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+### Statement
+
+$2.1.59.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$2.1.59.$ A weight of mass $m$, attached by a spring of stiffness $k$ to an axis, moves around this axis in a circle of radius $R$ with an angular velocity $\omega$. What is the length of the undeformed spring?
+
+### Solution
+
+We write Newton's second law for the weight, considering the elastic force of the spring $F_{\text{el}}$:
+$$m a_{\text{cp}} = F_{\text{el}} = k\Delta x$$
+
+Let's express $\Delta x$ through the length of the undeformed spring $l$:
+$$m a_{\text{cp}} = F_{\text{el}} = k(R - l) \quad (1)$$
+
+Let's find the centripetal acceleration through the angular velocity of rotation $\omega$:
+$$a_{\text{cp}} = \omega^2 R$$
+
+Substitute this into (1):
+$$m \omega^2 R = k(R - l)$$
+$$\frac{k}{m}(R - l) = \omega^2 R$$
+
+From which we find $l$:
+$$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$
+
+#### Answer
+$$l = R \left( 1 - \frac{m\omega^2}{k} \right)$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]