Question

The moment of inertia of a circular ring of mass $ M $ and diameter $ r $ about a tangential axis lying in the plane of the ring is:

• A. \( \frac{1}{2} M r^2 \)
• B. \( \frac{3}{8} M r^2 \)
• C. \( \frac{3}{2} M r^2 \)
• D. \( 2 M r^2 \)

Hint:

For a circular ring, the moment of inertia about an axis tangent to the ring is derived from the parallel axis theorem by adjusting for the offset from the center.

Answer 1

The Correct Option is B

The diameter is \( R \). The radius is therefore \(\frac{R}{2}\). The formula for the moment of inertia about a tangential axis is \(I_{\text{tangent}} = \frac{3}{2} m \left( \frac{R}{2} \right)^2 = \frac{3}{8} m R^2\). Consequently, the moment of inertia is \(I_{\text{tangent}} = \frac{3}{8} M r^2\).