:.t.-: circulo:· rr.()hon du�� rh is . The r c1cr u�; F rc)��":t: :·.,,-1 cf

f:i" l()

the

The m agnitude of the fictitious cent rifugal fo rce acting on an object on the surface of the Earth is, thus, Fcentrifugal,Earth

=

mm!RE

sine

(11-5)

Hence, the magnitude of this fictitious cent rifugal fo rce d epends upon the latitude of the object and is largest at the equator.

Example 11-6: Problem: A 10 kg object is at rest on the surface of the Earth at the equator. Wh at is the m agnitu de of the fictitious cent rifugal fo rce acting on the object? Solution: We first must solve fo r the angular velocity of the Earth. -s rad 1min = ) � m8 7_3x10 m = 2n rad (1day)( 1hr )( s 8 60s day 24hr 60min

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the Since the object is at the Equator, the radius of its circular motion is equal to the radius of Earth. From Equation 11-5, we have Fcentrifugal . ,Earth =(10kg)(7.3x10-

5

rad S

YJ (6.37x10

6

° m )sin(90 )

Fcentrifullal ,Earth =0.34N The magnitude of this force is small compared to the magnitude of the force of gravity acting on the object. F =mg � F =(10kg)(9.8 o

o

m

s2

lJ �

F =980N o

Thus, the fictitious centrifugal force can be neglected, and the object can be considered to be in an inertial reference frame15 •

The fictitious centrifugal force is also apparent when you drive your car around a turn, as dis­ cussed in Section 8-9. You feel yourself being pushed to the outside of the turn and interpret this sensation as resulting from a fictitious centrifugal force. Of course, this fictitious force "appears" only because you are describing your motion using a non-inertial reference frame; specifically, a refer­ ence frame fixed within a car experiencing centripetal acceleration. Naturally, in order for the car to experience the centripetal acceleration necessary to drive safely around a turn, a net force with a component directed toward the center of the circle must act on the car. In general, this component of the net force is the force of static friction acting between the tires and the road. It is a force of static friction since this force is acting to prevent the motion of the car relative to the road. In other words, this force is acting to keep the radius of the car's trajectory the same as the radius of the turn. Since there is an upper limit to the magnitude of the static fric­ tion force (depending upon the tire material and the condition of the road), turns are often also banked so that a component of the normal force acting on the car also can contribute to the net radially directed force acting on the car, as shown Figure 1 l. 19: The forces acting on a car d1 iving around a banked circle. The normal force is in Figure 11.19. denoted by n, the magnitude of the force of static Something similar happens when a pilot needs friction as (, and the magnitude of the force of to turn an airplane. She will rotate the airplane gravity as F . The direction of the centripetal acceleration fequired for the car to drive around the furn is indicated in blue. Components of both n and f, will give rise to the net force required for the centripetal acceleration.

15

However; it is, nevertheless, true that your apparent weight would be smallest at the equator.

CHAPTER ELEVEN

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around its longitudinal axis so that a component of the lift 1 6 is oriented in the direction she wants to turn the airplane. Finally, we n ow understand why astronauts are weightless (i.e., have an apparently weigh t of zero) when they are in spacecraft orbiting around the Earth. These spacecraft (and the astronauts inside them) are movin g in circular orbits around the Earth and thus must experience centripetal acceleration. The force associated with this centripetal acceleration is the force of gravity and thus the free body diagram for the astronauts in orbit can be represented by Figure 11.13; the negative direction for the y-axis in this free body diagram points toward the center of the Earth. Applying Newton's second law to this free body diagram gives us the following equation: (Fnety ) =may

(

v2 � n-F =m __t B R

) �

v2 n=F -m....!... B R

Substitution of the magnitude of the force of gravity from Equation 7-3 yields

m(

2)

M m v: M Em ME E F =G� n=G---m� n=- G--v 2 2 B t R R R R R

The variable ME in this equation is the mass of the Earth. In order for the astronaut to be weight­ less, the magnitude of the normal force must be zero.

m(

M 2 2 E n=O � 0=- G--v ) � v = G -E � v ,= + t t R R R i M

r

The (±) sign in this equation indicates that the result is valid for either direction of the angular velocity, as discussed previously. This equation indicates that for each distance R above the center of the Earth 17, there is a particular speed corresponding to weightlessness. This speed is, of course, also the speed required to maintain a circular orbit around the Earth.

• A force-based approach is necessary to describe the mechanics of systems in static equilibrium. - .t = 0 and rne - t= o · • The two conditions for a system to be in static equilibrium are F,, always be should • The axis of rotation for the fnet = 0 condition for static equilibrium located where one or more forces act on the system. . . 16 Lift . a irpl ane's wings. 1s a force tha t ac ts perpend'1cu1 a r to th e surface of an . atIOn r th E th b ilds upon the discussion of center of mass 17 The justification for using a point pa rticle appr?xm fo : ;; n � � ou his for physical pendula in Sec tion 10-5. You should feel inspired to ta ke d io a physics courses to lear n more ab t t .

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• The mechanical advantage of a lever is equal to the ratio of the moment arm of the effort to the moment arm of the load. • Any object moving in a circle, or through part of a circle, must experience centripetal acceleration. The direction of this acceleration is always toward the center of the circle, and the magnitude of the acceleration is a =Ro/ = C

v

2

.....L

R

In this equation, R is the radius of the circle. • Any object moving in a circle, or through part of a circle, must experience a net force that has a component directed toward the center of the circle. • A fictitious centrifugal force appears to act inside non-inertial reference frames under­ going circular or rotational motion.

Problems 1. A block is held in place by a massless rope that passes over a pulley as shown in Figure 11.20. The pulley is a uniformly dense solid cylinder that rotates around an axis through its center of mass. Is the tension T 1 greater than, equal to, or less than the tension T2 ? 2. A block is held in place by a Massless rope that passes over a pulley as shown in Figure 11.21. The pulley is a uniformly dense solid cylinder that rotates around an axis offset from its center. Is the tension T 1 greater than, equal to, or less than the tension T2 ? 3. The two cantilevers in Figure 11.22 have the same mass and length, and both are uniformly dense. The angle 0 is also the same for both cantilevers. Is the tension T 1 greater than, equal to, or less than the tension T2 ? 4. A rigid beam (i.e., a rigid rod) rests on two supports positioned at each end of the rod as shown in Figure 11.23. Because the density of the beam is non-uniform, the center of mass of the beam is twice as far from support B as it is from support A. Which support exerts a greater force on the beam?

hinge

Figure 1 1. 22

hinge

l

Fpuu

Figure l 1 . 20

l

Fpuu Figure 1 1.21

CHAPTER ELEVEN 5. A block is released from rest at the top of an oval track as shown in Figure 11.24. ! L/3 Jil :c The block then slides down the track and then back up the other side. Is the appar­ }1.'ir;,,t./i1:i'¥i ent weight of the block at the bottom of the track greater than, equal to, or less than the Support A magnitude of the force of gravity acting on Figure I 1.23 -- -�·--·�·---·-� the block? 6. A block and a uniformly dense solid sphere are sliding down identical semicircular tracks as shown in Figure 11.25. The block and the sphere have the same mass and the sphere is rolling without slipping. No energy is lost by the block or the sphere due to friction or drag. Each object will fly off its track at a characteristic angle 0max. Which object will have a larger value of 0max? F igure

I 297

L

I I

is,;

I I I I

: 19

I I

'0 ,

, ,

l l . 24

,, /

I / I , I / I , I/

I / , I I , I , I,

.,,

,/

Figure 11 25-------------------- - --------·--- - · -- �- -- ---

Support B

--- ··------------·-- -----------·-- ·- --- - - -

Quantitative Section 1 1-2

as 7. A uniformly dense rod with a mass of 3 kg is supported horizontally by two vertical ropes shown in Figure 11.26. What is the tension in rope A? What Rope B Rope A is the tension in rope B? 8. A rigid beam (i.e., a rigid rod) with a mass of 15 kg rests on U3 two supports positioned at each end of the rod as shown in Figure 11.23. Because the density of the beam is non­ uniform, the center of mass of the beam is twice as far from L support B as it is from support A. What is the magnitude the is t 11.26 Figure of the force exerted by support A on the beam? Wha ---------? beam the on B ort magnitude of the force exerted by supp

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1m 9. A rigid and uniformly dense beam (i.e., a rigid rod) with a mass of 10 kg and a length of 3 m rests on two supports as shown in Figure 11.27. The supports are positioned at each end of the rod. A 4 kg block is resting on the beam a distance of 1 m away from the center of mass Support A of the beam. What is the magnitude of the force Figure l l .27 exerted by support B on the beam? 10. A rigid and uniformly dense beam (i.e., a rigid rod) with a mass of 10 kg and a length of 3 m rests on two supports as shown in Figure 11.28. A 20 kg block is resting on the beam. What is the maximum distance that this block can be from support A without the beam tip­ ping? That is, what is the maximum value of x Support A in Figure 11.28 such that the beam remains in static equilibrium? Figure 11.28 11. A piece of modern art consists of four blocks supported by massless rods as shown in Figure 2d d 11.29. The system is in static equilibrium. What is the mass m? 12. The cantilever in Figure 11.30 is a triangle with a base of 0.5 m, a length of 4 m, and a mass of 500 kg. This cantilever is held in place by a steel cable that is attached to the far end of the cantilever at an angle of 0 = 30° with respect to the cantilever. What is the tension in the cable? The center of mass of this cantilever is located 4/3 m away from the hinge.

Support B

Support B --- ----·-- --- --

2d

d

3d

Figure 11.29 -------

hinge

Figure 11.30 .

-�--------·---�--------- ----·-�----�--------

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13. A 3 m long uniformly dense cantilever with a mass of 250 kg is attached to a hinge and also held in place by a steel cable that is attached to the far end of the cantilever at an angle of 0 = 30 ° with respect to the cantilever as shown in Figure 11.31. A 25 kg mass also hangs from the far end of the cantilever. What is the tension in the cable? What is the magnitude hinge ..._......,;.........,.....;.........,.....;.......;.......,...;;.;;:.....,.......:, of the force that the hinge exerts on the cantilever? 14. A 3 m long uniformly dense cantilever with a mass of 250 kg is attached to a hinge and also held in place by a steel cable that is attached to the far end of the cantilever at an angle of 0 = 30 ° with respect to the cantilever as shown in Figure 11.32. A 25 kg mass also hangs from the far end of the cantilever. What is the tension in the cable? 15. You must build the cantilever shown in Figure 11.5. The maximum length of cable you have available is 5 m and the maximum tension the cable can support is 245 N. The material used for the cantilever has a density of 40 kg per meter length of the cantilever. What is the maximum hinge length of the cantilever that you can build? 16. Two blocks are connected to each other by a massless rope that passes over an asymmetric cylindrical pulley as shown in Figure 11.33; the red circle denotes the axis ,�i���_:1_1_1 _:,}�-------- __ of rotation of the pulley. The pulley is massless and the distancedis equal to 5 cm. What is the mass of block B, if the mass of block A is 6 kg and the entire system is in static equilibrium? 17. Two blocks are connected to each other by a massless rope that passes over an asymmetric cylindrical pulley as shown in Figure 11.33; the red circle denotes the axis of rotation of the pulley. The pulley is uniformly dense and has a total mass of 4 kg. The distance dis equal to 5 cm. What is the mass of block B, if the mass of block A is 6 kg and the entire system is in static equilibrium? 18. A uniformly dense rigid rod with a mass of 10 kg and a length L = 2 m is leaning against a vertical surface as shown in Figure 11.34. The vertical surface is fric­ tionless, but there is friction between the rod and the horizontal surface. The coef­ ficient of static friction between the rod and the horizontal surface is 0.4. What is the minimum value for the angle 0 such Figure 11.34 Figure 11 33 that the system is in static equilibrium? -· " ------- --- -

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Section 11-4 19. An airplane is flying in a horizontal circle at a speed of 400 km/hr. If its wings are titled at a 30 ° angle relative to the horizontal, what is the radius of the circle in which the plane is flying? - - -r - - _____) 20. A 75 mg bob attached to a 1.5 m long string moves in a hori­ zontal circle as shown in Figure 11.35 to form a horizontal I --� ° I pendulum. If the angle is 0 = 30 what is the period of the I pendulum? That is, how much time is required for the bob �ig u :e 1 i_ .3 5 __________________________ _ to complete one revolution? 21. A 3 kg bob is attached to a 0.5 m long massless string and swings as a pendulum as shown in Figure 11.14. The pendulum is released from rest with an angle 0 = 60° relative to the vertical. What is the tension in the h string when 0 = 0° ? 22. A small block with a mass of 1.5 kg is released from rest on a ramp as shown in Figure 11.36; the angle 0 = 50°. The block slides down the ramp and then around - ___________:::,-.___;::a...,....::.___ the inside of a loop-the-loop track. The radius of the F���r�- �-] :._�-�-------- __ __ _ _ __ loop-the-loop is 0.25 m. The coefficient of kinetic friction on the ramp is µk = 0.15, but the rest of the track is frictionless. What is the minimum height h from which the block must be released to prevent it from falling off the track at the top of the loop-the-loop? 23. A common amusement park ride is the rotating cylinder. People step in­ side a vertically oriented cylinder and stand with their backs against the curved wall. The cylinder spins very rapidly around a vertical axis passing through its center, and at some angular velocity, the floor is pulled away. The riders then "stick" to the wall. If the radius of the cylinder is 5 m and the coefficient of static friction between the people and the wall is µs = 0.4, what is the maximum period of rotation of the cylinder for the floor to be removed safely? 24. Future spacecraft and space stations may create "artificial gravity" by rotating. Consider a cylindrical space station with a radius of 894 m that !� i ����--� 1 :._�_?____ _ rotates about its longitudinal axis. The inside surface of the cylinder is the floor of the space station upon which the astronauts live and work. What period of rotation is necessary for a 70 kg astronaut to have the same weight as she has on Earth? 25. A point particle with a mass of 2 kg is attached by a massless rope to a massless cylinder that is rotating with a constant angular speed as shown in Figure 11.3 7. The length of the rope is d = 50 cm. What is the rotational kinetic energy of this system when 8 = 30° ? 26. Two identical point particles, each with a mass of 4 kg, are attached by massless ropes to a mass­ less cylinder that is rotating with a constant angular speed as shown in Figure 11.38. The length of each rope is d = 40 cm. What is the rotational kinetic energy of this system when 8 = 40° ?

·--

--

CHAPTER ELEVEN 27. Two point particles, A and B, are connected by a thin mass­ less rod of length L = 2 m and are freely rotating in space with an angular speed of 3 rad/s as shown in Figure 11.16. The mass of particle A is 4 kg and the mass of particle B is 2 kg. What is the tension in the rod? 28. Two point particles, A and B, are connected by a thin mass­ less rod of length L = 3 m and are able to freely rotate. The mass of particle A is 4 kg and the mass of particle B is 2 kg. The system is initially at rest, but then a net torque of 6 Nm is applied to the system causing it to rotate. How many revolutions will be completed when the tension in the rod is 62.8 N? Figure 11 .38 29. Two identical point particles are connected by a thin mass- ---- ------less rod of length L = 4 m and are freely rotating in space under the influence of the applied forces showed in Figure 11.39. The L mass of each particle is 4 kg and the magnitude of each force is 2 N. The system is initially at rest. What is the tension in the rod after the forces have been applied for 5 s? 30. A block is sliding down a frictionless semicircular track as shown in Figure 11.25. If the block started from rest at the top of the track, at what angle 0 will the box "fly off" the track? F�£l-�r� l_l��-?__ _

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CHAPTER TWELVE Newtonian Mechanics for Systems of Moving Obiects Science is a way of thinking much more than it is a body of knowledge. -Car/Sagan

12-1 Introduction

T

he third of Newton's three laws of motion describes the relationship between the forces that interact:ng objects_ (or syst�ms of obje_cts) exert on each other. The correct application of Newton s 3rd law 1s essential when usmg forces to describe the kinematics of a system of interacting moving objects. However, as we shall see in examples in this chapter, the forces act­ ing between objects within the same system frequently do not contribute to the kinematics of the system as a whole. Thus, it is often more efficient to apply Newton's laws to the entire system rather than to each object in the system individually in order to describe the motion of the objects or the system.

12-2 Newton's 3rcl Law Newton's 3rd law can be expressed as Every force is always accompanied by a separate force of the same magnitude, but oriented in the opposite direction. Thus, the forces that interacting objects (or systems of objects) exert on each other are always equal in magnitude and opposite in direction.

Newton's 3rd law is often colloquially expressed as ''for every action there is an equal and opposite reaction." When a child runs across her backyard, her feet push backward on the Earth, and the Earth pushes forward on her. It is the force of the Earth pushing on her, together with the other forces that act on her, that determine her acceleration. She cannot make herself accelerate by exerting a force on the Earth since this is not a force that acts on her (Section 8-4). However, it follows from Newton's 3rd law that she can increase the magnitude of the force that the Earth exerts on her by increasing the magnitude of the force she exerts on the Earth (i.e., by pushing "harder" on the Earth). Similarly; when a person climbs a rope, it is the tension in the rope pulling up on the person that influences his 303

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•r)DV�>,j/\t;1cc; T1r·R· 'A"'l'-r AN') '' -'-' 1' • , .- .t "fvi, ' THE ENERGY OF PHYSICS, PART I: CLASSiCAL N\f::(h 1" '-.)

acceleration. However, we recognize from Newton's 3rd law that the tension in the rope (i.e., the force of the rope pulling up on the person) results from the person pulling down on the rope. While, at first glance, this may seem like nothing more than semantics, f - I ·.j U i ,2 -; 2 . l _· /, Newton's 3rd law does have an immediate practical use in constructing c..' s1�J!nq ot t'--"'1 F, =(0.04kg)( 9.8;) -> F, =0.39N

In Equation 13-3, Newton's 2nd law is expressed as a relationship between a force and a change in linear momentum. Although changes in linear momentum can be readily measured (i.e., both mass and velocity can be easily determined to high levels of accuracy), it is not always straightforward to determine the time dependence of the forces acting to cause that change in linear momentum. However, since we can typically also measure with high precision the time of an interaction we can, nevertheless, estimate the average net force associated with a change in linear momentum2 •

dp=fFnetdt � dp=(Fnet )avg fdt dp=(Fnet )avg M

(13-4)

The average net force is clearly not the same as the instantaneous force. Indeed, as shown in Example 13-2, the instantaneous force acting to cause a change in linear momentum may vary during the interval over which the change in linear momentum occurs. Nevertheless, Equation 13-4 does allow us to estimate the magnitude of the net force associated with the change in linear momentum.

