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When we used our
energy principle,

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suppose we have an object
that starts at a height h0.

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And this object is dropped,
and when it gets to the ground,

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it has some final velocity.

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And when we applied
our energy principle,

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assuming that there was
no air resistance, what

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we saw was that the change in
kinetic energy plus the change

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of potential energy was 0, so
we had 0 was equal to delta

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k plus delta u.

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And we saw that
the kinetic energy

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changed by 1/2 mv squared, and
the potential energy changed

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by minus mg h0.

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So we can compute the
velocity of the object

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as it's falling given
by square root of 2g h0.

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Now that we're considering
kinetic energy of rotation,

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recall that we show
that the kinetic energy

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of a pure rotation
about a fixed axis

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was 1/2 the moment
of inertia about

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that axis times the
angular speed squared.

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We now would like to
apply our energy principle

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to include rotational
kinetic energy along

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with the translational
kinetic energy.

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And the example that
we want to look at

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is something very simple.

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Suppose we have a pulley.

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Now our pulley has
a mass p, and we'll

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say it has a moment of inertia
of the pulley about the fixed

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axis passing through the
center of the pulley.

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And we have a mass 1
and another block 2.

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And let's suppose that
we release this system.

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And for the moment let's make
this surface frictionless.

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And suppose that block 2
falls down a certain distance.

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So in the final
state, block 2 will

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have dropped a
distance each final

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from its initial position.

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And what we like
to consider is find

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the velocity final of block 2.

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So we begin in the same way
that we've done this before,

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by considering our
energy diagrams.

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And so we'll have
an initial state.

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And in our initial
state, what we'll do

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is we'll just have
the initial 1, 2.

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And I'm going to choose u
equals 0, here's my pulley.

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And everything is at rest.

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And so in the initial
state, the initial energy, i

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initial, k initial,
plus u initial is 0.

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And in our final state,
we have the pulley

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is rotating with omega final.

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Block 1 is moving with
a velocity v final 1.

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And block 2 is also
moving with v2 final.

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And let's just suppose that this
was our u equals 0 position.

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And although it's not so
clear in the diagram, u final,

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it has moved down
to height h final.

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So what is the energy
in our final state?

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Well, we have to consider
all the different pieces.

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We have block 1, 1/2
m1, v1 final squared.

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We have the motion of block
2, 1/2 m2 v2 final squared.

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And we also have
the kinetic energy

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of the pulley, which is given
by 1/2 I about the pulley

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omega final squared.

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And what about our
final potential energy?

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Well, we've dropped
the height, h final.

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So we have block 2 has
moved minus m2 gh final.

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And now we have our
two energy states.

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And what we'd like to consider
is apply the energy principle

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just like we applied it
for this simple case.

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But before we do that, there
is a constraint condition

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that because the rope is fixed
in length, as block 1 moves,

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the pulley is rotating
and block 2 is moving.

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What we'll have is
fixed and not slipping.

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So as the rope moves
around the pulley,

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the pulley is moving with
the same motion as the rope,

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and the rope is moving with
the speeds of block 1 and 2,

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so we see that v1 final
is equal to v2 final.

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And now what about the pulley?

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If this is radius r, we
know that the velocity

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of a point on the
rim of a disk is

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moving with the speed
of the rope, which

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is the speed of
block 1 and block 2,

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So that's our omega final.

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And so that makes our final
energy, let's now gather terms.

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The velocities are the same.

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So we have 1/2 and 1 plus m2.

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And we'll just call this
the final for simplicity.

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1/2 m2 times v final squared.

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That accounts for
these two terms.

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And we have the moment of
inertia, kinetic energy

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associated with the wheel, which
is 1/2i omega final squared,

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which is v final squared over
r squared minus m2 gh final.

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And so now we can solve
our energy principle, which

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is because we're assuming
everything's frictionless,

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we have 0 equals E
final minus E initial,

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implies that E final
equals E initial.

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And we chose our initial
potential energy to be 0.

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So of course that's just
a choice of constant.

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So I can now solve this equation
by setting E final equal to 0.

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That's the same statement.

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And then you can
see algebraically I

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can solve for V final.

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And what I get is I'm just going
to write all these terms over.

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I get m2gh final.

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I'm going to divide by this
common coefficient, 1/2 and 1

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plus m2 plus the
moment of inertia

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divided by radius squared.

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And I now have to take the
square root of the whole thing.

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And that's how I can find
the velocity of block 2

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when it's dropped down a
certain distance to h final.

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So here we've generalized
our energy approach

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to include rotational
kinetic energy.