A ring ‘A’ of mass ‘m’ is attached to a stretched spring of force constant K, which is fixed at C on a smooth vertical circular track of radius R. Points A and C are diametrically opposite. When the ring slips from rest on the track to point B, making an angle of 30° with AC. (∠ACB = 30°) spring becomes unstretched. Find the velocity of the ring at B.

Question

G1 Physics · Accepted Answer

At A the spring length is 2R; at B it is unstretched and CB=√3R, so initial extension is R(2-√3). The ring drops by R/2. Energy conservation gives ½mv²=½K R²(2-√3)²+mg(R/2), so v=√{[KR²/m](2-√3)²+gR}. Hence B.