Example 13-3: Problem: A 2 kg block moving along a horizontal frictionless surface bounces off of a wall, as shown in the Figure 13.2. The speed of the block before and after the collision is 7 m/s and 4 m/s, respectively. What is the average force exerted by the wall on the block if the block was in contact with the wall for 1 ms?

Before

After

Figure ·--------�--------�-13.2: The system in Example 13-3. ··-

Solution: We will solve this problem using Equation 13-4. Since both the average net force and the change in linear momentum in this equation are vector quantities, the first step in our solution must be to specify the reference 2 This is analogous to the discussion in Section 7-6 in which we related the change in the kinetic energy of an object to the work done by the average net force acting on an object (Equation 7-9).

-- ----- -------

-

Positive direction

F wall on block

F block on wall

Figure 13.3 The direction of the aver­ age net force of the wall on the block �� Exompie_l 3-3 .

CHAPTER THIRTEEN

I 341

frame for defining the�e vectors. Let's define the horizontal direction to be the x-axis in the r fere�ce frame f�r th1s sys�em. The positive direction for the x-axis will point in the same : d1rect1on as th� _1mt1.� velocity of the block (i.e., to the right in Figure 13.2). The change m the Im ear momentum of the block resulting from its contact with the wall is !ip=mv1 -mv; � b.p=m(v1 -v;) = dp (Zkg)((-4:

)+: )) �

,\p = -22�m

Substitution of this change in linear momentum into Equation 13-4 gives us b.

(Fnet )avg = ___E_ b.t

�

(Fnet )avg --

-22

kg m

s 0.001s

(Fnet) avg = -22,000N The negative sign in our solution indicates that the direction of the average force is opposite the direction of the initial velocity of the block. In other words, as shown in Figure 13.3, the wall pushes the block to the left, as expected.

The relationship between average net force, interaction time, and change in linear momentum expressed in Equation 13-4 is also the motivation for crumple zones in cars. When a car experi­ ences a collision, nothing can be done about the associated change in the linear momentum of the car. However, by increasing the time interval over which the collision occurs (i.e., by increasing the time over which the linear momentum of the car changes), the magnitude of the average net force exerted on the car and, therefore, on the driver of the car (assuming she used her seat belt) can be red uced. By increasing the time over which the car slows to a stop following a collision, the crumple zone decreases the magnitude of the forces acting on the driver and, thus, helps to protect the driver. Finally, it should be noted that Equation 13-3 is often referred to as the impulse-momentum theo­ rem with the vector quantity impulse J defined in Equation 13-5.

J = f Fdt

(13-5)

As shown in Figure 13.4, since the impulse associated with a force is the integral of that force with respect to time, it is the same as the area under the curve of force versus time. In this way we can think of impulse as being similar to work. Work is the integral of force with respect to distance (Equation 9-2) and impulse is the integral of force with respect to time (Equation 13-5).

�

Time 13.4: The relationship between Fioure ,, ro! -1mpu I force, "tirne, onu se ... _ _ ___________ _

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

As defined in Equation 13-5, impulse is a vector quantity and, thus, we must specify both its magnitude and direction. As always, when dealing with 1-dimensional motion, we will rely upon (±) signs to denote the direction of a vector. For example, the impulses of the force of gravity and the force of the floor in Example 13-2 would be -

]gravity

=-0.0024

kg m S

kg m

J-flooronball =0.5064S

The net impulse acting on the ball in Example 13-2 is the sum of these two impulses and is thus in the positive direction, consistent with the direction of the change of momentum of the ball resulting from the collision with the wall.

· , · , · , ··· : · -11�1 milif•n rficm ma sfi� tions � "

.

-

.

.

'

-

The concept of relative descriptions of motion and veloc­ XA,external ity was first introduced in Section 2-6. The formal name given to this process of switching between reference frames for describing kinematics is a Galilean trans­ formation. Let's consider the system consisting of two Xs,external blocks shown in Figure 13.5. The positions of the blocks on the x-axis can be measured using a variety of reference frames (e.g., dif­ ferent origins). The reference frame xexternai is external to this system of blocks and has an origin on the x-axis of Figure 13.5: Measurements of the positions xexte = 0, as indicated by the dashed line on the far left of 1'No blocks relative to each other and to on rnal external reference frame (dashed line). in Figure 13.5. This external reference frame is constant regardless of the motion of the blocks. We can also use the location of block 8 to define the reference frame x8 whose origin is the location of block 8 on the x-axis.As block 8 moves, the location of x8 = 0 will also move. It follows from Figure 13.5 that we can relate the x-axis position of each of the blocks in the system by X A,external

= XA,B + XB,external

As introduced previously in Section 2-6, in this equation, we use subscripts to denote both the block and the reference frame. Thus, xA,B is the position of block A in the reference frame of block B (i.e., x A is the position of blockA relative to block 8). Similarly, x A x e ,8 ,e t rnal is the position of blockA relative to an external reference frame, and x B,external is the position of block 8 relative to an external referen ce frame. The velocities of the two blocks in our system can also be measured relative to each other or to an external reference frame, and these descriptions are related by VA ,external

= VA,B + VB ,external

I

CHAPTER THIRTEEN

343

Of course, we could extend these definitions of relative position and velocity to systems consisting of more than two objects and to multidimensional descriptions of these vectors.

13-4 Linear Momentum Conservation Since linear momentum is an extensive quantity, the linear momentum of a system of moving objects is the sum of the linear momenta of those objects (Equation 13-6). For this calculation, all of the velocity vectors are defined with respect to a reference frame that is external to the system. psystem =ml vl ,externa,+m2v2 ,externa,+ ... mn Vn ,externaI

�

-

""

-

psystem = £.. mi V; ,external

(13-6)

i=l

Consider the system shown in Figure 13.6 that consists of two blocks moving along a frictionless horizontal surface. The velocities of the blocks relative to the horizontal surface are also shown in Figure 13.6. The first step in determining the linear momentum of this system is to specify the reference frame with respect to which we will measure the velocities (and, hence, momenta) of the blocks that constitute the system. Let's choose the horizontal surface as the external reference frame for our calculation of the momentum of our system. Let's denote this as the x-axis with the positive direction pointing in the same direction as the velocity of the 4 kg block (Figure 13. 7). Since the speeds given for the blocks are relative to the horizontal surface, we have kg m

P-system =18 s

Figure 13.6: A system consisting of two blocks moving across a frictionless hori­ zontal surface. The indicated veiocities ore relative to the horizontal surface.

--�---�---------·----

+x 6 m/s -+

3 m/s +--

Figure 1 3 .7: Definition of the refer­ ence frame for calculating the mo­ mentum of the system in Figure 13.6.

The momentum of the system can also be expressed in terms of the motion of th center f mass of � � the system. Recall from Section 10-2 that we defined the center of mass of a collection ofobJects to be

i=l

ion Let's now choose a reference frame external to our system fior these measurements of posit

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THE ENERGY OF PHYSICS, PART I: CiASSICAL MECHA��!CS AND THERtv\ODYNAMICS

�m.f 1 1,extema

,£...

r

CM,external

= ..!.:.;-�1 ____

(13-7)

n

Lm ; i=l

If we assume that the masses of the objects in our system are constant in time, then differentiation of Equation 13-7 with respect to time gives us d �m.r. ,

n

n

d -r dt

CM ,external

,

t .£.J , d

externa

1

=..!.:.·-�1 _____

d _ � -rCM, dt

external

=

�m1.V1,. t_

extema l

i=l

i=l

This equation can then be simplified using Equation 13-6 to give us P d _ stem -rCM, xterna =-n t d ; sy

e

l

Lm i=l

We can now define a new vector quantity called the velocity of the center of mass derivative of the center of mass with respect to time. -

vcM

as the first

d dt

VCM =-rCM

With this definition, we have -

VCM,external

P system

= -n-- � p-system =

L

m;

(n ) �

1=1

mi vCM ,extern l a

i=l

p

stem

sy

=mtota vCM ,extern l

a

1

(13-8)

The linear momentum of the system is simply the product of the total mass of the system and the velocity of the center of mass. Equation 13-8 also indicates that we can treat a system of moving objects as though all of the mass of the objects were concentrated at the center of mass of the system. This is the natural extension of the definition of center of mass introduced in Section 10-5.

CHAPTER THIRTEEN

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345

Example 13-4: Problem: What is the velocity of the center of mass of the system of bloc ks shown m 16 F.1gure 13.. Solution: Starting with Equation 13-8 we have -

� VCM ,

psystem = mtota VCM ,extern l

al

extern al

P�stem =-m tota l

From our previous calculation, we know that

_

Psystem

gm =l8k �

v

cM ,external

S

V CM ,external

= ____.s'-6 kg

m = 3S

From the definition of the external reference frame for this problem (Figure 13. 7), we know that the direction of the velocity of the center of mass of this system is to the right in Figure 13.6.

We can now build upon the system approach to using Newton's 2nd law introduced in Section 12-6 to rewrite Newton's 2nd law as

(ft

)

external net

d p-system = dt

(13-9)

It follows from Equation 13-9 that since no net external force acts on an isolated system, the linear momentum of an isolated system is constant.

(ft

) -o

externa l net -

�

-0 d dt Psystem -

If a system's linear momentum is constant, we say that linear momentum is conserved for that system.It follows from Equation 13-9 that if the total mass of an isolated system remains constant, we have

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I

THE ENERGY OF PHYSICS, PART i: dv_ d_ o -0 � !!_vCM,extemal =0 / -psystem = � mtotal dt dt C',.," ,extema dt

Therefore, the velocity of the center of mass of an isolated system will be constant. m

n m

N'!lti5i\V&17Ni5' a

in ;mr:mrr,at;1w·�rm11mre�illl�---------------•

Example 1

5:

Problem: Block A has a mass of 2 kg and is moving with a speed of v0 = 2.8 m/s when it slides onto the top of block B, as shown in Figure 13.8. The mass of block B is 8 kg and the coefficient of kinetic friction between block __ A v"-A and block B is µk = 0.16. There is no friction ..Fl:::.., --------, between block B and the horizontal surface on 8 which it rests. What is the acceleration of block ; '/ 0 B relative to this horizontal surface as block A 1 .__,.o· slows to a stop relative to block B?

I

Solution: We can consider the two blocks together to form a single system. This is an iso­ lated system since there is no interaction between anything outside the system and either block (the horizontal surface along which block B slides is frictionless, e.g.). Therefore the linear momentum of the system will be conserved as the blocks move. ,1P

system

=0 � P-

system,J

= P-system,i

When writing expressions for the linear momenta of two blocks, we will use the horizontal surface along which block B slides as our external reference frame; we will refer to this sur­ face as the ground. We will also define a one-dimensional axis for the motion of the blocks, with a positive direction pointing toward the right in Figure 13.8. With these defections, the equation for linear momentum conservation for this system can be written as

The variables vA,o and vB,g in this equation denote the final velocities of block A and block B relative to the ground, respectively; we have dropped the subscript "/' for simplicity. Differentiating this equation with respect to time gives us

We can then use a Galilean velocity transformation to determine another equation for the the acceleration of block B relative to the ground. aA,g =aA,B +aB,

g

CHAPTER THIRTEEN

I

Combining these two equations then gives us

a

B

,g

-

(

--

m+

A

A

m

m

B

)

a A,

B

The acceleration ofblock A relative to block B can be determined using Newton's second law. The net force acting parallel to the velocity of block A is kinetic friction. Using our definition for the positive direction of the horizontal axis in Figure 13.8 then gives us F,,et =ma

�

- µkmAg

AaA, B

=m

�

QA, B

=

-µkg

Thus as.u

=-(

A

m

mA + mB

)(-µ g) � k

Substitution gives us a

B ,g

(21

2kg+8kg

m as.u =0.31s2

Since our answer is positive, the direction of the acceleration of block B is to the right in Figure 13.8. In other words, when block A slides on top ofblock B, block B will begin to move relative to the ground in the same direction that block A is moving relative to the ground. This forward motion of block B results from the force of friction acting between block A and block 8. Specifically, we know from Newton's third law that if this force of friction pushes back on block A (to the left in Figure 13.8) to oppose the continued motion of block A, it must push forward on block B (to the right in Figure 13.8). Note that this result predicts a8.u = 0 when m 8 = oo, as expected.

Example 13-6: Problem: Emily has a mass of SO kg and is standing on the left end ofa 15 m long and 575 kg perfectly level cart that has frictionless wheels and rolls on a horizontal and frictionless track; both Emily and the cart are initially at rest. Emily then starts to run along the cart toward the right end ofthe cart at a constant speed of 5 m/s relative to the cart. How far will Emily have run relative to the ground when she reaches the right end of the cart?

347

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

Solution: We can consider Emily and the cart together as a single system. This is an isolated system since there is no interaction between anything outside the system and either Emily or the cart (the cart is rolling along frictionless rails, e.g.). Therefore the momentum of the system will be conserved as Emily runs. /lpsystem -

O

4

Psystem,f

= Psystem,i

When writing expressions for the momenta of Emily and the cart, we will use the ground as our external reference frame. Since Emily and the cart were initially at rest, we have . vEmil.y,ground +mcart vcart,ground =0 P-system,/ =0 � mEmily In writing this equation we have ignored the subscript ''f" to denote these linear momenta in order to simplify our nomenclature. Since we were given Emily's velocity relative to the cart, we need to use a Galilean transformation to determine Emily's velocity relative to the ground. V Emily ,ground

= VEmily,cart + Vcart ,ground

4

V cart ,ground

= VEmily ,ground - VEmi/y,cart

Substitution yields

- . )

.

. ground + mcart (vEmily ground vEm1/y,cart = 0 . v.Emily, m Erm/y

VEmily ,ground

cart

= ( m m+ m ) VEmi/y ,cart Emily cart

Since velocity is a vector, we must specify a reference frame for its measurement. All of the motion occurs along 1-dimension, the horizontal direction, and we can define the positive direction for this axis to be the same as the direction of Emily's velocity relative to the cart. With this definition, we have -

VEmily,ground -

V

(

575kg 57Skg+S0kg

Emi/y,ground

)(s m ) -;

=4.6-

m S

The positive sign for our answer indicates that the direction of vEmil.y,ground is the same as the · · d Irection O f VEml/y,cart, as expected. The magnitude of VEmily.ground is less than the magnitude of • V smce l'mear momentum conservation for this system requires that the cart moves Emily, rt backward as Emily moves forward. As before, we can also interpret these directions in terms of Newton's 2nd and 3rd laws. The force exerted by Emily on the cart must be in the opposite ca

CHAPTER THIRTEEN

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349

dire ction of the force exerted by the cart on Em ily· Thus, the accelerat·10ns an ( d veloc1t1es · · and displacem ents) of E m ily and the cart m ust be in opposite directions. Since E m ily runs at a constant speed, we can use the constant acceleration kine m atics equations to relate her velocity to her displacem ent.

8XEmily, art = VEmily ,c rt .1.t � .1.t = 8X Emily, c

a

rt

ca

V Emily,cart

�

15m

M=

S

m

� .1t=3s

s

Therefore,

llxEm1!Y ·1 ,ground

= v Emily . ,ground.1.t � L\x Em11Y,groun .1 d

L\x Emily,ground

=

=(4 6 '

m S

)(3s)

13.8m

So, although Emily m oved 15 m with respect to the cart, she moved only 13.8 m with respect to the ground since the cart was also m oving with respect to the ground but in the opposite direction.

The solution for Exa m ple 13-6 can also be interpreted in terms of the center of m ass of the system . Since Emily and the cart were initially at rest, the initial velocity of the center of m ass of the system was VCM.external

=0

Since this is an isolated system, the velocity of the center of m ass must be constant as E mily runs. Therefore, the velocity of the center of mass is always zero regardless of the m otion of Emily and the cart. Furthermore, since the velocity of the center of m ass is zero, the position of the center of m ass will not change as Em ily m oves. The expression for the center of m ass of the syste m is _

XCM -

mEmily. x Emily. ,ground +mcart x ort,ground c

�

L\x cM

L\x = mEmily llxEmily,ground +mc rt cart,ground a

mEmily +mcart

Hence, VCM ,external

=0 �

A .. LM CM

To test this prediction, let's determine relative to the ground.

