Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown in figure. Calculate the moment of in inertia of the loop about the axis $XX$.

Answer 1

We know that moment of inertia of the ring about the line passing through centre and in the plane of the ring is

$I_{XX}=\frac{{MR}^2}{2}=\frac{\left(\rho L\right){\left(\frac{L}{2\pi }\right)}^2}{2}=\frac{\rho L^3}{8{\pi }^2}$
So by parallel axis thermo, $l_{XX}=l_{XX}+M{d}^2$
$=\frac{\rho L^3}{8{\pi }^2}+\rho L\times R^2=\frac{\rho L^3}{8{\pi }^2}+\rho L\times {\left(\frac{L}{2\pi }\right)}^2=\frac{\rho L^3}{8{\pi }^2}+\frac{\rho L^3}{4{\pi }^2}=\frac{3\rho L^3}{8{\pi }^2}$