We're going to talk
about a whole new concept, which is the concept
of momentum. We've all heard the expression, "We have a lot
of momentum going." Well, in physics,
momentum is a vector, and it is a product
between the mass of a particle and its velocity. And so the units would be kilograms times meters
per second. F = ma. That equals m dv/dt. That equals dmv/dt and that equals,
therefore, dp/dt. So what you see--
the force is dp/dt, and what that means is if a
particle changes its momentum, a force must have acted upon it. It also means if a force
acts on a particle, it will change its momentum. Let us now envision that we have
a whole set of particles which are interacting
with each other, and the interaction could be
gravitational interaction, could be electrical interaction, but they're interacting
with each other. Zillions of them--
a whole star cluster. I pick one here,
which I call mi, and I pick another here,
which I call mj. And there is an external force
on them, because they happen
to be exposed to forces from the outside world, and so on this one
is a force Fi external, and on this one is Fj external. But they're interacting
with each other, either attracting or repelling, and so in addition
to these external force, there is a force that j feels
due to the presence of i. And let's suppose
they are attracting each other. This would be F that j
experience in the presence of i. Actions equals minus reactions,
according to Newton's Third Law, so this force, Fij, will be
exactly the same as Fji, except in
the opposite direction. We call these internal forces. That's the interaction
between the particles. If these were
the only two forces, then the net force on this
object, on this particle i, would be... would be a force
in this direction. This would be F net. Now, you can do the same here. This would be F net
on particle j, and this would be F net
on particle i. But since there are zillions
of objects here, there are many of these
interacting forces, and so I can't tell you
what the net force will be. The net force is ultimately
the sum of the external force and all the internal forces. What is the total momentum
of these zillions of particles? Well, that's the sum
of the individual momenta. So that is p1 plus p2... pi...
and you have to add them all up. I take the derivative of this--
dp total/dt. That is p1/dt. Well, that's the force
on number one. It's the total force
on number one. So it is F1, but it is
the net force on F1. And F... and dp/dt for this
particle equals F2 net force, and the force on particle number
i equals Fi net, and so on. And so when we add up
all these forces, obviously that is the total
force on the entire system. Now comes the miracle. The miracle is that
all these internal forces eat each other up. This ji cancels this one if you
look at the system as a whole-- not if you look
at the individual particles, but on the system as a whole,
they all cancel each other out. And so the total force
on the system is simply
the total force external. And all the internal forces--
you can forget about it. And this means, then, that
we come to a key conclusion-- that dp total/dt... in fact, I have it
written down there. It's so important
that I want you to look at it this whole lecture. Look at this. You see that dp/dt
is the total force externally-- forget all the internal forces-- and what it means...
that if the external force, the total external force
on the whole system is zero, it means that the momentum
cannot change. Momentum is conserved, and that's called
the conservation of momentum. It only holds if the sum
of all external forces are zero. We could have hundreds of stars,
like a globular cluster, and they could collide
with each other, they could explode,
they could break apart-- all those forces are internal,
they don't count. The momentum of the cluster
as a whole would not change if there are no external forces
on the system, if the net sum, the total
external force, is zero. So the individual stars will change their momentum
all the time, because the individual particles
in the individual stars will experience, of course,
the internal forces. I'm not saying that
the individual particles do not continuously experience
momentum changes. It's just the system as a whole for which the momentum
is then conserved. We can do a very simple example so that you get
a feeling... numerical idea. I would have a...
