Physics - Rotation Question with Solution | TestHub

Let the density of the material of the cylinder be $\rho$.

The mass of the original solid cylinder is $M_1=\rho \pi R^{2}L$.
The moment of inertia of the original solid cylinder about its axis is:
$I_1=\frac{1}{2}{M}_1{R}^2=\frac{1}{2}(\rho \pi R^{2}L)R^2=\frac{1}{2}\rho \pi R^{4}L$

The carved out portion is a solid cylinder of radius $R/2$ and length $L/2$. Its mass is:
$M_2=\rho \pi {\left(\frac{R}{2}\right)}^2\left(\frac{L}{2}\right)=\frac{1}{8}\rho \pi R^{2}L$

The moment of inertia of this carved out portion about the same axis is:
$I_2=\frac{1}{2}{M}_2{\left(\frac{R}{2}\right)}^2=\frac{1}{2}\left(\frac{1}{8}\rho \pi R^{2}L\right)\left(\frac{R^2}{4}\right)=\frac{1}{64}\rho \pi R^{4}L$

Taking the ratio of $I_1$ to $I_2$:
$\frac{I_1}{I_2}=\frac{\frac{1}{2}\rho \pi R^{4}L}{\frac{1}{64}\rho \pi R^{4}L}=32$

Thus, the ratio is $32:1$.