Q.
A thin wire of length $l$ and mass $m$ is bent in the form of semicircle. Its moment of inertia about an axis joining its free ends will be
A) $\ m\ l^{2}$
B) Zero
C) $m\ l^{2}/\pi^{2}$
D) None of these
Explanation
(d) $\pi r = l\ \ \therefore r = l/\pi$
Moment of inertia of a ring about its diameter $= \frac{1}{2}Mr^{2}$
$\therefore$ Moment of inertia of semicircle $= \frac{1}{2}\left\lbrack m\left( \frac{l}{\pi} \right)^{2} \right\rbrack = \frac{ml^{2}}{2\pi^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.