Physics - Rotation Question with Solution | TestHub

Mass of the ring, $M=\lambda L$

Length of the wire is equal to the circumference of the ring, so $L=2\pi R\Rightarrow R=\frac{L}{2\pi }$

Moment of inertia of the ring about its diametric axis is $I_d=\frac{1}{2}M{R}^2$

Using the theorem of parallel axes, the moment of inertia about a tangential axis in its plane is $I_t=I_d+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Substituting the values of $M$ and $R$:

$I_t=\frac{3}{2}(\lambda L){\left(\frac{L}{2\pi }\right)}^2$

$I_t=\frac{3}{2}\lambda L\left(\frac{L^2}{4{\pi }^2}\right)=\frac{3\lambda L^3}{8{\pi }^2}$

Answer: $\frac{3\lambda L^3}{8{\pi }^2}$