A bob of mass m is suspended by a light string of length L . It is imparted a minimum horizontal velocity at the lowest point A such that it just completes half circle reaching the top most position B . The ratio of kinetic energies ( K . E . ) A ( K . E . ) B is :

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Aicharya · Accepted Answer

The velocity given is minimum, just enough to complete verticle circle. At the top most point, tension of the string will be zero and gravitational force will provide the required centripetal force.

Therefore,

$m g = \frac{m {(V_{H})}^{2}}{L} \Rightarrow \frac{1}{2} m {(V_{H})}^{2} = \frac{1}{2} g L$

Apply energy conservation between point $A$ and $B$ , we get

$\frac{1}{2} m V_{L}^{2} = \frac{1}{2} m V_{H}^{2} + m g (2 L)$

$\Rightarrow V_{L} = \sqrt{5 g L}$

Also, $V_{H} = \sqrt{g L}$

Required ratio, $\frac{(K . E)_{A}}{(K . E)_{B}} = \frac{\frac{1}{2} m (\sqrt{5 g L})^{2}}{\frac{1}{2} m (\sqrt{g L})^{2}} = \frac{5}{1}$

A bob of mass m is suspended by a light string of length L . It is imparted a minimum horizontal velocity at…

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