Question

Radius of gyration of a thin circular ring of mass m and radius r about a tangent in the plane of ring is

Answer 1

Moment of inertia at the centre and perpendicular to the plane of the ring.
So, about a point on the rim of the ring and the axis ⊥ to the plane of the ring.

The moment of inertia are given as
= mR² + mR² = 2mR² (parallel axis theorem)
 mK² = 2mR² (K = radius of the gyration)

 K = √2R² = √2 R.