Question

The moment of inertia of a solid cylinder about its own axis is the same as its moment of inertia about an axis passing through its centre of gravity and perpendicular to its length. The relation between its length L and radius R is

Answer 1

$L^2=3R^2$

as MI of a solid cylinder about its own axis is $\frac{1}{2}$ $M{R}^2$

and MI about an axis passing through its center of gravity and perpendicular

$M(\frac{L^2}{12}+\frac{R^2}{4})$

$\frac{1}{2}M{R}^2=M(\frac{L^2}{12}+\frac{R^2}{4})$

$\frac{R^2}{2}=\frac{L^2}{12}+\frac{R^2}{4}$

$\frac{R^2}{2}-\frac{R^2}{4}=\frac{L^2}{12}$

$L^2=3R^2$

$L^{\frac{1}{2}}=3^{\frac{1}{2}}R$