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phd in physics

INTERMEDIATE DYNAMICS Second Edition Patrick Hamill Contents Part 1. Principles of Mechanics 7 Chapter 1. A Brief Review of Introductory Concepts 9 1.1. Kinematics 9 1.2. Newton's Second Law 11 1.3. Work and Energy 13 1.4. Momentum 14 1.5. Rotational Motion 15 1.6. Summary 22 1.7. Problems 23 Chapter 2. Kinematics 29 2.1. Galileo Galilei (Optional Historical Note) 30 2.2. The Principle of Inertia 31 2.3. Basic Concepts in Kinematics 32 2.4. The Position of a Particle on a Plane 43 2.5. Unit Vectors 44 2.6. Kinematics in Two Dimensions 47 2.7. Kinematics in Three Dimensions 51 2.8. Summary 60 2.9. Problems 61 Chapter 3. Newton's Laws: Determining the Motion 69 3.1. Isaac Newton(Optional Historical Note) 70 3.2. The Law of Inertia 71 3.3. Newton's Second Law and the Equation of Motion 74 3.4. Newton's Third Law: Action Equals Reaction 78 3.5. Is Rotation Absolute or Relative? 81 3.6. Determining the Motion 83 3.7. Numerical Method to Determine the Motion (Optional) 94 iii iv 3.8. 3.9. CONTENTS Summary 98 Problems 99 Chapter 4. The Lagrangian Method 109 4.1. The Equation of Motion by Inspection 109 4.2. The Lagrangian 111 4.3. Lagrange's Equations 117 4.4. Degrees of Freedom 121 4.5. Generalized Momentum 123 4.6. Generalized Force (Optional) 126 4.7. The Calculus of Variations (Optional) 128 4.8. Hamilton's Equations (Optional) 134 4.9. Summary 138 4.10. Problems 140 Chapter 5. The Conservation of Energy 149 5.1. The Work-Energy Theorem 149 5.2. Work Along a Path. The Line Integral 151 5.3. Potential Energy 155 5.4. Force, Work, and Potential Energy 167 5.5. The Conservation of Energy 172 5.6. Energy Diagrams 176 5.7. Solving for the Motion: The Energy Integral 178 5.8. The Kinetic Energy of a System of Particles 181 5.9. Work on an Extended Body. Pseudowork 183 5.10. Summary 184 5.11. Problems 187 Chapter 6. Conservation of Linear Momentum 195 6.1. The Law of Conservation of Momentum 195 6.2. The Motion of a Rocket 196 6.3. Collisions 200 6.4. Inelastic Collisions. The Coefficient of Restitution 210 6.5. Impulse 211 6.6. Momentum of a System of Particles 212 6.7. Relative Motion and the Reduced Mass 214 6.8. Collisions in Center of Mass Coordinates (Optional) 216 6.9. Summary 221 6.10. Problems 223 Chapter 7. Conservation of Angular Momentum 233 7.1. Definition of Angular Momentum 233 7.2. Conservation of Angular Momentum 235 CONTENTS v 7.3. Angular Momentum of a System of Particles 237 7.4. Rotation of a Rigid Body about a Fixed Axis 244 7.5. The Moment of Inertia 247 7.6. The Gyroscope 250 7.7. Angular Momentum is an Axial Vector 253 7.8. Summary 255 7.9. Problems 257 Chapter 8. Conservation Laws and Symmetries 265 8.1. 8.2. 8.3. 8.4. 8.5. 8.6. 8.7. 8.8. 9.1. 9.2. 9.3. 9.4. 9.5. 9.6. 9.7. 9.8. 9.9. 9.10. Part 3. Symmetry 265 Symmetry and the Laws of Physics 267 Symmetries and Conserved Physical Quantities 268 Are the Laws of Physics Symmetrical? 270 Strangeness 272 Symmetry Breaking 274 Summary 274 Problems 274 The Gravitational Field 277 Chapter 9. The Gravitational Field 279 Part 2. Newton's Law of Universal Gravitation 280 The Gravitational Field 282 The Gravitational Field of an Extended Body 286 The Gravitational Potential 289 Field Lines and Equipotential Surfaces 293 The Newtonian Gravitational Field Equations 294 The Equations of Poisson and Laplace 298 Einstein's Theory of Gravitation (Optional) 299 Summary 306 Problems 308 The Mechanics of Particles 315 Chapter 10. Central Force Motion: The Kepler Problem 317 10.1. Johannes Kepler (Optional Historical Note) 318 10.2. Kepler's Laws 321 10.3. Central Forces 321 10.4. The Equation of Motion 328 10.5. Energy and the Effective Potential Energy 331 10.6. Solving the Radial Equation of Motion 336 10.7. The Equation of the Orbit 337 10.8. The Equation of an Ellipse 342 vi 10.9. 10.10. 10.11. 10.12. 10.13. Chapter 11.1. 11.2. 11.3. 11.4. 11.5. 11.6. 11.7. Chapter 12.1. 12.2. 12.3. 12.4. 12.5. 12.6. CONTENTS Kepler's Laws Revisited 348 A Perturbed Circular Orbit 355 Resonances 361 Summary 362 Problems 363 11. Harmonic Motion 373 Springs and Pendulums 373 Solving the Differential Equation (Optional) 376 The Damped Harmonic Oscillator 383 The Forced Harmonic Oscillator 389 Coupled Oscillators 402 Summary 409 Problems 410 12. The Pendulum 417 A Simple Pendulum with Arbitrary Amplitude 418 The Physical Pendulum 424 The Center of Percussion 429 The Spherical Pendulum 434 Summary 445 Problems 447 Chapter 13. Accelerated Reference Frames 453 13.1. 13.2. 13.3. 13.4. 13.5. 13.6. 13.7. 13.8. Part 4. A Linearly Accelerating Reference Frame 453 A Rotating Coordinate Frame 455 Fictitious Forces 457 Centrifugal Force and the Plumb Bob 460 The Coriolis Force 462 The Foucault Pendulum 469 Summary 476 Problems 476 The Mechanics of Extended Bodies 481 Chapter 14. Statics (Optional) 483 14.1. Basic Concepts 483 14.2. Couples, Resultants and Equilibrants 487 14.3. Reduction to the Simplest Set of Forces 489 14.4. The Hanging Cable 489 14.5. Stress and Strain 494 14.6. The Centroid (Optional) 497 14.7. The Center of Gravity (Optional) 499 CONTENTS vii 14.8. Equilibrium of Fluids 501 14.9. D'Alembert's Principle and Virtual Work (Optional) 507 14.10. Summary 511 14.11. Problems 513 Chapter 15. Rotational Kinematics 519 15.1. Orientation of a Rigid Body 520 15.2. Orthogonal Transformations 522 15.3. The Euler Angles 530 15.4. Euler's Theorem 534 15.5. Infinitesimal Rotations 545 15.6. Summary 547 15.7. Problems 549 Chapter 16. Rotational Dynamics 551 16.1. Angular Momentum 552 16.2. Kinetic Energy 556 16.3. Properties of the Inertia Tensor 558 16.4. The Euler Equations of Motion 571 16.5. Torque-Free Motion 572 16.6. The Spinning Top 574 16.7. Summary 582 16.8. Problems 585 Chapter 17. Waves 589 17.1. A Wave in a Stretched String 590 17.2. Direct Solution of the Wave Equation 593 17.3. Standing Waves 595 17.4. Traveling Waves 597 17.5. Standing Waves as a Special Case of Traveling Waves 600 17.6. Energy 603 17.7. Momentum 607 17.8. Summary 609 17.9. Problems 610 Chapter 18. Small Oscillations (Optional) 615 18.1. Introduction 615 18.2. Statement of the Problem 615 18.3. Normal Modes 621 18.4. Matrix Formulation 629 18.5. Normal Coordinates 632 18.6. Coupled Pendulums: An Example 634 18.7. Many Degrees of Freedom 638 viii CONTENTS 18.8. Transition to Continuous Systems 643 18.9. Summary 647 18.10. Problems 650 Part 5. Chapter 19.1. 19.2. 19.3. 19.4. 19.5. 19.6. 19.7. 19.8. 19.9. 19.10. 19.11. 19.12. Chapter 20.1. 20.2. 20.3. 20.4. 20.5. 20.6. 20.7. 20.8. Special Topics 653 19. The Special Theory of Relativity 655 Albert Einstein (Optional Historical Note) 656 Experimental Background 657 The Postulates of Special Relativity 659 The Lorentz Transformations 660 The Addition of Velocities 668 Simultaneity and Causality 670 The Twin Paradox 673 Minkowski Space-Time Diagrams 675 4-Vectors 679 Relativistic Dynamics 685 Summary 688 Problems 689 20. Classical Chaos (Optional) 695 Configuration Space and Phase Space 696 Periodic Motion 698 Attractors 700 Chaotic Trajectories and Liapunov Exponents 702 Poincar ́e Maps 703 The Henon Heiles Hamiltonian 705 Summary 707 Problems 708 Chapter A. Formulas and Constants 713 Chapter B. Answers to Selected Problems 717 Preface Although this book begins at an introductory level, by the end of the book the student will have been exposed to all of the subject matter usually found in an intermediate mechanics course as well as a few less common advanced topics. Organization This book is divided into five parts, preceded by a short chapter consisting of a review of a few essential introductory concepts. This chapter can be skipped by well prepared students, or assigned as read- ing for students who only need a quick refresher. Part I (Chapters 2 through 8) is called "Principles of Mechanics." This part covers all the basic concepts in mechanics. The topics of kine- matics, dynamics and the conservation principles are treated in depth. The first chapter in this part is called "Kinematics." This is the tradi- tional starting point for courses in intermediate mechanics. Here, the student is exposed to relations between