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Question: A uniform rod of length 2lis placed with one end in contact with the horizontal table and is then in...

A uniform rod of length 2lis placed with one end in contact with the horizontal table and is then inclined at an angle α to the horizontal and allowed to fall. When it becomes

horizontal, its angular velocity will be

A

ω=(3gsin⁡α2l)\omega = \sqrt{\left( \frac{3g\sin\alpha}{2l} \right)}

B

ω=(2l3gsin⁡α)\omega = \sqrt{\left( \frac{2l}{3g\sin\alpha} \right)}

C

ω=(gsin⁡αl)\omega = \sqrt{\left( \frac{g\sin\alpha}{l} \right)}

D

ω=(lgsin⁡α)\omega = \sqrt{\left( \frac{l}{g\sin\alpha} \right)}

Answer

ω=(3gsin⁡α2l)\omega = \sqrt{\left( \frac{3g\sin\alpha}{2l} \right)}

Explanation

Solution

Let the rod be inclined at an angle θ with upward vertical. The moment of inertia of the rod about the fixed axis through fixed end and perpendicular to rod.

= I=M(2l)23=4Ml23I = \frac{M(2l)^{2}}{3} = \frac{4Ml^{2}}{3}

KE = 12l(dθdt)2=12×(4ml23)(dθdt)2\frac{1}{2}l\left( \frac{d\theta}{dt} \right)^{2} = \frac{1}{2} \times \left( \frac{4ml^{2}}{3} \right)\left( \frac{d\theta}{dt} \right)^{2}

= 2Ml23(dθdt)2\frac{2Ml^{2}}{3}\left( \frac{d\theta}{dt} \right)^{2}

Let dθdt=ω,\frac{d\theta}{dt} = \omega, when rod is horizontal K. E. =2Ml23×ω2\frac{2Ml^{2}}{3} \times \omega^{2}

When the rod is brought down from an inclination (π2−α)\left( \frac{\pi}{2} - \alpha \right) to an inclination π2\frac{\pi}{2} with upward vertical, the workdone by the weight of rod is given by

Work done = mg × vertical height through which C.G. is moved

=mg[lcos⁡(π2−α)−lcos⁡π2]=Mglsin⁡αmg\left\lbrack l\cos\left( \frac{\pi}{2} - \alpha \right) - l\cos\frac{\pi}{2} \right\rbrack = Mgl\sin\alpha

∴2Ml23×ω2=Mglsin⁡α\therefore\frac{2Ml^{2}}{3} \times \omega^{2} = Mgl\sin\alphaSolving for ω, get

ω = (3gsin⁡α2l)\sqrt{\left( \frac{3g\sin\alpha}{2l} \right)}