PHYS 200: Fundamentals of Physics I

Lecture 8

 - Dynamics of a Multiple-Body System and Law of Conservation of Momentum


The dynamics of a many-body system is examined. Through a variety of examples, the professor demonstrates how to locate the center of mass and how to evaluate it for a number of objects. Finally, the Law of Conservation of Momentum is introduced and discussed. The lecture ends with problems of collision in one dimension focusing on the totally elastic and totally inelastic cases.

Transcript Audio Low Bandwidth Video High Bandwidth Video

Fundamentals of Physics I

PHYS 200 - Lecture 8 - Dynamics of a Multiple-Body System and Law of Conservation of Momentum

Chapter 1. Multibody Dynamics — The Two-Body System [00:00:00]

Professor Ramamurti Shankar: So, today we are going to do something different from what’s happened so far, in that we are going to study the dynamics of more than one body. Well, you might say “Look, we already did this last week, when I studied the Solar System,” where there are planets moving around the Sun so that makes at least two bodies, the Sun and the planet. But actually, the Sun was not doing anything interesting in our analysis. The Sun just stood there as a source of the gravitational force. It’s the planet that did all the orbiting and that was the problem in two dimensions, but of only one body. So now, we are going to enlarge our domain to more than one body obeying Newton’s laws.

So, let me start with the simplest possible case: two bodies. And I will again start with the simplest case of their moving in one dimension; then, we’ll put in more. So, here is the one-dimensional world in which these bodies are going to move, and this is my origin, x = 0, and I’m going to imagine one point mass m1, located at x1, another point mass m2, located at x2. Now, we know everything there is to know about these masses from the laws of Newton which I’m going to write down. The first mass obeys this equation, m1d2x1/ dt2. That’s ma, right? But I’m going to write this in a notation that’s more succinct. I’m tired of writing the second derivative in this fashion. I’m going to write it as follows, m1x1, with two dots on it; the two dots telling you it’s two derivatives, because one dot is one derivative, three dots is three derivatives. Of course, at some point this notation becomes unwieldy, but you never deal with more than two derivatives, so this works. This is ma, so don’t forget the dots, okay? This is not some foreign alphabet. Every dot is a derivative. You should remember that when I do subsequent manipulation. That’s ma, and that’s equal to force on body 1, F1.

Now, look at the body one and ask, “What are the forces acting on it?” Well, it could be the whole universe. But we’re going to divide that into two parts. The first part is going to be the force on body 1 due to body 2, which I’m going to denote by F12; that’s our notation. You and I agree that it’s the force on 1 due to 2. Then, there’s the force on 1 due to the external world; e stands for external. That means everything outside these two. So, the universe has many bodies; I have just picked these two guys. They’re 1 and 2, and the force on 1 is–some of it’s due to 2, and some of it’s due to everything else.

Similarly, I have another equation, m2x2 double dot is the force on 2 due to 1 plus the force on 2 due to the outside world. What do you mean by “outside world”? Maybe these two guys are next to some planet, and the planet’s way over to the right, it’s pulling all of them towards the planet with some gravitational force. So, everything else is called “external.” And I have 1 and 2, for example, are connected by a spring. The spring is not that important. It’s a way of transmitting force from one body to the other. If you compress the spring and let it go, these two masses will vibrate back and forth under the influence of the other person’s force. That’s an example of F12 and F21. For example, if the spring is compressed at this instant, it’s trying to push them out; that means, really, 1 is trying to push 2 out, whereas that way 2 is trying to push 1 to the left, this way. That’s an example of F12 and F21.

The external force could be due to something extraneous to these two bodies. So, one example is at the surface of the Earth. I take these two masses connected by a spring. Here is mass 1 and here’s mass 2. I squash the spring. If there is no gravity, they will just go vibrating up and down but let them fall into the field of gravity, so they’re also experiencing the mg due to gravity. So, they will both fall down and also oscillate a little with each other. They’re all described by this equation. This will be the spring force transmitted from 1 to 2. This will be the force of gravity, or it could be an electric force or any other force due to anything else; we are not interested.

Now, here is the interesting manipulation we’re going to perform. We are going to add the two things on the left-hand side and equate them to whatever I get on the right-hand side. Then, m1x1 double dot plus m2x2 double dot, and that’s going to be–I’m going to write it in a particular way, F1e + F2e + F12 + F21. I think you have some idea of where I’m headed now. So, what’s the next thing we could say? Yes?

Student: F12 + F21 = 0.

Chapter 2. The Center of Mass [00:05:41]

Professor Ramamurti Shankar: Yes, because that’s the Third Law of Newton. Whatever the underlying force, gravity, spring, anything, force on 1 due to 2, and force on 2 due to 1 will cancel, and everything I get today, the whole lecture is mainly about this one simple result, this cancellation. Then, this whole thing, I’m going to write as Fe, meaning the total external force on this two-body system. So, I have this peculiar equation. I’m going to rewrite it in a way that brings it to the form in which it’s most useful. I’m going to introduce a new guy, capital X. As you know, that’s called a center of mass coordinate, and it’s defined as m1x1 + m2x2 divided by capital M. Capital M is just the total mass, m1 + m2. If I do that, this is a definition. Then, we can write this equation as follows. I will write it and then we can take our time seeing that it is correct. So, this is really the big equation. Why don’t you guys try to fill in the blanks in your head? This is really correct. On the left-hand side, I have M times X double dot, so I have really m1 + m2 times X double dot. If you take the double dot of this guy, it’s m1x1 double dot plus m2x2 double dot, divided by m1 + m2. So, the left-hand side is indeed this; that’s all I want you to check. So, take this expression, divide by the total mass and multiply by the total mass. Well, the multiplying by the total mass is here, and when you divide by the total mass you get the second derivative of the center of mass coordinate.