acceleration, velocity and po- sition in Cartesian, plane polar, cylindrical, and spherical coordinates. A few simple concepts from vector analysis are introduced. A number of reasonably difficult projectile problems are included in the problems. The next chapter considers Newton's laws. It includes a discussion on "Determining the Motion" in which the student learns techniques for integrating Newton's second law to obtain the position as a function of time. This is done for constant forces, and for forces that are functions of time, of velocity, and of position. A short section called "Solving for the Motion by Numerical Methods" gives a flavor for the use of compu- tational techniques in physics. The role of computers in physics is not emphasized in this course. However, I realize that many instructors want to expose their students to computational methods, so I have in- cluded a few discussions of numerical techniques. Furthermore, nearly 1 2 PREFACE every chapter has a number of "Computational Projects." However, this text does not emphasize the role of computers in physics because I find that teaching the traditional material of intermediate mechanics takes most of two semesters and does not give enough time to delve into computational physics. Furthermore, many universities have in- cluded a computational physics course in the undergraduate physics curriculum. Chapter 4 is called "The Lagrangian Approach." I think it is im- portant for physics students to be exposed to the Lagrangian early on in mechanics. The Lagrangian is presented, at first, as a simple tech- nique for generating the equation of motion. Later in the chapter, I go through a derivation of the Lagrange equations using the Calculus of Variations. This section need not be covered if the instructor feels it is too advanced. The chapter ends with an optional discussion of the Hamiltonian and Hamilton's equations. There are several reasons why I chose to present the Lagrangian early in the course, perhaps the most important is that it gives the student a simple (almost "cookbook") technique for obtaining the equations of motion for a complicated dy- namical system. For example, the Lagrangian technique allows one to easily determine the equations of motion for a double pendulum, for a spherical pendulum, or for coupled oscillators. More importantly, it allows one to introduce the concepts of generalized momentum and ignorable coordinates and leads to the relation between conservation laws and symmetries. Furthermores, it lets the student know that this course is not simply a rehash of concepts learned in introductory physics. The following four chapters cover the conservations of energy, linear momentum, and angular momentum, followed by a short chapter on the relation between symmetry and conservation laws. The chapter on the conservation of energy discusses potential energy and the concept of effective potential. Potential energy naturally leads to a discussion of the gradient of a scalar field. There is a section on the way the "Del" operator can be expressed in cylindrical and spherical coordinates; this allows one to discuss coordinate transformations in general. (I be- lieve that introducing concepts from vector calculus as required by the physics is much more effective than stuffing all of the vector concepts into a single introductory chapter.) The chapters on the conservation of linear and angular momentum cover the usual topics (rockets, col- lisions, etc.) as well as some less usual topics such as the fact that angular momentum is an axial vector. Part II is a very short introduction to field theory. Since field theory is treated exhaustively in courses on electromagnetism, this study is PREFACE 3 limited to a single chapter on the gravitational field. Additional vector concepts are introduced here and the student is exposed to Gauss's law and the equations of Poisson and Laplace. Next is Part III (Chapters 10 - 13) called "The Mechanics of Par- ticles." This important section covers central force motion, harmonic motion and motion in accelerated reference frames. The student is ex- posed to the techniques of expansions in power series and the idea of using successive approximations to solve otherwise intractable physics problems. Chapter 10 deals with central force motion, as illustrated by the Kepler problem. Also considered is the stability of circular orbits which shows the student how to deal with small perturbations. Specif- ically, we imagine a comet striking a planet in a previously perfectly circular orbit and analyze the subsequent motion to determine stability and the frequency of radial oscillations. The next chapter is called "Simple Harmonic Motion." Damped and driven harmonic oscillators are treated in depth. A rather thorough dis- cussion on how to solve second order differential equations is included here. Coupled oscillators and normal modes are touched upon. The third chapter in this part (Chapter 12) is on the motion of a pendulum. We begin with the motion of a simple pendulum of arbi- trary amplitude and introduce elliptic integrals. Next we consider the physical pendulum, centers of oscillation and percussion, the spherical pendulum, and the conical pendulum. To spend a whole chapter on the pendulum may seem excessive, but it is a simple, easily visualized physical system that allows one to introduce many useful mathematical techniques without having to spend time explaining the motion. The fourth chapter in this part is called "Accelerated Reference Frames" in which we (mainly) consider motion on the surface of the rotating Earth. Coriolis forces and the Foucault pendulum are treated. Once again, perturbation theory is used to solve these problems. Part IV (Chapters 14-18) is called "The Mechanics of Extended Bodies." Portions of the material in this part may be too advanced for some classes, but the instructor should try to cover the chapters on rotational kinematics, rotational dynamics, and waves. This part begins with an optional chapter on statics that includes a discussion of d'Alembert's principle and virtual work. The next chapter (on rota- tional kinematics) introduces orthogonal transformations. We obtain the Euler angles and consider Euler's theorem. The following chapter (Chapter 16) on rotational dynamics introduces the inertia tensor and some simple methods from tensor analysis. The next chapter in this part is an introduction to wave motion. This topic is not considered 4 PREFACE in great detail since it is treated extensively in the undergraduate elec- tromagnetic theory class, nevertheless it does show the student how to solve a partial differential equation by separation of variables. The last chapter in this part treats small oscillations. It is rather advanced and the chapter is denoted as optional. The last part of the book is called "Special Topics" and consists of two chapters. The first of these is on special relativity. The second chapter is an introduction to classical chaos. Neither of these chapters is at all exhaustive; they are simply intended to give the student a flavor of these interesting subjects. A One-Semester Course Many instructors will find that the intermediate mechanics course in their department has been reduced to one semester. If such is the situation at your institution, you may wish to cover the material in Chapters 2, 4, 6, 7, 11, 13, 15, 16, 17 and include some material from the skipped chapters, in particular, Sections 3.6, 5.6, and 12.2 Exercises and Problems Learning physics requires doing physics, so I have included a large number of "exercises." These are found at the end of nearly every section. Most of them are fairly easy. Some are merely "plug-ins" to get the student to look at a formula and (hopefully) to think about it. Others ask the student to fill in the missing steps in a derivation. A few require a bit of clever thinking. Nearly all have answers given. I hope that students studying this book will solve every one of these exercises. At the end of each chapter is a collection of problems that are of the degree of difficulty to be expected from a course at this level. Most of these will require