Question

A circular portion of radius $R/4$ has been removed from the uniform disc of radius $R$, centred at $A$ as shown in figure. Then, centre of mass of the remaining portion of the uniform disc is:

• A. $\frac{R}{20}$ to the left of $A$
• B. $\frac{R}{12}$ to the left of $A$
• C. $\frac{R}{20}$ to the right of $A$
• D. $\frac{R}{12}$ to the right of $A$

Answer 1

The correct option is A $\frac{R}{20}$ to the left of $A$


Let origin $(0,0)$ be at the centre of complete disc i.e $A$
Let $\text{C.O.M}$ of remaining portion of mass $M$ be at $(-d,0)$
Due to symmetry about $x-$ axis, the $\text{COM}$ will lie along $x-$ axis only.
$\therefore \text{COM}$ of removed portion of mass $m$ is at $(R-\frac{R}{4},0)$

Let surface mass density of disc be $\rho {\text{kg/m}}^2$
$\therefore m=\rho \times \pi \times {\left(\frac{R}{4}\right)}^2=\frac{\rho \pi R^2}{16}\text{kg}$
and $M=\rho \pi R^2-m=\rho \pi R^2-\frac{\rho \pi R^2}{16}$
$\therefore M=\frac{15}{16}\rho \pi R^2$
$\text{COM}$ of combined $(M+m)$ will lie at geometric centre $A$ i.e $(0,0)$.
i.e $x_{CM}=\frac{M{x}_1+m{x}_2}{M+m}$
$0=\frac{M\times (-d)+m\times \frac{3R}{4}}{M+m}$
$\Rightarrow d=\frac{m\times \frac{3R}{4}}{M}=\frac{\left(\frac{\rho \pi R^2}{16}\right)\times \frac{3R}{4}}{\left(\frac{15\rho \pi R^2}{16}\right)}$
$\therefore d=\frac{3R}{4\times 15}=\frac{R}{20}$
(+ve sign indicated COM will lie to the left of origin)
Hence, $\text{COM}$ of remaining portion of disc is at distance $\frac{R}{20}$ to the left of $A$.