Physics - Rotation Question with Solution | TestHub

Let the mass of the circular loop be $M$ and its radius be $R$.

Given that the wire has length $L$ and linear mass density $ho$, the total mass of the loop is:
$M=hoL$

Since the wire is bent into a circular loop, its circumference is equal to $L$:
$2\pi R=L\Rightarrow R=\frac{L}{2\pi }$

The moment of inertia of a circular ring about its diameter is:
$I_d=\frac{1}{2}M{R}^2$

The axis $X{X}^{\prime }$ is a tangent to the circular loop in its plane, which is parallel to the diameter at a distance $R$. Using the parallel axis theorem, the moment of inertia about the axis $X{X}^{\prime }$ is:
$I_{X{X}^{\prime }}=I_d+M{R}^2$
$I_{X{X}^{\prime }}=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$

Substituting the values of $M$ and $R$ into the equation:
$I_{X{X}^{\prime }}=\frac{3}{2}(\rho L){\left(\frac{L}{2\pi }\right)}^2$
$I_{X{X}^{\prime }}=\frac{3}{2}(\rho L)\left(\frac{L^2}{4{\pi }^2}\right)$
$I_{X{X}^{\prime }}=\frac{3\rho L^3}{8{\pi }^2}$

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