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