Question

Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is :

• A. $\frac{1}{2}\frac{M{L}^2}{{\pi }^2}$
• B. $\frac{2}{5}\frac{M{L}^2}{{\pi }^2}$
• C. $\frac{1}{4}\frac{M{L}^2}{{\pi }^2}$
• D. $\frac{M{L}^2}{{\pi }^2}$

Answer 1

The correct option is D $\frac{M{L}^2}{{\pi }^2}$

When the wire is bent into the shape of a semi-circle, the length of the wire will become the circumference of the semi-circle.

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

Now, the moment of inertia of the semi-circular wire will be,

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