# 8. Dynamics of Multiple-Body System and Law of

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.
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
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
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
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
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_12X12 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,
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
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,
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.
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,
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?
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,
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?
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
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
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_2v2′. 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
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)v2. These are no great secrets;
you’ll find them in any textbook.
If you cannot follow my handwriting or you’re running
out of time, just what you should be understanding now is
that there are formulae for the final velocity when the
collision is totally elastic, or totally inelastic.
If they’re totally inelastic, it’s what I wrote there,
v^(‘) is something. The totally elastic you have a
formula like this one. So, here you just replace
everywhere; you saw an 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.”
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