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4.4. EXAMPLES 85 K′ cm = 1 2(m1 + m2)(v′ cm)2 = 1 2 (170 kg) & 0.353 m s −1.5 m s '2 = 111.8 J (4.26) K′ conv,i = K′ sys,i −K′ cm = 641.3 J −111.8 J = 529.5 J ≃529 J (4.27) This shows explicitly that the convertible energy, as I pointed out earlier in this chapter, is the same in every reference frame! (The equality is exact, if you keep enough decimals in the calculation.) Knowing this, we can simplify the calculation of the final kinetic energy, after the explosive sepa- ration: the convertible energy, K′ conv,f, will be the same as in the earth reference frame, that is to say, 21.2 J, and the total kinetic energy will be K′ sys,f = K′ cm + K′ conv,f = 111.8 J +21.2 J = 133 J. So, in this frame of reference, we have (to three significant figures): K′ sys,i = 641 J, K′ cm = 112 J, K′ conv,i = 529 J (before the collision) K′ sys = K′ cm = 112 J, K′ conv = 0 (right after the collision) K′ sys,f = 133 J, K′ cm = 112 J, K′ conv,i = 21.2 J (after the separation) So, even though the total kinetic energy is different in the two reference frames, all the (inertial) observers will agree as to the amount of kinetic energy “lost” in the collision, as well as the amount of kinetic energy put back into the system by the players’ pushing on each other. | University Physics I Classical Mechanics_Page_103_Chunk4301 |
86 CHAPTER 4. KINETIC ENERGY 4.5 Problems Problem 1 A 71-kg man can throw a 1-kg ball with a maximum speed of 6 m/s relative to himself. Imagine that one day he decides to try to do that on roller skates. Starting from rest, he throws the ball as hard as he can, so it ends up moving at 6 m/s relative to him, but he himself is recoiling as a result of the throw. (a) Assuming conservation of momentum, find the velocities of the man and the ball relative to the ground. (b) What is the kinetic energy of the system right after the throw? (By the system here we mean the man and the ball throughout.) Where did this kinetic energy come from? (c) Is the man’s reference frame inertial throughout this process? Why or why not? (d) Does the center of mass of the system move at all throughout this process? Problem 2 Analyze Problem 1 from Chapter 3 from the point of view of the system’s kinetic energy. In particular, answer the following questions: (a) What is the total kinetic energy of the system before and after the collision? How much of this energy is translational (that is, center-of-mass kinetic energy), and how much is convertible? (b) What kind of collision is this? (Elastic, inelastic, etc.) What is the coefficient of restitution? Problem 3 Analyze Problem 2 from Chapter 3 from the point of view of the system’s kinetic energy. In particular, answer the following questions: (a) What is the coefficient of restitution for the collision described in part (a) of the problem, and how much kinetic energy is “lost” in that collision? (b) What is the coefficient of restitution for the collision described in part (b) of the problem, and how much kinetic energy is “lost” in that collision? Problem 4 A 0.012-kg bullet, traveling at 850 m/s, hits a 2-kg block of wood that is initially at rest, and goes straight through it. Assume that the final velocity of the bullet relative to the block is 400 m/s, and that the system is isolated. (a) What is the coefficient of restitution for this collision? (b) How much kinetic energy is “lost” in the collision? (c) What is the final velocity of the block? Problem 5 A 2-kg object, moving at 1 m/s, collides with a 1-kg object that is initially at rest. Assume they form an isolated system. | University Physics I Classical Mechanics_Page_104_Chunk4302 |
4.5. PROBLEMS 87 (a) What is the initial kinetic energy of the system? How much of this is center of mass energy, and how much is convertible? (b) What is the maximum amount of kinetic energy that could be “lost” (converted to other forms of energy) in this collision? (c) If 60% of the amount you calculated in part (b) is in fact converted into other forms of energy in the collision, what are the final velocities of the two objects? | University Physics I Classical Mechanics_Page_105_Chunk4303 |
88 CHAPTER 4. KINETIC ENERGY | University Physics I Classical Mechanics_Page_106_Chunk4304 |
Chapter 5 Interactions and energy 5.1 Conservative interactions Let me summarize the physical concepts and principles we have encountered so far in our study of classical mechanics. We have “discovered” one important quantity, the inertia or inertial mass of an object, and introduced two different quantities based on that concept, the momentum m⃗v and the kinetic energy 1 2mv2. We found that these quantities have different but equally intriguing properties. The total momentum of a system is insensitive to the interactions between the parts that make up the system, and therefore it stays constant in the absence of external influences (a more general statement of the law of inertia, the first important principle we encountered). The total kinetic energy, on the other hand, changes while any sort of interaction is taking place, but in some cases it may actually return to its original value afterwards. In this chapter, we will continue to explore this intriguing behavior of the kinetic energy, and use it to gain some important insights into the kinds of interactions we encounter in classical physics. In the next chapter, on the other hand, we will return to the momentum perspective and use it to formally introduce the concept of force. Hence, we can say that this chapter deals with interactions from an energy point of view, whereas next chapter will deal with them from a force point of view. In the previous chapter I suggested that what was going on in an elastic collision could be inter- preted, or described (perhaps in a figurative way) more or less as follows: as the objects come together, the total kinetic energy goes down, but it is as if it was being temporarily stored away somewhere, and as the objects separate, that “stored energy” is fully recovered as kinetic energy. Whether this does happen or not in any particular collision (that is, whether the collision is elastic or not) depends, as we have seen, on the kind of interaction (“bouncy” or “sticky,” for instance) that takes place between the objects. 89 | University Physics I Classical Mechanics_Page_107_Chunk4305 |
90 CHAPTER 5. INTERACTIONS AND ENERGY We are going to take the above description literally, and use the name conservative interaction for any interaction that can “store and restore” kinetic energy in this way. The “stored energy” itself—which is not actually kinetic energy while it remains stored, since it is not given by the value of 1 2mv2 at that time—we are going to call potential energy. Thus, conservative interactions will be those that have a “potential energy” associated with them, and vice-versa. 5.1.1 Potential energy Perhaps the simplest and clearest example of the storage and recovery of kinetic energy is what happens when you throw an object straight upwards, as it rises and eventually falls back down. The object leaves your hand with some kinetic energy; as it rises it slows down, so its kinetic energy goes down, down. . . all the way down to zero, eventually, as it momentarily stops at the top of its rise. Then it comes down, and its kinetic energy starts to increase again, until eventually, as it comes back to your hand, it has very nearly the same kinetic energy it started out with (exactly the same, actually, if you neglect air resistance). The interaction responsible for this change in the object’s kinetic energy is, of course, the gravi- tational interaction between it and the Earth, so we are going to say that the “missing” kinetic energy is temporarily stored as gravitational potential energy of the system formed by the Earth and the object. We even have a way to describe what is going on mathematically. Recall the equation v2 f −v2 i = 2a∆x for motion under constant acceleration. Let us use y instead of x, for the vertical motion; let a = −g, and let vf just be the generic velocity, v, at some arbitrary height y. We have v2 −v2 i = −2g(y −yi) Now multiply both sides of this equation by 1 2m: 1 2mv2 −1 2mv2 i = −mg(y −yi) (5.1) The left-hand side of (5.1) is just the change in kinetic energy (from its initial value when the object was launched). We will interpret the right-hand side as the negative of the change in gravitational potential energy. To make this clearer, rearrange Eq. (5.1) by moving all the “initial” quantities to one side: 1 2mv2 + mgy = 1 2mv2 i + mgyi (5.2) We see, then, that the quantity 1 2mv2 + mgy stays constant (always equal to its initial value) as the object goes up and down. Let us define the gravitational potential energy of a system formed by the Earth and an object a height y above the Earth’s surface as the following simple function of y: U G(y) = mgy (5.3) | University Physics I Classical Mechanics_Page_108_Chunk4306 |
5.1. CONSERVATIVE INTERACTIONS 91 Then we see from Eq. (5.2) that K + U G = constant (5.4) This is a statement of conservation of energy under the gravitational interaction. For any interaction that has a potential energy associated with it, the quantity K + U is called the (total) mechanical energy. Figure 5.1 shows how the kinetic and potential energies of an object thrown straight up change with time. To calculate K I have used the equation v = vi −gt (taking ti = 0); to calculate U G = mgy, I have used y = yi + vit −1 2gt2. I have arbitrarily assumed that the object has a mass of 1 kg and an initial velocity of 2 m/s, and it is thrown from an initial height of 0.5 m above the ground. Note how the change in potential energy exactly mirrors the change in kinetic energy (so ∆U G = −∆K, as indicated by Eq. (5.1)), and the total mechanical energy remains equal to its initial value of 6.9 J throughout. 0 1 2 3 4 5 6 7 0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 K UG energy (J) t (s) Figure 5.1: Potential and kinetic energy as a function of time for a system consisting of the earth and a 1-kg object sent upwards with vi = 2 m/s from a height of 0.5 m. There is something about potential energy that probably needs to be mentioned at this point. Because I have chosen to launch the object from 0.5 m above the ground, and I have chosen to measure y from the ground, I started out with a potential energy of mgyi = 4.9 J. This makes sense, in a way: it tells you that if you simply dropped the object from this height, it would have picked up an amount of kinetic energy equal to 4.9 J by the time it reached the ground. But, actually, where I choose the vertical origin of coordinates is arbitrary. I could start measuring y | University Physics I Classical Mechanics_Page_109_Chunk4307 |
92 CHAPTER 5. INTERACTIONS AND ENERGY from any height I wanted to—for instance, taking the initial height of my hand to correspond to y = 0. This would shift the blue curve in Fig. 5.1 down by 4.9 J, but it would not change any of the physics. The only important thing I really want the potential energy for is to calculate the kinetic energy the object will lose or gain as it moves from one height to another, and for that only changes in potential energy matter. I can always add or subtract any (constant) number1 to or from U, and it will still be true that ∆K = −∆U. What about potential energy in the context in which we first encountered it, that of elastic collisions in one dimension? Imagine that we have two carts collide on an air track, and one of them, let us say cart 2, is fitted with a spring. As the carts come together, they compress the spring, and some of their kinetic energy is “stored” in it as elastic potential energy. In physics, we use the following expression for the potential energy stored in what we call an ideal spring2: U spr(x) = 1 2k(x −x0)2 (5.5) where k is something called the spring constant; x0 is the “equilibrium length” of the spring (when it is neither compressed nor stretched); and x its actual length, so x > x0 means the spring is stretched, and x < x0 means it is compressed. For the system of the two carts colliding, we can take the potential energy to be given by Eq. (5.5) if the distance between the carts is less than x0, and 0 (corresponding to a relaxed spring) otherwise. If we put cart 1 on the left and cart 2 on the right, then the distance between them is x2 −x1, and so we can write, for the whole interaction U(x2 −x1) =1 2k(x2 −x1 −x0)2 if x2 −x1 < x0 0 otherwise (5.6) This is enough to solve for the motion of the two carts, given the initial conditions. To see how, look in the “Examples” section at the end of this chapter. Here, I will just give you the result. For the calculation, shown in Fig. 5.2 below, I have chosen cart 1 to have a mass of 1 kg, an initial position (at t = 0) of x1i = −5 cm and an initial velocity of 1 m/s, whereas cart 2 has a mass of 2 kg and starts at rest at x2i = 0. I have assumed the spring has a length of x0 = 2 cm and a spring constant k = 1000 J/m2 (which sounds like a lot but isn’t really). The collision begins at tc = (x2i −x0 −x1i)/v1i = 0.03 s, which is the time it takes cart 1 to travel the 3 cm separating it from the end of the spring. Prior to that point, the total kinetic energy Ksys = 0.5 J, and the total potential energy U = 0. 1Of course, some choices may result in the potential energy, and even the total energy, being negative sometimes! If this notion of a negative total energy bothers you a bit, wait until the chapter on gravity (Chapter 10), where we will try to make some sense out of it. . . 2An “ideal spring” is basically defined, mathematically, by this expression, or by the corresponding force equation (6.21) (which we will study in the next chapter, and which goes by the name of Hooke’s law); usually, we also require that the spring be “massless” (by which we mean that its mass should be negligible compared to all the other masses involved in any given problem). Of course, for Eq. (5.5) to hold for x < x0, it must be possible to compress the spring as well as stretch it, which is not always possible with some springs. | University Physics I Classical Mechanics_Page_110_Chunk4308 |
5.1. CONSERVATIVE INTERACTIONS 93 As a result of the collision, the spring compresses and undergoes “half a cycle” of oscillation with an “angular frequency” ω = ! k/µ (where µ is, as in previous chapters, the “reduced mass” of the system, µ = m1m2/(m1 + m2)). That is, the spring is compressed and then pushes out until it gets back to its equilibrium length3. This lasts from t = tc until t = tc + π/ω, during which time the potential and kinetic energies of the system can be written as U(t) = 1 2µv2 12,i sin2[ω(t −tc)] K(t) = Kcm + 1 2µv2 12,i cos2[ω(t −tc)] (5.7) (don’t worry, all this will make a lot more sense after we get to Chapter 11 on simple harmonic motion, I promise!). After t = tc + π/ω, the interaction is over, and K and U go back to their initial values. 0 0.1 0.2 0.3 0.4 0.5 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 K U energy (J) t (s) Figure 5.2: Potential and kinetic energy as a function of time for a system of two carts colliding and compressing a spring in the process. If you compare Figure 5.2 with Figure 4.5 of Chapter 4, you’ll see that the kinetic energy curve looks very similar, except for the time scale, which here is hundredths of a second and over there was taken to be milliseconds. The quantity that determines the time scale here is the “half period” of oscillation, π/ω = π ! µ/k = 0.081 s for the values of k and µ assumed here. We could make this smaller by making the spring stiffer (increasing k), or the blocks lighter (reducing µ), but there’s not much point in trying, since the collisions in Chapters 3 and 4 were all just made up in any case. 3As noted earlier, we shall always assume our springs to be “massless,” that is, that their inertia is negligible. In turn, negligible inertia means that the spring does not “keep going”: it stops stretching as soon as it is back to its original length. | University Physics I Classical Mechanics_Page_111_Chunk4309 |
94 CHAPTER 5. INTERACTIONS AND ENERGY The main point is that this kind of physical setup (a cart fitted with a spring) would indeed give us an elastic collision, and a kinetic energy curve very much like the ones I used, for illustration purposes, in Chapter 4; only now we also have a potential energy curve to go with it, and to show where the energy is “hiding” while the collision lasts. (You might wonder, anyway, what kind of potential energy function would actually produce the made-up elastic collision curves in Chapters 3 and 4? The (perhaps surprising) answer is, I do not really know, and I have no way to find out! If you are curious about why, again look at the “Examples” section at the end of the chapter.) 5.1.2 Potential energy functions and “energy landscapes” The potential energy function of a system, as illustrated in the above examples, serves to let us know how much energy can be stored in, or extracted from, the system by changing its configuration, that is to say, the positions of its parts relative to each other. We have seen this in the case of the gravitational force (the “configuration” in this case being the distance between the object and the earth), and just now in the case of a spring (how stretched or compressed it is). In all these cases we should think of the potential energy as being a property of the system as a whole, not any individual part; it is, very loosely speaking, something akin to a “stress” in the system that can be turned into motion under the right conditions. It is a consequence of the principle of conservation of momentum that, if the interaction between two particles can be described by a potential energy function, this should be a function only of their relative position, that is, the quantity x1 −x2 (or x2 −x1), and not of the individual coordinates, x1 and x2, separately4. The example of the spring in the previous section illustrates this, whereas the gravitational potential energy example shows how this can be simplified in an important case: in Eq. (5.3), the height y of the object above the ground is really a measure of the distance between the object and the earth, something that we could write, in full generality, as |⃗ro −⃗rE| (where ⃗ro and ⃗rE are the position vectors of the Earth and the object, respectively). However, since we do not expect the Earth to move very much as a result of the interaction, we can take its position to be constant, and only include the position of the object explicitly in our potential energy function, as we did above5. Generally speaking, then, we can identify a large class of problems where a “small” object or “particle” interacts with a much more massive one, and it is a good approximation to write the potential energy of the whole system as a function of only the position of the particle. In one 4We will see why in the next chapter! But, if you want to peek ahead, nothing’s preventing you from reading sections 6.1 and 6.2 right now. Basically, to conserve momentum we need Eq. (6.6) to hold, and as you can see from Eq. (6.18), having the potential energy depend only on x1 −x2 ensures that. 5This will change in Chapter 10, when we get to study gravity over a planetary scale. | University Physics I Classical Mechanics_Page_112_Chunk4310 |
5.1. CONSERVATIVE INTERACTIONS 95 dimension, then, we have a situation where, once the initial conditions (the particle’s initial position and velocity) are known, the motion of the particle can be completely determined from the function U(x), where x is the particle’s position at any given time. This can be done, using calculus, essentially by the method illustrated in Example 5.6.3 at the end of this chapter (namely, let v = ± ! 2m(E −U(x)) and solve the resulting differential equation); but it is also possible to get some pretty valuable insights into the particle’s motion without using any calculus at all, through a mostly graphical approach that I would like to show you next. 0 5 10 15 20 25 30 -4 -3 -2 -1 0 1 2 3 4 energy (J) U(x) E = U+K x (m) Figure 5.3: A hypothetical potential energy curve for a particle in one dimension. The horizontal red line shows the total mechanical energy under the assumption that the particle starts out at x = −2 m with Ki = 8 J. The green line assumes the particle starts instead from rest at x = 1 m. In Figure 5.3 above I have assumed, as an example, that the potential energy of the system, as a function of the position of the particle, is given by the function U(x) = −x4/4 + 9x2/2 + 2x + 1 (in joules, if x is given in meters). Consider then what happens if the particle has a mass m = 4 kg and is found initially at xi = −2 m, with a velocity vi = 2 m/s. (This scenario goes with the red lines in Fig. 5.3, so please ignore the green lines for the time being.) Its kinetic energy will then be Ki = 8 J, whereas the potential energy will be U(−2) = 11 J. The total mechanical energy is then E = 19 J, as indicated by the red horizontal line. Now, as the particle moves, the total energy remains constant, so as it moves to the right, its poten- tial energy goes down at first, and consequently its kinetic energy goes up—that is, it accelerates. At some point, however (around x = −0.22 m) the potential energy starts to go up, and so the particle starts to slow down, although it keeps going, because K = E −U is still nonzero. However, when the particle eventually reaches the point x = 2 m, the potential energy U(2) = 19 J, and the kinetic energy becomes zero. | University Physics I Classical Mechanics_Page_113_Chunk4311 |
96 CHAPTER 5. INTERACTIONS AND ENERGY At that point, the particle stops and turns around, just like an object thrown vertically upwards. As it moves “down the potential energy hill,” it recovers the kinetic energy it used to have, so that when it again reaches the starting point x = −2 m, its speed is again 2 m/s, but now it is moving in the opposite direction, so it just passes through and over the next “hill” (since it has enough total energy to do so), and eventually moves outside the region shown in the figure. As another example, consider what would have happened if the particle had been released at, say, x = 1 m, but with zero velocity. (This is illustrated by the green lines in Fig. 5.3.) Then the total energy would be just the potential energy U(1) = 7.25 J. The particle could not possibly move to the right, since that would require the total energy to go up. It can only move to the left, since in that direction U(x) decreases (initially, at first), and that means K can increase (recall K is always positive as long as the particle is in motion). So the particle speeds up to the left until, past the point x = −0.22 m, U(x) starts to increase again and K has to go down. Eventually, as the figure shows, we reach a point (which we can calculate to be x = −1.548 m) where U(x) is once again equal to 7.25 J. This leaves no room for any kinetic energy, so the particle has to stop and turn back. The resulting motion consists of the particle oscillating back and forth forever between x = −1.548 m and x = 1 m. At this point, you may have noticed that the motion I have described as following from the U(x) function in Figure 5.3 resembles very much the motion of a car on a roller-coaster having the shape shown, or maybe a ball rolling up and down hills like the ones shown in the picture. In fact, the correspondence can be made exact—if we substitute sliding for rolling, since rolling motion has complications of its own. Given an arbitrary potential energy function U(x) for a particle of mass m, imagine that you build a “landscape” of hills and valleys whose height y above the horizontal, for a given value of the horizontal coordinate x, is given by the function y(x) = U(x)/mg. (Note that mg is just a constant scaling factor that does not change the shape of the curve.) Then, for an object of mass m sliding without friction over that landscape, under the influence of gravity, the gravitational potential energy at any point x would be U G(x) = mgy = U(x), and therefore its speed at any point will be precisely the same as that of the original particle, if it starts at the same point with the same velocity. This notion of an “energy landscape” can be extended to more than one dimension (although they are hard to visualize in three!), or generalized to deal with configuration parameters other than a single particle’s position. It can be very useful in a number of disciplines (not just physics), to predict the ways in which the configuration of a system may be likely to change. 5.2 Dissipation of energy and thermal energy From all the foregoing, it is clear that when an interaction can be completely described by a potential energy function we can define a quantity, which we have called the total mechanical | University Physics I Classical Mechanics_Page_114_Chunk4312 |
