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Well, we've been calculating
angular momentum about a point.

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Recall our definition
of angular momentum.

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We've been looking at
cases where the particle is

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moving linearly.

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Now let's look at a
case where the particles

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are undergoing circular motion.

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So suppose we have
a particle that's

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undergoing circular motion m.

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And let's choose some axes.

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And our particle, we
denote that it's rotating

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about the vertical axis.

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I'm going to call
that axis k hat.

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And I'll make that omega
squared, omega z positive.

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So it's rotating about
the k-axis, the z-axis.

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Now, when we calculate
the angular momentum,

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we know because it's
rotating about the z-axis,

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this is our z-axis, that
the particle has a velocity

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tangential to the circle.

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So its momentum is
tangential to the circle.

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And we draw our vector rs
to where the object is.

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Now, you could solve this
in Cartesian coordinates

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but then you might have to
do some vector decomposition.

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But because there's a central
point to this motion, whenever

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there's a central
point we like to choose

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cylindrical coordinates.

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And the way we'll do that is
we'll define some angle theta.

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If this were my plus
x and my plus y-axis,

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then that's
consistent with k hat

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being up in our
definition of omega.

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And I'll define an
r hat unit vector,

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which is pointing radially
outward from the center

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of the circle.

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And a theta hat vector, which
is tangent to the circle.

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And now I can calculate--

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I can write down, say the
radius of this circle is r.

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Then l of s, the vector rs
has radius r pointing outward.

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And the momentum vector
is pointing tangential.

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And that's a very easy
cross-product to make r hat

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cross theta hat.

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That's what we're defining
to be k hat, maintaining

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the cyclic order.

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And so, we get rp k hat.

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Now for circular motion
the momentum magnitude

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is m, the magnitude of
the theta component.

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We've made that positive,
which is mr omega z.

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And so ls is--

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There's an r here
and another r there.

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So we get mr squared
times omega z k hat,

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This is our vector omega.

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Now this turns out to be the
moment of inertia of a point

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particle located at the center.

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And so we conclude that
the angular momentum

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is proportional to
the angular velocity.

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Now, that's no surprise
in this particular case.

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Why?

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Because the vectors rs and p
are in the plane of motion.

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And whenever you
take a cross-product,

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l is perpendicular to
both of those vectors

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so it's perpendicular
to the plane of motion.

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And therefore, l has to
point in the z direction.