A mass $M$ is supported by a massless string wound around a uniform cylinder of mass $M$ and radius $R$. On releasing the mass from rest, it will fall with acceleration

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Aicharya · Accepted Answer

The given situation is shown in the figure,

We know that, torque, $\tau=I \alpha$ If $T$ is the tension in the string, then $ \tau=T R $ From Eqs. (i) and (ii), we get $ I \alpha=T R $ Under no slipping condition, $ \alpha=\frac{a}{R} $ Hence, $\quad I \cdot \frac{a}{R}=T R \Rightarrow T=\frac{I \cdot a}{R^2}$ For cylinder, $I=\frac{M R^2}{2}$ $ \therefore \quad T=\frac{M R^2}{2} \cdot \frac{a}{R^2} \Rightarrow T=\frac{m a}{2} $ For the motion of the block, $ \begin{aligned} & M g-T=M a \\ & \Rightarrow \quad M g-\frac{M a}{2}=M a \\ & \Rightarrow \quad M g=\frac{3 M a}{2} \Rightarrow a=\frac{2 g}{3} \\ & \end{aligned} $

A mass $M$ is supported by a massless string wound around a uniform cylinder of mass $M$ and radius $R$. On…

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