=0 � L\x

c

m

Emily

. ro art,g und

Au

UA

Emily,ground

+m

art

c

L\x cart.ground

=0

We begin by calculating the velocity of the cart

350

I

THE ENERGY OF PI-IYSICS ! Pfa.RT I· '-'-"'""',,_,,

-

Psystem,i

=0

�

VEmily,ground =VEmily . ,cart +vcart,ground __ Vcart,ground -

mEmilyv Emily,ground +mca vc rt

�

mEmily ( m . m ) VEmily, + ca Emily

art

mEmily (vEmily ,c +vca

cart

rt

art

�

V

cart,ground

rt

V cart,ground

=0

,ground

,ground

)+mcartvca

rt

,ground

=0

m 50kg ) =- ( -----'---)(5S + 575kg 50kg

m =-0.4S

As expected, the direction of Vcart ,ground is opposite the direction of VEmily,ground • The distance travelled by the cart can be determined using our constant acceleration kinematics equations. m �cart,ground =vc rt ,ground lit � �cart,ground =(-o.4 )(3s) S

a

�cart,ground =-1.2m We can now solve for the change in the position of the center of mass of the system. m Emily �Emily,ground +mca � a rt

c rt

,ground

=(50kg)(13.8m)+(575kg)(-1.2m)

mEmily� Emily ,ground + mca

rt

�

cart ,ground

=O

As predicted, the center of mass of the system has not changed since the system is isolated. We can now, at last, return to the our discussion of the gravitational potential energy shared between the Earth and a small block that is initially held in place a short height above the ground (Section 3-5). Let's consider the Earth and the block to form an isolated system. Then, when the block is released and falls down toward the Earth, the center of mass of the system will not change, so the Earth will also move up towards the block. However, since the mass of the Earth is so much larger than the mass of the block, the displacement of the Earth will be much smaller than the displacement of the block 3 • As the Earth and the block move toward each other, the force of gravity between the Earth and the block will do work on both the Earth and the block4 • It follows from the differences in the magnitudes of the displacements of the Earth and the block that the work done on the block by the force of gravity is larger than the work done on the Earth by the force of gravity (Section 7-5 and Section 9-4). Thus, the change in the kinetic energy of the Earth will be much smaller than the change in the kinetic energy

3 We could have obtained the same conclusion using a different application of Newton's 3rd law (Section 12-2). The magnitude of the force of gravity acting on the block is the same as the magnitude of the force of gravity acting on the Earth. However, since the mass of the Earth is so much larger than the mass of the block, the magnitude of the acceleration of the Earth will be much less than the magnitude of the acceleration of the block. Since the block has a larger acceleration, we know that it will cover a larger distance than the Earth during the time the block and the Earth come together. 4 From Newton's 3rd law (Section 12-2), we know that the magnitude of the gravitational force of the Earth on the block is equal to the magnitude of the gravitational force of the block on the Earth.

CHAPTER THIRTEEN

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351

of th� block5 ; theref�re, as proposed in Section 3-5, we can safely consider that all of the change in the _ grav1tat1onal potential energy of the system is converted into the kinetic energy of the block alone.

Example 13-7: Problem: A woman initially at rest starts to run across a horizontal field. Compare the changes in linear momentum and kinetic energy for the woman and the Earth. Solution: From Newton's 3rd law, we know that the magnitude of the force of the woman on the Earth is the same as the magnitude of the force of the Earth on the woman and that these forces point in opposite directions. It follows that the change in the linear momentum of the woman is equal in magnitude but opposite in direction to the change in the linear momentum of the Earth. Because the Earth is so much more massive than the woman, the magnitudes of the Earth's velocity and acceleration are much smaller than the magnitudes of the woman's velocity and acceleration. Therefore, the woman will always move a larger distance than the Earth will move. It follows that the work done by the Earth on the woman is larger than the work done by the woman on the Earth and, hence, that the change in the kinetic energy of the woman is much larger than the change in the kinetic energy of the Earth. As discussed in Example 13-7, running is fairly efficient since the energy6 expended by the runner to push on the Earth is converted almost entirely into the kinetic energy of the runner. We can use similar reasoning to understand why we cannot swim as fast as we can run. When swimming, we push ourselves forward by pushing backwards on the water. Unlike the Earth, however, the water will have a signficant displacement when we push on it. It follows that the work done by a swimmer pushing on the water is converted into both the kinetic energy of the water and the kinetic energy of the swimmer. This partioning of the energy results in swimming being less energy efficient that running. In other words, not all of the energy expended by the swimmer is converted into the kinetic energy of the swimmer. Similarly, it is technically incorrect to say that the Earth and the other planets orbit the Sun. Rather, the planets and the Sun are all orbiting around the center of mass of the solar system. Since the Sun is so much more massive than any of the planets, the center of mass of the solar system is very close to the geometric center of the Sun, and, thus, the motion of the Sun is very small. Nevertheless, it is the detection of such small wobbles in the position and/or velocity of other stars that has enabled astronomers to detect many exosolar planets. So far we have limited our discussion to systems with constant mass. It is possible, however, that across a system 's mass may change with time. For example, let's imagine an open containe� sliding _ a horizontal frozen lake in a snowstorm. The ice on the lake can be assumed to be fr1ct1onless, and, 5 Newton's 2nd law (Equation 7-8). 6 The energy obtained from digesting food or burning fat, for example.

352

I

lv\FCHA!'·JiCS A��D THERMODYNJ\N\ICS

THE ENERGY OF PHYSiCS 1 Pt,R-1 I:

thus, the container is an isolated system. The mass of the container increases as snow accumulates in it. Since linear momentum must be conserved7, for this isolated system we have d -0 � dt Psystem -

Thus, the speed of the container must decrease as its mass increases. Furthermore, the acceleration of the container is also a function of the mass of the container. _

dv

dm 1 v. dt J I

a =-1- � a =-(!!!L 2 )( f 1

dt

m

f

The magnitude of the acceleration decreases as the mass of the container increases. Plots of the mass, velocity, and acceleration of an isolated object whose mass increases linearly in time are shown in Figure 13.9. For this system, m i= 5 kg, vi= 10 m/s, and dm/dt = 0.1 kg/s. As shown in Figure 13.9, the magnitude of the object's acceleration is largest when the object's mass first starts to increase. The distance traveled by this object is determined through the integration of the velocity of the object with respect to time. For simplicity, let's assume that all motion occurs along a 1-dimensional axis, denoted as the x-axis, and define the object to be at position x at time t and at position x = 0 at time t = 0. mf =m.I+(

x

fo

dmt t � vt dt ) =

dx1 =

m.IvI.

. dmt m+ ( dt Jt I

mv i ; dt -a x-m, v ,f dm 0 (dm o (' m.+ -- t m.+ --t)t I I dt dt t

f

1)

m.v. dm1

x=-' -' In

dt

m.+ dmt t ( dt J I

m.I

7 The linear momentum ofthe snow is directed along the vertical axis, which is perpendicular to the horizontal motion of _ the contam�r. Thus, for the con�ervati�n oflinear momentum along the horizontal axis, there is only the linear momentum ofthe container. The conservat10n oflmear momentum associated with the vertical axis includes the linear momentum of the snow as well as the linear momentum ofthe Earth. But that's a separate problem to solve.

CHAPTER THIRTEEN

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353

Substitution g ives us

X __

(skg}(107) g 0.1 k s

1n

g Skg+(o.1� }

k � x=(soom)ln 1+(0.02 g)t s ] [

5k g

As shown in Figure 13.9, the position of the object does not increase linearly with time, as would be the case if the object's mass were constant; the dashed line in Fig ure 13.9 denotes the position of the object if its mass were constant. Furthermore, the object's position has decreased relative to what it would be with a constant mass (and thus constant velocity) since the magnitude of the object's velocity decreases with time as the object's mass increases with time. Similarly, the position of the object would be lar ger than what it would be with a constant mass if the object's mass were to decrease with time. 0

16 14

.....en

E

12

0

10

:ii:

� Cl>

8

u u

6 0

20

40

60

Time (s)

80

�

7

0

6

�

-E )C

5

400

0

20

40

60

Time (s)

80

100

· 1 3. 9: The dependence of acceI erat1on, ve I oc1· 1y, creases lineo�r _':'�h_t_i�::... _____ -- --- --

0

•• •• • 0

. .. . . . . • . •. . ••

600

20

..

• ••

...•

40

60

Time (s)

100

80

60

Time (s)

800

200

4

f= igure

40

20

1000

9

8

-0.2

0

10

E

-0.15

100

11

..... -!!

-0.1

• .. . .••

80

.•

100

1 placement for a system whose mass in an d d's

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

13..5 Collisions and Explosions In a perfectly inelastic collision, the colliding objects stick together and after the collision move together with a com­ mon final velocity. For example, consider the system shown in Figure 13.10 that consists of two blocks moving towards each other across a horizontal and frictionless surface. In this system, the magnitude of vA,I is larger than the magnitude of v8,1 , so, eventually. the two blocks collide. The two blocks stick together after the collision and, subsequently. move with a common velocity. How is the final velocity of the combined object after the collision related to the velocities of the blocks before the collision? Figure 13.10: A perfectly inelastic Since the surface across which the blocks slide is friction­ collision between two objects. The less, the blocks form an isolated system, and, therefore, linear indicated velocities ore relative to the horizontal surface aiong which momentum must be conserved during the collision. We could the blocks slide. also conclude that linear momentum is conserved based upon Newton's 3rd law. When the two blocks come into contact with each other, they will exert forces on each other, which will, in turn, create impulses that can change the momenta of the blocks. However, according to Newton's 3rd law, these forces and the associated impulses have the same magnitude but point in opposite directions. Therefore, the net impulse re­ sulting from the collision is zero, and the linear momentum of the system will not change. Similarly, we could argue that since the forces that the blocks exert on each other are internal forces, they cannot affect the motion of the system (Section 12-6). As before, we can express this requirement for linear momentum conservation mathematically as dPsyst.em

-

-

= 0 � Psyst.em.f = Psyst.em,i

Let's define the 1-dimensional external reference frame for our measurements of velocity (and linear momentum) to be the horizontal surface with a positive direction pointing in the same direction as vAJ

Therefore,

CHAPTER THIRTEEN

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355

Example 13-8: Problem consider th system shown in Figure 13.6 that consists of two blocks moving � '. _ onzontal surface. Assuming that the blocks undergo a perfectly inelas­ a long �nct1onless � � tic_ colhs10n, determme the velocity of combined object after the collision. Solution: We know that linear momentum must be conserved during the collision. Hence, �psystem =0 � psystem,[ =p-system,;

Let's adopt the same definition for a reference frame for this system that is shown in Figure 13. 7. Since the speeds of the blocks in Figure 13.6 are relative to the horizontal sur­ face (i.e., to the external reference frame), we have kg m - -p-system,i -18 S

Hence,

The combined object created by the inelastic collision of the two blocks is moving in the positive direction (to the right in Figure 13.6). The final velocity of the combined blocks is the same as the velocity of the center of mass before the collision (Example 13-4), as it should be since linear momentum is conserved for this collision.

Example 13-8. What happens to the energy in an inelastic collision? Let's consider the collision in 8 the The only energy of interest to this system is the translational kinetic energy of the blocks • Before collision, the translational kinetic energy of the system is

After the collision, the translational kinetic energy of the system is

8 The blocks are moving along a horizontal surface, so there is no change in their gravitational potential energies.

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The translational kinetic energy after the collision is less than the translational kinetic energy before the collision. This result can be generalized as Inelastic collisions result in energy being dissipated from the system. But where does the energy go? Some of the kinetic energy is converted into the potential energy to deform the shapes _a. . -,. .· _• va, 1 • of the objects and/or to make the objects stick together. � -� ._- ......__ ____ ..,..• < , c:�-- "· J,,.______� Furthermore, although the system might be instantaneously isolated during the collision9, eventually, there will be interac- Figure 13. l 1: A perfectly elastic co!!i­ sion between two objects. Ail of the tions with the outside environment that will dissipate energy velocities are defined relative to the from the system. For example, some of the energy will be dissi- �o_r_i!-_CJ��-s-�r!�c�--- __ �---- _____ _ __ pated as the kinetic energy of the air being pushed away from the collision (as sound, e.g.), and some is dissipated as heat, which we will discuss in Chapter 15. A collision in which objects bounce off each other, such as a rubber ball bouncing off of the floor, is an elastic collision. A perfectly elastic collision is a collision for which no energy is dissipated from the system. Let's consider the system shown in Figure 13.11 that consists of two blocks moving towards each other on a horizontal and frictionless surface. What is the final velocity of each block if their collision is perfectly elastic? The same reasoning that leads us to conclude that linear momentum is conserved during inelastic collisions indicates that linear momentum is also conserved during elastic collisions. Let's define the 1-dimensional reference frame for our measurements of velocity (and linear momentum) to be the horizontal surface with a positive direction pointing in the same direction as v A,/ ..

.:

rn

For perfectly elastic collisions, we also know that the energy of the system will remain constant. For our system, this requires that the initial and final kinetic energies are equal. Kf =KI

1 21 2 1 21 � 2mA vA .f + 2mB vB .f = 2mA vA .i + zmB vB2 .I

Solving these two equations allows us to determine the relationships between the initial and final velocities of the objects in the collision. It is often more convenient to express these relationships in terms of the initial and final linear momenta of the objects.

9 The time inte�l over which the collision occurs is so short that there is simply no time for energy to be exchanged _ environment . the outside with

CHAPTER THIRTEEN

PA = (mA -mB)pA _

,f

mA +m s

,

2mB

i

+(

I

357

2mA

) -s mA +ms P ,i

(13-10)

m m P =( )pA 1 + ( B- A ) -s i , m , mA +ms P , A +ms B/

As expected, all of the momenta in Equation 13-10 must be measured relative to an external refer­ ence frame.

Example 13-9: Problem: Consider the system shown in Figure 13.6 that consists of two blocks moving along a frictionless ho r izontal sur face. Assuming that the blocks undergo a perfectly elastic collision d etermine the velocity of each block after the collision.

Solution: We can solve this problem using Equation 13-10. Let's adopt the same definition for a reference frame fo r this system, as shown in Figure 13. 7. Since the speeds of the blocks in Figure 13.6are relative to the horizontal surface, we have 2(4kg) ( m 4kg -2kg ( m ( ( + k 4 ) J 6 ) Z kg)( -3 J J ( g p4kg.f = 4kg+2kg 4k g+2kg

7

7

k m ( g J (a kgm J � P P4kg,f =(a 4kg,/ =0 � 4kg)v4kg,f =0 s s

v4kg ,f = 0

2(2kg) m (4kg)( 6mJ+(2kg-4kgJ(z kg)( -3 J ) P2kg,f = 4kg+2kg 4 kg+2kg (

7

7

kg k gm kgm kgm (2kg )v =18-m � =18 � . -+2,/ 2k =16 p g 2kg f p2 kg.f 5 s S S

m v 2kg,/ = 97

ed,

The speed and the direction of the velocity of 2 kg block have chang has stopped moving.

and the 4 kg block

fo rces only, and• theref� re, linear An explosion is another process that results· from internal · think of an explosion as the momentum is also conserved durmg an explosion. 1 ndeed , one can

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opposite of an inelastic collision. Rather than objects colliding and sticking together, the objects in an explosion are initially together and then fly apart.

Example 13-10: Problem: A two-stage rocket is traveling at 1200 m/s with respect to the Earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 360 m/s relative to the second stage. The first stage is three times as massive as the second stage. What is the speed of the second stage with respect to the Earth after the separation? Solution: Let's denote the masses of the two stages of the rocket as m 1 and m2 • The Galilean velocity transformation equation for the velocities of this system is

In this expression, vi.Earth and vz, arth are the velocities of the 1st and 2nd stage relative to the Earth, respectively; and v1 _2 is the velocity of the first stage relative to the second state. The two stages of the rocket form our system, and we can consider the Earth as the external reference frame for our calculations. Since linear momentum is conserved during the explo­ sion, we have E

�P

system

=0 �

-

Psystem,f

= Psystem,i

Let's define the positive direction for the 1-dimensional motion of the rocket to point in the same direction as the initial velocity of the rocket.

Substitution of our equation for the Galilean velocity transformation yields

m, (v., + v,.,..,)+m, v ,,,,,. =(m, +m,)( 1200:)

V 2, Earth

m (

=1200-S

m1 ml

+ mz

)

V1, 2

Since the mass of the first stage is three times the mass of the second stage, we have

v2,

Ea rt h

=1200

3m2 -( s 3m2 + m2 ) Vl ' 2

m

�

V2 E rth

'

a

m 3 = 1200-- (-) V1,2 s 4

Finally, the direction of the v ,2 must be negative, according to the reference fra m e of this 1 system. V2, Ea rt h

m 3 m =1200-- - -360� ( )( S 4 S)

VZ, Earth

V2, Earth

m

m

S

S

=1200-+270-

m = 1470�

Since any forces involved in an explosion can be treated as internal forces, we know that the velocity of the center of m ass is not affected by an explosion. ;;:.:c,�:.-".:�-::..--.:::';".��i�t'lilll'tttlolillill'l QADc ' thus, indicating that heat is also an inexact differential •

Example 15-5: Problem: An ideal gas with energy 14 E = jNk8 moves quasi-statically (i.e., reversibly) from state A to state C as shown in Figure 15.8. What is the work done by the gas as it moves from state B to state C? What is the heat absorbed by the gas as it moves from state B to state C? T

B

� 6-1--------,----

! u, u,

!

C 2 .L---�---+------ A

D..

1 13 See Appendix F. 14 As mentioned previously, we will frequently drop the suffix "av�" when describing the energy (and pressure) of macroscopic systems. It is simply understood that these quantities are average quantities for these systems.

2

Volume (m 3)

3

Figure 15.8: The quasi-static process in Exai:1pJe_l 5-5..:_ _ _ ____ __ _ __ --�

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Solution: The work done by the gas is the area under the curve. Since the volume of the gas is being compressed as it moves from state B to state C, the work done by the gas will be negative.

W ={�(3m' -lm')(6P, -2P,)+(3m' -lm')(2P, -OP,)] W=-BJ The change in the energy of the gas as it moves from state B to state C can be found using Equation 15-9 and the equation for the energy of the gas.

E=�Nk T � E=�PV � M=��(Pv) � M=�(P1V1 2 B 2 2 2

-�v;)

The heat absorbed by the gas is then found using Equation 15-5.

M=Q-W � Q=M+W � Q=-24J+(-8J) � Q=-32J The gas released 32 J of heat as it moved from state B to state C.

As shown in the solution to Example 15-5, if the equation for the energy of the gas and the equa­ tion of state for the gas are known then the heat absorbed/released by a gas during a change of state can be found using the following procedure: • First determine the work from the area under the curve of pressure versus volume for the transition. • Then determine the change in energy from the equation of state and the equation for the energy of the gas. • Finally, determine the heat absorbed/released from the work and the change in energy using Equation 15-5. In Section 15-10 we will learn an additional method for determining the heat absorbed/release by a gas during a change of its state. However, the above procedure is often the simplest approach.

