A circular disk of moment of inertia I_t is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed $ω_{i}$ . 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 $ω_{f}$ . The energy lost by the initially rotating disc to friction is :

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$\frac{1}{2} \frac{I_{b} I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{b}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{1}{2} \frac{I_{t}^{2}}{(I_{t} + I_{b})} ω_{i}^{2}$

$\frac{I_{b} - I_{t}}{(I_{t} + I_{b})} ω_{i}^{2}$

answer is A.

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Detailed Solution

By COAM, $(I_{t} + I_{b}) ω_{f} = I_{t} ω_{i} \Rightarrow ω_{f} = \frac{I_{t} ω_{i}}{I_{t} + I_{b}}$
Energy lost by initially rotating disc
$= \frac{1}{2} I_{t} ω_{i}^{2} - \frac{1}{2} (I_{t} + I_{b}) ω_{f}^{2} = \frac{I_{b} I_{t}}{2 (I_{t} + I_{b})} ω_{i}^{2}$