A ball is released from height h as shown in fig. Find the condition for the particle to complete the circular path.

Explanation

According to law of conservation of energy
(K.E. + P.E) at A = (K.E. + P.E) at B
`rArr` 0 + mgh = `(1)/(2) mv^(2) + 0`
`rArr v = sqrt(2gh)`
But velocity at the lowest point of circle,
`v ge sqrt(5gR) rArr sqrt(2gh) ge sqrt(5gR) rArr h rArr (5R)/(2)`