So what have I done? I have introduced a fictitious entity, the center of mass. The center of mass is a location X, some kind of a weighted average of x1 and x2. By weighted, I mean if m1 and m2 are equal, then capital M will be two times that mass and you’ll just get x1 + x2 over 2. The center of mass will sit right in between. But if m1 is heavier, it’ll be tilted towards m1; if m2 is heavier, it’ll be tilted towards m2. It’s a weighted sum that gives a certain coordinate. There is nothing present at that location. There’s nobody there. All the stuff is either here or there. The center of mass is the location of a mathematical entity. It’s not a physical entity. If you go there and say, “What’s at the center of mass?” you typically won’t find anything. And it behaves like a body. After all, if you just said, “I’ve learned Newton’s laws,” and I walk into this room and I say this, you will say “Well, this guy’s talking about a body of mass, capital M, undergoing some acceleration due to the force.” So, the center of mass is the body whose mass seems to be the total mass of these two particles, whose acceleration is controlled by the same as Newton’s law, but the right-hand side contains only the external forces; this is the key. All the internal forces have canceled out, and what remains is the external force.

Now it turns, if you’ve got three bodies, you can do a similar manipulation. And again, you’ll have F12 and F23 and F32 and so on. They will cancel and what will remain will be a similar thing, but this is the total external force. So, if I can say this in words, what we have learned is that the advantage of introducing a quantity called “center of mass” is that it responds only to the total force; it doesn’t care about internal forces.

So, I’ll give an example. Here is some airplane, right? It’s in flight, and a couple of guys are having a fight, punching each other and so on. The rest of the passengers say “enough is enough” and they throw them out. So, they’re just floating around, affecting each other’s dynamics, and of course this person will feel a force due to that person, that person will feel a force due to this person, but what I’m telling you is the center of mass is going to drop like a rock. It’s going to accelerate with the force mgh; it’s going to accelerate with g. So, at one point this person may be having the upper hand and may be here, and the other person may be down here, but follow the center of mass and you’ll find it simply falls under gravity. So, the mutual forces do not affect the dynamics of the center of mass. Or, for example, suppose at some point, this person blows the other one up into, say, it’s a samurai bat, make it simple, cuts him up in two pieces, so now we’ve got three bodies now: the first protagonist and the other two now, unfortunately somewhat decimated. Now, you can take these three bodies, find their center of mass; it’ll be the same thing; it’ll just keep falling down. So, even though the system is becoming more and more complicated, you cannot change the dynamics of the center of mass. It responds only to the external force. If this fight was taking place in outer space where there’s no gravity, then as this fight continues and people are flying and parts are flying everywhere, the center of mass will just be in one location, not doing anything. Yes?

Student: Couldn’t the internal forces change the center of mass’s location?

Professor Ramamurti Shankar: No, that’s what I’m saying. The center of mass, if it changes it can certainly– No one says the center of mass cannot accelerate. It can accelerate due to external forces. But if there were no external forces, then the center of mass will behave like a particle with no force. If it’s not moving to begin with, it won’t move later. Or if it is moving to begin with, it’ll maintain a constant velocity.

So, here’s another example. Now you can obviously generalize this to more than one dimension. If you’re living in two dimensions, you will introduce an x coordinate and introduce a y coordinate and then you will have the center of mass as MR double dot equals F, and R would be m1r1 + m2r2 divided by the total mass; r1 and r2 are just the location now in two dimensions of these two masses. So, here is m1 and here is m2 and the center of mass you can check will be somewhere in between the line joining the two points, but it’ll now be a vector.

So here’s another example. You take a complicated object; it’s got masses and it’s got springs; it’s connected with thread, and chains and everything. You throw the whole mess into the air. All the different parts of it are jiggling and doing complicated movements, but if you follow the center of mass, in other words at every instant you take the m1r1 + m2r2 + m3r3 and so on, divided by the total mass, that coordinate will simply be following the parabolic path of a body curving under gravity. If at some point this complicated object fragments into two chunks, one will land here and one will land there. But at every instant, if you follow the center of mass, it’ll go as if nothing happened and it will land here. The center of mass does not care about internal forces, only about external forces. That’s the main point. And it was designed in such a way that external [correction: should have said “internal”] forces canceled in these dynamics. So, everything I’m going to do today is to take that equation, MX..= F or MR..= F, in vector form, and deduce some of the consequences.

Now, first of all, you should realize that if you’ve got several bodies, say three bodies, then I will define the center of mass to be mixi divided by the sum of mi. This is a shorthand, I’m going to write it only once, but you should know what the notation means. If there is i from 1 to 3, it really means m1x1 + m2x2 + m3x3 divided by m1 + m2 + m3. This summation is the notation mathematicians have introduced where the index i will go over a range from 1 to 3. Every term you let i take three different values and you do the sum. An exercise I’ve given to you guys to pursue at home is the following: if I got three bodies, 1, 2 and 3, you’ve defined the center of mass, you can either go to this formula, do all the m1x1’s and add them up, or you also have the following option. You can pick any two of them, say the first two – forget the third one – take these two, find their center of mass, let’s call it x1 and x2 and with that total mass M12, which is just m1 + m2, trade these two for a new fictitious object and put that here, and forget these two. But on that one point it deposits the mass of these two; now, you take the center of mass of this object with the third object, by the same weighting process, m3x3 + M12X12 divided by the total mass; you’ll get the same answer as here.

What I’m telling you is, if you’ve got many bodies and you want the center of mass of all of them, you can take a subset of them, replace them by their center of mass, namely, all their mass sitting at their center of mass, replace the other half by their mass sitting at their center of mass, and finally find the center of mass of these two centers of mass, properly weighted, and that’ll give you this result.

Okay, so, before I exploit that equation and find all the consequences, we have to get used to finding center of mass for a variety of things, as long as they give you ten masses, or a countable number of masses, we’ve just got to plug it in here; it’s a very trivial exercise. Things become more interesting if I give you not a set of discrete masses but discrete locations, namely, a countable set of masses, but I give you a rod like this. This is a rod of mass M and length L, and I say: “Where is the center of mass?” So, we have to adapt the definition that we have for this problem.

