Solution

Statement

$2.1.60^\ast$. A ring is made from a thin rubber cord of mass $m$ and stiffness $k$. This ring is spun around its axis. Find the new radius of the ring if its angular velocity of rotation is $\omega$, and its initial radius is $R_0$.

Solution

Consider a small segment of the cord corresponding to a small angle $2\alpha$. The length of this segment is $dl = 2\alpha R$.
Due to the uniformity of the cord, the mass of this small segment is:
$$dm = m \frac{2\alpha}{2\pi} = m \frac{\alpha}{\pi}$$

When the ring expands from radius $R_0$ to $R$, the total elongation of the cord is $\Delta L = 2\pi R - 2\pi R_0$.
The tension force $T$ in the cord is determined by Hooke's law:
$$T = k \Delta L = k(2\pi R - 2\pi R_0) = 2\pi k(R - R_0)$$

The tension forces act tangentially at both ends of our small segment. The resultant of these two forces is directed radially inward towards the center. For small angles ($\sin\alpha \approx \alpha$), this resultant force is:
$$F_{\text{res}} = 2T \sin\alpha \approx 2T\alpha$$

According to Newton's second law, this resultant force provides the centripetal acceleration $a_c = \omega^2 R$ for the segment:
$$dm \cdot a_c = 2T\alpha$$
$$m \frac{\alpha}{\pi} \omega^2 R = 2 \left[ 2\pi k(R - R_0) \right] \alpha$$

Cancel $\alpha$ from both sides and simplify:
$$m \frac{\omega^2 R}{\pi} = 4\pi k(R - R_0)$$
$$m \omega^2 R = 4\pi^2 k R - 4\pi^2 k R_0$$
$$R(4\pi^2 k - m \omega^2) = 4\pi^2 k R_0$$

From this, we find the new radius $R$:
$$R = \frac{4\pi^2 k R_0}{4\pi^2 k - m \omega^2} = \frac{R_0}{1 - \frac{m\omega^2}{4\pi^2 k}}$$

Analyzing the resulting expression: if $\omega \ge 2\pi\sqrt{k/m}$, the denominator becomes zero or negative, meaning the tension can no longer compensate for the centrifugal effect. The ring will stretch infinitely and eventually break.