5.2. DISSIPATION OF ENERGY AND THERMAL ENERGY 97 energy of the system, Emech = K + U, that is constant throughout the interaction. However, we already know from our study of inelastic collisions that this is rarely the case. Essential to the concept of potential energy is the idea of “storage and retrieval” of the kinetic energy of the system during the interaction process. When kinetic energy simply disappears from the system and does not come back, a full description of the process in terms of a potential energy is not possible. Processes in which some amount of mechanical energy disappears (that is, it cannot be found any- where anymore as either macroscopic kinetic or potential energy) are called dissipative. Mysterious as they may appear at first sight, there is actually a simple, intuitive explanation for them. All macroscopic systems consist of a great number of small parts that enjoy, at the microscopic level, some degree of independence from each other and from the body to which they belong. Macro- scopic motion of an object requires all these parts to move together as a whole, at least on average; however, a collision with another object may very well “rattle” all these parts and leave them in a more or less disorganized state. If the total energy is conserved, then after the collision the object’s atoms or molecules may be, on average, vibrating faster or banging against each other more often than before, but they will do so in random directions, so this increased “agitation” will not be perceived as macroscopic motion of the object as a whole. This kind of random agitation at the microscopic level that I have just introduced is what we know today as thermal energy, and it is by far the most common “sink” or reservoir where macroscopic mechanical energy is “dissipated.” In our example of an inelastic collision, the energy the objects had is not gone from the universe, in fact it is still right there inside the objects themselves; it is just in a disorganized or incoherent state from which, as you can imagine, it would be pretty much impossible to retrieve it, since we would have to somehow get all the randomly-moving parts to get back to moving in the same direction again. We will have a lot more to say about thermal energy in a later chapter, but for the moment you may want to think of it as essentially noise: it is what is left (the residual motional or configurational energy, at the microscopic level) after you remove the average, macroscopically-observable kinetic or potential energy. So, for example, for a solid object moving with a velocity vcm, the kinetic part of its thermal energy would be the sum of the kinetic energies of all its microscopic parts, calculated in its center of mass (or zero-momentum) reference frame; that way you remove from every molecule’s velocity the quantity vcm, which they all must have in common—on average (since the body as a whole is moving with that velocity)6. In order to establish conservation of energy as a fact (which was one of the greatest scientific triumphs of the 19th century) it was clearly necessary to show experimentally that a certain amount 6Note that thermal energy is not necessarily just kinetic; it may have a configurational component to it as well. For example, imagine a collection of vibrating diatomic molecules. You may think of each one as two atoms connected by a spring. The length of the “spring” at rest determines the molecule’s nominal chemical energy; thermal vibrations cause this length to change, resulting in a net increase in energy that—as for two masses connected by a spring—has both a kinetic and a configurational (or “potential”) component. | University Physics I Classical Mechanics_Page_115_Chunk4313 |
98 CHAPTER 5. INTERACTIONS AND ENERGY of mechanical energy lost always resulted in the same predictable increase in the system’s thermal energy. Thermal energy is largely “invisible” at the macroscopic level, but we detect it indirectly through an object’s temperature. The crucial experiments to establish what at the time was called the “mechanical equivalent of heat” were carried out by James Prescott Joule in the 1850’s, and required exceedingly precise measurements of temperature (in fact, getting the experiments done was only half the struggle; the other half was getting the scientific establishment to believe that he could measure changes in temperature so accurately!) 5.3 Fundamental interactions, and other forms of energy At the most fundamental (microscopic) level, physicists today believe that there are only four (or three, depending on your perspective) basic interactions: gravity, electromagnetism, the strong nu- clear interaction (responsible for holding atomic nuclei together), and the weak nuclear interaction (responsible for certain nuclear processes, such as the transmutation of a proton into a neutron7 and vice-versa). In a technical sense, at the quantum level, electromagnetism and the weak nuclear interactions can be regarded as separate manifestations of a single, consistent quantum field theory, so they are sometimes referred to as “the electroweak interaction.” All of these interactions are conservative, in the sense that for all of them one can define the equiv- alent of a “potential energy function” (generalized, as necessary, to conform to the requirements of quantum mechanics and relativity), so that for a system of elementary particles interacting via any one of these interactions the total kinetic plus potential energy is a constant of the motion. For gravity (which we do not really know how to “quantize” anyway!), this function immediately carries over to the macroscopic domain without any changes, as we shall see in a later chapter, and the gravitational potential energy function I introduced earlier in this chapter is an approximation to it valid near the surface of the earth (gravity is such a weak force that the gravitational interac- tion between any two earth-bound objects is virtually negligible, so we only have to worry about gravitational energy when one of the objects involved is the earth itself). As for the strong and weak nuclear interactions, they are only appreciable over the scale of an atomic nucleus, so there is no question of them directly affecting any macroscopic mechanical processes. They are responsible, however, for various nuclear reactions in the course of which nuclear energy is, most commonly, transformed into electromagnetic energy (X- or gamma rays) and thermal energy. All the other forms of energy one encounters at the microscopic, and even the macroscopic, level have their origin in electromagnetism. Some of them, like the electrostatic energy in a capacitor or the magnetic interaction between two permanent magnets, are straightforward enough scale-ups of their microscopic counterparts, and may allow for a potential energy description at the macroscopic 7Plus a positron and a neutrino. | University Physics I Classical Mechanics_Page_116_Chunk4314 |
5.4. CONSERVATION OF ENERGY 99 level (and you will learn more about them next semester!). Many others, however, are more subtle and involve quantum mechanical effects (such as the exclusion principle) in a fundamental way. Among the most important of these is chemical energy, which is an extremely important source of energy for all kinds of macroscopic processes: combustion (and explosions!), the production of electrical energy in batteries, and all the biochemical processes that power our own bodies. However, the conversion of chemical energy into macroscopic mechanical energy is almost always a dissipative process (that is, one in which some of the initial chemical energy ends up irreversibly converted into thermal energy), so it is generally impossible to describe them using a (macroscopic) potential energy function (except, possibly, for electrochemical processes, with which we will not be concerned here). For instance, consider a chemical reaction in which some amount of chemical energy is converted into kinetic energy of the molecules forming the reaction products. Even when care is taken to “channel” the motion of the reaction products in a particular direction (for example, to push a cylinder in a combustion engine), a lot of the individual molecules will end up flying in the “wrong” direction, striking the sides of the container, etc. In other words, we end up with a lot of the chemical energy being converted into disorganized microscopic agitation—which is to say, thermal energy. Electrostatic and quantum effects are also responsible for the elastic properties of materials, which can sometimes be described by macroscopic potential energy functions, at least to a first approxi- mation (like the spring we studied earlier in the chapter). They are also responsible for the adhesive forces between surfaces that play an important role in friction, and various other kinds of what might be called “structural energies,” most of which play only a relatively small part in the energy balance where macroscopic objects are involved. 5.4 Conservation of energy Today, physics is pretty much founded on the belief that the energy of a closed system (defined as one that does not exchange energy with its surroundings—more on this in a minute) is always conserved: that is, internal processes and interactions will only cause energy to be “converted” from one form into another, but the total, after all the forms of energy available to the system have been carefully accounted for, will not change. This belief is based on countless experiments, on the one hand, and, on the other, on the fact that all the fundamental interactions that we are aware of do conserve a system’s total energy. Of course, recognizing whether a system is “closed” or not depends on having first a complete catalogue of all the ways in which energy can be stored and exchanged—to make sure that there is, in fact, no exchange of energy going on with the surroundings. Note, incidentally, that a “closed” | University Physics I Classical Mechanics_Page_117_Chunk4315 |
100 CHAPTER 5. INTERACTIONS AND ENERGY system is not necessarily the same thing as an “isolated” system: the former relates to the total energy, the latter to the total momentum. A parked car getting hotter in the sun is not a closed system (it is absorbing energy all the time) but, as far as its total momentum is concerned, it is certainly fair to call it “isolated.” (And as you keep this in mind, make sure you also do not mistake “isolated” for “insulated”!) Hopefully all these concepts will be further clarified when we introduce the additional auxiliary concepts of force, work, and heat (although the latter will not come until the end of the semester). For a closed system, we can state the principle of conservation of energy (somewhat symbolically) in the form K + U + Esource + Ediss = constant (5.8) where K is the total, macroscopic, kinetic energy; U the sum of all the applicable potential energies associated with the system’s internal interactions; Esource is any kind of internal energy (such as chemical energy) that is not described by a potential energy function, but can increase the system’s mechanical energy; and Ediss stands for the contents of the “dissipated energy reservoir”—typically thermal energy. As with the potential energy U, the absolute value of Esource and Ediss does not (usually) really matter: all we are interested in is how much they change in the course of the process under consideration. (a) (c) K Ediss Ugravity K Ediss Ugravity K Ediss Ugravity K Ediss UgravityUelastic Uelastic Uelastic Uelastic (b) (d) Figure 5.4: Energy bar diagrams for a system formed by the earth and a ball thrown downwards. (a) As the ball leaves the hand. (b) Just before it hits the ground. (c) During the collision, at the time of maximum compression. (d) At the top of the first bounce. The total number of energy “units” is the same in all the diagrams, as required by the principle of conservation of energy. From the diagrams you can tell that the coefficient of restitution e = ! 7/9. Figure 5.4 above is an example of this kind of “energy accounting” for a ball bouncing on the ground. If the ball is thrown down, the system formed by the ball and the earth initially has both | University Physics I Classical Mechanics_Page_118_Chunk4316 |
5.5. IN SUMMARY 101 gravitational potential energy, and kinetic energy (diagram (a)). Note that we could write the total kinetic energy as Kcm + Kconv, as we did in the previous chapter, but because of the large mass of the earth, the center of mass of the system is essentially the center of the earth, which, in our earth-bound coordinate system, does not move at all, so Kcm is, to an excellent approximation, zero. Then, the reduced mass of the system, µ = mbME/(mb + ME) is, also to an excellent approximation, just equal to the mass of the ball, so Kconv = 1 2µv2 12 = 1 2mb(vb −ve)2 = 1 2mbv2 b (again, because the earth does not move). So all the kinetic energy that we have is the kinetic energy of the ball, and it is all, in principle, convertible (as you can see if you replace the ball, for instance, with a bean bag). As the ball falls, gravitational potential energy is being converted into kinetic energy, and the ball speeds up. As it is about to hit the ground (diagram (b)), the potential energy is zero and the kinetic energy is maximum. During the collision with the ground, all the kinetic energy is temporarily converted into other forms of energy, which are essentially elastic energy of deformation (like the energy in a spring) and some thermal energy (diagram (c)). When it bounces back, its kinetic energy will only be a fraction e2 of what it had before the collision (where e is the coefficient of restitution). This kinetic energy is all converted into gravitational potential energy as the ball reaches the top of its bounce (diagram (d)). Note there is more dissipated energy in diagram (d) than in (c); this is because I have assumed that dissipation of energy takes place both during the compression and the subsequent expansion of the ball. 5.5 In summary 1. For conservative interactions one can define a potential energy U, such that that in the course of the interaction the total mechanical energy E = U + K of the system remains constant, even as K and U separately change. The function U is a measure of the energy stored in the configuration of the system, that is, the relative position of all its parts. 2. The potential energy function for a system of two particles must be a function of their relative position only: U(x1 −x2). However, if one of the objects is very massive, so it does not move during the interaction, its position may be taken to be the origin of coordinates, and U written as a function of the lighter object’s coordinate alone. 3. For a system formed by the earth and an object of mass m at a height y above the ground, the gravitational potential energy can be written as U G = mgy (approximately, as long as y is much smaller than the radius of the earth). 4. The elastic potential energy stored in an ideal spring of spring constant k and relaxed length x0, when stretched or compressed to an actual length x, is U spr = 1 2k(x −x0)2. 5. For an object in one dimension, with position coordinate x, which is part of a system with potential energy U(x), the motion can be predicted from the “energy landscape” formed by | University Physics I Classical Mechanics_Page_119_Chunk4317 |
102 CHAPTER 5. INTERACTIONS AND ENERGY the graph of the function U(x). The idea, elaborated in Section 5.1.2 above, is to imagine the equivalent motion of an object sliding without friction over the same landscape, under the influence of gravity. 6. The fundamental interactions currently known in physics are gravity, the strong nuclear in- teraction and the electroweak interaction (which includes all electromagnetic phenomena). These are all conservative. 7. At a macroscopic level, one finds a number of interactions and associated energies that are de- rived from electromagnetism and quantum mechanics. Two important examples are chemical energy, and elastic energy (which is energy associated with the elasticity or “springiness” of a body). Elastic energy can often be described approximately by a potential energy function, and as such be included in calculations of the total mechanical energy of a system. 8. Interactions between macroscopic objects almost always involve the conversion of some type of energy into another. Typically, some of the total mechanical energy is always lost in the conversion process, because it is impossible to keep at least some of the energy from spreading itself randomly among the microscopic parts that make up the interacting objects. This is an intrinsically irreversible process known as dissipation of energy. 9. Most of the time the dissipated energy ends up as thermal energy, which is energy associated with a random agitation at the atomic or molecular level. 10. A closed system is one that does not exchange energy with its surroundings. This is not necessarily the same thing as an isolated system (which is one that does not exchange mo- mentum with its surroundings). For a closed system, the sum of its macroscopic mechanical energy (kinetic + potential) and all its other “internal” energies (chemical, thermal), must be a constant. | University Physics I Classical Mechanics_Page_120_Chunk4318 |
5.6. EXAMPLES 103 5.6 Examples 5.6.1 Inelastic collision in the middle of a swing Tarzan swings on a vine to rescue a helpless explorer (as usual) from some attacking animal or another. He begins his swing from a branch a height of 15 m above the ground, grabs the explorer at the bottom of his swing, and continues the swing, upwards this time, until they both land safely on another branch. Suppose that Tarzan weighs 90 kg and the explorer weighs 70 , and that Tarzan doesn’t just drop from the branch, but pushes himself offso that he starts the swing with a speed of 5 m/s. How high a branch can he and the explorer reach? Solution Let us break this down into parts. The first part of the swing involves the conversion of some amount of initial gravitational potential energy into kinetic energy. Then comes the collision with the explorer, which is completely inelastic and we can analyze using conservation of momentum (assuming Tarzan and the explorer form an isolated system for the brief time the collision lasts). After that, the second half of their swing involves the complete conversion of their kinetic energy into gravitational potential energy. Let m1 be Tarzan’s mass, m2 the explorer’s mass, hi the initial height, and hf the final height. We also have three velocities to worry about (or, more properly in this case, speeds, since their direction is of no concern, as long as they all point the way they are supposed to): Tarzan’s initial velocity at the beginning of the swing, which we may call vtop; his velocity at the bottom of the swing, just before he grabs the explorer, which we may call vbot1, and his velocity just after he grabs the explorer, which we may call vbot2. (If you find those subscripts confusing, I am sorry, they are the best I could do; please feel free to make up your own.) • First part: the downswing. We apply conservation of energy, in the form (5.8), to the first part of the swing. The system we consider consists of Tarzan and the earth, and it has kinetic energy as well as gravitational potential energy. We ignore the source energy and the dissipated energy terms, and consider the system closed despite the fact that Tarzan is holding onto a vine (as we shall see in a couple of chapters, the vine does no “work” on Tarzan—meaning, it does not change his energy, only his direction of motion—because the force it exerts on Tarzan is always perpendicular to his displacement): Ktop + U G top = Kbot1 + U G bot (5.9) In terms of the quantities I introduced above, this equation becomes: 1 2m1v2 top + m1ghi = 1 2m1v2 bot1 + 0 which can be solved to give v2 bot1 = v2 top + 2ghi (5.10) | University Physics I Classical Mechanics_Page_121_Chunk4319 |
104 CHAPTER 5. INTERACTIONS AND ENERGY (note that this is just the familiar result (2.10) for free fall! This is because, as I pointed out above, the vine does no work on the system.). Substituting, we get vbot1 = *& 5 m s '2 + 2 & 9.8 m s2 ' × (15 m) = 17.9 m s • Second part: the completely inelastic collision. The explorer is initially at rest (we assume he has not seen the wild beast ready to pounce yet, or he has seen it and he is paralyzed by fear!). After Tarzan grabs him they are moving together with a speed vbot2. Conservation of momentum gives m1vbot1 = (m1 + m2)vbot2 (5.11) which we can solve to get vbot2 = m1vbot1 m1 + m2 = (90 kg) × (17.9 m/s) 160 kg = 10 m s • Third part: the upswing. Here we use again conservation of energy in the form Kbot2 + U G bot = Kf + U G f (5.12) where the subscript f refers to the very end of the swing, when they both safely reach their new branch, and all their kinetic energy has been converted to gravitational potential energy, so Kf = 0 (which means that is as high as they can go, unless they start climbing the vine!). This equation can be rewritten as 1 2(m1 + m2)v2 bot2 + 0 = 0 + (m1 + m2)ghf and solving for hf we get hf = v2 bot2 2g = (10 m/s)2 2 × 9.8 m/s2 = 5.15 m 5.6.2 Kinetic energy to spring potential energy: block collides with spring A block of mass m is sliding on a frictionless, horizontal surface, with a velocity vi. It hits an ideal spring, of spring constant k, which is attached to the wall. The spring compresses until the block momentarily stops, and then starts expanding again, so the block ultimately bounces off. (a) In the absence of dissipation, what is the block’s final speed? (b) By how much is the spring compressed? Solution This is a simpler version of the problem considered in Section 5.1.1, and in the next example. The | University Physics I Classical Mechanics_Page_122_Chunk4320 |
5.6. EXAMPLES 105 problem involves the conversion of kinetic energy into elastic potential energy, and back. In the absence of dissipation, Eq. (5.8), specialized to this system (the spring and the block) reads: K + U spr = constant (5.13) For part (a), we consider the whole process where the spring starts relaxed and ends relaxed, so U spr i = U spr f = 0. Therefore, we must also have Kf = Ki, which means the block’s final speed is the same as its initial speed. As explained in the chapter, this is characteristic of a conservative interaction. For part (b), we take the final state to be the instant where the spring is maximally compressed and the block is momentarily at rest, so all the energy in the system is spring (which is to say, elastic) potential energy. If the spring is compressed a distance d (that is, x −x0 = −d in Eq. (5.5)), this potential energy is 1 2kd2, so setting that equal to the system’s initial energy we get: Ki + 0 = 0 + 1 2kd2 (5.14) or 1 2mv2 i = 1 2kd2 which can be solved to get d = *m k vi | University Physics I Classical Mechanics_Page_123_Chunk4321 |
106 CHAPTER 5. INTERACTIONS AND ENERGY 5.7 Advanced Topics 5.7.1 Two carts colliding and compressing a spring Unlike the example 5.6.2, which considered a stationary spring and asked only questions about initial and final states, this example is intended to show you how one can use “energy methods” to solve for the actual motion of a relatively complicated system as a function of time. The system is the two carts colliding, one of them fitted with a spring, considered in Section 5.1.1. Although all the physics involved is straightforward, the math is at a higher level than you will be using this semester, so I’m presenting this here as a “curiosity” only. First, recall the total kinetic energy for a collision problem can be written as K = Kcm + Kconv, where (if the system is isolated) Kcm remains constant throughout. Then, the total mechanical energy E = K + U = Kcm + Kconv + U. This is also constant, and before the interaction happens, when U = 0, we have E = Kcm + Kconv,i, so setting these two things equal and canceling out Kcm we get Kconv = Kconv,i −U (5.15) where the potential energy function is given by Eq. (5.6). Introducing the relative coordinate x12 = x2 −x1, and the relative velocity v12, Eq. (5.15) becomes 1 2µv2 12 = 1 2µv2 12,i −1 2k(x12 −x0)2 (5.16) an equation that must hold while the interaction is going on. We can solve this for v12, and then notice that both x12 and v12 are functions of time, and the latter is the derivative with respect to time of the former, so v12 = ± # v2 12,i −(k/µ) (x12 −x0)2 (5.17) dx12 dt = ± # v2 12,i −(k/µ) (x12(t) −x0)2 (5.18) (The “±” sign means that the quantity on the right-hand side has to be negative at first, when the carts are coming together, and positive later, when they are coming apart. This is because I have assumed cart 1 starts to the left of cart 2, so going in cart 2, as seen from cart 1, appears to be moving to the left.) Equation (5.18) is what is known, in calculus, as a differential equation. The problem is to find a function of t, x12(t), such that when you take its derivative you get the expression on the right-hand side. If you know how to calculate derivatives, you can check that the solution is in fact x12(t) = x0 −v12,i ω sin[ω(t −tc)] for tc ≤t ≤tc + π/ω (5.19) | University Physics I Classical Mechanics_Page_124_Chunk4322 |
5.7. ADVANCED TOPICS 107 where the quantity ω = ! k/µ, and the time tc is the time cart 1 first makes contact with the spring: tc = (x2i −x0 −x1i)/v1i. The solution (5.19) is valid for as long as the spring is compressed, which is to say, for as long as x12(t) < x0, or sin[ω(t −tc)] > 0, which translates to the condition on t shown above. Having a solution for x12, we could now obtain explicit results for x1(t) and x2(t) separately, using the fact that x1 = xcm −m2x12/(m1 +m2), and x2 = xcm+m1x12/(m1 +m2) (compare Eqs. (4.10), in chapter 4), and finding the position of the center of mass as a function of time is a trivial problem, since it just moves with constant velocity. We do not, however, need to do any of this in order to generate the plots of the kinetic and potential energy shown in Fig. 5.2: the potential energy depends only on x2 −x1, which is given explicitly by Eq. (5.19), and the kinetic energy is equal to Kcm + Kconv, where Kcm is constant and Kconv is given by Eq. (5.16), which can also be easily rewritten in terms of Eq. (5.19). The results are Eqs. (5.7) in the text. 5.7.2 Getting the potential energy function from collision data Consider the collision illustrated in Figure 3.4 (back in Chapter 3). Can we tell what the potential energy function is for the interaction between the two carts? At first sight, it may seem that all the information necessary to “reconstruct” the function U(x1−x2) is available already, at least in graphical form: From Figure 3.4 you could get the value of x2 −x1 at any time t; then from Figure 4.5 you can get the value of K (in the elastic-collision scenario) for the same value of t; and then you could plot U = E −K (where E is the total energy) as a function of x2 −x1. But there is a catch: Figure 3.4 shows that the colliding objects never get any closer than x2 −x1 ≃ 0.28 mm, so we have no way to tell what U(x2−x1) is for smaller values of x2−x1. This is essentially the problem faced by particle physicists when they use collisions (which they do regularly) to try to determine the precise nature of the interactions between the particles they study! You can check this for yourself. The functions I used for x1(t) and x2(t) in figure 3.4 are x1(t) = 1 3 + (2t −10) erf(10 −2t) + 10 erf(10) + t −e−4(t−5)2 √π , −5 x2(t) = 1 3 + (5 −t) erf(10 −2t) −5 erf(10) + t + e−4(t−5)2 2√π , (5.20) Here, “erf” is the so-called “error function,” which you can find in any decent library of mathemat- | University Physics I Classical Mechanics_Page_125_Chunk4323 |
108 CHAPTER 5. INTERACTIONS AND ENERGY ical functions. This looks complicated, but it just gives you the shapes you want for the velocity curves. The derivative of the above is v1(t) = 1 3 (1 + 2 erf(10 −2t)) v2(t) = 1 3 (1 −erf(10 −2t)) (5.21) and you may want to try plotting these for yourself; the result should be Figure 3.1. Now, assume (as I did for figure 4.5) that m1 = 1 kg, and m2 = 2 kg, and use these values and the results (5.21) (assumed to be in m/s) to calculate Ksys as a function of t. Then U = Esys −Ksys, with Esys = 1/2 J: U = 1 2 −1 2m1v2 1(t) −1 2m2v2 2(t) = 1 3 ( 1 −erf2(10 −2t) ) (5.22) and now do a parametric plot of U versus x2 −x1, using t as a parameter. You will end up with a figure like the one below: U (J) x2 - x1 (mm) Figure 5.5: The potential energy function reconstructed from the information available for the collision shown in Figs. 3.1, 3.4, 4.5. No information can be gathered from those figures (nor from the explicit expressions (5.20) and (5.21) above) on the values of U for x2 −x1 < 0.28 mm, the distance of closest approach of the two carts. | University Physics I Classical Mechanics_Page_126_Chunk4324 |
