Question

The moment of inertia of a uniform rod of length $2l$ and mass $m$ about an axis $(X-X)$ passing through its centre and inclined at an angle $\alpha ={30}^{\circ }$ is

• A. $\frac{m{l}^2}{12}$
• B. $\frac{m{l}^2}{3}$
• C. $\frac{m{l}^2}{6}$
• D. $\frac{m{l}^2}{4}$

Answer 1

The correct option is A $\frac{m{l}^2}{12}$



Mass per unit length of the rod $\lambda =\frac{m}{2l}$,
$dm=\lambda dx$
The perpendicular distance of the element from the given axis is $x\sin \alpha$

The desired moment of inertia is

$dI=\left(dm\right)(x\sin \alpha {)}^2$
$I=\underset{x=-l}{\overset{x=+l}{\int }}dI$
$=\underset{-l}{\overset{l}{\int }}(\frac{m}{2l}.dx)(x\sin \alpha {)}^2$
$=\frac{m{l}^2}{3}{\sin }^2\alpha$, Putting the value of $\alpha ={30}^{\text{o}}$ we get,

$I=\frac{m{l}^2}{12}$

Hence option A is the correct answer