Question

Find the distance of centre of mass of a uniform plate having semicircular inner and outer boundaries of radii $a$ and $b$ from the centre $\text{O}$.

• A. $\frac{2(b^2+a^2+ab)}{3\pi (a+b)}$
• B. $\frac{4(b^2+a^2+ab)}{3\pi (a+b)}$
• C. $\frac{3\pi (b^2+a^2+ab)}{4(a+b)}$
• D. $\frac{4(b^2+a^2+ab)}{3\pi (a+b)}$

Answer 1

The correct option is B $\frac{4(b^2+a^2+ab)}{3\pi (a+b)}$

Here, we can assume that a semicircular disc of radius $a$ is removed from the semicircular disc of radius $b$.

We know that coordinates of COM of semicircular disc of radius $r$ is given by

$(x,y)=(0,\frac{4r}{3\pi })$

Here,
Area of plate of radius $a=A_2=\frac{\pi a^2}{2}$

Area of plate of radius $b=A_1=\frac{\pi b^2}{2}$

$\therefore y_{com}$ of given plate is given by

$y_{com}=\frac{A_1{x}_1-A_2{x}_2}{A_1-A_2}$

$y_{com}=\frac{(\frac{\pi b^2}{2}\times \frac{4b}{3\pi })-(\frac{\pi a^2}{2}\times \frac{4a}{3\pi })}{(\frac{\pi b^2}{2}-\frac{\pi a^2}{2})}$

$y_{com}=\frac{4(b^3-a^3)}{3\pi (b^2-a^2)}=\frac{4(b-a)(b^2+a^2+ba)}{3\pi (b-a)(b+a)}$

$\therefore y_{com}=\frac{4(b^2+a^2+ab)}{3\pi (b+a)}$

From symmetry, we can say that $x_{com}=0$

$\therefore$Distance of COM of the given surface from centre
$\text{O}$ is $\frac{4(b^2+a^2+ab)}{3\pi (a+b)}$

Therefore, option $\left(B\right)$ is correct.