Question

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

Answer: (D)

Length of the wire $=L$
Density of the wire $=\rho$
Mass of the wire and hence of the loop is, $M=\rho L$
and radius of the loop is, $r=\frac{L}{2\pi }$

Moment of inertia of a ring about an axis passing through its center and lying in its plane, $I_1=\frac{M{r}^2}{2}$
Using parallel axis theorem, we can say that
$I_2=I_1+M{r}^2$
$\Rightarrow I_2=\frac{M{r}^2}{2}+M{r}^2$
$\Rightarrow I_2=\frac{3M{r}^2}{2}$
$\Rightarrow I_2=\frac{3}{2}\left(\rho L\right){\left(\frac{L}{2\pi }\right)}^2$
$\Rightarrow I_2=\frac{3\rho L^3}{8{\pi }^2}$