So, what should I do? Well, this is my origin of coordinates. If I had a set of masses with definite locations I know how to do it, but this is a continuum. The trick then is to say, I take a distance x from the left hand, and I cut myself a very thin sliver of taking this dx. That sliver has got a certain mass, and I argue it’s at a definite distance x from the coordinate’s origin. And if you’re nitpicking you will say, “What do you mean by definite distance?” It’s got a width dx, so one part of it is at x, the other part is at x + dx so it doesn’t have a definite coordinate. But if dx is going to 0, this argument will eventually be invalid. So dx goes to 0, sliver has a definite location, which is just the x coordinate of where I put it. So, to find the center of mass, which consists of multiplying every mass by its location and adding–Let me first find how much mass is sitting here. Let me call it Δm. How much mass is sitting there? Well, I do the following. I take the total mass and divide it by L; that’s the mass per unit length. And this fellow has a width dx, so the mass of this little sliver is (M/L) dx.

Therefore, the center of mass that I want is found by taking this sliver of that mass, multiplying by its coordinate and summing over all the slivers, which is what we do by the integral from 0 to L. Then, I should divide by the total mass, which is just M. You can see now if you do this calculation. I get 1/L; then, I get integral xdx from 0 to L, and that’s going to be L2 over 2. So, if you do that you’ll get L/2. I’m not doing every step in the calculus because at this point we should be able to do this without every detail. So, the center of mass of this rod, to nobody’s surprise, is right in the midpoint, but it assumes that the rod is uniform, whereas [if its] like a baseball bat, thicker at one end and thinner at the other end, of course no one is saying that, but we have assumed the mass per unit length, namely M or is this a fixed number, M/L, and then this is the answer.

But there are other ways to get this result without doing all the work, okay? So, we would like to learn that other method because it’ll save you a lot of time. It must be clear to most people that the center of mass of this rod is at the center. But how do we argue that? How do you make it official? If you do the integral you will get the answer, but I want to short circuit the integral. And here is a trick. It’s not going to work for arbitrary bodies. If I give you some crazy object like this, you cannot do anything. But this is a very symmetric object; you can sort of tell if I take the midpoint. There is as much stuff to the right as to the left and somehow you want to make that argument formal, and you do the following.

Suppose I have a bunch of masses, and the object’s really not even regular, and this is my origin of coordinates. If I replace every x by -x, okay–sorry, I should change this object, so this really looks like this. Here’s the object. Suppose I replace every x by -x that’s reflecting the body around this axis, it will look like this; it’ll be jutting to the right instead of the left. So, don’t go by my diagrams. You know what I’m trying to do. I’m trying to draw the mirror image of this object the other way. Then, I think it’s clear to everybody, if I do that, X will go to -X, because in this averaging mixi, sum of all the masses, if every x goes to -x the center of mass will go to -X. But now, take this rod and transport every particle to its negative coordinate. And you take that guy and put it here, the rod looks the same. If the rod looks the same, the center of mass must look the same. That means -xX has to be equal to x itself and the only answer is X = 0. So, without doing any detailed calculation, you argue that the answer is X = 0.

To do this, of course, you must cleverly pick your coordinates so that the symmetry of the body is evident. If you took the body like this and you took a reflection around this point, it goes into body flipped over. There is not much you can say about it. So, what you really want to do is to pick a point of reflection so that upon reflection the body looks the same; the body looks the same, the answer must be the same. But if you argue the answer must be minus of itself, therefore the answer is 0. This is how the center of mass of symmetric bodies can be found. So, we know the answer for this rod.

Suppose I give you not a thin rod, but a rod of non-0 width. We want to know its center of mass; we’re not going to do any more work now. By symmetry, I can argue that this has to be the center of mass, because I can take every point here and turn it into the point there by changing y to -y; therefore, capital Y will become minus capital Y, but the body looks the same after this mapping. So, capital Y is 0, and similarly capital X is 0, and that’s the center of mass.

Okay, now, what if I give you this object, anybody want to try that? Would you guys like to try this? Yes, can you tell me where the center of mass should be, yeah? No, no, the guy in front of you, yes.

Student: The center of mass, you can’t get the center of mass with two objects, and then this is where you said over in the curve, find the center between those two masses.

Professor Ramamurti Shankar: Okay, is that what your answer was? Okay, so the correct answer is, replace this mass by all of its mass, whatever it is, sitting here. Replace this one by all of its mass sitting here; then, forget the big bodies, replace them by points. Then, he’s got two masses located here, and you can find the weighted average. It may be somewhere there.

Okay, now let’s take one more object. Then I’m pretty much done with finding these centers of mass. The object I’m going to pick is a triangle that looks like this; it’s supposed to be symmetric, even though I’ve drawn it this way. That is b and that is b, and that distance is h; let the mass be M. Where is the center of mass of this object? Again, by symmetry, you can tell that the y coordinate of the center of mass must lie on this line, because if I take y to -y, it maps onto itself so it looks the same. But it’s supposed to reverse capital Y; therefore, Y is -Y and therefore it’s 0; so, it’s evidently lying somewhere on this line. I cannot do further calculations of this type by saying where it is on the line, because it has no longer a symmetry in the x direction; it’s symmetric on the y after flipping y, but I cannot take x to -x. In fact, if I take x to -x this one looks like this, it doesn’t map into itself. But I can pick any point here; if I take x to -x, the object looks different. It looks like that object and relating one object to another object is what I’m trying to do. I want to relate it to the same object. That cannot be done for x. It can be done for y; for x you’ve got to do some honest work.

The honest work we will do then is to take this thing, take a strip here, with location x and width dx, and the height of that strip here is y; y of course varies with x. So, I’m going to argue that to find the center of mass of this triangle I can divide it into vertical strips which are parallel to each other, and find the center of mass by adding the weighted average of all these things. For that, I need to know what’s the mass of the shaded region. So, let’s call it Δm. The mass of the shaded region is the mass per unit area. I’ll find the area later on in terms of b and h, but this is mass per unit area. Then, I need to know the area of the strip. I’m going to give the answer because I don’t have time to probe it. But you should think about what the answer for the area of the shaded region is. It’s got a height 2y, and it’s got the width dx. It’s not quite a rectangle because the edges are slightly tapered, but when Δx goes to 0, it’s going to look like a rectangle. So, the area is 2y Δx. But I don’t want to write everything in terms of y. I want to write it in terms of x; then, I do similar triangles. Similar triangles tell me that y/b is x/h, namely, that triangle compared to that triangle tells me y/b is x/h. Therefore, the y here can be replaced by bx/h. So, that is the mass of the sliver here, and its center is obviously here, so that is a mass there, there’s a mass there; I’ve got to do the weighted average of all of them. So remember, I don’t just integrate this over x; that would just give me the mass of the body. I should multiply it by a further x and then do the integral. So, what I really want to do to find the center of mass x, is to take that mass I got, M/A, there is a 2, there’s an h, there is a b, there is an x from there, and another x, because you have to multiply this by the x coordinate of this thing, because that’s the coordinate of the center of mass of this. There are two xs, that’s what you’ve got to remember. So, that should be integrated from 0 to h, that’ll give me h3/3. So, I get (2Mb/Ah) times h3 over 3.

