Question

The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{12} ML ^{2}$ (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. $\frac{1}{48}ML^{2}$
• B. $\frac{1}{12}ML^{2}$
• C. $\frac{1}{24}ML^{2}$
• D. $\frac{ML^{2}}{8\sqrt{3}}$

Answer 1

Answer: (B)

$\frac{1}{12}ML^{2}$

Answer 2

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 $I =\frac{1}{3}\left(\frac{ M }{2}\right)\left(\frac{ L }{2}\right)^{2}+\frac{1}{3}\left(\frac{ M }{2}\right)\left(\frac{ L }{2}\right)^{2}$ $=\frac{1}{3}\left[\frac{ ML ^{2}}{8}+\frac{ ML ^{2}}{8}\right]=\frac{ ML ^{2}}{12}$