Question

A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time period for oscillations of small amplitude is

(a) 2π√(l/g)

(b) 2π√(2l/3g)

(c) 2π√(3l/2g)

(d) 2π√(l/2g)

Answer 1

Correct Answer is: (b) 2π√(2l/3g)

T = mg(l/2 sin θ) ≃ 1/2 mglθ = - Iα

or 1/2 mglθ = - 1/3 mgl²α

or α = - (3g/2l) θ.

Put Ω² = 3g82l.

∴ α = -Ω²θ.

This represents an angular SHM with the time period 

2π/Ω = 2π√(2l/3g).