15-9 Heat Capacity Redux The heat capacity for a system relates energy absorbed or released by a system with a change in that system's temperature (Equation 15-6). As demonstrated previously, the change in temperature depends upon how the absorbed/released energy is partitioned among the available kinetic and potential energies of the system (Section 15-6). Thus, the relationship between heat absorbed/

CHAPTER FIFTEEN

I

429

released and a change in temperature will depend upon whether the heat is absorbed/released at constant volume (in which case the system can do no work) or at constant pressure (in which case the system can do work). We define the heat capacity at constant volume, CV' and the heat capacity at constant pressure, CP' using the following equations: CV

=(dQT ) d

V

CP

=(ddTQ)

p

The subscript in these equations indicates which variable (volume or pressure) is held constant during the absorption/release of the heat. According to the first law of thermodynamics, the change in the energy of a system is equal to the difference between the heat absorbed by the system and the work done by the system; we can use Equation 15-10 to calculate the work done by an ideal gas. dE=dQ-d W

� dQ=dE+dW � dQ=dE+PdV dV dQ = dE +P dT dT dT

It follows that this expression that the heat capacity at constant pressure is C = ( dQ) _ C = dE + p dV P P dT p dT dT

(15-11)

If the volume of the system is constant during the change in temperature associ�ted w'.th the absorp­ tion/release of heat, then the heat capacity at constant volume, denoted by CV' is apph cable. dV =O

dE � Cv =­ dT

(15-12)

Example 15-6: f the ato� � Problem: Let's model a solid as a 3-dimensional collection of N atoms. Each o . 1 l these osc1 l d e mo an c in the solid is free to vibrate around its eqm· ·1brmm l O cation and we · lations as simple harmonic motion (Section 6-8). What is the'heat capacity of this system? . ° f the atoms as simple harmonic motion, our . the motwn Solution: Since we can describe . system consists of a collection of N 3-dimensional har�o�ic osci·nators The equation for the energy of each oscillator (i.e., for each atom in the sohd) is

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THE ENERGY OF PHYSICS, PART i: CLASStCAL 1\1\l:CH/lNIC:S ,Ai'JD THERMODYNAMICS

According to the equipartition theorem, the average energy of each atom would be

{E,)=6(½v)

-> (E,)=3k, T

Substitution of this result into Equation 15-2 gives us

The heat capacity of the solid, which is a heat capacity at constant volume is, therefore, found using Equation 15-12.

This result is the Dulong-Petit law.

For an ideal gas, the only energy of the system is the translational kinetic energy of the molecules of the gas. We can determine the temperature dependence of the average value of these energies using the equipartition theorem (Example 15-4). Hence, for an ideal gas, the heat capacity at constant volume is

We can relate the heat capacity at constant pressure to the heat capacity at constant volume using the equation of state for the system. For an ideal gas, we have

PV=Nk,T -> V=[

N;,}

->

dV = NkB P dT

CHAPTER FIFTEEN

I

431

When a system absorbs heat, the energy flowing into the system can be converted into work, an increase in the temperature of the system, or both. The larger the fraction of the energy that is par­ titioned into work the less the amount of energy that is available to increase the temperature of the system 15• Thus, a system absorbing heat at constant pressure will have a smaller temperature change than if it was absorbing heat at constant volume. In other words, more heat must be absorbed by a system at constant pressure than at constant volume for the same change in temperature. It follows that CP must be greater than CV' as we have shown for the ideal gas. It also follows from this line of reasoning that

dV > 0 for an ideal gas. In other words, the volume of an ideal gas must increase as its dT

temperature increases. Similar equations for Cv and CP can be determined for non-ideal gases. The major difference in the derivation of these expressions is that while the energy of an ideal gas is a function of temperature only, the energy of a non-ideal gas can also depend upon volume (or other parameters). To account for this, we need to replace the total derivatives in the expressions for C and C derived previously v P with partial derivatives and add an extra term for the dependence of the energy on volume. Cv

=(aarE)

V

Cp =(aE)

dV +(P+(aE) ) ar v av r dT

(15-13)

We will discuss the properties of real gases in Section 17-7.

15-10 Cyclic Processes will an lcieal Geis We now turn our attention to an ideal gas that is under­ going a repeated cycle of pressure and volume changes. Let's consider the cycle shown in Figure 15.9; this style o f representation is referred to as a PV diagram since it denotes the relationship between the pressure (on the abscissa) and the volume (on the ordinate) of the gas. The gas starts at point A corresponding to pressure P1, volume V1, and temperature TA . The gas then undergoes a quasi-static isovolumetric increase in its pressure to arrive at point 8, which corresponds to pressure P , volume V1 , and temperature 'f_e· Sin�e 2 the volume of the gas remained constant durmg this process, the gas did no work. The heat absorbed by the gas during this process is

f::I

C

B

P2 AB

Oco

Q --+

ti) ti)

f

P1

--+

A V1

!

D

QDA

V2

Volume

.Figure 15.9: A cyclic process for an ideal g the area gos. The net work done by the as is ___ __ __ _ .__ enclosed by the path

QAB =CV (rB -T) A . ·tational potential energy into different kinetic 15 This is analogous to how changing the parti·r wm·ng of a change m grav1 energies affects the acceleration of a system.

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

Since the process of moving from point A to point B was quasi-static we can relate TA and T8 to the P1, P2, and V1 using the equation of state of an ideal gas. PV=Nk B T

-"7

p2 � -�� =Nk BTB -Nk BTA

Hence,

QAB --cV (�-�)� NkB

The system then undergoes a quasi-static isobaric expansion to point C. The work done by the system during this process is simply the area under the curve.

The heat absorbed by the gas during this process is

We can, again, use the equation of state to relate temperature to pressure and volume since this expansion was quasi-static. " '

A similar set of calculations allows us to determine the heat and work associated with the other processes in the cycle.

CHAPTER FIFTEEN

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433

The net work done by the system during this cycle is, thus,

As expected, the net work done by the gas is also the area enclosed by the path (Figure 15.9). Since the expansion of the gas (from state B to state C) occurred at a higher pressure than the compression of the gas (from state D to state A), the net work is positive. It follows that when displayed using a PV diagram, all clockwise cyclic processes have positive net work and all counter-clockwise cyclic processes have negative net work Similarly, the net heat absorbed by the system during this cycle is

Since our system is an ideal gas, we have CP CV +NkB � CV -CP =-NkB =

Substitution of this relationship gives us

Qnet

-w -

net

The temperature of the gas at each point during the cycle is related to the pressure and volume of the gas at that point, according to the equation of state. Thus, the te�perature 0� the gas at point A is the same before and after the system completes the cycle. Smee energy is a sta te function for a system, the energy of the gas at the end f the cycle must equal the energy of the · · 15 - 5), 1 gas a t the start of the cycle. This, of course, reqmres that Q = W &1or the eye 1 e (Equat'on which we calculated.

°

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THE ENERGY OF PHYSICS, PART i: CLASSICAL IV\ECH/>,N!CS AND THERMODYNAMICS

Example 15-7: Problem: An ideal gas is taken through the quasi-static cyclic process shown in Figure 15.10. The system begins in state A, then experiences an isovolumetric increase in pressure to state B, then an isothermal expansion to state C, and, finally, an isobaric compression back to state A. What is the heat absorbed or released during each process of the cycle? What is the net work done by the gas during the cycle? Solution: Let's begin by defining the two tem­ peratures of the system during this cycle. We will denote the temperature of the gas at state A to be T and the temperature of the gas at state 8 and state C to be T2 •

f

B

P2

-1-----•

P1

+---tt-�---, C

::::, tn tn

f D.

Figure l 5, 10. The 157.

•

-

I ic process 1n txan1p1e

-- ---·-------- -�-----�- ··------------�- ��-

1

There is no work done by the gas during the isovolumetric transition from state A to state 8 since the volume of the gas is constant during this process.

The heat associated with this process is

The work done by the gas during the isothermal transition from state B to state C is

Since this process is isothermal, the temperature of the gas is not changing, and, therefore, the energy of the gas is not changing. Thus, the gas must absorb energy from the environ­ ment to do the work associated with the increase in its volume.

CHAPTER FIFTEEN

Finally, for the isobaric compression of the

I

gas from state C to stateA, we have

Thus, heat is absorbed by the system when it tran sitions between state A and state B and between stateBand state C, and heat is released by the system during the transition from state C to state A. Similarly, work is done by the syst em during the transition from state B to state C, and work is done on the system (negative work done by the system) during the transition from state C to state A The net heat absorbed by the gas during a complete cycle is

Q-

�c, (r,-r,)+Nk,T,

Qnet =(cp -cV )(r1 -r2 )+Nk

r ln(v;J V

B 2

1

l

n( �

}c,(r,-r,)

� Qnet =Nk0(½ -I;)+Nk0J;ln(v;J V

1

Similarly, the net work done by the system during a complete cycle is

wnet =WAB +WBC +WCD

In summary,

w

Process

Q

AtoB

Cv (J;-½)

Bto C C toA

Nk,T, l ( n

�J

CP(½-I'i)

0 Nk,T, l ( n

�J

Nk8(½ -J;)

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THE ENERGY OF PHYSICS 1 PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

Entire Cycle As expected' Qnet = Wnet for a complete cycle, so the temperature of the gas is the same at the start and end of a complete cycle. This is, of course, expected since energy is a state function and, for an ideal gas, depends upon only the temperature of the gas. Since the energy of the ideal gas must be the same at the start and end of a complete cycle, the temperature of the ideal gas must be the same at the start and end of a complete cycle.

Example 15-8: Problem: An ideal gas with energy E = � Nk T is sub2 B jected to a cyclic, quasi-static (i.e. reversible) process shown in the PV diagram in Figure 15.11. The number of molecules in the gas is N = 1 x 10 24• What is the net work done by the gas during one complete cycle? What is the net heat absorbed/released during one complete cycle? What is the change in temperature of the gas between state B and state C? Solution: The net work done by the gas is the area enclosed by the path.

-

B

:. 6 -------------�----

1

2

3

Figure 15.11: The cyclic process 1n Example 15-8.

3 -1m 3 )(6Pa -2P) Wnet =!(3m � Wnet =4J 2 a

The net work is positive since the cyclic process in Figure 15.11 follows a clockwise path. In other words, the expansion of the gas occurs at a higher average pressure than the compres­ sion of the gas. The change in energy must be zero for one complete cycle. Thus, the net heat absorbed/released is equal to the net work done by the gas. M'=O � Qnet =Wnet � Qnet =4J

The gas absorbs a net 4 J as heat and does a net 4 J of work during one complete cycle. We can solve for the change in temperature directly from the equation of state for the gas. AT T T AT = pc Ve u B�c � = c - B � u B-tC Nk B

-

PB VB Nk B

1 ( � ATB-tC = -- PC VC -PB VB ) Nk B

CHAPTER FIFTEEN

I 437

Substitution yields �T

-

1

,�, - (1x10" }( 1.38 x 10-"

!)

{( 2Pa)(3m')-( 6Pa)(3m'))

· · · · It makes sense that the net work done b th a ::: 1: :: :ple 15-8 si�ce the expansion of :::; l: :�� :ta the gas (from state A to state B) occurs e e h t a 0 state C to state A). If the cycle in Figure 15.11 were run ba ckwards ;fr::�;:t::�0°:t:�: � :hc: : ::::; a o sta te A ), th� net work done by the gas would be negative, specifida lly -4 J, :: !:::::;� : 4 g e se J of heat m one complete cycle .

15-11 Free Expansion �o_n�i�er the system shown in Figure 15.12 in which an ideal gas 1s 1mtially restricted to half of the volume of a therma lly and meP=O chan�cally isolated container; a pa rtition keeps the gas in only the lefts1de of the container while the right side of the container is at vacuum. The initia l tempera ture, pressure, and volume of the g as are T, P0, and V01 respectively. When the partition is removed, the Partition Removed gas expands to fill the entire volume of the containe r. We refer to this process as a free expansion. Since the system is therma lly a nd m echanically isola ted, Q = 0 ..---------and W = 0 for this expansion. It follows that �E = O for this expan- Pof2, 2Vo, T sion. Consequently, if the energy of the gas has not changed, then the temperature of the gas has not cha nged. Eventually, the gas will come to thermal equilibrium aga in at which point the volume of Fig ure 15 · 12: Free exp ansion of the gas has doubled to 2 V0, and the pressure has decreased to P 0 /2. an ideal gas. --------------- ------- --- ------- -Why did the gas expand? There is no change in the potential energy of the gas associated with the expansion, so w e cannot argue th at expansion occurred to satisfy the principle of minimum potentia l energy. Furthermore, this expansion is irreversible since the g as will neve r spontaneously revert back to its original volume and pressure. Thus, we ca nnot a rgue that the expa nsion represents an equilibrium change in the state of the ga s. Of course, we can very easily explain why the gas expands in terms of the motion of the molecules of the g as. When the partition is removed, there is nothing preventing the molecules of gas from moving into the right-hand side of the containe r, and, thus, they will eventually fill the entire volume

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THE ENERGY OF PHYSICS, PART i CLA.SS!O\L 1v�ECHANICS t,,i'-1D THERMODYNAMICS

of the container. From the discussion of the statistical description of systems in Section 15-3, we can say that if given enough time, the molecules of the gas will eventually be evenly distributed among all possible accessible states (positions and velocities) of the entire container. We can, therefore, make a probability-based argument that the gas expanded simply because it is more likely to spread out among the entire accessible volume of the container rather than be confined to only a fraction of that volume. This probability-based explanation for the behavior of thermodynamic systems is the basis of a quantity known as entropy, which will be the subject of the next chapter.

• If the number of objects in a system is large enough, then all possible positions and kinetic energies accessible to the objects in the system will be associated with at least one object in the system. • Thermodynamic state: A configuration of the microscopic constituent objects in a macro­ scopic system. Thermodynamic states can be characterized using the associated macroscopic parameters (e.g., temperature, volume, pressure) of the entire system. • Accessible state: A thermodynamic state that is compatible with the macroscopic parameters of the system. • State variables: The set of parameters required to define a thermodynamic state. • Equation of state: An expression of the relationship between the state variables for a system. For example, the equation of state for an ideal gas relates the pressure, volume, number of molecules, and temperature of the gas.

The equation of state for an ideal gas can also be expressed as an energy density. PV=�E

3

• In thermal equilibrium, the probability of finding a system in any of its accessible states is independent of time. • In thermal equilibrium, all accessible states are equally probable. • Thermal equilibrium: Two systems are said to be in thermal equilibrium if they have the same temperature. No net heat is exchanged between systems in thermal equilibrium. • Heat: A change in energy associated with a non-mechanical interaction. Heat is denoted by the variable Q. • The infinitesimal work dW associated with an infinitesimal change in volume dV is dW=PdV

• The 1st law of thermodynamics: The change in the energy of a system is equal to difference of the heat absorbed by the system and the work done by the system. LiE=Q-W

CHAPTER FIFTEEN

I 439

• The equipartition theorem: A mean value of ½kB T is associate . d with each quadratic term in the expressi?n for the energy of a system at equilibrium. • Heat capacity: The proportionality between infinites1m . al heat absorbed or released by a system and the infinitesimal change in the temperature of the system. C=dQ dT The heat capacities at constant volume and constant pressure 1or c an 1'dea l gas are C

V

= dE dT

C

p

= dE + dV p dT dT

ProB�ms Conceptual

1. Which ideal gas will have a larger heat capacity: a monatomic ideal gas or a diatomic ideal gas? 2. Does the average energy of an ideal gas depend upon the volume of the gas? 3. An ideal gas is subject to the reversible (i.e., quasi-static) cyclic process shown in the PV diagram in Figure 15.13. Is the net work done by the gas during one complete cycle positive, negative, or zero? Is the net heat ab­ sorbed by the gas during one complete cycle positive, negative, or zero?

Section 15-5

-

B

� S+---1•�,--�----ac �

f

A

2+---t1£---+--1

3

2

Volume

(m 3 )

Quantitative

4. Through the interaction with its environment, a system releases 6 J of heat while doing 4 J of work. What is the change in the system's energy? 5. Through the interaction with its environment, a system's energy increases by 12 J while it does 4 J of work. What is the heat absorbed/released by the system?

Section 15-6

J while it does 6. Through the interaction with its environment, a system's energy increases by 12 What is the change in the 4 J of work. The heat capacity of the system is a constant 0.5 J/K. temperature of the system? act with each other. The 7. Two identical objects initially at 200 K and 300 K are brought into cont of heat the two objects heat capacity of the objects is a constant 2 J/K. Through the exchange

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eventually obtain a common final temperature. What is this final temperature? How much heat was exchanged during the process of obtaining the final temperature? 8. Two objects, A and B, initially at 200 K and 400 K, respectively, are brought into contact with each other. The heat capacity of object A is a constant 2 J/K and the heat capacity of object Bis a constant 3 J/K. Through the exchange of heat the two objects eventually obtain a common final temperature. What is this final temperature? How much heat was exchanged during the process of obtaining the final temperature? 9. Two objects, A and B, initially at 200 K and 400 K, respectively, are brought into contact with each other. Over this range of temperatures, the heat capacity of object A and object B can be approximated by the following equations:

cB = 4 lK +(o.02-1 )r K2

Through the exchange of heat the two objects eventually obtain a common final tempera­ ture. What is this final temperature? How much heat was exchanged during the process of obtaining the final temperature? 10. What is the total translational kinetic energy of 1 x 10 24 molecules of a monatomic ideal gas at 20 °C? Section 15-8 11. The pressure and volume of a gas are related by the following equation:

t

-

B

&+-----�--

How much work is done by the gas as it expands from a volume of 2 m3 to a volume of 3 m3 ? 12. An ideal gas with energy E=¾Nk8 T moves quasi-statically

(i.e., reversibly) from state A to state C as shown in the PV dia­ gram in Figure 15.14. What is the work done by the gas as it moves from state A to state B? What is the work done by the gas as it moves from state B to state C? 13. An ideal gas with energy E = Nk8 T moves quasi-statically (i.e.,

¾

reversibly) from state A to state C as shown in the PV diagram in Figure 15.15. What is the work done by the gas as it moves from state A to state B? What is the work done by the gas as it moves from state B to state C?