an object here as mass, m1, and here I have one, m2, and let this one have
a velocity, v1, and this one has a velocity, v2. One-dimensional problem,
just as a warm-up. They plow into each other
and they stick together. I put some glue on one,
and so they stick together. That's a given. You have to accept that. Let this be the direction
of increasing x, and so before the collision I have a certain amount
of momentum. I can do this on a horizontal
table, which is frictionless. I can do it on an ice surface,
which is nearly frictionless and I ignore air drag. So this is the situation before, and so the momentum that
I have before equals m1 v1-- this is my plus direction-- I could put vectors over there
but that's really not necessary, because it's
a one-dimensional problem-- plus m2v2. If you like that, be my guest. That's the momentum before. Now they stick together and so when they stick together,
their total mass is m1 + m2. And then their velocity
happens to be v prime. We often give a prime
to the situation after the collision occurs. And so now I can apply,
for the first time, the conservation of momentum. The momentum before must be the
same as the momentum afterwards because there are no
external forces on the system. When they collide,
you better believe it that there are internal forces. You better believe that there
is glue and it goes "plunk" and they feel those forces,
both of them. The momentum of each one
individually is changing, but not the momentum
of the system as a whole, and so this equals
(m1 + m2) v prime. And so, if you put in
some numbers-- suppose m1 equals one kilogram, and v1 is five meters
per second, let m2 be two kilograms and let
v2 be three meters per second. And you see, since they are both
in the same direction momentum here is 5 + 6 is 11,
so this is 11. And this is 1 + 2 is 3,
so that is 3 times v1 prime, and so v1 prime is eleven-thirds
meters per second. So the conservation of momentum tells you what the speed is
after the collision. When we deal with collisions, the velocities of the objects
before the collision are unprimed. You see here v1,
and you see here v2. Now, the velocities
after the collisions, by convention, are primed. You see here v prime. Now if after the collision object one and object two
have a different velocity, we will call it v1 prime
and v2 prime. Now in this case, we're dealing
with an inelastic collision, so m1 and m2 are stuck together. They are merged, so it is sufficient
to just call it v prime. You could have called it
v1 prime, because v1 prime is the same as
v2 prime is the same as v prime. And during
the next four minutes, you will see that a few times
do I call it v1 prime. I wish I really hadn't. It would be better
if I had just used v prime. That is unique and sufficient. Now, there was a certain amount
of kinetic energy before the collision, and of course, we can
all calculate that-- kinetic energy
before the collision. That is one-half m1 v1 squared
plus one-half m2 v2 squared-- one-half m1 v1 squared,
one-half m2 v2 squared. How much is it? You can do that
as well as I can. If you add that up,
you will find 21.5 joules. Trivial to stick in the numbers. Now I want to hear from you. What do you think the situation
is with the kinetic energy after the collision? Don't look at the numbers. Just use your intuition. It may be wrong. That's okay, my intuition
is often wrong. So there was a certain amount
of kinetic energy. We had this collision,
and I want you to tell me whether you think
that the kinetic energy has perhaps increased
or decreased or maybe the kinetic energy
hasn't changed. Who thinks the kinetic energy
has not changed? Good for you. I see some parents even. Who thinks the kinetic energy
has decreased? Even better for you. Hey, there's a professor
of physics there. Can't go wrong there. And who thinks the kinetic
energy has increased? No one thinks that. Hey, it's amazing. Let's take a look. Kinetic energy
after the collision. That would be one-half m1
plus m2 times v prime squared. Would we agree? Let me write this
at a different location. So this was the 21½ joules. Well, you can do this
as well as I can. We know that v1 prime is--
that is eleven-thirds. And you will find that this
number equals 20.2 joules. So the kinetic energy went down. Now, you may say,
"Well, big deal. "Little bit of kinetic energy--
1.3 joules. Who cares about 1.3 joules?" Well, it is a real big deal
in physics, let me tell you. And I can appre...
I can make you appreciate that it is a real big deal
by doing the following. I don't change the masses, but I just am going to change
the direction of the impact. Here is m1 and here is m2. The speeds remain the same--
v1 and this is v2. No change in the numbers except that they go now
(whooshes ) head-on. What is the momentum
of particle number one? Well, that is mv. 1 x 5 is plus 5. Remember, this is the increasing
direction of X. So the momentum is plus 5. What is the other one? That is 2 x 3,
that is 6, backwards. This is minus 6. So what is the total...
