$2.1.58.$ A heavy ball suspended on a thread of length $l$ has a second heavy ball suspended from it on a thread of the same length. When the balls rotate about a vertical axis passing through the top suspension point, both threads lie in the same plane and make constant angles $\alpha$ and $\beta$ with the vertical. Find the angular velocity of rotation of the balls.
Solution
Projecting the tension force $T$ for the lower ball, we write Newton's second law on the vertical and horizontal axes: $$T \sin\beta = m a_{\text{cp}}$$ $$mg - T \cos\beta = 0$$
From this: $$mg \tan\beta = m a_{\text{cp}} \quad (1)$$
Let's find the centripetal acceleration through the angular velocity of rotation $\omega$: $$a_{\text{cp}} = \omega^2 r = \omega^2(l_1 + l_2)$$ $$a_{\text{cp}} = \omega^2 l(\sin\alpha + \sin\beta)$$
Substitute this into (1): $$g \tan\beta = \omega^2 l(\sin\alpha + \sin\beta)$$
From which we find $\omega$: $$\omega = \sqrt{\frac{g \tan\beta}{l(\sin\beta + \sin\alpha)}}$$
