A thin uniform rod of length l is pivoted at its upper end. It is free to swing in a vertical plane. Its time ** for oscillations of small amplitude is
(a) 2π√(l/g)
(b) 2π√(2l/3g)
(c) 2π√(3l/2g)
(d) 2π√(l/2g)
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 **
2π/Ω = 2π√(2l/3g).
