Physics · Rotational Dynamics

JEE Advanced 2019 — Paper 2 — Question 3

A thin and uniform rod of mass $M$ and length $L$ is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle $60^{\circ}$ with vertical ? [ $g$ is the acceleration due to gravity]

Option A:
The radial acceleration of the rod's center of mass will be $\frac{3 g}{4}$
Correct
Option B:
The angular speed of the rod will be $\sqrt{\frac{3 g}{2 L}}$
Correct
Option C:
The angular acceleration of the rod will be $\frac{2 g}{L}$
Option D:
The normal reaction force from the floor on the rod will be $\frac{\mathrm{Mg}}{16}$
Correct

Answer: A, B, D

Step-by-step solution

To find $\omega$ , we apply COE : $\frac{1}{2} \sigma_{3}^{1} M L^{2} E \omega^{2}=M g \frac{L}{2}\left(1-\cos 60^{\circ}\right)$

$\omega=\sqrt{\frac{3 g}{2 L}}$

$\mathrm{a}_{\mathrm{CM}, \text { radial }}=\omega^{2} \frac{\mathrm{~L}}{2}=\frac{3 g}{4}$ .

To find $\alpha$ , we take torque about A :

$\mathrm{Mg} \frac{\mathrm{L}}{2} \sin 60^{\circ}=\frac{1}{3} \mathrm{ML}^{2} \alpha$

so, $\alpha=\frac{3 g}{2 L} \sin 60^{\circ}$ .

$a_{c m, \operatorname{tang}}=\alpha \frac{L}{2}=\frac{3 g}{4} \sin 60^{\circ}$

$\therefore \mathrm{Mg}-\mathrm{N}=\left(\mathrm{a}_{\mathrm{cm} ; \text { radial }} \cos 60^{\circ}+\mathrm{a}_{\mathrm{cm}, \text { tang }} \sin 60^{\circ}\right)$

$=M-\frac{3 g}{8}+\frac{9 g}{16}=$

$\therefore \mathrm{N}=\frac{\mathrm{Mg}}{16}$

$\therefore$ Correct options are A, B, D

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2019
Paper: Paper 2
Subject: Physics
Chapter: Rotational Dynamics
Topic: Kinematics of General Motion of a Rigid Body