8. Dynamics of Multiple-Body System and Law of
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01:11:48
Computer animation
Transcript: English(auto-generated)
00:00
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. 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
00:20
bodies, the Sun and the planet. But actually, the Sun was not doing anything interesting in our analysis. The Sun just stood there as the source of the gravitational force. It's a planet that did all the orbiting, and that was a problem in two dimensions, but of only one body. So now, we are going to enlarge our domain to more than
00:41
one body obeying Newton's laws. So, let me start with the simplest possible case of two bodies, and I will again start with the simplest case of two bodies. They're moving in one dimension. Then, we'll put in more.
01:03
So, here is the one-dimensional world in these bodies that are going to move. And this is my origin, x equal to 0, and I'm going to imagine one point mass m1 located at x1,
01:20
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, m d2x1 over
01:44
dt squared. 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 in the notation. The two dots telling you it's
02:01
two derivatives. If it's one dot, it's one derivative, three dots, it's 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 the subsequent manipulation.
02:22
So, this is ma, and that's ma, and that's equal to force on body one. 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
02:41
be the force on body one due to body two, which I'm going to denote by F12. That's our notation. You and I agree that's the force on one due to two. Then, there's the force on one due to the external world. E stands for external, that means everything outside
03:00
these two. So, the universe has many bodies. I've just picked these two guys. They're one and two, and the force on one is, some of it's due to two, and some of it's due to everything else. Similarly, I have another equation,
03:20
m2x2 double dot is the force on two due to one plus the force on two 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 is pulling all of them towards the planet with some gravitational force. So, everything else is called external.
03:41
Now, one and two, 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.
04:00
For example, if the spring is compressed at this instant, it's trying to push them out, that means really one is trying to push two outwards that way, two is trying to push one 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,
04:22
at the surface of the earth, I take these two masses connected by a spring. Here is mass one and here is mass two. I squash the spring. If there is no gravity, they will just go vibrating up and down, but let them fall under the field of gravity. So, they're also experiencing the mg due to gravity.
04:41
So, they will both fall down and also oscillate relative to each other. They're all described by this equation. This will be the spring force transmitted from one to two. This will be the force of gravity. Or it could be electric force or any other force due to anything else. I'm not interested. Now, here is the interesting manipulation we are going to perform.
05:01
We're going to add the two things on the left-hand side and equate them to whatever I get on the right-hand side. Then, I get m1x1 double dot plus m2x2 double dot, and that's going to be, I'm going to write it in a particular way, F1 external plus F2 external
05:24
plus F12 plus F21. I think you have some idea where I'm headed now. So, what's the next thing we could say? Yes?
05:41
Yes, because that's the third law of Newton. Whatever the underlying force, gravity, spring, anything, force on one due to two and force on two due to one 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
06:03
going to write as F external, 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.
06:24
As you know, that's called the center of mass coordinate, and it's defined as m1x1 plus m2x2 divided by capital M. Capital M is just the total mass, m1 plus m2. If I do that, this is a definition,
06:46
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
07:03
fill in the blanks in your head? This is really correct. On the left-hand side, I have capital M times x double dot. So, I have really m1 plus m2 times x double dot. If you take the double dot of this guy, it's m1x1 double
07:23
dot plus m2x2 double dot divided by m1 plus 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
07:41
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.
08:00
The center of mass is a location, capital X, some kind of a weighted average of x1 and x2. By weighted, I mean if m1 and m2 were equal, then capital M will be two times that mass, and you'll just get x1 plus x2 over 2. So, the center of mass will sit right in between, but if m1 is heavier, it'll be tilted towards m1. If m2 is heavier,
08:21
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 this 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
08:40
anything. And it behaves as a body. 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, I would say, well, this guy is talking about a body of mass, capital M, undergoing some acceleration due to the force. So, the center of mass is a body whose mass seems to be the
09:04
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 of the external
09:21
force? Now, if it turns out, 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 all cancel. And what will remain will be a similar thing where this is the total external force. So, if I can say this in
09:41
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 and doesn't care about internal forces. So, I'll give you an example. So, here is some airplane, right? It's in flight and a couple of guys are having a fight,
10:02
punching each other and so on. The rest of the passenger says, enough is enough, and they throw them out. So, they're just floating around, affecting each other's forces. 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
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the force mgh, or it's going to accelerate with g. So, at one point, this person may be having the upper hand and may be here, the other person may be down here, but following the center of mass, you'll find it simply falls under gravity. So, the mutual forces do not affect the dynamics of the plane, but they affect the center of mass.