¾

14. An ideal gas with energy E = Nk8 T moves quasi-statically (i.e., reversibly) from state A to state C as shown in the PV diagram in Figure 15.16. What is the work done by the gas as it moves from state A to state C?

1

3

2

Volume

(m l)

Figure 15. 14

-----2

�

---1

Volume Figure 15. 15

3

2

(ml )

CHAPTER FIFTEEN 15. An ideal gas with energy E = � Nk T is sub1· ected to a cycl· 1c, quas·i2 n static (i.e., re�ersible) process shown in the PV diagram in Figure 15.17. What is the work done by the gas as it moves from state A to state B? What is the work done by the gas as it moves from state C to state A?

Section 15-1 0 16. An ideal gas with energy E = � Nk T moves quasi-statically (i.e., 2 8 reversibly) from state A to state C as shown in the PV diagram in Figure 15.15. What is the change in the energy of the gas as it moves from state A to state B? 17. An ideal gas with energy E = �Nk T moves quasi-statically (i.e., 2 B reversibly) from state A to state C as shown in Figure 15.18. What is the change in the energy of the gas as it moves from state A to state C? � 18. An ideal gas with energy E = Nk T moves quasi-statically (i.e., 2 B reversibly) from state A to state C as shown in the PV diagram in Figure 15.19. What is the change in energy of the gas as it moves from state B to state C? 19. An ideal gas with energy E = � Nk Tis subjected to a cyclic, quasi-

2 B static (i.e., reversible) process shown in the PV diagram in Figure 15.11. What is the heat absorbed by the gas in going from A to C via the path ABC? 20. An ideal gas with energy E = � Nk T is subjected to a cyclic, 2 B quasi-static (i.e., reversible) process shown in the PV diagram in Figure 15.20. What is the heat absorbed by the gas during one complete cycle? If N = 1.5 x 1022, what is the change in tempera­ ture of the gas at it moves from state A to state B? 3 . IIY ('1.e., · 21. An ideal gas with energy E = -Nk T moves quas1-stat1ca 2 B reversibly) from state A to state C as shown in the PV diagram in Figure 15.21. What is the change in the energy of the gas as it moves from state A to state C? If N = 2 x 10 22, what is the change in temperature of the gas at it moves from state B to state C? ii

e:. Ill Ill

a.

5

2

r

A

� 0.. 1

2

Volume (m3)

Figure 15.20

3

2

./�c

------

t________

_________

3

2

1

Volume

1 S.21

I

(m 3)

44 l

6+-----.ac

I!!

�

2 __A

2

:g

f

2

A

----------1----4,C.

2

1

3

Volume (ml)

Fi(1ure 15 17 B

3 +-------,,,....-

I!! ::I

2

A

----------1-----

C

0..

3 2 Volume (m 3) Figure 15. 18

'ii e:,_

A

5 ----------�-�---- C

Ill

B � 2 -------�---+---

1

2

Volume (m3)

B

3

Volume (m3)

Figure i5.19

B

'ii

C

l

I

3

B

442

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

¾

22. An ideal gas with energy E= Nk8 Tis subjected to a cyclic, quasi­ static (i.e., reversible) process shown in the PV diagram in Figure 15.22. What is the heat absorbed by the gas for one full cycle? What is the change in the energy of the gas as it moves from state A to state C? 23. An ideal gas with energy E=�Nk8 T is subjected to a cyclic, 2 quasi-static (i.e., reversible) process shown in the PV diagram in Figure 15.23. If N = 1.45 x 1022, what is the change in tempera­ ture of the gas at it moves from state A to state C? What is the change in the energy of the gas during one complete cycle? What is the work done by the gas during one complete cycle? 24. An ideal gas with energy E= �Nk8 T is subjected to a cyclic, 2 quasi-static (i.e., reversible) process shown in the PV diagram in Figure 15.24. What is the net work done by the gas during one cycle? What is the difference in the energy of the gas between state C and state A? What is the heat absorbed by the gas in going from A to C via the path ABC?

i I!!

:II Ill Ill

I!!

B 6 _,______

2 .....

A_/�_C

i

i

2

Volume

3

(m3)

Figure 15.22

1

2

Volume Figure

3

(m3)

15.23

25. An ideal gas with energy E=§._Nk8 T is subjected to a cyclic, quasi-static (i.e., reversible) process 2 shown in the PV diagram in Figure 15.25. What is the heat absorbed by the gas during one complete cycle? If N = 2.9 x 1022, what is the change in temperature of the gas at it moves from state A to state C? B 3

A/

1

D 2

Volume (m 3) Figure

15.24

�

D

2

3

/

Volume (m3) Figure 15.25

3

C

CHAPTER SIXTEEN Entropy and the 2nd Law of Thermodynamics Everything is physics and mathematics. -Katherine Johnson

W

16-1 Introduction

e began this book with the principle of minimum potential energy: systems will always arra�g� themselves so as to minimize their total potential energy. We further indicated that It 1s the 2nd law of thermodynamics that justifies why systems behave this way. Indeed, it is the 2nd law ofthermodynamics that ultimately explains why systems behave as they do even when there is no change in the energy of the system (such as in the free expansion ofthe gas in Section 15-11). As we shall see, rather than changes in the energy ofa system, it is actually changes in the entropy ofa system that affect the behavior ofthe system.

16-2 Entropy 1 The statistical definition ofthe entropy ofa thermodynamic state is given in Equation 16-1 .

(16-1)

In Equation 16-1, the entropy (.5) is equal to the product of the Boltzmann constant (k8) and the natural log ofthe multiplicity ofthe state (.Q). The multiplicity of a state is the number ofconfigura­ tions of the components of the system that all correspond to the same thermodynamic state (i.e., to the same macroscopic parameters, such as temperature, pressure, and volume). As indicated in Equation 16-1, the units of entropy are J/K, and the largest entropy is associated with the state with the largest multiplicity. Entropy is an intrinsic property ofa system in the same way that energy and momenta are intrin y sic properties ofsystems. Furthermore, since the entropy ofa state is determined by the multiplicit of the state (Equation 16-1), it follows that entropy is a state function just like energy. I� other words, the the change in the entropy of a system associated with a change in the thermodynamic state of 1 This same equation can be found on Ludwig Boltzmann's tombstone in Vienna.

443

444

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

system is independent of the path (i.e., process) by which the change in entropy occurred. Finally, entropy is also an extensive parameter just like energy (Section 1-6). To give this discussion more context, let's consider all the possible outcomes of flipping a coin 4 times; for example, the first flip might result in heads, but the next 3 might all be tails. In Table 16-1, the 16 possible outcomes are grouped together by the number of heads that occurred. Table 16-1: The possible outcomes (heads is denoted by H, and tails is denoted by T) of flipping a coin 4 times. Number of Reads

:Associated Outcomes fmicrostotes)

Toto] Number of Ovtcomes (multiplicity}

0

TTTT

1

1

HTTT THTT TTHT TTTH

4

2

HHTT THHT HTHT THTH HTTHTTHH

6

3

THHH HTHH HHTH HHHT

4

4

HHHH

1

It's clear from Table 16-1 that the largest number of outcomes is associated with 2 of the 4 flips being heads. We would, therefore, conclude that getting 2 heads is the most probable outcome2 • This is the basis of the statistical definition of entropy. It is most probable to obtain 2 heads when flipping a fair coin 4 times since the largest fraction of the total outcomes corresponds to this re­ sult; i.e., the multiplicity associated with this result is the largest. When flipping a fair coin 4 times, the multiplicity for obtaining 2 heads (.Q = 6) is larger than the multiplicity for obtaining 4 heads (.Q = 1). Therefore, it is more likely that 2 heads will be obtained rather than 4 heads. Furthermore, the probability of obtaining one outcome relative to the probability of obtaining another outcome is the ratio of the multiplicities of the outcomes. PA QA -=-

PB

QB

(16-2)

In Equation 16-2, the subscripts A and B denote the possible outcomes or, more generally, accessible states of the system. For example, when flipping a fair coin 4 times, the ratio of the probability of getting 2 heads to the probability of getting 4 heads is

2 We assume that the coin is fair. That is, we assume that the probability of getting heads is equal to the probability of getting tails. In other words, each accessible state (each combination of heads and tails) is equally probable.

CHAPTER SIXTEEN

p_

�_

Q2heads

p4heads - Q4heads

�

I

445

p_2heads 6 -=- � p_2heads -6P 4heads P4heads 1

. When flipping a coin 4 times, you are 6 times more likelY to obtam 2 heads than to obtain 4 heads. . · Furthermore, according to Equation 16 -1, we can say that the entropy associated with obtaining . 2 heads is larger than the entropy associated WI'th gettmg 1 head.

Example 16-1: Problem: You flip a fair coin 3 t·imes. H ow much more hkely are you to obtain 2 heads than 3 heads.7 Solution: The possible outcomes are shown in the table below. Number of Heads

Associated Outcomes

Toto! Number of Outcomes

0

TTT

1

1

HTT THT TTH

3

HHT THH HHT

3

HHH

1

3

The ratio of the probability of getting 2 heads to the probability of getting 3 heads is �heads Q2heads �heads -=-3 � -=-�heads

Q3heads

�heads

l

P2heads =3P3heads

You are 3 times more likely to get 2 heads than to get 3 heads.

entropy with a measure of Lastly, it is worth mentioning that people often incorrectly associate e from Equation 16-1, en­ disorder. This cannot be ·true, of course, since disorder is subjective. Asid system can be arranged given tropy is best considered to be a measure of the number of ways that a em ( e.g., volume, temperature, the constraint of a particular set of macroscopic properties for that syst

446

I

THE ENERGY OF PHYSICS, PART i: CLASSIC.I.\L MECH/-\NiC:S 1\ND THERMODYNAMICS

or pressure). Ofcourse, such calculations become intractable for large systems. Indeed, determining the multiplicity for each thermodynamic state for 1 liter of water (i.e., enumerating the configura­ tions of all possible positions and velocities of the water molecules for a given temperature of the water) would be nearly impossible. Therefore, we rely upon statistical definitions and calculations similar to those used in Chapter 15. We can nevertheless rely upon the large number ofmicrostates accessible to macroscopic systems to argue, based upon maximum entropy, that such systems are far more likely to be found with an energy corresponding to their average energy than with any other energy when in thermodynamic equilibrium. Thus, we can refer to the average energy ofthe system as the energy of the system.

Although direct calculations ofthe entropy ofa system are often difficult (ifnot impossible), it is nev­ ertheless straightforward to calculate changes in the entropy of a system. According to the 2nd law of thermodynamics, if a system undergoes a quasi-static process at constant temperature in which it absorbs an infinitesimal amount of heat, the infinitesimal change in the entropy of the system is dS=dQ T

(16-3)

If the temperature of the system is constant as the heat is absorbed, then Equation 16-3 simplifies to Equation 16-4.

(16-4)

Example 16-2: Problem: One liter of water at 273 K is brought into contact with a large heat reservoir3 at 373 K. As a result of the interaction between the water and the reservoir, the temperature of the water will increase (i.e., heat will flow from the reservoir to the water, thereby increas­ ing the temperature of the water4). However, we will assume that the temperature of the reservoir r�mains constant. What has been the change in entropy of the water (Afwate,), of the reservoir (Af,eservo;), and ofthe entire system consisting ofboth water and heat reservoir (Afsystem) when the water has reached 373 K? The heat capacity ofthe water has a constant a value of 4180 J/K during this process. 3 The ten:iperature of a heat �eservoir remains constant regardless of the heat it exchanges with other systems. This can be accomplished, for example, 1f the energy of the heat reservoir is much, much larger than the energy of the system with which it exchanges heat. 4 See Equation 15-6.

CHAPTER SIXTEEN

I 447

Solution: The changes in entropy can be calculated using Equation 16-3 and Equation 15-6. Since the heat capacity of the water is constant, we have dQwater

= ter

Afwa

= Cwater

373 K

f

2 73 K

dT

dT

cwater T

-+

-+

S

d

r

wate

=

=c r

Afwate

C

water T

dT

373K r

wate

f

2 73 K

-+

dT

T

f

S water

d

=

f

dT C water T

-+ Afwater = Cwater (1nrl

373K

znK

)

373 llSwa r =(4rno l )1n( K) K 273 K -+ Afwa r =13051 K te

te

The entropy of the water has increased because the water has absorbed heat (Qwater > 0). Since the temperature of the reservoir remains constant, the change in the entropy of the reservoir can be found using Equation 16-4 and Equation 15-7 and the fact that any Q,eservolr = -Qwater·

-+

= reservoir Af reservoir T reservoir

Q

Af reservoir

.=

Af reservoir

= -Qwa

r

te

T reservoir

-( 4180½ )( 100K)

-+

Af reservoir

=

-C

wa

ter

l!!..Twater

_ reservoir

T

J -+ Afreservoir. =-1121K

373K

The entropy of the heat reservoir has decreased because it has released heat (Qreservoir < 0). Since entropy is an extensive parameter, the total change in the entropy of the entire system is the sum of the change in the entropy of the reservoir and the change in the entropy of the water. Afrotnl =llSreservoir. +Afwa

r

te

J J -+ llSrota/ =-1121-+1305K K

J llSrotnl =184_ K The total entropy of the system has increased.

448

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANiCS AND THERMODYNAMICS

In addition to the definition of entropy given in Equation 16-3, the 2nd law of thermodynamics also states that the entropy of an isolated system never decreases. Thus, for an isolated system, the change in entropy associated with any spontaneous process must be greater than or equal to zero. (16-5)

In Example 16-2, the change in the entropy of the system was positive indicating that if the water was brought into contact with the reservoir, heat would flow spontaneously from the reservoir to the water, and the temperature of the water would increase. Indeed, the 2nd law of thermodynamics explains why heat always flows from an object at higher temperature to an object at lower tem­ perature. Consider two objects, A and B, which are in thermal contact with each other but otherwise isolated from the outside environment. If an infinitesimal amount of heat flowed from object A to object B, the change in the entropy of the entire system consisting of both objects would be

The change in the entropy of object A is negative since it released heat, and the change in the entropy of object B is positive since it absorbed heat. In order for heat to flow spontaneously from object A to object B, the total change in the entropy of this system must be positive.

Hence, in order for heat to flow spontaneously from object A to object B, the temperature of object A must be larger than the temperature of object B. There is no change in entropy for a reversible process (i.e., there is no change in entropy for any process occurring at equilibrium). Afreversible =0

�

Afequilibrium

=O

(16-6)

Now suppose the water in Example 16-2 had been heated by first bringing it into contact with a heat reservoir at 323 K and then, after its temperature had reached 323 K, with a heat reservoir at 373 K. What is the change in the entropy of the entire system during this process? Since entropy is a state function, the change in the entropy of the water is the same regardless of the path over which the heat was absorbed. l:!.Swater

=13osl K

CHAPTER SIXTEEN

I

449

The change in the entropy of the 323 K reservoir is Af323Kreservoir

!J.S

323Kreservoir

=_

Q 323Kres ervoir

T

323K reservoir

� Af323Kreservoir

- (41Bo¾){soK) 323K

�

= Cwater dTwater T

323Kreservoir

Af323Kreservoir =-6471 K

The change in the entropy of the 373 K reservoir is Af373Kreservoir

Af373Kreservoir

= _ Q373Kreservoir T 373Kreservoir

C dT � Af373Kreservoir = _ water water T373Kreservoir

( 4180 n(soK) = ---'--�-- � Af 373Kreservoir 373K

=-560 l K

The total change in the entropy of the system is, thus,

Aftotal =981 K The change in the entropy for the system is still positive; this positive change in entropy indicates that this process of heating the water would also happen spontaneously. However, the change in entropy for this process of heating the water is closer to zero since this process more closely approximates a reversible process5 • Indeed, in the limit that the temperature of the water were raised using an infinite number of heat reservoirs, each of which was at an infinitesimally different temperature, the total change in entropy would be zero. In such a process the water would always be in equilibrium as its temperature was slowly raised or, to be it another way, the heating of the water would be a completely reversible process. Finally, according to the 2nd law of thermodynamics, the entropy of an isolated system at equilib­ rium must be a maximum. This is the equivalent of saying that a system at equilibrium is most likely to be found in the accessible state6 with the largest multiplicity (Equation 16-1).

5 That is, a process occurring at equilibrium. 6 See Section 15-3.

450

I

THE ENERGY OF PHYSICS, !)ART I CLASSICAL 1\t\ECHANICS AND THERMODYNAMICS

A heat engine is a device that absorbs heat from a high temperature reservoir, performs work, and then releases heat to a low temperature reservoir, as shown in Figure 16.1. Q

L '..••·•.·.·••. ··.··•...··...· •.· ··.··. ,.,·•··· \·/�If . · .,· ·. . ·.,··. _....

High Temperature Reservoir

�·>.' >;,'··,:·,-,· ,,._

Low Temperature Reservoir

Figure 16. l: /\ heat engine absorbs heat Q" frorn a reservoir at temperature T, 1, performs work \//, and then releases heat Q, to a reservoir at temperature L

We know from the first law of thermodynamics (Equation 15-5) that the work done by the heat engine is the difference between the net heat absorbed by the engine and the change in the (internal) energy of the engine. M=Q-W � W=Q -M The net heat in this equation is the difference between the heat absorbed from the high temperature reservoir and the heat released to the low temperature reservoir.