the total momentum... let's give it a magnitude
equals minus 1? The magnitude is one, of course and the momentum,
if I leave this off we would call that minus one because it's
a one-dimensional problem. The momentum before
is in this direction. It's minus one. So what now is v prime? They stick together. Here they are. And afterwards,
I'm not even sure whether they go this way
or this way. Yes, I am sure, because the
momentum is in this direction, so I predict that v1 prime
will be in this direction. And so this now equals m1
plus m2 times the new v1 prime. Well, m1 plus m2 is 1 + 2,
is 3 kilograms and so you see that v1 prime equals minus one-third meters
per second. So the whole system now goes off with one-third meters per second
in this direction. The kinetic energy before
has not changed-- those 21½ joules, right? That is independent of how
I collide them with each other. What now is the kinetic energy
afterwards? This is a great tragedy, because now you get one-half
the sum of the masses, which is three times
this small number squared. Goes with V squared, right? And now there
is only 0.17 joules left. Almost all kinetic
energy has been destroyed. So what you see here
in front of your eyes-- that kinetic energy
can be destroyed, but momentum cannot be destroyed in the absence
of external forces. Kinetic energy and momentum
is a totally different thing. The momentum of the individual
particles gets changed, but the net momentum
did not change but the kinetic energy
was destroyed. I can destroy the kinetic energy
completely if I want that. I can arrange a collision so that all kinetic energy
has been removed. Suppose this particle has a mass
five and the velocity is one. This is my shorthand notation. And this particle has a mass one
but it has a velocity five. The momentum of this system
is zero-- plus five in this direction,
minus five in this direction. But you bet your life
there is kinetic energy. After they collision...
they hit each other, we agreed that they would stick,
there was glue on it. Momentum afterwards must
therefore still be zero. Internal forces don't matter
no matter what happens. Kinetic energy is zero. So the whole system collides--
(bing)-- And afterwards you have here
the sum of the total masses and it just stands still, because I told you that they are
going to stick together. I used glue. If you have a car collision,
and two cars hit each other, and I compare the situation
just before the collision with just after the collision-- here are the two cars and one
has a speed in this direction, the other has a speed
in this direction, and they hit each other. And they stick together,
it's a given. You go (whooshes)--
one big clump. So this is before. And I can give them
some velocities. This is the speed of one, v1, and this is the speed
of the other, v2. Afterwards you see something
like this: v prime. Just a wreck! The impact time is so short that the change in momentum
due to friction with the road-- that would be an external force,
friction with the road. But that can be ignored,
is negligibly small. The cars hit. There is a huge internal force
going on. One slams on the other
and the other slams on one. There is even fireworks--
metal scrapes over metal. Friction, but that's
internal friction, not friction with the road. So momentum is approximately
conserved if we can ignore the friction
from the road during the impact because the impact time
is so short. So let here be car with mass m1 and here is the other car
with mass m2. And we'll make it
a two-dimensional problem, because we have seen only
one-dimensional problems now. Let's make it
a two-dimensional problem. So this is the direction
for this car, and let's say
this is the direction in which the other car is going,
velocity v2. And tragedy has it
that this is the place where they're going to collide. In what direction
and what will be the speed after the collision? I only compare just the moment
before they hit with just the moment
after they hit. What comes later
is a different story. When you have formed the wreck, clearly it's going to slide
on the road, and then there is an external
force which is friction, which will slow it down. It's just during the impact
I claim that to a reasonable
approximation, momentum will have
to be conserved. And so, what is the momentum
of this one? Well, this may be
the momentum of this one. It may have a very small mass,
small speed. And what is the momentum
of this one? Well, this may be the momentum
of number one and this could be the momentum
of number two. The net momentum is
the vectorial sum of these two, which is this. That is p total of the system. That is never going to change. That's before and after
the collision exactly the same. And therefore if you knew
this angle theta and you know p1 and you know p2,
then you can calculate in what direction the objects
are going to slide, but of course you can
also calculate then the velocity after the impact,
because this total momentum must be the sum of the total
of the two cars-- the mass of the two cars--
times v prime. And so you can calculate
