A uniform spring has an initial length l_0, linear density \lambda_0, stiffness coefficient k, and total mass M. Now, in a zero-gravity environment, it rotates around one of its ends with a constant angular velocity \omega. What will its length become under these conditions?
This is a question proposed by the user "Universe as Home" a few days ago. During this period, I attempted to solve it several times, made a few mistakes, and after various corrections, I arrived at the final answer. I would like to thank "Universe as Home" for the many reminders. If there are still errors in the following solution, I hope readers will discover and point them out.
First, the problem must be clearly understood. The spring is uniform before rotation, but after rotation, because different points experience different "inertial centrifugal forces," every part of the spring will elongate non-uniformly, causing the density to no longer be constant. Since the spring becomes non-uniform, using the "length ratio" to measure the stiffness coefficient ratio is no longer reliable; it should be expressed using the "mass ratio." Let the density function after rotation be \lambda = \lambda(r), where r is the distance from a cross-section of the spring to the center of rotation O. The inertial centrifugal force on each point of the spring is dF_c = r \omega^2 dm = \lambda r \omega^2 dr. Thus, the centrifugal force experienced by the cross-section of the spring at distance r from the center is F_c = \int_r^{l} \lambda r \omega^2 dr (Note: F_c stands for Centrifugal force, used here to avoid confusion with the F used below). The stiffness coefficient of a segment of spring with length dr is \frac{M}{dm}k. Due to the action of the centrifugal force, the elongation is \frac{\int_r^{l} \lambda r \omega^2 dr}{Mk/dm} = \frac{\omega^2 dm}{Mk} \int_r^{l} \lambda r dr. The length of this segment before rotation was \frac{dm}{\lambda_0}. Obviously, the density of this segment is: \lambda = \frac{dm}{\frac{dm}{\lambda_0} + \frac{\omega^2 dm}{Mk} \int_r^{l} \lambda r dr} = \frac{1}{\frac{1}{\lambda_0} + \frac{\omega^2}{Mk} \int_r^{l} \lambda r dr}
Look at the last step; we have already set up an "integral equation" regarding the density function: \lambda = \frac{1}{\frac{1}{\lambda_0} + \frac{\omega^2}{Mk} \int_r^{l} \lambda r dr}
This is the key to BoJone’s solution to this problem. For BoJone, solving integral equations is difficult. Through transformation, it can be turned into a differential equation (actually, for BoJone, solving differential equations isn’t easy either).
Let F = \int_r^{l} \lambda r dr, then \lambda = -\frac{dF}{r dr} = -\frac{\dot{F}}{r}. Substituting this into the original equation, we get: \begin{aligned} -\frac{\dot{F}}{r} &= \frac{1}{\frac{1}{\lambda_0} + \frac{\omega^2}{Mk}F} \\ -r &= \frac{\dot{F}}{\lambda_0} + \frac{\omega^2}{Mk}F\dot{F} \end{aligned}
This differential equation appears quite simple. Integrating once yields: C - \frac{1}{2} r^2 = \frac{F}{\lambda_0} + \frac{\omega^2}{2Mk}F^2
C is the integration constant. According to the definition of F, when r=l, we should have F=0, which implies C = \frac{1}{2} l^2. Solving for F from this, we get: F = \pm \sqrt{\frac{Mk}{\omega^2}(l^2-r^2) + \left(\frac{Mk}{\omega^2\lambda_0}\right)^2} - \frac{Mk}{\omega^2\lambda_0} Discard the negative value (think about why yourself?).
We can then calculate: \lambda = -\frac{dF}{r dr} = \frac{Mk}{\omega^2 \sqrt{\frac{M^2 k^2}{\omega^4 \lambda_0^2} + \frac{Mk (l^2-r^2)}{\omega^2}}} = \frac{1}{\sqrt{\frac{1}{\lambda_0^2} + \frac{\omega^2(l^2-r^2)}{Mk}}}
According to the definition of the density function, we must have: \int_0^l \lambda dr = \int_0^l \frac{dr}{\sqrt{\frac{1}{\lambda_0^2} + \frac{\omega^2(l^2-r^2)}{Mk}}} = M The result of this integral is: \frac{\sqrt{Mk}}{\omega} \arcsin\left( \frac{r\omega}{\sqrt{l^2\omega^2 + \frac{Mk}{\lambda_0^2}}} \right)
Thus we have: M = \frac{\sqrt{Mk}}{\omega} \arcsin\left( \frac{l\omega}{\sqrt{l^2\omega^2 + \frac{Mk}{\lambda_0^2}}} \right)
For ease of calculation, using the identity \arcsin(\frac{a}{b}) = \arctan\frac{a}{\sqrt{b^2-a^2}}, the above equation can be transformed into: M = \frac{\sqrt{Mk}}{\omega} \arctan\left(\frac{l\omega\lambda_0}{\sqrt{Mk}}\right)
Solving for l: l = \frac{\sqrt{Mk}}{\lambda_0 \omega} \tan(\omega\sqrt{M/k}) = l_0 \cdot \frac{1}{\omega}\sqrt{k/M} \tan(\omega\sqrt{M/k})
This answer is very beautiful! First, \frac{1}{\omega}\sqrt{k/M} and \omega\sqrt{M/k} are reciprocals of each other; these two possess a symmetrical beauty. Furthermore, the dimension of M is [M], and the dimension of k is [M/T^2], so the dimension of \sqrt{M/k} is [T]. Since the dimension of \omega is [1/T], both \frac{1}{\omega}\sqrt{k/M} and \omega\sqrt{M/k} are dimensionless quantities, which is self-consistent with the derivation process! Moreover, when \omega\sqrt{M/k} \to \pi/2, l \to \infty. In other words, \omega\sqrt{M/k} cannot exceed \pi/2. This provides a parameter to measure the performance of the spring: \omega = \frac{\pi}{2} \sqrt{k/M}. The larger this \omega is, the stronger the spring’s ability to "withstand stress," meaning its performance is better!
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