Question

A thin wire of length $l$ and mass $m$ is bent in the form of a semicircle as shown. Its moment of inertia about an axis joining its free ends will be,

• A. $ml^{2}$
• B. zero
• C. $\frac{m l^{2}}{\pi ^{2}}$
• D. $\frac{m l^{2}}{2 \pi ^{2}}$

Answer 1

Answer: (D)

The moment of inertia of semicircular portion is half of the total ring.
So, $I=\frac{1}{2}M_{\text{ring}}\frac{R^{2}}{2}=\frac{1}{2}\left(2 m\right)\frac{R^{2}}{2}=\frac{m R^{2}}{2}$
and $\pi R=l$ .
So, $R=\frac{l}{\pi }$
$l=\frac{m l^{2}}{2 \pi ^{2}}$ .