Question

Disc of radius $\frac{a}{2}$ is cut out from a disc of radius $a$. Find $x$ co-ordinate of centre of mass from origin.

• A. $\frac{5a}{6}$
• B. $\frac{a}{6}$
• C. $\frac{a}{3}$
• D. $\frac{a}{2}$

Answer 1

The correct option is A $\frac{5a}{6}$


Centre of mass of disc before removal of small disc from it,
${\overrightarrow{r}}_1=a\hat{i}+0\hat{j}$
Centre of mass of small disc removed,
${\overrightarrow{r}}_2=\frac{3a}{2}\hat{i}+0\hat{j}$
We know that for laminar bodies,
$\overrightarrow{r}=\frac{A_1{\overrightarrow{r}}_1-A_2{\overrightarrow{r}}_2}{A_1-A_2}$
$\Rightarrow \overrightarrow{r}=\frac{\pi a^2(a\hat{i}+0\hat{j})-\frac{\pi a^2}{4}(\frac{3a}{2}\hat{i}+0\hat{j})}{\pi a^2-\frac{\pi a^2}{4}}=\frac{5a}{6}\hat{i}+0\hat{j}$
Hence, centre of mass of the disc when small disc is removed is at $x=\frac{5a}{6}$

$\begin{array}{c}\text{Why this question ?}\\ \text{Concept: When some portion of a solid body is removed,}\\ \text{the centre of mass shifts to direction where more mass is concentrated.}\\ \text{Caution: Above used formula is only applicable for laminar bodies.}\end{array}$