Question

A semicircular disc is cut out from a square sheet of side $20\text{cm}$ as shown in the figure. What is the distance of the centre of mass of the new shape from $y-$ axis ? Consider the material to be of uniform density and the origin to be at point $O$.

• A. $9\text{cm}$
• B. $3.28\text{cm}$
• C. $6.28\text{cm}$
• D. $12\text{cm}$

Answer 1

The correct option is C $6.28\text{cm}$


Consider the semicircular part of radius $(R=10\text{cm})$ as a negative mass superimposed on the whole square sheet.

Area of square shape $A_1={20}^2$
Area of semicircular shape $A_2=\frac{\pi \times {10}^2}{2}$

COM of the semicircle $C$ lies at a distance of $\left(\frac{4R}{3\pi }\right)$ in -ve $x$ direction from the centre of semicircle.

$\Rightarrow$ $x$ coordinate of COM of semicircular part
$x_2=20-\frac{4R}{3\pi }=15.75\text{cm}$
and $x$ coordinate of COM of square sheet
$x_1=\frac{20}{2}=10\text{cm}$

For COM of new shape :
$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$=\frac{({20}^2\times 10-\frac{\pi \times {10}^2}{2}\times 15.75)}{({20}^2-\frac{\pi \times {10}^2}{2})}$

$\therefore x_{\text{CM}}=6.28\text{cm}$

$y_{\text{CM}}=10\text{cm}$ due to symmetry.

So, the COM of new shape is at a distance of $6.28\text{cm}$ from the Y axis.