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 at a distance of $\alpha R$ from the centre of the bigger disc. The value of $\alpha$ is _______.

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

Answer 1

The correct option is A $\frac{1}{3}$

Let the mass per unit area of the disc be $\sigma$

radius is $2R$

then mass of the original disc is $M=\pi (2R{)}^2\sigma =4\pi R^2\sigma$

Radius of the disc cut is R

them mass of the smaller disc $M^{{}^{\prime }}=\pi R^2\sigma =\frac{M}{4}$

Let $O$ and $O^{{}^{\prime }}$ be the centers of the original disc and the disc cut off from the original.

It is given that $O{O}^{{}^{\prime }}=R$

After the smaller disc has been cut from the original, the remaining portion is considered to be system of two masses. The two masses are M and $-M^{{}^{\prime }}=-\frac{M}{4}$

The negative sign indicates the mass has been removed from the original disc.

Let x be the distance through which the center of mass of the remaining portion shifts from point O

the relation between the center of masses of two masses as

$x=\frac{m_1{r}_1+m_2{r}_2}{m_1+m_2}$

$x=\frac{M\times 0-M^{{}^{\prime }}\times R}{M-M^{{}^{\prime }}}$

$x=\frac{M\times 0-\frac{M}{4}\times R}{M-\frac{M}{4}}$

$x=\frac{-\frac{MR}{4}}{\frac{3M}{4}}=\frac{-R}{3}$

therefore $\alpha =\frac{1}{3}$