When the mass leaves the contact from the hemisphere, then R = 0, so, mgcos = .

cos = Â —– (i)

We draw a perpendicular QS on PO. Therefore, SO = h = rcos .

Using conservation of energy at point P and Q we get, + mg(PS) =

Or, Â + mg(r- rcos) =

Or, Â = + 2gr(1- cos)

u =

Putting the value of u in equation (i) we get, cos = Â

Or, cos = + 2 – 2cos

Or, 3cos Â = Â + 2 Â

= .