In the following figure,a body of mass m is t | vedclass

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(D) For the falling mass $m$,the equation of motion is: $mg - T = ma$ (where $T$ is the tension in the string and $a$ is the linear acceleration).For

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$\frac{2mgt}{R(M + 2m)}$

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(D) For the falling mass $m$,the equation of motion is: $mg - T = ma$ (where $T$ is the tension in the string and $a$ is the linear acceleration).For the rotating cylinder,the torque equation is: $	au = I\alpha = TR$,where $I = \frac{1}{2}MR^2$ is the moment of inertia and $\alpha = \frac{a}{R}$ is the angular acceleration.Substituting $\alpha = \frac{a}{R}$ into the torque equation: $TR = (\frac{1}{2}MR^2)(\frac{a}{R}) \implies T = \frac{1}{2}Ma$.Substituting $T$ into the first equation: $mg - \frac{1}{2}Ma = ma \implies mg = a(m + \frac{M}{2}) \implies a = \frac{mg}{m + M/2} = \frac{2mg}{2m + M}$.Since the system starts from rest $(u = 0)$,the linear velocity at time $t$ is $v = at = \frac{2mgt}{2m + M}$.The angular velocity $\omega$ of the cylinder is given by $\omega = \frac{v}{R} = \frac{2mgt}{R(M + 2m)}$.

In the following figure,a body of mass $m$ is tied at one end of a light string and this string is wrapped around a solid cylinder of mass $M$ and radius $R$. At the moment $t = 0$,the system starts moving. If the friction is negligible,the angular velocity at time $t$ would be

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