everything. And that's what
the police is doing when they find wrecks
on the road. They actually use
the... the track, the... the skidding tracks of
the wreck to calculate v prime, and then they can try
to reconstruct the situation as it was before the collision. Now, in all these cases,
the objects stuck together. We've only considered cases
where they stick together, in which case we call those
in physics "completely inelastic
collisions." Now, next lecture I will
also deal with situations whereby during the collision, the particles bounce off
each other. In completely
inelastic collisions, you always lose kinetic energy-- sometimes a little,
as you see there; sometimes a lot,
as you see there; sometimes everything,
as you see there. Always in inelastic collisions
do you lose kinetic energy. Can we have, in a collision,
an increase of kinetic energy? Well, depends on how you define
the word "collision." The answer is yes. And in fact,
I will show you an example and even do a demonstration. In the most simple case
that I can think of, I have here a block
which has a certain mass, m. But there is an explosive inside and all of a sudden,
it goes "Boom!" It explodes. Well, before, the speed is zero
and so the momentum is zero. And there comes the bang--
(whooshes) and one piece flies
in this direction with a certain velocity,
v2 prime. This is m2. And another piece flies off
in this direction with mass m1,
with velocity v1 prime. Clearly, momentum
must be conserved. This explosion
is only internal forces, and so you can write down... if you call this
the increasing direction of x, so this momentum is positive
and this momentum is negative, the total momentum
will never change. It's the same before
as it is afterwards, so this must be m2 times v2
prime minus m1 times v1 prime. What happened
with the kinetic energy? Well, kinetic energy
has clearly increased. There was zero kinetic energy
to start with. This one now has kinetic energy
and this one has kinetic energy. Where did that come from? Well, it was the chemical
reaction of the explosion. Momentum, however,
was conserved. See, momentum really doesn't
care about these explosions. That's all internal forces. So be very careful. Never confuse momentum
with energy. Energy can change
or cannot change, can increase or decrease or
remain the same kinetic energy, but if there are no external
forces on the system, momentum is always conserved. And this is what I can show you. I have a demonstration set up
on this air track, and we do it on an air track so
there is a minimum of friction. And we're going to push
two cars together and we hold them together
with a spring. The spring is like
the explosion, like the dynamite
that can push these apart. So here is one car
on the air track, then there is a spring
and there is another car. Let this mass be m1
and this mass be m2, and this surface
is nearly frictionless. And I hold them together
with a string so that I can pull
the spring in. So this spring is compressed, and the whole system is at rest,
v equals zero. There is potential energy
in that spring. I take a burner (whooshes),
and I burn away this thread. This is the situation before. Momentum before is zero. Total momentum afterwards
would also be zero. It's identical
to what I did there. And so this object will fly
in this direction, and this object will fly
in that direction. And you already can tell what the ratio
of the velocities will be, because you will see that
v2 prime divided by v1 prime-- you see it right there--
equals m1 divided by m2. And so I will make
some predictions about the speed
of these two objects. The largest mass, by the way,
would get the smallest velocity, and the smallest mass would get
the largest velocity, the largest speed. What you see in here
are really speeds. We don't have the information
on the direction anymore. How do we measure the speed? Well, we actually measure
the time for the cars
to move ten centimeters. Each car has
a little metal plate which is ten centimeters long, and it blanks out
a light-emitting diode. And the moment that the light-
emitting diode is occulted [occluded], the timer starts, and the moment that the light-
emitting diode emerges again, the timer stops. So that's the way
we will measure the time for each car to move
by ten centimeters. The first experiment
that I will be doing, m1 is 244 plus or minus one gram and m2 is also 244
plus or minus one gram. In other words,
the uncertainty in the masses is something like 0.4%. That's just part of life. I can't do much better. The time that it takes
for object number one to go ten centimeters
is obviously ten divided by v1, which is the velocity that
we want to compare them with. Now, the ten centimeters
is really only known to an accuracy
of about one millimeters. So that is an uncertainty of 1%
in my timing in my velocity, so this is a 1% uncertainty. So if we measure the times that object number one
and object number two will go through
this timing device-- a measurement
of ten centimeters-- you would expect the times
to be the same to within roughly 1 + .5...
