Question

A circular lamina is having uniform thickness and radius $a$. A hole of dimeter $a$ is cut out from it as shown in the diagram. The position of centre of mass, of the remining portion will be,

• A. $(\frac{-a}{6},0)$
• B. $(\frac{-a}{3},0)$
• C. $(\frac{-a}{6},\frac{a}{2})$
• D. $(\frac{a}{3},\frac{a}{6})$

Answer 1

The correct option is A $(\frac{-a}{6},0)$

Let, $\rho$ be the density of the material and $t$ be it's thickness.

Now,

$A_1\to$area of the original lamina $=\pi a^2$

$A_2\to$ area of the removed portion

$=\pi {\left(\frac{a}{2}\right)}^2=\frac{\pi a^2}{4}$

Mass of the original lamina = $m_1=\rho V_1=\rho \pi a^{2}t$

Mass of the removed portion = $m_2=\rho V_2=\rho \frac{\pi a^2}{4}t$

Now,

$x_1,y_1$ = coordinates of centre of mass of original lamina $=(0,0)$

$x_2,y_2$ = coordinates of centre of mass of the removed portion $=(\frac{a}{2},0)$

$X_{COM}=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

Considering the removed portion has negative mass, the system now consists of the original lamina (mass $m_1$) of radius $a$ and the negative mass ($-m_2$), of radius $\frac{a}{2}$.

We get, $X_{COM}=\frac{\left(\rho \pi a^{2}t\right)\left(0\right)-\left(\rho \frac{\pi a^2}{4}t\right)\left(\frac{a}{2}\right)}{\rho \pi a^{2}t-\rho \frac{\pi a^2}{4}t}$

= $\frac{\left(0\right)-\left(\frac{a^2}{4}\right)\left(\frac{a}{2}\right)}{a^2-\frac{a^2}{4}}$

$=\frac{-\left(\frac{1}{8}\right)}{\left(\frac{3}{4}\right)}\left(a\right)=-\frac{a}{6}$

And,

$Y_{COM}=0$, as $y_1$ and $y_2$ both are zero.

Therefore, coordinates of $\text{COM}$ of the lamina shown in figure are,

$(-\frac{a}{6},0)$

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

Alternate method:

The problem can also be solved directly using the formula of $\text{COM}$ of remaining portion, which is given by:

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

Similarly,

$y_{COM}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}$

Why this Question? Tip: This is a very important format of problems which is very frequently asked in NEET. These problems are commonly referred as Cavity problems.