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