Question

A rod of linear mass density $ \lambda $ and length $ L $ is bent to form a ring of radius $ R $. Moment of inertia of the ring about any of its diameter is:

• A. \( \frac{\lambda L^3}{8\pi^2} \)
• B. \( \frac{\lambda L^3}{4\pi^2} \)
• C. \( \frac{\lambda L^3}{16\pi^2} \)
• D. \( \frac{\lambda L^3}{12} \)

Hint:

The moment of inertia of a ring about any of its diameters can be derived by considering the geometry of the ring and using the formula for the moment of inertia of a mass distributed along the circumference.

Answer 1

The Correct Option is A

For a rod of length \( L \) and linear mass density \( \lambda \), 
the total mass \( m \) of the rod is: \( m = \lambda L \) 
When the rod is bent into a ring of radius \( R \), the mass is uniformly distributed along the circumference of the ring.
The moment of inertia of the ring about any of its diameters is given by the formula for a ring: \( I = \frac{1}{2} m R^2 \) 

Substituting 
\( m = \lambda L \) into this expression: \( I = \frac{1}{2} \lambda L R^2 \) 
We know that the circumference of the ring is \( 2\pi R \), and the length of the rod is equal to the circumference of the ring, so: \( L = 2\pi R \) 

Thus, the radius \( R \) can be written as: \( R = \frac{L}{2\pi} \) 
Substituting this into the equation for the moment of inertia: \( I = \frac{1}{2} \lambda L \left( \frac{L}{2\pi} \right)^2 = \frac{1}{2} \lambda L \times \frac{L^2}{4\pi^2} \) 

Simplifying: 
\( I = \frac{\lambda L^3}{8\pi^2} \) 

Thus, the moment of inertia of the ring about any of its diameters is \( \frac{\lambda L^3}{8\pi^2} \). 
Therefore, the correct answer is Option (1).