Question

The moment of inertia of a uniform rod of length $2l$ and mass $m$ about an axis $x{x}^{\prime }$ passing through its centre and inclined at an angle $\alpha$ is

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

Answer 1

The correct option is A $\frac{m{l}^2}{3}{\sin }^2\alpha$

The desired moment of inertia is,
$I=\underset{x=-l}{\overset{x=+l}{\int }}dI=\underset{-l}{\overset{+l}{\int }}\left(\frac{m}{2l}dx\right)(x\sin \alpha {)}^2=\frac{m{l}^2}{3}{\sin }^2\alpha$
Hence, the correct answer is option (a).