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 semi-circular trajectory in the vertical plane with the string becoming slack on reaching the topmost point C, figure, Obtain an expression for (i) `v_(0)` (ii) the speeds at points B and C, (ii) the ration of kinetic energies `(K_(B)//K_(C))` at B and C.
Comment on the nature of the trajectory of the bob after it reahes the poing C.

Explanation

There are two external forces on the bob : gravity and the tension (T) in the string. The latter does no work since the displacement of the bob is always normal to the string. The potential energy of the bob is thus associated with the gravitational force only. The total mechanical energy E of the system is conserved. We take the potential energy of the system to be zero at the lowest point A. Thus, at A ,
`E= (1)/(2)mv_(0)^(2)" "(6.12)`
`T_(A)-mg= (mv_(0)^(2))/(L)` [Newton.s Second Law]
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.
Thus, at C `E= (1)/(2)mv_(c )^(2)+ 2mgL" "(6.13)`
`mg= (mv_(c )^2)/(L)` [Newton.s Second Law] (6.14)
where `v_(c )` is the speed at C. From Eqs. (6.13) and (6.14)
`E= (5)/(2)mgL`
Equatting this to the energy at A `(5)/(2)mgL= (m)/(2)v_(0)^(2)`
or, `v_(0)= sqrt(5gL)`.