Physics · Rotational Dynamics

JEE Advanced 2021 — Paper 1 — Question 10

A thin rod of mass M and length aa is free to rotate in horizontal plane about a fixed vertical axis passing through point O . A thin circular disc of mass M and of radius

a/4a / 4 is pivoted on this rod with its center at a distance a/4a / 4 from the free end so that it can rotate freely about its vertical axis, as shown in the figure.

Assume that both the rod and the disc have uniform density and they remain horizontal during the motion.

An outside stationary observer finds the rod rotating with an angular velocity Ω\Omega and the disc rotating about its vertical axis with angular velocity

4Ω4 \Omega. The total angular momentum of the system about the point O is (Ma2Ω48)n\left(\frac{\mathrm{Ma}^{2} \Omega}{48}\right) \mathrm{n}.

The value of n is \qquad .

Question figure

Answer: 49

Numerical answer — enter this value.

Step-by-step solution

Angular momentum of disc about O is LDO =M(3a4)(3a4)Ω+M2(a4)2(4Ω)L_{\text {DO }}=M\left(\frac{3 \mathrm{a}}{4}\right)\left(\frac{3 \mathrm{a}}{4}\right) \Omega+\frac{\mathrm{M}}{2}\left(\frac{\mathrm{a}}{4}\right)^{2}(4 \Omega)

Angular momentum of rod about O is

LRO=Ma23Ω\mathrm{L}_{\mathrm{RO}}=\frac{\mathrm{Ma}^{2}}{3} \Omega

So, L0=LDO+LRO=4948(Ma2Ω)\mathrm{L}_{0}=\mathrm{L}_{\mathrm{DO}}+\mathrm{L}_{\mathrm{RO}}=\frac{49}{48}\left(\mathrm{Ma}^{2} \Omega\right) So, n=49n=49

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2021
Paper
Paper 1
Subject
Physics
Chapter
Rotational Dynamics
Topic
Angular Momentum and its Conservation
A thin rod of mass M and length a is free to rotate in horizontal… | JEE Advanced 2021 PYQ with Solution · DhiX AI