Let's imagine that the heat engine functions ideally so that there is no change in the internal energy of the engine during its process of doing work (�E = 0); e.g., there is no change in the temperature of the engine and no energy dissipated to the environment. For such an ideal heat engine we have

If the temperature of the engine is between the temperatures of the reservoirs, heat will spon­ taneously flow from the high temperature reservoir to the engine and from the engine to the low temperature reservoir. The total change in the entropy of the system is a Aftotl

=

Afh 19. h temperature reservoir. +Aflow temperature reservoir +Afengine

The change in the entropy for the engine is zero (Mengme . = 0) because the temperature of the engine does not change (i.e., the engine is not absorbing any energy). Since the temperatures of the heat reservoirs remain constant, we can further simplify the expression for the total change in the entropy of the system using Equation 16-4. � Aftota1

=- QH QL + TH TL

CHAPTER SIXTEEN

I

451

The change in the entropy of the high temperature reservoir is negative since it is releasing heat to the engine and the change in the entropy of the low temperature reservoir is positive since it is absorbing heat from the engine. According to the second law of thermodynamics (Equation 16-5)

Hence, there is a limit to the work that can be done by the engine.

It is, therefore, the 2nd law of thermodynamics (i.e., the concept of entropy) that not only explains how heat engines function but also places a limit on their efficiency.

Example 16-3: Problem: An ideal heat engine absorbs heat from a 400 K reservoir and releases heat to a 100 K reservoir. If the engine absorbs 300 J of heat from the 400 K reservoir, what is the maximum work that can be done by the engine? Solution: The maximum work done by the heat engine is

w,;Q"(1- �) -"

w_=Qn(1-

�) -" w_=(3ooi)(1- ����)

Wmax =225J

Example 16-4: . s of an · deal heat engine are initially at Problem: The high and low temperature reservo1r i . 1Y· As a result ofthe operat_10n ofthe heat engine the high temperatures TH and TL' respective . heats until both have reached . cools and the low temperature reserv01r temperature reserv01r . · t the engme stops worki'ng· The heat capacities of a common temperature, TF' at wh'ICh pom . constant as the temperature of the . I , denoted bY C, and remam . the reservoirs are 1dent1ca

452

I

THE ENERGY OF PHYSICS PART I: CLASSICAL MECHANICS AND THERMODYNAMICS 1

reservoirs change. What is the maximum amount of work that can be obtained from this heat engine? Solution: The total amount of work done by the engine can be determined from the 1st law of thermodynamics. The heat released by the high temperature reservoir is Qhigh temperature reservoir = C ( TH - TF ) Similarly, the heat absorbed by the low temperature reservoir is Qlow temperature reservoir

=c(r-r L) F

Thus, the total work done by the engine is W=Qhigh temperature reservoir -Qlow temperature reservoir

We can now use the 2nd law of thermodynamics to determine the relationship between the initial and final temperatures of the reservoirs. As before, we will assume that the change in the entropy of the engine is zero. !!Stotal �O � Af.high temperature reservoir +Aflow temperature reservoir >O Since the heat capacities of the reservoirs are constant we have

r,

T

Afhigh temperature reservoir =Cf d T TH

TF�

(T J

�Af.high temperature reservoir =Cln..L TH

(

T !!Slow temperature reservoir. =C - �Afhigh temperature reservoir = C In i J T

f

�

�

CHAPTER SIXTEEN

I

453

Hence,

c1n(;)+c1n(TF)�o � Cln( T}. )>o H TT H L -

TL

�

Therefore, the largest amount of work that can be obtained from the engine is

w

TH +T-l C > --..J 2

lrr l

l

H L

The efficiency of a heat engine is defined as the ratio of the work done by the engine to the heat absorbed by the engine.

w

(16-7)

E=- � W=EQH QH

As derived above, the maximum efficiency obtained by an ideal engine is a simple function of the temperatures of the two reservoirs connected to the engine. All processes within the thermody­ namic cycle of an ideal engine (absorbing heat, doing work, and releasing heat) are reversible 7, and the efficiency is

Therefore, even an ideal engine will not be 100% efficient8 • Unfortunately, a real engine is never as efficient as an ideal engine. Indeed, the efficiencies of most real engines are less than 40% since some of the energy absorbed by the engine is dissipated through friction forces or absorbed by 7 l:!S = 0 for the thermodynamic cycle of the ideal heat engine. 8 If connected to a low temperatu re reservoir at O Kor a high temperature reservoir at efficient. However, such an engine would be impractical to construct.

00

K, the engine would be 100%

454 I THE ENERGY OF PHYSICS, P/,RT !: CLASSICAL Ml::CHAN!CS AND THERMODYNAMICS

the engine (thereby increasing both the temperature and the entropy of the engine). Indeed, the dissipation of energy by friction decreases the efficiency of the engine in the same way that the dissipation of energy by friction reduces the magnitude of the acceleration of a system of moving objects.

The 1st law of thermodynamics is a statement of energy conservation. We can use the 2nd law of thermodynamics (Equation 16-3) and Equation 15-10 to express the 1st law of thermodynamics as a relationship between changes in energy and changes in entropy9 • dE=dQ-dW � dE=TdS-PdV

(16-8)

As discussed in Section 15-7, the energy of an ideal gas is a function of the temperature of the gas only and can, therefore, be expressed in terms of the heat capacity of the gas at constant volume. dT P dE=CV dT � CvdT=TdS-PdV � dS=Cv-+-dV T T

Intuitively, it makes sense that increasing the volume of a gas would increase the entropy of the gas. Specifically, an increase in the volume of the gas provides more potential positions for the molecules of the gas to be located. Indeed, there are more ways to distribute the molecules of a gas if the volume accessible to the molecules of gas increases. The increase in the entropy of the gas associated with an increase in temperature follows directly from the 2nd law of thermodynamics (Equation 16-3) and can also be understood in terms of the kinetic energy of the molecules of gas. We know from the equipartition theorem (Section 15-6) that the average kinetic energy of the molecules of gas increases with increasing temperature. Thus, an increase in the temperature of the gas results in a larger distribution of potential velocities for the molecules of gas and hence, a larger entropy. We can now integrate this equation over a change in temperature and volume to determine the change in the entropy of the gas. Let's choose as the path of this integration (i.e., the path over which the temperature and volume of the gas change) to be a reversible process. Specifically, let's assume that the changes in temperature and volume occur quasi-statically so that the overall process is reversible. Since the changes in pressure and volume occur quasi-statically, the gas is always in thermodynamic equilibrium during these processes. Under these conditions, we can use the equation of state of the gas (Equation 15-9) to relate the temperature, pressure, and volume of the gas.

: In �quation 16-8, the only -:ork that is considered is work associated with a change in volume. Furthermore, it is mterestmg to note that all terms m Equation 16-8 are exact differentials.

CHAPTER SIXTEEN

Therefore, since Cv for an ideal gas is independent of temperature

I

455

(Section 15-9), we have

fdS= f C v dT + fNk dV � Af=C fdT +Nk dV s f T V V T B V M=Cvln(� ]+Nk ln(�] 8

(16-9)

Of course, since entropy is a state function, Equation 16-9 is valid whether or not the path taken by the gas is quasi-static 10 • We can also express Equation 16-9 in terms of the average energy of the gas using the equipartition theorem (Section 15-7). 2E1

E=¾Nk8

T

--> M=C v ln

3

:: +Nk8 ln(�]

3NkB M=Cvln(;, }Nk8 ln( �] Therefore, the entropy of the ideal gas can be considered to be a function of the energy and the volume of the gas. S=S(E,V) We will discuss this representation of entropy more in Section 16-7.

Example 16-5:

f

Problem: An ideal gas with a heat capacity cv = Nk, experiences the changes in volume and pressure shown in the PV diagram in Figure 16.2. The gas starts in state A, undergoes an isovolumetric increase in pressure to state B, and then an isothermal expansion to state C. What is the change in the entropy of the gas between state A and state C? Solution: The change in the entropy of the gas can be determined using Equation 16-9.

ca

ie

�

0.

B

f\

&O

20

TI

lsotherrm

J � •:

---------

-1

2000 1000 Volume (cm3 ) Figure 16.2: The PV diograrn for Example 16-5�-------- _ ----- _ --- -- -- _

functions only, as 10 This also follows from the fact that we can write the 1st law Of thermodYnamics in terms of state shown in Equation 16-8.

456

1

THE ENERGY OF PHYSICS, PART 1: CLASSICAL MECHANICS AND THERMODYNAMICS

The pressure at state C is the same as the pressure at state A.

5 Af=-Nk8 ln (Ve VA 2

J

The change in the entropy of the gas, thus, depends upon the ratio of the volume of the gas at the two states. Hence, 5 Af=-Nk8 ln 2

3 (2000cm )(

lm ) 100cm

lm (1000cm )( ) 100cm 3

3

3

Let's now return to the free expansion of an ideal gas introduced in Section 15-11. The change in the entropy of the gas experiencing the free expansion in Figure 15.12 would be

Free expansion occurs spontaneously because the change in entropy associated with that process is positive. Substitution of the heat capacity of an ideal monatomic gas (Section 15-9) into Equation 16-9 gives us

CHAPTER SIXTEEN

I

457

3 T V Cv =- NkB � Af=�Nk8 ln( 1)+Nk 1n 1 8 ( V) TI 2 2 i If we denote the reference point (i.e., the initial state for LlS) in this equation by the su · t 0, we bscnp · th"1s equation as can rewrite

Rearranging this expression yields -�Nk InT -Nk Inv) S=�NkBlnT+NkBinv+(s0 2 2 BOB o We can now define everything within the parenthesis to be the product Nk8 cr0, where the variable 0'0 is independent of T, V, and N. Rather, a0 depends upon the state of the system at the reference point only.

Let's consider a system that consists of N identical particles of an ideal gas in a container with volume Vat temperature T. A partition is in place that divides the container into two equal volumes (V/2) each of which contains an equal number (N/2) of particles of the gas; we can refer to these systems as A and B. The entropies of each of these systems are identical.

If the partition is removed, all of the particles of the gas can diffuse throughout the entire volume of the container. The entropy of the system after the partition is removed is thus 3 Szystem =-NkB 1nT+Nk8 1nV+Nk8 0"0 2 The change in entropy associated with removing the partition is thus Af =Szystem -(sA +sB ) � Af=Szystem -2sA

Af=Nk8 ln2

458

I

THE ENERGY OF PHYSICS, PART I: CLASSICAL MECHANICS AND THERMODYNAMICS

However, removing the partition (or re-inserting the partition) is a reversible process since it doesn't affect the distribution of accessible states to the system. In other words, we could easily recreate the initial distribution ofparticles in the system by re-inserting the partition. Therefore, the associated change in entropy should be zero, and not the positive number we calculated. This result is known as the Gibbs paradox, after Josiah Gibbs11• Removing the partition increases the volume accessible to each gas molecule by a factor of2. Ifthe two initial systems, A and B, contained different (i.e., distinguishable) gases then the act ofremoving the partition would allow each ofthe gases to expand and mix with the other gas. This would clearly be an irreversible process since we could not recover the initial distribution of gas particles by re­ inserting the partition. Therefore, the previously derived equation for the entropy of an ideal gas is valid only when the particles ofthe gas are distinguishable from one another. The correct expression for the entropy of an ideal gas when the particles of the gas are indistinguishable is12 (16-10)

Applying Equation 16-10 to determine the change in entropy associated with removing a partition dividing identical gasses yields

AS=(Nk,( ¾lnT+ln(; )+G, ))-{: k,[¾lnT+ln[

I

}a,]]

AS=( Nk,( }nr+In(; )+u, ))-( Nk,( ¾lnT+ln(; )+u,)) Li5=0

The change in entropy for the process is now zero, as required for this reversible process.

16-6 Equilibrium Conditions Let's imagine the situation shown in Figure 16.3 in which two systems, denoted as A and B, can interact with each other but are otherwise isolated from the outside environment. The total energy and volume ofthe systems are

11 Gibbs was an American theoretical scientist whose work in thermodynamics contributed to chem1·stry, physics' mathematics, and biology. 12 Yo� should feel inspired to take additional courses in thermodynamics and statistical mechan ics to learn how to _ derive this equation.

The two systems can exchange energy with each other through heat and work; however, since the systems are isolated from the outside environment, the total energy and volume of the systems must be con­ stant. Thus, the sum of any infinitesimal exchange of energy or change of volume between the systems must equal zero.

f-;qur"·� I(,:� 1\v'·1 ,',_\!er;,-, /\ u1:d H, (Ht: 1j1 ( 1.J(1 h; P8. We can, therefore, state that since PA is larger than P8, the molecules of dye will eventually diffuse to fill the entire volume of the container. Interestingly, we do not argue that it is impossible for the molecules of the dye to remain in their initial location or to eventually all move back there after having diffused through the entire volume of the container; rather, we accept that the probability of such an occurrence is very small relative to other outcomes. Now, let's imagine a different situation in which a system is in contact with a large heat reservoir. The system and the heat reservoir can exchange energy with each other, but both are unable to exchange any energy or do any work with the outside environment. Thus, the sum of the energy of the heat reservoir and the energy of the system are constant. Ereservmr. + Esystem = constant M reservoir +Msystem =0 � Mreservoir =-Msystem The temperature of the heat reservoir remains constant regardless of the heat it exchanges with the system. As discussed previously (Example 16-2), this can be accomplished if the energy of the heat reservoir is much, much larger than the energy of the system. Therefore, since entropy is a function of energy (Section 16-5), the total entropy of the system and the heat reservoir is approximately equal to the entropy of the heat reservoir. Stotal =Sreservoir. +Ssystem S reservoir. >>Ssystem � Sreservoir a:::Stotal

Let's now consider two separate accessible states for the system, denoted as A and 8, that have total energies EA and E8, respectively. The ratio of the probability of finding the system in state A to the probability of finding the system in state 8 depends upon the difference in the total entropy of each state; the total entropy is the sum of the entropy of the system and the entropy of the reservoir. According to Equation 16-15, we have

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From the 1st law of thermodynamics, we can reI ate the d'fti 1 erence m the entropy of the reservoir to . a change m the energy of the reservoir 17 • Since Tis constant, we have = TAf !:iE reservoir reservoir

---+

Af reservoir -

llE,...,,.,,r T

---+

Finally, since the system and the reservoir are isolated from the outside environment, we have = -1:iE !:iE reservoir �tem

---+

p � T .....:1.. == e k8 p B

---+

p .....:1.. == e

-(E.-E,) k,r

PB

Therefore the probability of finding a system in a particular state is proportional to both the energy of the state and the temperature of the system. (16-16)

This exponential factor is referred to as the Boltzmann factor for the state. The temperature in Equation 16-16 is the temperature of the heat reservoir, which is equal to the temperature of the system since the system and the reservoir are in thermal equilibrium. The relative probability of finding a system in two different states is, thus, proportional to the ratio of the Boltzmann factors of the two states. Therefore this relative probability scales with the temperature of the system.

Example 16-8: Problem: A system is in contact with a heat reservoir at 200 K. What is the ratio of the probability that the system is in a configuration with an energy of 10 pJ to the probabili that the system is in a configuration with an energy of 20 pJ7 Solution: The ratio of the probabilities can be found using Equation 16-16. 2 12 (1ox10- 1-2ox10-' 1)

p

1op1 =

p20pJ

e

-� k8 T

---+

p

1op1

p20pJ

=

e

(1.3Bx10-

23

¾J

),,ZOOK)

---+

� pJ _

O --

e 3.6xl0'

PZOpJ

e er state to the probability The ratio of the probability of the system being in the lo�er � � ely mfi mte. In other words, the _ of the system being in the higher energy state is effect1� e 1s effectively zero. probability of the system being in the higher energy stat

17

The reservoir does no work, so from E quat.ion 16-8

we have that dE = TdS.

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THE ENERGY OF PHYSICS, PART I: CLASSICAL MECH.AN!CS AND THE.RMODYf'.,JAMiCS

It follows from Equation 16-16 that systems are always most likely to be found in the configura­ tion with the lowest energy. Thus, it is the 2nd law of thermodynamics that justifies the principle of minimum potential energy introduced in Section 1-4.

It's clear from Equation 15-1 and Equation 15-2 that the energy of a system will change if the num­ ber of particles in the system change. Including this contribution to the first law of thermodynamics (Equation 16-8) gives us dE = TdS - PdV + µdN

(16-17)

The variableµ in Equation 16-17 is the chemical potential for the system. Chemical potential: A form of potential energy for a system. The unit ofchemical potential is the joule(]) for Equation 16-17. In other formulations of the first law ofthermodynamics, the units of chemical potential are ]/kg or ]/mole. We can obtain a mathematical definition for chemical potential by rearranging Equation 16-17.

p 1 µ dS=-dE+-dV--dN T T T It follows from this expression that entropy is a function of energy, volume, and the number of particles.

The total derivative of the entropy can therefore be written in terms of its partial derivatives with respect to energy, volume, and the number of particles.

Comparing these two expressions for dS gives us the following equation:

µ=-r(as) aN

E,v

(16-18)

It follows from Equation 16-18 that since the entropy of a system usually increases as the number of objects in the system increases, the chemical potential of most objects (or substances) is negative.

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467

Example 16-9 Problem: What is the chemical potential of an ideal

gas with E = � Nk T? 2 B •

Solu�on: We can determine the chemical potential by substituti ng Equation 16_10 into Equation 16-18. However, we must first express Equation 16-10 · terms o f the energy m ofthe gas. E = �Nk8 2

T

� S = Nk (�1n(�)+1n(�)+cr) B 2 3NkB N °

Substitution into Equation 16-18 yields

Expressing our result in terms of the temperature, rather than the energy, gives us ln(�)+cr -�] µ = -kB r[�lnT+ 2 N ° 2

Let's now extend the derivation in Section 16-6 to include the possibility that the two systems can exchange particles with one another. Ntotal

=

N A + N8

�

dN A +dN8 = 0 � dNA = -dN 8

When the two systems are in thermodynamic equilibrium any change in the total entropy of the sys­ tems resulting from the systems exchanging energy with each other must be zero (Equation 16-6). Thus, for any process occurring at equilibrium O µA _µB B A v (_!_ _ _!_)dEB +(PT - PT )d B +( T T )dN B = TB TA B A B A

n 16-6). In equilibrium the pressure and the temperature are the same for each system (Sectio Thus, the new condition for equilibrium is

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THE ENERGY OF PHYSICS PART I: CLASSICAL MECHANICS AND THERMODYNAMICS 1

.!. T (µA -µB )dNB =0 Since this equation must be valid for any change in the number of particles it follows that at equilib­ rium the chemical potentials of the two systems must be equal.

