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+### Statement
+
+$2.1.58.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.1.58.$ 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 $\omega$:
+$$\omega = \sqrt{\frac{g \tan\beta}{l(\sin\beta + \sin\alpha)}}$$
+
+#### Answer
+$$\omega = \sqrt{\frac{g \tan\beta}{l(\sin\beta + \sin\alpha)}}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]