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We are now considering the
motion of a rigid body.

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And we'd like to talk about
kinetic energy of rotation

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and properties of
that rigid body.

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Let's consider a
rigid body, and let's

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say that our rigid body is
rotating about an axis passing

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through a point S.

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If we look overhead
at our rigid body,

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then what we're going to do is
introduce a coordinate system.

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So this is our overhead view.

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And suppose that we have a
small element of our rigid body

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here, which I'm going
to write as delta m j.

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And that's a distance
r j from the center.

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Our rigid body has
an angle theta.

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And what we'd like to do is
describe a coordinate system,

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r hat, theta hat, and k
hat pointing like that.

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Now, with this rigid
body that's rotating,

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we describe the angular
velocity as the rate

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that the angle is changing
with respect to k hat.

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And so we can draw
the vector omega,

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and this establishes
our coordinate system

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for our rigid body and a small
element of mass, delta m j.

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Now, what we'd
like to consider is

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what does this mass element do?

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Imagine that it's over there.

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Because every point
in the rigid body

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has the same angular
velocity, this object

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is going in a circle with
angular velocity omega.

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Recall that omega is
perpendicular to the plane

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of motion.

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But the object itself
has a velocity,

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which I'm going to write v j.

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And that velocity vector, v j,
is in the tangential direction.

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And it's related to the z
component of our angular

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velocity by the
following relationship--

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it's how far we are
from the center point,

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S, times the z component
of the angular velocity.

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And it's pointing in the
tangential direction.

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So remember that omega z
is equal to d theta dt.

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And this describes
our coordinate system

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for the rigid body.

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Now what we'd like to discuss
is the kinetic energy.

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And the way we're going
to consider kinetic energy

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is, we're going to sum up
the rotational kinetic energy

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of every single mass element.

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So we begin by writing
k j rotational.

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And we know that that is just
1/2 times the mass element

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times the velocity of
that element squared.

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Now we can use our relationship
for the tangential velocity

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element related to
the angular velocity.

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And we have 1/2 delta m j r j
squared times omega z squared.

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Now keep in mind that omega z is
the same for every single mass

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element, but the distance
of the mass elements

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are all different by r j.

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So the total rotational
kinetic energy

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is the sum over j
from 1 to n of, let's

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put the 1/2 outside, times
delta m j r j squared.

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Now again, recall that every
element has the same omega z.

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So I can write parentheses
omega z squared.

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And that's our rotational
kinetic energy.

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So what we want to do
now is look at the limit

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as our delta m j becomes
very small, because we

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have a continuous body.

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And we'll write a
definition, which

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is going to be the
moment of inertia passing

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through this point, S, about
the axis passing through S

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is equal to the limit
as delta m j goes to 0

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or n goes to infinity of this
sum-- delta m j r j squared,

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j goes from 1 to n.

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Now because this is a limit
for the continuous body,

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we'll define it as the
integral over the body

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of a small mass element dm.

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That's a distance r squared.

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Now here, what is
the meaning of the r?

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For our continuous body
here, if we call this dm

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and we define the distance
from S to the body r-- and now

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I'm going to just put a
little notation in here.

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It's the distance from S, the
axis we're calculating about,

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to where the body is dm.

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So I'll write S dm.

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This is what we call
the moment of inertia

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of a continuous body.

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Now again, what's very
important to realize--

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it's a moment about
a particular axis.

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So this is about an axis
passing perpendicular

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to the plane of rotation
and through S, point S.

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So it's an axis that's
passing perpendicular

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to the plane of rotation
passing through the point S.

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And this is what we call the
moment of inertia of a body.

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Now, we'll see that the moment
can be expressed in terms

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of other physical quantities.

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As the course
develops, you'll see

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two or three more
fundamental relations

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for moment of inertia.

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But what we'd like to do now
is summarize our results--

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that the kinetic energy of
rotation is 1/2 for rotation

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about this axis, I'm just to
indicate passing through S,

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of I s times omega z squared.

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Now keep in mind, because
omega z is a component,

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it can be positive,
0, or negative.

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But the square is always a
positive definite quantity.

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And that's our kinetic
energy of rotation.

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Let's contrast that with our
translational kinetic energy.

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And we remember there that
was 1/2 times the total mass

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of the object
times v squared cm,

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where we're looking
at all of the objects

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at the center of mass.

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And this is the total
mass of the object.