Question

Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii $R_1$ and $R_2$ (figure 9 - E5)

• A. $=\frac{4}{\pi }\frac{(R_2^2+R_1^2+R_1{R}_2)}{R_1+R_2}$ above the centre
• B. $=\frac{4}{3\pi }\frac{(R_2^2+R_1^2+R_1{R}_2)}{R_1+R_2}$ above the centre
• C. $=\frac{2}{3\pi }\frac{(R_2^2+R_1^2+2R_1{R}_2)}{R_1+R_2}$ above the centre
• D. None of these

Answer 1

The correct option is B $=\frac{4}{3\pi }\frac{(R_2^2+R_1^2+R_1{R}_2)}{R_1+R_2}$ above the centre

The centre of mass of the plate will be on the symmetrical axis.
$\Rightarrow \overline{Y}cm=\frac{\left(\frac{\pi R_2^2}{2}\right)\left(\frac{4R_2}{3\pi }\right)-\left(\frac{\pi R_1^2}{2}\right)\left(\frac{4R_1}{3\pi }\right)}{\frac{\pi R_2^2}{2}-\frac{\pi R_1^2}{2}}$
$=\frac{\left(\frac{2}{3}\right)R_2^3-\left(\frac{2}{3}\right)R-1^3}{\frac{\pi }{2}2(r_2^2-R_1^2)}=\frac{4}{3\pi }\frac{(R_2-R_1)(R_2^2+R_1^2+R_1{R}_2)}{(R_2-R_1)(R_2+R_1)}$
$=\frac{4}{3\pi }\frac{(R_2^2+R_1^2+R_1{R}_2)}{R_1+R_2}$ above the centre.