Verified Solution • 2025–26 Session moderate

QJEEM21Ph0bd7: JEE Main 2021 Physics - Class 11 JEEM21Ph0bd7 Solution: Two discs have moments of inertia I 1 and I 2 about their…

Two discs have moments of inertia I1 and I2 about their respective axes perpendicular to the plane and passing through the centre. They are rotating with angular speeds, ω\omega 1 and ω\omega 2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :

Multiple Choice Options

  • A I1I2(I1+I2)(ω1−ω2)2{{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}
  • B (I1−I2)2ω1ω22(I1+I2){{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}
  • C I1I22(I1+I2)(ω1−ω2)2{{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}
  • D (ω1−ω2)22(I1+I2){{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}

Step-by-Step Verified Solution

From conservation of angular momentum we get

I1ω1+I2ω2=(I1+I2)ω{I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega

ω=I1ω1+I2ω2I1+I2\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}

ki=12I1ω12+12I2ω22{k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2

kf=12(I1+I2)ω2{k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}

ki−kf=12[I1ω12+I2ω22−(I1ω1+I2ω2)2I1+I2]{k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]

Solving above we get

ki−kf=12(I1I2I1+I2)(ω1−ω2)2{k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}

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