$1.5.17.$ The log, resting its lower end in the corner between the wall and the ground, touches the bottom of the truck at a height $H$ from the ground. Find the angular velocity of the log as a function of the angle $\alpha$ between it and the horizontal if the truck is moving away from the wall at speed $v$.
NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution 1.5.16
At time $t$ the car will travel the distance $vt$ and the bottom will have horizontal coordinate $x$
$$ x = H \tan \alpha\;(1) $$
Let's consider the change in angle $\alpha = \arctan \frac{H}{x}$ over a small period of time $dt$
$$ d\alpha = d\left(\arctan \frac{H}{x}\right) $$
$$ d\alpha = \frac{1}{1+\frac{H^2}{x^2}} \frac{H}{x^2} $$
$$ d\alpha = dx\frac{H}{x^2+H^2} $$
Considering that $\omega = \frac{d \alpha}{dt}$ and $v = \frac{dx}{dt}$
$$ \omega = \frac{vH}{x^2+H^2} $$
Substitute $(1)$
$$ \omega = \frac{v}{H(1 + \tan^2 \alpha)} $$
$$ \fbox{$\omega = \frac{v \sin^2 \alpha}{H}$} $$
In time $\Delta t$ the truck will travel a distance
$$ \Delta l=\upsilon \Delta t $$
The log will drop slightly, turning at a small angle $\Delta \alpha$. If the length of the arc of rotation is $\Delta x$, then by definition of the radian measure of angle
$$ \Delta \alpha=\frac{\Delta x }{R} $$
Angular velocity is defined as
$$ \omega=\frac{\Delta \alpha }{\Delta t }=\frac{\Delta x }{R\Delta t } $$
Но $\Delta x =\upsilon \Delta t\sin \alpha$.
$$ \omega=\frac{\upsilon \Delta t\sin \alpha }{R\Delta t }=\frac{\upsilon\sin \alpha }{R} $$
We can also write that
$$ H=R\sin \alpha $$
With this in mind
$$ \omega=\frac{\upsilon\sin^2 \alpha}{H} $$
$$\omega = \frac{v \sin^2 \alpha}{H}$$