5.8. PROBLEMS 109 5.8 Problems Problem 1 A particle is in a region where the potential energy has the form U = 5/x (in joules, if x is in meters). (a) Sketch this potential energy function for x > 0. (b) Assuming the particle starts at rest at x = 0.5 m, which way will it go if released? Why? (c) Under the assumption in part (b), what will be the particle’s kinetic energy after it has moved 0.1 m from its original position? (d) Now assume that initially the particle is at x = 1 m, moving towards the left with an initial velocity vi = 2 m/s. If the mass of the particle is 1 kg, how close to the origin can it get before it stops? Problem 2 A “ballistic pendulum” is a device (now largely obsolete, but very useful in its day) to measure the speed of a bullet as it hits a target. Let the target be a block of wood suspended from a string, as in the figure below. When the bullet hits, it is embedded in the wood, and together they swing, like a pendulum, to some maximum height h. The question is, how do you find the initial speed of the bullet (vi) if you know the mass of the bullet (m1), the mass of the block (m2), and the height h? h (a) (b) Figure 5.6: Ballistic pendulum. (a) Before the bullet hits. (b) After the bullet hits and is embedded in the block, at the maximum height of the swing. Problem 3 You drop a 0.5 kg ball from a height of 2 m, and it bounces back to a height of 1.5 m. Consider the system formed by the ball and the Earth, so we can speak properly of its gravitational potential | University Physics I Classical Mechanics_Page_127_Chunk4325 |
110 CHAPTER 5. INTERACTIONS AND ENERGY energy. (a) What is the kinetic energy of the ball just before it hits the ground? (b) What is the kinetic energy of the ball just after it bounces up? (c) What is the coefficient of restitution for this collision? (d) What kind of collision is this (elastic, inelastic, etc.)? Why? (e) If the coefficient of restitution does not change, how high would the ball rise on a second bounce? (f) On the graphs below, draw the energy bar diagrams for the system: (1) as the ball leaves your hand; (2) just before it hits the ground (assume h = 0 for practical purposes); (3) just after it leaves the ground on its way up (h = 0 still), and (4) at the top of its (first) bounce. Make sure to do this to scale, consistent with the values for the energies you have calculated above. Esource K U Ediss Esource K U Ediss Esource K U Ediss Esource K U Ediss (1) As the ball leaves your hand (2) Just before it hits the ground (3) Just after it leaves the ground on its way up (4) At the top of its bounce Problem 4 A 60-kg skydiver jumps from an airplane 4000 m above the earth. After falling 450 m, he reaches a terminal speed of 55 m/s (about 120 mph). This means that after this time his speed does not increase any more. (a) At the moment of the jump, what is the initial (gravitational) potential energy of the system formed by the earth and the skydiver? (Take U G = 0 at ground level.) (b) After the skydiver has fallen 450 m, what is the (gravitational) potential energy of the system? (Call this the “final” potential energy.) (c) What is the final kinetic energy of the diver at that time? (d) Assume the initial kinetic energy of the skydiver is zero. Is ∆K = −∆U for this system? If not, explain what happened to the “missing” energy. (e) Can the skydiver and the earth below (excluding the atmosphere!) be considered a closed system | University Physics I Classical Mechanics_Page_128_Chunk4326 |
5.8. PROBLEMS 111 here? Explain. (f) After the skydiver reaches terminal speed (and before he opens his parachute), he falls for a while at constant speed. What kind of energy conversion is taking place during this time? (Consider the system to be the earth, the skydiver, and the air around him). Problem 5 You shoot a 1-kg projectile straight up from a spring toy gun, and find that it reaches a height of 5 m. (How do you figure out the height? From the time of flight, of course! See problem 2 from Chapter 2.) You also measure that when you load the gun, the spring compresses a distance 10 cm. What is the value of the spring constant? | University Physics I Classical Mechanics_Page_129_Chunk4327 |
112 CHAPTER 5. INTERACTIONS AND ENERGY | University Physics I Classical Mechanics_Page_130_Chunk4328 |
Chapter 6 Interactions, part 2: Forces 6.1 Force As we saw in the previous chapter, when an interaction can be described by a potential energy function, it is possible to use this to get a full solution for the motion of the objects involved, at least in one dimension. In fact, energy-based methods (known as the Lagrangian and Hamiltonian methods) can be also generalized to deal with problems in three dimensions, and they also provide the most direct pathway to quantum mechanics and quantum field theory. It might be possible to write an advanced textbook on classical mechanics without mentioning the concept of force at all. On the other hand, as you may have also gathered from the example I worked out at the end of the previous chapter (section 5.6.3), solving for the equation of motion using energy-based methods may involve somewhat advanced math, even in just one dimension, and it only gets more complicated in higher dimensions. There is also the question of how to deal with interactions that are not conservative (at the macroscopic level) and therefore cannot be described by a potential energy function of the macroscopic coordinates. And, finally, there are specialized problem areas (such as the entire field of statics) where you actually want to know the forces acting on the various objects involved. For all these reasons, the concept of force will be introduced here, and the next few chapters will illustrate how it may be used to solve a variety of elementary problems in classical mechanics. This does not mean, however, that we are going to forget about energy from now on: as we will see, energy methods will continue to provide useful shortcuts in a variety of situations as well. We start, as usual, by considering two objects that form an isolated system, so they interact with each other and with nothing else. As we have seen, under these circumstances their individual momenta change, but the total momentum remains constant. We are going to take the rate of 113 | University Physics I Classical Mechanics_Page_131_Chunk4329 |
114 CHAPTER 6. INTERACTIONS, PART 2: FORCES change of each object’s momentum as a measure of the force exerted on it by the other object. Mathematically, this means we will write for the average force exerted by 1 on 2 over the time interval ∆t the expression (F12)av = ∆p2 ∆t (6.1) Please observe the notation we are going to use: the subscripts on the symbol F are in the order “by,on”, as in “force exerted by” (object identified by first subscript) “on” (object identified by second subscript). (The comma is more or less optional.) You can also see from Eq. (6.1) that the SI units of force are kg·m/s2. This combination of units has the special name “newton,” and it’s abbreviated by an uppercase N. In the same way as above, we can write the average force exerted by object 2 on object 1: (F21)av = ∆p1 ∆t (6.2) and we know, by conservation of momentum, that we must have ∆p1 = −∆p2, so we get our first important result, (F12)av = −(F21)av (6.3) That is, whenever two objects interact, they always exert equal (in magnitude) and opposite (in direction) forces on each other. This is most often called Newton’s third law of motion, or informally “the law of action and reaction.” We might as well now proceed along familiar lines and take the limit of Eqs. (6.1) and (6.2) above, as ∆t goes to zero, in order to introduce the more general concept of the instantaneous force (or just the “force,” without any further qualifiers). We then get F12 = dp2 dt F21 = dp1 dt (6.4) and, since Eq. (6.3) should hold for a time interval of any size, F12 = −F21 (6.5) Now, under most circumstances the mass of, say, object 2 will not change during the interaction, so we can write F12 = d dt(m2v2) = m2 dv2 dt = m2a2 (6.6) This is the result that we often refer to as “F = ma”, also known as Newton’s second law of motion: the (net) force acting on an object is equal to the product of its inertial mass and its acceleration. The formulation in terms of the rate of change of momentum, as in Eqs. (6.4), is, | University Physics I Classical Mechanics_Page_132_Chunk4330 |
6.1. FORCE 115 however, somewhat more general, so it is technically preferred, even though this semester we will directly use F = ma throughout. If you want an example of a physical situation where F = dp/dt is not equivalent to F = ma, consider a system where object 1 is a rocket, including its fuel, and “object” 2 are the gases ejected by the rocket. In this case, the mass of both “objects” is constantly changing, as the fuel is burned and more gases are ejected, and so the more general form F = dp/dt needs to be used to calculate the force on the rocket (the thrust) at any given time. At this point you may be wondering just what is Newton’s first law? It is just the law of inertia: an object on which no force acts will stay at rest if it is initially at rest, or will move with constant velocity. 6.1.1 Forces and systems of particles What if you had, say, three objects (let us make them “particles,” for simplicity), all interacting with one another? In physics we find that all our interactions are pairwise additive, that is, we can write the total potential energy of the system as the sum of the potential energies associated with each pair of particles separately. As we will see in a moment, this means that the corresponding forces are additive too, so that, for instance, the total force on particle 1 could be written as Fall,1 = F21 + F31 = dp1 dt (6.7) Consider now the most general case of a system that has an arbitrary number of particles, and is not isolated; that is, there are other objects, outside the system, that exert forces on some or all of the particles that make up the system. We will call these external forces. The sum of all the forces (both internal and external) acting on all the particles will take a form like this: Ftotal = Fext,1 + F21 + F31 + . . . + Fext,2 + F12 + F32 + . . . + . . . = dp1 dt + dp2 dt + . . . (6.8) where Fext,1 is the sum of all the external forces acting on particle 1, and so on. But now, observe that because of Newton’s third law, Eq. (6.5), for every term of the form Fij appearing in the sum (6.8), there is a corresponding term Fji = −Fij (you can see this explicitly already in Eq. (6.8) with F12 and F21), so all those terms (which represent all the internal forces) are going to cancel out, and we will be left only with the sum of the external forces: Fext,1 + Fext,2 + . . . = dp1 dt + dp2 dt + . . . (6.9) The left-hand side of this equation is the sum of all the external forces; the right-hand side is the rate of change of the total momentum of the system. But the total momentum of the system is | University Physics I Classical Mechanics_Page_133_Chunk4331 |
116 CHAPTER 6. INTERACTIONS, PART 2: FORCES just equal to Mvcm (compare Eq. (3.11), in the “Momentum” chapter). So we have Fext,all = dpsys dt = d dt(Mvcm) (6.10) This extends a previous result. We already knew that in the absence of external forces, the mo- mentum of a system remained constant. Now we see that the system’s momentum responds to the net external force as if the whole system was a single particle of mass equal to the total mass M and moving at the center of mass velocity vcm. In fact, assuming that M does not change we can rewrite Eq. (6.10) in the form Fext,all = Macm (6.11) where acm is the acceleration of the center of mass. This is the key result that allows us to treat extended objects as if they were particles: as far as the motion of the center of mass is concerned, all the internal forces cancel out (as we already saw in our study of collisions), and the point representing the center of mass responds to the sum of the external forces as if it were just a particle of mass M subject to Newton’s second law, F = ma. The result (6.11) applies equally well to an extended solid object that we choose to mentally break up into a collection of particles, as to an actual collection of separate particles, or even to a collection of separate extended objects; in the latter case, we would just have each object’s motion represented by the motion of its own center of mass. Finally, note that all the results above generalize to more than one dimension. In fact, forces are vectors (just like velocity, acceleration and momentum), and all of the above equations, in 3 dimensions, apply separately to each vector component. In one dimension, we just need to be aware of the sign of the forces, whenever we add several of them together. 6.2 Forces and potential energy In the last chapter I mentioned a special case that we encounter often, in which a lighter object is interacting with a much more massive one, so that the massive one essentially does not move at all as a result of the interaction. Note that this does not contradict Newton’s 3rd law, Eq. (6.5): the forces the two objects exert on each other are the same in magnitude, but the acceleration of each object is inversely proportional to its mass, so F12 = −F21 implies m2a2 = −m1a1 (6.12) and so if, for instance, m2 ≫m1, we get |a2| = |a1|m1/m2 ≪|a1|. In words, the more massive object is less responsive than the less massive one to a force of the same magnitude. This is just how we came up with the concept of inertial mass in the first place! Anyway, you’ll recall that in this situation I could just write the potential energy function of the whole system as a function of only the lighter object’s coordinate, U(x). I am going to use this | University Physics I Classical Mechanics_Page_134_Chunk4332 |
6.2. FORCES AND POTENTIAL ENERGY 117 simplified setup to show you a very interesting relationship between potential energies and forces. Suppose this is a closed system in which no dissipation of energy is taking place. Then the total mechanical energy is a constant: Emech = 1 2mv2 + U(x) = constant (6.13) (Here, m is the mass of the lighter object, and v its velocity; the more massive object does not contribute to the total kinetic energy, since it does not move!) As the lighter object moves, both x and v in Eq. (6.13) change with time (recall, for instance, our study of “energy landscapes” in the previous chapter, section 5.1.2). So I can take the derivative of Eq. (6.13) with respect to time, using the chain rule, and noting that, since the whole thing is a constant, the total value of the derivative must be zero: 0 = d dt $1 2m ( v(t) )2 + U ( x(t) )% = mv(t) dv dt + dU dx dx dt (6.14) But note that dx/dt is just the same as v(t). So I can cancel that on both terms, and then I am left with m dv dt = −dU dx (6.15) But dv/dt is just the acceleration a, and F = ma. So this tells me that F = −dU dx (6.16) and this is how you can always get the force from a potential energy function. Let us check it right away for the force of gravity: we know that U G = mgy, so F G = −dU G dy = −d dy(mgy) = −mg (6.17) Is this right? It seems to be! Recall all objects fall with the same acceleration, −g (assuming the upwards direction to be positive), so if F = ma, we must have F G = −mg. So the gravitational force exerted by the earth on any object (which I would denote in full by F G E,o) is proportional to the inertial mass of the object—in fact, it is what we call the object’s weight—but since to get the acceleration you have to divide the force by the inertial mass, that cancels out, and a ends up being the same for all objects, regardless of how heavy they are. Now that we have this result under our belt, we can move on to the slightly more challenging case of two objects of comparable masses interacting through a potential energy function that must be, as I pointed out in the previous chapter, a function of just the relative coordinate x12 = x2 −x!. | University Physics I Classical Mechanics_Page_135_Chunk4333 |
118 CHAPTER 6. INTERACTIONS, PART 2: FORCES I claim that in that case you can again get the force on object 1, F21, by taking the derivative of U(x2 −x1) with respect to x1 (leaving x2 alone), and reciprocally, you get F12 by taking the derivative of U(x2 −x1) with respect to x2. Here is how it works, again using the chain rule: F21 = −d dx1 U(x12) = −dU dx12 d dx1 (x2 −x1) = dU dx12 F12 = −d dx2 U(x12) = −dU dx12 d dx2 (x2 −x1) = −dU dx12 (6.18) and you can see that this automatically ensures that F21 = −F12. In fact, it was in order to ensure this that I required that U should depend only on the difference of x1 and x2, rather than on each one separately. Since we got the condition F21 = −F12 originally from conservation of momentum, you can see now how the two things are related1. The only example we have seen so far of this kind of potential energy function was in last chapter’s Section 5.1.1, for two carts interacting through an “ideal” spring. I told you there that the potential energy of the system could be written as 1 2k(x2 −x1 −x0)2, where k was the “spring constant” and x0 the relaxed length of the spring. If you apply Eqs. (6.18) to this function, you will find that the force exerted (through the spring) by cart 2 on cart 1 is F21 = k(x2 −x1 −x0) (6.19) Note that this force will be negative under the assumptions we made last chapter, namely, that cart 2 is on the right, cart 1 on the left, and the spring is compressed, so that x2 −x1 < x0. Similarly, F12 = −k(x2 −x1 −x0) (6.20) and this one, as it should, is positive. The results (6.19) and (6.20) basically tell you what we mean by an “ideal spring” in physics: it is a spring that pulls (if stretched) or pushes (if compressed) with a force that is proportional to the change from its equilibrium length. Thus, if you fasten one end of the spring at x = 0, and stretch it or compress it so that the other end is at x, the spring will respond by exerting a force F spr = −k(x −x0) (6.21) As you can see, this is negative if x > x0 > 0 (spring stretched, pulling force) and positive if x < x0 (spring compressed, pushing force). In fact, the spring exerts an equal (in magnitude) and opposite (in direction) force at the other end (the one attached to the wall), so Eq. (6.21) only gives the correct sign of the force at the end that is denoted by the coordinate value x. Equations (6.19) and (6.20) are a bit clearer in this respect: Eq. (6.19) gives the correct sign of the force at point x1, and Eq. (6.20) the correct sign at point x2. 1The result (6.18) generalizes to more dimensions, but to do it properly you need vectors and partial derivative notation, and I’m already bending the notational rules a little bit here. . . | University Physics I Classical Mechanics_Page_136_Chunk4334 |
6.2. FORCES AND POTENTIAL ENERGY 119 Figure 6.1 shows, in black, all the forces exerted by a spring with one fixed end, according as to whether it is relaxed, compressed, or stretched. I have assumed that it is pushed or pulled by a hand (not shown) at the “free” end, hence the subscript “h”, whereas the subscript “w” stands for “wall.” Note that the wall and the hand, in turn, exert equal and opposite forces on the spring, shown in red in the figure. x x0 Fs,w spr Fs,w spr Fs,h spr Fs,h spr Fh,s c Fh,s c Fw,s c Fw,s c Figure 6.1: Forces (in black) exerted by a spring with one end attached to a wall and the other pushed or pulled by a hand (not shown). In every case the force is proportional to the change in the length of the spring from its equilibrium, or relaxed, value, shown here as x0. For this figure I have set the proportionality constant k = 1. The forces exerted on the spring, by the wall and by the hand, are shown in red. Equation (6.21) is generally referred to as Hooke’s law, after the British scientist Robert Hooke (a contemporary of Newton). Of course, it is not a “law” at all, merely a useful approximation to the way most springs behave as long as you do not stretch them or compress them too much2. A note on the way the forces have been labeled in Figure 6.1. I have used the generic symbol “c”, which stands for “contact,” to indicate the type of force exerted by the wall and the hand on the spring. In fact, since each pair of forces (by the hand on the spring and by the spring on the hand, for instance), at the point of contact, arises from one and the same interaction, I should have used the same “type” notation for both, but it is widespread practice to use a superscript like “spr” to denote a force whose origin is, ultimately, a spring’s elasticity. This does not change the fact that the spring force, at the point where it is applied, is indeed a contact force. So, next, a word on “contact” forces. Basically, what we mean by that is forces that arise where objects “touch,” and we mean this by opposition to what are called instead “field” forces (such as gravity, or magnetic or electrostatic forces) which “act at a distance.” The distinction is actually 2Assuming that you can compress them! Some springs, such as slinkies, actually cannot be compressed because their coils are already in contact when they are relaxed. Nevertheless, Eq. (6.21) will still apply approximately to such a spring when it is stretched, that is, when x > x0. | University Physics I Classical Mechanics_Page_137_Chunk4335 |
120 CHAPTER 6. INTERACTIONS, PART 2: FORCES only meaningful at the macroscopic level, since at the microscopic level objects never really touch, and all forces are field forces, it is just that some are “long range” and some are “short range.” For our purposes, really, the word “contact” will just be a convenient, catch-all sort of moniker that we will use to label the force vectors when nothing more specific will do. 6.3 Forces not derived from a potential energy As we have seen in the previous section, for interactions that are associated with a potential energy, we are always able to determine the forces from the potential energy by simple differentiation. This means that we do not have to rely exclusively on an equation of the type F = ma, like (6.4) or (6.6), to infer the value of a force from the observed acceleration; rather, we can work in reverse, and predict the value of the acceleration (and from it all the subsequent motion) from our knowledge of the force. I have said before that, on a microscopic level, all the interactions can be derived from potential energies, yet at the macroscopic level this is not generally true: we have many kinds of interactions for which the associated “stored” or converted energy cannot, in general, be written as a function of the macroscopic position variables for the objects making up the system (by which I mean, typically, the positions of their centers of mass). So what do we do in those cases? The forces of this type with which we shall deal this semester actually fall into two different categories: the ones that do not dissipate energy, and that we could, in fact, associate with a potential energy if we wanted to3, and the ones that definitely dissipate energy and need special handling. The former category includes the normal force, tension, and the static friction force; the second category includes the force of kinetic (or sliding) friction, and air resistance. A brief description of all these forces, and the methods to deal with them, follows. 6.3.1 Tensions Tension is the force exerted by a stretched spring, and, similarly, by objects such as cables, ropes, and strings in response to a stretching force (or load) applied to them. It is ultimately an elastic force, so, as I said above, we could in principle describe it by a potential energy, but in practice cables, strings and the like are so stiffthat it is often all right to neglect their change in length altogether and assume that no potential energy is, in fact, stored in them. The price we pay for this simplification (and it is a simplification) is that we are left without an independent way to determine the value of the tension in any specific case; we just have to infer it from the acceleration 3If we wanted to complicate our life, that is. . . | University Physics I Classical Mechanics_Page_138_Chunk4336 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 121 of the object on which it acts (since it is a reaction force, it can assume any value as required to adjust to any circumstance—up to the point where the rope snaps, anyway). Thus, for instance, in the picture below, which shows two blocks connected by a rope over a pulley, the tension force exerted by the rope on block 1 must equal m1a1, where a1 is the acceleration of that block, provided there are no other horizontal forces (such as friction) acting on it. For the hanging block, on the other hand, the net force is the sum of the tension on the other end of the rope (pulling up) and gravity, pulling down. If we choose the upward direction as positive, we can write Newton’s second law for the second block as F t r,2 −m2g = m2a2 (6.22) Two things need to be realized now. First, if the rope is inextensible, both blocks travel the same distance in the same time, so their speeds are always the same, and hence the magnitude of their accelerations will always be the same as well; only the sign may be different depending on which direction we choose as positive. If we take to the right to be positive for the horizontal motion, we will have a2 = −a1. I’m just going to call a1 = a, so then a2 = −a. 1 2 Fr,1 t Fr,2 t FE,2 G Fs,1 fr a1 a2 Figure 6.2: Two blocks joined by a massless, inextensible strength threaded over a massless pulley. An optional friction force (in red, where fr could be either s or k) is shown for use later, in the discussion in subsection 3.3. In this subsection, however, it is assumed to be zero. The second thing to note is that, if the rope’s mass is negligible, it will, like an ideal spring, pull with a force with the same magnitude on both ends. With our specific choices (up and to the right is positive), we then have F t r,2 = F t r,1, and I’m just going to call this quantity F t. All this yields, | University Physics I Classical Mechanics_Page_139_Chunk4337 |
