# Motion in a vertical circle

A 0.8 kg ball is whirled on a string r = 0.4 meters long in a vertical circular path. At the bottom of the circle, the ball is h = 0.45 meters from the ground. At the top of the circle, the ball has a speed of 3 m/s. Assume that the total energy of the ball is kept constant.

a. Calculate total energy of the ball (ground is zero potential energy)
I chose the top most point of the circle to do ths since PE is maximum there.
ME = KE + PE
Ke = mv^2/2 = (.8kg)*(3m/s)^2/2 = 3.6 J
PE = mgh = (.8kg)*(10m/s^2)*(1.25m) =10 J
Total energy = 13.6 J

Im not sure if i calculated the PE correctly. I know that PE at the top of the circle is mg2r but then i have to consider the height from the ground to the top of the circle. So i figured i should add the distance from the ground up and multiply by mg. is that right?

Next part of the question asks me to calculate velocity at the bottom of the circle. Here KE is larger than PE.
ME = KE + PE
13.6 J = mv^2/2 + mgh
13.6 J = (.8kg * v^2)/2 + (.8kg)*(10m/s^2)*(.45m)
When i solve for velocity i get 5m/s. how can this be? Shouldn't it be 3m/s as it is on the top? What am i doing wrong?