The kinetic energy of the circular disc in initial state is,
[tex]K^{1r} = \frac{1}{2}I^{t}ω^{ \frac{2}{i} } [/tex]
The initial kinetic energy of the other disc dropped on the first disc is,
[tex]K^{2r} = \frac{1}{2}I^{b}(0)^{2}[/tex]
[tex] = > K^{2r} = 0J[/tex]
As both the disc in the final state are moving with the same angular velocity, thus, the net kinetic energy in the final state is,
[tex]E^{f} = \frac{1}{2}(I^{t} + I^{b})ω^{ \frac{2}{f} }[/tex]
The net kinetic energy in the initial state is,
[tex]E^{i} = K^{1r} + K^{2r}[/tex]
[tex] = > E^{i} = \frac{1}{2} I^{t}ω^{ \frac{2}{i} } + 0[/tex]
[tex] = > E^{i} = \frac{1}{2} I^{t}ω^{ \frac{2}{i} }[/tex]
Thus, the change in the kinetic energy during the change of state is,
[tex]dE = E^{f} - E^{i}[/tex]
[tex] = > dE = \frac{1}{2} (I^{t} + I^{b})ω^{ \frac{2}{f} } - \frac{1}{2}I^{t}ω^{ \frac{2}{i} } [/tex]
[tex] = > dE = \frac{1}{2} I^{t}(ω^{ \frac{2}{f} } - ω^{ \frac{2}{i} }) + \frac{1}{2}I^{b}ω^{ \frac{2}{f} } [/tex]
This change in value of energy is the energy lost initially rotating the disk to friction.
Hence, the energy lost in the given case is [tex]dE = \frac{1}{2} I^{t}(ω^{ \frac{2}{f} } - ω^{ \frac{2}{i} }) + \frac{1}{2}I^{b}ω^{ \frac{2}{f} } [/tex]
