Question

One quadrant of a circular disc is removed from the original disc of radius $5\text{cm}$. Take reference axes and origin $O$ as shown in the figure. Find the sum of the magnitudes (in $\text{cm}$) of the $x$ and $y$ coordinates of the $\text{COM}$ of the new shape. Consider the material to be of uniform density.

• A. $6\text{cm}$
• B. $5.12\text{cm}$
• C. $4.24\text{cm}$
• D. $1.41\text{cm}$

Answer 1

The correct option is D $1.41\text{cm}$

Considering the removed quadrant as the negative mass superimposed on the whole circular disc.

Area of circular disc $A_1=\pi \times 5^2{\text{cm}}^2$
and Area of disc quadrant $A_2=\frac{\pi \times 5^2}{4}{\text{cm}}^2$


COM of the circular disc $(x_1,y_1)=(0,0)$

COM of a quadrant of radius $r$ is at distance $\frac{4r}{3\pi }$ from both axis.

$\therefore$ COM of the quadrant $(x_2,y_2)=(\frac{4\times 5}{3\pi },\frac{4\times 5}{3\pi })$

Applying the formula for coordinates of $\text{COM}$ of system:

$x_{\text{CM}}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$
$=\frac{(\pi \times 5^2)\times 0-\frac{\pi \times 5^2}{4}\times \frac{4\times 5}{3\pi }}{\pi \times 5^2-\left(\frac{\pi \times 5^2}{4}\right)}$
$\therefore x_{\text{CM}}=-0.707\text{cm}$

Simillarly,
$y_{\text{CM}}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}=-0.707\text{cm}$

Hence sum of magnitudes of the coordinates of centre of mass of the new shape $=0.707+0.707=1.414\text{cm}$