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

To determine the moment of inertia of a ring formed from a rod, we first establish the mass distribution and geometric relationships:

1. A rod with linear mass density \(\lambda\) and length \(L\) is bent into a ring of radius \(R\).
2. The rod's length equals the ring's circumference: \(L = 2\pi R\). Consequently, \(R = \frac{L}{2\pi}\).
3. The total mass \(M\) of the rod, and thus the ring, is \(M = \lambda L\).
4. The moment of inertia of a ring about its diameter is given by the formula \(I = \frac{1}{2} M R^2\).
5. Substituting the expressions for \(M\) and \(R\) yields:

\[I = \frac{1}{2} \cdot \lambda L \cdot \left(\frac{L}{2\pi}\right)^2\]

1. Simplifying the expression results in:

\[I = \frac{\lambda L}{2} \cdot \frac{L^2}{4\pi^2} = \frac{\lambda L^3}{8\pi^2}\]

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