A bead slides without friction around a loop-the-loop. The bead is released from a height of 10.1 m from the bottom of the loop-the-loop which has a radius 3 m. The acceleration of gravity is 9.8 m/s^2 .
What is its speed at the top of the loop? Answer in units of m/s.
The Attempt at a Solution
My first thought was to find the PE of the ball at the top, since the KE and the PE are always going to add up to that number, no matter where the ball is on the track. I did mgh = 1/2mv^2 = m(9.8)(10.1) = 98.98m
So the PE = mass times 98.98
then i did 98.98m=1/2mv^2 to get v=14.06982587, then i divided that by two since the ball is at the top of the loop, it's halfway through it i thought to get 7.034912935, which was wrong...
So what am I doing wrong? Any help would be greatly appreciated