Dynamics & Relativity (Part 1) 2024-2025 PDF

Dynamics & Relativity (part 1) Judd Harrison University of Glasgow September 22, 2024 ii For 2024-25 These notes cover the first part of the Dynamics and Relativity course. They are essentially the same as the notes for the previous year, kindly provided by Chris B...

Dynamics & Relativity (part 1) Judd Harrison University of Glasgow September 22, 2024 ii For 2024-25 These notes cover the first part of the Dynamics and Relativity course. They are essentially the same as the notes for the previous year, kindly provided by Chris Bouchard, with some minor alterations. I’ve set up a Padlet page for us to use in an effort to encourage discussion. Post your questions about the course, or physics more generally (anonymously if you like) in Padlet and I’ll answer them there. Figure 1: To access our Padlet page either scan this QR code or visit https://padlet.com/juddharrison/uofg-physics-1-d-r1-2024-2025-4jj938krdpta3or8. The password is "D&R1padlet" iii iv FOR 2024-25 Notation and Units 0.1 Vectors and components We will indicate a vector using boldface and an arrow above, e.g., \myvec v, with x, y, and z components given by v_x,\ v_y,\ \text {and}\ v_z. The components of the position vector ⃗ r will simply be x, y, and z. The unit vectors along these directions are \ihat ,\ \jhat ,\ \text {and}\ \khat , respectively. We will discuss vectors representing things like position and velocity, which will be functions of time, ⃗ r(t) and ⃗v(t). A space-saving notation makes time-dependence implicit, as in \myvec r = \myvec r(t). When a specific time is used for the evaluation, e.g., t = t0 , a subscript will be used as in \myvec r_0 = \myvec r(t_0), \qquad \myvec r_1 = \myvec r(t_1), \qquad {\rm etc.} This means that the components of a vector will potentially have two subscripts, one specifying which component and another for the time, e.g., \text {$x$ component of }\, \myvec v(t_0) &= v_{x0} = v_{0x}. The order of the subscripts is irrelevant. 0.2 Uncertainty Uncertainty is an inherent part of any experiment and without experiment, sci- ence is just math. You can’t do science without addressing uncertainty. We’ll v vi NOTATION AND UNITS indicate uncertainty in either of the following two ways, here giving the mass of the Earth, M_\oplus &= (5.9726 \pm 0.0007) \times 10^{24} \ {\rm kg}, \\ &= 5.9726(7) \times 10^{24} \ {\rm kg}. If we don’t need to be precise and give the uncertainty, but still want to indicate that we aren’t using exact values, we will use the approximately equals sign as in M_\oplus \approx 6\times 10^{24}\ {\rm kg}. We can be even less precise and use an order of magnitude for the value, which means the nearest power of 10, M_\oplus \sim 10^{25}\ {\rm kg}. 0.3 Units We will work exclusively in the SI (meters, kilograms, seconds,... ) system of units discussed in Appendix A. Contents For 2024-25 iii Notation and Units v 0.1 Vectors and components....................... v 0.2 Uncertainty............................... v 0.3 Units.................................. vi 1 Introduction 1 1.1 Applicability.............................. 2 1.2 An aside................................ 3 2 Non-Circular Motion 5 2.1 Definitions: position, velocity, and acceleration.......... 5 2.2 One-dimensional motion....................... 9 2.2.1 Uniform acceleration..................... 9 2.2.2 Freely falling body...................... 11 2.2.3 What if acceleration is NOT uniform?........... 13 2.3 Two-dimensional motion....................... 13 2.3.1 Projectile motion....................... 13 2.3.2 Relative velocity....................... 16 2.4 Plotting position and velocity versus time............. 19 3 Circular Motion 21 3.1 Uniform circular motion....................... 22 3.2 Non-uniform circular motion..................... 27 4 Force and Newton’s Laws 29 4.1 Newton’s First Law.......................... 31 4.2 Newton’s Second Law........................ 31 4.3 Newton’s Third Law......................... 33 4.4 Free body diagrams.......................... 33 4.5 Friction................................. 35 5 Gravity 37 vii viii CONTENTS 6 Simple Harmonic Motion 39 6.1 Definitions............................... 39 6.2 Mass on a Spring........................... 40 6.3 The simple pendulum......................... 42 Index 44 A SI Units 47 B Vector Notation and Operations 49 C Non-circular motion problems 51 D Circular motion problems 65 E Friction problems 73 Chapter 1 Introduction These are notes for lectures given as part of the first half of the Dynamics & Relativity portion of the first year course, Physics 1. Their purpose is to fo- cus our discussion on those aspects of dynamics you are expected to learn as part of your studies. They represent a minimum amount of physics you should digest over the course of our lectures. You are encouraged, in fact you are expected, to dig deeper and follow your own curiosity beyond the outline they provide. For this, you may find it useful to consult the text for this course, University Physics. To help with this, when we discuss a topic in these notes, I’ll try to point you to the relevant bits in the textbook. An occasion- ally irreverent take on the physics discussed throughout your first couple of years at university can be found in The Feynman Lectures on Physics. I highly recommend these, especially the audio version which is freely available at feynmanlectures.caltech.edu/flptapes.html! Figure 1.1: Newton contemplates the fall of an apple and develops a description of motion still in use today to understand and explore the cosmos. Images from The College of William & Mary (left) and NASA’s JPL (right). 1 2 CHAPTER 1. INTRODUCTION Figure 1.2: Range of known distance scales in our universe. For an interactive and more detailed tour, visit scaleofuniverse.com. 1.1 Applicability We will study the motion of objects as first developed by Newton , famously inspired by a falling apple. From this romantic beginning the study has flourished and found application well beyond what Newton was likely able to imagine, Fig. 1.1! In the first half of this course we’ll limit ourselves to studying the motion of objects that move slowly — by this I mean motion at speeds that are much less than the speed of light , c = 299\,792\,458\ {\rm m/s}. (1.1) We’ll also restrict ourselves to the study of distance scales that are large — by this I mean that the objects we study will be larger, and the distances they travel will be further, than roughly the size of an atom, a distance characterized by the so-called Bohr radius , a_0 \sim 10^{-10}\ {\rm m}. (1.2) We also need to avoid situations where objects are too massive. Mass warps space and time, inflicting Newton’s equations with small discrepancies. Earth is sufficiently massive that these effects (google GPS to see a practical example) are measurable as very small corrections to what Newton would’ve predicted! We should limit ourselves to masses that are less than the mass of the Earth, M_{\oplus } = 5.9726(7) \times 10^{24} \ {\rm kg}, (1.3) where the number in parentheses is the error in the last digit. 1.2. AN ASIDE 3 Technically, we don’t need to limit ourselves in these ways. The physics of fast moving objects (special relativity) and the physics of objects over small distances (quantum mechanics) are both known, you’ll learn about them each during your time at the University of Glasgow. The combined effects of both fast and small (special relativity plus quantum mechanics) is called quantum field theory, and you may learn about it too, depending on your course of study. We routinely use quantum field theory to successfully describe the physics of fast and small things in collisions at the Large Hadron Collider. We also understand the physics of very massive objects1. However, special relativity, quantum mechanics, and general relativity are too difficult to use as a starting point, so we will start with the physics of relatively slow, big, and not too massive objects. This may seem like we’re boxed into a small corner, but don’t despair! Light is fast, atoms are small, and Earth is massive! Most of what we observe in our everyday lives easily obeys these conditions. You may therefore find that you have an intuition for many of our discussions. This is convenient, but don’t let it lull you into a false sense of security! Nature, it turns out, has some incredibly interesting and rather unintuitive aspects to her behavior for speeds, distances, and masses with which we are unfamiliar. While we won’t deal with these topics in the first half of this course, keep this warning in mind as you continue learning physics. Many great physicists have fallen victim to the bravado that we know all there is to know, only to be embarrasingly disproven when experiment provided a deeper look into the workings of Nature. 