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\sqrt3} $

Answer 1

Answer: (B)

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