Isn't centripetal force equal to mv^2 over r?

Question

or is it mv^2*r?

Answer 1

It's mv^2/r.

Here is how you can stop doubting:

That formula is a SPECIAL CASE for circular morion of the general formula [for any 3D motion of a point-mass] which gives the normal component of the force (i.e., the projection of the whole force on the plane perpendicular to the trajectory) provided "r" is understood to be the "radius of curvature" of the trajectory at that particular point.  See link below for a nice proof.

One way to be sure that the other formula is WRONG is to analyze the "dimension" of the quantity mv^2*r... It's like a mass multiplied by the cube of a distance divided by the square of a time (because v^2 is the square of a distance divided by the square of a time).  On the other hand, a force should be a mass multiplied by a distance divided by the square of a time (since it's mass times acceleration).  The formula mv^2/r matches that.

Make good use of both parts of the above answer.

You do not need to memorize such things in detail if you just analyze dimensions (which can be done very quickly) and KNOW that no bizarre coefficients are involved in the formula.  Better, if you have understood the [beautiful] proof for the expression of the normal acceleration given in the link, you'll be able to reproduce it quickly ten years from now (trust me, I have been there) and will never doubt again!

Enjoy.