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.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.5.17.$ 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 $v$.

    <h3>Solution</h3>
    <p>
        <p>NO: Before viewing the solution to this problem, I recommend that you familiarize yourself with the solution <a href="/1.5.16">1.5.16</a></p>

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 $(1)$

$$\omega = \frac{v}{H(1 + \tan^2 \alpha)}$$

$$\fbox{$\omega = \frac{v \sin^2 \alpha}{H}$}$$

Alternative solution

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

$$\omega=\frac{\upsilon\sin^2 \alpha}{H}$$

    <h4>Answer</h4>
    <p>
        $$\omega = \frac{v \sin^2 \alpha}{H}$$
    </p>


<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>