Question

A ball is tied to one end of a string. The other end of the string is fixed. The ball is set in motion around a vertical circle without friction. At the top of the circle, the ball has a speed of v = sqrt(Rg). At what angle θ should the string be cut so that the ball will travel through the center of the circle?

Answer 1

To get started, picture the ball at the top of the circle, I like it going to the left with velocity v. [Yes, -v should work.]

At the top of the circle its KE is (1/2) m v^2 = (1/2) m g R

and if we take the center of the circle as h=0 for PE, we have

KE + PE = M g R (1 + 1/2)

Let theta =0 be a horizontal line drawn from the center of circle.

We need ball to be somewhere in theta (@) = 0 to pi/2 to move horizontally back toward the center and upward for awhile.

The ball that hits the center will start at y = R sin @ and have a velocity u, with x component -u cos(@) and y component v sin(@).
Then the distance traveled back toward x=0 will be - t u cos(@).

I am not sure this is how to solve the problem, but I hope this is helpful.

Need to find the time for it to go from y to y=0 under the acceleration of gravity, y=0 = y + t u sin(@) - (1/2) g t^2.