A track is mounted on a large wheel that is free to tum with negligible friction about a vertical axis (Fig). A toy train of mass Mis placed on the track and, with the system initially at rest, the train's electrical power is turned on. The train reaches speed v with respect to the track. What is the wheel's angular speed if its mass is m and its radius is r? (Treat it as a hoop, and neglect the mass of the spokes and hub.)

No external torque acts on the system consisting of the train and wheel, so the total angular momentum of the system (which is initially zero) remains zero. 

Let I = ${\mathrm{mR}}^2$ be the rotational inertia of the wheel (which we treat as a hoop). Its angular momentum is 

$\overrightarrow{{\mathrm{L}}_{\mathrm{wheel}}} = \left(\mathrm{I\omega }\right)\hat{\mathrm{k}} = -{\mathrm{mR}}^2\left|\mathrm{\omega }\right|\hat{\mathrm{k}}$

where $\hat{\mathrm{k}}$ is up in figure and that last step (with the minus sign) is done in recognition that the wheel's clockwise rotation implies a negative value for $\mathrm{\omega }$. The linear speed of a point on the track is -|$\mathrm{\omega }$|R  and the speed of the train (going counterclockwise in figure with speed v'relative to an outside observer) is therefore ${\mathrm{v}}^{\text{'}} = \mathrm{v}-\left|\mathrm{\omega }\right|\mathrm{R}$ where v is its speed relative to the tracks. Consequently, the angular momentum of
the train is $\overrightarrow{{\mathrm{L}}_{\mathrm{train}}} = \mathrm{M}(\mathrm{v}-|\mathrm{\omega }\left|\mathrm{R}\right)\mathrm{R}\hat{\mathrm{k}}$. Conservation of angular momentum yields

$0 = \overrightarrow{{\mathrm{L}}_{\mathrm{wheel}}}+\overrightarrow{{\mathrm{L}}_{\mathrm{train}}} = -{\mathrm{mR}}^2\left|\mathrm{\omega }\right|\hat{\mathrm{k}}+ \mathrm{M}(\mathrm{v}-|\mathrm{\omega }\left|\mathrm{R}\right)\mathrm{R}\hat{\mathrm{k}}$

which we can use to solve for |$\mathrm{\omega }$|.
Solving for the angular speed, the result is

$\left|\mathrm{\omega }\right| = \frac{\mathrm{MvR}}{(\mathrm{M}+\mathrm{m}){\mathrm{R}}^2}= \frac{\mathrm{v}}{({\displaystyle \frac{\mathrm{m}}{\mathrm{M}}}+1)\mathrm{R}}$