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We'd like to now calculate
the angular momentum

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about an axis rotation of a
symmetric object like a ring.

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00:00:12,110 --> 00:00:14,270
So here is our axis of rotation.

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We're going to have
omega pointing up.

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That means omega's z will
be positive, because that's

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going to be our k hat axis.

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And this is a ring of
mass m and radius r.

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So here's our radius.

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Now, and we're going
to choose a point that

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lies somewhere on the
symmetry axis of the ring.

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And the object is rotating
about that symmetry axis.

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So it's passing through
the center of mass.

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Now, the way we'll
do this is we'll

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divide our ring up into
pairs of symmetric objects.

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So we have an m here and
a little delta m there.

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And we're going to calculate
the angular momentum

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first of that pair.

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Well, we've already
made that calculation

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that the angular momentum
about s of the pair

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is twice the mass
of each object.

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That's the mass of the
pair times r squared

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and it points in
the omega direction.

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And so to get the angular
momentum of the ring,

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we'll just sum up a
set of symmetric pairs

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until we expand out
over the entire ring.

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And so L of the
ring about s is just

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the sum over the pairs
of the mass of the pair

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times r squared omega.

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And r squared and omega
are all constants.

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The total mass of
the ring is just

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the sum over the pairs
of the mass of the pair.

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And that's the total
mass m of the ring.

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And so what we get is m r
squared omega L of the ring s.

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This is the moment of inertia
about the axis for the ring,

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because all the mass is
distributed a distance r away.

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And so we see for a continuous
symmetric body like a ring,

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again, because it's symmetric,
the angular momentum

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only has a component along
the axis of rotation.

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Now, what if we had an
extended symmetric object?

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And let's see if I can possibly
draw something like that.

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So we'll draw a
pear shaped object.

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What is the moment of inertia of
this pear shaped object, which

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is intended to be
symmetric, about the z-axis?

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Well, I can think of
that pear shaped object

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as consisting of a ring
and it's a solid object.

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This is a solid
symmetric object.

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And then inside the
ring is just more rings.

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So the whole disk here is
just a superposition of rings.

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And so the moment of inertia--

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so the angular momentum
of just this disk

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is the moment of the
disk about the axis.

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And now I'll just add more
disks that are symmetric.

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And so by exactly
the same calculation

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as I have here, the moment of
this symmetric object is just--

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the moment of inertia is more
complicated for this object.

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And it points in
the omega direction.

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So again, in conclusion,
a symmetric object

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about the axis of
rotation only has

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a component of L pointing
in the direction of omega.

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The constant of proportionality
between the angular momentum

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and the angle the
velocity is the moment

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of inertia about that axis.

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This we need to calculate
for each symmetric object.