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$2 respectively and are brought into contact face to face with their axes of rotation coaxial. The loss in kinetic energy of the system in the process is given by :
Solution
From conservation of angular momentum we get<br><br>${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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