Question

A solid cylinder of uniform density of radius $2cm$ has mass of $50g$ . If its length is $10cm$ . Calculate its moment of inertia about.
(i) its own axis of rotation passing through the centre,
(ii) an axis passing through its centre and perpendicular to its length.

Answer 1

(i)

The moment of inertia of the cylinder about its own axis passing through its centre is given as,

$I=\frac{M{R}^2}{2}$

$=\frac{50\times {10}^{-3}\times {\left(2\times {10}^{-2}\right)}^2}{2}$

$={10}^{-5}kg{m}^2$

(ii)

The moment of inertia of the cylinder about an axis passing through its centre and perpendicular to the length is given as,

$I=\frac{M{R}^2}{4}+\frac{M{l}^2}{12}$

$=\frac{50\times {10}^{-3}\times {\left(2\times {10}^{-2}\right)}^2}{4}+\frac{50\times {10}^{-3}\times {\left(10\times {10}^{-2}\right)}^2}{12}$

$=4.67\times {10}^{-5}kg{m}^2$