Solution

Statement

$2.1.63.$ A horizontal disk begins to spin around its axis with an angular velocity that increases linearly with time $\omega = \varepsilon t$. At what angular velocity will a body located at a distance $r$ from the disk's axis begin to slide off it, if the coefficient of friction between them is $\mu$?

Solution

Because the angular velocity is constantly changing, the body experiences two components of acceleration on the plane of the disk:

1. Centripetal (normal) acceleration, directed towards the center: $a_n = \omega^2 r$
2. Tangential acceleration, directed perpendicular to the radius: $a_\tau = \varepsilon r$

Since these two vectors are strictly perpendicular, we find the total magnitude of acceleration $a$ using the Pythagorean theorem:
$$a = \sqrt{a_n^2 + a_\tau^2} = \sqrt{(\omega^2 r)^2 + (\varepsilon r)^2} = r \sqrt{\omega^4 + \varepsilon^2}$$

The body starts to slide at the exact moment when the required force to maintain this total acceleration exceeds the maximum possible static friction force $F_{\text{fr}} = \mu N = \mu mg$.
Writing Newton's second law for the threshold of slipping:
$$ma = \mu mg \implies a = \mu g$$

Substitute the expression for total acceleration:
$$r \sqrt{\omega^4 + \varepsilon^2} = \mu g$$

Square both sides to solve for $\omega$:
$$r^2 (\omega^4 + \varepsilon^2) = \mu^2 g^2$$
$$\omega^4 + \varepsilon^2 = \frac{\mu^2 g^2}{r^2}$$
$$\omega^4 = \frac{\mu^2 g^2}{r^2} - \varepsilon^2$$

Taking the fourth root, we find the angular velocity $\omega$:
$$\omega = \left( \frac{\mu^2 g^2}{r^2} - \varepsilon^2 \right)^{1/4}$$

Note: This solution is physically meaningful only if the expression under the root is positive, i.e., $\varepsilon < \mu g / r$. If the angular acceleration $\varepsilon$ is extremely large ($\varepsilon \ge \mu g / r$), the tangential acceleration alone immediately overcomes friction, and the body slips at $t=0$ when $\omega = 0$.