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.
Solution
Direction of the torque of the rotational force Direction of the torque of the rotational force
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: $$u=v\cos(\frac{\pi}{2}-\alpha )=v\sin\alpha\tag{1}$$ 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 $$\boxed{\omega =\frac{u}{R}=v \frac{h}{R^2}}$$
