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 of mass \( M \) and radius \( R \) about an axis passing through the center of the ring and perpendicular to its plane is given by: \[ I_{\text{center}} = MR^2 \] However, the question asks for the moment of inertia about an axis passing through a tangential point. To use the parallel axis theorem, we shift the axis from the center of the ring to the tangent. The parallel axis theorem states: \[ I_{\text{tangent}} = I_{\text{center}} + Md^2 \] where \( d \) is the distance between the center and the tangent (which is \( R \) for a ring). Therefore: \[ I_{\text{tangent}} = MR^2 + MR^2 = \frac{5MR^2}{2} \] Thus, the correct answer is \( \frac{5MR^2}{2} \).