A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical motion (circular) as shown in figure. Initially, by applying horizontal velocity $v_0$ at the point ' A ', the string becomes slack when, the bob reaches at the point ' D '. The ratio of the kinetic energy of the bob at the points $B$ and $C$ is ______ -.

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Aicharya · Accepted Answer

$\begin{aligned} & \frac{1}{2} \mathrm{mv}_{\mathrm{A}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mgh} \ & \Rightarrow \frac{1}{2} \mathrm{~m}(5 \mathrm{~g} \ell)=\frac{1}{2} \mathrm{mv}_{\mathrm{B}}^2+\mathrm{mg} \frac{\ell}{2} \ & \Rightarrow \frac{5 \mathrm{mg} \ell}{2}-\frac{\mathrm{mg} \ell}{2}=\mathrm{KE}_{\mathrm{B}} \ & \Rightarrow \mathrm{KE}_{\mathrm{B}}=2 \mathrm{mg} \ell \ & \frac{1}{2} \mathrm{mv}_{\mathrm{C}}^2=\frac{1}{2} \mathrm{mv}_{\mathrm{D}}^2+\mathrm{mg} \frac{\ell}{2} \ & \Rightarrow \mathrm{KE}_{\mathrm{C}}=\frac{1}{2} \mathrm{mg} \ell+\mathrm{mg} \frac{\ell}{2}=\mathrm{mg} \ell \ & \Rightarrow \frac{\mathrm{KE}_{\mathrm{B}}}{\mathrm{KE}_{\mathrm{C}}}=2\end{aligned}$

A bob of mass $m$ is suspended at a point $O$ by a light string of length $l$ and left to perform vertical…

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