Q.
A rod of length $L$ and mass $M$ is bent to form a semi-circular ring as shown in figure. The moment of inertia about $XY$ is
A) $\frac{ML^{2}}{2\pi^{2}}$
B) $\frac{ML^{2}}{\pi^{2}}$
C) $\frac{ML^{2}}{4\pi^{2}}$
D) $\frac{2ML^{2}}{\pi^{2}}$
Explanation
(a) M.I. of ring about diameter $I = \frac{MR^{2}}{2}$ …(i)
$\because L = \pi R\ \therefore R = L/\pi$
From equation (i), $I = \frac{ML^{2}}{2\pi^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.