JEE MAIN - Physics (2021 - 27th August Evening Shift - No. 4)
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 :
$${{{I_1}{I_2}} \over {({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$
$${{{{({I_1} - {I_2})}^2}{\omega _1}{\omega _2}} \over {2({I_1} + {I_2})}}$$
$${{{I_1}{I_2}} \over {2({I_1} + {I_2})}}{({\omega _1} - {\omega _2})^2}$$
$${{{{({\omega _1} - {\omega _2})}^2}} \over {2({I_1} + {I_2})}}$$
Explanation
From conservation of angular momentum we get
$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega $$
$$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$$
$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$
$${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$$
$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$
Solving above we get
$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$
$${I_1}{\omega _1} + {I_2}{\omega _2} = ({I_1} + {I_2})\omega $$
$$\omega = {{{I_1}{\omega _1} + {I_2}{\omega _2}} \over {{I_1} + {I_2}}}$$
$${k_i} = {1 \over 2}{I_1}\omega _1^2 + {1 \over 2}{I_2}\omega _2^2$$
$${k_f} = {1 \over 2}({I_1} + {I_2}){\omega ^2}$$
$${k_i} - {k_f} = {1 \over 2}\left[ {{I_1}\omega _1^2 + {I_2}\omega _2^2 - {{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}} \over {{I_1} + {I_2}}}} \right]$$
Solving above we get
$${k_i} - {k_f} = {1 \over 2}\left( {{{{I_1}{I_2}} \over {{I_1} + {I_2}}}} \right){({\omega _1} - {\omega _2})^2}$$
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