Question

The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12}M{L}^2$ (where $M$ is the mass and $L$, the length of the rod). The rod is bent in the middle so that the two halves make an angle of $60^{\circ }$. The moment of inertia of the bent rod about the same axis would be :

• A. $\frac{1}{48}M{L}^2$
• B. $\frac{1}{12}M{L}^2$
• C. $\frac{1}{24}M{L}^2$
• D. $\frac{M{L}^2}{8\sqrt{3}}$

Answer 1

Answer: (B)

Since, rod is bent at the middle, so each part of it will have same length $\left(\frac{L}{2}\right)$ and mass $\left(\frac{M}{2}\right)$ as shown.

Moment of inertia of each part through its one end
$=\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2$
Hence, net moment of inertia through its middle point $O$ is
$I=\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2+\frac{1}{3}\left(\frac{M}{2}\right){\left(\frac{L}{2}\right)}^2$
$=\frac{1}{3}\left[\frac{M{L}^2}{8}+\frac{M{L}^2}{8}\right]=\frac{M{L}^2}{12}$