As shown in the figure, a bob of mass m is tied to a massless string whose other end portion is wound on a fly wheel (disc) of radius r and mass m. When released from rest the bob starts falling vertically. When it has covered a distance of h, the angular speed of the wheel will be:

Moment of inertia of the flywheel (disc): $$I = \frac{1}{2}mr^2$$

No-slip string unwinding condition: $$v = r\omega$$

Total initial mechanical energy of the system: $$E_i = 0$$

Total final mechanical energy after bob drops by height $$h$$: $$E_f = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2 - mgh$$

From $$E_i = E_f$$: $$mgh = \frac{1}{2}m(r\omega)^2 + \frac{1}{2}\left(\frac{1}{2}mr^2\right)\omega^2$$

$$mgh = \frac{1}{2}mr^2\omega^2 + \frac{1}{4}mr^2\omega^2 = \frac{3}{4}mr^2\omega^2$$

$$gh = \frac{3}{4}r^2\omega^2 \implies \omega^2 = \frac{4gh}{3r^2}$$

$$\omega = \frac{1}{r}\sqrt{\frac{4gh}{3}}$$