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

$ I_1=\frac{m_1 R^2}{2} I_2=\frac{m_2(R / 2)^2}{2}$
$ \frac{I_1}{I_2}=\frac{4 m _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$