Question

A rod of linear mass density $ \lambda $ and length $ L $ is bent into the form of a ring of radius $ R $. The moment of inertia of the ring about any of its diameters is:

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

Hint:

For a ring, the moment of inertia about its diameter can be derived using the formula for a uniform mass distribution and the relationship between radius and length.

Answer 1

The Correct Option is D

Step 1: Moment of inertia of a ring.
For a ring, the moment of inertia about any of its diameters is: \[ I = \frac{1}{2} M R^2 \] where: - \( M = \lambda L \) is the mass of the ring, - \( R \) is the radius of the ring.
Step 2: Substitute mass into the formula.
Substituting \( M = \lambda L \) into the moment of inertia formula: \[ I = \frac{1}{2} \lambda L R^2 \]
Step 3: Relate radius and length.
The length of the rod is the circumference of the ring, so: \[ L = 2\pi R \quad \Rightarrow \quad R = \frac{L}{2\pi} \]
Step 4: Substitute for \( R \).
Substitute \( R = \frac{L}{2\pi} \) into the moment of inertia formula: \[ I = \frac{1}{2} \lambda L \left( \frac{L}{2\pi} \right)^2 = \frac{1}{2} \lambda L \times \frac{L^2}{4\pi^2} = \frac{\lambda L^3}{8\pi^2} \]
Final Answer:
Thus, the moment of inertia of the ring about any of its diameters is \( \frac{\lambda L^3}{8\pi^2} \), corresponding to option (4).