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