Question

A thin wire of length L and uniform linear density $\rho$ used to form a ring. The M.I. of this ring about tangential axis laying in its planes is:

• A. $\frac{\rho L^3}{8{\pi }^2}$
• B. $\frac{\rho L^3}{16{\pi }^2}$
• C. $\frac{3\rho L^3}{12{\pi }^2}$
• D. $\frac{3\rho L^3}{8{\pi }^2}$

Answer 1

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

REF.Image

Mass of ring = $\rho \times L$

Then,

MOI about $O{O}^{\prime }=\left(\rho L\right)(\frac{L}{2\pi }{)}^2$

$=\frac{\rho L^3}{4{\pi }^2}$ $2\pi r=L$, density = $\rho$ $r=\frac{L}{2\pi }$

Then,

MOI about $N{N}^{\prime }=\frac{\rho L^3}{4{\pi }^2}\times \frac{1}{2}=\frac{\rho L^3}{8{\pi }^2}$

(By perpendicular axis theorem)

Finally by parallel axis theorem

MOI about $M{M}^{\prime }=\frac{\rho L^3}{8{\pi }^2}+\left(\rho L\right)(\frac{L}{2\pi }{)}^2$

$=\frac{\rho L^3}{8{\pi }^2}+\frac{\rho L^3}{4{\pi }^2}=\frac{\rho L^3+2\rho L^3}{8{\pi }^2}=\frac{3\rho L^3}{8{\pi }^2}$