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Two discs of moment of inertia $I_1$ and $I_2$ and angular speeds ${\omega _1}\,{\rm{and }}{\omega _2}$ are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational $KE$ of system will be
$\frac{{{I_1}{\omega _1} + {I_2}{\omega _2}}}{{2({I_1} + {I_2})}}$
$\frac{{({I_1} + {I_2})\,{{({\omega _1} + {\omega _2})}^2}}}{2}$
$\frac{{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}}}{{2({I_1} + {I_2})}}$
None of these
Solution
By the law of conservation of angular momentum ${I_1}{\omega _1} + {I_2}{\omega _2} = \left( {{I_1} + {I_2}} \right)\,\omega $
Angular velocity of system $\omega = \frac{{{I_1}{\omega _1} + {I_2}{\omega _2}}}{{{I_1} + {I_2}}}$
Rotational kinetic energy $ = \frac{1}{2}\left( {{I_1} + {I_2}} \right)\;{\omega ^2}$ $ = \frac{1}{2}\left( {{I_1} + {I_2}} \right)\,{\left( {\frac{{{I_1}{\omega _1} + {I_2}{\omega _2}}}{{{I_1} + {I_2}}}} \right)^2}$$ = \frac{{{{({I_1}{\omega _1} + {I_2}{\omega _2})}^2}}}{{2({I_1} + {I_2})}}$.
