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Question

(D) According to the law of conservation of energy,the potential energy lost by the mass $m$ is converted into the kinetic energy of the mass $m$ and

Vedclass · Accepted Answer

$\sqrt{\frac{4mgh}{2m + M}}$

Vedclass · Answer

(D) According to the law of conservation of energy,the potential energy lost by the mass $m$ is converted into the kinetic energy of the mass $m$ and the rotational kinetic energy of the disc.$mgh = \frac{1}{2} mv^{2} + \frac{1}{2} I\omega^{2}$Since the string is wrapped around the disc,the moment of inertia of the disc is $I = \frac{1}{2} MR^{2}$ and the angular velocity is $\omega = \frac{v}{R}$.Substituting these values:$mgh = \frac{1}{2} mv^{2} + \frac{1}{2} \left( \frac{1}{2} MR^{2} \right) \left( \frac{v}{R} \right)^{2}$$mgh = \frac{1}{2} mv^{2} + \frac{1}{4} Mv^{2}$$mgh = v^{2} \left( \frac{2m + M}{4} \right)$$v^{2} = \frac{4mgh}{2m + M}$$v = \sqrt{\frac{4mgh}{2m + M}}$

$A$ massless string is wrapped around a disc of mass $M$ and radius $R$. Another end is tied to a mass $m$ which is initially at height $h$ from the ground level as shown in the figure. If the mass is released,then its velocity while touching the ground level will be:

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