Last time, I discussed the problem of the elongation of a rotating spring from the perspective of density. Since I was new to topics like elastic deformation, it took a lot of effort. Over the past few days, I have re-examined Hooke’s Law and derived a relatively simpler solution to this problem from another perspective. I am recording it here, along with some of my superficial views on elastic deformation.
In the process of solving this, I once again experienced the feeling that all roads lead to Rome. Science is wonderful in this way; a single destination often has more than one path. Different paths provide us with different scientific perspectives, ultimately allowing us to appreciate different scientific beauties. Walking multiple paths allows us to appreciate the breathtaking beauty of science from various angles, which is also why many travel enthusiasts travel thousands of miles to see beautiful scenery!
Magnificent Dividing Line
Previously, we wrote Hooke’s Law as F = k \cdot \Delta l, where \Delta l is the elongation of the spring. This notation is simple but not perfect because although k is a constant, it is influenced by too many factors, such as the length and thickness of the spring, which affect the stiffness coefficient k. In university mechanics textbooks, Hooke’s Law is written as: \frac{F_n}{S} = E \frac{\Delta l}{l_0}
Here, F_n is the force acting perpendicular to the cross-section of the spring; S is the cross-sectional area; E is called the elastic modulus (also known as Young’s modulus), which is a quantity related to the material. For tension and compression, E is generally different, but the difference is not large. \frac{\Delta l}{l_0} is called the (normal) strain. Since the overall strain of the spring may not be uniform, the rigorous strain should be written as \frac{d(\Delta l)}{dl}. Therefore, Hooke’s Law should be written as: \frac{F_n}{S} = E \frac{d(\Delta l)}{dl} \tag{1}
If the force is uniform, then we have: \frac{F_n}{S \cdot E} dl = d(\Delta l). Integrating this yields: \frac{F_n}{S \cdot E} l_0 = \Delta l \Rightarrow F_n = \frac{E \cdot S}{l_0} \Delta l Comparing this with F = k \Delta l, we can conclude: k = \frac{E \cdot S}{l_0}. This equation clearly explains which quantities k is related to (E, S, l_0 represent the material, thickness, and length of the spring, respectively).
Magnificent Dividing Line
The above is just BoJone’s superficial view of Hooke’s Law. Now, let’s get to the main topic and provide another method for solving the elongation of a rotating spring.
Select the center of rotation of the spring as the reference point. Suppose a mass element dm was originally located at coordinates (x, x+dx) on the spring. After rotation, its coordinates are (y, y+dy). Then the elongation is dy - dx, and the strain is \frac{dy - dx}{dx} = \dot{y} - 1 (where \dot{y} = \frac{dy}{dx}). According to (1), we have: F_n = E \cdot S \cdot (\dot{y} - 1) = k l_0 (\dot{y} - 1) \tag{2}
F_n is the inertial centrifugal force acting on the cross-section, which can be expressed as F_n = F_c = \int_y^l y \omega^2 dm, where l is the length of the spring after rotation. From this, and combined with equation (2), we have the boundary condition: at y=l, \dot{y}=1. Now we differentiate both sides of equation (2) with respect to x: \begin{aligned} d(F_n) &= k l_0 \ddot{y} dx \\ d(F_n) &= -y \omega^2 dm = -\lambda_0 \omega^2 y dx \\ \ddot{y} &= -\frac{\lambda_0 \omega^2}{k l_0} y \end{aligned} \tag{3}
This is a second-order linear homogeneous differential equation. To obtain a solution in a relatively simple form, we do not use the general solution method. Instead, as with problems we have handled before, we let \ddot{y} = \dot{y} \frac{d\dot{y}}{dy}. Substituting this into (3), separating variables, and integrating, we get: \dot{y}^2 = -\frac{\lambda_0 \omega^2}{k l_0} y^2 + C_1 According to the initial conditions y=l, \dot{y}=1, we find C_1 = \frac{\lambda_0 \omega^2}{k l_0} l^2 + 1. Substituting this back into the equation and continuing with the separation of variables, we get: dx = \frac{dy}{\sqrt{\frac{\lambda_0 \omega^2}{k l_0} l^2 + 1 - \frac{\lambda_0 \omega^2}{k l_0} y^2}} Integrating this yields: x = \frac{1}{\omega} \sqrt{\frac{k l_0}{\lambda_0}} \arcsin\left(\frac{\omega \sqrt{\frac{\lambda_0}{k l_0}} y}{\sqrt{\frac{\lambda_0 \omega^2}{k l_0} l^2 + 1}}\right) \tag{4} (The integration constant is 0 because the initial condition is y=0, x=0).
Equation (4) is the result we desired. We want to find the value of l, so we can substitute x=l_0 and y=l into (4). Using the transformation \arcsin(a/b) = \arctan\frac{a}{\sqrt{b^2-a^2}}, it becomes: l_0 = \frac{1}{\omega} \sqrt{\frac{k l_0}{\lambda_0}} \arctan\left(\omega \sqrt{\frac{\lambda_0}{k l_0}} l\right)
That is: l = l_0 \cdot \frac{1}{\omega} \sqrt{\frac{k}{\lambda_0 l_0}} \tan\left(\omega \sqrt{\frac{\lambda_0 l_0}{k}}\right) = l_0 \cdot \frac{1}{\omega} \sqrt{\frac{k}{M}} \tan\left(\omega \sqrt{\frac{M}{k}}\right)
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