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JEE Mains · Physics · STD 11 - System of particles and rotational motion
- JEE Mains 2018
- Hard
- MCQ
A thin rod \(MN\), free to rotate in the vertical plane about the fixed end \(N\), is held horizontal . When the end \(M\) is released the speed of this end, when the rod makes an angle \(\alpha \) with the horizontal, will be proportional to ( see figure)

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\sqrt {\cos \alpha } \)
Step-by-step Solution
Detailed explanation
When the rod makes an angle \(\alpha \)
Displacement of center of mass \( = \frac{1}{2}\cos \,\alpha \)
\(mg\frac{1}{2}\cos \alpha = \frac{l}{2}I{\omega ^2}\,\)
\(mg\frac{1}{2}\cos \alpha = \frac{{m{l^2}}}{6}{\omega ^2}\) (M.I. of thin uniform rod about an axis passing through its center of mass and perpendicular to the rod \(I = \frac{{m{l^2}}}{{12}}\))
\( \Rightarrow \omega = \sqrt {\frac{{3g\cos \alpha }}{l}} \)
speed of end \( = \omega \times l = \sqrt {3g\cos \alpha l} \)
i.e., Speed of end, \(\omega \, \propto \,\sqrt {\cos \alpha } \)
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