1.2 An aside Before we get down to the business of studying in detail the not-too-fast motion of not-too-small and not-too-massive objects over not-too-short distances, a subject often referred to as Classical (or Newtonian) Mechanics, let me first address something that might be bothering you at this point — it bothered me. Because what we’re about to discuss must eventually be corrected to allow for very fast moving objects, very massive objects, or motion over atomic or shorter distances, that means what we are about to study is merely an approximation of the truth. Why would we do this? Out of some respect for the past and the order in which things were figured out? Wouldn’t it be better to just learn things correctly from the beginning? The answers to these question are important and I think it’s worth pointing them out right here in the beginning. First, we don’t (and may never) know “the truth!” Human beings have yet to develop a coherent description of Nature valid for all speeds, masses, and dis- tances. Maybe the best we can realistically hope for are a set of approximations 1 Interestingly, what is not yet known is how to combine the physics of fast, small, and massive objects. In fact, we can’t even combine the physics of short distances and large masses into a quantum theory of gravity. 4 CHAPTER 1. INTRODUCTION valid over the range of distances, speeds, and masses for which we are capable of probing Nature, never knowing what we may find if only we could look a little deeper inside the microscopic structure of matter or space-time, or gaze further into the universe at ever larger structures of the cosmos. With this possibility in mind, it makes sense to learn first the simplest approximations, those valid in our everyday lives. Second, the approximations we are about to learn are very good ones — provided they’re applied when the conditions outlined above are met. Note that when you learn how to improve upon these approximations, which will begin in the second half of this semester with the study of special relativity, you should take care to note the uncertainties introduced when the approximations are made. Approximations are an integral part of doing physics, but unless you know how to assess the accuracy of predictions when approximations are used, the predictions aren’t very useful. I will try to make the third point by way of analogy. Imagine we wanted to predict what would happen to the temperature in our classroom if the pressure were to increase. We could first realize that the physics at play involves a staggering number of interacting air molecules, each of which is built of atoms. Each atom, in turn, consists of a cloud of electrons surrounding a nucleus, with the electrons within each atom, and those within different atoms, interacting with one another by exchanging photons. Each nucleus is itself composed of even smaller constituents, called neutrons and protons, and these are built of quarks and gluons. An ultimate description of the temperature in our classroom would then be cast in terms of the quarks, gluons, electrons, and photons. However, even with the help of the most powerful supercomputers in existence today, we are incapable of manipulating the fundamental theory describing these interactions, the Standard Model of particle physics, to reproduce the equivalent of the ideal-gas law, developed by Clapeyron in 1834! The ability to identify the relevant degrees of freedom for the problem at hand is an incredibly powerful simplifying tool! Details of the interactions among the quarks, gluons, electrons, and photons are unnecessary if our goal is to describe macroscopic properties of the gas in the classroom, like temperature. We need only consider an effective theory written in terms of the relevant degrees of freedom, those at play over distance scales commensurate with the quantity we wish to calculate. In this case, the effective theory is a statistical description of the gas in the classroom written in terms of idealized, point-like molecules that bounce off one another like little billiard balls! Molecular structure, and all the detailed complications that entails, results in only small corrections to the ideal-gas equation. Chapter 2 Non-Circular Motion We begin with a study of problems where particles aren’t constrained to move in a circle. These problems are most conveniently discussed using cartesian coordinates: x, y and z. 2.1 Definitions: position, velocity, and acceleration We define the position of an object at point P by the vector ⃗ r that begins at the origin and extends to P. In three dimensions and cartesian coordinates this is represented by, \myvec r = x_P\, \ihat + y_P\, \jhat + z_P\, \khat , \label {eq:r} (2.1) and is depicted in Fig. 2.1. For a brief summary of vector notation and operations, see Appendix B. Because our goal is to study the motion of objects as a function of time we treat ⃗ r as an unknown function of time, ⃗ r(t). Many of the problems we solve in this class will involve finding ⃗ r at a particular time. This means, of course, Figure 2.1: The position ⃗ r of point P is defined relative to the origin. 5 6 CHAPTER 2. NON-CIRCULAR MOTION Figure 2.2: The position vector ⃗r(t) varies with time as the object moves along the red trajectory from time t0 to a later time t. that the components of ⃗ r are functions of time as well, and we will in general have xP (t), yP (t), and zP (t). The static picture of Fig. 2.1 is replaced by the dynamical situation depicted in Fig. 2.2. To characterize this unknown function r(t), we can Taylor series1 expand about an initial time t 0 , ⃗ \myvec {r}(t) = \myvec {r}(t_0) + \frac {d\myvec {r}(t_0)}{dt}\, (t-t_0) + \frac {1}{2} \frac {d^2 \myvec {r}(t_0)}{dt^2}\, (t-t_0)^2 + \frac {1}{6} \frac {d^3 \myvec {r}(t_0)}{dt^3}\, (t-t_0)^3 + \dots , \label {eq:expansion} (2.2) where I’ve used the shorthand notation, d⃗ r(t 0 )/dt = d⃗r(t)/dt¯ t= t0 , and similarly ¯ for higher-order derivatives. The unknown function ⃗ r(t) is then determined by the times, t and t 0 , its initial value ⃗ r(t 0 ), and the derivatives d⃗ r(t 0 )/dt, d 2⃗ r(t 0 )/dt2 , d 3⃗ r(t 0 )/dt3 ,... We define velocity ⃗ v to be the first of these derivatives of position with respect to time ,2 d⃗ r ⃗ v = (2.3) dt dxP d yP dzP = ı̂ + ȷ̂ + k̂. (2.4) dt dt dt 1 The Taylor series expansion of f (x) about point x = a, is given by f(x) &= \sum _{n=0}^\infty \frac {1}{n!} \left.\frac {d^nf(x)}{dx^n}\right |_{x=a} (x-a)^n \\ &= f(a) + \frac {1}{1!}\left.\frac {df(x)}{dx}\right |_{x=a} (x-a) + \frac {1}{2!}\left.\frac {d^2f(x)}{dx^2}\right |_{x=a} (x-a)^2 + \dots , where 0! = 1 and d 0 f (x)/dx0 = f (x). For the special case a = 0, this is known as the Maclaurin series. This expansion exists provided the function f (x) is differentiable. 2 Note that you may occasionally see the derivative with respect to time denoted by a dot,... i.e. ⃗ r = d⃗ r = d 2⃗ r/dt, ⃗ r/dt2 , and so on. 2.1. DEFINITIONS: POSITION, VELOCITY, AND ACCELERATION 7 Velocity is a derivative, which means it’s obtained by taking the limit of a ratio of differences. Using the positions shown in Fig. 2.2, the velocity at time t0 would be given by, \myvec v(t_0) = \lim _{t \to t_0} \frac {\myvec r(t) - \myvec r(t_0)}{t - t_0}. \label {eq:vderiv} (2.5) Note that it is only a function of t0. That is, it is an instantaneous quantity, meaning that it’s defined for a single instant of time. If you knew the positions at times t and t0 , you could approximate the velocity from the differences in Eqn. (2.5), without taking the limit, \myvec v(t, t_0) \approx \frac {\myvec r(t) - \myvec r(t_0)}{t - t_0} = \frac {\Delta \myvec r}{\Delta t}, \label {eq:vav1} (2.6) but this would only be approximate. It depends on two times, t0 and t, and is sometimes written using the shorthand notation with ∆ indicating a difference. We will occasionally discuss an average, as opposed to instantaneous, velocity. If you knew the instantaneous velocities ⃗ v(t 0 ) and ⃗ v(t), the average velocity can be calculated exactly the way you’d think to calculate it, \myvec v_{\rm av} = \frac {1}{2}\big (\myvec v(t_0) + \myvec v(t)\big ). \label {eq:vav2} (2.7) We often speak of speed , referring to the magnitude of the velocity, v = \sqrt {v_x^2 + v_y^2 + v_z^2}. \label {eq:speed} (2.8) Acceleration is defined as the second-order time derivative of position, d 2⃗r ⃗ a = 2 (2.9) dt d 2 xP d 2 yP d 2 zP = ı̂ + ȷ̂ + k̂. (2.10) dt2 dt2 dt2 Note that acceleration can equivalently be written as the first-order time-derivative of ⃗ v, d⃗v ⃗ a = (2.11) dt dv x dv y dv z = ı̂ + ȷ̂ + k̂. (2.12) dt dt dt The derivative in Eqn. (2.11) is defined by the limit, \myvec {a}(t_0) = \lim _{t \to t_0} \frac {\myvec v(t) - \myvec v(t_0)}{t-t_0}. (2.13) 8 CHAPTER 2. NON-CIRCULAR MOTION Just as we did with velocity, we can also speak of an approximate acceleration, defined by the difference of velocities, \myvec a(t, t_0) \approx \frac {\myvec v(t) - \myvec v(t_0)}{t-t_0} = \frac {\Delta \myvec v}{\Delta t}, \label {eq:avg_a} (2.14) and an average acceleration, calculable if you know ⃗ a(t 0 ) and ⃗ a(t), \myvec a_{\rm av} = \frac {1}{2}\big (\myvec a(t_0) + \myvec a(t)\big ). (2.15) We should, in principle, continue this description of ⃗ r(t) by defining higher and higher order time-derivatives, ad infinitum. For example, the third-order derivative of ⃗ r(t) is called the jerk 3 ! In this course (unless