Question

A uniform square sheet has a side length of $2R$. A circular plate of radius $\left(\frac{R}{2}\right)$ is cut off from the square sheet as shown in the figure. Find the center of mass of the remaining portion when $R=8\text{cm}$.

• A. $4.6\text{cm}$
• B. $5.8\text{cm}$
• C. $3.95\text{cm}$
• D. $0.97\text{cm}$

Answer 1

The correct option is D $0.97\text{cm}$

Let $A_1\to$ mass of the square plate.
$A_2\to$ mass of the removed circular plate.
$A_1=(2R{)}^2=4R^2$
$A_2=\pi \left({\frac{R}{2}}^2\right)=\frac{\pi R^2}{4}$

Taking origin at centre of square plate,
x coordinate of COM of square plate $x_1=0$ and
x coordinate of COM of circular plate $x_2=\frac{R}{2}$

x coordinate of $\text{COM}$ of remaining portion:-
$X_{COM}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$=\frac{\left(4R^2\right)\left(0\right)-\left(\frac{\pi R^2}{4}\right)\left(\frac{R}{2}\right)}{4R^2-\frac{\pi R^2}{4}}$
$=\frac{-\pi R}{2(16-\pi )}$

Putting $R=8\text{cm}$ (given)
$X_{COM}=\frac{-\pi \times 8}{2(16-\pi )}=-0.97\text{cm}.$
$Y_{COM}$ will be zero due to symmetry.

COM of the remaining portion will be $0.97\text{cm}$ left of the center of the square.