WEBVTT
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Let's now consider how
to find the velocity
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of the center of mass of a wheel
that's rolling without slipping
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down an inclined plane.
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So let's draw a picture.
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Here is the wheel that
starts at t equals 0.
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And this is an angle phi.
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And at some later time
t final, the wheel
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has dropped a distance h
along the incline plane.
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And now let's figure
out what is the velocity
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of the center of
mass of the wheel
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when it gets to the bottom.
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And we're now going to
apply our energy arguments.
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Now the first thing we have to
realize is a very subtle point
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here is that there
can be friction.
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And we saw that when a wheel
is rolling without slipping,
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there can be static
friction at the contact
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point between the
wheel and the ground.
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There is a gravitational force.
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And a normal force.
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Now static friction does no
work because the point is always
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instantaneously at
rest and f dot ds is 0
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because the object is at rest.
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And so the ds is 0.
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So our static
friction does no work.
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And therefore, from
our energy principle
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that the external work equals
E final minus E initial,
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we have no external work and
so our energy is constant.
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Now how do we
analyze our energy?
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Well, I've set things up.
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So I'm going to choose my
potential energy to be 0
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when it's at the bottom.
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And that our initial energy is
only equal to potential energy.
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If the wheel has a mass m,
gravitational force down,
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the potential energy
initially is mgh.
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The final energy is
only kinetic energy.
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The wheel is rolling
with omega final.
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Its center of mass is moving
with v center of mass final.
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And we just saw that
the kinetic energy
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has two contributions; the
translational kinetic energy,
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1/2 m--
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we'll just call
this final squared.
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And we'll make that the
final just to make it easier.
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And it has the kinetic
energy of rotation.
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Now the rolling without
slipping condition
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is that the translational
center of mass speed
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is equal to the
radius of the wheel,
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r, times the angular
speed, omega final.
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So I can write E final
as 1/2 of m plus I
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cm over r squared times the
final squared, where I'm just
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replacing omega final
equals v final over r.
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And therefore, that the kinetic
energy by the energy principle
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tells us that E
initial equals E final.
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So mgh equals 1/2
times m plus I cn
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over r squared v final square.
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So we can find that v final is
equal to the square root of 2
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mgh divided by m
plus 1/2 cm r square.
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And that is the final
speed of the center
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of mass translational's
velocity of the wheel
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as it dropped a height h.