Physics - Rotation Question with Solution | TestHub

Let $R$ be the radius of the semicircular ring.

The length of the wire is bent into a semicircle, so the arc length is equal to $L$.

$\pi R=L\Rightarrow R=\frac{L}{\pi }$

The axis $X{X}^1$ passes through the two ends of the semicircular wire, which corresponds to the diametric axis of the full circle.

The moment of inertia of a semicircular ring of mass $M$ and radius $R$ about its diametric base is the same as that of a full ring of the same mass and radius about its diameter.

$I=\frac{1}{2}M{R}^2$

Substituting the value of $R$ in the above equation, we get:

$I=\frac{1}{2}M{\left(\frac{L}{\pi }\right)}^2$

$I=\frac{M{L}^2}{2{\pi }^2}$

Answer: $\frac{M{L}^2}{2{\pi }^2}$