Question

A thin rod of mass M and length L is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

• A. $\frac{ML^2}{2\pi^2}$
• B. $\frac{ML^2}{4\pi^2}$
• C. $\frac{ML^2}{8\pi^2}$
• D. $\frac{ML^2}{\pi^2}$

Answer 1

Answer: (C)

$I = \frac{MR^2}{2}$
but $2 \pi R = L \, \Rightarrow \, R =\frac{L}{2\pi}$
$\therefore \, I = \frac{ML^2}{8 \pi^2}$