Question

A thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular coil. Moment of inertia of the coil about the tangential axis in its plane is __________.

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

Answer 1

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

Given: A thin wire of length L and uniform linear mass density ρ is bent into a circular coil.

To find the moment of inertia of the coil about the tangential axis in its plane

Solution:

The moment of inertia of the thin wire coil about its diameter

$I=\frac{1}{2}M{R}^2$

Here $M=V\times \rho$ (mass=volume $\times$ density)

and we know $L=2\pi R⟹R=\frac{L}{2\pi }$ (as thin wire of length L is bent into a circular coil)

$I=\frac{1}{2}M{R}^2⟹I=\frac{1}{2}L\rho \times {\left(\frac{L}{2\pi }\right)}^2⟹I=\frac{L^3\rho }{8{\pi }^2}$

Now using parallel axis theorem, we get

$I_{x{x}^{\prime }}=I+M{R}^2⟹I_{x{x}^{\prime }}=\frac{L^3\rho }{8{\pi }^2}+L\rho {\left(\frac{L}{2\pi }\right)}^2⟹I_{x{x}^{\prime }}=\frac{L^3\rho +2L^3\rho }{8{\pi }^2}⟹I_{x{x}^{\prime }}=\frac{3L^3\rho }{8{\pi }^2}$

is the moment of inertia of the coil about the tangential axis in its plane