From a hemisphere of radius $\mathrm{R}$, a cone of base radius $\frac{\mathrm{R}}{2}$ and height $\mathrm{R}$ is removed as shown in figure. Calculate the height of centre of mass of the remaining object in cm if $\mathrm{R}=28 \mathrm{cm}$.

Let $\rho$ be the mass density, then mass of the removed cone is

${\mathrm{m}}_1=\rho \left(\frac{1}{3}\pi {\left(\frac{\mathrm{R}}{2}\right)}^2\mathrm{R}\right)=\frac{\pi {\mathrm{R}}^3\rho }{12}$

Also, centre of mass of cone is at height $\frac{\mathrm{R}}{4}$

Mass of remaining hemisphere is

${\mathrm{m}}_2=\left(\frac{2}{3}\pi {\mathrm{R}}^3-\frac{\pi {\mathrm{R}}^3}{12}\right)\rho =\frac{7\pi {\mathrm{R}}^3\rho }{12}$

Let its centre of mass be at a height y Since common centre of mass of ${\mathrm{m}}_1+{\mathrm{m}}_2$ (i.e., a hemisphere) is at a height  $\frac{3R}{8}$ so we get

$\frac{3R}{8}=\frac{\rho \left(\frac{\pi R^3}{12}\right)\frac{R}{4}+\rho \left(\frac{7}{12}\pi R^3\right)y}{\rho \left(\frac{2}{3}\pi R^3\right)}$

$\Rightarrow \frac{2}{3}\left(\frac{3\mathrm{R}}{8}\right)=\frac{\mathrm{R}}{48}+\frac{7\mathrm{y}}{12}\Rightarrow \frac{7\mathrm{y}}{12}=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{48}=\frac{11\mathrm{R}}{48}$

$\Rightarrow \mathrm{y}=\frac{11\mathrm{R}}{28}=\frac{11}{28}\left(28\right)=11\mathrm{cm}$