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

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Question

(A) Let $a$ be the linear acceleration of the mass $m$ and $\alpha$ be the angular acceleration of the cylinder. Since the string does not slip,$a = R

Vedclass · Accepted Answer

$\frac{g}{2}$

Vedclass · Answer

(A) Let $a$ be the linear acceleration of the mass $m$ and $\alpha$ be the angular acceleration of the cylinder. Since the string does not slip,$a = R\alpha$,which implies $\alpha = \frac{a}{R}$.For the falling mass $m$,the equation of motion is: $mg - T = ma$ . . . $(i)$For the rotation of the hollow cylinder,the torque $	au = I\alpha$ is provided by the tension $T$:$T 	imes R = I\alpha$Since the moment of inertia $I$ of a hollow cylinder about its axis is $mR^2$,we have:$T 	imes R = (mR^2) 	imes \left( \frac{a}{R} ight)$$T = ma$ . . . (ii)Substituting equation (ii) into equation $(i)$:$mg - ma = ma$$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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