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I would now like to show
you how to calculate

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the moment of inertia of
a typical continuous body.

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Let's consider a
rigid rod, very thin.

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And what we want
to do is calculate

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the moment of inertia of this
body about the center of mass.

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Let's say the body is of length
L, and it has total mass M.

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Now, recall that the
moment of inertia

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about the center
of mass we defined

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as an integral of dm r-squared
integrated over the body.

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Now, our challenge
today is to understand

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exactly what all these terms
are in this expression.

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What is dm?

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What is r?

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And what do we mean by an
integral over the body?

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So now let's do a stepwise
interpretation of each term.

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The crucial thing is to
introduce coordinates systems.

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So let's choose a
coordinate system.

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And let's put the origin
at the center of mass.

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Now, the most important
thing is that we're

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going to do an integral.

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So we need to introduce
the integration variable.

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And that's the hardest part
of setting up intervals.

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So what we want to do
is arbitrarily choose

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an element dm.

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So there's our elements dm.

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And here's our
integration variable,

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it's a distance x
from the origin.

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So that's the
integration variable.

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Now, there's two things
when we set up the integral.

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r is equal to that
integration variable.

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r is abstract in
this expression,

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but in this concrete realization
it is the integration variable.

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The second place the integration
variables shows up is in dm.

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dm is a mass in
this small element.

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But if we want to express that
in terms of our integration

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variable, we have to express
it in terms of the differential

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length dx.

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So dm mass is equal to the
total mass per unit length.

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We're assuming the
rod is uniform times

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the length dx of
our small piece.

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And now we've set up the
two pieces that are crucial

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and all we have
to think about now

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is what does an integral mean.

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Well, an integral means that
we're dividing up the piece

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into a bunch of small
elements and we're

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adding the contribution
of each small element.

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So in particular, when we
write Icm equals-- now,

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we can write it as m over Ldx.

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That was our dm.

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And the distance of dm from the
point we're computing the axis

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is x-squared.

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Now, the question is what is
our integration variable doing?

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Well, x is going from minus L
over 2-- that's at this end--

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to x equals plus L over
2 on the other end.

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So we have x minus L over
2x equals plus L over 2.

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And now we've set
up the integral

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for the moment of
inertia, and the rest

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is just doing an integral.

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Recall that the integral of dx
x-squared is x-cubed over 3.

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And so this integral
then simply becomes

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Icm equals m over L
x-cubed over 3-- evaluating

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from the limits minus L over
2 to x equals plus L over 2.

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Again when you evaluate
the limits, what

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we get is m over L we have to
put in L over 2 cubed divided

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by 3 minus L over 2
cubed divided by 3,

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and that's 1 over 2 cubed
is an eighth-- divide

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by third, that's the 24.

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24 minus 24 is a 12.

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And so what we get
for Icm is m over L,

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12 L-cubed, or
1/12 m L-squared is

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the moment of inertia
about the center of mass

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of our rigid rod.

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And this is a measure
of how the mass is

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distributed about this axis.