$2.8.37^*.$ A wheel of radius $R$ can rotate freely around its axis. Drive belts moving at a speed $v$ are pressed against the side surface of the wheel at a distance $h$ from the axis of rotation. Determine the steady-state angular velocity of the wheel if its contact with the drive belt occurs only along the rim.
In order for the wheel to move it is necessary that a moment of force associated with the frictional force from drive belts is transmitted, which is directed along the axis of rotation.
Thus, the momentum will make an angle with velocity $v$ equal to $\pi-\alpha$
Projecting onto the linear rotation speed of the wheel: $$u=v\cos(\frac{\pi}{2}-\alpha)=v\sin\alpha\quad(1)$$ From the drawing $$\sin\alpha = \frac{h}{R}$$ Substituting into $(1)$ $$u=v \frac{h}{R}$$ From where we find the linear velocity, given that the wheel is circular $$\boxed{\omega=\frac{u}{R}=v \frac{h}{R^2}}$$
$$\omega=\frac{hv}{R^2}$$