Centripetal force: if radius decreases, does ANGULAR or TANGENTIAL velocity change? Having conceptual trouble with this aspect of centripetal force. Say we have a puck on a frictionless table attached to a string that I am holding through a small hole, so that the puck moves in a circular path. So $F=m\;\dfrac{v^{2}}{r}=m\;\omega^{2}\; r$. If I pull on the string to change the radius of the puck's path, does angular velocity change, or linear? The math makes it look like it could be either, or both, which seems like it can't be right.  
 A: In your scenario, angular momentum $m v r$ is preserved (because your pulling force is radial, with no tangential component). So if you reduce $r$ by half, $v$ must double, and since $\omega = v/r$, it increases by a factor of four.
Note this means in a small amount of time that the area swept out by the string is proportional to $v$ and $r$. Since they change by opposite factors the area remains the same.
It's a basic rule of orbits:

Note also that the kinetic energy increases by a factor of four, because $v$ doubles.
That energy comes from the work done in pulling the string.
If the pulling had not been radial, but had a tangential component, that would have confused the issue.
By the way: Doing work against centripetal force is (to me) an interesting topic.
For example, if you have a pendulum hanging quietly from a string, making the string shorter or longer has no interesting effect.
But if the pendulum is already swinging, you can pull it up when it is in the center of the swing (against large centripetal force), and drop it down when it is at either end of the swing (small centripetal force), moving the weight in a figure-8.
This imparts energy to the pendulum so that it swings higher.
I suspect this is also a possible way to look at other forms of oscillatory motion, such as traveling uphill on a skateboard in a series of S-turns.
It might also be relevant to flapping wings, where the downstroke works against a higher force because the bird is in a slight upward curve, and the upstroke works against lesser force because the bird is in a slight downward curve.
The net work comes out as kinetic energy.
I'm sure this explanation will get flak, but I think it could be one way to understand what's going on.
A: As the puck is moving in uniform circular motion the centripetal force is normal to the velocity vector, no work is produced and the total energy of the puck, which is kinetic $\mathcal K=(1/2)mv^{2}$, remains constant.
Pulling the string, even slowly, the string tension becomes oblique to the velocity vector during the transition, work is produced transfered to the puck as kinetic energy. So the speed $v$ is increased, the radius is decreased and from $\omega=v/r$ the angular velocity is increased also.
