Moment of inertia is the name given to rotational inertia, the rotational analogue of mass for linear motion.
When the mass is left to descend after winding, there is a loss in potential energy P_loss. $$P_{loss}=mgh$$ This is converted into three energies: The flywheel’s kinetic energy, K_f, given by: $$K_f=\frac{1}{2}Iω^2$$ The vertical kinetic energy of falling mass, K_m: $$K_m=\frac{1}{2}mv^2$$ Loss due to frictional torque, W_f: $$W_f=\frac{n}{2N}Iω^2$$ Equating energies, $$P_{loss}=K_f+K_m+W_f$$ $$⇒mgh=\frac{1}{2}Iω^2+\frac{1}{2}mv^2+\frac{n}{2N}Iω^2$$ $$⇒mgh= \frac{1}{2}Iω^2+\frac{1}{2}mω^2r^2+\frac{n}{2N}Iω^2$$ $$⇒mgh= \frac{1}{2}Iω^2+\frac{1}{2}mω^2r^2+\frac{n}{2N}Iω^2$$ $$I=\frac{Nm}{N+n}\left[\frac{2gh}{ω^2}-r^2\right]$$