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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Detailed Solution

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