Medium MCQ +4 / -1 PYQ · JEE Mains 2021

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, $\omega$1 and $\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 :

  1. A ${{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$
  2. B ${{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}$
  3. C ${{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$ Correct answer
  4. D ${{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}$

Solution

From conservation of angular momentum we get<br><br>${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega$<br><br>$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$<br><br>${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$<br><br>${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$<br><br>$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$<br><br>Solving above we get<br><br>$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$

About this question

Subject: Physics · Chapter: Rotational Motion · Topic: Centre of Mass

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