$A$ 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 circular trajectory in the vertical plane with the string becoming slack only on reaching the topmost point,$C$. This is shown in the figure. Obtain an expression for: $(i) v_{o}$; $(ii)$ the speeds at points $B$ and $C$; $(iii)$ the ratio of the kinetic energies $(K_{B} / K_{C})$ at $B$ and $C$. Comment on the nature of the trajectory of the bob after it reaches the point $C$.

(N/A) 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 $A$.
At $A$,the total energy is $E = \frac{1}{2} m v_{o}^{2}$.
At the highest point $C$,the string slackens,meaning the tension $T_{C} = 0$. Applying Newton's Second Law at $C$: $mg = \frac{m v_{C}^{2}}{L}$,which gives $v_{C} = \sqrt{gL}$.
The total energy at $C$ is $E = \frac{1}{2} m v_{C}^{2} + mg(2L) = \frac{1}{2} m(gL) + 2mgL = \frac{5}{2} mgL$.
Equating energy at $A$ and $C$: $\frac{1}{2} m v_{o}^{2} = \frac{5}{2} mgL \implies v_{o} = \sqrt{5gL}$.
At point $B$ (horizontal level),the height is $L$. By conservation of energy: $\frac{1}{2} m v_{o}^{2} = \frac{1}{2} m v_{B}^{2} + mgL$.
Substituting $v_{o}^{2} = 5gL$: $\frac{5}{2} mgL = \frac{1}{2} m v_{B}^{2} + mgL \implies \frac{1}{2} m v_{B}^{2} = \frac{3}{2} mgL \implies v_{B} = \sqrt{3gL}$.
$(iii)$ The ratio of kinetic energies: $\frac{K_{B}}{K_{C}} = \frac{\frac{1}{2} m v_{B}^{2}}{\frac{1}{2} m v_{C}^{2}} = \frac{3gL}{gL} = \frac{3}{1}$.
After reaching point $C$,if the string becomes slack,the bob is under the influence of gravity only. It will follow a parabolic projectile trajectory starting from point $C$ with a horizontal velocity $v_{C} = \sqrt{gL}$.