122 CHAPTER 6. INTERACTIONS, PART 2: FORCES then, the following two equations: F t = m1a F t −m2g = −m2a (6.23) The system (6.23) can be easily solved to get a = m2g m1 + m2 F t = m1m2g m1 + m2 (6.24) 6.3.2 Normal forces Normal force is the reaction force with which a surface pushes back when it is being pushed on. Again, this works very much like an extremely stiffspring, this time under compression instead of tension. And, again, we will eschew the potential energy treatment by assuming that the surface’s actual displacement is entirely negligible, and we will just calculate the value of F n as whatever is needed in order to make Newton’s second law work. Note that this force will always be perpendicular to the surface, by definition (the word “normal” means “perpendicular” here); the task of dealing with a sideways push on the surface will be delegated to the static friction force, to be covered next. If I am just standing on the floor and not falling through it, the net vertical force acting on me must be zero. The force of gravity on me is mg downwards, and so the upwards normal force must match this value, so for this situation F n = mg. But don’t get too attached to the notion that the normal force must always be equal to mg, since this will often not be the case. Imagine, for instance, a person standing inside an elevator at the time it is accelerating upwards. With the upwards direction as positive, Newton’s second law for the person reads F n −mg = ma (6.25) and therefore for this situation F n = mg + ma (6.26) If you were weighing yourself on a bathroom scale in the elevator, this is the upwards force that the bathroom scale would have to exert on you, and it would do that by compressing a spring inside, and it would record the “extra” compression (beyond that required by your actual weight, mg) as extra weight. Conversely, if the elevator were accelerating downward, the scale would record you as being lighter. In the extreme case in which the cable of the elevator broke and you, the elevator and the scale ended up (briefly, before the emergency brake caught on) in free fall, you would all be falling with the same acceleration, you would not be pushing down on the scale at all, and its normal force as well as your recorded weight would be zero. This is ultimately the reason | University Physics I Classical Mechanics_Page_140_Chunk4338 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 123 for the apparent weightlessness experienced by the astronauts in the space station, where the force of gravity is, in fact, not very much smaller than on the surface of the earth. (We will return to this effect after we have a good grip on two-dimensional, and in particular circular, motion.) 6.3.3 Static and kinetic friction forces The static friction force is a force that prevents two surfaces in contact from slipping relative to each other. It is an extremely useful force, since we would not be able to drive a car, or ride a bicycle, or even walk, without it—as we know from experience, if we have ever tried to do any of those things on a low-friction surface (such as a sheet of ice). The science behind friction (known technically as tribology) is actually not very simple at all, and it is of great current interest for many reasons—whether the ultimate goal is to develop ways to reduce friction or to increase it. On an elementary level, we are all aware of the fact that even a surface that looks smooth on a macroscopic scale will actually exhibit irregularities, such as ridges and valleys, under a microscope. It makes sense, then, that when two such surfaces are pressed together, the bumps on one of them will hit, and be held in place by, the bumps on the other one, and that will prevent sliding until and unless a sufficient force is applied to temporarily “flatten” the bumps enough to allow the thing to move4. As long as this does not happen, that is, as long as the surfaces do not slide relative to each other, we say we are dealing with the static friction force, which is, at least approximately, an elastic force that does not dissipate energy: the small distortion of the “bumps” on the surfaces that takes place when you push on them typically happens slowly enough, and is small enough, to be reversible, so that when you stop pushing the two surfaces just go back to their initial state. This is no longer the case once the surfaces start sliding relative to each other. At that point the character of the friction force changes, and we have to deal with the sliding, or kinetic friction force, as I will explain below. The static friction force is also, like tension and the normal force, a reaction force that will adjust itself, within limits, to take any value required to prevent slippage in a given circumstance. Hence, its actual value in a particular situation cannot really be ascertained until the other relevant forces— the other forces pushing or pulling on the object—are known. For instance, for the system in Figure 6.2, imagine there is a force of static friction between block 1 and the surface on which it rests, sufficiently large to keep it from sliding altogether. How large 4This picture based, essentially, on classical physics, leaves out an atomic-scale effect that may be important in some cases, which is the formation of weak bonds between the atoms of both surfaces, resulting in an actual “adhesive” force. This is, for instance, how geckos can run up vertical walls. For our purposes, however, the classical picture (of small ridges and valleys bumping into each other) will suffice to qualitatively understand all the examples we will cover this semester. | University Physics I Classical Mechanics_Page_141_Chunk4339 |
124 CHAPTER 6. INTERACTIONS, PART 2: FORCES does this have to be? If there is no acceleration (a = 0), the equivalent of system (6.23) will be F s s,1 + F t = 0 F t −m2g = 0 (6.27) where F s s,1 is the force of static friction exerted by the surface on block 1, and we are going to let the math tell us what sign it is supposed to have. Solving the system (6.27) we just get the condition F s s,1 = −m2g (6.28) so this is how large F s s,1 has to be in order to keep the whole system from moving in this case. There is an empirical formula that tells us approximately how large the force of static friction can get in a given situation. The idea behind it is that, microscopically, the surfaces are in contact only near the top of their respective ridges. If you press them together harder, some of the ridges get flattened and the effective contact area increases; this in turn makes the surfaces more resistant to slippage. A direct measure of how strongly the two surfaces press against each other is, actually, just the normal force they exert on each other. So, in general, we expect the maximum force that static friction will be able to exert to be proportional to the normal force between the surfaces: --F s s1,s2 -- max = µs --F n s1,s2 -- (6.29) where s1 and s2 just mean “surface 1” and “surface 2,” respectively, and the number µs is known as the coefficient of static friction: it is a tabulated quantity that is determined experimentally, by testing the slippage of different surfaces against each other under different loads. In our example, the normal force exerted by the surface on block 1 has to be equal to m1g, since there is no vertical acceleration for that block, and so the maximum value that F s may have in this case is µsm1g, whatever µs might happen to be. In fact, this setup would give us a way to determine µs for these two surfaces: start with a small value of m2, and gradually increase it until the system starts moving. At that point we will know that m2g has just exceeded the maximum possible value of |F s 12|, namely, µsm1g, and so µs = (m2)max/m1, where (m2)max is the largest mass we can hang before the system starts moving. By contrast with all of the above, the kinetic friction force, which always acts so as to oppose the relative motion of the two surfaces when they are actually slipping, is not elastic, it is definitely dissipative, and, most interestingly, it is also not much of a reactive force, meaning that its value can be approximately predicted for any given circumstance, and does not depend much on things such as how fast the surfaces are actually moving relative to each other. It does depend on how hard the surfaces are pressing against each other, as quantified by the normal force, and on another tabulated quantity known as the coefficient of kinetic friction: ---F k s1,s2 --- = µk --F n s1,s2 -- (6.30) | University Physics I Classical Mechanics_Page_142_Chunk4340 |
6.3. FORCES NOT DERIVED FROM A POTENTIAL ENERGY 125 Note that, unlike for static friction, this is not the maximum possible value of |F k|, but its actual value; so if we know F n (and µk) we know F k without having to solve any other equations (its sign does depend on the direction of motion, of course). The coefficient µk is typically a little smaller than µs, reflecting the fact that once you get something you have been pushing on to move, keeping it in motion with constant velocity usually does not require the same amount of force. To finish offwith our example in Figure 2, suppose the system is moving, and there is a kinetic friction force F k s,1 between block 1 and the surface. The equations (6.23) then have to be changed to F t −µkm1g = m1a F t −m2g = −m2a (6.31) and the solution now is a = m2 −µkm1 m1 + m2 g F t = m1m2(1 + µk) m1 + m2 g (6.32) You may ask, why does kinetic friction dissipate energy? A qualitative answer is that, as the surfaces slide past each other, their small (sometimes microscopic) ridges are constantly “bumping” into each other; so you have lots of microscopic collisions happening all the time, and they cannot all be perfectly elastic. So mechanical energy is being “lost.” In fact, it is primarily being converted to thermal energy, as you can verify experimentally: this is why you rub your hands together to get warm, for instance. More dramatically, this is how some people (those who really know what they are doing!) can actually start a fire by rubbing sticks together. 6.3.4 Air resistance Air resistance is an instance of fluid resistance or drag, a force that opposes the motion of an object through a fluid. Microscopically, you can think of it as being due to the constant collisions of the object with the air molecules, as it cleaves its way through the air. As a result of these collisions, some of its momentum is transferred to the air, as well as some of its kinetic energy, which ends up as thermal energy (as in the case of kinetic friction discussed above). The very high temperatures that air resistance can generate can be seen, in a particularly dramatic way, on the re-entry of spacecraft into the atmosphere. Unlike kinetic friction between solid surfaces, the fluid drag force does depend on the velocity of the object (relative to the fluid), as well as on a number of other factors having to do with the object’s shape and the fluid’s density and viscosity. Very roughly speaking, for low velocities the | University Physics I Classical Mechanics_Page_143_Chunk4341 |
126 CHAPTER 6. INTERACTIONS, PART 2: FORCES drag force is proportional to the object’s speed, whereas for high velocities it is proportional to the square of the speed. In principle, one can use the appropriate drag formula together with Newton’s second law to calculate the effect of air resistance on a simple object thrown or dropped; in practice, this requires a somewhat more advanced math than we will be using this course, and the formulas themselves are complicated, so I will not introduce them here. One aspect of air resistance that deserves to be mentioned is what is known as “terminal velocity” (which I already introduced briefly in Section 2.3). Since air resistance increases with speed, if you drop an object from a sufficiently great height, the upwards drag force on it will increase as it accelerates, until at some point it will become as large as the downward force of gravity. At that point, the net force on the object is zero, so it stops accelerating, and from that point on it continues to fall with constant velocity. When the Greek philosopher Aristotle was trying to figure out the motion of falling bodies, he reasoned that, since air was just another fluid, he could slow down the fall (in order to study it better) without changing the physics by dropping objects in liquids instead of air. The problem with this approach, though, is that terminal velocity is reached much faster in a liquid than in air, so Aristotle missed entirely the early stage of approximately constant acceleration, and concluded (wrongly) that the natural way all objects fell was with constant velocity. It took almost two thousand years until Galileo disproved that notion by coming up with a better method to slow down the falling motion—namely, by using inclined planes. 6.4 Free-body diagrams As Figure 6.1 shows, trying to draw every single force acting on every single object can very quickly become pretty messy. And anyway, this is not usually what we need: what we need is to separate cleanly all the forces acting on any given object, one object at a time, so we can apply Newton’s second law, Fnet = ma, to each object individually. In order to accomplish this, we use what are known as free-body diagrams. In a free-body diagram, a potentially very complicated object is replaced symbolically by a dot or a small circle, and all the forces acting on the object are drawn (approximately to scale and properly labeled) as acting on the dot. Regardless of whether a force is a pulling or pushing force, the convention is to always draw it as a vector that originates at the dot. If the system is accelerating, it is also a good idea to indicate the acceleration’s direction also somewhere on the diagram. The figure below (next page) shows, as an example, a free-body diagram for block 1 in Figure 6.2, in the presence of both a nonzero acceleration and a kinetic friction force. The diagram includes all the forces, even gravity and the normal force, which were left out of the picture in Figure 6.2. | University Physics I Classical Mechanics_Page_144_Chunk4342 |
6.5. IN SUMMARY 127 Fr,1 t Fs,1 n FE,1 G Fs,1 k a Figure 6.3: Free-body diagram for block 1 in Figure 6.2, with the friction force adjusted so as to be compatible with a nonzero acceleration to the right. Note that I have drawn F n and the force of gravity F G E,1 as having the same magnitude, since there is no vertical acceleration for that block. If I know the value of µk, I should also try to draw F k = µkF n approximately to scale with the other two forces. Then, since I know that there is an acceleration to the right, I need to draw F t greater than F k, since the net force on the block must be to the right as well. And, if I were drawing a free-body diagram for block 2, I would have to make sure that I drew its weight, F G E,2, as being greater in magnitude than F t, since the net force on that block needs to be downwards. 6.5 In summary 1. Whenever two objects interact, they exert forces on each other that are equal in magnitude and opposite in direction (Newton’s 3rd law). 2. Forces are vectors, and they are additive. The total (or net) force on an object or system is equal to the rate of change of its total momentum (Newton’s 2nd law). If the system’s mass is constant, this can be written as Fext,all = Macm, where M is the system’s total mass and acm is the acceleration of its center of mass. Only the external forces contribute to this equation; the internal forces cancel out because of point 1 above. 3. For any interaction that can be derived from a potential energy function U(x1 −x2), the force exerted by object 2 on object 1 is equal to −dU/dx1 (where the derivative is calculated treating x2 as a constant), and vice-versa. 4. The force of gravity on an object near the surface of the earth is known as the object’s weight, and it is equal (in magnitude) to mg, where m is the object’s inertial mass. 5. An ideal spring whose relaxed length is x0, when stretched or compressed to a length x, exerts a pulling or pushing force, respectively, at both ends, with magnitude k|x −x0|, where k is called the spring constant. | University Physics I Classical Mechanics_Page_145_Chunk4343 |
128 CHAPTER 6. INTERACTIONS, PART 2: FORCES 6. When dealing with macroscopic objects we introduce several “constraint” forces whose values need to be determined from the accelerations through Newton’s second law: the tension F t in ropes, strings or cables; the normal force F n exerted by a surface in response to applied pressure; and the static friction force F s that prevents surfaces from slipping past each other. 7. The maximum possible value of the static friction force is µs|F n|, where µs is the coefficient of static friction. 8. The force of sliding or kinetic friction, F k, appears when two surfaces are sliding past each other. Its magnitude is µk|F n| (µk is the coefficient of static friction), and its sign is such as to oppose the sliding motion. Unlike the forces in 6 above, it is a dissipative force. 9. A free-body diagram is a way to depict all (and only) the forces acting on an object. The object should be represented as a small circle or dot. The forces should all be drawn as vectors originating on the dot, with their directions correctly shown and their lengths approximately to scale. The acceleration of the object should also be indicated elsewhere in the picture. The forces should be labeled like this: F type by,on. | University Physics I Classical Mechanics_Page_146_Chunk4344 |
6.6. EXAMPLES 129 6.6 Examples 6.6.1 Dropping an object on a weighing scale (Short version) Suppose you drop a 5-kg object on a spring scale from a height of 1 m. If the spring constant is k = 20, 000 N/m, what will the scale read? (Long version) OK, let’s break that up into parts. Suppose that a spring scale is just a platform (of negligible mass) sitting on top of a spring. If you put an object of mass m on top of it, the spring compresses so that (in equilibrium) it exerts an upwards force that matches that of gravity. (a) If the spring constant is k and the object’s mass is m and the whole system is at rest, what distance is the spring compressed? (b) If you drop the object from a height h, what is the (instantaneous) maximum compression of the spring as the object is brought to a momentary rest? (This part is an energy problem! Assume that h is much greater than the actual compression of the spring, so you can neglect that when calculating the change in gravitational potential energy.) (c) What mass would give you that same compression, if you were to place it gently on the scale, and wait until all the oscillations died down? (d) OK, now answer the question at the top! Solution (a) The forces acting on the object sitting at rest on the platform are the force of gravity, F G E,o = −mg, and the normal force due to the platform, F n p,o. This last force is equal, in magnitude, to the force exerted on the platform by the spring (it has to be, because the platform itself is being pushed down by a force F n o,p = −F n p,o, and this has to be balanced by the spring force). This means we can, for practical purposes, pretend the platform is not there and just set the upwards force on the object equal to the spring force, F spr s,p = −k(x −x0). So, Newton’s second law gives Fnet = F G E,o + F spr s,p = ma = 0 (6.33) For a compressed spring, x −x0 is negative, and we can just let d = x0 −x be the distance the spring is compressed. Then Eq. (6.33) gives −mg + kd = 0 so d = mg/k (6.34) when you just set an object on the scale and let it come to rest. (b) This part, as the problem says, is a conservation of energy situation. The system formed by the spring, the object and the earth starts out with some gravitational potential energy, and ends | University Physics I Classical Mechanics_Page_147_Chunk4345 |
130 CHAPTER 6. INTERACTIONS, PART 2: FORCES up, with the object momentarily at rest, with only spring potential energy: U G i + U spr i = U G f + U spr f mgyi + 0 = mgyf + 1 2kd2 max (6.35) where I have used the subscript “max” on the compression distance to distinguish it from what I calculated in part (a) (this kind of makes sense also because the scale is going to swing up and down, and we want only the maximum compression, which will give us the largest reading). The problem said to ignore the compression when calculating the change in U G, meaning that, if we measure height from the top of the scale, yi = h and yf = 0. Then, solving Eq. (6.35) for dmax, we get dmax = * 2mgh k (6.36) (c) For this part, let us rewrite Eq. (6.34) as meq = kdmax/g, where meq is the “equivalent” mass that you would have to place on the scale (gently) to get the same reading as in part (b). Using then Eq. (6.36), meq = k g * 2mgh k = . 2mkh g (6.37) (d) Now we can substitute the values given: m = 5 kg, h = 1 m, k = 20, 000 N/m. The result is meq = 143 kg. (Note: if you found the purely algebraic treatment above confusing, try substituting numerical values in Eqs. (6.34) and (6.36). The first equation tells you that if you just place the 5-kg mass on the scale it will compress a distance d = 2.45 mm. The second tells you that if you drop it it will compress the spring a distance dmax = 70 mm, about 28.6 times more, which corresponds to an “equivalent mass” 28.6 times greater than 5 kg, which is to say, 143 kg. Note also that 143 kg is an equivalent weight of 309 pounds, so if you want to try this on a bathroom scale I’d advise you to use smaller weights and drop them from a much smaller height!) 6.6.2 Speeding up and slowing down (a) A 1400-kg car, starting from rest, accelerates to a speed of 30 mph in 10 s. What is the force on the car (assumed constant) over this period of time? (b) Where does this force comes from? That is, what is the (external) object that exerts this force on the car, and what is the nature of this force? (c) Draw a free-body diagram for the car. Indicate the direction of motion, and the direction of the acceleration. (d) Now assume that the driver, traveling at 30 mph, sees a red light ahead and | University Physics I Classical Mechanics_Page_148_Chunk4346 |
6.6. EXAMPLES 131 pushes on the brake pedal. Assume that the coefficient of static friction between the tires and the road is µs = 0.7, and that the wheels don’t “lock”: that is to say, they continue rolling without slipping on the road as they slow down. What is the car’s minimum stopping distance? (e) Draw a free-body diagram of the car for the situation in (d). Again indicate the direction of motion, and the direction of the acceleration. (f) Now assume that the driver again wants to stop as in part (c), but he presses on the brakes too hard, so the wheels lock, and, moreover, the road is wet, and the coefficient of kinetic friction is only µk = 0.2. What is the distance the car travels now before coming to a stop? Solution (a) First, let us convert 30 mph to meters per second. There are 1, 609 meters to a mile, and 3, 600 seconds to an hour, so 30 mph = 10 × 1609/3600 m/s = 13.4 m/s. Next, for constant acceleration, we can use Eq. (2.4): ∆v = a∆t. Solving for a, a = ∆v ∆t = 13.4 m/s 10 s = 1.34 m s2 Finally, since F = ma, we have F = ma = 1400 kg × 1.34 m s2 = 1880 N (b) The force must be provided by the road, which is the only thing external to the car that is in contact with it. The force is, in fact, the force of static friction between the car and the tires. As explained in the chapter, this is a reaction force (the tires push on the road, and the road pushes back). It is static friction because the tires are not slipping relative to the road. In fact, we will see in Chapter 9 that the point of the tire in contact with the road has an instantaneous velocity of zero (see Figure 9.8). (c) This is the free-body diagram. Note the force of static friction pointing forward, in the direction of the acceleration. The forces have been drawn to scale. Fr,c s Fr,c n FE,c G a v | University Physics I Classical Mechanics_Page_149_Chunk4347 |
132 CHAPTER 6. INTERACTIONS, PART 2: FORCES (d) This is the opposite of part (a): the driver now relies on the force of static friction to slow down the car. The shortest stopping distance will correspond to the largest (in magnitude) acceleration, as per our old friend, Eq. (2.10): v2 f −v2 i = 2a∆x (6.38) In turn, the largest acceleration will correspond to the largest force. As explained in the chapter, the static friction force cannot exceed µsF n (Eq. (6.29)). So, we have F s max = µsF n = µsmg since, in this case, we expect the normal force to be equal to the force of gravity. Then |amax| = F s max m = µsmg m = µsg We can substitute this into Eq. (6.38) with a negative sign, since the acceleration acts in the opposite direction to the motion (and we are implicitly taking the direction of motion to be positive). Also note that the final velocity we want is zero, vf = 0. We get −v2 i = 2a∆x = −2µsg∆x From here, we can solve for ∆x: ∆x = v2 i 2µsg = (13.4 m/s)2 2 × 0.7 × 9.81 m/s2 = 13.1 m (e) Here is the free-body diagram. The interesting feature is that the force of static friction has re- versed direction relative to parts (a)–(c). It is also much larger than before. (The forces are again to scale.) Fr,c s Fr,c n FE,c G a v (f) The math for this part is basically identical to that in part (d). The difference, physically, is that now you are dealing with the force of kinetic (or “sliding”) friction, and that is always given by F k = µkF n (this is not an upper limit, it’s just what F k is). So we have a = −F k/m = −µkg, and, just as before (but with µk replacing µs), ∆x = v2 i 2µkg = (13.4 m/s)2 2 × 0.2 × 9.81 m/s2 = 45.8 m | University Physics I Classical Mechanics_Page_150_Chunk4348 |
6.6. EXAMPLES 133 This is a huge distance, close to half a football field! If these numbers are accurate, you can see that locking your brakes in the rain can have some pretty bad consequences. | University Physics I Classical Mechanics_Page_151_Chunk4349 |