explicitly stated otherwise) we will consider only situations with uniform acceleration, meaning the acceleration is constant, not a function of time, 0 = \frac {d\myvec {a}}{dt} = \frac {d^3\myvec {r}}{dt^3}. (2.16) There is no jerk in this class. This also ensures all higher-order time derivatives of ⃗ r in Eqn. (2.2) are zero, since they can be written as a time derivatives of ⃗a. That is, for n > 2, \frac {d^n \myvec r}{dt^n} = \frac {d^{n-2} }{dt^{n-2}}\ \frac {d^2 \myvec r}{dt^2} = \frac {d^{n-2} \myvec a}{dt^{n-2}} = 0. (2.17) Therefore, using our definitions of velocity and acceleration, and our assumption that acceleration is uniform, we arrive at the following expression for ⃗ r(t), \myvec {r}(t) = \myvec {r}(t_0) + \myvec {v}(t_0)\cdot (t-t_0) + \frac {1}{2}\cdot \myvec {a}\cdot (t-t_0)^2. \label {eq:3dmotion} (2.18) I’ve inserted dots in Eqn. (2.18) for multiplication only to keep you from thinking, for example, that ⃗ a is a function of (t − t 0 ). A less explicit notation, but one that we will use, writes Eqn. (2.18) as, \myvec {r} = \myvec {r}_0 + \myvec {v}_0 (t-t_0) + \frac {1}{2} \myvec {a} (t-t_0)^2, \label {eq:3dmotion_simple} (2.19) where time-dependence is just not written, so there’s no confusing what is meant by ⃗ a(t − t 0 ). In this notation, when we evaluate something at a particular time, we’ll indicate this with a subscript, for example, ⃗ v(t 0 ). v0 = ⃗ 3 Some higher order derivatives are snap (fourth), crackle (fifth), pop (sixth), lock (sev- enth), and drop (eighth). 2.2. ONE-DIMENSIONAL MOTION 9 2.2 One-dimensional motion After introducing the main characters (⃗ r, ⃗ v, and ⃗ a) in their full three-dimensional glory, we simplify now to one dimension to study relations between these quanti- ties and perform some calculations without the added difficulty of three dimen- sions. 2.2.1 Uniform acceleration Let’s assume for the moment that we have motion only along the x direction: \text {position: } &\myvec r = x\, \ihat , \\ \text {velocity: } &\myvec v = v_x\, \ihat = \frac {dx}{dt}\, \ihat , \label {eq:vx} \\ \text {acceleration: } &\myvec a = a_x\, \ihat = \frac {dv_x}{dt}\, \ihat = \frac {d^2 x}{dt^2}\, \ihat. \label {eq:ax} (2.22) Note that we could have just as easily assumed motion along the y or z direction and the analogous expressions (with ȷ̂ s and y subscripts or k̂s and z subscripts) would apply. You’ll see problems with motion along these other directions as well. We assume uniform acceleration (remember, we’ll almost always assume this, and when we don’t I’ll be sure to say so). This lets us do something pretty simple in Eqn. (2.22) (which is nothing but the one-dimensional version of the definition of acceleration), a_x &= \frac {dv_x}{dt}, \label {eq:approach} \\ \text {constant $a_x$}\ \ &\Longrightarrow \ \ a_x \int _{t_0}^t dt' = \int _{v_{0x}}^{v_x} dv'_x, \label {eq:constanta} \\ \Longrightarrow \ \ \Aboxed { v_x &= v_{0x} + a_x (t-t_0)\. } \label {eq:1deom1} (2.25) Eqn. (2.25) (see Eqn. (2.8) and the surrounding discussion in the textbook) is one of a handful of particularly useful relations that we’ll use for one-dimensional motion with uniform acceleration.4 I put a box around these expressions. Given the time-dependence for v x in Eqn. (2.25), we can carry out a similar game by integrating the expression within Eqn. (2.21) (which is nothing but the 4 A note on notation; v is being used here to stand for v (t ), similar to the notation we 0x x 0 used in Eqn. (2.19). 10 CHAPTER 2. NON-CIRCULAR MOTION one-dimensional version of the definition of velocity), v_x &= \frac {dx}{dt}, \\ \text {Eqn.~(\ref {eq:1deom1}) for $v_x$}\ \ &\Longrightarrow \ \ \int _{t_0}^t \big (v_{0x} + a_x(t'-t_0)\big ) dt' = \int _{x_0}^x dx',\\ \text {constant $v_{0x}$, $a_x$, and $t_0$}\ \ &\Longrightarrow \ \ v_{0x}(t-t_0) + a_x \left [\frac {1}{2} (t^2 - t_0^2) - t_0 (t-t_0)\right ] = x-x_0 \\ \Longrightarrow \ \ \Aboxed { x &= x_0 + v_{0x}(t-t_0) + \frac {1}{2} a_x (t-t_0)^2 \ , } \label {eq:1deom2} (2.29) which corresponds to Eqn. (2.13) in the textbook. Notice that we have simply recovered the one-dimensional version of the series expansion for ⃗r in Eqn. (2.18). This shouldn’t be too surprising, since all we’ve done is play around with the same definitions of velocity and acceleration that were used to write down Eqn. (2.18). If we had ended up with something different, there would’ve been trouble! Eqns. (2.25) and (2.29) contain basically all the information we’ll need to solve all the problems of this section, but we can make our life a little easier by combining them in a couple more ways. For example, we can combine them to eliminate any dependence on t − t0 , which could be useful in problems where times aren’t given to us. To do this, we square Eqn. (2.25), v_x^2 &= \big ( v_{0x} + a_x(t-t_0) \big )^2, \\ &= v_{0x}^2 + 2a_x \left ( v_{0x} (t-t_0) + \frac {1}{2} a_x (t-t_0)^2 \right ), \\ \Longrightarrow \ \ \Aboxed { v_x^2 &= v_{0x}^2 + 2a_x (x-x_0) \ ,} \label {eq:1deom3} (2.32) where the last step uses Eqn. (2.29). This expression is given in the textbook in Eqn. (2.13). We can also combine Eqns. (2.25) and (2.29) to eliminate acceleration, for problems that don’t give us acceleration. To do this we start with Eqn. (2.25), x-x_0 &= v_{0x}(t-t_0) + \frac {1}{2} a_x (t-t_0)^2, \\ &= \big ( v_{0x} + \frac {1}{2} a_x (t-t_0) \big ) (t-t_0), \\ \text {using Eqn.~(\ref {eq:1deom1})}\ \ &\Longrightarrow \ \ x-x0 = \big ( v_{0x} + \frac {1}{2}(v_x-v_{0x}) \big ) (t-t_0) \\ \Longrightarrow \ \ \Aboxed { x-x_0 &= \frac {1}{2}(v_{0x} + v_x) \,(t-t_0) \. }\label {eq:1deom4} (2.36) Notice that (v0 x + v x )/2 is just the average x-component of the velocity over the time range from t0 to t, i.e. the one-dimensional version of Eqn. (2.7), and we 2.2. ONE-DIMENSIONAL MOTION 11 see that Eqn. (2.36) is nothing but the one-dimensional version of Eqn. (2.6), i.e. v x, av = ∆ x/∆ t. I point out that we keep arriving at expressions that are equivalent to what we started with just to give you a warm fuzzy feeling that what we’re doing all holds together. From a practical point of view it’s not necessary that you recognize all these equivalences. Though of course, it should help your mental organization if you realize that there are only a few definitions at play (⃗ r, ⃗ v, and a) and one assumption (uniform acceleration). From the practical perspective, ⃗ we need to be able to apply the boxed equations to problems. To help with this let’s look at an example, a freely falling body. 2.2.2 Freely falling body We start by making approximations to make the problems easier. First, we assume the only acceleration around is that due to Earth’s gravity.5 This means we’ll ignore gravitational acceleration coming from everything else in the universe (the moon, the sun, Pluto,...). It turns out, because of the large distances to this other stuff, that this is a pretty good approximation. It also means we’ll ignore other non-gravitational sources of acceleration, including the rotation of the Earth and friction from air resistance. This last approximation, ignoring friction, is actually not a very good one, so we’ll come back to it later and figure out how to account for it. Second, we’ll assume the acceleration from Earth’s gravity is constant. In reality, as you move further from the center of the Earth, the gravitational attraction becomes weaker and the resultant acceleration is smaller. However, because the Earth is so large compared to the distances our cannonballs and the like will be falling, any change in g is so small that we neglect it. The Earth’s gravitational acceleration is g \approx 9.8\ {\rm m/s}^2 \ \ \text {(pointing toward the surface of the Earth)}. (2.37) If you’re working a problem in the book, you may be given a slightly more precise value for g. If this happens be sure to use the value given in the problem. In order to see a decrease of g from 9.8 m/s2 to 9.7 m/s2 , you’d have to travel about 20 miles away from the surface of the Earth, into the stratosphere! Imagine Galileo dropping an apple from the top of the leaning tower of Pisa. Let’s define the y axis to point up, away from the surface of the Earth, and define the initial time, initial y position, and initial velocity to all be zero. The problem, shown in Fig. 2.3, is for us to calculate the y position of the apple, and the y component of the apple’s velocity at a few different times during its fall. We start by realizing that this is one-dimensional motion, albeit along the y axis. This means we can use, with the simple change x → y, any of the boxed equations from the last section. What is the acceleration of the apple? We 5 We’ll discuss the relation between forces like gravity and the acceleration they produce when we talk about Newton’s Laws a few lectures from now. 12 CHAPTER 2. NON-CIRCULAR MOTION know the apple is undergoing free fall, which means it’s only being accelerated by gravity. But notice the direction in which the apple is accelerating (which is also the direction it’s being pulled) is downward, which means in the negative y direction, so a_y = -g. (2.38) We want to know y and v y and