Q. Four thin rods of same mass $M$ and same length $l$, form a square as shown in figure. Moment of inertia of this system about an axis through centre $O$ and perpendicular to its plane is
A) $\frac{4}{3}\ Ml^{2}$
B) $\frac{Ml^{2}}{3}$
C) $\frac{Ml^{2}}{6}$
D) $\frac{2}{3}\ Ml^{2}$
Explanation
(a) Moment of inertia of rod $AB$ about point $P$ and perpendicular to the plane $= \frac{Ml^{2}}{12}$
M.I. of rod $AB$ about point $'O^{'} = \frac{Ml^{2}}{12} + M\left( \frac{l}{2} \right)^{2} = \frac{Ml^{2}}{3}$
(By using parallel axis theorem)
But the system consists of four rods of similar type so by but the symmetry $I_{System} = 4\left( \frac{Ml^{2}}{3} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.