significant effort on the part of the student. However, I believe that a student who has read the chapter and worked the exercises will be prepared to attack the problems. Acknowledgements PREFACE 5 I thank my colleagues at San Jose State University and NASA Ames Research Center, particularly Drs. Alejandro Garcia and Michael Kauf- man. I am especially indebted to the many students in my mechanics courses whose influence on this book cannot be overestimated. Part 1 Principles of Mechanics Chapter 1 A Brief Review of Introductory Concepts This introductory chapter is intended to remind you of some of the basic concepts from your introductory physics course. You should probably read through this chapter quickly, work out some of the ex- ercises and try a few of the problems at the end of the chapter. If you remember all of this material, go on to Chapter 2, but if you find that you have forgotten some of these basic concepts, you should go back to your introductory physics textbook and review the material there. Most of the standard introductory physics texts are well written and contain many instructive figures and diagrams. It is a good idea to go back to that text whenever you are exposed to the same material on a more advanced level. As you may recall, the mechanics section of your introductory physics book covered the following topics: • Kinematics • Newton's Second Law • Work and Energy • Momentum • Rotational Motion We now very briefly review some concepts from each of these items. 1.1. Kinematics Kinematics is defined as the study of motion. Essentially, kine- matics involves determing the relationships between position, velocity, acceleration, and time. 9 10 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Position is denoted by the vector r and the change in position (or displacement) can be written as ∆r. Velocity is defined as the displace- ment with respect to time, so the average velocity is given by < v >= ∆r ∆t where ∆t is the time during which the object moved a distance ∆r. As the time interval becomes very small, we replace the difference (represented by ∆) with the derivative and write v = dr. (1.1) dt The change in velocity with respect to time is called the acceleration and is given by a = dv. (1.2) dt Clearly, we can use the definitions of acceleration and velocity to write the inverse relations: and dv = adt, dr = vdt. 1.1.1. Motion in a Straight Line at Constant Acceleration. For motion in a straight line, we do not need to use vector notation. Letting the position to be represented by x, if the acceleration is con- stant the integrals above are easy to evaluate, leading to the familiar relations and v(t) = at + v0, x(t) = 1at2 + v0t + x0, (1.3) (1.4) You probably memorized Equations (1.3) and (1.4) in your intro- ductory physics course. You should, however, keep in mind that they are only valid if the acceleration is constant. In this course, we shall frequently be concerned with non-constant accelerations and these re- lations cannot be used. Two other useful equations can be obtained from (1.3) and (1.4) as follows. Solve one equation for t and substitute it in the other to obtain 2a(x−x0)=v2 −v02, 2 where v0 and x0 are the initial velocity and the initial position. 1.2. NEWTON'S SECOND LAW 11 or solve one equation for a and substitute it in the other equation to obtain x − x0 = 1(v + v0)t. 2 Comment 1. In this book you will find a large number of "exercises." They are not difficult. They are intended to give you a chance to review and understand the concepts in the preceding section. It would be a good idea to work out the solution for each exercise as you read through the text. At the end of each chapter you will find a set of problems that are significantly more difficult than the exercises. However, if you solve the exercises, you will be well prepared for solving the problems. Many students find working the exercises to be a good preparation for the examinations. Exercise 1.1. You were driving your new Ferrari at 62 mph (= 100 km/hr) when you spotted a police car. Naturally, you hit the brakes. You slowed to 31 mph, covering a distance of 50 m. (a) What is your constant acceleration? (b) How much time did it take to slow to 31 mph? Answers: (a) -5.79 m/s2 (b) 2.4 sec. 1.2. Newton's Second Law The study of the relation between the forces acting on a body and the motion of the body is called dynamics. In your introductory course you were exposed to dynamics in the form of Newton's Second Law. This law states that a body of mass m that is acted upon by a force F will accelerate at a = F/m. This relationship is usually remembered as F = ma. (1.5) As you shall see, this form of Newton's second law is only valid if the mass is constant. 12 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Figure 1.1. A block sliding down an inclined plane with friction. Sketch (a) shows the forces acting on the block. Sketch (b) shows the free-body diagram. Sketch (c) shows the x-and y-axes inclined so that they are par- allel and perpendicular to the plane. WorkedExample1.1. Determinetheaccelerationofablock of mass m sliding down an inclined plane of angle θ. Assume the coefficient of sliding friction is μ. See Figure 1.1. Solution: The forces acting on the block are gravity mg (down- wards), the normal force N (perpendicular to the plane), and the frictional force μN (parallel to the plane). These forces are illus- trated in Figure 1.1(a). The free-body diagram with all of the forces acting at the center of mass of the block is shown in Figure 1.1(b). In a problem such as this one, it is convenient to assume the axes are parallel to and perpendicular to the surface of the plane as indicated in Figure 1.1(c). There is no acceleration perpendicular to the plane so the net force in that direction must be zero. Consequently, N = mg cos θ. The net force down the plane is Fd = mg sin θ − μN . The acceler- ation of the block is a= Fd =gsinθ−μgcosθ. m Exercise 1.2. Two blocks of mass M1 and M2 are tied together. They are sitting on a smooth frictionless surface as shown in Figure 1.2. A force F is applied to the free string attached to M1. What is the tension in the string between the two blocks? Answer: T = M2F/(M1 + M2). Exercise1.3. Ablockofmass25kgisheldinplaceonaninclined plane of angle 30◦ as shown in Figure 1.3. The coefficient of static 1.3. WORK AND ENERGY 13 Figure 1.2. Two blocks on a smooth frictionless surface connected by a massless string. friction is 0.4. a) Draw the free body diagram. What forces act on the block? b) What is the tension in the string? c) If the string is cut, what is the acceleration of the block? Answers (b) T = 37.63 N (c) a = 1.51 m/s2. m 30o Figure 1.3. A block on an inclined plane. What is the acceleration of the block if the string is cut? 1.3. Work and Energy Imagine pushing a box along a horizontal surface, such as a table- top. The force you are applying can be denoted by F. As you might expect, there will be opposing forces such as friction, air resistance, etc., but for now we are only interested in the force you are exerting. Ifyoupushtheboxfromx1 tox2,thework youdoonitis x2 W = F · dx. (1.6) x1 When we do work on an object, we often change its energy. The energy of an object is either kinetic energy (energy of motion) or po- tential energy (energy of position). For a particle of mass m moving with speed v the kinetic energy (denoted by T) is given by T = 1mv2. (1.7) 2 14 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS The potential energy of an object of mass m raised a height h above the surface of the Earth is given by V = mgh. (1.8) We shall often be interested in the potential energy of a stretched or compressed spring. The potential energy of such a system is given by V = 1kx2, (1.9) 2 where k is the spring constant and x represents the amount the spring is stretched or compressed. Power is defined as the work done per unit time. Suppose a force F acts on a particle for an infinitesimal time interval, from t to t+dt. The work done during this interval is dW = F · dx. Therefore,the power is P = dW = F·dx. dt dt Since dx/dt ≡ v, another expression for power is P = F · v. Exercise 1.4. A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back past its original point, (b) When it is 15 m above the street, and (c) Just before it hits the street. Answer: (a) -5 m/s, (c)24.76 m/s. Exercise 1.5. A horse drags a 100 kg sled a distance of 4 km in 20 minutes. The horse exerts one horsepower, of course. What is the coefficient of sliding friction between the sled and the ground? Answer: μk = 0.23. 