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A uniform rod of length l can rotate freely in a vertical plane about a fixed horizontal axis passing through O. Starting from rest in its unstable equilibrium position, the rod turns through an angle θ. Its angular speed ω is expressed by

1. √(6g/l) sin θ
2. √(6g/l) sin(θ/2)
3. √(6g/l) cos(θ/2)
4. √(6g/l) cos θ

Correct answer: √(6g/l) sin(θ/2)

Solution

With I = ml^2/3 about the end and the centre of mass falling (l/2)(1-cos th) from the inverted position: (1/2)(ml^2/3)w^2 = mg(l/2)(1-cos th). Thus w^2 = (3g/l)(1-cos th) = (6g/l)sin^2(th/2), so w = sqrt(6g/l) sin(th/2).

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