134 CHAPTER 6. INTERACTIONS, PART 2: FORCES 6.7 Problems Problem 1 (a) Draw a free-body diagram for the skydiver in Problem 4 of Chapter 5. (b) What is the magnitude of the air drag force on the skydiver, after he reaches terminal speed? Problem 2 A book is sent sliding along a table with an initial velocity of 2 m/s. It slides for 1.5 m before coming to a stop. What is the coefficient of kinetic friction between the book and the table? Problem 3 You are pulling on a block of mass 4 kg that is attached, via a rope of negligible mass, to another block, of mass 6 kg. The coefficient of kinetic friction between the blocks and the surface on which they are sliding is µk. You find that when you apply a force of 20 N, the whole thing moves at constant velocity. (a) Draw a free-body diagram for each of the two blocks (b) What is the coefficient of kinetic friction between the blocks and the surface? (c) What is the tension in the rope? Problem 4 A box of mass 2 kg is sitting on top of a sled of mass 5 kg, which is resting on top of a frictionless surface (ice). (a) What is the normal force exerted by the box on the sled? (And by the sled back on the box.) (b) If you pull on the sled with a force of 35 N, how large does the coefficient of static friction, µs, between the box and the sled have to be, in order for the box to move with the sled? Draw free-body diagrams for the box and for the sled under this assumption (that they move together). (c) Suppose that µs is less than the value you got in part (b), so the box starts to slide back (relative to the sled). If the coefficient of kinetic friction µk between the box and the sled is 0.15, what is the acceleration of the sled, and what is the acceleration of the box, while they are sliding relative to each other (so, before the box falls off, and while you are still pulling with a 35-N force)? Draw again the free-body diagrams appropriate to this situation. Problem 5 You stick two objects together, one with a mass of 10 kg and one with a mass of 5 kg, using a glue that is supposed to be able to provide up to 19 N of force before it fails. Suppose you then pull on the 10 kg block with a force of 30 N. (a) What is the acceleration of the whole system? (b) What is the force exerted on the 5 kg block, and where does it come from? Does the glue hold? (c) Now suppose you pull on the 5 kg block instead with the same force. Does the glue hold this time? | University Physics I Classical Mechanics_Page_152_Chunk4350 |
6.7. PROBLEMS 135 Problem 6 Draw a free-body diagram for a 70-kg person standing in an elevator carrying a 15-kg backpack (do not consider the backpack a part of the person!). (a) if the elevator is not moving, and (b) if the elevator is accelerating downwards at 2 m/s2. In each case, what is the magnitude of the normal force exerted on the person by the floor? | University Physics I Classical Mechanics_Page_153_Chunk4351 |
136 CHAPTER 6. INTERACTIONS, PART 2: FORCES | University Physics I Classical Mechanics_Page_154_Chunk4352 |
Chapter 7 Impulse, Work and Power 7.1 Introduction: work and impulse In physics, “work” (or “doing work”) is what we call the process through which a force changes the energy of an object it acts on (or the energy of a system to which the object belongs). It is, therefore, a very technical term with a very specific meaning that may seem counterintuitive at times. For instance, as it turns out, in order to change the energy of an object on which it acts, the force needs to be at least partly in line with the displacement of the object during the time it is acting. A force that is perpendicular to the displacement does no work—it does not change the object’s energy. Imagine a satellite in a circular orbit around the earth. The earth is constantly pulling on it with a force (gravity) directed towards the center of the orbit at any given time. This force is always perpendicular to the displacement, which is along the orbit, and so it does no work: the satellite moves always at the same speed, so its kinetic energy does not change. The force does change the satellite’s momentum, however: it keeps bending the trajectory, and therefore changing the direction (albeit not the magnitude) of the satellite’s momentum vector. Of course, it is obvious that a force must change an object’s momentum, because that is pretty much how we defined force anyway. Recall Eq. (6.1) for the average force on an object: ⃗Fav = ∆⃗p/∆t. We can rearrange this to read ∆⃗p = ⃗Fav∆t (7.1) For a constant force, the product of the force and the time over which it is acting is called the 137 | University Physics I Classical Mechanics_Page_155_Chunk4353 |
138 CHAPTER 7. IMPULSE, WORK AND POWER impulse, usually denoted as ⃗J ⃗J = ⃗F ∆t (7.2) Clearly, the impulse given by a force to an object is equal to the change in the object’s momentum (by Eq. (7.1)), as long as it is the only force (or, alternatively, the net force) acting on it. If the force is not constant, we break up the time interval ∆t into smaller subintervals and add all the pieces, pretty much as we did with Figure 1.5 in Chapter 1 in order to calculate the displacement for a variable velocity. Formally this results in an integral: ⃗J = " tf ti ⃗F(t) dt (7.3) Graphically, the x component of the impulse is equal to the area under the curve of Fx versus time, and similarly for the other components. You will get to see how it works in a lab experiment this semester. There is not a whole lot more to be said about impulse. The main lesson to be learned from Eq. (7.1) is that one can get a desired change in momentum—bring an object to a stop, for instance—either by using a large force over a short time, or a smaller force over a longer time. It is easy to see how different circumstances may call for different strategies: sometimes you may want to make the force as small as possible, if the object on which you are acting is particularly fragile; other times you may just need to make the time as short as possible instead. Of course, to bring something to a stop you not only need to remove its momentum, but also its (kinetic) energy. If the former task takes time, the latter, it turns out, takes distance. Work is a much richer subject than impulse, not only because, as I have indicated above, the actual work done depends on the relative orientation of the force and displacement vectors, but also because there is only one kind of momentum, but many different kinds of energy, and one of the things that typically happens when work is done is the conversion of one type of energy into another. So there is a lot of ground to cover, but we’ll start small, in the next section, with the simplest kind of system, and the simplest kind of energy. 7.2 Work on a single particle Consider a particle that undergoes a displacement ∆x while a constant force F acts on it. In one dimension, the work done by the force on the particle is defined by W = F∆x (constant force) (7.4) and it is positive if the force and the displacement have the same sign (that is, if they point in the same direction), and negative otherwise. | University Physics I Classical Mechanics_Page_156_Chunk4354 |
7.2. WORK ON A SINGLE PARTICLE 139 In three dimensions, the force will be a vector ⃗F with components (Fx, Fy, Fz), and the displacement, likewise, will be a vector ∆⃗r with components (∆x, ∆y, ∆z). The work will be defined then as W = Fx∆x + Fy∆y + Fz∆z (7.5) This expression is an instance of what is known as the dot product (or inner product, or scalar product) of two vectors. Given two vectors ⃗A and ⃗B, their dot product is defined, in terms of their components, as ⃗A · ⃗B = AxBx + AyBy + AzBz (7.6) This can also be expressed in terms of the vectors’ magnitudes, | ⃗A| and | ⃗B|, and the angle they make, in the following form: ⃗A · ⃗B = | ⃗A|| ⃗B| cos φ (7.7) A ¶ B Figure 7.1: Illustrating the angle φ to be used when calculating the dot product of two vectors by the formula (7.7). One way to think of this formula is that you take the projection of vector ⃗A onto vector ⃗B (indicated here by the blue lines), which is equal to | ⃗A| cos φ, then multiply that by the length of ⃗B (or vice-versa, of course). Figure 7.1 shows what I mean by the angle φ in this expression. The equality of the two definitions, Eqs. (7.6) and (7.7), is proved in mathematics textbooks. The advantage of Eq. (7.7) is that it is independent of the choice of a system of coordinates. Using the dot product notation, the work done by a constant force can be written as W = ⃗F · ∆⃗r (7.8) Equation (7.7) then shows that, as I mentioned in the introduction, when the force is perpendicular to the displacement (φ = 90◦) the work it does is zero. You can also see this directly from Eq. (7.5), by choosing the x axis to point in the direction of the force (so Fy = Fz = 0), and the displacement to point along any of the other two axes (so ∆x = 0): the result is W = 0. If the force is not constant, again we follow the standard procedure of breaking up the total displacement into pieces that are short enough that the force may be taken to be constant over | University Physics I Classical Mechanics_Page_157_Chunk4355 |
140 CHAPTER 7. IMPULSE, WORK AND POWER each of them, calculating all those (possibly very small) “pieces of work,” and adding them all together. In one dimension, the final result can be expressed as the integral W = " xf xi F(x)dx (variable force) (7.9) So the work is given by the “area” under the F-vs-x curve. In more dimensions, we have to write a kind of multivariable integral known as a line integral. That is advanced calculus, so we will not go there this semester. 7.2.1 Work done by the net force, and the Work-Energy Theorem So much for the math and the definitions. Where does the energy come in? Let us suppose that F is either the only force or the net force on the particle—the sum of all the forces acting on the particle. Again, for simplicity we will assume that it is constant (does not change) while the particle undergoes the displacement ∆x. However, now ∆x and Fnet are related: a constant net force means a constant acceleration, a = Fnet/m, and for constant acceleration we know the formula v2 f −v2 i = 2a∆x applies. Therefore, we can write Wnet = Fnet∆x = ma∆x = m1 2 ( v2 f −v2 i ) (7.10) which is to say Wnet = ∆K (7.11) In words, the work done by the net force acting on a particle as it moves equals the change in the particle’s kinetic energy in the course of its displacement. This result is often referred to as the Work-Energy Theorem. As you may have guessed from our calling it a “theorem,” the result (7.11) is very general. It holds in three dimensions, and it holds also when the force isn’t constant throughout the displacement— you just have to use the correct equation to calculate the work in those cases. It would apply to the work done by the net force on an extended object, also, provided it is OK to treat the extended object as a particle—so basically, a rigid object that is moving as a whole and not doing anything fancy such as spinning while doing so. Another possible direction in which to generalize (7.11) might be as follows. By definition, a “particle” has no other kind of energy, besides (translational) kinetic energy. Also, and for the same reason (namely, the absence of internal structure), it has no “internal” forces—all the forces acting on it are external. So—for this very simple system—we could rephrase the result (7.11) by saying that the work done by the net external force acting on the system (the particle in this case) is equal to the change in its total energy. It is in fact in this form that we will ultimately generalize (7.11) to deal with arbitrary systems. | University Physics I Classical Mechanics_Page_158_Chunk4356 |
7.3. THE “CENTER OF MASS WORK” 141 Before we go there, however, I would like to take a little detour to explore another “reasonable” extension of the result (7.11), as well as its limitations. 7.3 The “center of mass work” All the physics I used in order to derive the result (7.11) was F = ma, and the expression v2 f −v2 i = 2a∆x, which applies whenever we have motion with constant acceleration. Now, we know that for an arbitrary system, of total mass M, Fext,net = Macm [see Eq. (6.11)]. That is enough, then, to ensure that, if Fext,net is a constant, we will have Fext,net∆xcm = ∆Kcm (7.12) where Kcm, the translational kinetic energy, is, as usual, Kcm = 1 2Mv2 cm, and ∆xcm is the displace- ment of the center of mass. The result (7.12) holds for an arbitrary system, as long as Fext,net is constant, and can be generalized by means of an integral (as in Eq. (7.9)) when it is variable. So it seems that we could define the left-hand side of Eq. (7.12) as “the work done on the center of mass,” and take that as the natural generalization to a system of the result (7.11) for a particle. Most physicists would, in fact, be OK with that, but educators nowadays frown on that idea, for a couple of reasons. First, it seems that it is essential to the notion of work that one should multiply the force by the displacement of the object on which it is acting. More precisely, in the definition (7.4), we want the displacement of the point of application of the force1. But there are many examples of systems where there is nothing at the precise location of the center of mass, and certainly no force acting precisely there. This is not necessarily a problem in the case of a rigid object which is not doing anything funny, just moving as a whole so that every part has the same displacement, because then the displacement of the center of mass would simply stand for the displacement of any point at which an external force might actually be applied. But for many deformable systems, this would not be case. In fact, for such systems one can usually show that Fext,net∆xcm is actually not the work done on the system by the net external force. A simple example of such a system is shown below, in Figure 7.2. 1As the name implies, this is the precise point at which the force is applied. For contact forces (other than friction; see later), this is easily identified. For gravity, a sum over all the forces exerted on all the particles that make up the object may be shown to be equivalent to a single resultant force acting at a point called the center of gravity, which, for our purposes (objects in uniform or near-uniform gravitational fields) will be the same as the center of mass. | University Physics I Classical Mechanics_Page_159_Chunk4357 |
142 CHAPTER 7. IMPULSE, WORK AND POWER Fs,2 spr Fs,1 spr Fh,2 c Fh,2 c ∆x1 ∆x2 ∆xcm 1 1 2 2 Figure 7.2: A system of two blocks connected by a spring. A constant external force, ⃗F c h,2, is applied to the block on the right. Initially the spring is relaxed, but as soon as block 2 starts to move it stretches, pulling back on block 2 and pulling forward on block 1. Because of the stretching of the spring, the displacements ∆x1, ∆xcm and ∆x2 are all different, and the work done by the external force, F c h,2∆x2, is different from the “center of mass work” F c h,2∆xcm. In this figure, the two blocks are connected by a spring, and the external force is applied to the block on the right (block 2). If the blocks have the same mass, the center of mass of the system is a point exactly halfway between them. If the spring starts in its relaxed state, it will stretch at first, so that the center of mass will lag behind block 2, and F c h,2∆x2, which is the quantity that we should properly call the “work done by the net external force” will not be equal to F c h,2∆xcm. We find ourselves, therefore, with a very general and potentially rather useful result, Eq. (7.12), that looks a lot like it should be “the work done on the system by the net external force” but, in fact, is that only sometimes. On the other hand, the result is so useful that simply referring to it all the time by “Eq. (7.12)” will not do. I propose, therefore, to call it the “center of mass work,” in between quotation marks, just so we all know what we are talking about, and remember the caveats that go with it. We can now move to the real theorem relating the work of the external forces on a system to the change in its energy. What we have seen so far are really just straightforward applications of Newton’s second law. The main result coming up is deeper than that, since it involves also, ultimately, the principle of conservation of energy. 7.4 Work done on a system by all the external forces Consider the most general possible system, one that might contain any number of particles, with possibly many forces, both internal and external, acting on each of them. I will again, for simplicity, | University Physics I Classical Mechanics_Page_160_Chunk4358 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 143 start by considering what happens over a time interval so short that all the forces are approximately constant (the final result will hold for arbitrarily long time intervals, just by adding, or integrating, over many such short intervals). I will also work explicitly only the one-dimensional case, although again that turns out to not be a real restriction. Let then Wall,1 be the work done on particle 1 by all the forces acting on it, Wall,2 the work done on particle 2, and so on. The total work is the sum Wall,sys = Wall,1 + Wall,2 + . . .. However, by the results of section 7.2, we have Wall,1 = ∆K1 (the change in kinetic energy of particle 1), Wall,2 = ∆K2, and so on, so adding all these up we get Wall,sys = ∆Ksys (7.13) where ∆Ksys is the change in kinetic energy of the whole system. So far, of course, this is nothing new. To learn something else we need to look next at the work done by the internal forces. It is helpful here to start by considering the “no-dissipation case” in which all the internal forces can be derived from a potential energy2. We will consider the case where dissipative processes happen inside the system after we have gained a full understanding of the result we will obtain for this simpler case. 7.4.1 The no-dissipation case The internal forces are, by definition, forces that arise from the interactions between pairs of particles that are both inside the system. Because of Newton’s 3rd law, the force F12 (we will omit the “type” superscript for now) exerted by particle 1 on particle 2 must be the negative of F21, the force exerted by particle 2 on particle 1. Hence, the work associated with this interaction for this pair of particles can be written W(1, 2) = F12∆x2 + F21∆x1 = F12 (∆x2 −∆x1) (7.14) Notice that ∆x2 −∆x1 can be rewritten as x2,f −x2,i −x1,f + x1,i = x12,f −x12,i = ∆x12, where x12 = x2 −x1 is the relative position coordinate of the two particles. Therefore, W(1, 2) = F12∆x12 (7.15) Now, if the interaction in question is associated with a potential energy, as I showed in section 6.2, F12 = −dU/dx12. Assume the displacement ∆x12 is so small that we can replace the derivative by just the ratio ∆U/∆x12 (which is consistent with our assumption that the force is approximately constant over the time interval considered); the result will then be W(1, 2) = F12∆x12 ≃−∆U ∆x12 ∆x12 = −∆U (7.16) 2Or else they do no work at all: the magnetic force between moving charges is an example of the latter. | University Physics I Classical Mechanics_Page_161_Chunk4359 |
144 CHAPTER 7. IMPULSE, WORK AND POWER Adding up very many such “infinitesimal” displacements will lead to the same final result, where ∆U will be the change in the potential energy over the whole process. This can also be proved using calculus, without any approximations: W(1, 2) = " x12,f x12,i F12dx12 = − " x12,f x12,i dU dx12 dx12 = −∆U. (7.17) We can apply this to every pair of particles and every internal interaction, and then add up all the results. On one side, we will get the total work done on the system by all the internal forces; on the other side, we will get the negative of the change in the system’s total internal energy: Wint,sys = −∆Usys. (7.18) In words, the work done by all the (conservative) internal forces is equal to the change in the system’s potential energy. Let us now put Eqs. (7.13) and (7.18) together: the difference between the work done by all the forces and the work done by the internal forces is, of course, the work done by the external forces, but according to Eqs. (7.13) and (7.18), this is equal to Wext,sys = Wall,sys −Wint,sys = ∆Ksys + ∆Usys (7.19) which is the change in the total mechanical (kinetic plus potential) energy of the system. If we further assume that the system, in the absence of the external forces, is closed, then there are no other processes (such as the absorption of heat) by which the total energy of the system might change, and we get the simple result that the work done by the external forces equals the change in the system’s total energy: Wext,sys = ∆Esys (7.20) As a first application of the result (7.20), consider again the blocks connected by a spring shown in Fig. 2. You can see now why the work done by the external force F c h,2 has to be different, and in fact larger, than the “center of mass work”: the latter only gives us the change in the translational energy, but the former has to give us the change in the total energy—translational, convertible, and potential: F c h,2∆xcm = ∆Kcm F c h,2∆x2 = ∆Kcm + ∆Kconv + ∆U spr (7.21) As another example, imagine you throw a ball of mass m upwards (see Figure 7.3, next page), and it reaches a maximum height h above the point where your hand started to move. Let us define the system to be the ball and the earth, so the force exerted by your hand is an external force. Then you do work on the system during the throw, which in the figure is the interval, from A to B, during which your hand is on contact with the ball. The bar diagram on the side shows that some of this work goes into increasing the system’s (gravitational) potential energy (because the | University Physics I Classical Mechanics_Page_162_Chunk4360 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 145 ball goes up a little while in contact with your hand), and the rest, which is typically most of it, goes into increasing the system’s kinetic energy (in this case, just the ball’s; the earth’s kinetic energy does not change in any measurable way!). A B C rising, from A to C ∆K ∆U Wext falling, from C to B ∆K ∆U Wext the throw, from A to B ∆K ∆U Wext h Figure 7.3: Tossing a ball into the air. We consider the system formed by the ball and the earth. The force exerted by the hand (which is in contact with the ball from point A to point B) is therefore an external force. The diagrams show the system’s energy balance over three different intervals. So how much work did you actually do? If we knew the distance from A to B, and the magnitude of the force you exerted, and if we could assume that your force was constant throughout, we could calculate W from the definition (7.4). But in this case, and many others like it, it is actually easier to find out how much total energy the system gained and just use Eq. (7.20). To find ∆E in | University Physics I Classical Mechanics_Page_163_Chunk4361 |
146 CHAPTER 7. IMPULSE, WORK AND POWER practice, all we have to do is see how high the ball rises. At the ball’s maximum height (point C), as the second diagram shows, all the energy in the system is gravitational potential energy, and (as long as the system stays closed), all that energy is still equal to the work you did initially, so if the distance from A to C is h you must have done an amount of work Wyou = ∆U G = mgh (7.22) The third diagram in Figure 7.3 shows the work-energy balance for another time interval, during which the ball falls from C to B. Over this time, no external forces act on the ball (recall we have taken the system to be the ball and the earth, so gravity is an internal force). Then, the work done by the external forces is zero, and the change in the total energy of the system is also zero. The diagram just shows an increase in kinetic energy at the expense of an equal decrease in potential energy. What about the work done by the internal forces? Eq. (7.18) tells us that this work is equal to the negative of the change in potential energy. In this case, the internal force is gravity, and the corresponding energy is gravitational potential energy. This change in potential energy is clearly visible in all the diagrams; however, when you add to it the change in kinetic energy, the result is always equal to the work done by the external force only. Put otherwise, the internal forces do not change the system’s total energy, they just “redistribute” it among different kinds—as in, for instance, the last diagram in Fig. 7.3, where you can clearly see that gravity is causing the kinetic energy of the system to increase at the expense of the potential energy. We will use diagrams like the ones in Figure 7.3 to look at the work-energy balance for different systems. The idea is that the sum of all the columns on the left (the change in the system’s total energy) has to equal the result on the far-right column (the work done by the net external force): that is the content of the theorem (7.20). Note that, unlike the energy diagrams we used in Chapter 5, these columns represent changes in the energy, so they could be positive or negative. Just as for the earlier energy diagrams, the picture we get will be different, even for the same physical situation, depending on the choice of system. This is illustrated in Figure 7.4 below (next page), where I have taken the same throw shown in Fig. 7.3, but now the system I’m looking at is the ball only. This means gravity is now an external force, as is the force of the hand, and the ball only has kinetic energy. Normally one would show the sum of the work done by the two external forces on a single column, but here I have chosen to break it up into two columns for clarity. As you can see, during the throw the hand does positive work, whereas gravity does a comparatively small amount of negative work, and the change in kinetic energy is the sum of the two. For the longer interval from A to C (second diagram), gravity continues to do negative work until all the kinetic energy of the ball is gone. For the interval from C to B, the only external force is gravity, which now does positive work, equal to the increase in the ball’s kinetic energy. | University Physics I Classical Mechanics_Page_164_Chunk4362 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 147 the throw, from A to B ∆K Wgrav Whand rising, from A to C ∆K Whand Wgrav } Wext } Wext ∆K Whand Wgrav } Wext falling, from C to B Figure 7.4: Work-energy balance diagrams for the same toss illustrated in Fig. 7.3, but now the system is taken to be the ball only. Of course, the numerical value of the actual work done by any particular force does not depend on our choice of system: in each case, gravity does the same amount of work in the processes illustrated in Fig. 7.4 as in those illustrated in Fig. 7.3. The difference, however, is that for the system in Fig. 7.4, gravity is an external force, and now the work it does actually changes the system’s total energy, because the gravitational potential energy is now not included in that total. Formally, it works like this: in the case shown in Fig. 7.3, where the system is the ball and the earth, we have ∆K + ∆U G = Whand. By the result (7.18), however, we have ∆U G = −Wgrav, and so this equation can be rearranged to read ∆K = Wgrav + Whand, which is just the result (7.20) when the system is the ball alone. Ultimately, the reason we emphasize the importance of the choice of system is to prevent double counting: if you want to count the work done by gravity as contributing to the change in the system’s total energy, it means that you are, implicitly, treating gravity as an external force, and therefore your system must be something that does not have, by itself, gravitational potential energy (the case of the ball in Figure 7.4); conversely, if you insist on counting gravitational potential energy as contributing to the system’s total energy, then you must treat gravity as an internal force, and leave it out of the calculation of the work done on the system by the external forces, which are the only ones that can change the system’s total energy. 7.4.2 The general case: systems with dissipation We are now ready to consider what happens when some of the internal interactions in a system are not conservative. There are two key observations to keep in mind: first, of course, that energy will always be conserved in a closed system, regardless of whether the internal forces are “conservative” or not: if they are not, it merely means that they will convert some of the “organized,” mechanical energy, into disorganized (primarily thermal) energy. | University Physics I Classical Mechanics_Page_165_Chunk4363 |
