As shown in the figure,a bob of mass m is tie | vedclass

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(C) By the law of conservation of energy,the loss in potential energy of the bob is equal to the gain in the total kinetic energy (translational kinet

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$\frac{1}{r} \sqrt{\frac{4 g h}{3}}$

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(C) By the law of conservation of energy,the loss in potential energy of the bob is equal to the gain in the total kinetic energy (translational kinetic energy of the bob + rotational kinetic energy of the disc).Loss in potential energy = $mgh$Gain in kinetic energy = $\frac{1}{2} mv^2 + \frac{1}{2} I \omega^2$Since the string is wound on the disc,the linear velocity of the bob $v$ is related to the angular velocity $\omega$ of the disc by $v = r\omega$.The moment of inertia of the disc is $I = \frac{1}{2} mr^2$.Equating the energies: $mgh = \frac{1}{2} mv^2 + \frac{1}{2} (\frac{1}{2} mr^2) \omega^2$Substituting $v = r\omega$: $mgh = \frac{1}{2} m(r\omega)^2 + \frac{1}{4} mr^2 \omega^2$$mgh = \frac{1}{2} mr^2 \omega^2 + \frac{1}{4} mr^2 \omega^2 = \frac{3}{4} mr^2 \omega^2$Solving for $\omega$: $\omega^2 = \frac{4gh}{3r^2}$$\omega = \frac{1}{r} \sqrt{\frac{4gh}{3}}$

As shown in the figure,a bob of mass $m$ is tied by a massless string whose other end portion is wound on a flywheel (disc) of radius $r$ and mass $m$. When released from rest,the bob starts falling vertically. When it has covered a distance of $h$,the angular speed of the wheel will be

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