We refer to this equilibrium as diffusive equilibrium since it corresponds to the condition when the net flow of particles between the two systems is zero. The change in associated with any spontaneous process must be greater than or equal to zero (Equation 16-5). The corresponding condition for particle flow between two systems at the same temperature is therefore

Thus, if dN8 is positive (i.e., if the number of particles in state B is increasing) then µ8 is less than µA. In other words, particles will spontaneously move toward lower chemical potential. If we re-express the equation for the chemical potential of an ideal gas (Example 16-9) in terms of the density of the gas (denoted by n), we have

Thus, chemical potential increases as density increases. Because of this, an ideal gas will always diffuse toward lower density (Section 16-8). Similarly, we can show that the chemical potential of an ideal gas is proportional to the pressure of the gas. PV = Nk, T ---> µ = - ,T[ ¾lnT +ln( kn+a,-�] k

µ =-k, T [�ln T +ln(k, )-lnP+a, -�] ---; µ oc k, T ln(P) Thus, ideal gases always diffuse toward lower pressure. In general, chemical potential is a function of temperature and pressure, which we can express as a series expansion. 2

µ = µ 0 +a1 (Lir)+a2 (Lir) +...+,B1 (M)+,az(,iP)2 + ... µ = µ o + Ia i (Lirr + L.B;(Mr I

i

(16-19)

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In Equation 16-19, the variable µo 1·s the chem1ca . 1 potential at a re�erence temperature and pressure; the variables AT and AP are changes m . temperature and pressure, respectively, to that refer. e to ence temperature and pre ssure ; and the constants a; and /3 are umqu each substance/system. 1

Example 16-1 O . Problem: Determine the melting t re O f ice . usmg the following expressions for the chemical potential of solid and liq:i:::::;

so1,d

=-236590-J-+(-44.77-J -)(T-298K) mol molK

(µwater tquid

=-23718Q _J_ +(-69.91-J )(T-298K) molK mo }

(µwater)

.

Solution: Soli� water (i.e , ice) will melt when the chemical potential ofliquid water is less : tha the chemical potential of solid wate r. The melting temperature (frequently denoted as � Tm) 1s the temperature at which these two chemical potentials are equal.

J )(r -298K) -236590-J-+(-44.77-J-)(rm-298K)=-237180-J-+(-69.91-molK m mol molK mol -23.47K - � (rm -298K)= • (25.14-J -)(rm-298K)=-590-J mol molK ° Tm = 274.53K � Tm = 1.37 C

The difference between this calculation of Tm and the empirical value of O °C results from the truncation of the s eries expansions in Equation 16-19 and neglecting the effects ofpressure on chemical potential.

sing pressure, but The chemical potential of both solid and liquid water increases with increa words, the /3; other In water. the magnitud e of this increase is larger for solid water than for liquid se ofthis, it is coefficients in Equation 16-19 are larger for solid water than for liquid water. Becau possible to melt ice through the application of pressure. phase ofwater (the gas phase) Solid and liquid water are two different phases of water; the third change in entropy for the is water vapor. Because transitions between phases are associated with a se are also associated with an system, we would anticipate from Equation 16-4 that changes in pha

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THE ENERGY OF PHYSICS, PART i: CLASSICAL N\ECHAN1C:S ,t,.ND THERMOD'{NAMICS

phase transition� occur at exchange of heat with the environment outside the system. Indeed, since _ 0n 16-6), it follows (Sect1 rature tempe nt consta a at thus and s phase two en betwe the equilibrium of the corresponding from Equation 16-4 that the heat associated with a phase change is the product temperature and change in entropy. L=TAf

(16-20)

The variable L in Equation 16-20 is referred to as the latent heat. The latent heat of fusion is associ­ ated with liquid-solid phase transitions, and the latent heat of vaporization is associated with the liquid-gas phase transition. A more commonly used variable is the specific latent heat, denoted as I in Equation 16-21, which is the latent heat per unit of mass of the substance.

I= !:_ � I= !]_ m m

(16-21)

Example 16-11 Problem: The specific latent heat of fusion for water is 334 kJ/kg. What is the heat required to melt 2 kg of ice? What is the associated change in entropy? Solution: Starting with Equation 16-21, we have l=g_ � Q=lm � Q=(334 _!g_ )(2kg) � Q=668kJ kg m . The change in entropy for this process can be determined using Equation 16-4; this process occurs at the melting temperature of ice. 6�� Q � Af=-J Af=� Af=2446T 273K K

The change in entropy is positive since the system absorbed heat to melt the ice.

. , · ·

I 6-1 D l.ooking Ahead

In our energy-based approach to kinematics, we determined a description of the kinematics of a system (e.g., the acceleration of the system) by differentiating an equation for the energy of the system with respect to the position of an object in the system (or with respect to time, etc.). In this chapter, we have shown that the equation for the entropy of a system can be differentiated to determine a description of the thermodynamics of a system. Similarly; just as it is changes in energy that are responsible for the

CHAPTER SIXTEEN

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471

kinematics of a system18, ultimately, it is 7._Nk T j----------- ::ao----: changes in entropy that drive the behavior 2 B of a system. Furthermore, although it is � ·c;ca often too difficult to determine an equation Q. for the entropy of a system using Equation � 16· 1, it is, nevertheless, straightforward ni to calculate changes in the entropy of a INkBT +----=------2 system using the 2nd law of thermodynamics (Equation 16-3). Typically, all that is needed for these calculations is heat Temperature capacity of the system, which can be read­ Figure 16.4: The heat capacity at constant volume ol oyy ily measured. gen as a function of temperature. Interestingly, despite our frequent as­ sumptions to the contrary, heat capacity is always a function of temperature. Indeed, heat capacity increases with increasing temperature. Cons ider, for example, the heat capacity at constant vo lume of molecular oxygen (0 2 ) as a function of temperature (Figure 16-4). At low temperatures, the heat capacity at constant volume o f oxygen is �Nk8 , as expected from the equipartition theorem (Sectio n 15-6). As the temperature increases, hotvever, the heat capac­ ity at constant volume suddenly increases to �Nk8 • We would argue, based upon the equ i partition theorem, that this increase in the heat capacity is associated with the presence of additional terms in the equation for the energy o f the gas. For example, ifwe included the energy associated with the rotational motion of the atoms within the molecule of oxygen, we would have.

:!

1 1 1 1 1 2 2 OJ K. =-mv 2 +-mv 2 +-mv z +-/ my +-/ zz z ' 2 x z Y 2 z zy

�

() K.

' avg

5 =-kB T 2

In thi s equation, we have defined the x-axis to be the longitudinal axis o f the molecule. The mo ment of in ertia for this axis of rotation is much smaller than the moments o f inertia for rotations around the perpen d icular y-axis and z-axis. Thus, we igno re contributions to the kinetic energy for rotations around the x-axis. The average energy o f a system o f N molecules of oxyge·n and the associated heat capa c ity would, therefore, be Eavg =N(Ki)avg their equil�br ium s�para�ion Similarly, ifwe included the vibrational motion of the atoms around pot�nt1al energ1_es. Smee within the molecule, we would need to include the associated vibrational · funct1· 0ns, the heat capaci·ty of the gas would mcrease agam. these energies are also quadratic 7 Eavg =-NkB T z

-?

7 Cv =z NkB

· ·mize their potential 1 8 For example, systems always arrange themse Ives t o mini

energy.

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THE ENERGY OF PHYSICS, PART I: ClASSICAL MECHANICS AND THERMODYNAMICS

However, although the equipartition theorem can provide an energy-based interpretation of oxygen's different heat capacities, it cannot explain the temperature dependence of these heat capacities. That is, why do the rotational and vibrational energies contribute to the heat capacity only at higher temperatures? Our first attempt to answer this question would likely involve using Boltzmann factors (Equation 16-15) to justify the temperature dependence of the probability of observing a mol­ ecule of oxygen with rotational or vibrational energy. However, this is ultimately unsatisfying as it cannot account for the abrupt transitions between the observed heat capacities shown in Figure 16.4. The ultimate answer to this question is that the rotational and vibrational motions of the oxygen molecule require a finite minimum amount of energy in order to be "switched on." This was one of the first discoveries that energy is quantized. Indeed, rather than having a continuous distribution of possible values, the energy of a system comes in discrete bunches only. This, in turn, brought about the development of quantum mechanics. These concepts also led to the development of statistical mechanics, which provides a framework for determining an equation for the entropy of a system from knowledge of the energies accessible to the system (e.g., using Boltzmann factors). In this way, we end up back where we started again:. with energy. If you can characterize the energy and entropy of a system, you can describe the system completely.

Summary • Entropy: An intrinsic property of a system. Entropy is best considered to be a measure of the number of ways that a system can be arranged given the constraint of a particular set of mac­ roscopic properties for that system (e.g., volume, temperature, or pressure). The mathematical definition for the entropy of a system is S=k8 lnQ

• The ratio of the multiplicities of two states is equal to the probability of obtaining (or observing) those two states.

• The 2nd law of thermodynamics: The infinitesimal change in the entropy of a system associ­ ated with the absorption of an infinitesimal amount of heat through a quasi-static process is dS=aQ T

Furthermore, the entropy of an isolated system never decreases. Thus, for an isolated system, the change in entropy associated with any spontaneous process must be greater than or equal to zero.

CHMTER SIXTEEN

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473

It follows that the entropy of a system must be a maximum when the system is at equilibrium. Finally, there is no change in entropy for a reversible process (i.e., there is no change in entropy for any process occurring at equilibrium). �reversible . = 0 � �equilibrium =0

• Heat engine: A device that absorbs heat from a high temperature reservoir, performs work, and then releases heat to a low temperature reservoir. The efficiency of a heat engine is the ratio of the work done to the heat absorbed.

According to the 2nd law of thermodynamics, the maximum efficiency of an ideal heat engine is

• The 1st law of thermodynamics: The entropy of a system can be written in terms of the energy and volume of the system.

ds=(as) dE+(as) a

aE

v

v

dV E

This form of the 1st law also provides the definitions of temperature and pressure for a system.

1 = as (a )

r

E

as

P= r ( V

av )

E

• Heat capacity: A measure of the dependence of the entropy of a system the system.

s

cv =r(aar)

V

cp =r(aars)

p

on the temperature of

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THE ENERGY OF PHYSICS, PART i: CLASSICAL tV\ECHANiCS J\ND THERMODY:'-JAMICS

• The change in the entropy of an ideal gas associated with a change in the temperature _ _ (or equivalently energy) and/ or volume of the gas is given by the followmg equation

Af=C, ln (;, )+Nk, In (�) Af=C, In (;, )+Nk, In (�) • The probability of finding a system in a particular state is proportional to both the energy of the state and the temperature of the system.

This exponential factor is referred to as the Boltzmann factor for the state. • Chemical potential: a form of potential energy for a system. Chemical potential can be deter­ mined from the dependence of the entropy of a system on the number of objects in the system. µ = -T(a!)

a

E,V

Objects will spontaneously move toward lower chemical potential to maximize entropy. • Latent heat: the energy associated with a phase transition. It is determined from the tempera­ ture of the transition and the change in entropy associated with the transition. L=TAS

Conceptual 1. System A contains N molecules of an ideal gas in a 1 m3 closed container. System B contains N molecules of the same ideal gas in a 2 m3 closed container. The temperatures of the gases are the same. The entropy of which system is larger? B 2. Can you lower the temperature of your kitchen by open- S!_ 6 +------• ing the door of the refrigerator? ::::, 3. You allow an ideal gas to expand freely from volume A_ _ V to volume 2V. Later you allow the same ideal gas to _C � 2 "I" T freely expand from volume 2 V to volume 3 V. Is the total o.. i I 1 1 change in entropy for these processes greater than, less 1 3 2 than, or equal to the total change in entropy that would Volume (m 3 ) have occurred if the gas had been allowed to expand from volume Vto volume 3Vin one process? Figure 16.5

i e

/�

CHAPTER SIXTEEN

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475

4. An ideal gas is subjected to the cyclic, quasi-static ( i· e·, reversible) process shown in the PV diagram in Figure 16.5. What is the change in entrop y of the gas during one complete cycle?

Quantitative Section 16-2

5. If you flip a fair coin N times, the probability that you obtam · n heads Is · given . by the following equation:

If you flip a fair coin 200 times, how much more likely are you to obtain 100 heads than 0 heads? Report the natural log of the probability ratio. 6. If you flip a fair coin N times, the probability that you obtain n heads is given by the following _ equat10n:

If you flip a fair coin 200 times, how much more likely are you to obtain 100 heads than 50 heads? Report the natural log of the probability ratio. 7. What is the total number of unique ways to distribute two identical objects in three separate containers such that there is no more than one object per container? 8. What is the total number of unique ways to distribute two non-identical objects in three separate containers such that there is no more than one object per container? 9. What is the total number of unique ways to distribute two identical objects in three separate containers when there is no limit on the number of objects that can be in a container? Section 16-3

10. Two identical objects initially at 200 Kand 300 K are brought into contact with each other. The heat capacity of the objects is a constant 2 J/K. Through the exchange of heat the two objects eventually obtain a common final temperature. What is the change in entropy associated with this process? 11. Two objects, A and B, initially at 200 Kand 400 K respectively, are brought into contact with each t B is a other. The heat capacity of the object A is a constant 2 J/K and the heat capacity of objec common final constant 3 J/K. Through the exchange of heat the two objects eventually obtain a temperature . What is the change in entropy associated with this process? l2. The heat capacity of an object is given by the following equation: C-14000! +(200�,

Jr+�, )r'

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THE ENERGY OF PHYSICS, PART I: CI.ASSICAl MECH.AN!CS AND THERMODYNAMICS

What is the change in the entropy of the object associated with raising its temperature from 290 K to 380 K? 13. Two objects, A and B, initially at 200 K and 400 K respectively, are brought into contact with each other. Over this range of temperatures, the heat capacity of object A and object B can be approximated by the following equations: 1 )r cB = 4 lK +(o.02K 2

Through the exchange of heat the two objects eventually obtain a common final tempera­ ture. What is the change in entropy associated with this process? Section 16-4 14. An ideal heat engine absorbs 100 J of heat from a 400 K reservoir and releases 75 J of heat to a 200 K reservoir. How much work is done by the engine? 15. The low temperature reservoir for an ideal heat engine is at 200 K. If the heat engine must do 200 J of work and can release a maximum of 200 J to the low temperature reservoir, what is the minimum temperature for the high temperature reservoir? 16. An ideal heat engine absorbs heat from a 500 K reservoir and releases heat to a 100 K reservoir. If the engine releases 200 J of heat to the 100 K reservoir, what is the maximum amount of heat that can be absorbed from the 500 K reservoir? 17. An ideal heat engine absorbs heat from a 500 K reservoir and releases heat to a 100 K reservoir. If the engine releases 200 J of heat to the 100 K reservoir, what is the maximum amount of work that can be done by the engine? 18. The high temperature reservoir for an ideal heat engine is at 400 K. If the heat engine must do 150 J of work and can release a minimum of 250 J from the low temperature reservoir, what is the maximum required temperature for the low temperature reservoir? 19. A refrigerator is a heat engine running in reverse, as shown in Figure 16.6. Work that is done on the refrigerator enables it to absorb heat from a low temperature reservoir and release heat to a high temperature reservoir. Consider an ideal refrigerator connected to a 200 K and 400 K reser­ voir. What is the minimum amount of work required for this refrigerator to absorb 300 J of energy from the low temperature reservoir?

High Temperature Reservoir Figure 16.6

Low Temperature Reservoir

I

CHAPTER SIXTEEN

Section 16-5 20. 5 x 10 23 molecules ofan ideal gas with energy E = � Nk r 2 B is subjected to a cyclic, quasi-static (i.e., reversible) process shown in the PV diagram in Figure 16. 7. What is the change in the entropy of the gas during one complete cycle? 21. 1 x 10 24 molecules ofan ideal gas with energy E = � Nk T 2 B is subjected to a cyclic, quasi-static (i.e., reversible) pro­ cess shown in the PV diagram in Figure 16.8. What is the change in the entropy of the gas as it moves from state A to state C? 5 22. 1 x 10 24 molecules ofan ideal gas with energy E = Nk T 2 B is subjected to a cyclic, quasi-static (i.e., reversible) pro­ cess shown in the PV diagram in Figure 16.9. What is the change in the entropy of the gas as it moves from state A to state C? 5 23. 2 x 10 24 molecules ofan ideal gas with energy E = Nk8 T

-"'

Q.

5

B

!

II) II)

!

Q.

2

A 1

is subjected to a cyclic, quasi-static (i.e., reversible) pro­ cess shown in the PV diagram in Figure 16.11. What is the change in the entropy of the gas as it moves from state A to state D?

-"' Q.

3 Cl.

!

2

---

�

A

Volume

--"' Q.

5

a., :::s

II)

C

!

1D

! 1

2

Volume (m 3) Figure 16. 11

(m 3)

A

B

2

II 3

2

Volume (m

A

1

3

2

1

1

II) II)

Cl.

(m3 )

B

a., Q.

B

3

2

Volume

2

is subjected to a cyclic, quasi-static (i.e., reversible) pro­ cess shown in the PV diagram in Figure 16.10. What is the change in the entropy of the gas as it moves from state A to state B? 3 24. 1 x 1024 molecules ofan ideal gas with energy E = 2 Nk8 T

-Q.

3

477

e e

Q.

B

6

:::s

II) II)

2

3)

A_/

:

!