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Or, for example, suppose at some point this person blows the other one up into, say it's a samurai bat, makes it simple, cuts him up into two pieces. So, now we've got three bodies now. First, the protagonist, and the other two now, unfortunately, somewhat decimated. Now, you can take these three bodies, find their center of
11:02
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,
11:22
the center of mass will just be in one location, not doing anything. Yes? 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
11:44
will behave like a particle with no force. If it is not moving to begin with, it won't move later. Or if it is moving to begin with, it will 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,
12:03
you introduce a y-coordinate, and then you will have the center of mass as mr double dot equal to F. And r would be m1r1 plus m2r2 divided by the total mass.
12:23
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 will now be a vector.
12:42
So, here's another example. So, 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 plus
13:01
m2r2 plus 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,
13:21
it'll go as if nothing happened and it will land here. The center of mass does not care about the internal forces, but only about external forces. That's the main point. And it was designed in such a way that external forces canceled in its dynamics. So, everything I'm going to do today is to take that equation, mx double dot equal
13:42
to F or mr double dot equal to F in vector form and deduce all the consequence. 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 sum over mi.
14:01
Okay, this is a shorthand. I'm going to write it only once, but you should know what the notation means. If it's i from 1 to 3, it really means m1x1 plus m2x2 plus m3x3 divided by m1 plus m2 plus m3. This summation is the notation mathematicians have introduced
14:23
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,
14:40
1, 2, and 3, to find the center of mass, you can either go to this formula, do all the m1x1s 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,
15:02
which is just m1 plus m2. Trade these two for a new fictitious object and put that here and forget these two. But on that one point, you deposit 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 plus this
15:25
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
15:42
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 will give you this result. Okay, so before I exploit that equation and find all the
16:03
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
16:22
with 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 its center of mass? So, we have to adapt the definition that we have for this problem.
16:43
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 continual. The trick then is to say, I take a distance x from the left hand and I cut myself a
17:02
very thin sliver of thickness dx. That sliver has got a certain mass, and I argue it's at a definite distance x from the coordinate's origin. If you're nitpicking, you will say, what do you mean by definite distance?
17:21
It's got a width dx, so one part of it is at x, the other part is at x plus dx, so it doesn't have a definite coordinate. But if dx is going to zero, this argument will eventually be invalid. So, if dx goes to zero, that 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
17:42
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? But I do the following. I take the total mass and divide by l. That's the mass per unit length. And this fellow has a width dx, so the mass of this little
18:05
sliver is m over 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
18:22
slivers, which is what we do by the integral from zero 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 over l, then I get integral x dx from zero to l, and that's going to be
18:42
l squared over 2. So, if you do that, you'll get l over 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
19:04
uniform. It was 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, mo is a fixed number, m over l, and then this is the answer. But there are other ways to get this result without doing all
19:21
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.
19:41
And here is the 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, that it's 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
20:01
masses, and the object's really not even regular. This is my origin of coordinates. If I replace every x by minus x, okay? Sorry, I should change this object so this is really, looks like this. Here's the object. Suppose I replace every x by minus x.
