Question

A solid cylinder of mass 20 kg has length 1 m and radius 0.2 m. Then its moment of inertia ( in kg- $m^2$ about an axis passing through its centre of gravity and perpendicular to its length is.

Answer 1

The moment of inertia of cylinder is :-

$I=\frac{M{L}^2}{12}+\frac{m{r}^2}{4}$
$=\frac{20\times 1^2}{12}+\frac{20\times (0.2{)}^2}{4}$

$\Rightarrow I=\frac{28}{15}kg{m}^2$