Question

A uniform circular disc of radius 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 a distance c 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. $\frac{b^{2}c}{\left(a^2-b^2\right)}$
• C. $\frac{\pi c^2}{a^2-b^2}$
• D. $\frac{c{a}^2}{c^2-b^2}$

Answer 1

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

$x_{cm}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$

$A_1=\pi \times a^2$ {area of whole disc}

$A_2=\pi b^2$ {area of remove d disc}

$x_1=0$ (centre of mass of whole disc)

$x_2=c$ (centre of mass of removed disc)

$x_{cm}=\frac{\left(x{a}^2\right)\left(0\right)-\left(\pi b^2\right)\left(c\right)}{(\pi a^2-\pi b^2)}=\frac{-\pi b^{2}c}{\pi (a^2-b^2)}$

$x_{cm}=\frac{-b^{2}c}{(a^2-b^2)}$

From point 0, centre of mass l is at left at a distance $=\frac{b^{2}c}{(a^2-b^2)}$

Hence, option $\left(B\right)$ is correct answer.