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JEE Main 20205 SeptemberMorning Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Morning Shift
Subject
Physics
Chapter
Rotational Motion

JEE Main 2020 · 5 September · Morning Shift · Physics · Rotational Motion

A wheel is rotating freely with an angular speed ω on a shaft. The...

A wheel is rotating freely with an angular speed ω\omega on a shaft. The moment of inertia of the wheel is II and the moment of inertia of the shaft is negligible. Another wheel of moment of inertia 3I3I initially at rest is suddenly coupled to the same shaft. The resultant fractional loss in the kinetic energy of the system is:

Options

  1. A56\dfrac{5}{6}
  2. B14\dfrac{1}{4}
  3. C00
  4. D34\dfrac{3}{4}

Correct answer

D. 34\dfrac{3}{4}

Step-by-step solution

From angular momentum conservation
Iω+0=IωC+3IωCI\omega + 0 = I{\omega}_{C} + 3I{\omega}_{C}
\omegac=ω4\omegac = \dfrac{\omega}{4}

Loss of kinetic energy =12Iω2−12(I+3I)(ω4)2= \dfrac{1}{2}I{\omega}^{2} - \dfrac{1}{2} ( I + 3I ) {\left(\dfrac{\omega}{4}\right)}^{2}

=12Iω2−12Iω24= \dfrac{1}{2}I{\omega}^{2} - \dfrac{1}{2}I\dfrac{{\omega}^{2}}{4}

=38Iω2= \dfrac{3}{8}I{\omega}^{2}

Fractional loss =34= \dfrac{3}{4}

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