20:22
That's reflecting the body around this axis. So, the answer is, it will look like this. It will be jutting to the right instead of to 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 minus x. Because in this averaging,
20:43
mi xi summed over all the masses, if every x goes to minus x, the center of mass will go to minus x. But now take this rod and take the same 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,
21:01
the center of mass must look the same. That means minus x has to be equal to x itself, and the only answer is x equal to zero. So, without doing any detailed calculation, you argue that the answer is x equal to zero. To do this, of course, you must cleverly pick your coordinates so that the symmetry of the body is
21:20
evident. If you took the body like this, and you took a reflection around this point, it goes into a 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
21:41
itself, therefore the answer is zero. 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 symmetric rod, but a rod of non-zero 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
22:04
mass, because I can take every point here, turn it into the point there by changing y to minus y. Therefore, capital Y will become minus capital Y. But the body looks the same after this mapping, so capital Y is zero. Similarly, capital X is zero and
22:23
that's the center of mass. Okay, now what if I give you this object? Anybody want to try that?
22:42
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. Okay.
23:14
Is that what your answer was? Okay, so the correct answer is replace this mass by all of its mass, whatever it is sitting
23:22
here, replace this one by all of its mass sitting here, then forget the big bodies, replace them by points, then you've got two masses located here and you can find the weighted average which may be somewhere there. Okay, now let's take one more object and then I'm pretty
23:44
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.
24:01
Let the mass be m, where it's 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 minus y, it maps onto itself so it looks the same,
24:22
but it's supposed to reverse the capital Y, therefore y is minus y and therefore it's zero, 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 in the y,
24:41
under flipping y, but I cannot say take x to minus x. In fact, if I take x to minus x, this one looks like this. It doesn't map into itself, but I can pick any point here. So, if I take x to minus x, the object looks different. It looks like that object, and relating one object to another object is what I'm trying to do.
25:01
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
25:22
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.
25:41
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. But then, I need to know the area of this strip.
26:01
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 zero, it's going to look like a rectangle.
26:20
So, the area is 2yΔx. 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 tells me that y over B is x over H. Namely, that triangle compared to that triangle tells me
26:44
that y divided by B is x divided by H. Therefore, the y here can be replaced by Bx over H. So, that is the mass of the sliver here,
27:04
and its center is obviously here. So, there 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'll 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
27:20
to find the center of mass, x, is to take that mass. I got m over A. There is a 2. There's an H. There is a B. There's an x from there and another x because you have to multiply this by the x coordinate of this thing,
27:41
and I'm going to multiply it by 2mvA because that's the coordinate of the center of mass of this. There are two x's. That's what you've got to remember. So, that should be integrated from 0 to H. That'll give me H³ over 3. So, I get 2mvAH
28:01
H³ over 3. Now, you know there's one more thing we have to do. We must replace the area of the triangle by ½ base times altitude, which is BH. Well, I also forgot to divide everything by the mass because the center of mass is
28:22
this weighted average divided by the total mass. So, I got to divide by 1 over m. Well, I claim if you do this and cancel everything, you will get the answer of 2 thirds H.
28:40
Okay, so not surprisingly, the center of mass of this is not half way to the other end, but two thirds of the weight 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. You'll be able to take some body, slice it up in some fashion, and find the location
29:01
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.
29:22
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 the center of mass when it imagines all the mass
29:40
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,
30:03
but let me now take a thorough analysis of this basic equation, MR double dot equals F. We're just going to analyze the consequence. So, there are several cases you can consider.
30:22
Case one is F external not equal to zero. So, these are two bodies subject to mutual force and to an outside force. So, the simplest example I
30:42
have already given to you, but I will 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 that if you 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,
31:00
it'll do all kinds of durations and jiggling as it moves around. But if you found its center of mass, its center of mass will follow simply a parabola, because the external force on it is just mg. So, it'll be MR double dot equals mg, and that's just motion with constant acceleration in the y direction, and it's just a projectile problem.
31:21
And I repeat once more for emphasis that if this object broke into two objects, typically what will 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
31:40
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. So, 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
32:03
two. Case two, if you want case 2a, is F external equal to zero? What does it mean if F external is zero? That means this is zero. That means MR double dot is
32:26
a constant, because it's not changing. Who is this MR dot? What does it stand for?
