Question

A uniform solid cylinder with radius $R$ and length $L$ has moment of inertia $I_1$, about the axis of the cylinder. A concentric solid cylinder of radius $R^{\prime }=\frac{R}{2}$ and length $L^{\prime }=\frac{L}{2}$ is carved out of the original cylinder. If $I_2$ is the moment of inertia of the carved out portion of the cylinder then $\frac{I_1}{I_2}$ $=$_______
(Both $I_1$ and $I_2$ are about the axis of the cylinder)

Answer 1

Answer: 32

$I_1=\frac{m_1{R}^2}{2}{I}_2=\frac{m_2(R/2{)}^2}{2}$
$\frac{I_1}{I_2}=\frac{4m_1}{m_2}=\frac{4\cdot \rho \pi R^2\ell }{\rho \cdot \frac{\pi R^2}{4}\times \frac{\ell }{2}}\Rightarrow \frac{I_1}{I_2}=32$