| A circular disk of moment of inertia \[{{I}_{t}}\] is rotating in a horizontal plane, about its symmetry axis, with a constant angular speed \[{{\omega }_{t}}\]. 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 due to friction is [AIPMT (S) 2010] |
A) \[\frac{1}{2}\frac{I_{b}^{2}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]
B) \[\frac{1}{2}\frac{I_{t}^{2}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]
C) \[\frac{1}{2}\frac{{{I}_{b}}-{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]
D) \[\frac{1}{2}\frac{{{I}_{b}}{{I}_{t}}}{({{I}_{t}}+{{I}_{b}})}\omega _{i}^{2}\]
Correct Answer: D
Solution :
| Loss of energy, \[\Delta E=\frac{1}{2}{{I}_{t}}\omega _{i}^{2}-\frac{1}{2}\frac{I_{t}^{2}\omega _{i}^{2}}{2({{I}_{t}}+{{I}_{b}})}\] |
| \[=\frac{1}{2}\frac{I_{b}^{{}}{{I}_{t}}\omega _{i}^{2}}{({{I}_{t}}+{{I}_{b}})}\] |
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