What is the minimum speed the rock must have at the top of the circle if it is to always stay in contact with the bottom of the bucket?

 First lets determine the equation. Well at the top of the circle both the normal force and the weight are in the same direction. So we have Fnet=N+mg. Since this is a circular path the Fnet is also = to (mv^2)/r. 

We convert the situation where the rock is no longer in contact with the bottom to terms relevant to the equation. So, what is a requirement for normal force? The object must be in contact with the surface, meaning it can't be in free fall. Realizing this means that the instant when the object does not touch the bucket is where the normal force = 0. 

Now we have N+mg=(mv^2)/r where N=0 is the case we are interested in. This leaves 0+mg=(mv^2)/r 
Solve for v: 
v=(gr)^(1/2) or v=3.28m/s 

Ibelieve the answer is b