1.4. Momentum A moving particle is characterized by having a particular momen- tum. When we use the term "momentum" we usually are referring to the linear momentum, not to be confused with the angular momentum, which we will define in a few moments. 1.5. ROTATIONAL MOTION 15 The momentum is simply defined as mass times velocity and is denoted by the letter p. Thus, p = mv. (1.10) If the mass of a body is constant, the time derivative of the mo- mentum is dp = d(mv) =mdv =ma dt dt dt But according to Newton's second law, F = ma, so Newton's second law can be expressed as F = dp (1.11) dt Here F is the net or total vector sum of all forces acting on the body. Consequently, we appreciate that if the net force is zero, the time de- rivative of the momentum is zero. That is, the momentum of the body is constant. This is called the law of conservation of momentum. Exercise 1.6. A 1500 kg car traveling east at 40 km/hr turns a corner and speeds up to a velocity of 50 km/hr due north. What is the change in the car's momentum? Answer: 26, 700 kg m/s at 38.7◦ West of North. 1.5. Rotational Motion The motion of a body rotating about a fixed axis is mathematically identical to one dimensional linear motion. Recall that kinematics studies the relationship between position, velocity, and acceleration. Rotational kinematics deals with angular position (θ), angular velocity (ω), and angular acceleration (α), where and ω ≡ dθ, dt α ≡ dω = d2θ. dt dt2 You will discover that rotational motion can be very complicated. To keep things simple for the moment, consider the special case of a symmetrical body rotating about a fixed axis, such as the wheel illustrated on the left side of Figure 1.4. 16 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Figure 1.4. A wheel mounted on a fixed axis. The wheel is allowed to rotate but not to translate. For a fixed, stationary axis, the center of the wheel is at rest. All other points are moving in circles around it. If you looked straight down the axis you would see the circle shown on the right side of Figure 1.4. Point P is on the rim of the wheel. The angular position of P is given by the angle between some fixed line and the radius vector to P. If the wheel is turning, after a time dt point P will have moved a distance ds to P′. Recall from geometry that ds = rdθ where dθ (in radians of course!) is the angle subtended by the arc P P ′ = ds. Point P moves with speed v = ds = rdθ = rω. dt dt The speed of point P is called the "tangential speed" because instan- taneously P is moving tangent to the rim of the wheel. It is sometimes convenient to write the tangential speed as vT . Then vT =rω. (1.12) Taking the time derivative of vT yields the tangential acceleration aT, dv d dθ d2θ aT=dt=dt rdt =rdt2, where we used the fact that r is constant. But dω = α, so dt aT =rα. (1.13) Exercise 1.7. A wheel initially spinning at ω0 = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and the number of revolutions it makes before stopping. Answers: -2.5 rad/s2; 79.6 rev. 1.5. ROTATIONAL MOTION 17 1.5.1. Rotational Dynamics. Rotational dynamics is the anal- ysis of the rotational motion of a body subjected to external torques. Consider a body constrained to rotate about a fixed axis as shown in Figure 1.5. I drew the body in the shape of a plane lamina for simplicity. Let a force F act on the body at a point on its rim. Let us assume (again for simplicity) that F is perpendicular to the axis of rotation. The point of application of F is specified by the vector r whose origin is at the axis of rotation. The body cannot accelerate linearly because the axis is fixed. The applied force causes the body to rotate about the axis. The tendency of a force to cause a rotation is called the moment of the force, or more commonly, the torque. Just as you can think of a force as a pull, you can think of a torque as a twist. The ability of a force to produce a rotation depends not only on the magnitude of the force, but also on its direction and on the location of the point where the force is applied to the body. To define torque draw a line having the direction of the force and passing through the point of application. This is called the "line of action of the force." (See Figure 1.6.) Next draw a line that starts at the axis of rotation and intersects the line of action at a 90◦ angle. This line is called the lever arm. The lever arm is the shortest distance from the axis of rotation to the line of action of the force. Its length is r sin θ where θ is the angle between r and F. The magnitude of the Axis of rotation r Figure 1.5. Illustration of torque. The laminar body is free to rotate around the fixed axis of rotation. The vector r, with origin at the axis, specifies the point of application of the force. F 18 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Figure 1.6. Definition of lever arm. The axis of rota- tion is perpendicular to the plane of the figure. torque is given by the relation: Torque = Force times Lever Arm. This elementary definition of torque is complicated and cumber- some. Fortunately, vector concepts allow us to simplify the definition. Torque (N) is the cross product of r and F, thus: N ≡ r×F. (1.14) In Equation (1.14) r is the vector from the axis of rotation to the point of application of the force. The direction of the torque is perpen- dicular to the plane defined by r and F and it is usually represented as a vector along the axis of rotation. Exercise 1.8. Show that a given force applied at any point along the line of action yields the same torque. Exercise 1.9. An athlete is holding a 2.5 meter pole by one end. The pole makes an angle of 60◦ with the horizontal. The mass of the pole is 4 kg. Determine the torque exerted by the pole on the athlete's hand. (The mass of the pole can be assumed to be concentrated at the center of mass.) Answer: Torque = 24.5 N m. 1.5. ROTATIONAL MOTION 19 1.5.2. Statics. This section is an overview of some of the basic concepts of statics. Newton's first law tells us that if no external force acts on a body, it will move at constant velocity. That is, zero force implies zero ac- celeration. (This is also a consequence of the Newton's second law.) Consequently, a body that is not accelerating must have no net exter- nal force acting on it. Similarly, a body that is rotating at a constant angular velocity has no angular acceleration. Such a body has no net external torque acting on it. A body that has no linear acceleration and no angular acceleration is said to be in equilibrium. The conditions for equilibrium are: and Fi = 0, (1.15) Ni = 0, (1.16) where Fi and Ni are the net external forces and torques acting on the system. A body in equilibrium has constant linear velocity and constant angular velocity. A particular but important special case occurs when both the linear and angular velocity are zero. This is called static equilibrium. The general conditions for equilibrium are given by Equations (1.15) and (1.16). I will state these conditions in words for a situation in which all of the forces applied to a body lie in the same plane (but it is not difficult to generalize). If a body is in equilibrium then: 1. The algebraic sum of the components of the forces in each of two mutually perpendicular directions is zero, and, 2. The algebraic sum of the torques about any point in the plane of the forces is zero. The first condition is obvious. The second condition will be proved in Chapter 7. It is very useful for solving simple statics problems because it allows us to take the torques about any convenient point. This point does not even have to be in the body. Worked Example 1.2. The captain of the swimming team (who weighs 600 N) is posing on the end of a 4 m diving board that weighs 150 N. The board is bolted to two pillars that are 1.5 m apart. The first pillar is at the end of the board. Determine the tension (or compression) acting on the bolts. See Figure 1.7. 