say 1½%. There is, however,
an additional problem that is very difficult
for me to evaluate and that's the following. If here is
the light-emitting diode, and here is that metal plate
that I mentioned to you, which comes over this diode, occults [occludes] the diode,
and says "start the timer," and then the other end,
of course, moves off
and then the timer stops-- that criterion for on and off
for one diode may not be exactly the same
as for the other. Here is one system. One car will fly
in this direction. And here is another system, and the car will fly
in that direction. And that's not so easy
to evaluate unless you do
a lot of experiments. So I would roughly guess-- I'm always on
the conservative side-- that this may add an uncertainty
in the criterion of the diode of another millimeter. And so effectively, I really don't know any better
than two millimeters what the ten centimeters is. So I would say we really
have to allow here for two percent plus .4%, so I would predict that
the times of these two cars will be the same
to within roughly 5%. That would be my prediction. So we're going to write down
here the times, and then we're going
to measure them. So time one and time two, and we will see whether
these two numbers are the same within the uncertainty
of our measurements. Here is the air track. Have to turn on the airflow. Oh, let me tell,
for those of you... For the parents in my audience
who haven't seen this one, this is a marvelous device. Um... it is a track
which is well-constructed, so that these cars
fit beautifully on this V-shaped track, and then
we blow air out of this track, which lifts these cars up
so that they float, and so we can move them
in horizontal direction with extremely little friction. There is no metal
touching metal. The only friction you have
is the air drag. That you can never avoid. When you go through the air
with a car, there is wind-- the same wind that you feel
when you drive your bicycle. But there's very little, and that's why these experiments
are done on these air tracks. These are not cheap, by the way,
these air tracks-- they're very expensive. Well, you... you pay
$25,000 tuition, so you might as well get
something for it, right? So I hope that these timers
are on. They are. And now we have here one car, and notice how frictionless
it goes when I touch it. Very nice, almost frictionless. And here we have another one, and they have equal mass
to within one gram. And there is a spring
between them. Some of you may be able
to see it. And I'm going to attach now that
string to hold them together. And so I put potential energy
in the spring by pushing them together,
and they are lined up now here. They have no momentum. The timer's not on? Has to be reset. Thank you. They're now both zero? Yeah, you're sure? Okay. So now I'm going
to burn off that wire. The momentum of each one
of those cars will change. One will go in this direction,
the other one in this direction. Kinetic energy will increase. That's the potential energy
from the spring. But the momentum
of the system as a whole after I burn the wire
will remain zero. Ready for this? Make me happy. What do you see? You're going to make me unhappy? I better walk around. I can't wait. 220... oh, fantastic,
absolutely fantastic! 223, 219 milliseconds. That is well
within the uncertainties. What was the number, 219? 219 milliseconds
and 223 milliseconds. If I repeat the experiment
and I make the string shorter, I will get different times. I can never predict the times, because if I make the spring
more squeezed, then obviously I will have
more potential energy. And so the speeds of the cars
will be higher, but the times will be the same within the uncertainty
of the measurements. So I can never predict the time. Now we're going
to do an experiment whereby I'm going to make m2
twice the mass. m2 is 488 plus or minus
one gram. Now, that's going
to be interesting, because now you really
begin to test the conservation of momentum. The momentum before is zero. I burn the wire, momentum
afterwards is also zero. But the velocities...
oh, I erased this-- v2 prime divided by v1 prime
equals m1 divided by m2. And so twice the mass
will get half the speed. And so now you're going to see that one car will go twice
as slow as the other. That's the only way that nature
can conserve momentum. Nature has no choice. Nature can deal with kinetic
energy one way or another, but it cannot finagle momentum. So it must give
the more massive car half the speed
than it gives the other car. And so we're going
to redo this experiment now with a car here
which is twice the mass. And this one is one mass,
so this is the 244, and the one on the...
on your right is the 488. So I have to bring them together
again with a string. So I... I do the work now. I always do the work here
in 26 100. I do the work. I squeeze these springs. I get paid for that, by the way. And that's potential energy
that goes into the spring. There it is. Momentum is zero. Kinetic energy is zero. I have to zero the timers. Oh, and I would like
to make a prediction. I would like to see... we'll measure T1
and then we'll measure T2, and then we get a number
and we get a number, and if we multiply this number
by 2.00, then I would like
to get this number within the uncertainties
of the measurements. There is always an uncertainty
in any measurement that you do. All right? Make sure that...
are the timers zero? Okay, there we go. What do you see? Ah! 406, 193-- whoo! On the button! 406 and 193. 0,406-- is that what you see? Yeah? And 0,193. Now, I don't even need
my calculator to multiply this by two. I can even do that by heart. 9-3-- , .396, .406. Close enough. The two are
in excellent agreement within the 2½% uncertainty
in each measurement. So, you see, that's exactly
how you can demonstrate the conservation of momentum. They picked up kinetic energy, but after the wire was burned,
the momentum remained zero. Now I change the topic. It... it appears
that I change the topic, but you will see in a few weeks
that it's really related-- not even a few weeks-- you will see
that it's really related to the conservation of momentum. I'm going to explain to you now what we physicists mean
by center of mass of a system. The center of mass of a system
is defined as follows: I have here some kind
of an object-- not just a point mass,
but it has a finite size. It could be a hammer. It could be something
like this-- a squash racquet. And let the center of mass
be, for instance, here. This is the center of mass. I pick any origin I want to. You're totally free
to choose this origin-- doesn't matter
where you take it. And this, now, is the position
vector of the center of mass. And now I carve out a zillion
little mass elements, m of i, covering the entire object, and then this is
the position vector, r of i. And the center of mass
is defined as follows: the total mass of this object times the position vector
of the center of mass is the sum, summed over i-- over all these
little elements, i-- times the position vectors of
the individual little particles that make up this object. v center of mass-- I find that by taking the derivative
of this equation-- equals one over the mass total. I bring this on this side,
I take the derivative here, and so here I get the sum
of m of i times the dri/dt. And so this is the sum
of i... of mi vi, simply taking
the first derivative, which changes positions
into velocities. Now, this is also 1 over m total
times the total momentum, because I've all these momenta
of these individual particles. They have a...
