A bob of mass m is suspended by a light string of length L. It is imparted a horizontal velocity `v_(o)` at the lowest point A such that it completes a semi-circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point, C. This is shown in Fig. 6.6. Obtain an expression for
The speeds at points B and C.

Explanation

It is clear from Eq. (6.14)
`v_(c )= sqrt(gL)`
At B, the energy is `E=(1)/(2)mv_(B)^(2)+mgL`
Equatting this to the energy at A and employing the result from (i), namely `v_(0)^(2)= 5gL`,
`(1)/(2)mv_(B)^(2)+mgL= (1)/(2)mv_(0)^(2)`
`(5)/(2)mgL`
`v_(B)= sqrt(3gL)`.