WEBVTT
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I would now like to calculate
the moment of inertia
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for a very thin disk.
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So we have a thin disk.
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And the radius of
that disk is r.
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And it has a mass m.
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And I would like to calculate
the moment of inertia
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for this disk.
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Now, let's just
remind what point
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we're calculating it about,
about the center of mass.
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So our definition
of moment of inertia
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was take a small element,
mass element to the disk.
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In fact, we're going to see
it doesn't have to be small.
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Take a mass element to
the disk that's useful,
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and multiply it by the
perpendicular distance squared
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from the point we're
calculating it.
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So the way I'll do it
is I will choose a ring.
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I'm gonna choose a
ring of radius r.
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And now I'll make the
ring a certain thickness.
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And this thickness is dr.
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Now, in this
calculation, we're going
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to take a limit as
dr goes to zero.
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So even though the ring
has some finite thickness,
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its radius-- we'll
eventually treat
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treated as all of
the mass element
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a distance r from the center.
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So r will be our
integration variable.
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And that will be equal to
rcm, what we're calling
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rcm in the abstract result.
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Now, the dm is the tricky part.
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So what is the mass that's
contained in this area disk
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of radius r and thickness dr?
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Well, one way to think
about that is it's-- here we
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didn't say this, but our
disk is going to be uniform.
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And so we can describe
the mass per unit area
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as the total mass divided by
the area of the whole disk.
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And then we can say that
the mass in that ring
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is equal to sigma
mass per area times
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the area of the outer ring minus
the area of the inner ring.
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Now, when we expand this
out, dm, m over pi r squared,
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we get pi r squared plus
2rdr plus dr quantity
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squared minus pi r squared.
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And you can see
those terms cancel.
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And so what I get is
m times pi r squared.
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And in here I have
2 pi r dr. Now,
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this is only order
dr, plus a second term
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that goes like pi dr squared.
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And so, when I take this
limit as dr goes to 0,
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this term is much, much
smaller than that term.
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And so I can say my mass
element is m pi r squared times
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2 pi r dr.
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Now, let's think about this
term, why it makes sense.
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When we're shrinking our
ring, so taking a limit as dr
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goes to 0, and the ring just
becomes an extremely thin ring
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at radius r, then this
piece is a circumference,
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and this piece is
just the width.
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And so it's no surprise
that area is 2 pi r
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times d pi r dr in the limit.
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And now that enables us to
write the moment of inertia
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about the center of mass, icm.
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Let's pull out these
constants, m pi r squared.
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Now we're integrating
over the body.
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Let's hold off on the
limits for the moment,
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and put our values for dm.
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That's 2 pi r dr. And we
have our distance squared,
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which was, again, the
radius of r squared.
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And so the pis will cancel.
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I have 2m over r-squared times
the integral of r cubed dr.
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Now, we're supposedly
integrating over the body,
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but what does that body
integral actually mean?
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Well, what we're doing is
we're taking a series of rings
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and adding them up as
we go from the origin
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out to the radius
of the whole disk.
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So the limits of our body
integral with respect
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to our integration variable,
we start with rings
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that essentially have no width.
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And we're integrating
these, we're adding up
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the contribution of
every ring until we
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get to rings of radius r.
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And our integration
variable, r cubed, dr.
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Now, this is an integral
that's easy to do.
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That's r to the forth over
4 between 0 and r equals r.
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And when we put that
in, the 2 cancels the 4.
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And oh, the pi we lost.
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So let's make sure
this pi should be in m.
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So we have the 2 over the 4
is one half, and r squared.
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And that is the moment
of inertia of it
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does about an axis
passing through the center
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perpendicular to the
plane of the disk.