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Let's determine the moment
of inertia of a big wheel.

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Here we have a big wheel and
on the side it's attached here

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at the center .

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And there's a string going
around here, and on that string

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is a little mass hanging.

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And our disk here
has a radius R.

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And we now want to in
a little experiment

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what the moment of
inertia is of this disk.

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And what we have
to do is we have

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to drop this mass to
the ground, and we

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need to measure
what height is here,

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and we need to measure
how long it takes to drop.

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So we need to measure t.

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So how are we going
to go about that?

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Well the moment of
inertia, that probably

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has something to do with
the torque of this wheel.

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So we have to start
doing a torque analysis.

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So torque about the center of
mass equals moment of inertia

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about the center of mass times
the angular acceleration.

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And we need to recall
that the torque is

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the product of the radius
times the perpendicular force,

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so we're going to
have R here, and then

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the force that's
acting on this wheel

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here is actually this
tension force here.

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So we are going to
have our RT here.

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And now we need to look at
how things actually moving

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with respect to
coordinate system.

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So if I let this mass
drop, to the disk

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it's going to spin clockwise.

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So we're going to have
theta hat going this way.

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And if theta hat
is going clockwise,

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my k hat vector is going
to go into the board.

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And if k hat is
going to go there,

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and it's rotating clockwise,
the angular acceleration

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is going to go in
the k hat direction.

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And if that is the
case, then the torque

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is going to follow suit.

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So torque also goes
into the board.

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And so that means
here we're going

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to deal with k hat direction.

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And we're going to
have Icm and then

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alpha Z in the k hat direction.

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OK, so we can solve this
I RT over alpha Z. OK,

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well that's pretty good, but
we have two unknowns-- the T

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and the alpha Z. And we can use
some other concepts to actually

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get information on those.

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The T, as you can guess
already, plays a role here

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in this massless string, so
we can do a quick free body

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diagram, an F equals ma analysis
to get to that tension force.

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So we have a little mass, m,
here gravity is acting on it.

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And we have this
tension force here.

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And we're going
to put j hat down.

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So we're going to get
mg minus t, equals ma.

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It's only going to
go in y direction,

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so we can just leave it here.

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And we can solve this for t.

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And then we have mg minus a.

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OK, good, so we have that.

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Now about the
angular acceleration.

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And whenever there was a string
going around a disk then,

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we of course, have a
constrained condition,

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because the linear acceleration
of this little mass going down

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is related to the radius of
the disk, times the angular

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acceleration.

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So we can solve this for alpha.

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And then we have a over R.
So let's put that in here.

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R, and then we have
mg minus a over a,

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and then we get another R here.

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And we can write that a little
bit more compact, mR squared g

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over a minus 1.

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Good, so now we have one last
hurdle, namely that a here.

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That a we can't measure.

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I said in the beginning, we
want to make an experiment.

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Actually we need experiment,
because we can't otherwise

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get to this a, so
what we need is

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a relation that connects what we
can measure, which is the time.

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It falls down the height here to
the acceleration of this block.

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And of course, that comes from
one dimensional kinematics.

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And we know that h equals 1/2 at
squared, so we can solve for a.

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2h over t squared.

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And now we can
stick that in here.

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And we have mR squared, gt
squared over 2h minus 1.

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And let's just write
this here again.

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And that is our final solution.

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So now we have only
measurable quantities here.

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The t we can measure.

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We just need a stopwatch.

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And the h we can
measure, as well.

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And this actually
already resembles,

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if you know the
theoretical, this already

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resembles the
theoretical solution,

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which of course is for a disk.

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1/2 mR squared, so that's
what one expects for a disk.

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And you see that we're very
close, so this term here

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is probably something
like 1.5, or should better

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come out to be 1.5,
because if we subtract 1,

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we get to that 1/2 here.

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And you can use it to
in return-- in return,

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you can also use it
to predict the time

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it takes to fall down if
you know what the height is

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or vice versa.