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(C) For the translational motion of the hanging mass $m$:$mg - T = ma$  --- $(1)$For the rotational motion of the pulley of mass $m$ and radius $R$:$T

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$\frac{2}{3}g$

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(C) For the translational motion of the hanging mass $m$:$mg - T = ma$  --- $(1)$For the rotational motion of the pulley of mass $m$ and radius $R$:$T \cdot R = I \alpha$Since the string does not slip,the linear acceleration $a$ of the mass is related to the angular acceleration $\alpha$ of the pulley by $a = \alpha R$,so $\alpha = \frac{a}{R}$.The moment of inertia $I$ of a uniform circular disc is $\frac{1}{2}mR^2$.Substituting these into the torque equation:$T \cdot R = (\frac{1}{2}mR^2) \cdot (\frac{a}{R})$$T = \frac{1}{2}ma$  --- $(2)$Substituting $(2)$ into $(1)$:$mg - \frac{1}{2}ma = ma$$mg = ma + \frac{1}{2}ma = \frac{3}{2}ma$$a = \frac{2}{3}g$

$A$ mass $m$ hangs with the help of a string wrapped around a pulley on a frictionless bearing. The pulley has mass $m$ and radius $R$. Assuming the pulley to be a perfect uniform circular disc,the acceleration of the mass $m$,if the string does not slip on the pulley,is

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