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When we analyzed how the
position vector changed,
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we know that the velocity
for circular motion
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is given by the
radius times the rate
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that the angle is changing.
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And it points tangential
to the circle.
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So let's draw a few
characteristic arrows
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to show that.
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At this point, we'll draw
these pictures with d theta dt
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positive.
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So the velocity
points like that.
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It points like this.
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It points like that.
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And these are all the velocity
vectors at different times.
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Notice that if we make--
consider the special case
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in which d theta dt is a
constant, in that instance,
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the magnitude of
the velocity, v,
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is given by r magnitude
of d theta dt.
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And that is also a constant.
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But the velocity vector
is changing direction.
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And we know by definition
that the acceleration
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is the derivative of velocity.
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And so what we see
here is where we
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have a vector that's constant
in magnitude but changing
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direction.
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And we now want to
calculate the derivative
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in this special case.
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We refer to this case as
uniform circular motion.
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So this special case is often
called uniform circular motion.
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OK.
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How do we calculate the
derivative of the velocity?
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Well, recall that the
velocity vector, r d theta
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dt-- those are all constants--
because it's in the theta hat
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direction, once again,
will decompose theta hat
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into its Cartesian components.
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You see it has a minus i hat
component and a plus j hat
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component.
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The i hat component
is opposite the angle.
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So we have minus sine theta
of t i hat plus cosine
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theta of t j hat.
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So when I differentiate
the velocity in time,
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this piece is constant, so I'm
only again applying the chain
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rule to these two functions.
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So I have r, d theta dt.
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And I differentiate sine.
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I get cosine with a minus sign.
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So I have minus cosine theta.
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I'll keep the function of
t, just so that you can
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see that-- d theta dt i hat.
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Over here, the
derivative of cosine
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is minus sine d theta dt.
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That's the chain rule-- sign
of theta dt, d theta dt, j hat.
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And now I have this
common d theta dt term,
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and I can pull it out.
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And I'll square it.
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Now whether do you think that
dt is positive or negative,
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the square is always
positive, so this quantity
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is always positive.
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And inside I have--
I'm also going
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to pull the minus sign out.
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And I have cosine theta of t i
hat plus sine theta of t j hat.
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Now what we have here is
the unit vector r hat t.
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r hat has a cosine adjacent in
the i hat direction and a sine
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component in the
H Hut direction.
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So our acceleration--