Question

A solid cylinder is made of radius R and height 3R having mass density $\rho$ . Now two half spheres of radius R are removed from both ends. The moment of inertia of remaining portion about axis ZZ' can be calculated as $\frac{29}{6K}\pi R^5\rho$ . Find K.

Answer 1

Answer: 5

Effectively we have removed a solid sphere out of the solid cylinder.
The mass and moment of inertia of the entire cylinder are
$M_{cylinder}=\left(\pi R^2\times 3R\right)\rho$
$I_{cylinder}=\frac{1}{2}\left(3\pi R^3\rho \right)R^2$
Similarly, the mass and moment of inertia of the solid sphere are
$M_{sphere}=\frac{4}{3}\pi R^3\rho$
$I_{sphere}=\frac{2}{5}\left(\frac{4}{3}\pi R^3\rho \right)R^2$
$I_{remaining}=I_{cylinder}-I_{sphere}=\frac{29}{30}\pi R^5\rho$
$I_{remaining}=\frac{29}{6\times 5}\pi R^5\rho$
$\Rightarrow K=5$