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A circular disk of moment of inertia $I_t$ is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $\omega_i$. Another disk of moment of inertia $I_b$ is dropped coaxially onto the rotating disk. Initially the second disk has zero angular speed. Eventually both the disks rotate with a constant angular speed $\omega_{\mathrm{f}}$. The energy lost by the initially rotating disc due to friction is
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The correct answer is:
$\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}}}{\left(\mathrm{L}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)} \omega_{\mathrm{i}}^2$
Loss of energy, $\Delta \mathrm{E}=\frac{1}{2} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2-\frac{1}{2} \frac{\mathrm{I}_{\mathrm{t}}^2 \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}$
$$
=\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}
$$
$$
=\frac{1}{2} \frac{\mathrm{I}_{\mathrm{b}} \mathrm{I}_{\mathrm{t}} \omega_{\mathrm{i}}^2}{\left(\mathrm{I}_{\mathrm{t}}+\mathrm{I}_{\mathrm{b}}\right)}
$$
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