Question

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

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

$\frac{1}{3}$

Answer 2

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 $CM$ of new disc is at point $O _{2}$ where, $O_{1} O _{2}=\alpha R$ (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) $x_{ CM }=\alpha R =\frac{M \times 0-\left(\frac{M}{4}\right) \times R}{M-\frac{M}{4}}$ $=-\frac{R}{3}$ $\Rightarrow \alpha=\frac{1}{3}$