Question

A circular disc of radius $R$ is removed from a bigger circular disc of radius $2R$, such that the circumferences of the discs coincide. The centre of mass of the new disc is $\alpha/R$ from the centre of the bigger disc. The value of $\alpha$ is

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

Answer 1

Answer: (B)

$\frac{1}{3}$

Answer 2

In figure,, O is the centre of circular disc of radius $2R$ and mass $M.C_1$ is centre of disc of radius R, which is removed. If $s$ is mass per unit area of disc, then $M=\pi\left(2R\right)^{2}\sigma$ Mass of disc removed, $M_{1}=\pi R^{2}\sigma=\frac{1}{4}M$ Mass of remaining disc, $M_{2}=M-M_{1}$ $=M-\frac{1}{4}M=\frac{3}{4}M$ Let centre of mass of remining disc be at $C_{2}$ where $OC_{2} = x$ As $M_{1}\times OC_{1}=M_{2}\times OC_{2}$ $\therefore \frac{M}{4}R=\frac{3M}{4}x$ $x=\frac{R}{3}=\alpha R \,\therefore \alpha=\frac{1}{3}$