As no external torque is applied to the system, the angular momentum of the system remains conserved.

$\therefore$ L_i = L_f

According to given problem,

${I_t}{\omega _i} = \left( {{I_t} + {I_b}} \right){\omega _f}$

$\Rightarrow {\omega _f} = {{{I_t}{\omega _i}} \over {\left( {{I_t} + {I_b}} \right)}}$    ...(i)

Initial energy, ${E_i} = {1 \over 2}{I_t}\omega _i^2$   ...(ii)

Final energy, ${E_f} = {1 \over 2}\left( {{I_t} + {I_b}} \right)\omega _f^2$   ...(iii)

Substituting the value of ${\omega _f}$ from equation (i) in equation (iii), we get

Final energy, ${E_f} = {1 \over 2}\left( {{I_t} + {I_b}} \right){\left( {{{{I_t}{\omega _i}} \over {\left( {{I_t} + {I_b}} \right)}}} \right)^2}$

$= {1 \over 2}{{I_t^2\omega _i^2} \over {{I_t} + {I_b}}}$   ...(iv)

Loss of energy, $\Delta$E = E_i – E_f

$= {1 \over 2}{I_t}\omega _i^2 - {1 \over 2}{{I_t^2\omega _i^2} \over {\left( {{I_t} + {I_b}} \right)}}$     (Using (ii) and (iv))

$= {{\omega _i^2} \over 2}\left( {{I_t} - {{I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}} \right) = {{\omega _i^2} \over 2}\left( {{{I_t^2 - {I_b}{I_t} - I_t^2} \over {\left( {{I_t} + {I_b}} \right)}}} \right)$

$= {1 \over 2}{{{I_b}{I_t}} \over {\left( {{I_t} + {I_b}} \right)}}\omega _i^2$