Question

As shown in the figure, a circular sheet is removed from a uniform rectangular sheet. Find the horizontal distance of centre of mass of the remaining sheet from the origin $\text{O}$, after the removal of circular sheet. Use $\pi =\frac{22}{7}$ for calculation.

• A. $10.6\text{cm}$
• B. $2.18\text{cm}$
• C. $15.2\text{cm}$
• D. $4.34\text{cm}$

Answer 1

The correct option is B $2.18\text{cm}$

Considering the circular sheet as the negative mass superimposed on the whole rectangular sheet, will give the new shape.


Area of rectangular sheet $A_1=80\text{cm}\times 40\text{cm}=3200{\text{cm}}^2$
Area of circular sheet $A_2=\pi r^2=\pi \times 100{\text{cm}}^2=100\pi {\text{cm}}^2$

COM of the rectangular sheet is at origin $O$:
$(x_1,y_1)=(0,0)$
COM of the circular sheet $(x_2,y_2)=(20,0)$

Replacing the respective shapes as point masses placed at their COM,
$x$ coordinate of COM of new shape is:
$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$=\frac{(3200\times 0)-(100\pi \times 20)}{\left(3200\right)-\left(100\pi \right)}=\frac{-2000\pi }{3200-100\pi }$

$\therefore x_{\text{CM}}=-\frac{\frac{44000}{7}}{\frac{20200}{7}}=-2.18\text{cm}$

Hence, COM of new shape is at a distance of $2.18\text{cm}$ along$-ve$ x-axis.
Also, the COM of system will lie along the x-axis only, because system is symmetrical about x-axis as shown in figure.