Now, you know there is one more thing we have to do. We must replace the area of the triangle by ½ base times altitude, which is bh. I also forgot to divide everything by the mass, because the center of mass is this weighted average divided by the total mass, so I’ve got to divide by 1/M. Well, I claim, if you do this and cancel everything you will get the answer of 2/3h. Okay, so not surprisingly the center of mass of this is not halfway to the other end, but two-thirds of the way because it’s top heavy; this side of it is heavier.

This is the level of calculus you should be able to do in this course, be able to take some body, slice it up in some fashion, and find the location of the center of mass. You combine symmetry arguments with actual calculation. For this sliver, by symmetry, you know the center of mass is at the center, you don’t waste your time, but then when you add these guys there is no further symmetry you can use; you have to do the actual work.

So, what have I done so far? What I’ve done is point out to you that when you work with extended bodies, or more than one body, we can now treat the entire body, replace the body by a single point for certain purposes. The single point is called a center of mass, where it imagines all the mass concentrated at the center of mass. So, you have created a brand new entity which is fictitious. It has a mass equal to total mass. It has a location equal to the center of mass, and it moves in response to the total force. And it’s not aware of internal forces, and that’s what we want to exploit. I already gave you a clue as to what the implications are, but let me now take a thorough analysis of this basic equation MR..= F. We’re just going to analyze the consequence. So there are several cases you can consider.

Case one: F external not equal to 0. So, these are two bodies subject to mutual force and to an outside force, but the simplest example I’ve already given to you – but I’ll repeat it – we’re not going to do this in great detail. We all know, if I fire a point mass like this, it will do that. What I’m now telling you is, if I take a complex body made of 20,000 parts, all connected to each other pushing and pulling, if you fling that crazy thing in the air it’ll do all kinds of gyrations and jiggling as it moves around. But if you found its center of mass, the center of mass will follow simply a parabola, because the external force on it is just Mg. So, it’ll be MR..= mg and that’s just motion with constant acceleration in the y direction, and it’s just a projectile problem. And I repeat once more, for emphasis, that if this object broke into two objects, typically what’ll happen is one will fly there and one will land here, but at every instant if you found their center of mass you will find it proceeds as if nothing happened. For example, if you have an explosive device that blows them apart, and the pieces are all flying, but that’s just coming from one part of the piece pushing on another part of the piece, but those forces are of no interest to us. As far as the external force goes, it is still gravity so the center of mass will continue traveling. Beyond that, I’m not going to do too much with this thing. So let me now go to Case II.

Chapter 3. Law of Conservation of Momentum — Examples and Applications [00:32:05]

Case II. If you want, case 2a. F external is equal to 0. What does it mean if F external is 0? That means this is 0. That means MR.. = 0, that means MR. is a constant because it’s not changing. Who is this MR.? What does it stand for? Well, it looks like the following. If you take a single particle of mass m, and velocity x., we use the symbol p, maybe I’ve never used it before in the course, and that’s called the momentum. The momentum of a body is this peculiar combination of mass and velocity. In fact, in terms of momentum we may write Newton’s law. Instead of saying it’s mdv/dt, which is ma, you can also write it as d by dt of mv, because m is a constant and you can take it inside the derivative, and that we can write as dp/dt. Sometimes, instead of saying force is mass times acceleration, people often say “force is the rate of change of momentum.” The rate at which the momentum of a body is changing is the applied force. So, if I’ve not introduced to you the notion of momentum, well, here it is. So, if you think about it that way, this looks like the momentum of the center of mass, and we are told the momentum of the center of mass does not change if there are no external forces. But the momentum of the center of mass has a very simple interpretation in terms of the parts that make up the center of mass; let’s see what it is.

Let’s go back here. Remember, let me just take two bodies and you will get the idea; it’s m1 + m2, which is total M, and let’s take just one dimension when it’s m1x1.+ m2x2. over m1 + m2; that is what Mx. is. So, m1 + m2 cancels here, and you find this is just p1 + p2. Let’s use the symbol capital P for momentum of the center of mass. So, the momentum of the center of mass is the sum of the momentum of the two parts, but what you’re learning is – so let me write it one more time – if F eternal is equal to 0, then p1 + p2 does not change with time. This is a very, very basic and fundamental property, and it’s in fact another result that survives all the revolutions of relativity and quantum mechanics, where what I’ve said for two bodies is true for ten bodies; you just do the summation over more terms.

So, let me say in words what I’m saying. Take a collection of bodies. At a given instant everything is moving; it’s got its own velocity and its momentum, add up all the momenta. If you had one dimension, just add the numbers. If in two dimensions, add the vectors; you get a total momentum. That total momentum does not change if there are no outside forces acting on it. So, a classic example is two people are standing on ice. Their total momentum is 0 to begin with, and the ice is incapable of any force along the plane. It’s going to support you vertically against gravity, but if it’s frictionless it cannot do anything in the plane. Then, the claim is that if you and I are standing and we push against each other and we fly apart, my momentum has to be exactly the opposite of your momentum, because initially yours plus mine was 0; that cannot change because there are no external forces. If two particles are pushing against each other, they can only do so without changing the total. Okay, so p1 + p2 does not change, and here’s another context in which it’s important.

