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A circular disc of radius $R$ is removed from a bigger circular disc of radius $2 R$ such that the circumferences of the discs touch. The centre of mass of the new disc is at a distance $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is
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The correct answer is:
$\frac{1}{3}$
Let centre $O_1$ of disc be the origin. Due to symmetry the centre of mass of remaining part will lie at $x$-axis. Let $\mathrm{CM}$ of new disc is at point $\mathrm{O}_2 \quad$ where, $\mathrm{O}_1 \mathrm{O}_2=\alpha \mathrm{R}$ (given).
here mass of cut off portion,
$m=\frac{\pi(R)^2 \cdot M}{\pi(2 R)^2}=\frac{M}{4}$
and position of its centre of mass, $O_1 O_1=R$ Hence, for remaining part (new disc)
$\begin{aligned}
x_{\mathrm{CM}}=\alpha R & =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} \\
& =-\frac{R}{3} \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$
here mass of cut off portion,
$m=\frac{\pi(R)^2 \cdot M}{\pi(2 R)^2}=\frac{M}{4}$
and position of its centre of mass, $O_1 O_1=R$ Hence, for remaining part (new disc)
$\begin{aligned}
x_{\mathrm{CM}}=\alpha R & =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}} \\
& =-\frac{R}{3} \Rightarrow \alpha=\frac{1}{3}
\end{aligned}$
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