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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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Verified Answer
The correct answer is:
$\frac{m}{m+\frac{M}{2}} g$
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
$$
\begin{aligned}
& m g-\frac{M R \alpha}{2}=m a \\
\Rightarrow & m g-\frac{M R}{2} \cdot \frac{a}{R}=m a \Rightarrow m g-\frac{m a}{2}=m a \\
\Rightarrow & m g=\frac{m a}{2}+m a \Rightarrow m g=\left(\frac{M}{2}+m\right) a \\
\Rightarrow & a=\frac{m g}{m+\frac{M}{2}}
\end{aligned}
$$
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
$$
\begin{aligned}
& m g-\frac{M R \alpha}{2}=m a \\
\Rightarrow & m g-\frac{M R}{2} \cdot \frac{a}{R}=m a \Rightarrow m g-\frac{m a}{2}=m a \\
\Rightarrow & m g=\frac{m a}{2}+m a \Rightarrow m g=\left(\frac{M}{2}+m\right) a \\
\Rightarrow & a=\frac{m g}{m+\frac{M}{2}}
\end{aligned}
$$
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