together, a total momentum. And so now you come to another very important
statement in physics, and that is that the total
momentum of an object-- which could be a hammer, which
could be a squash racquet-- is the total mass of that object times the velocity
of the center of mass. And if I take
the derivative of this, then dp/dt of
the total momentum-- of which we learned that
the total momentum, dp/dt, is the total external force, we already learned
that earlier-- that now is the total mass
of the system times the acceleration
of the center of mass, because I take the derivative
of this equation. That gives me dP/dt here, and the velocity changes
to acceleration. And look at this! This is really an amazing
statement. This says F = ma. But if I have here
this squash racquet and here is the center of mass, then if I throw this object up
in 26 100, as I will do later, then the center of mass behaves as if all the mass
of the entire squash racquet was right at that
center of mass. So the behavior of the center of
mass is extremely predictable, whereas the behavior of
the squash racquet is not. It may start tumbling. If the external force on
the squash racquet were zero, then it would continue to go
always with the same velocity, the center of mass. If I had here a hammer--
this is a hammer-- and there were
no external forces-- I would be somewhere
in outer space-- and it would have
a certain velocity. Then the center of mass,
but only the center of mass, would have a velocity
that never changes, because if there's
no external force then there is only
a constant velocity. There's no acceleration. But the hammer itself
may be tumbling. A little later in time, the
hammer may have this position. A little earlier in time, the hammer may have had
this position. But the center of mass
is just one smooth motion. That is very mysterious that
there is one and only one point in any one of us--
in you, in me, in any object-- that the center of mass behaves as if all the matter
were together in one point. And so this is another
quite important statement: For the center of mass,
the total momentum is the total mass times
the velocity of the center of mass. And you take the derivative
of that equation and you get F = ma. And that means if
the external force is zero-- you can go back
to the upper line again-- if the external total force
is zero, then the momentum of the system
is conserved, and so the center of mass will then keep
its velocity unchanged. Let's do one calculation
to give you some experience on how you derive
the center of mass. I'll take a simple case. I won't take a hammer. It's a little bit complicated. I will take three masses
which are held together by very unphysical...
by massless rods, say. Then I have three point masses and that makes my life
a little easier. Let this be the Y axis
and this the X axis, and here at zero
I have a mass m. At a distance l,
I have here a mass 2m. This is also l
and this is also l, so this is equilateral triangle. These are massless rods,
and here I have a mass m. And I'm asking you,
where is the center of mass? So you have three point masses, and they are connected
with massless rods. Well, for those of you who have
a good feeling for symmetry, they would say certainly it has
to lie somewhere on this line. And it's probably slanted
in the direction of the 2m, so it will probably
be somewhere here. And so this will be the position
vector r center of mass, and the individual
position vectors from the origin here
will be this, and this will be a position
vector to this object, and the position vector
to this object is zero. And so now we can calculate the position of the center
of mass as follows. We know that the total mass, which is 4m times the position
vector, r center of mass-- I go all the way up there
on the blackboard; there's my definition
for center of mass-- equals the sum
of the individual masses times their position vectors. So it is the sum of i
m of i times r i. And this i goes
from one to three. Now, this is
a vectorial equation, and whenever we have a vectorial
equation, it sometimes pays off to split it
into two components-- a y and to an x component. And so in the x direction... of course,
the same equation holds for the x component
of these vectors. So now I have that 4m times the x component
of the center of mass equals this mass times the x component
of its position vector, which is zero, plus this mass, which is 2m times
the x position, which is l-- so plus 2m times l--
plus this mass times the x component of this
mass, which is one-half l. So that gives me
plus one-half m times l. My m goes and so I get
that x center of mass equals 2½ divided by four. That is five-eighths l. So we were not too far off
where we put it. Now, in the y direction,
you can do exactly the same. You can split it up
into the position vector of this object, which is
one-half l square root three. This one has no y component
and this one has no y component, so this is very easy. So you're going to get that 4m
