Question

A uniform circular disc of mass M and radius a is given. A part of the radius b is removed from it. The distance between centre of original disc and circular hole's centre is c. The new position of C.M. from O is:

• A. $\frac{bc}{a-b}$
• B. $\frac{b^2}{b-a}$
• C. $\frac{-b^{2}c}{a^2-b^2}$
• D. $\frac{b^{2}c}{a^2+b^2}$

Answer 1

The correct option is C $\frac{-b^{2}c}{a^2-b^2}$

$\sigma =\frac{m}{\pi a^2-\pi b^2}$
$=\frac{m}{\pi (a^2-b^2)}$
$\therefore$ Mass of whole disc
$=\frac{m}{\pi (a^2-b^2)}\times \pi a^2$
$=\frac{m{a}^2}{(a^2-b^2)}$
and mass of hole $=\frac{m}{\pi (a^2-b^2)}\times \pi b^2$
$=\frac{m{b}^2}{(a^2-b^2)}$
$\therefore$ C.M. $\equiv \frac{\frac{m{a}^2}{(a^2-b^2)}\times 0-\frac{m{b}^2\times c}{a^2-b^2}}{m}=\frac{-b^{2}c}{(a^2-b^2)}$