20 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Figure 1.7. What forces are exerted on the bolts that hold the diving board on the pillars? Solution: The board is in equilibrium so forces up = forces down and torques clockwise (CW) = torques counterclockwise (CCW). The first condition leads to 600+150=F1 +F2. The second condition leads to 600×4+150× 2=F2 ×1.5. Therefore, F2 = 1800 N (up) and F1= -1050N (down). 1.5.3. Rotational Kinetic Energy. The rotational kinetic en- ergy of an extended rigid body is obtained (at least conceptually) by summing over the kinetic energy of each particle in the body. If the body is rotating about a fixed axis and if the body is symmetrical about this axis, the rotational kinetic energy is given by a fairly simple formula. Recall that every particle in the body is moving in a circle, and that the linear velocity v of the particle is related to the angular velocity of the body, ω, according to v = ωr⊥ where r⊥ is the per- pendicular distance from the axis of rotation to the particle. So the kinetic energy of the particle is T = 1mv2 = 1m(ωr⊥)2. The total 22 energy of the body will be the sum over all of the particles (molecules) in the body. If we denote the mass of the ith particle by mi and its 1.5. ROTATIONAL MOTION 21 perpendicular distance from the axis by ri⊥, we have T= 1mr2 ω2 i 2 i i⊥ The quantity mir2 is called the moment of inertia of the body and i i⊥ is denoted I. Therefore the rotational kinetic energy can be expressed as T = 1Iω2. (1.17) 2 Note the similarity with the translational kinetic energy T = 1mv2. 2 Exercise 1.10. A disk of mass M and radius R is initially at rest. It is acted upon by a constant torque N for a time T. Determine the final angular velocity of the disk and the work required to spin it up to this final state. If the energy was supplied by a motor, what was the average power output of the motor? Answer: Average Power = N2T/MR2. 1.5.4. Angular Momentum. In general, the equations for linear motion go over to the equations for rotational motion if we substitute I for m and ω for v. Thus, for example, the magnitude of the angular momentum of a single particle of mass m moving in a circle of radius r with speed v will be l = Iω, (1.18) This relation can also be expressed as l = mr2ω or as l = mvr. But angular momentum is a vector, so l = Iω where ω is the angular velocity vector, directed along the axis of rota- tion. For an extended rigid body the total angular momentum is usually denoted by L and, as you might expect, is given by where I = mr2. wherenowI= mir2 . i i⊥ L = Iω 22 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS The rotational analogue of Newton's second law is obtained by sub- stituting torque for force, moment of inertia for mass and angular ac- celeration for linear acceleration. This yields N = Iα. Recall that Newton's second law can be expressed as F = dp. dt Similarly, for rotational motion we can write N = dL. dt (1.19) Just as linear momentum is constant if the net force is zero, so too, angular momentum is constant if the net torque is zero. Exercise 1.11. A spinning ice skater speeds up by pulling her hands to the side of her body. Approximate the ice skater by a cylinder of radius 10 cm and mass 30 kg. Her hands are two point masses of 0.25 kg each and her (massless) arms extend 90 cm from her torso (or 1 m from the axis of rotation). If she was initially rotating at 2 rev/s, what is her final angular velocity? Answer: 8.4 rev/s. 1.6. Summary The position of a particle is given by the vector r. Velocity is defined by and the acceleration is v = dr, dt a = dv. dt Newton's second law for a particle with constant mass is F = ma. (1.20) Work is defined by the integral of the dot product of force and displacement, W = F · dr. (1.21) 1.7. PROBLEMS 23 Kinetic energy is given by for linear motion, and T = 1mv2, 2 T = 1Iω2, 2 for rotational motion. Potential energy of an object raised a height h above the Earth's surface is The potential energy stored in a spring is V = 1kx2. 2 The linear momentum of a particle of mass m moving at velocity v is defined as p = mv. (1.22) The relation between angular velocity and linear velocity is given by the cross product: v = ω × r, (1.23) where ω is a vector pointing along the axis of rotation whose magnitude is equal to the angular velocity. V = mgh. Torque is defined by N = r × F. The angular momentum of a particle can be written as l = r×p. The rotational version of Newton's Second Law is N = dl . dt 1.7. Problems (1.24) (1.25) (1.26) Problem 1.1. Two ships are sailing in a thick fog. Initially, ship A is 10 miles north of ship B. Ship A sails directly east at 30 miles per hour. Ship B sails due east at constant speed vB then turns and sails due north at the same speed. After two hours, the ships collide. Determine vB. 24 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Problem 1.2. You carefully observe an object moving along the x−axis and determine that its position as a function of time is given by x(t)=2t−3t2 +t3. (a) What is the position at time t = 2 seconds? (b) What is the velocity at time t = 2 seconds? (c) What is the acceleration at time t = 2 seconds? (d) How far did it travel between times t = 0 and t = 2 seconds? (Note: Distance, not displacement! It might be helpful to plot x vs t for 0 ≤ t ≤ 2.) Problem 1.3. The first train leaves the station and accelerates at a constant rate to its maximum speed of 100 km/hr, reaching this speed at a distance of 2 km from the station. Five minutes later, a second train leaves the station and accelerates to 100 km/hr in 4 km. What is the distance between the two trains when they both reach maximum speed? Problem 1.4. A rocket is trying to land on a barge in ocean. The rocket engines deliver an upward force of 36,000 N and the rocket is observed to be descending at a constant safe speed of 3 m/s when it is 100 m above the barge. But suddenly there is a malfunction and the rocket engine thrust is reduced to 30,000 N. (a) What is the mass of the rocket? (b) What is the acceleration of the rocket after the malfunction? (c) Assume this acceleration is maintained constant during the final descent. What is the speed of the rocket when it contacts the barge? Problem 1.5. You travel a distance d in time t. (a) If you travel at speed v1 for half the time and at v2 for the other half of the time, what is your average speed? (b) If you travel at speed v1 for half the distance and at speed v2 for the other half of the distance, what is your average speed? Problem 1.6. There is a long straight road out in the desert and it goes through a small town that has just one police car. The police car accelerates at 2 m/sec2 unitl it reaches a maximum speed of 200 km/hour. A car full of escaped criminals skpeeds through the town at its top speed which is 150 km/hour. The police car, starting from rest, gives chase. How far from the town do the police catch up to the criminals? Problem 1.7. A police car is at rest on the side of the road when a wild teenager comes speeding by at 75 miles per hour. The police 1.7. PROBLEMS 25 car starts immediately and accelerates at 8 miles per hour per second. At that same moment the teenager steps on the gas, but his car only accelerates at 2 miles per hour per second. (a) How far from the starting point does the police car overtake the speeder? (b) How fast are they going at that time? (c) Why is the speed you calculated for the police car unrealistic? Problem 1.8. A brick is on a wooden plank that is resting on a table. One end of the plank is slowly raised so that it forms an angle θ with the horizontal table top. When θ = 60◦ the brick starts to slide down the plane. (a) Draw the free body diagram. (b) Determine the coefficient of static friction between brick and plank. (c) If the coef- ficient of sliding friction is one half of the coefficient of static friction, determine the acceleration of the block. Problem 1.9. A block of mass M is on an inclined