32:42
Well, it looks like the following. If you take a single particle of mass M and velocity x dot, we use the symbol P. Maybe I've never used it before in the course, and that's called the momentum. The momentum of the body is
33:02
this peculiar combination of mass and velocity. So, the momentum of the body is the same as mass and velocity. In fact, in terms of momentum, we may write Newton's law. Instead of saying it's m dv 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
33:21
derivative. And that we can write as dp dt. Sometimes, instead of saying force is mass times acceleration, people often say that the rate of change of force is the rate of change of momentum. The rate of which the momentum of a body is changing is the applied force.
33:40
So, if I've not introduced you to 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 is a very simple interpretation in terms of the parts that make up the
34:01
momentum. So, 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 m one plus m two, which is total m. And let's take just one dimension when it's mx one dot plus m two x two dot divided by m one plus m
34:22
two. That is what mx dot is. So, m one plus m two cancels here, and you find this is just P one plus P two. Let's use the symbol capital P for momentum of the center of mass.
34:41
So, the momentum of the center of mass is the sum of the momenta of the two parts. But what you're learning is, so let me write it one more time. If F external is equal to zero, then P one plus P two does not change with time. This is a very,
35:04
very basic and fundamental property, and it's in fact one other result that survives all the revolutions of relativity and quantum mechanics. Well, 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
35:21
I'm saying. Take a collection of bodies. At a given instant, everything is moving. So, you've got a total momentum, it's got its own velocity and its own momentum. Add up all the momenta. If you're in one dimension, just add the numbers. If you're in two dimensions, add the vectors. You get a total momentum. That total momentum does not change if there is no outside force acting on it.
35:43
So, a classic example is two people are standing on ice. Their total momentum is zero to begin with, and the ice is incapable of any force along the plane. It can support you vertically against gravity, but if it's frictionless, it cannot do anything in the plane. Then, the claim is that if you
36:01
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 zero, and that cannot change because there are no external forces. So, if two particles push against each other, they can only do so without changing the total. Okay, so P1 plus P2 does
36:25
not change. And here's another context in which it's important. Suppose there is a mass, m1, going with a velocity v1. And here's a second mass, m2, going with some velocity v2. They collide. When they collide,
36:41
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. 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'. What I'm telling you is that m1v1 plus m2v2 will be equal
37:06
to m1v1' plus m2v2'. 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. That's the reason why, even though individually the
37:22
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
37:41
and this will start moving with the speed of the target, will move with the speed of the projectile. So, momenta of individual objects have changed. One was moving before it's not moving, one was at rest, it's moving, but when you add up the total, it doesn't change. And this is called the law of conservation of momentum. So, that's so important.
38:02
I'm just going to write it down here one more time. And the basic result is, if external forces are zero, then p1 plus p2 plus p3 and so on before will be
38:23
p1 prime plus p2 prime plus p3 prime and so on, where this means before and that means after. And what's it? When is before and when is after? Pick any two times in the life
38:41
of these particles. It's like law of conservation of energy, where we said E1 equals E2. There, one and two stood for before and after. Well, here we cannot use one and two for before and after, because one and two and three are labeling particles. So, the before quantities are written without a prime, the after quantities are written with a prime.
39:02
Everybody follow this? It's very important you follow the statement and follow the conditions under which it's written. So, 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.
39:21
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 the only force in the block is the one due to each other, then the total momentum will not change. So, the case that I considered
39:40
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, external forces are 0,
40:02
center of mass was 0. In other words, center of mass was at rest. Do you find these different cases complicated? Then I don't mind telling you one more time. Center of mass behaves like a
40:22
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
40:42
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 equal to ma to a body and there is 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 will go at the velocity. If it was at rest, it will remain at rest.
41:02
The same goes for the center of mass. If the center of mass was moving, it will preserve its momentum. That really means the sum of the momentum 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 have done them in
41:24
either order. I can take the case where there's no external force, there is no motion, or I can just, I chose to do it the opposite way, but 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, has to keep moving no matter
41:40
what. In the simplest cases, 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 that 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.
42:01
So, let's look at it a little later. That'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
42:22
in view of what I've said? Yes? Yeah, but 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 night? That's one way.
