A bob of heavy mass $m$ is suspended by a light string of length $I$. The bob is given a horizontal

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A bob of heavy mass $m$ is suspended by a light string of length $I$. The bob is given a horizontal velocity $v_{0}$ as shown in figure. If the string gets slack at some point $P$ making an angle $\theta$ from the horizontal, the ratio of the speed $v$ of the bob at point $P$ to its initial speed $v_{0}$ is:

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At Point $P,mgsin	heta = \frac{mv^{2}}{l}$By conservation of mechanical energy at point $P \in Q$$$\begin{matrix} & \frac{1}{2}mv_{0}^{2} = \frac{1}{2}mv^{2} + mg(I + Isin	heta) \ & \frac{v_{0}^{2}}{2} = \frac{v^{2}}{2} + gl(1 + sin	heta) \end{matrix}$$Put $gl = \frac{v^{2}}{sin	heta}$ using (1)$$\begin{matrix} & \frac{v_{0}^{2}}{2} = \frac{v^{2}}{2} + \frac{v^{2}}{sin	heta}(1 + sin	heta) \ & \frac{v_{0}^{2}}{2} = \frac{v^{2}}{2} + \frac{v^{2}}{sin	heta} + v^{2} \ & \frac{v_{0}^{2}}{2} = \frac{3}{2}v^{2} + \frac{2v^{2}}{2sin	heta} \ & v_{0}^{2} = v^{2}\left\lbrack 3 + \frac{2}{sin	heta} ightbrack \ & \frac{v}{v_{0}} = \left( \frac{sin	heta}{3sin	heta + 2} ight)^{\frac{1}{2}} \end{matrix}$$

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