1. Given

• Bob mass $=m$
• Flywheel is a disc of mass $m$ and radius $r$
• String does not slip, so $v=r\omega$
• Bob starts from rest and falls through height $h$

We need angular speed $\omega$ after falling distance $h$.

2. Use conservation of mechanical energy

Loss in gravitational potential energy of the bob: $mgh$

This becomes:

• Translational kinetic energy of bob
• Rotational kinetic energy of disc

So, $mgh=\frac12 mv^2+\frac12 I\omega^2$

3. Moment of inertia of the disc

For a solid disc about its central axis, $I=\frac12 mr^2$

Substitute into energy equation: $mgh=\frac12 mv^2+\frac12\left(\frac12 mr^2\right)\omega^2$

Using $v=r\omega$, $mgh=\frac12 m(r\omega)^2+\frac14 mr^2\omega^2$

$mgh=\left(\frac12+\frac14\right)mr^2\omega^2$

$mgh=\frac34 mr^2\omega^2$

4. Solve for $\omega$

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

Hence, $\omega=\frac1r\sqrt{\frac{4gh}{3}}$

5. Match with options

This corresponds to: $\boxed{\text{C: }\frac1r\sqrt{\frac{4gh}{3}}}$

6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.