42:42
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? It moves.
43:00
Maybe you can tell me, you cannot point out from there, but tell me which way you think it's moving, right? In the beginning, it's somewhere here, a little later it's there, a little later it's there. So, 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
43:23
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 a momentum that points this way now and points that way later. But look what I've said here. Take one and two to be the
43:43
Earth and the Sun, it doesn't add up to the same momentum. 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.
44:02
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,
44:22
but not as big as another 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.
44:41
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 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
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Sun and you get a trajectory for the Earth on a bigger circle, the center remains fixed. So, if you apply laws 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,
45:20
for example, then you will say the centripetal acceleration, mv² over 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 over r, the r you put will be the distance to the center of mass from where you are. That will be the mv²
45:40
over r. But when you equate that to the force of gravity, either gm1m2 over r squared, 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
46:00
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 is providing that force,
46:20
but 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 will have a circular orbit, and you can take some time calculating now the relation between time period and radius and whatnot. So, this is called a two-body problem.
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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
47:00
where the center of mass is not moving. So, I'll just give you a couple of examples, and then I'll stop. But I won't do the numbers. Here's one example. There is a carriage that contains a horse, and the horse is on this far end. And as they tell us to do,
47:21
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 end of the railway carriage is here. Now, you cannot see the horse, okay? The horse is inside. The horse decides to now,
47:41
instead 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
48:02
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,
48:21
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
48:40
given the length of the carriage, find out how much the carriage moves. So, you think you can do that? Give some numbers and plug in the things. For example, this guy is at a distance L over 2 from this. The horse is at a distance L. Make this your origin of coordinates.
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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 will be somewhere 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. Recompute the center of
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mass. When you do that, remember that the center of the carriage is L over 2 plus d from this origin. The horse is at the distance d from the origin. Equate the centers of mass and you will get an equation where the only unknown will be d. And you solve for d and it will tell you how much it moves.
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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
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boat. Maybe I'll show the boat like a boat. Here it is. Okay, that's the boat. Now, you are here.
50:24
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
50:41
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, very natural, that you want to come as far to the left as possible and then jump.
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Suppose it is true that d, which is say 3 meters, is the maximum you can possibly jump, whereas you cannot jump d plus x. So, you say, let me go to the end, then I'm safe because I can jump the distance d. And again, we know that's not going to be, it's not going to work. Because when you move,
51:22
look, it's very simple. If you move and nobody else moves, we've got a problem. Because if you found a center of mass with one location for you and you change your location and nothing else changes, the 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
51:40
horizontal force, but that's ignored in this calculation. There are no horizontal forces. If you move, everything else has to move. So, what will happen is that when you move, the boat would 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,
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delta, and you have to find that delta. You find it by the same trick. 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.
52:21
But it's convenient to pick the shore as your origin, find the weighted sum of your location and your mass, boat's location, and boat's mass. At the end of the day, put yourself on the left end of the boat, and let's say it has moved a distance, delta, so the real distance now is d plus delta. That's where you are. That plus L over 2 is where the center of the boat is.
52:43
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 plus delta. Everybody follow that? That's another example of the center of mass doesn't move. Now, let's ask what happens next.
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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? It'll be moving, and why do you say it'll be moving? Yeah, right.
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There's one way to say that. Center of mass cannot move, so if you move to the left, the boat will move to the right. That's the equivalent way to say that. Yes? Right. Momentum cannot change. Originally, momentum was 0, nobody was moving. If suddenly you're moving, the boat has to move the other way.
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Of course, it doesn't move with the same velocity or the same speed. It moves at 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, let's say, move with a big speed, the boat moves with a small speed, then these two numbers in magnitude will be equal.
54:02
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 technically speaking, it 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?
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No, your momentum is zero. Yes, 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
54:41
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. It's obviously pushing you to the right because you were flying to the left and you were stopped.
55:00
So, your combined system, you and the boat, have a rightward force acting for a 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 momentum. 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, it keeps going.