148 CHAPTER 7. IMPULSE, WORK AND POWER The second observation is that the work done by an external force on a system does not depend on where the force comes from—that is to say, what physical arrangement we use to produce the force. Only the value of the force at each step and the displacement of the point of application are involved in the definition (7.9). This means, in particular, that we can use a conservative interaction to do the work for us. It turns out, then, that the generalization of the result (7.20) to apply to all sorts of interactions becomes straightforward. To see the idea, consider, for example, the situation in Figure 7.5 below. This is essentially the same as Figure 6.2, which we analyzed in detail from the perspective of forces and accelerations in the previous chapter. Here I have broken it up into two systems. System A, outlined in blue, consists of block 1 and the surface on which it slides, and includes a dissipative interaction—namely, kinetic friction—between the block and the surface. The force doing work on this system is the tension force from the rope, ⃗F t r,1. 1 2 Fr,1 t Fr,2 t FE,2 G Fs,1 fr a1 a2 A B Figure 7.5: Block sliding on a surface, with friction, being pulled by a rope attached to a block falling under the action of gravity. The motion of this system was solved for in Section 6.3. Because the rope is assumed to have negligible mass, this force is the same in magnitude as the | University Physics I Classical Mechanics_Page_166_Chunk4364 |
7.4. WORK DONE ON A SYSTEM BY ALL THE EXTERNAL FORCES 149 force ⃗F t r,2 that is doing negative work on system B. System B, outlined in magenta, consists of block 2 and the earth and thus it includes only one internal interaction, namely gravity, which is conservative. This means that we can immediately apply the theorem (7.20) to it, and conclude that the work done on B by ⃗F t r,2 is equal to the change in system B’s total energy: Wr,B = ∆EB (7.23) However, since the rope is inextensible, the two blocks move the same distance in the same time, and the force exerted on each by the rope is the same in magnitude, so the work done by the rope on system A is equal in magnitude but opposite in sign to the work it does on system B: Wr,A = −Wr,B = −∆EB (7.24) Now consider the total system formed by A+B. Assuming it is a closed system, its total energy must be constant, and so any change in the total energy of B must be equal and opposite the corresponding change in the total energy of A: ∆EB = −∆EA. Therefore, Wr,A = −∆EB = ∆EA (7.25) So we conclude that the work done by the external force on system A must be equal to the total change in system A’s energy. In other words, Eq. (7.20) applies to system A as well, as it does to system B, even though the interaction between the parts that make up system A is dissipative. Although I have shown this to be true just for one specific example, the argument is quite general: if I use a conservative system B to do some work on another system A, two things happen: first, by virtue of (7.20), the work done by B comes at the expense of its total energy, so Wext,A = −∆EB. Second, if A and B together form a closed system, the change in A’s energy must be equal and opposite the change in B’s energy, so ∆EA = −∆EB = Wext,A. So the result (7.20) holds for A, regardless of whether its internal interactions are conservative or not. What is essential in the above reasoning is that A and B together should form a closed system, that is, one that does not exchange energy with its environment. It is very important, therefore, if we want to apply the theorem (7.20) to a general system—that is, one that includes dissipative interactions—that we draw the boundary of the system in such a way as to ensure that no dissipation is happening at the boundary. For example, in the situation illustrated in Fig. 7.5, if we want the result (7.25) to apply we must take system A to include both block 1 and the surface on which it slides. The reason for this is that the energy “dissipated” by kinetic friction when two objects rub together goes into both objects. So, as the block slides, kinetic friction is converting some of its kinetic energy into thermal energy, but not all this thermal energy stays inside block 1. Put otherwise, in the presence of friction, block 1 by itself is not a closed system: it is “leaking” energy to the surface. On the other hand, when you include (enough of) the surface in the system, you can be sure to have “caught” all the dissipated energy, and the result (7.20) applies. | University Physics I Classical Mechanics_Page_167_Chunk4365 |
150 CHAPTER 7. IMPULSE, WORK AND POWER 7.4.3 Energy dissipated by kinetic friction In the situation illustrated in Fig. 7.5, we might calculate the energy dissipated by kinetic friction by indirect means. For instance, we can use the fact that the energy of system A is of two kinds, kinetic and “dissipated,” and therefore, by theorem (7.20), we have ∆K + ∆Ediss = F t r,1∆x1 (7.26) Back in section 6.3, we used Newton’s laws to solve for the acceleration of this system and the tension in the rope; using those results, we can calculate the displacement ∆x1 over any time interval, and the corresponding change in K, and then we can solve Eq. (7.26) for ∆Ediss. If we do this, we will find out that, in fact, the following result holds, ∆Ediss = −F k s,1∆x1 (7.27) where F k s,1 is the force of kinetic friction exerted by the surface on block 1, and must be understood to be negative in this equation (so that ∆Ediss will come out positive, as it must be). It is tempting to think of the product F k s,1∆x as the work done by the force of kinetic friction on the block, and most of the time there is nothing wrong with that, but it is important to realize that the “point of application” of the friction force is not a single point: rather, the force is “distributed,” that is to say, spread over the whole contact area between the block and the surface. As a consequence of this, a more general expression for the energy dissipated by kinetic friction between an object o and a surface s should be ∆Ediss = |F k s,o||∆xso| (7.28) where I am using the Chapter 1 subscript notation xAB to refer to “the position of B in the frame of A” (or “relative to A”); in other words, ∆xso is the change in the position of the object relative to the surface or, more simply, the distance that the object and the surface slip past each other (while rubbing against each other, and hence dissipating energy). If the surfaces is at rest (relative to the Earth), ∆xso reduces to ∆xEo, the displacement of the object in the Earth reference frame, and we can remove the subscript E, as we typically do, for simplicity; however, in the rare cases when both the surface and the objet are moving (as in part (c) of Problem 3 in Chapter 6, the sled problem) what matters is how far they move relative to each other. In that case we have |∆xso| = |∆xo −∆xs| (with both ∆xo and ∆xs measured in the Earth reference frame). 7.5 Power By “power” we mean the rate at which work is done, which is to say, the rate at which energy is taken in, or given out, or converted from one form to another. The SI unit of power is the watt (W), | University Physics I Classical Mechanics_Page_168_Chunk4366 |
7.6. IN SUMMARY 151 which is equal to 1 J/s. The average power going into or coming out of a system by mechanical means, that is to say, through the action of a force applied at a point undergoing a displacement ∆x, will be Pav = ∆E ∆t = W ∆t = F ∆x ∆t (7.29) assuming the force is constant. Note that in the limit when ∆t goes to zero, this gives us the instantaneous power associated with the force F as P = Fv (7.30) where v is the (instantaneous) velocity of the point of application of the force. This one-dimensional result generalizes to three dimensions as P = ⃗F · ⃗v (7.31) using again the dot product notation. An important goal of this chapter has been to develop a set of tools that you may use to find out where power is spent, and how much: in any practical situation, which systems are giving energy and which are taking it in, what forms of energy conversion are taking place, and where and through which means are the exchanges and conversions happening. These are extremely important practical questions; the problems and exercises that you will see here will give you a feel for the variety of situations that can already be analyzed by this “systems-based” approach, but in a way they will still do little more than scratch the surface. 7.6 In summary 1. The change in the momentum of a system produced by a force ⃗F acting over a time ∆t is given the name of “impulse” and denoted by ⃗J. For a constant force, we have ⃗J = ∆⃗p = ⃗F∆t. 2. Work, or “doing work” is the name given in physics to the process by which an applied force brings about a change in the energy of an object, or of a system that contains the object on which the force is acting. 3. The work done by a constant force ⃗F acting on an object or system is given by W = ⃗F · ∆⃗r, where the dot represents the “dot” or “scalar” product of the two vectors, and ∆⃗r is the displacement undergone by the point of application of the force while the force is acting. If the force is perpendicular to the displacement, it does no work. 4. For a system that is otherwise closed, the net sum of the amounts of work done by all the external forces is equal to the change in the system’s total energy, when all the types of energy are included. Note that, for deformable systems, the displacement of the point of application may be different for different forces. | University Physics I Classical Mechanics_Page_169_Chunk4367 |
152 CHAPTER 7. IMPULSE, WORK AND POWER 5. The result in 4 above holds only provided the boundary of the system is not drawn at a physical surface on which dissipation occurs. Put otherwise, kinetic friction or other similar dissipative forces (drag, air resistance) must be included as internal, not external forces. 6. The work done by the internal forces in a closed system results only in the conversion of one type of energy into another, always keeping the total energy constant. 7. For a system with no internal energy, like a particle, the work done by all the external forces equals the change in kinetic energy. This result is sometimes called the Work-Energy theorem in a narrow sense. 8. For any system, if ⃗Fext,net (assumed constant) is the sum of all the external forces, the following result holds: ⃗Fext,net · ∆⃗rcm = ∆Kcm where Kcm is the translational (or “center of mass”) kinetic energy, and ∆⃗rcm is the displace- ment of the center of mass. This is only sometimes equal to the net work done on the system by the external forces. 9. For an object o sliding on a surface s, the energy dissipated by kinetic friction can be directly calculated as ∆Ediss = |F k s,o||∆xso| where |∆xso| = |∆xo −∆xs| is the distance that the two surfaces in contact slip past each other. This expression, with a negative sign, can be used to take the place of the “work done by friction” in applications of the results 7 and 8 above to systems involving kinetic friction forces. 10. The power of a system is the rate at which it does work, that is to say, takes in or gives up energy: Pav = ∆E/∆t. When this is done by means of an applied force F, the instantaneous power can be written as P = Fv, or, in three dimensions, ⃗F · ⃗v. | University Physics I Classical Mechanics_Page_170_Chunk4368 |
7.7. EXAMPLES 153 7.7 Examples 7.7.1 Braking Suppose you are riding your bicycle and hit the brakes to come to a stop. Assuming no slippage between the tire and the road: (a) Which force is responsible for removing your momentum? (By “you” I mean throughout “you and the bicycle.”) (b) Which force is responsible for removing your kinetic energy? Solution (a) According to what we saw in previous chapters, for example, Eq. (6.10) ∆psys ∆t = Fext,net (7.32) the total momentum of the system can only be changed by the action of an external force, and the only available external force is the force of static friction between the tire and the road (static, because we assume no slippage). So it is this force that removes the forward momentum from the system. The stopping distance, ∆xcm, and the force, can be related using Eq. (7.12): F s r,t∆xcm = ∆Kcm (7.33) (b) Now, here is an interesting fact: the force of static friction, although fully responsible for stopping your center of mass motion does no work in this case. That is because the point where it is applied—the point of the tire that is momentarily in contact with the road—is also momentarily at rest relative to the road: it is, precisely, not slipping, so ∆x in the equation W = F∆x is zero. By the time that bit of the tire has moved on, so you actually have a nonzero ∆x, you no longer have an F: the force of static friction is no longer acting on that bit of the tire, it is acting on a different bit—on which it will, again, do no work, for the same reason. So, as you bring your bicycle to a halt the work Wext,sys = 0, and it follows from Eq. (7.20) that the total energy of your system is, in fact, conserved: all your initial kinetic energy is converted to thermal energy by the brake pad rubbing on the wheel, and the internal force responsible for that conversion is the force of kinetic friction between the pad and the wheel. | University Physics I Classical Mechanics_Page_171_Chunk4369 |
154 CHAPTER 7. IMPULSE, WORK AND POWER 7.7.2 Work, energy and the choice of system: dissipative case Consider again the situation shown in Figure 7.5. Let m1 = 1 kg, m2 = 2 kg, and µk = 0.3. Use the solutions provided in Section 6.3 to calculate the work done by all the forces, and the changes in all energies, when the system undergoes a displacement of 0.5 m, and represent the changes graphically using bar diagrams like the ones in Figure 7.3 (for system A and B separately) Solution From Eq. (6.32), we have a = m2 −µkm1 m1 + m2 g = 5.55 m s2 F t = m1m2(1 + µk) m1 + m2 g = 8.49 N (7.34) We can use the acceleration to calculate the change in kinetic energy, since we have Eq. (2.10) for motion with constant acceleration: v2 f −v2 i = 2a∆x = 2 × & 5.55 m s2 ' × 0.5 m = 5.55 m2 s2 (7.35) so the change in kinetic energy of the two blocks is ∆K1 = 1 2m1 ( v2 f −v2 i ) = 2.78 J ∆K2 = 1 2m2 ( v2 f −v2 i ) = 5.55 J (7.36) We can also use the tension to calculate the work done by the external force on each system: Wext,A = F t r,1∆x = (8.49 N) × (0.5 m) = 4.25 J Wext,B = F t r,2∆y = (8.49 N) × (−0.5 m) = −4.25 J (7.37) Lastly, we need the change in the gravitational potential energy of system B: ∆U G B = m2g∆y = (2 kg) × & 9.8 m s2 ' × (−0.5 m) = −9.8 J (7.38) and the increase in dissipated energy in system A, which we can get from Eq. (7.28): ∆Ediss = −F k s,1∆x = µkF n s,1∆x = µkm1g∆x = 0.3 × (1 kg) × & 9.8 m s2 ' × (0.5 m) = 1.47 J (7.39) We can now put all this together to show that Eq. (7.20) indeed holds: Wext,A = ∆EA = ∆K1 + ∆Ediss = 2.78 J + 1.47 J = 4.25 J Wext,B = ∆EB = ∆K2 + ∆U G B = 5.55 J −9.8 J = −4.25 J (7.40) | University Physics I Classical Mechanics_Page_172_Chunk4370 |
7.7. EXAMPLES 155 To plot all this as energy bars, if you do not have access to a very precise drawing program, you typically have to make some approximations. In this case, we see that ∆K2 = 2∆K1 (exactly), whereas ∆K1 ≃2∆Ediss, so we can use one box to represent Ediss, two boxes for ∆K1, three for Wext,A, four for ∆K2, and so on. The result is shown in green in the picture below; the blue bars have been drawn more exactly to scale, and are shown for your information only. system A ∆K1 ∆Ediss Wext system B ∆K2 ∆UG Wext 7.7.3 Work, energy and the choice of system: non-dissipative case Suppose you hang a spring from the ceiling, then attach a block to the end of the spring and let go. The block starts swinging up and down on the spring. Consider the initial time just before you let go, and the final time when the block momentarily stops at the bottom of the swing. For each of the choices of a system listed below, find the net energy change of the system in this process, and relate it explicitly to the work done on the system by an external force (or forces) (a) System is the block and the spring. (b) System is the block alone. (c) System is the block and the earth. Solution (a) The block alone has kinetic energy, and the spring alone has (elastic) potential energy, so the total energy of this system is the sum of these two. For the interval considered, the change in kinetic energy is zero, because the block starts and ends (momentarily) at rest, so only the spring energy changes. This has to be equal to the work done by gravity, which is the only external force. So, if the spring stretches a distance d, its potential energy goes from zero to 1 2kd2, and the block falls the same distance, so gravity does an amount of work equal to mgd, and we have Wgrav = mgd = ∆Esys = ∆K + ∆U spr = 0 + 1 2kd2 (7.41) | University Physics I Classical Mechanics_Page_173_Chunk4371 |
156 CHAPTER 7. IMPULSE, WORK AND POWER (b) If the system is the block alone, the only energy it has is kinetic energy, which, as stated above, does not see a net change in this process. This means the net work done on the block by the external forces must be zero. The external forces in this case are the spring force and gravity, so we have Wspr + Wgrav = ∆K = 0 (7.42) We have calculated Wgrav above, so from this we get that the work done by the spring on the block, as it stretches, is −mgd, or (by Eq. (7.41)) −1 2kd2. Note that the force exerted by the spring is not constant as it stretches (or compresses) so we cannot just use Eq. (7.4) to calculate it; rather, we need to calculate it as an integral, as in Eq. (7.9), or derive it in some indirect way as we have just done here. (c) If the system is the block and the earth, it has kinetic energy and gravitational potential energy. The force exerted by the spring is an external force now, so we have: Wspr = ∆Esys = ∆K + ∆U G = 0 −mgd (7.43) so we end up again with the result that Wspr = −mgd = −1 2kd2. Note that both the work done by the spring and the work done by gravity are equal to the negative of the changes in their respective potential energies, as they should be. 7.7.4 Jumping For a standing jump, you start standing straight (A) so your body’s center of mass is at a height h1 above the ground. You then bend your knees so your center of mass is now at a (lower) height h2 (B). Finally, you straighten your legs, pushing hard on the ground, and take off, so your center of mass ends up achieving a maximum height, h3, above the ground (C). Answer the following questions in as much detail as you can. (a) Consider the system to be your body only. In going from (A) to (B), which external forces are acting on it? How do their magnitudes compare, as a function of time? (b) In going from (A) to (B), does any of the forces you identified in part (a) do work on your body? If so, which one, and by how much? Does your body’s energy increase or decrease as a result of this? Into what kind of energy do you think this work is primarily converted? (c) In going from (B) to (C), which external forces are acting on you? (Not all of them need to be acting all the time.) How do their magnitudes compare, as a function of time? (d) In going from (B) to (C), does any of the forces you identified do work on your body? If so, which one, and by how much? Does your body’s kinetic energy see a net change from (B) to (C)? What other energy change needs to take place in order for Eq. (7.20) (always with your body as the system) to be valid for this process? Solution | University Physics I Classical Mechanics_Page_174_Chunk4372 |
7.7. EXAMPLES 157 (a) The external forces on your body are gravity, pointing down, and the normal force from the floor, pointing up. Initially, as you start lowering your center of mass, the normal force has to be slightly smaller than gravity, since your center of mass acquires a small downward acceleration. However, eventually F n would have to exceed F G in order to stop the downward motion. (b) The normal force does no work, because its point of application (the soles of your feet) does not move, so ∆x in the expression W = F∆x (Eq. (7.4)) is zero. Gravity, on the other hand, does positive work, since you may always treat the center of mass as the point of application of gravity (see Section 7.3, footnote). We have F G y = −mg, and ∆y = h2 −h1, so Wgrav = F G y ∆y = −mg(h2 −h1) = mg(h1 −h2) Since this is the net work done by all the external forces on my body, and it is positive, the total energy in my body must have increased (by the theorem (7.20): Wext,sys = ∆Esys). In this case, it is clear that the main change has to be an increase in my body’s elastic potential energy, as my muscles tense for the jump. (An increase in thermal energy is always possible too.) (c) During the jump, the external forces acting on me are again gravity and the normal force, which together determine the acceleration of my center of mass. At the beginning of the jump, the normal force has to be much stronger than gravity, to give me a large upwards acceleration. Since the normal force is a reaction force, I accomplish this by pushing very hard with my feet on the ground, as I extend my leg’s muscles: by Newton’s third law, the ground responds with an equal and opposite force upwards. As my legs continue to stretch, and move upwards, the force they exert on the ground decreases, and so does F n, which eventually becomes less than F G. At that point (probably even before my feet leave the ground) the acceleration of my center of mass becomes negative (that is, pointing down). This ultimately causes my upwards motion to stop, and my body to come down. (d) The only force that does work on my body during the process described in (c) is gravity, since, again, the point of application of F n is the point of contact between my feet and the ground, and that point does not move up or down—it is always level with the ground. So Wext,sys = Wgrav, which in this case is actually negative: Wgrav = −mg(h3 −h2). In going from (B) to (C), there is no change in your kinetic energy, since you start at rest and end (momentarily) with zero velocity at the top of the jump. So the fact that there is a net negative work done on you means that the energy inside your body must have gone down. Clearly, some of this is just a decrease in elastic potential energy. However, since h3 (the final height of your center of mass) is greater than h1 (its initial height at (A), before crouching), there is a net loss of energy in your body as a result of the whole process. The most obvious place to look for this loss is in chemical energy: you “burned” some calories in the process, primarily when pushing hard against the ground. | University Physics I Classical Mechanics_Page_175_Chunk4373 |
158 CHAPTER 7. IMPULSE, WORK AND POWER 7.8 Problems Problem 1 In a mattress test, you drop a 7.0 kg bowling ball from a height of 1.5 m above a mattress, which as a result compresses 15 cm as the ball comes to a stop. (a) What is the kinetic energy of the ball just before it hits the mattress? (b) How much work does the gravitational force of the earth do on the ball as it falls, for the first part of the fall (from the moment you drop it to just before it hits the mattress)? (c) How much work does the gravitational force do on the ball while it is compressing the mattress? (d) How much work does the mattress do on the ball? (e) Now model the mattress as a single spring with an unknown spring constant k, and consider the whole system formed by the ball, the earth and the mattress. By how much does the potential energy of the mattress increase as it compresses? (f) What is the value of the spring constant k? Problem 2 A block of mass 1 kg is sitting on top of a compressed spring of spring constant k = 300 N/m and equilibrium length 20 cm. Initially the spring is compressed 10 cm, and the block is held in place by someone pushing down on it with his hand. At t = 0, the hand is removed (this involves no work), the spring expands and the block flies upwards. (a) Draw a free-body diagram for the block while the hand is still pressing down. Try to get the forces approximately to scale. The following question should help. (b) What must be the force (magnitude and direction) exerted by the hand on the block? (c) How much elastic potential energy was stored in the spring initially? (d) Taking the system formed by the block and the earth, how much total work is done on it by the spring, as it expands to its equilibrium length? (You do not need to do a new calculation here, just think of conservation of energy.) (e) How high does the block rise above its initial position? (f) Treating the block alone as the system, how much net work is done on it by the two external forces (the spring and gravity) from the time just before it starts moving to the time it reaches its maximum height? (Again, no calculation is necessary if you can justify your answer.) Problem 3 A crane is lifting a 500-kg object at a constant speed of 0.5 m/s. What is the power output of the crane? Problem 4 In a crash test, a car, initially moving at 30 m/s, hits a wall and crumples to a halt. In the process of crumpling, the center of mass of the car moves forward a distance of 1 m. (a) If the car has a mass of 1, 800 kg, what is the magnitude of the average force acting on it while it stops? What, physically, is this force? (b) Does the force you found in (a) actually do any work on the car? (Think carefully!) (c) What is the net change in the car’s kinetic energy? Where does all that kinetic energy go? | University Physics I Classical Mechanics_Page_176_Chunk4374 |
