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

Answer 1

Answer: (B)

The total length of the rod is L. Since, the rod is bent at the middle, so each part of it will have same length $\frac{L}{2}$ and mass $\frac{M}{2}$

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