Rotational Motion

JEE Main level, single correct, medium

121 of 280

Question 121 of 280


A uniform rod of length ' ℓ′{\ell }^{'} is pivoted at one of its ends on a vertical shaft of negligible radius. When the shaft rotates at angular speed ω\omega the rod makes an angle θ\thetawith it (see figure). To find θ\theta equate the rate of change of angular momentum (direction going into the paper) mℓ212ω2sin⁡θ\dfrac{m{\ell }^{2}}{12}{\omega }^{2}\sin \theta about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH{F}_{H}and Fv{F}_{v} about the CM. The value of θ\theta is then such that:

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Answer: (D) cos⁡θ=3g2ℓω2\cos \theta =\dfrac{3g}{2\ell {\omega }^{2}}

Torque of centrifugal force τcf=dm.xsin⁡θω2xcosϑ=mlω2sin⁡θcosθ∫0lx2dx{\tau }_{cf}=dm.x\sin {\text{θω}}^{2}xcosϑ=\dfrac{m}{l}{\omega }^{2}\sin \theta cos\theta {\int }_{0}^{l}{x}^{2}dx
τef=ml2ω2sinθcosθ3{\tau }_{ef}=\dfrac{m{l}^{2}{\omega }^{2}sin\theta cos\theta }{3}
τmg=τcf{\tau }_{mg}={\tau }_{cf}
mg.l2sin⁡θ=ml2ω2sinθcos⁡θ3mg.\dfrac{l}{2}\sin \theta =\dfrac{m{l}^{2}{\omega }^{2}sin\theta \cos \theta }{3}
cos⁡θ=3g2lω2\cos \theta =\dfrac{3g}{2l{\omega }^{2}}

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