Uploaded by YaleCourses on 22.09.2008

Transcript:

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 m_1, located at

x_1, another point mass

m_2, located at

x_2.

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_1

d^(2)x_1/ dt^(2). 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, m_1x

_1, 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, F_1.

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

F_12; 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, m_2x

_2 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 F_12 and

F_21. 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 F_12 and

F_21. 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, m_1x _1 double dot plus

m_2x _2 double dot,

and that's going to be--I'm going to write it in a

particular way, F_1e +

F_2e + F_12 + F_21.

I think you have some idea of where I'm headed now.

So, what's the next thing we could say?

Yes? Student:

F_12 + F_21 = 0.

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 F_e,

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

m_1x_1 + m_2x_2

divided by capital M. Capital M is just the

total mass, m_1 + m_2.

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

m_1 + m_2 times X

double dot. If you take the double dot of

this guy, it's m_1x _1 double dot plus

m_2x _2 double dot,

divided by m_1 + m_2.

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 x_1 and x_2.

By weighted, I mean if m_1

and m_2 are equal,

then capital M will be two times that mass and you'll

just get x_1 + x_2 over 2.

The center of mass will sit right in between.

But if m_1 is heavier, it'll be tilted towards

m_1; if m_2 is

heavier, it'll be tilted towards m_2.

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 F_12 and

F_23 and F_32 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

m_1r_1 + m_2r_2

divided by the total mass; r_1 and

r_2 are just the location now in two

dimensions of these two masses. So, here is

m_1 and here is m_2 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 m_1r_1 +

m_2r_2 + m_3r_3

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 m_ix _i divided by the

sum of m_i. 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 m_1x_1 +

m_2x_2 + m_3x_3

divided by m_1 + m_2 +

m_3. 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 m_1x _1'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 x_1 and

x_2 and with that total mass

M_12, which is just m_1

+ m_2, 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, m_3x_3 +

M_12X< sub>12 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

L^(2) 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

m_ix _i,

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 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 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 h^(3)/3.

So, I get (2Mb/Ah) times h^(3) 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. 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 m_1 +

m_2, which is total M,

and let's take just one dimension when it's

m_1x_1.+ m_2x_2.

over m_1 + m_2;

that is what Mx. is. So, m_1 +

m_2 cancels here, and you find this is just

p_1 + p_2.

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 p_1 + p_2 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 p_1 + p_2 does not

change, and here's another context in which it's important.

Suppose there is a mass m_1,

going with some velocity v_1,

and here's the second mass m_2,

going with some velocity v_2;

they collide. When they collide,

all kinds of things can happen. I mean, m_1

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 m_1 going with

a new velocity v_1',

and m_2 going with a new velocity

v_2'. But what I'm telling you is

that m_1 v_1 + m_2

v_2 will be equal to m_1

v_1' + m_2 v_2‘.

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 p_1 + p_2 +

p_3 and so on, "before" will be

p_1' + p_2' +

p_3', 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 E_1 = E_2,

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 prime,

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 mv^(2) /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 mv^(2)/r,

the r you put will be the distance to the center of

mass from where you are. That'll be the mv^(2)/r.

But when you equate that to the force of gravity,

the Gm_1m _2 over

r^(2), 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.

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

v_0. 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 v_0. 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 - v_0; 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 v_0,

but the rocket itself is going to the right at speed v.

So, the speed as seen from the ground will be v -

v_0.

So, the Law of Conservation of Momentum will say Mv = (M -

δ) (v + Δv) + δ(v - v_0).

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 -vδ. I don't want to call it

δv, I want to call it vδ + MΔv - (δ) (Δv ) + δv -

v_0δ. I want to call it vδ.

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 vδ and that vδ.

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

MΔv = v_0 times δ.

So, I'm going to write it as Δv/v_0 = 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 v_0,

will be the log of M initial over M final.

So, you will find final is v initial +

v_0 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

v_0 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.

So, we're going to take the collision of two bodies,

one body, another body, m_1v

_1, m_2v

_2, they collide.

At the end of the day, you can have the same two

bodies moving at some velocities v_1',

v_2'. 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, m_1v_1'

+ m_2v~~2'.~~

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 v_1' and

v_2' 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 (m_1 +

m_2) v', so you get v ‘=

(m_1v_1 + m_2v_2

)/(m_1 + m_2).

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

m_1v_1^(2) + ½

m_2v_2^(2) = ½

m_1v_1^('2) + ½ m_2v_2

^('2). You can, it turns out,

juggle this equation and that equation and solve for

v_1^(') and v_2^(').

Well, I'll tell you what the answer is.

I don't expect you to keep solving it.

The answer is that v_1^(') =

(m_1 - m_2)/(m_1 +

m_2)v_1 + (2 m_2/m_1 +

m_2)v< sub>2.

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 m_1 you put an m_2,

2m_1 over m_1 +

m_2v_1. 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 v_0 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 naive 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

m_0^(2). I equate these two guys and I

found v_0; 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.

