Let's look at the angular momentum
of two particles, one sitting here and one over there.
And they are circling each other.
And we want to determine the angular
momentum of a point on the ground right here underneath.
That's the whole point, s.
And here is the center.
So the circle has radius, R. And here we have a height, h.
and let's have the particles go counterclockwise.
Because then we can define our k hat vector to go up.
And if these particles are going this way,
then omega is also going to be going in the k hat direction--
in the positive k hat direction.
So we know that L equals r cross p.
So we need to determine where is r and what's going on with p.
So our r vector goes here from point s to our first particle,
rs1.
And here is number two.
And we have rs2 sitting over here.
And p, the momentum of the particles-- well,
if these guys going in the counterclockwise direction,
then the momentum of particle one
is going to go into the board following
the direction of the motion.
And here p2 is going to come out of the board.
If we consider this equation here,
we can already graphically determine
what the answer is actually going to be.
And so we do r cross p.
So r cross p, which means L. Ls1 is going
to point in this direction.
Ls1.
And alternatively, Ls2 is going to go
this direction, which means if we add these two components--
and we do want the total angular momentum of the system
of both particles.
So we have to add them.
We'll get here our L of the system.
So in order to calculate that, we
need to calculate first the Ls1 and then the Ls2.
And then we need to add them.
The crucial point here is to be careful
with the coordinate system.
For the first particle, k is going up.
And r is going outwards.
And theta hat is going into the board.
Then we can write up Ls1 equals rs1 cross p1.
And we can express this vector here with h and with r.
So we have R in the r hat direction
plus h in the k hat direction cross p.
And p is the mass of the particle times the angular
velocity, the component.
That is R omega z.
And that goes in the theta direction.
So we need to solve this cross product here.
And we're going to have mR squared omega z.
And then we have r hat cross theta hat.
That's k hat plus hmR omega z.
And then we have k hat cross theta hat.
That's anti-cyclic.
And we get an r hat.
And this is, by the way, an r1.
And this is another r1.
And we get a minus because it's anti-cyclic.
So we're going to turn this here into a minus.
And that is the angular momentum for our first particle.
Now we're going to do the same in an analogous
way for particle number two.
But again, we need to look at the coordinate system.
That's very crucial here.
So k hat still goes up.
r hat now goes into the other direction.
And accordingly, theta hat comes out of the board.
And then we can just write down our Ls2 here as the same thing,
mR squared omega z k hat.
And here we have minus hmR omega z r2 hat.
And we know from our setup here that r1 equals minus r2 hat.
r1 hat equals minus r2 hat, which
means when we now put it all together,
we want L equals Ls1 plus Ls2.
We can tally this up.
And from this, we will see that the r hat term actually
falls out.
And we are left with 2 times the first term.
So we have 2mR squared omega z k hat.
And that's exactly what we wanted.
We wanted an L pointing in the k hat direction.
That's exactly this here.
And what you can also see is that the 2m here
is the mass, the total mass of the system, namely
two particles.
We could add another--
we could do the same exercise again
for four points because we'll add two here.
Then we would get 4m here.
So this is the mass of the system.
And if we were to add many, many more points all along the rim
here, we get to a solid ring, which
means that either we know the total mass of the ring
that we could stick in here.
Or we can integrate along the rim
and then get to the total mass here.
And we'll always get to the same kind of form.