Q. The moment of inertia of a rod about an axis through its centre and perpendicular to it is $\frac{1}{2}{ML}^{2}$ (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 makes an angular of $60^{0}$. The same axis would be
A) $\frac{1}{48}{ML}^{2}$
B) $\frac{1}{12}{ML}^{2}$
C) $\frac{1}{24}{ML}^{2}$
D) $\frac{{ML}^{2}}{8\sqrt{3}}$
Explanation
(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.
[image:7_image180.emf]
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
$I = \frac{1}{3}\left( \frac{M}{2} \right)\left( \frac{L}{2} \right)^{2} + \frac{1}{3}\left( \frac{M}{2} \right)\left( \frac{L}{2} \right)^{2}$
$= \frac{1}{3}\left\lbrack \frac{ML^{2}}{8} + \frac{ML^{2}}{8} \right\rbrack = \frac{ML^{2}}{12}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.