A block of mass $m$ is hanging by a rope tied to a rotating solid disc of mass $M$ and radius $R$ as shown in the figure. If $\alpha$ is the angular acceleration of the disc, then the linear acceleration of the block is ( $\mathrm{g}=$ acceleration due to gravity)

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Aicharya · Accepted Answer

The given situation is shown below

If $a$ be the acceleration of block, then by equation of motion. $ m g-T=m a...(i) $ $\Rightarrow$ We know that, torque $(\tau)$ in the disc $ \tau=I \alpha $ $\Rightarrow T \cdot R=\frac{M R^2}{2} \times \alpha$ $\left(\because \tau=T \cdot R\right.$ and $\left.I=\frac{M R^2}{2}\right)$ $ T=\frac{M R \alpha}{2}...(ii) $ From Eqs. (i) and (ii), we get $ \begin{aligned} & m g-\frac{M R \alpha}{2}=m a \\ \Rightarrow & m g-\frac{M R}{2} \cdot \frac{a}{R}=m a \Rightarrow m g-\frac{m a}{2}=m a \\ \Rightarrow & m g=\frac{m a}{2}+m a \Rightarrow m g=\left(\frac{M}{2}+m\right) a \\ \Rightarrow & a=\frac{m g}{m+\frac{M}{2}} \end{aligned} $

A block of mass $m$ is hanging by a rope tied to a rotating solid disc of mass $M$ and radius $R$ as shown…

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