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

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(D) Let $a$ be the linear acceleration of the mass $m$ and $\alpha$ be the angular acceleration of the cylinder. The relationship between linear and a

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

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(D) Let $a$ be the linear acceleration of the mass $m$ and $\alpha$ be the angular acceleration of the cylinder. The relationship between linear and angular acceleration is $a = R\alpha$,so $\alpha = \frac{a}{R}$.For the falling mass $m$,the equation of motion is: $mg - T = ma$ ... $(1)$For the rotating cylinder,the torque equation is: $	au = I\alpha$,where $I = \frac{1}{2}MR^2$. Thus,$TR = \frac{1}{2}MR^2 \alpha$.Substituting $\alpha = \frac{a}{R}$,we get $TR = \frac{1}{2}MR^2 (\frac{a}{R}) = \frac{1}{2}MRa$,which simplifies to $T = \frac{1}{2}Ma$ ... $(2)$Substituting $T$ from $(2)$ into $(1)$:$mg - \frac{1}{2}Ma = ma$$mg = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$$a = \frac{2mg}{M + 2m}$Since $\alpha = \frac{a}{R}$,we have $\alpha = \frac{2mg}{R(M + 2m)}$.Starting from rest at $t=0$,the angular velocity $\omega$ at time $t$ is given by $\omega = \omega_0 + \alpha t = 0 + \alpha t = \frac{2mgt}{R(M + 2m)}$.

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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