Two discs have moments of intertia $I_{1}$ and $I_{2}$ 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:
Correct Option: , 3
From conservation of angular momentum we get
$\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}=\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega$
$\omega=\frac{\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}}{\mathrm{I}_{1}+\mathrm{I}_{2}}$
$\mathrm{k}_{\mathrm{i}}=\frac{1}{2} \mathrm{I}_{1} \omega_{1}^{2}+\frac{1}{2} \mathrm{I}_{2} \omega_{2}^{2}$
$\mathrm{k}_{\mathrm{f}}=\frac{1}{2}\left(\mathrm{I}_{1}+\mathrm{I}_{2}\right) \omega^{2}$
$\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{1}{2}\left[\mathrm{I}_{1} \omega_{1}^{2}+\mathrm{I}_{2} \omega_{2}^{2}-\frac{\left(\mathrm{I}_{1} \omega_{1}+\mathrm{I}_{2} \omega_{2}\right)^{2}}{\mathrm{I}_{1}+\mathrm{I}_{2}}\right]$
Solving above we get
$\mathrm{k}_{\mathrm{i}}-\mathrm{k}_{\mathrm{f}}=\frac{1}{2}\left(\frac{\mathrm{I}_{1} \mathrm{I}_{2}}{\mathrm{I}_{1}+\mathrm{I}_{2}}\right)\left(\omega_{1}-\omega_{2}\right)^{2}$
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