A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is :-
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Question

A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point at its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is :-

Options

G1 Physics · Accepted Answer

The ring radius is R=1 m. For oscillation about a tangent diameter perpendicular to the ring plane, I=I about CM+MR²=(1/2)MR²+MR²=(3/2)MR². Equivalent simple-pendulum length is Lₑ=I/(MR)=3R/2=1.5 m. Correct option: 3.