times y of the center of mass equals this mass m times the y component
of that position vector, and that is one-half l
square root of three. And so you lose your m, and so
you see that y center of mass then becomes the square root of
three divided by eight times l. And I think that's about 0.22L,
very roughly. Yes. And so you see that
we didn't put it in so badly. It's about one-fifth
of this distance. It's about one-fifth higher
than this distance. And so you can calculate
the center of mass. That's really not too hard,
if you have discrete points. If you have a car or if you have
an object like this, then it is, of course, much harder to calculate
the center of mass. I will teach you in a few weeks
a very easy way to determine experimentally where the center of mass
is located-- experimentally,
which is different from calculating
it analytically, as we just did. So I mentioned to you that
the motion of the center of mass is very uniform in the absence
of external forces, and that I can demonstrate for
you again with the air track. We have a system here
of two cars which I connected by a spring. I will turn on the air shortly because the air makes
a lot of noise. This is... these are two cars
connected by a spring, and I will give these cars
a certain motion and they will go
in this direction. And they will oscillate
in a weird way because they are connected
with a spring and I keep them connected. And it will be nearly impossible for us to evaluate the motion
of these two cars individually. But if I give the whole system
a certain velocity and then they go like this, and they keep going like this
and making crazy things, momentum of the center of mass
will not change-- only of the center of mass. Not of this car,
not of that car. That's the uniqueness
of the center of mass. And so the center of mass
will just laugh at us and ignore all these motions and will travel at
a constant speed very nicely. So if you concentrate
on that little object, you may be able to see that. You may need a lot of
imagination to see that, because you're going
to be distracted by the weird motion
of the other two objects. This is the center of mass. It's right in the middle. The objects have the same mass. There we go. You see how complicated
this motion is of the individual cars? Can you see that the center of
mass is moving very uniformly, or can you not see that? Oh, you think you can see that? You have a lot of imagination. But I will help you. I will turn on...
I will turn off all the lights and only turn on
ultraviolet light. And ultraviolet light
will interact with that little ball,
the center of mass, and when I then make it dark, you will only see the motion
of the center of mass. And then you can really see that the center of mass moves
in a very civilized way. I'll bring it out here again. Okay. So now I'm going to help you
to concentrate. Are the timers off? Yes, timers are off? Okay, dark. Let your eyes get used
to the darkness. Okay, I'm going to do
the same thing, and now look at
that center of mass. And we know that these cars
are doing crazy things. Hard to predict but the center
of mass behaves decently. Beautifully! Constant velocity. I can let it go backwards,
so you can enjoy this once more. Center of mass motion, in the absence of external
forces has a constant velocity. When I throw up a hammer, then the hammer will do
very weird things. The hammer will start
to tumble and rotate, but the center of mass will
behave in a civilized way. If I throw up a hammer, then the center of mass
and only the center of mass will just go along
a perfect parabola as if it were just
a tennis ball. Now at one point... I will do it
with a squash racquet-- at one point the squash racquet
may be like this, and at another point, the squash
racquet may be like this, and a little later,
it may be like this, but the center of mass
of the squash racquet will perfectly go
along a parabola. And so we have here
a squash racquet and we have here
the center of mass. I have also here a regular... well, it's not quite
a tennis ball, but close enough. This one, you would expect it to behave perfectly
like a parabola. From this one,
you would not expect it to behave like a parabola. Let me throw this one up
in light, and you will see that it has
very strange motion. If I show the whole thing in UV, then you will see the same kind
of beautiful parabola as you would see
with this ball-- something like this. Forget the fact
that it lights up. You remember it was
the last lecture that I wanted to remind you of. So now we're going
to turn off the lights and I want to show you the
motion of the center of mass. You see the center of mass here? Can you all see it? Okay, there we go. You ready? Concentrate only
on the center of mass. Nice parabola or not? I'll do it again. You see the center of mass? Can you see it? You can still see it, right? Wonderful parabola for me! All right. Enjoy the presence
of your parents. Have a good weekend. See you Monday.