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 m_1, located at

x_1, another point mass

m_2, located at

x_2.

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_1

d^(2)x_1/ dt^(2). 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, m_1x

_1, 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, F_1.

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

F_12; 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, m_2x

_2 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 F_12 and

F_21. 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 F_12 and

F_21. 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, m_1x _1 double dot plus

m_2x _2 double dot,

and that's going to be--I'm going to write it in a

particular way, F_1e +

F_2e + F_12 + F_21.

I think you have some idea of where I'm headed now.

So, what's the next thing we could say?

Yes? Student:

F_12 + F_21 = 0.

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 F_e,

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

m_1x_1 + m_2x_2

divided by capital M. Capital M is just the

total mass, m_1 + m_2.

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

m_1 + m_2 times X

double dot. If you take the double dot of

this guy, it's m_1x _1 double dot plus

m_2x _2 double dot,

divided by m_1 + m_2.

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 x_1 and x_2.

By weighted, I mean if m_1

and m_2 are equal,

then capital M will be two times that mass and you'll

just get x_1 + x_2 over 2.

The center of mass will sit right in between.

But if m_1 is heavier, it'll be tilted towards

m_1; if m_2 is

heavier, it'll be tilted towards m_2.

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 F_12 and

F_23 and F_32 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

m_1r_1 + m_2r_2

divided by the total mass; r_1 and

r_2 are just the location now in two

dimensions of these two masses. So, here is

m_1 and here is m_2 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 m_1r_1 +

m_2r_2 + m_3r_3

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 m_ix _i divided by the

sum of m_i. 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 m_1x_1 +

m_2x_2 + m_3x_3

divided by m_1 + m_2 +

m_3. 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 m_1x _1'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 x_1 and

x_2 and with that total mass

M_12, which is just m_1

+ m_2, 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, m_3x_3 +

M_12X< sub>12 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

L^(2) 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

m_ix _i,

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 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 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 h^(3)/3.

So, I get (2Mb/Ah) times h^(3) 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. 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 m_1 +

m_2, which is total M,

and let's take just one dimension when it's

m_1x_1.+ m_2x_2.

over m_1 + m_2;

that is what Mx. is. So, m_1 +

m_2 cancels here, and you find this is just

p_1 + p_2.

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 p_1 + p_2 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 p_1 + p_2 does not

change, and here's another context in which it's important.

Suppose there is a mass m_1,

going with some velocity v_1,

and here's the second mass m_2,

going with some velocity v_2;

they collide. When they collide,

all kinds of things can happen. I mean, m_1

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 m_1 going with

a new velocity v_1',

and m_2 going with a new velocity

v_2'. But what I'm telling you is

that m_1 v_1 + m_2

v_2 will be equal to m_1

v_1' + m_2 v_2‘.

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 p_1 + p_2 +

p_3 and so on, "before" will be

p_1' + p_2' +

p_3', 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 E_1 = E_2,

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 prime,

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 mv^(2) /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 mv^(2)/r,

the r you put will be the distance to the center of

mass from where you are. That'll be the mv^(2)/r.

But when you equate that to the force of gravity,

the Gm_1m _2 over

r^(2), 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.

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

v_0. 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 v_0. 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 - v_0; 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 v_0,

but the rocket itself is going to the right at speed v.

So, the speed as seen from the ground will be v -

v_0.

So, the Law of Conservation of Momentum will say Mv = (M -

δ) (v + Δv) + δ(v - v_0).

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 -vδ. I don't want to call it

δv, I want to call it vδ + MΔv - (δ) (Δv ) + δv -

v_0δ. I want to call it vδ.

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 vδ and that vδ.

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

MΔv = v_0 times δ.

So, I'm going to write it as Δv/v_0 = 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 v_0,

will be the log of M initial over M final.

So, you will find final is v initial +

v_0 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

v_0 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.

So, we're going to take the collision of two bodies,

one body, another body, m_1v

_1, m_2v

_2, they collide.

At the end of the day, you can have the same two

bodies moving at some velocities v_1',

v_2'. 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, m_1v_1'

+ m_2v

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 v_1' and

v_2' 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 (m_1 +

m_2) v', so you get v ‘=

(m_1v_1 + m_2v_2

)/(m_1 + m_2).

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

m_1v_1^(2) + ½

m_2v_2^(2) = ½

m_1v_1^('2) + ½ m_2v_2

^('2). You can, it turns out,

juggle this equation and that equation and solve for

v_1^(') and v_2^(').

Well, I'll tell you what the answer is.

I don't expect you to keep solving it.

The answer is that v_1^(') =

(m_1 - m_2)/(m_1 +

m_2)v_1 + (2 m_2/m_1 +

m_2)v< sub>2.

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 m_1 you put an m_2,

2m_1 over m_1 +

m_2v_1. 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 v_0 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 naive 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

m_0^(2). I equate these two guys and I

found v_0; 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.