Question

A rod of length $L$ and mass $M$ is bent to form a semicircular ring as shown in figure. The moment of inertia about $XY$ is

• A. $\frac{1}{4} \frac{M L^{2}}{\pi^{2}}$
• B. $\frac{2 M L^{2}}{3 \pi^{2}}$
• C. $\frac{M L^{2}}{4}$
• D. $\frac{M L^{2}}{2}$

Answer 1

Answer: (A)

Here,
$L=\pi R$ or $R=\frac{L}{\pi}$
$\therefore $ Moment of inertia about
$X Y=\frac{1}{2}\left(\frac{1}{2} M R^{2}\right)$
$=\frac{1}{4} \frac{M L^{2}}{--\pi^{2}}$