Suppose there is a mass m1, going with some velocity v1, and here’s the second mass m2, going with some velocity v2; they collide. When they collide, all kinds of things can happen. I mean, m1 may bump its head on that and come backwards, or it could be a heavy object that pushes everything in the forward direction, or at the end of the day you will have some m1 going with a new velocity v1, and m2 going with a new velocity v2. But what I’m telling you is that m1 v1 + m2 v2 will be equal to m1 v1’ + m2 v2. In a collision, of course, one block exerts a force on the other block, and the other block exerts an opposite force on this block, and that’s the reason why even though individually the momenta could be very different, finally the momenta will add up to the same total. Here’s a simple example; you can show that if this mass and that mass are equal, and say this one is at rest, that one comes and hits this. You can show under certain conditions this one will come to rest and this will start moving with the speed of the–the target will move at the speed of the projectile. So, momenta of individual objects have changed. One was moving before, it is not moving; one was at rest, it’s moving, but when you add up the total, it doesn’t change.

This is called the Law of Conservation of Momentum. So, that’s so important I’m just going to write it down here one more time. And the basic result is, if external forces are 0, then p1 + p2 + p3 and so on, “before” will be p1’ + p2’ + p3, and so on, where this means “before” and that means “after,” . When is “before” and when is “after.” Pick any two times in the life of these particles, it’s like Law of Conservation of Energy, where we said E1 = E2, there 1 and 2 stood for “before” and “after.” Well, here we cannot use 1 and 2 for “before” and “after” because 1 and 2 and 3 are labeling particles. So, the “before” quantities are written without a′, and the “after” quantities are written with a prime.

Everybody follow this? It’s very important you follow this statement and follow the conditions under which it’s valid. There cannot be external forces. For example, in the collision of these two masses, if there is friction between the blocks and the table, you can imagine they collide and they both come to rest after a while. Originally, they had momentum and finally they don’t. What happened? Well, here you have an explanation, namely, the force of friction was an external force acting on them. What I’m saying is that if the only force on each block is the one due to each other, then the total momentum will not change. So, the case that I considered, 2a, was external force equal to 0, but center of mass was moving, because it had a momentum. Then, the claim is that momentum will not change.

I’m coming to the last case, which is, if you want, case 2b. The external forces are 0, the center of mass was 0; in other words, center of mass was at rest. If you find these different cases complicated, then I don’t mind telling you one more time. The center of mass behaves like a single object responding to the external force. It’s clear that if the external force is non-zero, the center of mass will accelerate. If the external force is 0, the center of mass will not accelerate. There are cases 1 and 2. 2a and 2b are the following: if it does not accelerate, its velocity will not change. So then, you have the two cases, it had a velocity, which it maintained, or it had no velocity, in which case it does not even move. See, if you apply F = ma to a body and there are no forces, you cannot say the body will be at rest. You will say the body will maintain the velocity. So, if it had a velocity, it’ll go at the velocity, if it was at rest it’ll remain at rest; the same goes for the center of mass. If the center of mass was moving, it’ll preserve its momentum. That really means the sum of the momenta of the individual pieces will be preserved. If it was at rest, since the external force is 0, it will remain at rest. So, I want to look at the consequence of this one. I could’ve done them in either order. I can take the case where there’s no external force, there is no motion, or I chose to do with the opposite way, where I took the most complicated case, where external force is not 0. Then, I took the case where it is 0 but the center of mass was moving to begin with, and therefore it has to keep moving no matter what. And the simplest case is the center of mass was at rest; then, it’s not moving now and it will never move.

So, let me give an example of where the idea comes into play. We looked at the planetary motion of the Sun and the Earth, and I said the Earth goes around the Sun, so let’s look at it a little later; there’s the Earth and the only force between the Earth and the Sun is the mutual force of gravitation. Now, my question to you is, “Is this picture of the Sun sitting here and the Earth moving around acceptable or not in view of what I’ve said?” Yes?

Student: The Sun and Earth revolve around a mutual center of gravity.

Professor Ramamurti Shankar: Yes, that’s the answer to the problem. But what is wrong if I just say the Sun remains here and the Earth goes in a circle, which is what we accepted last time?

Student: The momentum of the Sun doesn’t change, but it changes in the momentum of the Earth.

Professor Ramamurti Shankar: That’s one way. Do you understand what he said? He said the momentum of the Sun is not changing. The momentum of the Earth is changing, so the total momentum is changing. The total momentum cannot change, so that’s not acceptable. But in terms of center of mass, you can say something else, yes?

Student: The center of mass moves.

Professor Ramamurti Shankar: It moves, maybe you can tell me, you cannot point out from there, but tell me which way you think it’s moving.

Student: In a circle.

Professor Ramamurti Shankar: Right, in the beginning it’s somewhere here. A little later it’s there, a little later it’s there, so the center of mass would do this, if the picture I gave you last time was actually correct. So, a Sun of finite mass staying at rest and a planet orbiting around it is simply not acceptable, because the center of mass is moving without external forces; that’s not allowed. Or as he said, the momentum is constantly changing, this guy has no momentum, that guy has the momentum that points this way now and points that way later. But look what I’ve said here: take 1 and 2 to be the Earth and the Sun, it doesn’t add up to the same number. So, we sort of know what the answer should be. We know that the thing that cannot move is not the Sun and it’s not the Earth. It’s the center of mass; that’s what cannot move. If originally it was at rest, it’ll remain at rest. So, the center of mass, if it cannot move, so let’s start off the Sun here, start off the Earth here, join them, the center of mass is somewhere here. Actually, the Sun is so much more massive than the Earth, the center of mass lies inside the Sun. But I’m taking another Solar System where the Sun is a lot bigger than the Earth but not as big as in our world, so I can show the center of mass here. That cannot move. So, what that means is, a little later, if the planet is here and I want to keep the center there, the Sun has to be here, and somewhat later when the planet’s there the Sun has to be here. So, what will happen is, the Sun will go around on a circle of smaller radius, the planet will go around on a circle of a bigger radius, always around the center of mass. So, you’ve got this picture now, it’s like a dumbbell, asymmetric, big guy here, small guy here, fix that and turn it.

