WEBVTT
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We've already talked
about kinetic energy
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for an object that's
rotating and translating.
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Let's take, again, a
simple object like a wheel.
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Center of mass is
moving, and we want
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to consider this is a bunch
of point-like particles,
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th Jth particle with mass MJ,
and a position vector RCMJ.
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And now what we'd like to do
is, what is the angular momentum
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of this about some point P?
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And we're going to do the
same type of decomposition
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that we've been
doing right along.
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Let's draw the vector RJ.
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This is the vector from
the point P to the object.
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Here's the vector capital R,
and there's the vector RCMJ.
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And our same vector
addition, R plus RCMJ.
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j Now let's calculate
the angular momentum
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about the point
P. Now just let's
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recall two facts about the
center of mass reference frame
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that we've already used.
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The first is that the sum of
the velocities, the momentum
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in the center of mass
reference frame is zero,
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and if you integrate this
equation in just exactly
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the same way, if you added
up the mass times each
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of the position
vectors, that's MJ,
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MJ, in the center of mass
frame, that's also zero.
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So we've used this when we
talked about kinetic energy.
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Now we're going to consider
both of these results.
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So recall that angular
momentum about a point
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is the sum of the vector from
the point, cross-product MJ,
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times the velocity BJ.
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Now we can also use our
law of addition velocities,
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VJ is V plus the
VCMJ, and so now
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we're going to have to
do both substitutions.
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That's why this calculation is
a little bit more complicated.
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We have R plus RCMJ,
cross MJ times capital
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D. That is the velocity
of the center of mass
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with respect to the
ground frame plus VCMJ.
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Now we have four terms
in the cross-product,
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and this time we'll
write them all out.
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So the first term is the sum
over J of R cross MJ capital D.
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The second term is,
we'll take this one,
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and we'll cross with that one.
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But because MJ is just a scalar,
I'm going to pull it in front.
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So that's MJ, RCMJ, and V is the
same for every single particle
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in the object.
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Remember, V is just the
velocity of one reference frame
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with respect to the other.
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So I can pull that out, cross
V. And you're already noticing
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that this term will be zero.
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The next term is
when I take R cross
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MJVJ, V center of
mass, J, summed over J,
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that's JR cross MJ, VCMJ.
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But again, remember that
R is the same vector.
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It's just from P to
the center of mass,
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so I didn't need to
put the sum here.
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I can pull R out of the sum
and I have my sum like that.
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And again, you're probably
noticing that this term is zero
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and our final term is
RCMJ cross MJ VCMJ.
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So our final term is sum
R, RCMJ cross MJ VCMJ.
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Now from these two
conditions, one and two,
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both of these two
terms are zero.
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And that's the power of using
the center of mass reference
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frame.
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So LP, now let's look
at the first term.
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Here, R is the same
for every particle,
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V is the same for
every particle,
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when we're summing
over the mass,
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we're just getting
the total mass,
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so the first piece
is R cross M total V.
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And this last piece is precisely
the definition of the angular
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momentum of the Jth particle
in the center of mass frame.
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So that's the sum
over J of LCMJ,
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because RCM cross MVJ R
cross, say whatever speed this
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is in the center of mass frame,
it could be that sum VCMJ.
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BJ minus capital V.
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This is exactly the
angular momentum
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of that Jth mass in the
center of mass frame,
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and when we total all of
this up, we have R cross
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and total V plus LCM.
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Now what this first term is,
if you treated the whole object
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as a point mass, M total, moving
with V, and here's our point P
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and there's R, this
is just what we
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call the translational
angular momentum.
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It's treating the whole
system as a point-like mass,
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and just calculate
moving with speed V,
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and that's just the
angular momentum.
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We've called this V.
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And this piece is the angular
momentum about center of mass.
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So sometimes, we can think
of the angular momentum
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as corresponding
to a translational,
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it has to do with the
actual orbit of the object.
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You could call this
orbital angular momentum.
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And this is the fact that
the object is rotating.
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You could think of this as
the spin angular momentum
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about the center of mass.
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This is the same
type of decomposition
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that we saw with kinetic energy.
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There was a translational
component of kinetic energy
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and a rotational component
of kinetic energy.
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This is the analogous case for
angular momentum about a point
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P.