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Question
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
\[m K^2 = 2m R^2 \]
K = Radius of the gyration
\[ \Rightarrow K = \sqrt{2 R^2} = \sqrt{2}R\]
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