Question

The moment of inertia of a ring of mass \( M \) and radius \( R \) about an axis passing through a tangential point in the plane of ring is:

• A. \( \frac{5MR^2}{2} \)
• B. \( \frac{3MR^2}{2} \)
• C. \( \frac{4MR^2}{3} \)
• D. \( \frac{2MR^2}{3} \)

Hint:

When calculating the moment of inertia about an axis that is not passing through the center, use the parallel axis theorem to shift the axis to the required point.

Answer 1

The Correct Option is A

The moment of inertia of a ring with mass \( M \) and radius \( R \) about an axis through its center and perpendicular to its plane is \( I_{\text{center}} = MR^2 \). The question requires the moment of inertia about a tangential axis. Applying the parallel axis theorem, we shift the axis from the center to the tangent. The theorem states \( I_{\text{tangent}} = I_{\text{center}} + Md^2 \), where \( d \) is the distance between the center and the tangent, equal to \( R \) for a ring. Therefore, \( I_{\text{tangent}} = MR^2 + MR^2 = \frac{5MR^2}{2} \). The correct answer is \( \frac{5MR^2}{2} \).