A uniform thin rod of length L and mass M is used as a physical pendulum. It is pivoted about a horizontal axis perpendicular to the rod at a point located a distance a from its centre of mass. The rod is given a small angular displacement in a vertical plane and released. What is the time period of small oscillations?
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Question

A uniform thin rod of length L and mass M is used as a physical pendulum. It is pivoted about a horizontal axis perpendicular to the rod at a point located a distance a from its centre of mass. The rod is given a small angular displacement in a vertical plane and released. What is the time period of small oscillations?

Options

G1 Physics · Accepted Answer

For a uniform rod, the moment of inertia about its centre is I₀ = ML²/12. The pivot is at distance a from the centre. By the parallel-axis theorem, Iₚ = ML²/12+Ma². For a physical pendulum, T = 2π√[Iₚ/(Mga)]. Substituting, T = 2π√[ (ML²/12+Ma²)/(Mga) ]. Cancel M: T = 2π√[ (L²/12+a²)/(ga) ]. Correct option: A.