Particle, P mass, m moving in a horizotal circle with uniform speed and the smooth inner surface of a hemisphere with base radius a. A light inextensible string is attached to P and passes through a small hole at the base of the hemisphere. A mass, m hangs at rest at the opposite end to P. The plane of motion is a/2 below the rim of the hemisphere. prove that v^2 = 3ga. I have so far concluded that the angle between normal reaction and the vertical is the same as between the string and the vertical and is 60 degrees. and that the radius of the horizontal circle is ((root 3/2)a) I have then tried to resolve horizntally so T-R = ma(v^2/ ((root 3/2)a)) and resolving vertically Tcos 60 + r cos 60 = 2mg. But I cant get it to work out!