Energy lost due to friction between disks – Rankers Physics

Energy lost due to friction between disks

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 into 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 due to friction is: (2010)

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

Solution:

By conservation of angular momentum, $I_{t}\omega_{i} = (I_{t}+I_{b})\omega_{f}$, yielding $\omega_{f} = \frac{I_{t}\omega_{i}}{I_{t}+I_{b}}$. The loss in kinetic energy is $\Delta K = \frac{1}{2}I_{t}\omega_{i}^{2} - \frac{1}{2}(I_{t}+I_{b})\omega_{f}^{2}$. Substituting $\omega_{f}$ simplifies to $\Delta K = \frac{1}{2}\frac{I_{t}I_{b}}{(I_{t}+I_{b})}\omega_{i}^{2}$.

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