You get a trajectory for the Sun, and you get a trajectory for the Earth on a bigger circle; the center remains fixed. So, if you apply loss of gravity, I’ve given you a homework problem, you’ve got to be careful about one thing. When you apply the Law of Gravity, you may apply it to the Earth for example. Then, you will say the centripetal acceleration mv2 /r is the force of gravity. When you do that calculation be careful; v is the velocity of the planet, and when you do the mv2/r, the r you put will be the distance to the center of mass from where you are. That’ll be the mv2/r. But when you equate that to the force of gravity, the Gm1m2 over r2, for that r it is the actual distance from the Earth to the Sun that you should keep, because the force of gravity is a function of the distance between the planets, not between the planet and the center of mass. The actual force on the Earth is really coming not from here but on the other side of this where the Sun is. But luckily, at every instant, the Sun is constantly pulling it towards the center of the circle. It’s a very clever solution. The planet moves around a circle. It has an acceleration towards the center, and somebody’s providing that force. ut that somebody is not at the center but always on the other side of the line joining you to the center, so you still experience a force towards the center. Under the action of that force, you can show it’ll have a circular orbit and you can take some time in calculating now the relation between time period and radius and whatnot. So, this is called a two-body problem. So, this is one example where you realize, “Hey, center of mass, if I follow it, it cannot be moving and therefore the actual motion of planets is more complicated than we thought.”

Okay, then, there is a whole slew of problems one can do, where the center of mass is not moving. So, I’ll just give you a couple of examples; then, I will stop, but I won’t do the numbers. Here is one example. That is a carriage that contains a horse, and the horse is on this far end. And as they tell us to do, we won’t draw the horse, we’ll just say it’s a point mass m, and the railway carriage is a big mass, capital M, and let’s say the left hand of the railway carriage is here. Now, you cannot see the horse, okay? The horse is inside; the horse decides to–;now, he said, “I’m tired of sitting on this side of this room, I’m going to the other side.” The horse goes to the other side. First of all, you will know something’s going on without looking in, because when the horse moves to the left the carriage has to move to the right. First, convince yourself the carriage has to move somewhere because originally the center of mass between these two objects – that one and that one – is something in between, somewhere here. If the horse came to that side, the center of mass is now the average of those two, which is somewhere over there; the center of mass has moved and that’s not allowed, the center of mass cannot move. So, if the center of mass is originally on that line, it has to remain in that line. So, what will happen in the end is that the horse will come here, the center of the carriage will be there, but the center of mass will come out the same way.

So, a typical problem, you guys will be expected to solve, will look like this. Given all these masses and given the length of the carriage, find out how much the carriage moves. Do you think you can do that? Give some numbers and plug in the things. For example, this guy is at a distance L/2 of this; the horse is at a distance L. Make this your origin of coordinates. Take the weighted average of those two, and get the x coordinate of the center of mass. You don’t have to worry about the y because there’s nothing happening in the y. So, the x coordinate of that and that’ll be somewhere in here. At the end of the day, let us say it has moved an unknown distance d, which is what you’re trying to calculate. Then, compute the center of mass. When you do that, remember that the center of the carriage is L/2 + d from this origin. The horse is at the distance d from the origin. Equate the center as a mass and you will get an equation that the only unknown will be d, and you solve for a d and it’ll tell you how much it moves. Anybody have a question about how you attack this problem? Find the center of mass before, find the center of mass after, equate them and that linear equation will have one unknown, which is the d by which the carriage has moved and you can solve for it.

Okay, here’s another problem. Here is a shore, and here is a boat; maybe I’ll show the boat like a boat, here it is, okay that’s the boat. Now, you are here. So, the boat has a certain mass, which we can pretend is concentrated there. You have a little mass m, and the boat is at a distance, say, d, from the shore, and you are at a certain distance x from the edge of the boat, and you want to get out, okay. You want to go to shore, so what do you do? So, if you’re Superman or Superwoman you just take off and you land where you want. But suppose you have limited jumping capabilities. It is very natural that you want to come as far to the left as possible and then jump. Suppose it is true that d, which is, say, three meters, is the maximum you can possibly jump, whereas you cannot jump d + x. So, you say, “Let me go to the end and I’m safe because I can jump the distance d.” And again, we know that’s not going to work because when you move–Look it’s very simple. If you move and nobody else moves we’ve got a problem, because if you found the center of mass with one location for you, and you change your location and nothing else changes, center of mass will change and that’s not allowed. Here, I’m assuming there are no horizontal forces. In real life, the water will exert a horizontal force, but that’s ignored in this calculation. There are no horizontal forces. If you move, everything else has to move.

So, what’ll happen is that, when you move, the boat will have moved from there maybe somewhere over to the right like this. You are certainly at the edge of the boat, but the boat has moved a little extra distance Δ, and you have to find that Δ. You find it by the same trick. You find the center of mass of you and the boat, preferably with this as the origin. You can use any origin you want for center of mass, it’s not going and it’s not going according to anybody, but it’s convenient to pick the shore as your origin, find the weighted sum of your location and your mass, the boat’s location and boat’s mass. At the end of the day, put yourself on the left hand of the boat, and let’s say it has moved a distance Δ, so the real distance now is d + Δ. That’s where you are. That plus L/2 is where the center of the boat is. Now, find the new center of mass and equate them and you will find how much the boat would have moved, and that means you have to jump a distance d + Δ. Everybody follow that? That’s another example where the center of mass doesn’t move. Now, let’s ask what happens next. So, you leap in the air, okay? Now you’re airborne. What do you think is happening when you’re airborne? Yes? Go ahead! What’s happening?

Student: The boat will move the other way.

Professor Ramamurti Shankar: It’ll be moving, and why do you say it’ll be moving?

Student: Why?

Professor Ramamurti Shankar: Yeah.

Student: Because the center of mass still won’t be the same thing.

Professor Ramamurti Shankar: Right, there’s one way to say that. The center of mass cannot move, so if you move to the left, the boat will move to the right. What’s the equivalent way to say that? Yes?

Student: The momentum can’t change.

