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In kinematics, if we
know acceleration,
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which we generally can get
from Newton's second law,
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then we'd like to
integrate the acceleration
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to get velocity and position.
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How does that work
for circular motion?
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Well, when you think
about circular motion
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and you think about the
tangential direction,
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then that, in some sense,
is a one-dimensional motion.
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And so what we'll see is that
the tangential description
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of this motion can be integrated
to get the tangential velocity
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and position, just
in a similar way
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that we did with
linear motion, where
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if we were given
the acceleration,
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we integrated
velocity and position.
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So let's look at a
particular example.
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Suppose we're given that
the tangential component
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of the acceleration is given
by some simple polynomial, A
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minus Bt.
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Now, in this case,
the acceleration
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is certainly non-constant.
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What we'd like to
do-- remember, this
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is equal to rd squared
theta dt squared.
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So if we integrate the
tangential component
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of the acceleration
with some variable
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from t prime going from
some initial value, which
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we can call 0, to
some final value,
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then that will
give us the change
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in the tangential velocity.
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So for this case,
V theta of t is
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equal to V theta t0 plus the
integral of A minus B t prime
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d t prime.
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Now, we needed an integration
variable because recall
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that our functional
dependence is the upper limit
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of the integral.
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We saw that in
one-dimensional kinematics.
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So we're going from some
t0 0 to some prime t prime.
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Well, this integral
is now-- we'll
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write this V theta 0, where,
again, t0 we're calling 0.
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And when we do this integral,
A minus B t prime and d
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t prime-- that's an integral
from 0 to t prime equals t.
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t prime equals 0.
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This integral is not hard to do.
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We get A of t minus
1/2 B of t squared.
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And there's that
constant initial term.
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And so the velocity at time
t of the tangential component
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of the velocity is given
by that expression.
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In exactly the same fashion,
the angle-- how much angle
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does this go through?
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Well, we have to
be careful there.
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So if we want to do
the arc length, s of t,
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we have to multiply the
angle by the radius.
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And that is just the integral
of the tangential component
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of the velocity from t prime
equal 0 to time t prime t.
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And so in this example, we
have a slightly complicated
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integral, V0 d t prime, 0 to t.
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I'm going to drop--
well, I'll write it
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in-- t prime plus the
integrals A t prime minus 1/2
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B t prime squared.
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I didn't need to split this up.
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But I'm splitting it up, anyway.
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And I have really one,
two, three integrals to do.
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And what I get is V
theta 0 t-- the result
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of the first integral.
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The result of the second
interval is 1/2 At squared.
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And the result of
the third integral
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is minus 1/6 B t cubed.
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And that gives me this.
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I have to be a little
careful on this side.
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It's how much has the arc length
changed as we go from time 0
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to time t.
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So that's the change
in arc length.
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And that's a typical
example of where we're
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given tangential acceleration.
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We integrate it to get
the tangential velocity.
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And then when we integrate
the tangential velocity,
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we're getting the change in
arc length around the circle.