The thread wound on the axis of the coil is pulled at a speed v at an angle \alpha to the horizon. The reel rolls along a horizontal plane without slipping. Find the speed of the axis and the angular velocity of the rotation of the coil. At what angles does the \alpha axis move to the right? to the left? The thread is so long that the angle \alpha does not change when moving.

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>
        $1.5.9^*.$ The thread wound on the axis of the coil is pulled at a speed $v$ at an angle $\alpha$ to the horizon. The reel rolls along a horizontal plane without slipping. Find the speed of the axis and the angular velocity of the rotation of the coil. At what angles does the $\alpha$ axis move to the right? to the left? The thread is so long that the angle $\alpha$ does not change when moving.

    <h3>Solution</h3>
    <p>
        <p>We denote the radius of the circle along which the coil rolls around the instantaneous axis by $ R^* $:</p>

$$R^* = R \cos \alpha - r$$

The angular velocity of the coil $\omega$ is defined as the ratio of the linear velocity to the radius:

$$\omega = \frac{v}{R \cos \alpha - r}$$

The velocity of the axis $v_0$ can be found by considering the rotation of the coil as a movement around the center $O$:

$$v_0 = \omega R$$

We substitute the found angular velocity:

$$v_0 = \frac{v R}{R \cos \alpha - r}$$

The coil will move to the right if the velocity $v_0$ is positive, and to the left if $v_0$ is negative. The condition for changing direction is:

$$R \cos \alpha = r$$

Accordingly, the critical angle $\alpha^*$ is determined from:

$$\cos \alpha^* = \frac{r}{R}$$

    </p>

    <h4>Answer</h4>
    <p>
        $u=\frac{v R}{R \cos \alpha-r} ;$ $ \omega=\frac{v}{R \cos \alpha-r}$ to the right at $\cos \alpha>r / R $, left at $\cos \alpha < r/R$
    </p>


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