plane with a coefficient of sliding friction μ. At a given instant of time the block is at some point P on the plane and is moving up the plane with a speed v0. (a) Obtain the time for the block to reach its highest point relative to P. (b) Obtain the time for the block to slide back down to point P. (c) Obtain an expression for the velocity of the block at the time it returns to P. Problem 1.10. A toy rocket burns fuel for 1.5 seconds. During that time, it accelerates upwards at 60 m/s2. It then coasts upwards to some maximum altitude before falling back down. Determine the maximum altitude reached. Problem 1.11. Atwood's machine consists of two weights (M1 and M2) suspended at the ends of a string that passes over a pulley. Assume massless, inextensible strings and a frictionless pulley. Let M1 =6 kg and M2 = 5.5 kg. The masses are released from rest. Determine the distance descended by the 6 kg mass when its velocity reaches 0.5 m/s. Problem 1.12. (a) Determine the rotational kinetic energy of a wheel of your bicycle when your linear speed is 20 km/hour. You may assume the wheel is a hoop of mass 1.5 kg and radius 30 cm. (b) Compare your result with the translational kinetic energy of the wheel. (c) Give a theoretical explanation for the relationship found in parts (a) and (b). Problem 1.13. The frictional force between water and seabed in shallow seas causes an increase in the day by about 1 ms/century. Determine the torque that causes this change. Assume the Earth is a uniform sphere. 26 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS Problem 1.14. A meter stick has a pivot at one end. It is found to be in static equilibrium when acted upon by three forces that act at different points along the meter stick and act in different directions. We conclude that the net torque about the pivot is zero. Prove that the net torque about any other (arbitrary) point is also zero. Problem 1.15. To build the pyramids it was necessary to pull heavy stones up inclined planes. Suppose a 2000 kg stone was dragged up a 20◦ incline at a speed of 0.25 m/s by a gang of 20 laborers. The coefficient of kinetic friction between the stone and the incline was 0.4. How much power was exerted by each laborer? Problem 1.16. Two wooden blocks, both of mass 3 kg, are sliding in the same direction at 2 m/s on a frictionless horizontal surface. Let M1 =M2 =3kgandassumetheirspeedis2m/s. Athirdblockof mass 1 kg is also sliding in the same direction at a speed of 10 m/s, and it collides with the trailing 3 kg block. The third block is covered with a sticky, gooey substance, so it sticks to the trailing block. This combination block catches up with and collides elastically with the leading 3 kg block. Determine the final speed of the leading block. (Note that this is a one-dimensional problem. You may find it easier to solve the problem if you insert numerical values sooner rather than later.) Problem 1.17. You are driving along at 60 mph (=100 km/hr). Your car's wheels have a radius of 35 cm. (a) Determine the angular velocity of the wheels. (b) What is the angular displacement of a wheel when you travel 1 km? Problem 1.18. Aeronautical engineers have developed "tip jet" helicopters in which small jet engines are attached to the tips of the rotor. One such helicopter was powered by two ramjets. For a ramjet to develop thrust, it needs to be moving through the air quite rapidly, and is not efficient until it is moving at about 1000 km/hr. Assume the rotor diameter is 10 meters. Determine the angular speed of the rotor when the ramjet is moving through the air at 1000 km/hr. Problem 1.19. Figure 1.8 shows an angle iron of mass M hang- ing from the pivot at point P. The three segments have equal length. Determine whether or not this is a stable equilibrium orientation. Problem 1.20. A solid ball of mass M and radius R rolls down an inclined plane. What is its translational speed when it has descended a vertical distance h? 1.7. PROBLEMS 27 P Figure 1.8. An angle iron. Determine equilibrium. Problem 1.21. A disk of mass 72 kg and radius 50 cm is rotating at 2000 rpm. (a) Determine its angular momentum. (b) If acted upon by a retarding force of 20 N acting tangent to the rim of the disk, determine the time required to stop the disk. COMPUTATIONAL PROJECTS: The following problems can be solved easily if you know a computer language such as MATLAB, Python, FORTRAN, or C++. A few can also be solved using a spread sheet program such as EXCEL. Although some of the computational projects in this book can be solved analytically, they will require a significant amount of brainless labor that can be done much more con- veniently using a computer. Computational Project 1.1. The position of a particle as a function of time is given by x = 5t3 − 2t (meters). Plot the position as a function of time for the interval t = −5 sec to t = +10 sec. Using the relationship v = ∆x/∆t, obtain the average velocity at one second intervals. Plot the average velocity as a function of time and on the same graph, plot the analytical expression for the velocity. Computational Project 1.2. This is a more realistic version of problem 1.4. In that problem a teenager driving at 33.5 m/s (∼75 mph) speeds past a parked policeman. The policeman and the teenager then accelerate at given constant rates and you are required to deter- mine how far from the starting point the policeman catches up to the teenager. To make the problem somewhat more realistic, assume the teenager accelerates at aT = kT e−bT t and the policeman accelerates at aP = kP e−bP t, where kT = 2.0 m/s2 and kP = 3.5 m/s2. By plotting thepositionsasfunctionsoftimeforvariousvaluesofbT andbP obtain reasonable values for these constants. (Make sure your answers are 28 1. A BRIEF REVIEW OF INTRODUCTORY CONCEPTS reasonable. Obviously, the policeman will catch the teenager, but your answer is not reasonable if the distance required is hundreds of miles!) Chapter 2 Kinematics Kinematics in one dimension is fairly simple because position, ve- locity and acceleration can be treated as scalars. Motion in a straight line and rotation about a fixed axis are examples of one-dimensional motion. The motion of a projectile in a constant gravitational field can be resolved into two linked one-dimensional problems. Going to two or three dimensions complicates the problem significantly because now you need to express the basic quantities as vectors. In this chapter you will learn the relations between the position, velocity and acceleration vectors in the three main coordinate systems used in physics: Cartesian coordinates, cylindrical coordinates and spherical coordinates. (Rota- tional motion in three dimensions is considerably more difficult than linear motion and will be left to Chapters 13 and 16.) Although some of the material presented in this chapter will be familiar to you, you will also find many new concepts. These concepts are used throughout the course, so please make sure you understand this chapter thoroughly. Be aware that many students find this material rather difficult. I think it is important for you as a physicist to know something about the people upon whose shoulders you are standing, so in this book I have included a few sections marked "Optional Historical Note." You will encounter very little physics in these sections, but you might find them interesting and helpful in placing a few famous physicists in historical context. The first Historical Note describes the life of the person who invented your profession. 