physics
posted by emath on .
A tennis ball is a hollow sphere with a thin wall. It is set rolling without slipping at 4.24 m/s on a horizontal section of a track. It rolls around the inside of a vertical circular loop 90.0 cm in diameter and finally leaves the track at a point 19.0 cm below the horizontal section.
(a) Find the speed of the ball at the top of the loop.
(b) Find its speed as it leaves the track.
I tried using energy conservation but I keep getting the wrong answer

Did you include the rotational kinetic energy along with the translational kinetic energy?
Total KE = (1/2) M V^2 + (1/2) I w^2
= (1/2) M V^2 + (1/2)(2/3)M R^2*(V/R)^2
= (5/6) M V^2
(a) Solve (5/6) M [V1^2  V2^2] = M g H
Solve for V2 at the top. (H = 0.9 m)
M cancels out
(b) I don't understand how the ball can be below the below the horizontal section. I thought the loop was above the track.