1

�- c

2

Volume (m 3 )

I 3

25. 5 x 10 20 molecules ofan ideal gas with energy E = _§_ Nk8 T is subjected to a cyclic, quasi-static (i.e., reversible) process shown in the PV diagram in Figure 16.12. This gas forms a heat engine that is connected to a high temperature reservoir at 900 K and a low temperature at 200 K. What is the change in entropy of this whole system (the gas and the reservoirs) during one complete cycle of the gas?

D.

5

C

G>

,n ,n D.

2

A

1

2

Volume

3

(m 3)

re l (). 1 L

Section 16-6

26. A 100 cm long rectangular container with a cross-sectional area of 5 cm 2 is divided into two compartments by a movable partition that slides through a frictionless mechanism. Both compartments are filled with the same ideal gas, but at different pressures. The partition is initially held in place exactly in the middle of the container (i.e., 50 cm from the right end of the container). In this initial configuration, the pressure of the gas in the right compartment is 6 x 10 6 Pa and the pressure of the gas in the left compartment is 2 x 10 6 Pa. The difference in the pressures of the two gases causes the partition to move when it is released and the final equilibrium position of the partition is different from its initial position. The entire system is thermally and mechanically isolated and you may ignore the heat capacities of the container and the partition. How far from the right end of the cylinder is the final equilibrium position of the partition? What is the increase of the total entropy of the system if the initial temperature of the system was 270 K? Section 16-7 27. The equation for the entropy of a system of N molecules of a gas is

In this equation cr is a constant that is independent of N, V, and T. What is the equation of state for this system? You may assume that the energy of the gas is a function of only the temperature of the gas. What is the heat capacity at constant pressure for this system? 28. The entropy of a system is given by the following equation:

S=µEV In this equation,µ is a constant. What is an equation of state for this system? 29. The entropy of a system is given by the following equation:

S=µE.Jv In this equation,µ is a constant. What is an equation of state for this system?

CHAPTEi< SIXTE:[N

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30. The entropy of a system is given by the following equation: II.S2 = NE+ yV 2 In this equation, N is the number of molec ules in the system, and ,1 and yare constants. What is an equation of state for this system? W hat is Cv for this system? 31. The heat capacity at constant volume of a system at temperature Tis given by CV

1

= }.,T 2

In this equation, ,1 is a constant. The entropy of this system is independent of the volume of the system and depends upon only the temperature of the system; the entropy of the system is s0 and T = 0 K. What is the equation describing the entropy of this system as a function of temperature?

Section 16-8 32. A system is in thermal equilibrium at temperature T. What is the ratio of the probability that the system is in a configuration with an energy of 3 k8T to the probability that the system is in a configuration with an energy of 2 k8 T? 33. A system is in contact with a heat reservoir at 4000 K. What is the ratio of the probability that the system is in a configuration with an energy of 5 pJ to the probability that the system is in a configuration with an energy of 10 pJ? What is this ratio if the temperature of the heat reservoir is 10 12 K? What is this ratio if the temperature of the heat reservoir is infinite?

Section 16-9 34. The chemical potentials of the solid and liquid phases of lead are given (approximately) by the following equations: J

(µ1ead )solid =(-64.8 moleK )(T-298K) J 1 )(T-298K) ( µ1ead ) liquid =2220---(71.7 moleK mole What is the melting temperature of lead? 35. A 0.5 kg block of ice is placed in 2 liters (i.e., 2 kg) of water that is initially �t 3oo K. The heat 1 334 k�/kg. What capacity of water is 4.19 kJ/kgK and the specific latent heat of fusion for water � · lated from the outside environment? · system 1'f 1·t 1s · iso is the final equilibrium temperature of this

CHAPTER SEVENTEEN Fluids I am always doing that which I cannot do, in order that I may learn how to do it. -Pablo Picasso

L

17- 1 Introduction iquids and gases are also both.fluids. Fluid: a substance that can flow.

The focus of this chapter will be on the static and dynamic properties of fluids, including both liq­ uids and gases. In our discussion of these properties, we will rely upon the principles of statistical mechanics and thermodynamics from Chapter 15 and Chapter 16. This chapter is therefore also an extension of applying those principles to other systems. The basic differences between the three states of matter-solid, liquid, and gas-result from the interactions between their molecules 1 • The molecules within a solid interact very strongly with one another, and these strong interactions are what give a solid its definite volume and shape. The interac­ tions between molecules in a liquid or gas are weaker than those of a solid and therefore liquids and gases do not have definite shapes. The intermolecular interactions of a liquid are slightly stronger than those of a gas and therefore liquids have a definite volume, whereas unconfined gases do not.

17-2 Definitions We begin our discussion of fluid statics and dynamics with several definitions. Two common charac­ teristics used to describe a fluid are compressibility and viscosity. Compressibility: a measurement of the change in the volume of a substance resulting from a change in the pressure applied to that substance. Viscosity: a measurement of the internal friction in a fluid.

1

Oratoms.

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THE ENERGY OF PHYSICS, PART i

We will confine our discussion to idea/fluids. An idea/fl uid is: • incompressible: the density is constant throughout the fluid. • nonviscous: the viscosity is zero. flow. Ideal fluidflow is: Similarly, we will restrict our discussion of fluid flow to ideal fluid • steady: all particles constituting the fluid have the same velocity. • irrotational: the flow is smooth and has no turbulence. be hydrostatic equilibrium Steady fluid flow is also called laminar fluid flow. Lastly, a fluid is said to when the velocity of its flow at each point is constant over time. As shown previously in our discussion of ideal gases (Section 15-7), another defining character­ istic of a fluid is its pressure. Besides the pascal, another common, although not SI, unit for pressure is the atmosphere2 • latm = 101.3x10 3 Pa We will typically assume that the atmospheric pressure is equal to 1 atm.

Example 17- 1 Problem: A massless circular suction cup with a radius of 4 cm is pushed flat against the ceiling. What is the force required to pull the suction cup off of the ceiling? Solution: If the suction cup is pushed flat against the ceiling, then all of the air has been pushed out from between the suction cup and the ceiling. As shown in Figure 17.1, the pressure inside the cup will be zero and the pressure outside the cup will be equal to atmo­ spheric pressure. Due to symmetry, the atmospheric pressure outside the cup will give rise to a net force pushing upwards on the cup. Let's define a reference frame for this system in which they-axis denotes the vertical axis with a positive y direction pointing up from the ground (i.e., the negative direction for the y-axis points down from the ceiling). The x-axis of this reference frame will be perpendicular -------x to the y-axis. In the free-body diagram for the suction n cup, we will denote the pressure of the air as Pa,r'. the p = Pair F cross-sectional area of the suction cup as A, the normal force acting on the suction cup (from the ceiling) as n, r 1 11 and the force pulling the suction cup off of the ceiling F��� � �7-� _! :__sy���'l1�i� E_;:;9}·n_e�_ 1_!_ 1 · as F. With these definitions and Equation 15-8, the freebody diagram for the suction cup is shown in Figure 17.1. Applying Newton's second law to each of the coordinate axes in this free-body diagram yields the following equations: (Fnett =max (Fnet)

2

y

� O=max � ax =0

= ma

y

� P.a,r A-n-F = may

This unit is supposed to correspond to the average atmospheric pressure at sea level at the latitude of Paris, France.

CHAPTER SEVENTEEN I 483 If th� suction cup remains attached to the ceiling the component ofits acceleration along the y-axis must be zero. Substitution of aY = O into the equation above yields ay =0 � P.= A-n-F= � n=P A-F As long as the suction cup is in contact with the ceiling, the magnitude of the normal force must be greater than zero. n � O � pa;,A-F � 0 � F � P. A � F � P. trR 2 mr mr Substitution gives us

A force greater than 509 N is required to pull the suction cup off of the celling.

All of us are subject to the pressure of the atmosphere pushing down on us. If we assume that a per­ son has a cross-sectional area of 0.1 m2 when viewed from above, then the force of the atmosphere pushing down on a person can be calculated using Equation 15-8.

That's the equivalent of having a 1030 kg block resting on the person. And this atmospheric pressure is exerted not only on the top ofthe person, but on all sides ofthe person. The human body is adapted to this air pressure, however, so we don't notice the associated force at all. Indeed, the internal pres­ sure within our bodies is equal to the atmospheric pressure so our bodies are not crushed by the force of the atmosphere pushing on us. In fact, we rely upon tiny differences between our internal pressure and the atmospheric pressure outside our bodies to breathe. When you inhale, your dia­ phragm contacts and the volume of your lungs expands, resulting in a decrease in the pressure of the air in your lungs (Equation 15-9). In this way, the pressure of the air in the lungs can be reduced below the outside air pressure and thereby allow air to flow into your lungs. When you exhale, the diaphragm relaxes and the volume of your lungs decreases, resulting in the pressure of the air in your lungs being higher than the outside air pressure. This then allows for air to flow from the lungs to the outside. In Section 15-7 we derived an equation for the pressure ofan ideal gas based upon the momentum transfer associated with collisions of the molecules of the gas with the walls of the container holding the gas. In this calculation we assumed that the pressure of the gas is constant throughout the entire volume of the gas. Specifically, we assumed that the average kinetic energies of the gas molecules were independent of the volume ofthe gas (i.e., the average kinetic energy ofa gas molecule was inde­ pendent of the position of the gas molecule within the container). This approximation is reasonably

484

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THE ENERGY OF fJHYS!CS,. P.ARl I: CASSiCAL MECHA��iCS AND THERMODYNAMICS

valid if the volume of the gas ( or fluid) is small. Consider, for example, a small volume of fluid within a much larger container of fluid as shown in Figure 17.2. If the size of the volume is small enough3, the pressure pushes on the entire surface of the volume with equal strength. Alternatively, the pressure in a fluid can vary over large sub-volumes of the entire fluid. Consider the volume of fluid shown in Figure 17.3, which extends from the surface of the fluid down a distance d. The top of the container of the fluid is open and exposed to the air. The pressure of the fluid at its surface (at the interface pre�sure is cor.st:v�,; 1 ne with the air) is equal to the pressure of the air at the surface uf ·:J:1 1::fi:-1iies; (lC! n�,-S since the fluid is in equilibrium with the air at this point srna:: (Section 16-6). In order to determine the pressure at the base of the volume, let's define a reference frame for this system in which the y-axis denotes the vertical axis with a positive direction pointing up from the bottom of the container holding the fluid. The x-axis of this reference frame will be perpendicular to the y-axis with a positive direction point­ ing to the right in Figure 17.3. In the free-body pair diagram for the volume of fluid, we will denote y the pressure of the fluid at the surface (at the --,II r-A I interface with the air) as P ;,, the pressure at the I Tl

0

base of the volume as P, the force of gravity act­ ing on the water within the volume as F, and the cross-sectional area of the top and bottom faces of the fluid as A. With these definitions, the free­ body diagram for the volume of fluid is shown in Figure 17.3. Applying Newton's second law to the y-axis of this free-body diagram gives us

d

F(s; . :"F; 1

;

I

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I I I I I

____I,

: A .,

PA

----4111,--- X

p

7 3 ,,i r��.::. presst.;:·e 'vvitl-,:�:

Ci

(Fnet ) =may � PA-P.a,r A-Fg =may y

Since the volume of fluid is in static equilibrium, the acceleration of the volume must be zero. Substitution of aY = 0 into the equation above yields aY =0 � PA-P. A-FB =0 � PA=P. A+ mn mr a,r c::J The mass of the flui� contai�ed �ithin the volume is the product of the density of the fluid, p, and the volume of the contamer, which 1s the product of the distance d and the cross-sect ional area A. m=pV � m=pAd � PA=P. A+p Adn t:::J a1r P=P.a1r +pl:Jnd 3

In other words, if the size of the volume is infinitesimally small.

(17-1)

CHAPTER SEVENTEEN

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485

The pressure of the fluid thus increases with depth below the surface (at the interface with the air). The pressure in Equation 17-1 is called the hydrostatic pressure because it is derived under the condition that the fluid is at rest. We see from Equation 17-1 that the pressure of a fluid increases with increasing depth beneath the surface of the fluid. This change in pressure for sea water, which has a density ofapproximately 1025 kg/m3 , would be atm pg = (1025 � )(9.8;) -; pg=10045: -; pg""0.1m Thus, the pressure of sea water increases by approximately 1 atm for each 10 m you descend. In contrast, the pressure of the atmosphere decreases with increasing height above the surface ofthe Earth. Ifwe assume the average density ofair to be 1.2 kg/m3 we have kg m Pa atm pg= ( 1.2-3 )( 9.8-2 ) � pg= 11.76- � pg"" O.lm m km s So the pressure of the air decreases by approximately 1 atm for each 10 km you ascend. Of course, since the density of a fluid is also often a function of temperature, Equation 17-1 is strictly valid for isothermal conditions only.

Example 17-2 Problem: A tube closed at one end is filled with mercury and then inverted into a dish of mercury, as shown in Figure 17.4, to make a barometer. The closed end of the barometer is approxi­ mately at vacuum while the surface of the mercury in the dish is at atmospheric pressure. What is the height of the column of mercury in the barometer? The density of mercury is 13.5 x 10 3 kg/m3 • Solution: The height of the column can be determined using Equation 17.1. The pressure at the height ofthe column, which is at vacuum, must increase to atmospheric pressure at the bottom ofthe column. p

patm =pgh � h= -EE!l. pg

The height ofthe column will be 760 mm.

P=O

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: / ,1 · Ti,c 1 /')

h=0.76m

h

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Example 17-3 Problem: Water (p = 1000 kg/m3) fills the u-shaped tube shown in Figure 17.5. The taller column is exposed to the air and the shorter column is closed. What is the pres­ sure at the top of the closed column? Solution: The pressure at the top of the closed column is found using Equation 171. Since the pressure at the top of the open tube is atmospheric pressure we have

3.1 m 1.7 m

Fif1ure _17 5:_The_systs1r1_::_,_E:,w1:p > · 7 3_

P=�;,+pgd � P=101.3x103 :2 (1000:� )(9.8; )(3.1m -1.7m) +

P=115020Pa � P=1.14atm

Since the pressure of the liquid is equal to the pressure of the air at all interfaces of the liquid and the air, liquids rise to the same height in all open regions of a container when the liquid is in hydrostatic equilibrium (Figure 17.6). Let's return to the system in Figure 17.3 and determine how the pressure of the fluid varies with depth when the entire fluid is accelerating. Solving for the pressure P in this case gives us

Figure 17.6: Liquids in hydrostmic equ,ld::irium rise to the some height in o!i open ,egion3:::,f CJ conto'._'.'72��-

PA-P.air A-mg= may � PA=Pa1r A+m(g +ay ) m= p Ad � PA=�;, A+p Ad(g+ay ) P=P.air +pd(g+ay ) The pressure increases when the system is accelerating upwards and decreases when the system is accelerating downwards. This is analogous to changes in the apparent weight of a person riding in an accelerating elevator (Section 8-9). Furthermore, when the system is in free-fall we have ay =-g � P=P.air

CHAPTER ScVENTEEN

-

During free-fall the pressure everywhere in the fluid is the same and equal to the pressure at the surface (at the interface with the air). m Finally, let's consider the system shown in Figure 17.7 in which two pistons with different cross-sectional A1 areas support blocks with different masses. The fluid enclosed in the system is ideal. Since the fluid is incom­ pressible and the two blocks in Figure 17.7 are at the same height, the pressure of the fluid pushing on each piston must be equal (Equation 7-1). Let's denote the pressure acting on the piston with cross-sectional area A 1 as P1 and the pressure acting on the piston with cross­ sectional area A 2 as P2 • If the system is in equilibrium (i.e., the blocks aren't moving) the forces associated with these pressures must equal the forces ofgravity acting on the blocks.

I 487

2m A2

p1 =P2 � mg= 2mg � A2 =2A1 A1

�

The cross-sectional area ofthe piston supporting the more massive block in Figure 17.7 must be twice the cross-sectional area ofthe other piston. By analogy to the simple machines in Section 7-7 and Section 11-3, we can consider the system shown in Figure 17.7 to be a hydraulic4 lever.

When the volume of an ideal gas changes, the average distance between molecules in the gas also changes. However, since we assume that interactions between the molecules of an ideal gas do not give rise to potential energy, this change in the average distance between molecules is not associated with a change in the average energy of the ideal gas. This is the reason why an ideal gas experiences no change in its average energy during free-expansion (Section 15-11). Indeed, we can determine the average energy ofan ideal gas from the kinetic energy of the molecules alone (Example 15-4). In contrast, since the energy ofa liquid does depend upon intermolecular interactions within the liquid (and thus on the average distance between the molecules in the liquid), the average energy of a liquid can change when its volume or shape changes. For ex­ ample, suppose that a thin liquid is stretched on a wire frame by a movable bar, as shown in Figure 17.8. 4 Hydraulic denotes a liquid moving in a confined space under pressure.

17,8:

488

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THE ENERGY OF PHYS!CS, PART I: ClASS!CJ,J ,'viECH1\NiCS Al' P2 and there is a drop in the pressure as the fluid moves through the constricted section of the tube. The Venturi effect is the phenomenon explaining how atomizers, carburetors, and spray guns work In each of the cases, the drop in pressure associated with moving air through a constricted section of a tube is used to draw up fluid from a reservoir.

CHAPTER SEVENTEEN

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505

Example 17-11 Problem: A hydraulic lift is created by connected two differ­ ent pistons with different cross-sectional areas, as shown in Figure 17.26; the system contains an ideal fluid. What is the ratio ofthe magnitudes ofthe forces F1 and F2 pushing down on the two pistons? What is the ratio of the displacements associated with moving the two pistons?

F1

A1

Solution: Let's use the subscripts 1 and 2 to denote the piston on the left and the piston on the right, respectively, in Figure 17.26. Since the fluid is incompressible we can use Equation 17-5 to relate the pressures exerted by the fluid on the two pistons.

--·· h·--................

! A2

r

Fig urn ; 7. 26 The hycJrnuli