Verified Solution • 2025–26 Session moderate

QJEEM20Ph3d03: JEE Main 2020 Physics - Class 11 JEEM20Ph3d03 Solution: A wheel is rotating freely with an angular speed on a shaft.…

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

Multiple Choice Options

  • A 0
  • B 56{5 \over 6}
  • C 14{1 \over 4}
  • D 34{3 \over 4}

Step-by-Step Verified Solution

Applying Angular Momentum conservation

Iω=(I+3I)ω′I\omega = (I + 3I)\omega '

⇒\Rightarrow ω′=Iω4I=ω4\omega ' = {{I\omega } \over {4I}} = {\omega \over 4}

ki=12Iω2{k_i} = {1 \over 2}I{\omega ^2}

kf=12(4I)(ω′)2{k_f} = {1 \over 2}(4I){(\omega ')^2}

=2I(ω4)2=18Iω2= 2I{\left( {{\omega \over 4}} \right)^2} = {1 \over 8}I{\omega ^2}

Fractional loss =Ki−KfKi=12Iω2−18Iω212Iω2= {{{K_i} - {K_f}} \over {{K_i}}} = {{{1 \over 2}I{\omega ^2} - {1 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}}

= 38Iω212Iω2=34{{{3 \over 8}I{\omega ^2}} \over {{1 \over 2}I{\omega ^2}}} = {3 \over 4}

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