Question

Find the radius of gyration of a circular ring of radius r about a line perpendicular to the plane of the ring and passing through one of its particles.

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
= mR² + mR² = 2mR² (parallel axis theorem)
 mK² = 2mR² (K = radius of the gyration)

 K = √2R² = √2 R.