we know the acceleration and times, so we can most easily use Eqns. (2.25) and (2.29). Noting that v0 y = 0 m/s, t0 = 0 s, and y0 = 0 m (initial conditions like this are often given implicitly in the statement of the problem), we have v_y &= -gt , \\ y &= -\frac {1}{2} g t^2. (2.40) Plugging in the times we’re interested in gives, y_1 = -4.9\ {\rm m}, \qquad & \qquad v_{1y} = -9.8\ {\rm m/s}, \nonumber \\ y_2 = -19.6\ {\rm m}, \qquad & \qquad v_{2y} = -19.6\ {\rm m/s}, \nonumber \\ y_3 = -44.1\ {\rm m}, \qquad & \qquad v_{3y} = -29.4\ {\rm m/s}. (2.41) After any calculation you should always make a few sanity checks, verifying that: numbers have the proper units, the signs of the numbers make sense, and the magnitudes of the numbers is reasonable. Figure 2.3: An application of a freely falling body. 2.3. TWO-DIMENSIONAL MOTION 13 2.2.3 What if acceleration is NOT uniform? This will be the exception and not the rule, but we should have some idea of how to handle a problem where acceleration changes with time. The approach, stick- ing with one dimension, starts similarly enough to our approach in Eqn. (2.23), with the definition of acceleration. The difference shows up in Eqn. (2.24), where we can no longer move a x outside of the integral over time, since it’s now a function of time. We can still write the equation in terms of the integral, giving a_x &= \frac {d v_x}{dt} , \\ \Longrightarrow \ \ \int _{v_{0x}}^{v_x} dv'_x &= \int _{t_0}^t a_x\, dt' , \\ \Longrightarrow \ \ \Aboxed {v_x - v_{0x} &= \int _{t_0}^t a_x\, dt' \. } \label {eq:genv} (2.44) We then do the same thing to arrive at an expression for the position, v_x &= \frac {dx}{dt} , \\ \Longrightarrow \ \ \int _{x_0}^x dx' &= \int _{t_0}^t v_x\, dt' , \\ \Longrightarrow \ \ \Aboxed {x - x_0 &= \int _{t_0}^t v_x\, dt' \. } \label {eq:genx} (2.47) If you’re given the time-dependence of a x , plugging it into Eqn. (2.44) and doing the integral will give you the time-dependence of v x , which you can then plug into the Eqn. (2.47) and, after doing one more integral, arrive at the time- dependence of x. This is exactly what we did in Section 2.2.1, where a x had a very simple time dependence (it was constant). 2.3 Two-dimensional motion Motion in two dimensions isn’t really any more difficult than motion in one dimension. We simply apply the expressions for uniform acceleration in one dimension to each of the dimensions in which the object travels. That is, we do the same thing as in one dimension, we just do it twice. 2.3.1 Projectile motion Projectile motion is a lot like the motion of freely falling bodies. The big dif- ference is that with projectile motion, our falling objects are given some initial velocity. This initial velocity can result in our falling body travelling not only 14 CHAPTER 2. NON-CIRCULAR MOTION downward, but also moving upward and horizontally — think cannonballs and take a look at Fig. 2.4. Let’s work out the expressions for x, v x , y, and v y in the case of projectile motion. We begin with the x and y versions of the boxed expressions for uniform acceleration, x &= x_0 + v_{0x} (t-t_0) + \frac {1}{2} a_x (t-t_0)^2, \\ v_x &= v_{0x} + a_x (t-t_0), \\ y &= y_0 + v_{0y} (t-t_0) + \frac {1}{2} a_y (t-t_0)^2, \\ v_y &= v_{0y} + a_y (t-t_0). (2.51) In projectile motion, as with freely falling bodies, the only acceleration involved is that from gravity. This means, again with the y-axis pointing up, a_x = 0\ {\rm m/s}^2 \qquad \text {and} \qquad a_y = -g, (2.52) and we have the generic expressions for projectile motion \boxed { \begin {aligned} x &= x_0 + v_{0x} (t-t_0), \\ v_x &= v_{0x}, \\ y &= y_0 + v_{0y} (t-t_0) - \frac {1}{2} g (t-t_0)^2, \\ v_y &= v_{0y} - g (t-t_0). \label {eq:projectile} \end {aligned} } (2.53) The corresponding expressions in the textbook are Eqns. (3.20), (3.21), (3.22), and (3.23). As an application of projectile motion, let’s calculate some properties of the cannonball’s trajectory. Before starting though, let’s define the initial conditions to simplify our expressions, and set t0 = 0 s, x0 = 0 m, and y0 = 0 m. The first aspect we discuss is the shape of the cannonball’s trajectory, which is described by the function y(x), depicted by the dotted curve in Fig. 2.4. To Figure 2.4: Projectile motion of a cannonball. 2.3. TWO-DIMENSIONAL MOTION 15 get y(x), we start with the expression for y in Eqn. (2.53) and replace t with x using the expression for x, which tells us that t = x/v0 x. The resulting expression for y(x) is then, y(x) = x \left ( \frac {v_{0y}}{v_{0x}} \right ) - x^2 \left ( \frac {g}{2v_{0x}^2} \right ), \label {eq:proj_shape} (2.54) which is a parabola. Projectile motion is often referred to as parabolic motion. Next, let’s use this to figure out how far the cannonball travels before it hits the ground, in other words, the distance labeled R in Fig. 2.4. One way to find this is to notice that when the cannonball hits the ground, y = 0. So we look at Eqn. (2.54) and ask what value(s) of x give us y = 0? One of these is x = 0, telling us the cannonball’s initial position is the origin. There’s another solution, one for which x is not zero. This is the solution we’re interested in, and gives us the value of R , 0 &= R\left ( \frac {v_{0y}}{v_{0x}} \right ) - R^2 \left ( \frac {g}{2v_{0x}^2} \right ), \\ \Longrightarrow \ \ R &= \frac {2v_{0x} v_{0y}}{g}. (2.56) Now, let’s find out how high the cannonball travelled. A slick way to do this is to realize that the cannonball reaches its highest point halfway through its flight. This is because the flight of the cannonball is symmetric. That is to say, the path represented by the dotted line is the same whether you fire the cannon from x = 0 pointed to the right (like we did) or whether you moved the cannon to x = R and fired it to the left. The cannonball feels no acceleration along the x axis, so it doesn’t have any way to distinguish right from left. The highest point, indicated by h in Fig. 2.4, is therefore6 h &= y(R/2), \\ \Longrightarrow \ \ h &= \frac {v_{0y}^2}{2g}. (2.58) As a final comment, note that Fig. 2.4 shows the angle α of the initial velocity of the cannonball relative to the x axis. Given this angle and the initial speed of the cannonball v0 , the x- and y-components of the velocity can be found with some simple trigonometry, v_{0x} &= v_0 \cos \alpha , \\ v_{0y} &= v_0 \sin \alpha. (2.60) 6 Another way to see this is to find the point where the slope of y(x) is zero, which is the inflection point where the cannonball stops moving up and starts moving down — the cannonball’s highest point. By setting d y/dx = 0, you can solve for the value of x where the slope is zero and you get the same answer. 16 CHAPTER 2. NON-CIRCULAR MOTION 2.3.2 Relative velocity Take two people, let’s call them A and B, and ask them each how far away and in which direction something is, let’s call the something point P. If you ask, you’ll get two different answers. This is because each person has their own reference frame. Zero meters away means right where they are. We’re going to relate measurements of distances and velocities made by A and B. To do this, we need to introduce a little bit of notation. We’ll call xP / A the x coordinate of P as measured by A , or equivalently the x coordinate of P in reference frame A. Fig. 2.5 shows our two reference frames and the distances to point P as measured in each. In this situation, A and B agree on the y coordinate of P , though in general they won’t necessarily agree on anything about P. If y is the height, then maybe A and B are both standing on the same level floor. The floor gives them both the same point of reference for heights, or values of y. In or notation, this means7 y_{B/A} = &\ y_{A/B} = 0. (2.61) Because they’re standing apart, however, they don’t agree on the x coordinate of P. We can relate xP /B and xP / A using the difference between where A and B are standing. In our notation, this difference is xB/ A (if you ask A ) or x A /B (if you ask B). Taking a look at Fig. 2.5, it’s not too hard to convince ourselves that x_{P/A} &= x_{P/B} + x_{B/A}, \\ y_{P/A} &= y_{P/B}. (2.63) Figure 2.5: The relative position of P as seen in reference frames A and B. 7 Note that, in general, y B/ A ̸= yA /B. If you’re 2 m from me to the East, then I’m also 2 m from you, but to the West. Same distance, opposite direction. We indicate this mathematically with a minus sign, yB/ A = − yA /B. 2.3. TWO-DIMENSIONAL MOTION 17 In general, of course, A and B won’t always agree on the value of y. B might climb on a desk or A might be on a different floor. We can easily make this generalization, and also add a third dimension, the z coordinate, see Fig. 2.6. The vector expression relating the position of an arbitrary point P , as measured in any two reference frames, A and B, is \myvec r_{P/A} = \myvec r_{P/B} + \myvec r_{B/A}. \label {eq:relpos} (2.64) Figure 2.6: The relative position of P according to A and B in full three- dimensional glory. 