29 30 2. KINEMATICS 2.1. Galileo Galilei (Optional Historical Note) Galileo Galilei (1564 - 1642) was a brilliant but difficult man whose studies of the physical world launched a scientific and cultural revolu- tion. This cantankerous Italian genius was the first modern scientist. He rejected authority and based his conclusions on observation, exper- iment, and rational analysis. His ideas were opposed by the powers of the Church, and he ended his days under house arrest. Neverthe- less, his view of the universe and his methods for discovering scientific truth eventually won out and today every physicist is an intellectual descendent of Galileo.1 Galileo did not invent the telescope, but he built one and was prob- ably the first person to make a scientific study of what he saw in the sky. In short order, he discovered that the Moon has mountains and valleys, that the Milky Way is made up of individual stars, that Jupiter has satellites orbiting it, and that the Sun rotates and has spots on its surface. He became the first "celebrity scientist." He got into trouble, however, because he was outspoken and had little patience with peo- ple who put faith above reason. He was particularly offensive towards those who believed the Earth was the center of the universe and that the Sun, planets and stars revolved around it. This "geocentric" uni- verse was based on the philosophy of Aristotle which had been adapted to Christianity by Thomas Aquinas and was subsequently accepted as the official philosophy of the Church. Therefore, Galileo's outspoken attacks on the geocentric theory were considered by many to be hereti- cal. Galileo was eventually called to Rome and ordered by Church authorities to neither teach nor defend the Copernican theory that the Sun is the center of the Universe. Shortly thereafter Galileo wrote a book in the form of a dialogue between three men. In this book he ridiculed the Aristotelians and even put some arguments used by the Pope in the mouth of a character named "Simplicio" who was pictured as a bit of a dunce. Galileo was called back to Rome by the Inquisition and was eventually condemned to life imprisonment, to take place in own home. He was 70 years old at the time. He continued to work, publishing his last book, "Discourses and Mathematical Demonstra- tions Relating to Two New Sciences" in 1638. This book describes much of the work in physics that he had carried out over the previous 30 years. He died at age seventy eight. Galileo's life is an instructive example of the conflict between a brilliant individual and the powers that rule society. There is no doubt 1An excellent biography of Galileo is "Galileo's Daughter" by Dava Sobel, Walker and Co., New York, 1999. 2.2. THE PRINCIPLE OF INERTIA 31 that Galileo changed the world and our understanding of it. Some people condemn the Church authorities for trying to stop scientific progress; others are more understanding of the authorities who saw their entire world view threatened by a revolutionary iconoclast. 2.2. The Principle of Inertia Among his many accomplishments, Galileo was arguably the first person to truly understand motion. He performed a series of experi- ments with pendulums of different lengths, as well as falling bodies and balls rolling down inclined planes. His experiments with these simple devices allowed him to grasp the essential aspects of motion, something that had eluded the greatest minds of previous ages. Prior to Galileo's work many respected philosophers (including Aristotle) described mo- tion in ways that modern scientists consider to be either meaningless or just plain silly. For example, they said, "Water flows downhill be- cause it is trying to return to its natural place, which is the ocean." Or they said, "Smoke rises because smoke is air and the natural place for air is up in the sky." They explained the motion of the stars and planets with the preposterous statement that, "The celestial bodies are perfect, therefore they move about the Earth in perfect circles." A crucial element in this erroneous theory of motion was the idea that an object will move to its "natural place" and then remain at rest in that place. This was called, of course, "natural motion." For something to move away from its natural place, it was necessary to give it a push or a pull. This kind of motion, requiring a force, they called "violent motion." Galileo's most significant insight was that objects do not tend to come to rest, but rather they tend to keep on moving. In fact they tend to keep on moving at a constant speed in a straight line. (We call this uniform motion.) The property of material bodies to maintain their state of motion is called inertia. As you recall from your introductory course in physics, the law of inertia states that: A body will remain in uniform motion as long as no net external force acts on it. This basic principle has come to be known as Newton's first law. We will return to it later. Although his equipment was crude, Galileo's experiments showed him that time is an essential component of motion. Mechanical clocks had not yet been invented, but Galileo devised a number of ingenious ways to estimate time intervals. He counted his pulse or watched the 32 2. KINEMATICS oscillations of a pendulum or weighed the amount of water dripping out of a container. Time is a very difficult concept to comprehend. Everyone has an intuitive, qualitative understanding of time. If you think about it, you will realize that your personal definition of time has some connection to the motion of material bodies. The length of a day is related to the rotation of the Earth, and a year is the Earth's orbital period. Most people describe an hour and a minute in terms of the motion of the hands on a clock. Scientists define the second in terms of the vibrations of certain molecules. Is it possible to define time without reference to motion? If all objects in the universe were perfectly still, would time exist? Newton avoided these philosophical questions by assuming time to be an absolute parameter that is continuously changing at a constant rate. Einstein gave an even simpler definition when he said, "Time is what you measure with a clock." We shall adopt Newton's point of view at present, then come back for a deeper look at the question of time when we consider Einstein's theory of relativity in Chapter 19.2 As mentioned above, kinematics is the study of the relationships between position, velocity, acceleration, and time. Galileo and his dis- ciples grappled with these concepts. Fortunately, we can use many powerful mathematical tools that were not available to Galileo. For ex- ample, the calculus was invented by Newton and Leibniz after Galileo's death, and vector analysis was not developed until about 150 years ago. 2.3. Basic Concepts in Kinematics Let me remind you that the general relations between position (r), velocity (v), and acceleration (a) are given by v = d r = r ̇ , dt a = d v = v ̇ = ̈r . dt One-dimensional motion refers to any kind of motion for which the position can be described in terms of a single parameter. For example, the position of a car driving down a curving road can be specified by its distance from the starting point. Similarly, the position of a bead sliding along a wire can be specified by the distance to the bead from some given point on the wire. A single number is enough to 2Some physicists believe that time, in the Newtonian sense of one event happen- ing before another event, does not apply in the realm of quantum mechanics. For an interesting discussion of this problem see the article by Charles Seife, "Quantum Physics: Spooky Twins Survive Einsteinian Torture," Science, 294, 1265, (2001). 2.3. BASIC CONCEPTS IN KINEMATICS 33 completely specify the position of the bead, no matter how the wire may be twisted or curled. However, the linear distance may not be the most convenient parameter to use. For example, the position of a bead sliding on a circular hoop might best be described in terms of an angle. The simplest example of one-dimensional motion is a particle mov- ing along a straight line, such as the x-axis. The acceleration is a=dv. dt Here I wrote a and v rather than a and v because if the motion is one- dimensional there is no need to use vectors. Multiplying both sides by dt gives the following very simple differential equation: dv = adt. To solve, you integrate both sides: vf tf dv = adt, v0 where the limits indicate that the time runs from t0 to tf (that is, "initial" time to "final" time), and the initial and final values of the velocity are v0 and vf . The quantity t0 represents the time when the particle was at the initial position x0. Since the starting time for any process is arbitrary (it just depends on when you push the button on the stopwatch) it is nearly always set equal to zero (t0 = 0). Now you can go no further. You cannot evaluate the integral adt because you do not have an expression for the acceleration as a function of time. But if you do know the acceleration as a function of time, a = a(t), then you can carry out the integration. When you do so, you will obtain an expression for the velocity as a function of time,3 v = v(t). Once you have determined the velocity, you can continue the analysis and write dx = v(t)dt, which follows immediately from the definition of velocity. Integrating once again, you will obtain an expression for the posi- tion as a function of time, or, more explicitly, x = x(t), x = x(t, v0, x0). t0 3To be precise, this will give you an expression for the velocity in terms of the time, the initial time and the initial velocity, v = v(t, t0, v0). 