A uniform rod of length 'l' is pivoted at one of its ends on a vertical shaft of negligible radius When the shaft rotates at angular speed the rod makes an angle θ with it (see figure). To find θ equate the rate of change of angular momentum (direction going into the paper) m 2 12 2 θ θ about the centre of mass (CM) to the torque provided by the horizontal and vertical forces FH and FV about the CM. The value of θ is then such that:

(A) \cos \theta=\frac{g}{2 \ell \omega^{2}}

(C) \cos \theta=\frac{2 g}{3 \ell \omega^{2}}

(D) \cos \theta=\frac{g}{\ell \omega^{2}}

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JEE Mains · Physics · STD 11 - System of particles and rotational motion

JEE Mains 2020
Hard
MCQ

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

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Correct Answer

(B) \(\cos \theta=\frac{3 g}{2 \ell \omega^{2}}\)

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\(F_{V}=m g\)

\(F_{H}=m \omega^{2} \frac{\ell}{2} \sin \theta\)

\(mg \frac{\ell}{2} \sin \theta- m \omega^{2} \frac{\ell}{2} \sin \theta \frac{\ell}{2} \cos \theta=\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta\)

\(\cos \theta=\frac{3}{2} \frac{g}{\omega^{2} \ell}\) \(....(ii)\)

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