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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Rotational Motion

JIPMER 2018 · Physics · Rotational Motion

A disc of moment of inertia 2 ~kg-m^2 revolving with 8 rad / s is...

A disc of moment of inertia 2 kg−m22 \mathrm{~kg}-\mathrm{m}^2 revolving with 8rad/s8 \mathrm{rad} / \mathrm{s} is placed on another disc of moment of inertia 4 kg−m24 \mathrm{~kg}-\mathrm{m}^2 revolving 4rad/s4 \mathrm{rad} / \mathrm{s}. What is the angular frequency of composite disc?

Options

  1. A4rad/s4 \mathrm{rad} / \mathrm{s}
  2. B316rad/s\frac{3}{16} \mathrm{rad} / \mathrm{s}
  3. C163rad/s\frac{16}{3} \mathrm{rad} / \mathrm{s}
  4. D165rad/s\frac{16}{5} \mathrm{rad} / \mathrm{s}

Correct answer

C. 163rad/s\frac{16}{3} \mathrm{rad} / \mathrm{s}

Step-by-step solution

Given that, moment of inertia of discs, I1=2 kg−m2,\mathrm{I}_1=2 \mathrm{~kg}-\mathrm{m}^2,\mathrm{I}2=4 kg−m2_2=4 \mathrm{~kg}-\mathrm{m}^2 Angular velocity of discs, ω1=8rad/s\omega_1=8 \mathrm{rad} / \mathrm{s}, ω2=4rad/s\omega_2=4 \mathrm{rad} / \mathrm{s} From angular momentum conservation principle, I1ω1+\mathrm{I}_1 \omega_1+\mathrm{I}_2 \omega_2=\left(\mathrm{I}1+_1+\mathrm{I}_2\right) \omega or 2×8+4×4=(2+4)ω\quad 2 \times 8+4 \times 4=(2+4) \omega or ω=16+166=326=163rad/s\quad \omega=\frac{16+16}{6}=\frac{32}{6}=\frac{16}{3} \mathrm{rad} / \mathrm{s}
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