Question

A solid cylinder of mass $20Kg$ has length $1m$ and radius $0.2m$. Then its moment of inertia $(inkg-m^2)$ about its geometrical axis is-

Answer 1

$\because I=\frac{M{R}^2}{2}forsolidcylinderI=\frac{20\times \left(0.2\right)\times \left(0.2\right)}{2}I=0.4kg{m}^2$