A wheel having a moment of inertia of 0.500 kg-m^2 and a radius of...

Initial moment of inertia of the wheel, $I_1 = 0.500 \text{ kg-m}^2$

Initial angular speed, $\omega_1 = 20.0 \text{ rad/s}$

Mass of the particle, $m = 200 \text{ g} = 0.200 \text{ kg}$

Radius of the wheel, $r = 20.0 \text{ cm} = 0.200 \text{ m}$

Moment of inertia of the particle about the axis of rotation, $I_p = m r^2 = 0.200 \times (0.200)^2 = 0.008 \text{ kg-m}^2$

Final moment of inertia of the system, $I_2 = I_1 + I_p = 0.500 + 0.008 = 0.508 \text{ kg-m}^2$

By the principle of conservation of angular momentum:

$I_1 \omega_1 = I_2 \omega_2$

$0.500 \times 20.0 = 0.508 \times \omega_2$

$\omega_2 = \dfrac{10.0}{0.508} \approx 19.7 \text{ rad/s}$