Question

What is the $y$ coordinate of the centre of mass of a semi-circular disc as shown in the figure? The smaller disc has been cut from the larger one. Consider the discs to have uniform density. Consider the axes and origin as shown in figure.

• A. $19.8\text{cm}$
• B. $25\text{cm}$
• C. $29.3\text{cm}$
• D. $8\text{cm}$

Answer 1

The correct option is A $19.8\text{cm}$

Considering the removed semicircular part $(R=20\text{cm})$ as the negative mass superimposed on the bigger semicircular part $(R=40\text{cm})$ will give the required shape of disc.


$A_1=\frac{\pi \times {40}^2}{2}$ &
$A_2=\frac{\pi \times {20}^2}{2}$

COM of the semicircular disc lies at a distance of $\left(\frac{4R}{3\pi }\right)$ from the centre along $y-$ axis.

Also, the semicircular disc is symmetrical about the y axis. So, the COM will lie on the y- axis. i.e $x_{CM}=0$

$\Rightarrow y_{\text{CM}}=\frac{A_1{y}_1-A_2{y}_2}{A_1-A_2}$
$=\frac{(\pi \times \frac{{40}^2}{2}\times \frac{4\times 40}{3\pi }-\pi \times \frac{{20}^2}{2}\times \frac{4\times 20}{3\pi })}{(\frac{\pi \times {40}^2}{2}-\pi \times \frac{{20}^2}{2})}$
$=\frac{42666.666-5333.33}{\pi (800-200)}$
$\therefore y_{\text{CM}}=19.8\text{cm}$