7.8. PROBLEMS 159 Problem 5 A block of mass 3 kg slides on a horizontal, rough surface towards a spring with k = 500 N/m. The kinetic friction coefficient between the block and the surface is µk = 0.6. If the block’s speed is 5 m/s at the instant it first makes contact with the spring, (a) Find the maximum compression of the spring. (b) Draw work-energy bar diagrams for the process of the block coming to a halt, taking the system to be the block and the surface only. | University Physics I Classical Mechanics_Page_177_Chunk4375 |
160 CHAPTER 7. IMPULSE, WORK AND POWER | University Physics I Classical Mechanics_Page_178_Chunk4376 |
Chapter 8 Motion in two dimensions 8.1 Dealing with forces in two dimensions We have been able to get a lot of physics from our study of (mostly) one-dimensional motion only, but it goes without saying that the real world is a lot richer than that, and there are a number of new and interesting phenomena that appear when one considers motion in two or three dimensions. The purpose of this chapter is to introduce you to some of the simplest two-dimensional situations of physical interest. A common feature to all these problems is that the forces acting on the objects under consideration will typically not line up with the displacements. This means, in practice, that we need to pay more attention to the vector nature of these quantities than we have done so far. This section will present a brief reminder of some basic properties of vectors, and introduce a couple of simple principles for the analysis of the systems that will follow. To begin with, recall that a vector is a quantity that has both a magnitude and a direction. The magnitude of the vector just tells us how big it is: the magnitude of the velocity vector, for instance, is the speed, that is, just how fast something is moving. When working with vectors in one dimension, we have typically assumed that the entire vector (whether it was a velocity, an acceleration or a force) lay along the line of motion of the system, and all we had to do to indicate the direction was to give the vector’s magnitude an appropriate sign. For the problems that follow, however, it will become essential to break up the vectors into their components along an appropriate set of axes. This involves very simple geometry, and follows the example of the position vector ⃗r, whose components are just the Cartesian coordinates of the point it locates in space (as shown in Figure 1.1). For a generic vector, for instance, a force, like the one shown in Figure 8.1 below, the components Fx and Fy may be obtained from a right triangle, as indicated there: 161 | University Physics I Classical Mechanics_Page_179_Chunk4377 |
162 CHAPTER 8. MOTION IN TWO DIMENSIONS F F Fx Fy x y θ θ 180° − θ Fx Figure 8.1: The components of a vector that makes an angle θ with the positive x axis. Two examples are shown, for θ < 90◦(in which case Fx > 0) and for 90◦< θ < 180◦(in which case Fx < 0). In both cases, Fy > 0. The triangle will always have the vector’s magnitude (|⃗F| in this case) as the hypothenuse. The two other sides should be parallel to the coordinate axes. Their lengths are the corresponding components, except for a sign that depends on the orientation of the vector. If we happen to know the angle θ that the vector makes with the positive x axis, the following relations will always hold: Fx = |⃗F| cos θ Fy = |⃗F| sin θ |⃗F| = # F 2x + F 2y θ = tan−1 Fy Fx (8.1) Note, however, that in general this angle θ may not be one of the interior angles of the triangle (as shown on the right diagram in Fig. 8.1), and that in that case it may just be simpler to calculate the magnitude of the components using trigonometry and an interior angle (such as 180◦−θ in the example), and give them the appropriate signs “by hand.” In the example on the right, the length of the horizontal side of the triangle is equal to |⃗F| cos(180◦−θ), which is a positive quantity; the correct value for Fx, however, is the negative number |⃗F| cos θ = −|⃗F| cos(180◦−θ). In any case, it is important not to get fixated on the notion that “the x component will always be proportional to the cosine of θ.” The symbol θ is just a convenient one to use for a generic angle. There are four sections in this chapter, and in every one there is a θ used with a different meaning. When in doubt, just draw the appropriate right triangle and remember from your trigonometry classes which side goes with the sine, and which with the cosine. For the problems that we are going to study in this chapter, the idea is to break up all the forces involved into components along properly-chosen coordinate axes, then add all the components along any given direction, and apply Fnet = ma along that direction: that is to say, we will write (and | University Physics I Classical Mechanics_Page_180_Chunk4378 |
8.1. DEALING WITH FORCES IN TWO DIMENSIONS 163 eventually solve) the equations Fnet,x = max Fnet,y = may (8.2) We can show that Eqs. (8.2) must hold for any choice of orthogonal x and y axes, based on the fact that we know ⃗Fnet = m⃗a holds along one particular direction, namely, the direction common to ⃗Fnet and ⃗a, and the fact that we have defined the projection procedure to be the same for any kind of vector. Figure 8.2 shows how this works. Along the dashed line you just have the situation that is by now familiar to us from one-dimensional problems, where ⃗a lies along ⃗F (assumed here to be the net force), and |⃗F| = m|⃗a|. However, in the figure I have chosen the axes to make an angle θ with this direction. Then, if you look at the projections of ⃗F and ⃗a along the x axis, you will find ax = |⃗a| cos θ Fx = |⃗F| cos θ = m|⃗a| cos θ = max (8.3) and similarly, Fy = may. In words, each component of the force vector is responsible for only the corresponding component of the acceleration. A force in the x direction does not cause any acceleration in the y direction, and vice-versa. a F ax Fx ay Fy x y θ Figure 8.2: If you take the familiar, one-dimensional (see the black dashed line) form of ⃗F = m⃗a, and project it onto orthogonal, rotated axes, you get the general two-dimensional case, showing that each orthogonal component of the acceleration is proportional, via the mass m, to only the corresponding component of the force (Eqs. (8.2)). In the rest of the chapter we shall see how to use Eqs. (8.2) in a number of examples. One thing I can anticipate is that, in general, we will try to choose our axes (unlike in Fig. 8.2 above) so that one of them does coincide with the direction of the acceleration, so the motion along the other direction is either nonexistent (v = 0) or trivial (constant velocity). | University Physics I Classical Mechanics_Page_181_Chunk4379 |
164 CHAPTER 8. MOTION IN TWO DIMENSIONS 8.2 Projectile motion Projectile motion is basically just free fall, only with the understanding that the object we are tracking was “projected,” or “shot,” with some initial velocity (as opposed to just dropped from rest). Unlike in the previous cases of free fall that we have studied so far, we will now assume that the initial velocity has a horizontal component, as a result of which, instead of just going straight up and/or down, the object will describe (ignoring air resistance, as usual) a parabola in a vertical plane. The plane in question is determined by the initial velocity (more precisely, the horizontal component of the initial velocity) and gravity. A generic trajectory is shown in Figure 8.3, showing the force and acceleration vectors (constant throughout) and the velocity vector at various points along the path. y x ymax height xmax height xrange vi yi vx,i vy,i FG a Figure 8.3: A typical projectile trajectory. The velocity vector (in green) is shown at the initial time, the point of maximum height, and the point where the projectile is back to its initial height. Conceptually, the problem turns out to be extremely simple if we apply the basic principle intro- duced in Section 8.1. The force is vertical throughout; so, after the throw, there is no horizontal acceleration, and the vertical acceleration is just −g, just as it always was in our earlier, one- dimensional free-fall problems: ax = Fx m = 0 ay = Fy m = −g (8.4) | University Physics I Classical Mechanics_Page_182_Chunk4380 |
8.2. PROJECTILE MOTION 165 The overall motion, then, is a combination of motion with constant velocity horizontally, and motion with constant acceleration vertically, and we can write down the corresponding equations of motion immediately: vx = vx,i vy = vy,i −gt x = xi + vx,it y = yi + vy,it −1 2gt2 (8.5) where (xi, yi) are the coordinates of the launching point (there is usually no reason to make xi anything other than zero, so we will do that below), and (vx,i, vy,i) the initial components of the velocity vector. By eliminating t in between the last two Eqs. (8.5), we get the equation of the trajectory in the x-y plane: y = yi + vy,i vx,i x − g 2v2 x,i x2 (8.6) which, as indicated earlier, and as shown in Fig. 8.3, is indeed the equation of a parabola. The apex of the parabola (highest point in the trajectory) is at xmax height = vx,ivy,i/g. We can get this result from calculus, or from a comparison of Eq. (8.6) with the canonical form of a parabola, or we can use some physics: the maximum height is reached, as usual, when the vertical velocity becomes momentarily zero, so solving the vy equation (8.5) for tmax height and substituting in the x equation, we get tmax height = vy,i g xmax height = vx,ivy,i g ymax height = yi + v2 y,i 2g (8.7) The last of these equations should look familiar. It is, indeed a variation on our old friend v2 f −v2 i = −2g∆y, only now instead of the full velocity ⃗v we have to use only the vertical velocity component vy. Just like for one-dimensional motion, this result follows again from conservation of energy: throughout the flight, we must have K + U G = constant, only now there is a component to the kinetic energy—the part associated with the horizontal motion—which remains constant on its own. In general, the kinetic energy of a particle will be 1 2m|⃗v|2, where |⃗v| is the magnitude of the velocity vector—that is to say, the speed. In two dimensions, this gives K = 1 2mv2 x + 1 2mv2 y (8.8) | University Physics I Classical Mechanics_Page_183_Chunk4381 |
166 CHAPTER 8. MOTION IN TWO DIMENSIONS For projectile motion, however, vx does not change, so any change in K will affect only the second term in Eq. (8.8). Conservation of energy between any two instants i and f gives K + U G = 1 2mv2 x,i + 1 2mv2 y,i + mgyi = 1 2mv2 x,i + 1 2mv2 y,f + mgyf (8.9) The 1 2mv2 x,i term cancels, and therefore v2 y,f −v2 y,i = −2g(yf −yi) (8.10) Another quantity of interest is the projectile’s range, or maximum horizontal distance traveled. We can calculate it from Eqs. (8.5), by setting y equal to the final height, then solving for t (which generally requires solving a quadratic equation), and then substituting the result in the equation for x. In the simple case when the final height is the same as the initial height, we can avoid the need for calculating altogether, and just reason, from the fact that the trajectory is symmetric, that the total horizontal distance traveled will be twice the distance to the point where the maximum height is reached, that is, xrange = 2xmax height: xrange = 2vx,ivy,i g (only if yf = yi) (8.11) As you can see, all these equations depend on the initial values of the components of the velocity vector ⃗vi. If ⃗vi makes an angle θ with the horizontal, and we simplify the notation by calling its magnitude vi, then we can write vx,i = vi cos θ vy,i = vi sin θ (8.12) In terms of vi and θ, the range equation (8.11) becomes xrange = v2 i sin(2θ) g (only if yf = yi) (8.13) since 2 sin θ cos θ = sin(2θ). This tells us that for any given launch speed, the maximum range is achieved when the launch angle θ = 45◦(always assuming the final height is the same as the initial height). In real life, of course, there will always be air resistance, and all these results will be modified somewhat. Mathematically, things become a lot more complicated: the drag force depends on the speed, which involves both components of the velocity, so the horizontal and vertical motions are no longer decoupled: not only is there now an Fx, but its value at any given time depends both on vx and vy. Physically, you may think of the drag force as doing negative work on the projectile, and hence removing kinetic energy from it. Less kinetic energy means, basically, that it will not travel quite as far either vertically or horizontally. Surprisingly, the optimum launch angle remains pretty close to 45◦, at least if the simulations at this link are accurate: https://phet.colorado.edu/en/simulation/projectile-motion | University Physics I Classical Mechanics_Page_184_Chunk4382 |
8.3. INCLINED PLANES 167 8.3 Inclined planes Back in Chapter 2, I stated without proof that the acceleration of an object sliding, without friction, down an inclined plane making an angle θ with the horizontal was g sin θ. I can show you now why this is so, and introduce friction as well. FG θ θ F n F k a θ F k FG Eb sb sb sb F n sb Eb a x y x y Figure 8.4: A block sliding down an inclined plane. The corresponding free-body diagram is shown on the right. Figure 8.4 above shows, on the left, a block sliding down an inclined plane and all the forces acting on it. These are more clearly seen on the free-body diagram on the right. I have labeled all the forces using the ⃗F type by,on convention introduced back in Chapter 6 (so, for instance, ⃗F k sb is the force of kinetic friction exerted by the surface on the block); however, later on, for algebraic manipulations, and especially where x and y components need to be taken, I will drop the “by, on” subscripts, and just let the “type” superscript identify the force in question. The diagrams also show the coordinate axes I have chosen: the x axis is along the plane, and the y is perpendicular to it. The advantage of this choice is obvious: the motion is entirely along one of the axes, and two of the forces (the normal force and the friction) already lie along the axes. The only force that does not is the block’s weight (that is, the force of gravity), so we need to decompose it into its x and y components. For this, we can make use of the fact, which follows from basic geometry, that the angle of the incline, θ, is also the angle between the vector ⃗F g and the negative y axis. This means we have F g x = F g sin θ F g y = −F g cos θ (8.14) Equations (8.14) also show another convention I will adopt from now, namely, that whenever the symbol for a vector is shown without an arrow on top or an x or y subscript, it will be understood to refer to the magnitude of the vector, which is always a positive number by definition. | University Physics I Classical Mechanics_Page_185_Chunk4383 |
168 CHAPTER 8. MOTION IN TWO DIMENSIONS Newton’s second law, as given by equations (8.4) applied to this system, then reads: F g x + F k x = max = F g sin θ −F k (8.15) for the motion along the plane, and F g y + F n y = may = −F g cos θ + F n (8.16) for the direction perpendicular to the plane. Of course, since there is no motion in this direction, ay is zero. This gives us immediately the value of the normal force: F n = F g cos θ = mg cos θ (8.17) since F g = mg. We can also use the result (8.17), together with Eq. (6.30), to get the magnitude of the friction force, assuming we know the coefficient of kinetic (or sliding) friction, µk: F k = µkF n = µkmg cos θ (8.18) Substituting this and F g = mg in Eq. (8.15), we get max = mg sin θ −µkmg cos θ (8.19) We can eliminate the mass to obtain finally ax = g (sin θ −µk cos θ) (8.20) which is the desired result. In the absence of friction (µk = 0) this gives a = g sin θ, as we had in Chapter 2. Note that, if you reduce the tilt of the surface (that is, make θ smaller), the cos θ term in Eq. (8.20) grows and the sin θ term gets smaller, so we must make sure that we do not use this equation when θ is too small or we would get the absurd result that ax < 0, that is, that the force of kinetic friction has overcome gravity and is accelerating the object upwards! Of course, we know from experience that what happens when θ is very small is that the block does not slide: it is held in place by the force of static friction. The diagram for such a situation looks the same as Fig. 8.4, except that ⃗a = 0, the force of friction is F s instead of F k, and of course its magnitude must match that of the x component of gravity. Equation (8.15) then becomes max = 0 = F g sin θ −F s (8.21) Recall from Chapter 6 that the force of static friction does not have a fixed value: rather, it will match the applied force up to a maximum value given by Eq. (6.29): F s max = µsF n = µsmg cos θ (8.22) where I have used Eq. (8.17), since clearly the equation (8.16) still applies along the vertical direction. So, on the one hand we have the requirement that F s = mg sin θ to keep the block from | University Physics I Classical Mechanics_Page_186_Chunk4384 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 169 sliding, and on the other hand the constraint F s ≤µsmg cos θ. Putting these together we conclude that the block will not slide as long as mg sin θ ≤µsmg cos θ (8.23) or tan θ ≤µs (8.24) In short, as long as θ is small enough to satisfy Eq. (8.24), the block will not move. Once θ exceeds the value tan−1 µs, we can apply the result (8.20) for the acceleration. Note that, since we always have µs ≥µk, the result (8.20) will always be positive if θ > tan−1 µs, that is, if sin θ > µs cos θ. What if we send the block sliding up the plane instead? The acceleration would still be pointing down (since the object would be slowing down all the while), but now the force of kinetic friction would point in the direction opposite that indicated in Figure 8.4, since it always must oppose the motion. If you go through the same analysis I carried out above, you will get that ax = g (sin θ + µk cos θ) in that case, since now friction and gravity are working together to slow the motion down. 8.4 Motion on a circle (or part of a circle) The last example of motion in two dimensions that I will consider in this chapter is motion on a circle. There are many examples of circular (or near-circular) motion in nature, particularly in astronomy (as we shall see in a later chapter, the orbits of most planets and many satellites are very nearly circular). There are also many devices that we use all the time that involve rotating or spinning objects (wheels, gears, turntables, turbines. . .). All of these can be mathematically described as collections of particles moving in circles. In this section, I will first introduce the concept of centripetal force, which is the force needed to bend an object’s trajectory into a circle (or an arc of a circle), and then I will also introduce a number of quantities that are useful for the description of circular motion in general, such as angular velocity and angular acceleration. The dynamics of rotational motion (questions having to do with rotational energy, and a new important quantity, angular momentum) will be the subject of the next chapter. 8.4.1 Centripetal acceleration and centripetal force As you know by now, the law of inertia states that, in the absence of external forces, an object will move with constant speed on a straight line. A circle is not a straight line, so an object will not naturally follow a circular path unless there is a force acting on it. | University Physics I Classical Mechanics_Page_187_Chunk4385 |
170 CHAPTER 8. MOTION IN TWO DIMENSIONS Another way to see this is to go back to the definition of acceleration. If an object has a velocity vector ⃗v(t) at the time t, and a different velocity vector ⃗v(t + ∆t) at the later time t + ∆t, then its average acceleration over the time interval ∆t is the quantity ⃗vav = (⃗v(t + ∆t) −⃗v(t))/∆t. This is nonzero even if the speed does not change (that is, even if the two velocity vectors have the same magnitude), as long as they have different directions, as you can see from Figure 8.5 below. Thus, motion on a circle (or an arc of a circle), even at constant speed, is accelerated motion, and, by Newton’s second law, accelerated motion requires a force to make it happen. θ R s v(t) v(t) v(t+Δt) v(t+Δt) Δv θ P Q Figure 8.5: A particle moving along an arc of a circle of radius R. The positions and velocities at the times t and t + ∆t are shown. The diagram on the right shows the velocity difference, ∆⃗v = ⃗v(t + ∆t) −⃗v(t). We can find out how large this acceleration, and the associated force, have to be, by applying a little geometry and trigonometry to the situation depicted in Figure 8.5. Here a particle is moving along an arc of a circle of radius r, so that at the time t it is at point P and at the later time t+∆t it is at point Q. The length of the arc between P and Q (the distance it has traveled) is s = Rθ, where the angle θ is understood to be in radians. I have assumed the speed to be constant, so the magnitude of the velocity vector, v, is just equal to the ratio of the distance traveled (along the circle), to the time elapsed: v = s/∆t. Despite the speed being constant, the motion is accelerated, as I just said above, because the direction of the velocity vector changes. The diagram on the right shows the velocity difference vector ∆⃗v = ⃗v(t + ∆t) −⃗v(t). We can get its length from trigonometry: if we split the angle θ in half, we get two right triangles, and for each of them |∆⃗v|/2 = v sin(θ/2). Thus, we have |⃗aav| = |∆⃗v| ∆t = 2v sin(θ/2) ∆t (8.25) for the magnitude of the average acceleration vector. The instantaneous acceleration is obtained by taking the limit of this expression as ∆t →0. In this limit, the angle θ = s/R = v∆t/R becomes | University Physics I Classical Mechanics_Page_188_Chunk4386 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 171 very small, and we can use the so-called “small angle approximation,” which states that sin x ≃x when x is small and expressed in radians. Therefore, by Eq. (8.25), |⃗aav| = 2v sin(θ/2) ∆t ≃vθ ∆t = v2∆t/R ∆t (8.26) This expression becomes exact as ∆t →0, and then ∆t cancels out, showing the instantaneous acceleration has magnitude |⃗a| = ac = v2 R (8.27) This acceleration is called the centripetal acceleration, which is why I have denoted it by the symbol ac. The reason for that name is that it is always pointing towards the center of the circle. You can kind of see this from Figure 8.5: if you take the vector ∆⃗v shown there, and move it (without changing its direction, so it stays ‘parallel to itself”) to the midpoint of the arc, halfway between points P and Q, you will see that it does point almost straight to the center of the circle. (A more mathematically rigorous proof of this fact, using calculus, will be presented in the next chapter, section 9.3.) The force ⃗Fc needed to provide this acceleration is called the centripetal force, and by Newton’s second law it has to satisfy ⃗Fc = m⃗ac. Thus, the centripetal force has magnitude Fc = mac = mv2 R (8.28) and, like the acceleration ⃗ac, is always directed towards the center of the circle. Physically, the centripetal force Fc, as given by Eq. (8.28), is what it takes to bend the trajectory so as to keep it precisely on an arc of a circle of radius R and with constant speed v. Note that, since ⃗Fc is always perpendicular to the displacement (which, over any short time interval, is essentially tangent to the circle), it does no work on the object, and therefore (by Eq. (7.11)) its kinetic energy does not change, so v does indeed stay constant when the centripetal force equals the net force. Note also that “centripetal” is just a job description: it is not a new type of force. In any given situation, the role of the centripetal force will be played by one of the forces we are already familiar with, such as the tension on a rope (or an appropriate component thereof) when you are swinging an object in a horizontal circle, or gravity in the case of the moon or any other satellite. At this point, if you have never heard about the centripetal force before, you may be feeling a little confused, since you almost certainly have heard, instead, about a so-called centrifugal force that tends to push spinning things away from the center of rotation. In fact, however, this “centrifugal force” does not really exist: the “force” that you may feel pushing you towards the outside of a curve when you ride in a vehicle that makes a sharp turn is really nothing but your own inertia—your body “wants” to keep moving on a straight line, but the car, by bending its trajectory, is preventing it from doing so. The impression that you get that you would fly radially out, as opposed to along a tangent, is also entirely due to the fact that the reference frame you are in (the car) is continuously | University Physics I Classical Mechanics_Page_189_Chunk4387 |
172 CHAPTER 8. MOTION IN TWO DIMENSIONS changing its direction of motion. You will find this effect illustrated in some detail in an example in the “Advanced Topics” section, if you want to look at it in more depth. On the other hand, getting a car to safely negotiate a turn is actually an important example of a situation that requires a definite centripetal force. On a flat surface (see the “Advanced Topics” section for a treatment of a banked curve!), you rely entirely on the force of static friction to keep you on the track, which can typically be modeled as an arc of a circle with some radius R. So, if you are traveling at a speed v, you need F s = mv2/R. Recalling that the force of static friction cannot exceed µsF n, and that on a flat surface you would just have F n = F g = mg, you see you need to keep mv2/R smaller than µsmg; or, canceling the mass, v2 R < µsg (8.29) This is the condition that has to hold in order to be able to make the turn safely. The maximum speed is then vmax = √µsgR, which, as you can see, will depend on the state of the road (for instance, if the road is wet the coefficient µs will be smaller). The posted, recommended speed will typically take this into consideration and will be as low as it has to be to keep you safe. Notice that the left-hand side of Eq. (8.29) increases as the square of the speed, so doubling your speed makes that term four times larger! Do not even think of taking a turn at 60 mph if the recommended speed is 30, and do not exceed the recommended speed at all if the road is wet or your tires are worn. 8.4.2 Kinematic angular variables Consider a particle moving on a circle, as in Figure 8.6 below (next page). Of course, we can just use the regular, cartesian coordinates, x and y, to describe its motion. But, in a way, this is carrying around more information than we typically need, and it is also not very transparent: a value of x and y does not immediately tell us how far the object has traveled along the circle itself. Instead, the most convenient way to describe the motion of the particle, if we know the radius of the circle, is to give the angle θ that the position vector makes with some reference axis at any given time, as shown in Fig. 8.6. If we choose the x axis as the reference, then the conversion from a description based on the radius R and the angle θ to a description in terms of x and y is simply x = R cos θ y = R sin θ (8.30) so knowing the function θ(t) we can immediately get x(t) and y(t), if we need them. (Note: in this section I am using an uppercase R for the magnitude of the position vector, to emphasize that it is a constant, equal to the radius of the circle.) | University Physics I Classical Mechanics_Page_190_Chunk4388 |
