Physics - Rotation Question with Solution | TestHub

Total mass of the ring, $M=mL$.

Since the wire of length $L$ is bent into a circular ring of radius $R$, its circumference is $L$.

$2\pi R=L\Rightarrow R=\frac{L}{2\pi }$

The axis $y{y}^{\prime }$ is a tangent to the ring in its plane. The moment of inertia of a ring about its diametric axis is $I_d=\frac{1}{2}M{R}^2$.

Using the parallel axis theorem, the moment of inertia of the ring about the tangent $y{y}^{\prime }$ is:

$I_{y{y}^{\prime }}=I_d+M{R}^2$

$I_{y{y}^{\prime }}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

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

$I_{y{y}^{\prime }}=\frac{3}{2}(mL){\left(\frac{L}{2\pi }\right)}^2$

$I_{y{y}^{\prime }}=\frac{3}{2}mL\left(\frac{L^2}{4{\pi }^2}\right)$

$I_{y{y}^{\prime }}=\frac{3m{L}^3}{8{\pi }^2}$

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