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Questions  

A thin horizontal uniform rod AB of mass m and length l can rotate freely about a vertical axis passing through its end A. At a certain moment, the end B starts experiencing a constant force F which is always perpendicular to the original position of the stationary rod and directed in a horizontal plane. The angular velocity of the rod as a function of its rotation angle θ measured relative to the initial position should be

a
6F sin θml
b
2F sin θml
c
3F sin θml
d
5F sin θml

detailed solution

Correct option is A

Work done by the torque∆W = ∫τdθ  = ∫0θFl cos θ dθ∆W = Fl sin θNow using work energy theorem,∆W = ∆k∴ Flsinθ = [12(ml23)ω2-0]Which gives, ω = 6F sin θml

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