$2.1.57.$ A small bead is threaded onto a smooth wire ring of radius $R$ positioned vertically. The ring rotates with an angular velocity $\omega$ about a vertical axis passing through its diameter. Where is the bead located?
Solution
Let's write Newton's second law for the bead on the vertical and horizontal axes, projecting the reaction force $T$ of the ring: $$T \sin\alpha = m a_{\text{cp}}$$ $$mg - T \cos\alpha = 0$$
From this, we obtain: $$mg \tan\alpha = m a_{\text{cp}} \quad (1)$$
Let's find the centripetal acceleration through the angular velocity of rotation $\omega$: $$a_{\text{cp}} = \omega^2 r = \omega^2 R \sin\alpha$$
Substitute this into (1): $$mg \frac{\sin\alpha}{\cos\alpha} = m \omega^2 R \sin\alpha$$ $$g = \omega^2 R \cos\alpha$$
From which we find $\cos\alpha$: $$\cos\alpha = \frac{g}{\omega^2 R}$$
Considering the domain of the cosine function: $$-1 \le \cos\alpha \le 1$$
If $g > \omega^2 R$, the angular velocity will not be enough to lift the body to any angle, and the body will take a stable position at the lowest point of its trajectory, corresponding to: $$\alpha = 0$$
Answer
$\cos\alpha = \frac{g}{\omega^2 R}$ for $\frac{g}{\omega^2 R} \le 1$; $\alpha = 0$ for $\frac{g}{\omega^2 R} > 1$
