A circular disk of moment of inertia $I_{t}$ is rotating in a horizontal plane, about its symmetry a

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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 $\omega_{f}$. The energy lost by the initially rotating disc to friction is\

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(d) : As no external torque is applied to the system, the angular momentum of the system remains conserved.\ $	herefore \quad L_{i}=L_{f}$\ According to given problem, $$ I_{t} \omega_{i}=\left(I_{t}+I_{b}ight) \omega_{f} \quad 	ext { or } \quad \omega_{f}=\frac{I_{t} \omega_{i}}{\left(I_{t}+I_{b}ight)} 	ag{i} $$ Initial energy, $E_{i}=\frac{1}{2} I_{t} \omega_{i}^{2}$\ Final energy, $E_{f}=\frac{1}{2}\left(I_{t}+I_{b}ight) \omega_{f}^{2}$\ Substituting the value of $\omega_{f}$ from equation (i) in equation (iii), we get\ Final energy, $$ E_{f}=\frac{1}{2}\left(I_{t}+I_{b}ight)\left(\frac{I_{t} \omega_{i}}{I_{t}+I_{b}}ight)^{2}=\frac{1}{2} \frac{I_{t}^{2} \omega_{i}^{2}}{\left(I_{t}+I_{b}ight)} 	ag{iv} $$ Loss of energy, $\Delta E=E_{i}-E_{f}$ $$ \begin{aligned} & \quad=\frac{1}{2} I_{t} \omega_{i}^{2}-\frac{1}{2} \frac{I_{t}^{2} \omega_{i}^{2}}{\left(I_{t}+I_{b}ight)} \quad 	ext { (Using (ii) and (iv)) } \ & =\frac{\omega_{i}^{2}}{2}\left(I_{t}-\frac{I_{t}^{2}}{\left(I_{t}+I_{b}ight)}ight)=\frac{\omega_{i}^{2}}{2}\left(\frac{I_{t}^{2}+I_{b} I_{t}-I_{t}^{2}}{\left(I_{t}+I_{b}ight)}ight) \ & =\frac{1}{2} \frac{I_{b} I_{t}}{\left(I_{t}+I_{b}ight)} \omega_{i}^{2} \end{aligned} $$