55:20
You can calculate. Can you calculate how fast the boat is moving? Can anybody tell me how to calculate how fast the boat is moving? Yes? But I am on the shore now,
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fallen on the shore. I'm asking you how fast the boat is moving. 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.
56:01
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,
56:21
the boat will go really fast the opposite way. Okay, that's the end of this family of center of mass problems. So, I'm going to another class of problems. This involves a rocket.
56:41
I'm just 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,
57:01
the rocket goes one way, the air goes the other way. 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,
57:23
right now. What rockets do is they emit negative gases and the gases have a certain exhaust velocity. That velocity is called v zero. In magnitude it's pointing away from the rocket and it has a fixed value relative to
57:40
the rocket, not relative to the ground. If you're riding with the rocket and you look at the fumes coming out of the back, they will be leaving you at that speed v naught. A short time later, what happens? A short time later, the rocket has a mass m minus delta because it's lost some of
58:02
its own body mass in the form of exhaust fumes. Exhaust fumes, I'm just showing them as a blob here, and the rocket's velocity now is not v, but v plus delta v. 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,
58:21
the velocity of the rocket fume with respect to the ground is v minus v zero. 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? It's the mass, delta, but what is its
58:42
velocity? Its velocity with respect to the rocket is pointing to the left at v zero, but the rocket itself is going to the right at speed v, so the speed as seen from the ground will be v minus v zero. So, the law of conservation of momentum will say
59:00
m times v is equal to m minus delta times v plus delta v plus delta times v minus v zero. This is momentum before and momentum after equal.
59:20
So, you open up this bracket, you get, sorry, my letters better be uniform. This is mv, then minus v delta. I don't want to call it delta v. I want to call it v delta, plus m delta v minus delta times delta v, plus delta v minus delta
59:45
times v zero. I want to call it v zero delta. The reason I want to put the delta on the right is you may get confused. Delta usually stands for the change of something, so I don't want, that's not what I mean. So, you cancel this mv and you cancel this mv. You cancel this v.
01:00:00
and that's vδ. This one you ignore because it's a product of two infinitesimals. One is the amount of gas in a small time, Δt, the other is the infinitesimal change in velocity. We keep things which are linear in this. Then you get the result m times Δv is equal to v0 times Δ.
01:00:26
So, I'm going to write it as Δv over v0 is equal to Δ over m.
01:00:42
This is the relation between the change in the velocity of the rocket, the velocity of the exhaust gas as seen by the rocket, the amount emitted in the small time divided by the mass at that instant. But what, in the sense of calculus, what is the change in the mass of the rocket? If m is the mass of the rocket, what would you call as the
01:01:03
change in the mass of the rocket in this short time? Yep? No, no, no. In terms of the symbols here, what's the change in the mass of the rocket? It's Δ, but it's really speaking minus Δ.
01:01:21
If you keep track of the sign, the change in the variable is really negative. And Δ here stands for a positive number. So, if you remember that, you'll write it as minus dm over m. Now, the rest is simple mathematics. I don't want to do this, but if you integrate this side
01:01:40
and you integrate that side, and you know dm over m is a logarithm, you'll find the result m, you'll find the result that the v at any time is v final, is equal to v initial plus, or maybe I might as well do this integral here.
01:02:00
So, this integral will be v final minus v initial over v zero will be the log of m initial over m final. So, you will find v final is v initial plus v zero log m initial over m final.
01:02:24
I'm doing it rather fast because I'm not that interested in following this equation any further. 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
01:02:40
conservation of momentum. I'm not going to hold you responsible in any great detail for the derivation, but there 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
01:03:01
before to the mass after times v naught is the change in the velocity. So, you've got more ammunition to do your homework problems. I'm going to discuss the last and final topic, which is the subject of collisions. So, we're going to take the
01:03:20
collision of two bodies, one body, another body, m one, v one, m two, v two. They collide. At the end of the day, you're going to have the same two bodies moving at some velocities, v one prime, v two prime. So, our goal is to find the final velocities.
