A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity `v_(0)` at the lowest point A such that it completes a semicircular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C this is shown in Fig. Obtain an expression for `v_(o)`

Explanation

Thus, at A :
`E (1)/(2) mv_(0)^(2). . . . .(1)`
`T_(A) - mg = (mv_(0)^(2))/( L)` [Newton.s Second Low]
where `T_(A)` is the tension in the string at A. At the highest point C, the string slackens, as the tension in the string `(T_(c))` becomes zero .
`E = (1)/(2) mv_(c)^(2) + 2 mgL . . . . (2)`
`mg = (mv_(c)^(2))/(L)` [ Newton.s Second Low ] . . . . (3)
where `v_(c )` is the speed at C. From equations (2) & (3)
`E = (5)/(2) mgL`
Equating this to the energy at A
`(5)/(2) mgL = (m)/(2) v_(0)^(2) or, v_(0) = sqrt(5 g L)`