18 CHAPTER 2. NON-CIRCULAR MOTION Figure 2.7: Relative velocity. Now imagine A and B are both moving, and so is the point P !8 We would like to be able to relate the velocities A and B measure for point P in the same way we’ve just related the positions. Though this seems a little bit harder, luckily we already have the positions figured out and we know that velocity is just the derivative of position with respect to time. All we have to do is differentiate Eqn. (2.64) with respect to time! The velocities are related by, \Aboxed { \myvec v_{P/A} &= \myvec v_{B/A} + \myvec v_{P/B}\. } \label {eq:relvel} (2.65) The expression for relative velocities was simple for us to arrive at, but to think about it requires a little more thought than when everything was static. To help with this, let’s work out an example. You (Y ) are in an airport with time to kill, reading a physics book, when you notice Usain Bolt (U ) sprinting along the moving walkway (W ) to try and catch a plane. The walkway moves at a speed of 1 m/s and Usain sprints at an incredible clip of 12.4 m/s. 1. How fast does Usain run relative to you? This is a one dimensional problem. We start by defining which direction is going to be positive, we choose the right and draw an axis accordingly. I’ve labeled it x. Since we’re working in one dimension there’s no need to keep track of “ x”, “ y”, or “ z” components, so I’ll leave off x subscripts. Usain is like the point we want to know about, and we want his speed in your reference frame, so we want to know vU /Y. The other reference frame in the problem is defined by the walkway. The one-dimensional version of Eqn. (2.65) for this problem is, v_{U/Y} = v_{U/W} + v_{W/Y}. (2.66) We’re given the speed of Usain relative to the walkway, vU /W = 12.4 m/s, and the speed of the walkway relative to you, vW /Y = 1 m/s. Both of these 8 Recall from our first discussion that none of the speeds involved can be too close to the speed of light, or else we’d need to consult Einstein’s special theory of relativity to accurately figure things out. 2.4. PLOTTING POSITION AND VELOCITY VERSUS TIME 19 are positive, since they point to the right. Plugging in the numbers, we find, v_{U/Y} = 13.4~{\rm m/s}. (2.67) 2. How fast are you moving relative to the walkway? We know how fast the walkway is moving relative to you, vW /Y. We also know that your motion relative to the walkway is equal and opposite, v_{Y/W} = -v_{W/Y}. (2.68) Since the walkway is moving 1 m/s away from you to the right (positive), that means v_{Y/W} = -1~{\rm m/s}. (2.69) 3. How fast is the walkway moving relative to Usain? We know how fast Usain is moving relative to the walkway, vU /W. We also know that the walkway’s motion relative to Usain is equal and opposite, v_{W/U} = -v_{U/W}. (2.70) Since Usain is moving 12.4 m/s away from the walkway to the right (pos- itive), that means v_{W/Y} = -12.4\ {\rm m/s}. (2.71) 2.4 Plotting position and velocity versus time We’ve discussed how to characterize the motion of freely falling bodies, projec- tiles by discussing how to calculate the position and components of velocity as functions of time or at specific points of the trajectory (e.g. the highest point attained by a cannonball). Now we discuss another way to characterize the motion, plotting it. Given that we’ve figured out how to arrive at expressions describing the motion, it’s natural to plot these expressions versus time to get a better feel for what the motion actually looks like. Plots of x(t) and v x (t) versus t, which we’ll refer to as x – t and v x – t graphs, are shown in Fig. 2.8 for the case of projectile motion. The x – t graph shows us that the cannonball fired in Fig. 2.4 moves from its initial x position at x0 , to ever-increasing values of x — that is, it moves along the positive x axis. The v x – t graph shows us that the speed with which the cannonball progresses along the x axis is constant. Note that this is just the slope of the x – t graph, which it has to be since we defined the velocity to be the derivative of the position, so v x = dx/dt. Note that although the problem might start at time t = t0 , we can still plot x(t) and v x (t) for earlier times (as I’ve done in Fig. 2.8). Just keep in mind that the only part of the plotted curves that apply to the problem are those for t ≥ t0. 20 CHAPTER 2. NON-CIRCULAR MOTION Figure 2.8: (Left) The x – t graph for projectile motion, described by x(t) = x0 + v0 x (t − t 0 ). The slope of x at time t 0 is v0 x , which I’ve drawn for the case v0 x > 0, though of course this can be negative depending on the direction of motion and how we decide to draw the axes. (Right) The v x – t graph for v x (t) = v0 x. Because the only acceleration in projectile motion is from the Earth, and because we decided to label the axis along which Earth’s acceleration points the y axis (see Fig. 2.4), there is no acceleration in the x direction for projectile motion. This means the x-component of velocity, which corresponds to the slope in our x – t graph, is constant. You can also look at y – t and v y – t graphs for two-dimensional motion, or even θ – t and ω – t graphs for circular motion (which is coming next). In all cases, a “something” – t graph simply means to plot “something” as a function of t. Chapter 3 Circular Motion Circular motion takes place in two dimensions, so we could have included it in the last section. However, it’s different enough in other ways that it deserves its own section. Let’s introduce the ideas by way of an example. Consider the piece of gum stuck to the spinning record in Fig. 3.1. The gum is stuck a distance r from the center of the record. To say exactly where it’s at we also need to give the angle θ , defined relative to wherever we decide to set θ = 0, I’ve chosen the horizontal dashed line. This is the first thing to notice about circular motion... to specify the position of something it’s more convenient to use polar coordinates, r and θ , instead of cartesian coordinates, x and y. The next thing to note is the way we define how fast the record is spinning. If you happen to be familiar with vinyl, you may know that this is typically specified in units of rpm, revolutions per minute. This rate of spinning is called the angular speed , and we’ll use the symbol ω for it. One revolution is 2π radians (which is 360 degrees), so ω tells us how many radians (or degrees) the record rotates through per unit time. Note that angular speed is a different thing than the (linear) speed we introduced in Eqn. (2.8). We didn’t call it linear back then because we didn’t need to distinguish it from circular, but in the current discussion I’ll Figure 3.1: An example of uniform circular motion: a record spins at angular speed of ω = 33 rpm, with a piece of gum stuck to it at r = 1/6 m. 21 22 CHAPTER 3. CIRCULAR MOTION add linear if that’s what I mean, to make sure there’s no confusion. 3.1 Uniform circular motion Uniform circular motion means the motion traces out a circle, so fixed r , with constant angular speed. (It also means the record player isn’t wobbling, or spinning about other axes.) Before we delve too deeply into the description of uniform circular motion in terms of polar coordinates ( r and θ ), we’ll first talk about linear velocity and linear acceleration. Question: If I tell you the gum’s linear speed v = 3.5 m/s, what would you say is it’s linear velocity ? Answer: While the linear speed may be constant, the direction the gum is travel- ing changes constantly. Since the linear velocity is both the linear speed and the direction, it changes constantly, too. For example, when the gum is at the top in Fig. 3.1 its linear velocity is 3.5 m/s to the left, and when it’s at the bottom its linear velocity is 3.5 m/s to the right. Question: What does this tell us about the linear acceleration? Answer: It’s not zero! We can figure out the direction of the linear acceleration by looking at Fig. 3.2 and considering the average linear acceleration, \myvec a_{\rm av} &= \frac {\myvec v_2 - \myvec v_1}{\Delta t}, \label {eq:avg_a_2} (3.1) as introduced in Eqn. (2.14). Note that ∆ t is the time difference between ⃗ v2 and ⃗v1. Because the linear speed in uniform circular motion is constant, the speeds v1 and v2 are the same. Because of this, when you calculate ⃗ v2 −⃗v1 , and turn it into ⃗ aav by dividing by ∆ t, the result points directly toward the center of the circle. Now imagine taking the limit ∆ t → 0, which will turn ⃗ aav into the instantaneous acceleration ⃗a. Look at the right-most image in Fig. 3.2 and picture what happens in this limit. ⃗ v2 and ⃗v1 get closer and closer together, but Figure 3.2: Direction of average linear acceleration for uniform circular motion. In the limit ∆ t → 0, ⃗ arad. aav → ⃗ 3.1. UNIFORM CIRCULAR MOTION 23 at any point of the limit, ⃗ aav continues to point toward the center of the circle. Therefore, linear acceleration ⃗ a points toward the center of the circle for uniform circular motion. Because of this, it is often called radial acceleration (it points along the radius of the circle) or centripetal acceleration (Greek for towards the center). In the book, see Eq. (3.28) and the surrounding text, they call it ⃗ arad , so we’ll use the same notation. We can figure out the magnitude a rad by analyzing