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Let's consider the
motion of a wheel that's

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rolling along the ground with
some center of mass velocity

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vcm.

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And because the
wheel is rotating

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it has an angular velocity.

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And you can see that
that vector is directed

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into the plane of the board.

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Now, what we'd like to do is
consider the kinetic energy

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of this continuous body.

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A little bit later on, that
body has moved some distance.

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And what we want to
consider is the fact

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that not only is every
point in the body moving

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with the center of
mass speed, but there's

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this additional
rotational energy

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that's associated with
the fact that every point

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in the center of
mass reference frame

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is undergoing circular motion.

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So how do we describe that?

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Well, we'll do that by choosing
some point in the body.

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So let's pick a point.

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We'll call that
mj, with mass mj.

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And the velocity of this point,
remember, has two components.

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To simplify it,
we'll give ourselves

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a little more picture here.

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Every single point in
the object has the vcm.

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But because this object is
undergoing circular motion,

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there is vcmj.

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That's the rotational
circular tangential velocity.

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And so the vector
sum of these two

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is the actual velocity
vj of the j-th object.

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vj is equal to the
center of mass velocity

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plus the tangential rotational
velocity that it has,

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because it's undergoing
circular motion.

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And now what we'd like to do
is calculate the kinetic energy

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of this object.

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Well, the kinetic energy
is the sum j from 1 to n

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of 1/2 mj times the velocity
of this j-th particle squared,

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which we can take
as a dot product.

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So we can write that as vcm
plus vcmj dot vcm plus vcmj.

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And that's just vj squared.

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So when we look at these terms,
it looks complicated at first.

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But there's some nice-- there's
going to be vcm dot vcm.

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There's two cross terms.

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They're identical.

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And vcmj dot vcmj.

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So let's write out
those three terms.

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We have 1/2 mj.

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vcm dot vcm is vcm squared.

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Now, every point in the
object has the same vcm.

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So we can pull that
one out of the sum.

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And now we'll take
these cross terms.

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So we have the sum
over j from 1 to n.

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There's two cross terms.

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So the 2's are going to cancel.

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And inside here,
we have to remember

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to keep our mass element.

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That's important.

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Now, I'm going to write
it as mj vcmj vector.

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Now, remember, when
you dot with vcm,

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every single point
has the same vcm.

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Every j-th element
has the same vm,

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so I can pull that vcm outside.

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And finally, I
have the last term,

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which is the sum over j from
1 to n of 1/2 mj vcmj squared.

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And that's just the dot
product of those two terms.

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And so our kinetic energy
looks rather complicated,

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but let's focus on
this term right here.

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Because recall from our
video on the center of mass

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that the definition of the
center of mass reference frame,

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so if you're moving
in the center of mass,

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that in the center of mass
reference frame, the sum of mj

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vcmj is equal to 0.

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So for instance, if you're
in the center of mass frame,

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you're moving with vcm.

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The only velocity is this.

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And in that frame, the
sum of mj vcmj is 0.

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And we did a video
on that one before.

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And that's exactly
what's in this term.

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So this term is 0.

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So this, remember, was
how we defined the center

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of mass reference frame.

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And therefore, our kinetic
energy consists of two pieces.

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This first piece is just 1/2 the
total mass times vcm squared.

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And our second piece over here,
we'll just write it out now--

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1/2 sum over j mj vcmj squared.

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Now, if you are moving
with the center of mass,

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then this j-th object is just
undergoing circular motion.

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And so we have our
result that we've

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used many times is
that the velocity,

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the tangential
rotational velocity,

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is just equal to
the radius rsj--

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so let's introduce that rsj--

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times the angular speed omega.

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And when we put
that into this term,

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we see our kinetic
energy has two pieces--

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m total v center of mass squared
plus 1/2 j goes from 1 to n--

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I didn't finish that sum there--

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mj rsj squared.

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Now, just remember that every
single point in the object

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has the same angular speed,
and so we can pull out

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the omega squared in there.

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And because this is a continuous
body and we take the limit,

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as we've done before,
as mj goes to 0,

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this quantity of
mass times distance

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squared is just the
moment of inertia

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about the center of
mass of that body.

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And in conclusion, K
is 1/2 m total v center

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of mass squared plus 1/2
the moment of inertia

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about the center of mass times
the angular speed squared.

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Now, this is the same
crucial decomposition

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that we've talked
about many times.

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This first piece is what we
call the translational kinetic

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energy, because it
just represents how

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the center of mass is moving.

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And the second piece
is what we call

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the rotational kinetic
energy, because it's

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a representation of just the
kinetic energy of rotation.

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For example, if you were in
the center of mass frame,

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there would be no
translational energy,

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and this would be the
only kinetic energy.