34 2. KINEMATICS 2.3.1. Motion in One Dimension with Constant Accelera- tion. As discussed in Section 1.1.1, if the acceleration is constant, we can integrate dv = adt to obtain and v(t) = v0 + at, (2.1) x(t) = x0 + v0t + 1at2. (2.2) 2 I am sure you are very familiar with Equations (2.1) and (2.2) from your introductory physics course. In fact, you probably memorized them. However, it is more important at this stage in your physics career to understand the process for obtaining these equations. Exercise 2.1. Starting with x = v0t + 1 at2 and v = v0 + at, obtain 2 x = 1(v + v0)t, 2 and Are these expressions valid if the acceleration is not constant? the relations 2ax = v2 − v02. (2.3) (2.4) I must emphasize that the formulas obtained above (Equations 2.1 through 2.4) are only valid for constant acceleration. If the acceleration is a function of time or velocity or position, then you must go through the entire procedure to obtain the correct expressions for v(t) and x(t). 2.3.2. Projectile Motion (Motion in a Plane). A projectile is any object that is launched with some initial velocity and then moves under the action of the gravitational force which imparts to it a con- stant downward acceleration. The horizontal motion is motion at constant velocity, and therefore x = x0 + v0xt, and the vertical motion is motion at constant acceleration, and there- fore y = y0 + v0yt − 1gt2, 2 where g = 9.8 m/s. It is helpful to remember that at the top of the trajectory the ver- tical velocity (vy) is zero. This gives us a useful way to determine 2.3. BASIC CONCEPTS IN KINEMATICS 35 the time for the projectile to reach the top of its trajectory. Since vy = v0y + ayt we have so 0 = v0y − gttop ttop = v0y/g. Worked Example 2.1. Consider the projectile motion of a cannon ball. Assume the initial speed is v0 and the cannon is aimed at an angle θ above the horizontal. Determine the range. Solution: The range (R) is the horizontal distance traveled by the projectile along a flat horizontal plane; it is the value of x when y = 0. The initial velocity components are v0x = v0 cos θ and v0y = v0 sin θ. Since the total time of flight is twice the time required to reach the top of the trajectory, the range is given by x(t = 2ttop). It is easy to obtain a formula for the range in terms of the initial velocity as follows: vy = v0y − gt. But vy = 0 when t = ttop so 0 = v0y −gttop = v0sinθ−gttop. Consequently, Therefore, ttop = (v0 sin θ) /g. R = v0x(2ttop) = 2v0 cos θ (v0 sin θ) /g, v02 R = 2 g sinθcosθ. (2.5) Our analysis of projectile motion made two assumptions: (1) The acceleration of gravity is constant, and (2) There are no horizontal forces acting on the projectile. These assumptions mean that we are assuming a flat, non-rotating Earth, and that there is no air resis- tance acting on the body. These seemingly preposterous assumptions (a flat, motionless, airless Earth!) illustrate an important technique in physics. When faced with a difficult problem, we solve a similar, simpler problem. Frequently, the solution to the simple problem is sufficiently accurate. The solution of the simple problem is often re- ferred to as the "zeroth order" approximation to the solution of the real problem. Later in this course you will learn techniques for includ- ing the factors that were ignored in the zeroth order approximation to get results that are nearer and nearer to the exact solution. You will appreciate more fully how this technique works when you study projec- tile motion on a rotating Earth in Chapter 13. As you can imagine, a 36 2. KINEMATICS working physicist is continually faced with very complicated problems. It is part of a physicist's education to learn what is fundamentally im- portant and what can be left until later as a refinement (or "higher order approximation"). For example, you can estimate the trajectory of a thrown baseball with reasonable accuracy by assuming a flat, non- rotating Earth, but if you want to calculate the trajectory of a rocket, those factors will have to be included. Exercise 2.2. Determine the elevation angle for a projectile such that the range will be maximized. (Hint: the maximum value of a function is determined by setting its derivative to zero.) Answer: 45◦. Exercise 2.3. An airplane traveling at 900 km/hr at 5,000 m altitude is directly over the target when it drops a bomb. How far from the target does the bomb hit? Answer: ≃ 8 km. In working problems involving projectile motion in a uniform grav- itational field, keep in mind that the horizontal motion has constant velocity (vx = constant) and the vertical motion has constant accel- eration (ay = constant). Projectile motion is a combination of two independent one-dimensional motions. These are related to one an- other through the time. If you eliminate the time from the equations, you obtain an equation for the trajectory of the projectile as shown in the following example. Worked Example 2.2. Prove that the path of a projectile in a uniform gravitational field is a parabola Solution: The equation of a parabola whose axis is parallel to the y axis in Cartesian coordinates is y = ax2 + bx + c. In section 2.3.2 we showed that for a projectile Therefore, y = −1gt2 + v0yt + y0, 2 x = v0xt + x0. t = x − x0 , v0x 2.3. BASIC CONCEPTS IN KINEMATICS 37 and consequently g(x−x0)2 x−x0 y = −2 v2 + v0y v + y0, 0x 0x y=− g (x2−2xx0+x20)+v0yx−v0yx0, 2v02x v0x v0x y=− g x2+[( g 2x0+v0y)]x+(−gx20 −v0yx0+y0). y=− g x2+v0yx. 2v0x v0x 2v02x 2v0x v0x 2v02x v0x This has the required form as can more easily be appreciated if we setx0 =0andy0 =0toobtain Although projectile problems tend to be straightforward, sometimes it is necessary to use a more subtle approach, as illustrated in the following example. Worked Example 2.3. A cannon that fires shells with a muz- zle speed v0 is mounted on the top of a cliff a height h above a level plain. Show that the angle at which the cannon should be aimed to give maximum range is θ = sin−1(v0/ 2(v02 + gh)). Solution: I am including this problem to show that a seem- ingly simple projectile problem can require considerable ingenuity to obtain a solution. The obvious approach would be to attempt to solve the problem by using the fact that if t is the time of flight (tf), then the value of x is the range (R) and the value of y is −h (where we assume that the coordinate origin (0,0) is at the cannon). That is R = (v0 cosθ)(tf), −h = (v0 sin θ) (tf ) − 1g(tf )2. 2 You can solve the first equation for tf , obtaining tf = R/(v0 cos θ). Plugging into the second equation leads to gR2 1 − h = R tan θ − 2v02 cos2 θ . (2.6) Now, you could solve this quadratic equation to obtain an expres- sion for R as a function of θ. Then taking the derivative of R with respect to θ and setting it equal to zero gives an expression for θ 38 2. KINEMATICS that maximizes R. However, the resultant expression is so compli- cated that you will find it extremely difficult to solve for θ. (You might want to give it a try.) However, the problem can be s

drag equation

R = ½ρCAv2

newtons second law

The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object

newtons first law

an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force

newtons third law

every action has an equll and opposite reaction.

newtons second law equation

f=ma

momentum

p = mv