01:03:43
That's the goal of physics. I tell you what's happening now, I'm asking you what's happening later. So, here is one. There are two conditions you need because you're trying to find two unknowns, right? If you want two unknowns, I need two equations. One equation, always true. So, let me write that down. Always true. Always true is the condition
01:04:06
that the momentum before is the momentum after, and the momentum after is m one, v one prime, plus m two, v two prime. You need a second equation to solve for the two unknowns.
01:04:21
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 v one prime and
01:04:42
v two prime are not two unknowns, but a single unknown, v prime. 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 m one plus m two times v prime.
01:05:03
So, you get v prime is m one v one plus m two v two over m one plus m two. That's a simple case. You can write it like this. Two things hit, stick together, and move at a common speed.
01:05:21
The common speed should be so that the total momentum agrees with what you have before. That's called totally inelastic. The other category is called totally elastic. In a totally elastic collision, kinetic energy is conserved. That you can write as the
01:05:44
following relation involving quadratic things. m one v one squared plus one half m two v two prime squared is equal to one half m one v one prime squared plus one half m two v two prime squared.
01:06:03
You can, it turns out, juggle this equation and that equation and solve for v one prime and v two prime. Well, I'll tell you what the answer is. I don't expect you to keep solving it. The answer is that v one prime is equal to m one minus
01:06:23
m two over m one plus m two times v one plus two m two over m one plus m two v two. These are no great secrets. You'll find them in any textbook. If you cannot follow my handwriting, you're running out of time. Just what you should be
01:06:41
understanding now is that there are formulas 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 one prime is something. If they're totally elastic, you have a formula like this one. So here, you just replace everywhere you saw an m one, you put an m two.
01:07:08
So, m one over m one plus m two v one. 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
01:07:22
you the two bodies. I'll tell you the masses, I'll tell you the numbers, I'll tell you the initial velocities, plug in the numbers, you get the final velocity. 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
01:07:41
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 opposite velocities. Total momentum is zero. They slam, they sit together as a lump. They've got no kinetic energy in the end.
01:08:00
In the beginning, they both had kinetic energy. So, kinetic energy is not conserved in a totally inelastic collision. Inelastic collision, it is. So, here is an example that tells you how to do this carefully. So, this is called the ballistic pendulum. So, if you have a pistol, you manufacture the pistol, the bullets coming out of the pistol at a certain speed,
01:08:22
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 to the minus 10 seconds. In the old days, this is the trick people had. You go and hang a chunk of wood from the ceiling, and then you fire the bullet
01:08:42
with some speed v zero, and you know its mass exactly. The bullet comes, and then it ramps, ramps into this chunk. So, I cannot draw one more picture, so you guys imagine now bullets embedded in this. I think you also know intuitively the minute it's embedded, the whole thing sets
01:09:01
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 condition. You can measure this thing. This is like a pendulum, so the pendulum rises up now to
01:09:22
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 are allowed to use in the two stages. So, pay attention and then we're done. In the first collision,
01:09:40
when the bullet rammed into this block, you cannot use law of conservation of energy. In other words, you might be naive and say, look, I don't care about what happened in between. Finally, I got a certain energy, m plus m times g times h. That's my potential energy, no kinetic. Initially, I had minus one-half m zero squared. I equate these two guys,
01:10:02
and I found v zero. 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's a totally inelastic collision in the middle, and you can't use the law of conservation of energy in the middle because what will 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
01:10:21
conservation of momentum all the time in the first collision to deduce that m plus 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 it starts moving,
01:10:42
it's like a pendulum with 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 one-half m plus m times this velocity squared,
01:11:02
you may in fact equate that to m plus mgh. So, let me summarize this last result. In every collision, no matter what, momentum is conserved. Energy may or may not be. And if I give you a problem like this, where in between
01:11:21
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 just equation. When can I use conservation of kinetic energy? When can I not? Bullet ramming into a chunk of wood, you better know you
01:11:41
cannot use conservation of kinetic energy. But once the combination is going up, trading kinetic for potential, you can.