Question

A square of side $\frac{a}{2}$ is removed from a disc having radius $a$. Find centre of mass of remaining portion.

• A. $x=-\frac{2a}{\pi }$
• B. $x=-\frac{a}{8\pi -2}$
• C. $x=-\frac{4a}{3\pi }$
• D. $x=-\frac{a}{3\pi -4}$

Answer 1

The correct option is B $x=-\frac{a}{8\pi -2}$

Before removal of square from the disc, the centre of mass will be at origin $\text{O}$.

$\therefore \overrightarrow{r_1}=0\hat{i}+0\hat{j}$

Now, centre of mass of the square will be its geometrical centre which lie on the $\text{x-axis}$ at $x=\frac{a}{2}$.

$\therefore \overrightarrow{r_2}=\frac{a}{2}\hat{i}+0\hat{j}$

The centre of mass of the remaining portion after removal of square,

$\overrightarrow{r}=\frac{A_1\overrightarrow{r_1}-A_2\overrightarrow{r_2}}{A_1-A_2}$

where, $A_1$ is area of the disc and $A_2$ is area of the square.

Substitituting the values in the above equation,

$\Rightarrow \overrightarrow{r}=\frac{\pi a^2(0\hat{i}+0\hat{j})-\frac{a^2}{4}(\frac{a}{2}\hat{i}+0\hat{j})}{\pi a^2-\frac{a^2}{4}}$

$\Rightarrow \overrightarrow{r}=\frac{\frac{-a^3}{8}\hat{i}}{\frac{4\pi a^2-a^2}{4}}$

$\Rightarrow \overrightarrow{r}=-\frac{a\hat{i}}{8\pi -2}$

Hence, the centre of mass will lie at $(x=-\frac{a}{8\pi -2},y=0)$