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

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Solution

Moment of inertia of the ring about a point on the rim of the ring and the axis perpendicular to the plane of the ring = mR² + mR² = 2mR² (from parallel axis theorem)
$We know that : m K^{2} = 2 m R^{2} (K = R adius of the gyration) \Rightarrow K = \sqrt{2 R^{2}} = \sqrt{2} R$

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