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Q.

A smooth cylindrical glass lies on the ground. A small particle of mass ‘m’ lies on the bottom surface touching the circumference as shown in the figure. The minimum value of $v_{0}$ (along the circumference) such that particle just reaches at the top of glass is $\sqrt{\frac{K g h}{3}}$ . Find the value of K.

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answer is 8.

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Detailed Solution

By the conservation of energy
$\frac{1}{2} m v_{0}^{2} = m g h + \frac{1}{2} m v^{2} ....... (i)$
By conservation of angular momentum
$\frac{m v_{0}^{} R}{2} = m v R$
From equation (i) and equation (ii)
$v_{0} = \sqrt{\frac{8 g h}{3}}$
$\Rightarrow k = 8$

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