Question

A uniform circular disc of radius $a$ is taken. A circular portion of radius $b$ has been removed from its as shown in the figure, If the centre of hole is at $c$ distance from the centre of the disc, the distance $x_2$ of the centre of mass of the remaining part from the initial centre of mass $O$ is given by:

• A. $\frac{\pi b^2}{(a^2-c^2)}$
• B. $\frac{c{b}^2}{(a^2-b^2)}$
• C. $\frac{{\pi }^2}{(a^2-b^2)}$
• D. $\frac{c{a}^2}{(c^2-b^2)}$

Answer 1

The correct option is B $\frac{c{b}^2}{(a^2-b^2)}$

Surface mass density $\sigma =\frac{M}{\pi a^2}$

Mass of Removed portion $M^{\prime }=\sigma \times \pi b^2$

$=\frac{M}{\pi a^2}\times \pi b^2$

$M^{\prime }=\frac{M{b}^2}{a^2}$

Taking 0 as origin

$\left[r_{com}\right]=\frac{M\times 0-M^{\prime }\times C}{M-M^{\prime }}=\frac{-\frac{M{b}^2}{a^2}\times C}{M-\frac{M{b}^2}{a^2}}$

$=\frac{-\frac{M{b}^{2}C}{a^2}}{\frac{M{a}^2-M{b}^2}{a^2}}$

$=\frac{-b^{2}C}{a^2-b^2}$

$x_2=|r_{com}|=\frac{b^{2}C}{a^2-b^2}$