Rotational Motion

JEE Main level, single correct, medium

180 of 280

Question 180 of 280

A thin rod MN\mathrm{MN}, free to rotate in the vertical plane about the fixed end N\mathrm{N}, is held horizontal. When the end M\mathrm{M} is released the speed of this end, when the rod makes an angle α\alpha with the horizontal, will be proportional to: (see figure)

Show answer and solution

Answer: (A) cos⁡α\sqrt{\cos \alpha}

When the rod makes an angle α\alpha
 Displacement of centre of mass =l2cos⁡αmgl2cos⁡α=l2Iω2mgl2cos⁡α=ml26ω2(∵ M.I. of thin \begin{aligned} &\text { Displacement of centre of mass }=\frac{l}{2} \cos \alpha \\ &m g \frac{l}{2} \cos \alpha=\frac{l}{2} I \omega^2 \\ &m g \frac{l}{2} \cos \alpha=\frac{m l^2}{6} \omega^2 \quad(\because \quad \text { M.I. of thin } \end{aligned}
uniform rod about an axis passing through its centre of mass and perpendicular to the
rod⁡I=ml212)⇒ω=3gcos⁡αl\begin{aligned} &\left.\operatorname{rod} I=\frac{m l^2}{12}\right) \\ &\Rightarrow \omega=\sqrt{\frac{3 g \cos \alpha}{l}} \end{aligned}
Speed of end =ω×l=3gcos⁡αl=\omega \times l=\sqrt{3 g \cos \alpha l}
i.e., Speed of end, ω∝cos⁡α\omega \propto \sqrt{\cos \alpha}

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Rotational dynamics and rolling simulationDrag the sliders and watch the numbers in this question change.Revise Rotational Motion firstFormula sheet, solved examples and the traps students fall into.

More Physics practice

JEE Main Physics practice question on Rotational Motion (Rotational kinematics), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.