A thin ring of radius 0.50 m is suspended from a point on its circumference and oscillates in its own vertical plane. Take g = 10 m/s². If ω² = N s⁻², find N.

Question

G1 Physics · Accepted Answer

For a thin ring about its centre, I₀ = MR². Using the parallel-axis theorem for a pivot on the circumference, Iₚ = MR²+MR² = 2MR². The centre of mass is at distance d = R from the pivot. Hence, ω² = Mgd/I = MgR/(2MR²) = g/(2R). With g = 10, R = 0.50, ω² = 10/(1) = 10 s⁻². Hence, N = 10.