8.4. MOTION ON A CIRCLE (OR PART OF A CIRCLE) 173 + r v P (x,y) x y R sin θ θ R cos θ Figure 8.6: A particle moving on a circle. The position vector has length R, so the x and y coordinates are R cos θ and R sin θ, respectively. The conventional positive direction of motion is indicated. The velocity vector is always, as usual, tangent to the trajectory. Although the angle θ itself is not a vector quantity, nor a component of a vector, it is convenient to allow for the possibility that it might be negative. The standard convention is that θ grows in the counterclockwise direction from the reference axis, and decreases in the clockwise direction. Of course, you can always get to any angle by coming from either direction, so the angle by itself does not tell you how the particle got there. Information on the direction of motion at any given time is best captured by the concept of the angular velocity, which we represent by the symbol ω and define in a manner analogous to the way we defined the ordinary velocity: if ∆θ = θ(t + ∆t) −θ(t) is the angular displacement over a time ∆t, then ω = lim ∆t→0 ∆θ ∆t = dθ dt (8.31) The standard convention is also to use radians as an angle measure in this context, so that the units of ω will be radians per second, or rad/s. Note that the radian is a dimensionless unit, so it “disappears” from a calculation when the final result does not call for it (as in Eq. (8.35) below). For motion with constant angular velocity, we clearly will have θ(t) = θi + ω(t −ti) or ∆θ = ω∆t (constant ω) (8.32) | University Physics I Classical Mechanics_Page_191_Chunk4389 |
174 CHAPTER 8. MOTION IN TWO DIMENSIONS where ω is positive for counterclockwise motion, and negative for clockwise. (There is a sense in which it is useful to think of ω as a vector, but, since it is not immediately obvious how or why, I will postpone discussion of this next chapter, after I have introduced angular momentum.) When ω changes with time, we can introduce an angular acceleration α, defined, again, in the obvious way: α = lim ∆t→0 ∆ω ∆t = dω dt (8.33) Then for motion with constant angular acceleration we have the formulas ω(t) = ωi + α(t −ti) or ∆ω = α∆t (constant α) θ(t) = θi + ωi(t −ti) + 1 2α(t −ti)2 or ∆θ = ωi∆t + 1 2α(∆t)2 (constant α) (8.34) Equations (8.32) and (8.34) completely parallel the corresponding equations for motion in one dimension that we saw in Chapter 1. In fact, of course, a circle is just a line that has been bent in a uniform way, so the distance traveled along the circle itself is simply proportional to the angle swept by the position vector ⃗r. As already pointed out in connection with Fig. 8.5, if we expressed θ in radians then the length of the arc corresponding to an angular displacement ∆θ would be s = R|∆θ| (8.35) so multiplying Eqs. (8.32) or (8.34) by R directly gives the distance traveled along the circle in each case. Δθ s r(t) r(t+∆t) ∆r Figure 8.7: A small angular displacement. The distance traveled along the circle, s = R∆θ, is almost identical to the straight-line distance |∆⃗r| between the initial and final positions; the two quantities become the same in the limit ∆t →0. Figure 8.7 shows that, for very small angular displacements, it does not matter whether the distance traveled is measured along the circle itself or on a straight line; that is, s ≃|∆⃗r|. Dividing by ∆t, using Eq. (8.35) and taking the ∆t →0 limit we get the following useful relationship between the angular velocity and the instantaneous speed v (defined in the ordinary way as the distance traveled per unit time, or the magnitude of the velocity vector): |⃗v| = R|ω| (8.36) | University Physics I Classical Mechanics_Page_192_Chunk4390 |
8.5. IN SUMMARY 175 As we shall see later, the product Rα is also a useful quantity. It is not, however, equal to the mag- nitude of the acceleration vector, but only one of its two components, the tangential acceleration: at = Rα (8.37) The sign convention here is that a positive at represents a vector that is tangent to the circle and points in the direction of increasing θ (that is, counterclockwise); the full acceleration vector is equal to the sum of this vector and the centripetal acceleration vector, introduced in the previous subsection, which always points towards the center of the circle and has magnitude ac = v2 R = Rω2 (8.38) (making use of Eqs. (8.29) and (8.36)). These results will be formally established in the next chapter, after we introduce the vector product, although you could also verify them right now—if you are familiar enough with derivatives at this point—by using the chain rule to take the derivatives with respect to time of the components of the position vector, as given in Eq. (8.30) (with θ = θ(t), an arbitrary function of time). The main thing to remember about the radial and tangential components of the acceleration is that the radial component (the centripetal acceleration) is always there for circular motion, whether the angular velocity is constant or not, whereas the tangential acceleration is only nonzero if the angular velocity is changing, that is to say, if the particle is slowing down or speeding up as it turns. 8.5 In summary 1. To solve problems involving motion in two dimensions, you should break up all the forces into their components along a suitable pair of orthogonal axes, then apply Newton’s second law to each direction separately: Fnet,x = max, Fnet,y = may. It is convenient to choose your axes so that at least one of either ax or ay will be zero. 2. An object thrown with some horizontal velocity component and moving under the influence of gravity alone (near the surface of the Earth) will follow a parabola in a vertical plane. This results from horizontal motion with constant velocity, and vertical motion with constant acceleration equal to −g, as described by equations (8.7). 3. To analyze motion up or down an inclined plane, it is convenient to choose your axes so that the x axis lies along the surface, and the y axis is perpendicular to the surface. Then, if θ is the angle the incline makes with the horizontal, the force of gravity on the object will also make an angle θ with the negative y axis. 4. Recall that the force of kinetic friction will always point in a direction opposite the motion, and will have magnitude F k = µkF n, whereas the force of static friction will always take | University Physics I Classical Mechanics_Page_193_Chunk4391 |
176 CHAPTER 8. MOTION IN TWO DIMENSIONS on whatever value is necessary to keep the object from moving, up to a maximum value of F s max = µsF n. 5. An object moving in an arc of a circle of radius R with a speed v experiences a centripetal acceleration of magnitude ac = v2/R. “Centripetal” means the corresponding vector points towards the center of the circle. Accordingly, to get an object of mass m to move on such a path requires a centripetal force Fc = mv2/R. 6. To describe the motion of a particle on a circle of radius R, we use an angular position variable θ(t), in terms of which we define angular displacement ∆θ, angular velocity ω = dθ/dt, and angular acceleration α = dω/dt. The equations for motion in one dimension with constant acceleration apply to circular motion with constant α with the changes x →θ, v →ω and a →α. 7. The displacement along the circle, s, corresponding to an angular displacement ∆θ, is (in magnitude) s = R|∆θ|. Similarly, the (linear) speed of the particle (magnitude of its velocity vector) is equal to v = R|ω|, and the tangential component of its acceleration vector has magnitude at = R|α|. In addition to this, the particle always has a radial acceleration component ar equal to the centripetal acceleration of point 5 above. | University Physics I Classical Mechanics_Page_194_Chunk4392 |
8.6. EXAMPLES 177 8.6 Examples You will work out a rather thorough example of projectile motion in the lab, and Section 8.3 above already has the problem of a block sliding down an inclined plane worked out for you. The following example will show you how to use the kinematic angular variables of section 8.4.2 to deal with motion in a circle, and to calculate the centripetal acceleration in a simple situation. The section on Advanced Topics deals with a few more challenging (but interesting) examples. 8.6.1 The penny on the turntable Suppose that you have a penny sitting on a turntable, a distance d = 10 cm from the axis of rotation. Assume the turntable starts moving, steadily spinning up from rest, in such a way that after 1.3 seconds it has reached its final rotation rate of 33.3 rpm (revolutions per minute). Answer the following questions: (a) What was the turntable’s angular acceleration over the time interval from t = 0 to t = 1.3 s? (b) How many turns (complete and fractional) did the turntable make before reaching its final velocity? (c) Assuming the penny has not slipped, what is its centripetal acceleration once the turntable reaches its final velocity? (d) How large does the static friction coefficient between the penny and the turntable have to be for the penny not to slip throughout this process? Solution (a) We are told that the turntable spins up “steadily” from t = 0 to t = 1.3 s. The word “steadily” here is a keyword that means the (angular) acceleration is constant (that is, the angular velocity increases at a constant rate). What is this rate? For constant α, we have, from Eq. (8.34), α = ∆ω/∆t. Here, the time interval ∆t = 1.3 , so we just need to find ∆ω. By definition, ∆ω = ωf −ωi, and since we start from rest, ωi = 0. So we just need ωf. We are told that “the final rotation rate” is 33.3 rpm (revolutions per minute). What does this tell us about the angular velocity? The angular velocity is the number of radians an object moving in a circle (such as the penny in this example) travels per second. A complete turn around the circle, or revolution, corresponds to 180◦, or equivalently 2π radians. So, 33.3 revolutions, or turns, per minute means 33.3×2π radians per 60 s, that is, ωf = 33.3 × 2π rad 60 s = 3.49 rad s (8.39) | University Physics I Classical Mechanics_Page_195_Chunk4393 |
178 CHAPTER 8. MOTION IN TWO DIMENSIONS The angular acceleration, therefore, is α = ∆ω ∆t = ωf −ωi ∆t = 3.49 rad/s 1.3 s = 2.68 rad s2 (8.40) (b) The way to answer this question is to find out the total angular displacement, ∆θ, of the penny over the time interval considered (from t = 0 to t = 1.3 s), and then convert this to a number of turns, using the relationship 2π rad = 1 turn. To get ∆θ, we should use the equation (8.34) for motion with constant angular acceleration: ∆θ = ωi∆t + 1 2α(∆t)2 (8.41) We start from rest, so ωi = 0, We know ∆t = 1.3 s, and we just calculated α = 2.68 rad/s2, so we have ∆θ = 1 2 × 2.68 rad s2 × (1.3 s)2 = 2.26 rad (8.42) This is less than 2π radians, so it takes the turntable less than one complete revolution to reach its final angular velocity. To be precise, since 2π radians is one turn, 2.26 rad will be 2.26/(2π) turns, which is to say, 0.36 turns—a little more than 1/3 of a turn. (c) For the questions above, the penny just served as a marker to keep track of the revolutions of the turntable. Now, we turn to the dynamics of the motion of the penny itself. First, to get its angular acceleration, we can just use Eq. (8.38), in the form ac = Rω2 = 0.1 m × $ 3.49 rad s %2 = 1.22 m s2 (8.43) noticing that R, the radius of the circle on which the penny moves, is just the distance d to the axis of rotation that we were given at the beginning of the problem, and ω, its angular velocity, is just the final angular velocity of the turntable (assuming, as we are told, that the penny has not slipped relative to the turntable). (d) Finally, how about the force needed to keep the penny from slipping—that is to say, to keep it moving with the turntable? This is just the centripetal force needed “bend” the trajectory of the penny into a circle of radius R, so Fc = mac, where m is the mass of the penny and ac is the centripetal acceleration we just calculated. Physically, we know that this force has to be provided by the static (as long as the penny does not slip!) friction force between the penny and the turntable. We know that F s ≤µsF n, and we have for the normal force, in this simple situation, just F n = mg. Therefore, setting F s = mac we have: mac = F s ≤µsF n = µsmg (8.44) This is equivalent to the single inequality mac ≤µsmg, where we can cancel out the mass of the penny to conclude that we must have ac ≤µsg, and therefore µs ≥ac g = 1.22 m/s2 9.8 m/s2 = 0.124 (8.45) | University Physics I Classical Mechanics_Page_196_Chunk4394 |
8.7. ADVANCED TOPICS 179 8.7 Advanced Topics 8.7.1 Staying on track (This example studies a situation that you could easily setup experimentally at home (you can use a whole sphere instead of a half-sphere!), although to get the numbers to work out you really need to make sure that the friction between the surface and the object you choose is truly negligible. Essentially the same mathematical approach could be used to study the problem of a skier going over a mogul, or a car losing contact with the road if it is going too fast over a hill.) A small object is placed at the top of a smooth (frictionless) dome shaped like a half-sphere of radius R, and given a small push so it starts sliding down the dome, initially moving very slowly (vi ≃0), but picking up speed as it goes, until at some point it flies offthe surface. (a) At that point, when the object loses contact with the surface, what is the angle that its position vector (with origin at the center of the sphere) makes with the vertical? (b) How far away from the sphere does the object land? θ θmax F n sb FG Eb θ θ (a) (b) Figure 8.8: An object (small block) sliding on a hemispherical dome. The drawing (a) shows the angle θmax at which the object flies off(red dashed line), and a smaller, generic angle θ. The drawing (b) shows the free-body diagram corresponding to the angle θ. Solution (a) As we saw in Section 8.4, in order to get an object to move along an arc of a circle, a centripetal force of magnitude mv2/r is required. As long as our object is in contact with the surface, the forces acting on it are the normal force (which points along the radial direction, so it makes a negative contribution to the centrifugal force) and gravity, which has a component mg cos θ along the radius, towards the center of the circle (see Figure 8.8(b), the dashed light blue line). So, the centripetal force equation reads mv2 R = mg cos θ −F n (8.46) | University Physics I Classical Mechanics_Page_197_Chunk4395 |
180 CHAPTER 8. MOTION IN TWO DIMENSIONS The next thing we need to do is find the value of the speed v for a given angle θ. If we treat the object as a particle, its only energy is kinetic energy, and ∆K = Wnet (Eq. (7.11)), where Wnet is the work done on the particle by the net force acting on it. The normal force is always perpendicular to the displacement, so it does no work, whereas gravity is always vertical and does work Wgrav = −mg∆y (taking upwards as positive, so ∆y is negative). In fact, from Figure 8.8(a) (follow the dashed blue line) you can see that for a given angle θ, the height of the object above the ground is R cos θ, so the vertical displacement from its initial position is ∆y = −(R −R cos θ) (8.47) Hence we have, for the change in kinetic energy, 1 2mv2 −1 2mv2 i = mgR −mgR cos θ (8.48) Assuming, as we are told in the text of the problem, that vi ≃0, we get v2 ≃2gR −2gR cos θ, and using this in Eq. (8.46) 2mg −2mg cos θ = mg cos θ −F n (8.49) or F n = 3mg cos θ−2mg. This shows that F n starts out (when θ = 0) having its usual value of mg, and then it becomes progressively smaller as the object slides down. The point where the object loses contact with the surface is when F n = 0, and that happens for 3 cos θmax = 2 (8.50) or θmax = cos−1(2/3) = 48.2◦. Recalling that ∆y = −(R −R cos θ), we see that when cos θ = 2/3, the object has fallen a distance R/3; put otherwise, its height above the ground at the time it flies offis 2R/3, or 2/3 of the initial height. (b) This is just a projectile problem now. We just have to find the values of the initial conditions (xi, yi, vx,i and vy,i) and substitute in the equations (8.5). By inspecting the figure, you can see that, at the time the object flies off, xi = R sin θmax = 0.745R yi = R cos θmax = 0.667R (8.51) Also, we found above that v2 ≃2gR −2gR cos θ, and when θ = θmax this gives v2 = 0.667gR, or v = 0.816√gR. The projection angle in this case is −θmax; that is, the initial velocity of the projectile (dashed red arrow in Figure 8.8(a)) is at an angle 48.2◦below the positive x axis, so we have: vx,i = vi cos θmax = 0.544 ! gR vy,i = −vi sin θmax = −0.609 ! gR (8.52) | University Physics I Classical Mechanics_Page_198_Chunk4396 |
8.7. ADVANCED TOPICS 181 Now we just use these results in Eqs. (8.5). Specifically, we want to know how long it takes for the object to reach the ground, so we use the last equation (8.5) with y = 0 and solve for t: 0 = yi + vy,it −1 2gt2 (8.53) The result is t = 0.697 ! R/g. (You do not need to carry the “g” throughout; it would be OK to substitute 9.8 m/s2 for it. I have just kept it in symbolic form so far to make it clear that the quantities we derive will have the right units.) Substituting this in the equation for x, we get x = xi + vx,it = 0.745R + 0.544 ! gR × 0.697 ! R/g = 1.125R (8.54) (Note how the g cancels, so we would get the same result on any planet!) Since the sphere has a radius R, the object falls a distance 0.125R away from the sphere. 8.7.2 Going around a banked curve Roadway engineers often bank a curve, especially if it is a very tight turn, so the cars will not have to rely on friction alone to provide the required centripetal force. The picture shows a car going around such a curve, which we can model as an arc of a circle of radius r. In terms of r, the bank angle θ, and the coefficient of static friction, find the maximum safe speed around the curve. F n sc FG Ec F s sc x y θ F n sc FG Ec F s sc x y a Figure 8.9: A car going around a banked curve (sketch and free-body diagram). The center of the circle is towards the right. The figure shows the appropriate choice of axes for this problem. The criterion is, again, to choose the axes so that one of them will coincide with the direction of the acceleration. In this case, the acceleration is all centripetal, that is to say, pointing, horizontally, towards the center of the circle on which the car is traveling. It may seem strange to see the force of static friction pointing down the slope, but recall that for a car turning on a flat surface it would have been pointing inwards (towards the center of the circle), | University Physics I Classical Mechanics_Page_199_Chunk4397 |
182 CHAPTER 8. MOTION IN TWO DIMENSIONS so this is the natural extension of that. In general, you should always try to imagine which way the object would slide if friction disappeared altogether: ⃗F s must point in the direction opposite that. Thus, for a car traveling at a reasonable speed, the direction in which it would skid is up the slope, and that means ⃗F s must point down the slope. But, for a car just sitting still on the tilted road, ⃗F s must point upwards, and we shall see in a moment that in general there is a minimum velocity required for the force of static friction to point in the direction we have chosen. Apart from this, the main difference with the flat surface case is that now the normal force has a component along the direction of the acceleration, so it helps to keep the car moving in a circle. On the other hand, note that we now lose (for centripetal purposes) a little bit of the friction force, since it is pointing slightly downwards. This, however, is more than compensated for by the fact that the normal force is greater now than it would be for a flat surface, since the car is now, so to speak, “driving into” the road somewhat. The dashed blue lines in the free-body diagram are meant to indicate that the angle θ of the bank is also the angle between the normal force and the positive y axis, as well as the angle that ⃗F s makes below the positive x axis. It follows that the components of these two forces along the axes shown are: F n x = F n sin θ F n y = F n cos θ (8.55) and F s x = F s cos θ F n y = −F s sin θ (8.56) The vertical force equation is then: 0 = may = F n y + F s y −F G = F n cos θ −F s sin θ −mg (8.57) This shows that F n = (mg + F s sin θ)/ cos θ is indeed greater than just mg for this problem, and must increase as the angle θ increases (since cos θ decreases with increasing θ). The horizontal equation is: max = F n x + F s x = F n sin θ + F s cos θ = mv2 r (8.58) where I have already substituted the value of the centripetal acceleration for ax. Equations (8.57) and (8.58) form a system that needs to be solved for the two unknowns F n and F s. The result is: F n = mg cos θ + mv2 r sin θ F s = −mg sin θ + mv2 r cos θ (8.59) | University Physics I Classical Mechanics_Page_200_Chunk4398 |
8.7. ADVANCED TOPICS 183 Note that the second equation would have F s becoming negative if v2 < gr tan θ. This means that below that speed, the force of static friction must actually point up the slope, as discussed above. We can call this particular speed, for which F s becomes zero, vno friction: vno friction = ! gr tan θ (8.60) What this means is that it is possible to arrange the banking angle so that a car going at a specific speed would not have to rely on friction at all in order to make the curve: the normal force would be just right to provide the required centripetal acceleration. A car going at that speed would not feel either pulled down or pushed up the slope. However, a car going faster than that would tend to “fly off”, and the static friction force would be required to pull it in and keep it on the curve, whereas a car moving more slowly would tend to slide down and would have to be pushed up by the friction force. Friction, therefore, provides a range of safe speeds to drive in this case, just as it did in the flat surface case. We can calculate the maximum safe speed as we did before, recalling that we must always have F s ≤µsF n. Substituting Eqs. (8.59) in this expression, and solving for v, we get the condition vmax = √gr . µs + tan θ 1 −µs tan θ (8.61) This reproduces our result (8.29) for θ = 0 (a flat road), as it should. To put some numbers into this, suppose the curve has a radius of 20 m, and the coefficient of static friction between the tires and the road is µs = 0.7. Then, for a flat surface, we get vmax = 11.7 m/s, or about 26 mph, whereas for a bank angle of θ = 10◦(the angle chosen for the figure above) we get vmax = 14 m/s, or about 31 mph. Equation (8.61) actually indicates that the maximum velocity would “become infinite” for a finite bank angle, namely, if 1 −µs tan θ = 0, or tan θ = 1/µs (if µs = 0.7, this corresponds to θ = 55◦). This is mathematically correct, but of course we cannot take it literally: it assumes that there is no limit to how large a normal force the roadway may exert without sustaining damage, and also that F s can become arbitrarily large as long as it stays below the bound F s ≤µsF n. Neither of these assumptions would hold in real life for very large speeds. Also, the angle θ = tan−1(1/µs) is much too steep: recall that, according to Eq. (8.24), the force of friction will only be able to keep an object (initially at rest) from sliding down the slope if tan θ ≤µs, which for µs = 0.7 means θ ≤35◦. So, with a bank angle of 55◦you might drive on the curve, provided you were going fast enough, but you could not park on it—the car would slide down! Bottom line, use Eq. (8.61) only for moderate values of θ . . . and do not exceed θ = tan−1 µs if you want a car to be able to drive around the curve slowly without sliding down into the ditch. | University Physics I Classical Mechanics_Page_201_Chunk4399 |
184 CHAPTER 8. MOTION IN TWO DIMENSIONS 8.7.3 Rotating frames of reference: Centrifugal force and Coriolis force Imagine you are inside a rotating cylindrical room of radius R. There is a metal puck on the floor, a distance r from the axis of rotation, held in place with an electromagnet. At some time you switch offthe electromagnet and the puck is free to slide without friction. Find where the puck strikes the wall, and show that, if it was not too far away from the wall to begin with, it appears as if it had moved straight for the wall as soon as it was released. Solution The picture looks as shown below, to an observer in an inertial frame, looking down. The puck starts at point A, with instantaneous velocity ωr pointing straight to the left at the moment it is released, so it just moves straight (in the inertial frame) until it hits the wall at point B. From the cyan-colored triangle shown, we can see that it travels a distance √ R2 −r2, which takes a time ∆t = √ R2 −r2 ωr (8.62) In this time, the room rotates counterclockwise through an angle ∆θroom = ω∆t: ∆θroom = √ R2 −r2 r (8.63) O A A‘ B B‘ r R Figure 8.10: The motion of the puck (cyan) and the wall (magenta) as seen by an inertial observer. This is the angle shown in magenta in the figure. As a result of this rotation, the point A’ that was initially on the wall straight across from the puck has moved (following the magenta dashed line) to the position B’, so to an observer in the rotating room, looking at things from the point O, the puck appears to head for the wall and drift a little to the right while doing so. | University Physics I Classical Mechanics_Page_202_Chunk4400 |
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