Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with center O. The moment of inertia of the loop about the axis which is tangent to this loop 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: (A), (D)

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

Mass of wire $m=\rho l$

Radius of ring $r=\frac{l}{2\pi }$

The moment of inertia of the loop about the axis which is tangent to this loop is $I=m{r}^2/2+m{r}^2=3m{r}^2/2=\frac{3\rho l^3}{8{\pi }^2}$