Solutions of Savchenko Problems in Physics
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<h3> Statement </h3>
<p>
$2.8.37^*.$ 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.
<h3>Solution</h3>
<p>
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.
<p>
Thus, the momentum will make an angle with velocity $v$ equal to $\pi-\alpha$
</p>
<p>
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
$$\boxed{\omega=\frac{u}{R}=v \frac{h}{R^2}}$$
</p>
<h4>Answer</h4>
<p>
$$\omega=\frac{hv}{R^2}$$
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