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:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\cos \theta=\frac{3 g}{2 \ell \omega^{2}}\)
Step-by-step Solution
Detailed explanation
\(F_{V}=m g\)
\(F_{H}=m \omega^{2} \frac{\ell}{2} \sin \theta\)
\(mg \frac{\ell}{2} \sin \theta- m \omega^{2} \frac{\ell}{2} \sin \theta \frac{\ell}{2} \cos \theta=\frac{ m \ell^{2}}{12} \omega^{2} \sin \theta \cos \theta\)
\(\cos \theta=\frac{3}{2} \frac{g}{\omega^{2} \ell}\) \(....(ii)\)
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