Professor Ramamurti Shankar: Right, the momentum cannot change. Originally, the momentum was 0, nobody was moving, but suddenly you’re moving, the boat has to move the other way. Of course, it doesn’t move with the same velocity, or the same speed; it moves with the same momentum. So, the big M of the boat times the small v of the boat, will equal your small m times your big V. In other words, you, unless you move with a big speed, the boat moves with a small speed; then, these two numbers in magnitude will be equal. So, if you’re going on one of the big cruise ships, you jump on the shore, you’re not going to notice the movement of the ship, but it technically speaking does move the other way. Okay, you’re airborne, okay. Then, a few seconds later you collapse on the shore; you’re right there. Now, what’s happening to the boat? Is it going to stop now? Your momentum is 0, yes?

Student: But you’ve been stopped by the force of the ground.

Professor Ramamurti Shankar: Yes. Everybody agree? I will repeat that answer, but you should all have figured this out. The boat will not stop just because you hit the shore. The boat will keep moving because there’s no force on the boat; it’s going to keep moving. The question is, “How come I suddenly have momentum in my system when I had no momentum before?” It’s because the F external has now come into play. Previously, it was just you and the boat and you couldn’t change your total momentum. But the ground is now pushing you, and it’s obviously pushing you to the right because you were flying to the left and you were stopped. So, your combined system, you and the boat, have a rightward force acting for the time it took to stop you; it’s that momentum that’s carried by the boat. A better way to say this is, you and the boat exchange momenta, you push the boat to the right, you move to the left, and your momentum is killed by the shore. The boat, no reason to change, and keeps going. So, can you calculate how fast the boat is moving? Can anybody tell me how to calculate how fast the boat is moving? Yes?

Student: [inaudible]

Professor Ramamurti Shankar: But I’m on the shore now. I’ve fallen on the shore. I’m asking how fast the boat is moving.

Student: The boat is moving as fast as [inaudible]

Professor Ramamurti Shankar: Right, I think he’s got the right answer. If I only told you that I jumped and landed on the shore, that’s not enough to predict how fast the boat is moving. But if I told you my velocity when I was airborne, then of course I know my momentum and you can find the boat momentum and that’s the velocity it will retain forever. So, you need more information than simply saying, “I jumped to the shore.” It depends with what velocity I left the boat and landed on the ground. If I leap really hard, the boat will go really fast the opposite way. Okay, that’s the end of this family of center of mass problems.

Chapter 4. The Rocket Equation [00:56:32]

So, I’m going to another class of problems. This involves a rocket, and it’s going to derive the rocket equation. A rocket is something everybody understands but it’s a little more complicated than you think. Everyone knows you blow up a balloon, you let it go, the balloon goes one way, the air goes the other way, action and reaction are equal, even lay people know that. Or, if you stand on a frozen lake and you take a gun and you fire something, but then the bullet goes one way and you go the opposite way, again, because of conservation of momentum. The rocket is a little more subtle and I just want to mention a few aspects of it. I don’t want to go into the rocket problem in any detail. It’s good for you to know how these things are done.

Here’s a rocket whose mass at this instant is M, whose velocity is v right now. What rockets do is they emit gasses, and the gasses have a certain exhaust velocity. That velocity is called v0. In magnitude, it’s pointing away from the rocket, and it has a fixed value relative to the rocket, not relative to the ground. If you are riding with the rocket and you look at the fumes coming out of the back, they will be leaving you at that speed v0. A short time later–What happens a short time later, the rocket has a mass M - Δ because it’s lost some of its own body mass in the form of exhaust fumes. The exhaust fumes, I’m just showing them as a blob here, and the rocket’s velocity now is not v, but v + Δv. And what’s the velocity of the fumes? Here’s where you’ve got to be careful. If your velocity was v at that instant, the velocity of the rocket fume, with respect to the ground is v - v0; that’s the part you’ve got to understand. The rocket has a smaller mass and bigger velocity; everyone understands that. But what’s the momentum of the gas leaving the rocket if the mass is Δ? But what is its velocity? Its velocity with respect to the rocket is pointing to the left of v0, but the rocket itself is going to the right at speed v. So, the speed as seen from the ground will be v - v0.

So, the Law of Conservation of Momentum will say Mv = (M - Δ) (v + Δv) + Δ(v - v0). This is, the momentum before and the momentum after are equal. So, you open out this bracket you get – sorry my letters better be uniform – this is Mv, then -v Δ. I don’t want to call it Δv, I want to call it v Δ + M Δv - ( Δ) ( Δv ) + Δv - v0 Δ. I want to call it v Δ. The reason I want to put the Δ on the right is you may get confused. Δ usually stands for the change of something, so that’s not what I mean. So, you cancel this Mv and you cancel this Mv. You cancel this v Δ and that v Δ. This one you ignore because it’s the product of two infinitesimals, one is the amount of gas in the small time Δt, the other is infinitesimal change in velocity; we keep things which are linear in this. Then, you get the result M Δv = v0 times Δ. So, I’m going to write it as Δv/v0 = Δ over M. This is the relation between the change in the velocity of the rocket; the velocity of the exhaust gases seen by the rocket, the amount emitted in the small time divided by the mass at that instant. But in the sense of calculus, what is the change [dM] in the mass of the rocket? If M is the mass of the rocket, what would you call this a change, the mass of the rocket, in this short time? Yep?

Student: [inaudible]

Professor Ramamurti Shankar: No, no, no, in terms of the symbols here what’s the change in the mass of the rocket?

Student: Δ

Professor Ramamurti Shankar: It’s Δ, but it’s really speaking - Δ. You could keep track of the sign, the change in the variable is really negative, and delta here stands for a positive number. So, if you remember that, you’ll write this -dM/M. Now, the rest is simple mathematics. I don’t want to do this, but if you integrate this side and you integrate that side, and you know dM/M is a logarithm; you will find the result that the v at any time is v final = v initial plus–or maybe I might as well do this integral here. This integral will be v final - v initial over v0, will be the log of M initial over M final. So, you will find final is v initial + v0 log M initial over M final. I’m doing it rather fast because I’m not that interested in following this equation any further. It’s not a key equation like what I’ve been talking about now. So, this is just to show you how we can apply the Law of Conservation of Momentum. I’m not going to hold you responsible in any great detail for the derivation, but that is a formula that tells you the velocity of the rocket at any instant, if you knew the mass at that instant. The rocket will pick up speed and its mass will keep going down, and the log of the mass before to the mass after times v0 is the change in the velocity of the rocket. Okay, so I have to give you some more ammunition to do your homework problems; so, I’m going to discuss the last and final topic, which is the subject of collisions.