the geometry of Fig. 3.2 in a bit more detail. We start with the definition of a rad , given by the limit of Eqn. (3.1), a_{\rm rad} = \lim _{\Delta t \to 0} \frac {|\myvec v_2 - \myvec v_1|}{\Delta t}, \label {eq:arad} (3.2) and then figure out what |⃗ v1 | is. We do this using Fig. 3.3, where we’ve v2 − ⃗ labeled a few more bits of information. First, we include the fact the speed is constant by writing ⃗ v1 = v û1 and ⃗ v2 = v û2 , where û1 and û2 are vectors of unit length that point in the direction of the velocity. We also label the radius of the circle r , the angle ∆θ , and the arc length ∆s. Next, we identify the two shapes shown to the right in Fig. 3.3. Because the sides labeled r and v are perpendicular, these shapes have a common angle, ∆θ. The shape with sides r and the segment of the circle ∆s is a sector , and obeys the relation \Delta \theta &= \frac {\Delta s}{r}. (3.3) The other shape is a triangle, and the analogous relation is only approximately true, \Delta \theta &\approx \frac {|\myvec v_2 - \myvec v_1|}{v}. (3.4) However, in the limit ∆θ → 0, which corresponds to the ∆ t → 0 limit in which we’re working, the difference between a sector and a triangle goes away and we can assert that, in this limit, |\myvec v_2 - \myvec v_1| &= \frac {v \Delta s}{r}. (3.5) Figure 3.3: Magnitude of radial acceleration for uniform circular motion. 24 CHAPTER 3. CIRCULAR MOTION Plugging this into Eqn. (3.2) gives a_{\rm rad} &= \lim _{\Delta t \to 0} \left ( \frac {v\, \Delta s}{r\, \Delta t} \right ), \\ \Longrightarrow \ \ \Aboxed {\myvec a_{\rm rad} &= \frac {v^2}{r},\ \text {towards center of circle} \. } \label {eq:arad_vec} (3.7) To get Eqn. (3.7), we’ve used the fact that v = \lim _{\Delta t \to 0} \frac {\Delta s}{\Delta t}. (3.8) Now let’s cast our discussion of uniform circular motion in terms of angular velocity and angular acceleration (see Section 9.1 in the textbook). We’ve already mentioned angular speed ω is the number of radians (or degrees) per unit time, \omega = \frac {d\theta }{dt}. (3.9) To make this a velocity, we need to specify a direction as well. Take a look at the record in Fig. 3.4, it can only spin in one of two ways... clockwise or counterclockwise. This is an okay description, but it’s not complete. Say the record player is sitting on a table and I tell you it’s spinning clockwise. How do you know I wasn’t describing the motion while lying on the floor under the table? Think about this, clockwise means something different depending on whether I’m above or below the table. To remove this ambiguity, we define the direction of rotation using the right hand rule. Take your right hand (you’ll get the wrong answer every time if you use your left hand!) and curl your fingers toward the palm of your hand in the direction of rotation. Now stick your thumb out. It’s pointing in the direction of the angular velocity. If you do this for the record in Fig. 3.1, your thumb should be pointing into the page. This might seem funny — the angular velocity is perpendicular to the direction of motion — but it’s just notation, and all that matters about notation is that it’s unambiguous, and the right hand rule is unambiguous. Angular velocity is, \Aboxed { \myvec \omega = \frac {d\theta }{dt}\, \text {, direction by right hand rule} \. } (3.10) Angular acceleration is defined to be the derivative of angular velocity with respect to time, \Aboxed { \myvec \alpha = \frac {d\myvec \omega }{dt} \ , } (3.11) which is zero for uniform circular motion, because the angular speed is constant and the direction from the right hand rule isn’t changing. This last point is why 3.1. UNIFORM CIRCULAR MOTION 25 we stated in the beginning of this section that the record wasn’t wobbling or spinning about its other axes. There is one more term we use for rotational motion, and that is the period T which is the time it takes for one revolution. There are 2π radians (or 360 degrees) in one revolution. This means we can write the angular speed of the gum stuck on the record as “the number of radians in one revolution divided by the time of one revolution”, or \Aboxed { \omega = \frac {2\pi }{T} \. } (3.12) We can do something similar for linear speed of the gum, where instead of the total number of radians in one revolution we use the total distance travelled by the gum in one revolution. This distance is the circumference of the circle with radius r. Therefore, \Aboxed { v = \frac {2\pi r}{T} \. } (3.13) From these two equations, we see that linear and angular speed are related. In fact, we can write the linear speed and the magnitude of linear acceleration [see Eqn. (3.7)] in terms of ω, \boxed { \begin {aligned} v &= \omega \, r, \\ a_{\rm rad} &= \omega ^2 r. \end {aligned} } (3.14) See Section 9.3 in the textbook for further discussion. For uniform circular motion we see that the angular speed (e.g. the record player’s rpm) is constant, but the linear speed and linear acceleration both depend how far you are from the center of the record. The farther out you are, the faster you’re going. This is probably intuitive if you recall playing on merry go rounds. Let’s work Problem 3.25 in the textbook to help cement some of these ideas, see Fig. 3.5. Figure 3.4: The right hand rule determines the direction of angular velocity. 26 CHAPTER 3. CIRCULAR MOTION Figure 3.5: Problem 3.25 in the texbook. (a) We’re asked to find the radial acceleration of an object on the Earth’s equator, expressed in units of m/s2 and as a fraction of the Earth’s accel- eration g. a_{\rm rad} &= \frac {v^2}{r} \\ &= \frac {4\pi ^2 r}{T^2} \ \text {, using}\ v=\frac {2\pi r}{T} \\ &= \frac {4 \pi ^2 \times 6380\times 10^3\ {\rm m}}{\big (24\ {\rm hr} \times \frac {60\ {\rm min}}{\rm hr} \times \frac {60\ {\rm s}}{\rm min}\big )^2} \\ &= 0.0337 \ {\rm m}/{\rm s}^2 \\ &= (3.4 \times 10^{-3}) \times g (3.19) (b) We’re asked what period of the Earth’s rotation would generate a rad > g. a_{\rm rad} &> g \\ \Longrightarrow \ \ \frac {4 \pi ^2 r}{T^2} &> g \\ \Longrightarrow \ \ T^2 &< \frac {4\pi ^2 r}{g} \\ \Longrightarrow \ \ T &< 2\pi \sqrt {\frac {r}{g}} \\ \Longrightarrow \ \ T &< 5069.6\ {\rm s}\ (\text {or 1.4 hours}) (3.24) 3.2. NON-UNIFORM CIRCULAR MOTION 27 Figure 3.6: Non-uniform circular motion results in two components of accelera- tion, ⃗ arad and ⃗ atan. 3.2 Non-uniform circular motion As an example of non-uniform circular motion, consider what happens to the gum on the record when you turn off the record player. The record, and therefore the gum, begins to slow because of friction, meaning the linear and angular speeds decrease. A decrease in the linear speed means a linear acceleration (it’s negative, which is often called deceleration). When a velocity vector is shrinking in magnitude, the acceleration points in the opposite direction. Therefore, like the linear velocity, this acceleration is tangent to the circle traced out by the gum. However, the record is still spinning, so the gum still has radial acceleration, too. This situation is depicted in Fig. 3.6. The total linear acceleration for non-uniform circular motion is then, \myvec a_{\rm total} & = \myvec a_{\rm rad} + \myvec a_{\rm tan}. (3.25) Because the radial and tangential components of ⃗ atotal are orthogonal, it has magnitude, a_{\rm total} &= \sqrt { a_{\rm rad}^2 + a_{\rm tan}^2 }. (3.26) 28 CHAPTER 3. CIRCULAR MOTION Chapter 4 Force and Newton’s Laws The motion we’ve been discussing is caused by forces. The connection between forces and the motion they cause is provided by Newton’s Laws of motion. Fig. 4.1 shows some examples of the forces we’ll consider. Here’s a brief descrip- tion of the forces acting in Fig. 4.1: The push applied by the blue person is a contact force, meaning it is applied directly to the box. The red person is pulling on the rope, and the rope transmits this pull to the box. By the time the pull reaches the box it might not be the same force applied by the red person — maybe the rope is stretchy. As a result of this, we call the force of the rope on the box tension. Tension is also a contact force. Friction opposes the motion of the box and results from molecular bonds formed between the box and the floor it’s sitting on and from the fact that the surface of the box and the surface of the floor are rough, see Fig. 4.2. Friction is a contact force. The weight of the box is the force applied to the box by the Earth, through gravitational attraction. It acts toward the center of the earth. Note in Fig. 4.1 the angle of the weight relative to the perpendicular to the surface on which the box is sitting. This angle is the same as the angle of elevation of the surface. Weight is a long distance force, meaning the Earth doesn’t have to touch the box in order to apply the force of gravity. The weight an object of mass m feels on Earth is \Aboxed { \myvec W = m g\ \text {, toward the center of the Earth.} } (4.1) Weight is a force and has units of Newtons. The weight and mass of an object are often, incorrectly, used interchangeably. Gravity is the ultimate long distance force, reaching throughout the universe. More on gravity in Ch. 5. 