Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at O. the moment of inertia of loop about axis $x-x^{{}^{\prime }}$ is

• A. $\frac{\rho L^3}{8{\pi }^2}$
• B. $\frac{\rho L^3}{16{\pi }^2}$
• C. $\frac{5\rho L^3}{16{\pi }^2}$
• D. $\frac{3\rho L^3}{8{\pi }^2}$

Answer 1

The correct option is D $\frac{3\rho L^3}{8{\pi }^2}$

We have $2\pi r=L$
or
$r=\frac{L}{2\pi }$
Moment of inertia of circular ring about centroidal axis perpendicular to its radius is $m{r}^2$.
Let moment of inertia about any diameter be I. Thus $2I=m{r}^2$ (using perpendicular axis theorem)
or
$I=\frac{m{r}^2}{2}$
Thus moment of inertia about any tangential axis as shown would be $\frac{m{r}^2}{2}+m{r}^2=\frac{3}{2}m{r}^2$
or
$\frac{3}{2}\rho L\frac{L^2}{4{\pi }^2}=\frac{3\rho L^3}{8{\pi }^2}$