A mass m is supported by a massless string wo | vedclass

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(D) For the falling mass $m$,the equation of motion is: $mg - T = ma$  ... $(1)$For the rotation of the hollow cylinder,the torque $	au$ is given by

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$\frac{g}{2}$

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(D) For the falling mass $m$,the equation of motion is: $mg - T = ma$  ... $(1)$For the rotation of the hollow cylinder,the torque $	au$ is given by $	au = I\alpha$. Since the cylinder is hollow,its moment of inertia is $I = mR^2$.The torque is provided by the tension $T$,so $T \cdot R = I\alpha = (mR^2) \cdot \frac{a}{R}$,where $a = R\alpha$ is the linear acceleration of the string.This simplifies to $T = ma$ ... $(2)$Substituting equation $(2)$ into equation $(1)$:$mg - ma = ma$$mg = 2ma$$a = \frac{g}{2}$

$A$ mass $m$ is supported by a massless string wound around a uniform hollow cylinder of mass $m$ and radius $R$. If the string does not slip on the cylinder,with what acceleration will the mass fall on release?

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