29 30 CHAPTER 4. FORCE AND NEWTON’S LAWS Figure 4.1: Examples of the types of forces we’ll be dealing with. The normal force is the response of the surface to the weight of the box. This force keeps the box from falling through the surface and accelerating toward the center of the earth. The normal force is applied perpendicular (or normal, hence the name) to the surface. If you know the weight of the box, and as long as the box is being supported by the surface and not crashing through it, then the normal is equal in magnitude to the component of the weight along the normal (i.e. along the dotted line in Fig. 4.1. Because forces have a direction and magnitude, we represent them using ⃗. The units of force are Newtons, N = kg · m / s2. When vectors, for example F multiple forces act on the same point of an object, the object responds to the resultant force, which is the vector sum of all the individual forces, \myvec R &= \Sigma \, \myvec F. (4.2) Note that it’s important the forces act on the same point. If they don’t, the object may rotate, which is a complication we won’t deal with in these lectures. We will treat all the forces acting on our objects as if they were acting on the Figure 4.2: Microscopic views of different surfaces. Image from academic.greensboroday.org. 4.1. NEWTON’S FIRST LAW 31 same point, the center of mass of the object. This is why I drew all the force vectors in Fig. 4.1 acting on the center of mass of the box. 4.1 Newton’s First Law Newton’s First Law of motion is: A body acted on by no net force moves with a constant velocity, which may be zero. A body is said to be in equilibrium when there is no net force acting on it, ⃗ = 0. Inertia, the tendency of an object to stay at rest or to keep moving i.e. Σ F once in motion, encodes the concept of Newton’s First Law. In fact, Newton’s First Law is sometimes called the Law of Inertia. Imagine you’re wearing spandex pants while sitting on the smooth leather bench seat (a frictionless combination!) in the back of your friend’s car while your friend drives with constant velocity along Byres Road. You are not wearing your seatbelt! Question: What forces are acting on you? Answer: Your weight (i.e. gravity) acts down, the normal force from the seat ⃗ = 0. acts up to cancel out your weight. There is no resultant force on you, Σ F You are in equilibrium. Without slowing down, your friend turns onto Great Western Road. Question: What happens to you? Answer: You slide across the seat and slam into the door of the car (should’ve worn your seatbelt). The only forces acting on you are your weight and the normal force from the seat. In the reference frame of your friend, you started moving with no net forces acting on you, in violation of Newton’s First Law! This is because Newton’s First Law doesn’t work in a reference frame if the reference frame is accelerating. An intertial reference frame is one in which Newton’s First Law holds. If refer- ence frame A is an inertial frame, then an other reference frame B moving with constant velocity relative to A, that is ⃗vB/ A = constant, is also inertial. Unless specifically stated, all our reference frames will be inertial. 4.2 Newton’s Second Law ⃗ acts on an object Newton’s Second Law of motion is: If a net external force Σ F of mass m, the object accelerates according to, \Sigma \, \myvec F = m \myvec a. \label {eq:N2} (4.3) 32 CHAPTER 4. FORCE AND NEWTON’S LAWS Figure 4.3: What force is required to swing this kid around? In a typical problem we’ll be given some individual forces acting on some object and asked to calculate the resulting trajectory of the object. The big- picture steps to solve this type of problem are: ⃗. 1. Add up all the forces, Σ F 2. Use Newton’s Second Law and the objects mass m to calculate the result- ing acceleration, ⃗ ⃗. a = Σ F/m 3. Take this acceleration and use the equations for uniform acceleration we’ve already studied to calculate the object’s trajectory. Though this is typical, it isn’t the only way a problem can be posed. For example, in Fig. 4.3 what force must the guy in the old-time swimsuit supply to keep the girl of mass m swinging around in a circle of radius r with constant linear speed v? To keep the girl from flying off into the bushes he has to keep changing her velocity by pulling on her feet while he spins around. The acceleration he has to generate is the radial acceleration, a rad = v2 /r. Since she has mass m, and we know the acceleration, we can use Newton’s Second Law the other way around to determine the force, \Aboxed { \myvec F = \frac {mv^2}{r}\ \text {, pointed to the center of the circle}. } (4.4) This problem arrives at in important result, the force required to maintain uniform circular motion. A subtle point, this force is not generated by uniform circular motion, rather it is the force required to keep circular motion going. If the force disappears (the guy lets go of the kid) uniform circular motion stops (and she flies off into the air... projectile motion). 4.3. NEWTON’S THIRD LAW 33 Figure 4.4: (Left) A frictionless pulley with inelastic cable connecting blocks of mass m 1 and m 2 , each of which is initially at rest. (Right) the free body diagrams (FBDs) for m 1 and m 2. 4.3 Newton’s Third Law Newton’s Third Law of motion states: If object A exerts force on object B (the “action”) then B exerts a force on A (the “reaction”) that is equal in magnitude but opposite in direction, \myvec F_{\text {A on B}} = - \myvec F_{\text {B on A}}. (4.5) Note that the two forces involved in this law, the action-reaction pair, do not act on the same object. If you find yourself thinking that since these two forces are equal but opposite and they should therefore cancel each other, remind yourself that they are acting on two different objects. This law is the idea behind speed boat, spacecraft, submarine, jet, and more more types of propulsion. For example, a controlled explosion and a nozzle act to provide a force, the action, on exhaust. The exhaust responds with an equal but opposite force, the reaction, applied to the jet. The exhaust is expelled from the jet with great force and in return, the jet receives a great force in the opposite direction. 4.4 Free body diagrams Free body diagrams (FBDs) are book-keeping tools to help us keep track of the forces acting in a problem, so that we can figure out exactly what goes into the left hand side of Newton’s Second Law, Eqn. (4.3). Let’s demonstrate how to use a free body diagram by way of an example. Fig. 4.4 shows a frictionless pulley hanging from the ceiling, with an inelastic cable connecting two blocks of mass m 1 and m s. We will calculate the acceleration of each block using free body diagrams, also shown in Fig. 4.4. 34 CHAPTER 4. FORCE AND NEWTON’S LAWS Let’s start with m 1 and analyze the FBD for the x-components of the forces acting on it. There are none! Given the x and y axes defined in Fig. 4.4, there are no forces acting on m 1 and m 2 in the x direction. The y components of the forces acting on m 1 , however, are not zero. We have the tension T from the cable acting up and the weight of m 1 acting down. Note that the tension on each block is the same because the cable is inelastic (i.e. not stretchy). \text {FBD for $x$-component of forces on $m_1$:\qquad \qquad }\sum F_{1x} &= m_1 a_{1x} \nonumber \\ \Longrightarrow 0 &= a_{1x} (4.6) \text {FBD for $y$-component of forces on $m_1$:\qquad \qquad }\sum F_{1y} &= m_1 a_{1y} \nonumber \\ \Longrightarrow T - m_1 g &= m_1 a_{1y} \label {eq:prob_2} (4.7) \text {FBD for $x$-component of forces on $m_2$:\qquad \qquad }\sum F_{2x} &= m_2 a_{2x} \nonumber \\ \Longrightarrow 0 &= a_{2x} (4.8) \text {FBD for $y$-component of forces on $m_2$:\qquad \qquad }\sum F_{2y} &= m_2 a_{2y} \nonumber \\ \Longrightarrow T - m_2 g &= m_2 a_{2y} \label {eq:prob_1} (4.9) Because a cable connects m 1 and m 2 , their motion is also connected. In fact, because the cable is not stretchy, the speed and magnitude of acceleration of the two blocks is the same. The direction of the motion, however, and the direction of the acceleration, is opposite — if m 1 moves or accelerates up, then m 2 moves or accelerates down. That is, a 2 y = −a 1 y. Using this in Eqn. (4.9), we have T - m_2g &= -m_2 a_{1y} \nonumber \\ \Longrightarrow T &= m_2\, (g - a_{1y}). (4.10) We can use this in Eqn. (4.7) to solve for a 1 y , m_2\, (g - a_{1y}) - m_1 g &= m_1 a_{1y} \\ \Rightarrow (m_2 - m_1)\, g &= (m_1 + m_2)\, a_{1y} \\ \Rightarrow a_{1y} &= \frac {m_2 - m_1}{m_1 + m_2}\, g, (4.13) and using the fact that a 2 y = −a 1 y , a_{2y} = \frac {m_1 - m_2}{m_1 + m_2} g. (4.14) 4.5. FRICTION 35 Figure 4.5: Transition from static to kinetic friction as a block on a surface is pushed harder and harder. The accelerations ⃗ a1 and ⃗ a2 are then, \myvec a_1 = \frac {m_2 - m_1}{m_1 + m_2} g\ \jhat \qquad \text {and}\qquad \myvec a_2 = \frac {m_1 - m_2}{m_1 + m_2} g\ \jhat. (4.15) This suggests, for example, that if block 2 is heavier it will accelerate downward, and if both blocks have equal mass, they will remain at rest. This makes sense. Appendix C provides a collection of worked problems using FBDs to solve problems with non-circular motion and without friction and Appendix D provides a collection for circular motion without friction. 