Chapter 5. Elastic and Inelastic Collisions [01:03:19]

So, we’re going to take the collision of two bodies, one body, another body, m1v1, m2v2, they collide. At the end of the day, you can have the same two bodies moving at some velocities v1, v2. Our goal is to find the final velocities; that’s a goal of physics. I tell you what’s happening now. I’m asking you what’s happening later. So here, there are two conditions you need because you’re trying to find two unknowns, right? We want two unknowns, I need two equations.

One equation always true, so let me write that down, always true. Always true is the condition that the momentum before is the momentum after, m1v1’ + m2v2. You need a second equation to solve for the two unknowns, and that’s where there are two extreme cases for which I can give you the second equation, the one extreme case is called “Totally inelastic.”

In a totally inelastic collision, the two masses stick together. That means v1 and v2 are not two unknowns, but a single unknown v’. Then, it’s very easy to solve for the momentum, because they stick together and move as a unit. So, you can write here that is equal to (m1 + m2) v’, so you get v’ = (m1v1 + m2v2)/(m1 + m2). That’s a simple case: two things hit, stick together, and move at a common speed. The common speed should be such that the total momentum agrees with what you had before. That’s called “total Inelastic.”

The other category is called “totally elastic.” In a totally elastic collision, the kinetic energy is conserved. That you can write as the following relation involving quadratic things, ½ m1v12 + ½ m2v22 = ½ m1v1‘2 + ½ m2v2‘2. You can, it turns out, juggle this equation and that equation and solve for v1 and v2. Well, I’ll tell you what the answer is. I don’t expect you to keep solving it. The answer is that v1 = (m1 - m2)/(m1 + m2)v1 + (2 m2/m1 + m2)v2. These are no great secrets; you’ll find them in any textbook. If you cannot follow my handwriting or you’re running out of time, just what you should be understanding now is that there are formulae for the final velocity when the collision is totally elastic, or totally inelastic. If they’re totally inelastic, it’s what I wrote there, v is something. The totally elastic you have a formula like this one. So, here you just replace everywhere; you saw an m1 you put an m2, 2m1 over m1 + m2v1. Don’t waste too much time writing this. I think you can go home and fill in the blanks; it’s in all the books. What you carry in your head is there’s enough data to solve this, because I will tell you the two bodies, I’ll tell you their masses, I’ll tell you the initial velocities; so plug in the numbers you get the final velocity. So, remember this, elastic, inelastic collision, this is in one dimension.

Now, I’ll give you a typical problem where you have to be very careful in using the Law of Conservation of Energy. You cannot use the Law of Conservation of Energy in an inelastic collision. In fact, I ask you to check if two bodies–Take two bodies identical with opposite velocities; the total momentum is 0. They slam, they sit together as a lump. They’ve got no kinetic energy in the end. In the beginning, they both had kinetic energy. So, kinetic energy is not conserved in a totally inelastic collision, in an elastic collision it is.

So, here is an example that tells you how to do this carefully. So, this is called a Ballistic Pendulum. So, if you have a pistol – you manufactured a pistol – the bullet’s coming out of the pistol at a certain speed, and you want to tell the customer what the speed is. How do you find it? Well, nowadays we can measure these things phenomenally well with all kinds of fancy techniques, down to 10-10 seconds. In the old days, this is the trick people had. You go and hang a chunk of wood from the ceiling. Then, you fire the bullet with some speed v0 and you know its mass exactly. The bullet comes, rams into this chunk. I cannot draw one more picture, so you guys imagine now. The bullet is embedded in this, and I think you also know intuitively the minute it’s embedded, the whole thing sets in motion. Now, you could put this on a table. Forget all the rope. If you can find the speed of the entire combination, then by using Conservation of Momentum, you can find out the speed of the bullet. But that’s hard to measure; people have a much cleverer idea. You should ram into this thing. This is like a pendulum. So, the pendulum rises up now to a certain maximum height that you can easily measure. And from that maximum height you can calculate the speed of the bullet.

So, I’m going to conclude by telling you what equations you’re allowed to use in the two stages; so pay attention and then we’re done. In the first collision, when the bullet rammed into this block, you cannot use Law of Conservation of Energy. In other words, you might be naïve and say, “Look, I don’t care about what happened in between; finally, I’ve got a certain energy, M + m times g times h, that’s my potential energy, not kinetic.” Initially, I had ½ m02. I equate these two guys and I found v0; that would be wrong. That’s wrong because you cannot use the Law of Conservation of Energy in this process when I tell you that it’s a totally inelastic collision in the middle. Because, what’ll happen is, some energy will go into heating up the block; it might even catch fire if the bullet’s going too fast.

But you can use the Law of Conservation of Momentum all the time in the first collision to deduce that M + m times some intermediate velocity is the incoming momentum. You understand that? From that, you can find the velocity v with which this composite thing, block and bullet, will start moving. Once they start moving, it’s like a pendulum with the initial momentum, or energy. It can climb up to the top and convert the potential to kinetic, or kinetic to potential. There is no loss of energy in that process. Therefore, if you extract this velocity and took ½ (M + m) times this velocity squared, you may in fact equate that to (M + m)gh.

So, let me summarize this last result. In every collision, no matter what, momentum is conserved; the energy may or may not be. And if I give you a problem like this where in between there’s some funny business going on which is not energy conserving, don’t use energy conservation from start to finish. Use momentum conservation, find the speed of the composite object. This is what you’ve got to understand in your head. It’s not this equation. When can I use Conservation of Mechanical Energy? When can I not? A bullet driving into a chunk of wood, you better know you cannot use Conservation of Kinetic Energy. But once the combination is going up, trading kinetic for potential, you can. All right, so let me stop now and remind you guys, no meeting Wednesday; we meet three times next week and on Wednesday you come for your homework exchange.

[end of transcript]

Back to Top
mp3 mov [100MB] mov [500MB]