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 (mg) and tension (T) in the string. The tension does no work as displacement is always perpendicular to the string. Total mechanical energy (E) of the system is conserved.
If we take P.E. of the system to be zero at the lowest point A, then
At A, `E=(1)/(2)mv_(0)^(2)` ...(i)
The necessary centripetal force `mv_(0)^(2)//L` is provided by `(T_(A)-mg)`.
`T_(A)-mg=(mv_(0)^(2))/(L)` ...(ii)
At the highest point C, the string slackens as the tension in the string `(T_(c))` becomes zero.
`:.` At C, `mg=(mv_(C)^(2))/(L)` ...(iii)
and total energy, `E=(1)/(2)mv_(C)^(2)+mg(2L)` ...(iv)
From (iii), `v_(C)=sqrt(gL)` ...(v)
From (iv), `E=(1)/(2)m(gL)+2mgL=(5)/(2)mgL`
Using (i), `(1)/(2)mv_(0)^(2)=(5)/(2)mgL, v_(0)=sqrt(5gL)` ...(vi)
At B, the energy is `E=(1)/(2)mv_(B)^(2)+mg(L)`
or `(1)/(2)mv_(B)^(2)=E-mg(L)=(5)/(2)mgL-mgL`
`=(3)/(2)mgL`
`v_(B)=sqrt(3gL)`
`(K_(B))/(K_(C))=((1)/(2)mv_(B)^(2))/((1)/(2)mv_(C)^(2))=(3gL)/(gL)=(3)/(1)`
At C, the string becomes slack and the velocity of the bob is horizontal and to the left. Therefore, the bob will continue on its circular path and complete the revolution.