4.5 Friction There are several types of friction: The force of kinetic friction acts on an object while the object slides on a surface. It’s magnitude, f K is given by, \Aboxed { f_K = \mu _K n, } (4.16) where µK is the coefficient of kinetic friction and n is the magnitude of the normal force. The force of kinetic friction points in the direction opposite the motion and is only present when the object is moving. The force of static friction acts on an object in response to other forces that would cause the object to slide on a surface. This is the force of friction when the object is not moving. Once the object starts moving, 36 CHAPTER 4. FORCE AND NEWTON’S LAWS kinetic friction takes over. The magnitude of the force of static friction, f S , is given by \Aboxed { f_S \leq \mu _S n, } (4.17) where µS is the coefficient of static friction and n is the magnitude of the normal force. The force of static friction points opposite to the force(s) that are trying to cause the object to slide and is given by an inequality because it matches the other forces that are trying to make the object slide, canceling them out, until they are bigger than µS n, at which point the object begins sliding. Fig. 4.5 explains the transition from static to kinetic friction as the applied force on an object is increased and Fig. 4.6 addresses the microscopic origin of static and kinetic friction. The force of rolling friction is treated in a manner similar to kinetic friction, f_r = \mu _r n, (4.18) where µr is the coefficient of rolling friction. We won’t make use of this, so I didn’t put a box around it. Fluid (or air) resistance is another type of friction, one whose opposing force to motion depends on the speed of the motion, the size and shape of the object of the body moving, and the properties of the fluid (or air) the body is moving through. We won’t get into the details of this. See Section 5.3 of the textbook for additional details. Appendix E provides a collection of worked problems that include the effects of friction. Figure 4.6: (Left) The roughness of the surfaces contributes to both static and kinetic friction. (Right) Static friction is also due, in part, to the formation of bonds between the molecules within the two surfaces. These bonds are broken when the surfaces begin moving relative to one another and are therefore not present in kinetic friction. Chapter 5 Gravity Figure 5.1: The gravitational force of attraction between two objects is propor- tional to the product of their masses and inversely proportional to the squared distance between their centers of mass. Newton’s law of gravity tells us that the gravitational force between the two objects depicted in Fig. 5.1 is proportional to the product of their masses and falls off with the inverse of the distance between them squared, \Aboxed { \myvec F_g = \frac {G m_1 m_2}{r_{12}^2}\ \text {(attractive)}, } \label {eq:NLOG} (5.1) where the constant of proportionality G is Newton’s constant , G = 6.67430(15) \times 10^{-11}\ {\rm m}^3\, {\rm kg}^{-1}\, {\rm s}^{-2}. (5.2) The direction of the force of gravity is easy to figure out, gravity always pulls objects towards each other. Because G is such a small number, the force of gravity is very weak in comparison to the other forces. Electricity and magnetism can defeat Earth’s gravity with a small refrigerator magnet. However, all the other forces (electricity and magnetism, the weak force, and the strong force) 37 38 CHAPTER 5. GRAVITY can be both attractive and repulsive. Over large distances and lots of matter involved, the positive and negative forces cancel each other out and only gravity remains. On distance scales of the solar system and larger, gravity dominates! Consider the force you feel from gravity when you are height h above the surface of the Earth, F_g = \frac {G m M_\oplus }{(R_\oplus + h)^2}, (5.3) where your mass is m and Earth’s mass and radius are M⊕ = 5.97217(13) × 1024 kg and R ⊕ ≈ 6.378 × 106 m, respectively. Newton’s second law lets us turn this into the acceleration you feel due to Earth’s gravity, g &= \frac {G M_\oplus }{(R_\oplus + h)^2} \nonumber \\ &= \frac {G M_\oplus }{R_\oplus ^2} \left ( 1 + \frac {h}{R_\oplus } \right )^{-2} \nonumber \\ &= \frac {G M_\oplus }{R_\oplus ^2} \left ( 1 - \frac {2h}{R_\oplus } + \mathcal O \left ( \frac {h}{R_\oplus }\right )^2 \right ). \label {eq:trueg} (5.4) The last line is an expansion valid for h/R⊕ < 1, which is valid for any reasonable height you’re likely to reach while jumping around, kicking footballs, or even shooting cannons on the surface of the Earth. Plugging in the values for G, M⊕ and R⊕ gives the value we’re expecting g = 9.8\ {\rm m}/{\rm s^2} \left ( 1 - \frac {2h}{R_\oplus } + \mathcal O \left ( \frac {h}{R_\oplus }\right )^2 \right ), (5.5) times a correction the reduces g the higher up we go. This correction is tiny, how- ever, and validates the uniform acceleration approximation we make for freely- falling body and projectile motion problems. For g in Eq. (5.4) to decrease to 9.7 m/s2 , you’d need to be at a height of about 20 miles, three times higher than the peak of Mt. Everest. Chapter 6 Simple Harmonic Motion Sections 14.1, 14.2, and 14.5 in the textbook. Simple harmonic motion (which we’ll abbreviate SHM) is an important topic in physics. You’ll see the concepts repeated in different guises throughout your physics studies, including graduate studies if you decide to pursue an advanced degree. 6.1 Definitions To help define some key terms, we use the frictionless block and spring setup shown in Fig. 6.1. The only relevant direction for this problem is the x direction, so we’ll treat it as a one-dimensional problem along the x axis. At x = 0, the block and spring are in equilibrium, that is, there no net force. As we pull the block to the right, the spring stretches. The distance from the equilibrium position is called the displacement. In the image on the right in Fig. 6.1, the displacement is A. In this stretched condition, the spring exerts a force on the block to the left. This is called a restoring force, because it always points toward the equilibrium position, attempting to restore the system to equilibrium. Figure 6.1: A frictionless block and spring setup (left) in equilibrium and (right) with the spring stretched and the block displaced a distance A from equilibrium. 39 40 CHAPTER 6. SIMPLE HARMONIC MOTION If we let go of the block, the displacement x will oscillate about the equi- librium position, 0, between the displacement extremes, ± A. The amplitude of the oscillation is the magnitude of the maximum displacement, A. The period T is the time for one complete cycle, that is, the time required to go from x = A to x = − A , and then back to x = A. Note the similarity between this definition of period and what we used for circular motion. This is not an accident and we’ll see that they are in fact the same thing. The frequency f is the number of cycles per unit time. Frequency has units of Hertz (Hz), where 1 Hz = 1 s−1. Period and frequency are related to each other, \Aboxed { f &= \frac {1}{T}. } (6.1) Angular frequency ω is related to frequency by \Aboxed { \omega = 2\pi f, } (6.2) which means ω = 2π/T , just like our definition of angular speed for circular motion. As with the period, the angular frequency is the same thing as the angular speed we discussed for circular motion. We’ll see this shortly, when we relate SHM to uniform circular motion. 6.2 Mass on a Spring An example of a system that undergoes simple harmonic motion is an ideal spring, which obeys Hooke’s Law, \Aboxed { F_x = -k x, } \label {eq:Hooke} (6.3) where k is the spring constant and the restoring force F x is proportional to the displacement. Whenever the restoring force is proportional to the displacement, we have Simple Harmonic Motion, or SHM for short. A body that undergoes SHM is called a harmonic oscillator. Applying Newton’s Second Law to the harmonic oscillator gives, \Sigma F_x &= m a_x \\ \Longrightarrow -k x &= m a_x \\ \Longrightarrow \frac {d^2 x}{dt^2} + \frac {k}{m} x &= 0 \label {eq:SHM_diffeq}. (6.6) This is a differential equation. You’ll learn how to solve them in your mathematics courses. Here, I’ll simply give you the solution. It is, \Aboxed { x &= A \cos (\omega t + \phi ), \label {eq:SHM_x}} (6.7) 6.2. MASS ON A SPRING 41 Figure 6.2: The x component of uniform circular motion is identical to the motion of a harmonic oscillator, x = A cos(ω t + φ). where φ is the phase angle, which is fixed by the initial condition, x_0 &= A\cos (\phi ). \label {eq:SHM_x0} (6.8) For example, if we started the clock ( t = 0) when we let go of the block at x = A , then we’d have x0 = A and φ = 0. For SHM, the angular frequency is given by \Aboxed { \omega = \sqrt {\frac {k}{m}}. } \label {eq:omega_SHM} (6.9) You should take Eqn. (6.7) and see for yourself that it satisfies Eqn. (6.6), with the identification of ω in Eqn. (6.9). Given the time-dependence