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(D) For the mass $m$,the equation of motion is: $mg - T = ma$ $(1)$For the cylinder of mass $M$ and radius $R$,the torque equation is: $	au = I\alpha

Vedclass · Accepted Answer

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

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(D) For the mass $m$,the equation of motion is: $mg - T = ma$ $(1)$For the cylinder of mass $M$ and radius $R$,the torque equation is: $	au = I\alpha$,where $I = \frac{1}{2}MR^2$ and $\alpha = \frac{a}{R}$.So,$TR = (\frac{1}{2}MR^2)(\frac{a}{R}) \implies T = \frac{1}{2}Ma$ $(2)$Substituting $(2)$ into $(1)$: $mg - \frac{1}{2}Ma = ma \implies mg = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$.Thus,the linear acceleration is $a = \frac{2mg}{M + 2m}$.The angular acceleration is $\alpha = \frac{a}{R} = \frac{2mg}{R(M + 2m)}$.Since the system starts from rest,the angular velocity at time $t$ is $\omega = \alpha t = \frac{2mgt}{R(M + 2m)}$.

In the figure shown,a mass $m$ is attached to a light string which is wrapped around a solid cylinder of mass $M$ and radius $R$. The system starts from rest at $t = 0$. If friction is negligible,what will be the angular velocity at time $t$?

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