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(D) Let the acceleration of the mass be $a$ and the angular acceleration of the disc be $\alpha$.Since the string moves without slipping on the disc,$

Vedclass · Accepted Answer

$\sqrt{\frac{4mgh}{2m + M}}$

Vedclass · Answer

(D) Let the acceleration of the mass be $a$ and the angular acceleration of the disc be $\alpha$.Since the string moves without slipping on the disc,$a = R\alpha$.For the mass $m$,the equation of motion is $mg - T = ma$,so $T = m(g - a)$.For the disc,the torque equation is $	au = I\alpha$,where $	au = TR$ and $I = \frac{1}{2}MR^2$.Thus,$TR = (\frac{1}{2}MR^2)(\frac{a}{R})$,which simplifies to $T = \frac{1}{2}Ma$.Equating the two expressions for $T$: $m(g - a) = \frac{1}{2}Ma$.$mg - ma = \frac{1}{2}Ma \implies mg = a(m + \frac{M}{2}) \implies a = \frac{2mg}{2m + M}$.Using the kinematic equation $v^2 = u^2 + 2as$ with $u = 0$ and $s = h$:$v^2 = 2(\frac{2mg}{2m + M})h = \frac{4mgh}{2m + M}$.Therefore,$v = \sqrt{\frac{4mgh}{2m + M}}$.

$A$ massless string is wrapped around a disc of mass $M$ and radius $R$. Another end is tied to a mass $m$ which is initially at height $h$ from ground level as shown in the figure. If the mass is released,then its velocity while touching the ground level will be

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