Find the centre of mass of a uniform plate having semicircular inner and outer boundaries of radii $R_1$ and $R_2$. 
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The centre of mass of the blate will be on the symmetrical axis.
$\Rightarrow\bar{\text{y}}_{\text{cm}}=\frac{\Big(\frac{\pi\text{R}_2^2}{2}\Big)\Big(\frac{\text{4R}_2}{3\pi}\Big)-\Big(\frac{\pi\text{R}_1^2}{2}\Big)\Big(\frac{4\text{R}_1}{3\pi}\Big)}{\frac{\pi\text{R}_2^2}{2}-\frac{\pi\text{R}_1^2}{2}}$
$=\frac{\Big(\frac{2}{3}\Big)\text{R}_2^3-\Big(\frac{2}{3}\Big)\text{R}_1^3}{\frac{\pi}{2}\big(\text{R}_2^2-\text{R}_1^2\big)}=\frac{4}{3\pi}\frac{(\text{R}_2-\text{R}_1)\big(\text{R}^2_2+\text{R}_1^2+\text{R}_1\text{R}_2\big)}{(\text{R}_2-\text{R}_1)(\text{R}_2+\text{R}_1)}$
$=\frac{4}{3\pi}\frac{\big(\text{R}_2^2+\text{R}_1^2+\text{R}_1\text{R}_2\big)}{\text{R}_1+\text{R}_2}$ above the centre.
$\Rightarrow\bar{\text{y}}_{\text{cm}}=\frac{\Big(\frac{\pi\text{R}_2^2}{2}\Big)\Big(\frac{\text{4R}_2}{3\pi}\Big)-\Big(\frac{\pi\text{R}_1^2}{2}\Big)\Big(\frac{4\text{R}_1}{3\pi}\Big)}{\frac{\pi\text{R}_2^2}{2}-\frac{\pi\text{R}_1^2}{2}}$
$=\frac{\Big(\frac{2}{3}\Big)\text{R}_2^3-\Big(\frac{2}{3}\Big)\text{R}_1^3}{\frac{\pi}{2}\big(\text{R}_2^2-\text{R}_1^2\big)}=\frac{4}{3\pi}\frac{(\text{R}_2-\text{R}_1)\big(\text{R}^2_2+\text{R}_1^2+\text{R}_1\text{R}_2\big)}{(\text{R}_2-\text{R}_1)(\text{R}_2+\text{R}_1)}$
$=\frac{4}{3\pi}\frac{\big(\text{R}_2^2+\text{R}_1^2+\text{R}_1\text{R}_2\big)}{\text{R}_1+\text{R}_2}$ above the centre.
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