A bob of mass m is tied by a massless string | vedclass

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(A) Let the angular velocity of the disc be $\omega$ and the linear velocity of the bob be $v$. Since there is no slipping,$v = R\omega$.Applying the

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$\frac{2}{R} \sqrt{\frac{gh}{3}}$

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(A) Let the angular velocity of the disc be $\omega$ and the linear velocity of the bob be $v$. Since there is no slipping,$v = R\omega$.Applying the law of conservation of energy: The loss in potential energy of the bob equals the gain in kinetic energy of the bob plus the gain in rotational kinetic energy of the disc.$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$For a disc,the moment of inertia $I = \frac{1}{2}mR^2$.Substituting $I$ and $v = R\omega$ into the equation:$mgh = \frac{1}{2}m(R\omega)^2 + \frac{1}{2}(\frac{1}{2}mR^2)\omega^2$$mgh = \frac{1}{2}mR^2\omega^2 + \frac{1}{4}mR^2\omega^2$$mgh = \frac{3}{4}mR^2\omega^2$$gh = \frac{3}{4}R^2\omega^2$$\omega^2 = \frac{4gh}{3R^2}$$\omega = \frac{2}{R} \sqrt{\frac{gh}{3}}$.

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