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en/2.1.60.md
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| + | ### Statement | ||
| + | |||
| + | $2.1.60.$ [Insert the problem statement] | ||
| + | |||
| + | ### 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 $R$: | ||
| + | $$R = \frac{4\pi^2 k R_0}{4\pi^2 k - m \omega^2} = \frac{R_0}{1 - \frac{m\omega^2}{4\pi^2 k}}$$ | ||
| + | |||
| + | <i>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.</i> | ||
| + | |||
| + | #### Answer | ||
| + | $R = \frac{R_0}{1 - m\omega^2 / (4\pi^2 k)}$ for $\omega < 2\pi\sqrt{k/m}$; | ||
| + | for $\omega \ge 2\pi\sqrt{k/m}$ the ring stretches infinitely. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $2.1.60.$ [Insert the problem statement] | |||
| ### 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 $R$: | |||
| $$R = \frac{4\pi^2 k R_0}{4\pi^2 k - m \omega^2} = \frac{R_0}{1 - \frac{m\omega^2}{4\pi^2 k}}$$ | |||
| <i>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.</i> | |||
| #### Answer | |||
| $R = \frac{R_0}{1 - m\omega^2 / (4\pi^2 k)}$ for $\omega < 2\pi\sqrt{k/m}$; | |||
| for $\omega \ge 2\pi\sqrt{k/m}$ the ring stretches infinitely. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||