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 :

Solution

Physics sir · 4 steps

Watch the trap word: r here is the diameter, so the radius is r/2. Then it is just parallel axes.

1. Radius first. The ring's diameter is r, so its radius is R = r/2. Work in R and switch to r only at the end.
2. About a diameter. A diameter is a centre axis lying in the plane of the ring. By the perpendicular axis theorem, two perpendicular diameters add up to the axis through the centre perpendicular to the plane: 2I_d = MR². So I_d = 1/2MR².
3. Shift to tangent. The tangent in the plane is parallel to a diameter and a distance R from it. By the parallel axis theorem, add MR².

   Iₜ = 1/2MR² + MR² = 3/2MR² = 3/2M(r/2)² = 3/8Mr²

4. So. The moment of inertia about the tangent in the plane of the ring is 3/8Mr², option (a).

The idea to keep. Always check whether the question gives the radius or the diameter. For a ring, start from 1/2MR² about a diameter and add MR² to reach a tangent in its plane.

Answer from the NTA final answer key. Solution by PrepWiser.