The circle is so harmonious and perfect that mathematicians and physicists hold it in high regard. Geometrically, a local segment of a curve can be viewed as an arc of a circle, allowing us to study it using the properties of circles (in mathematics, the reciprocal of the curvature radius is the curvature; the greater the curvature, the more curved the line). Physicists like to treat a local segment of a particle’s curved trajectory as circular motion, using the methods of circular motion to describe it. Both research methods tell us that two different types of "lines" can be equivalent within an infinitesimal range. This also provides a guiding principle for our scientific inquiries: transform the unknown into the known, and view the unknown through the known. In both the physical and mathematical treatments, there is one point where all roads lead to Rome: after treating the trajectory as a circle, what is the radius of that circle? We must first derive it.
In mathematical analysis, one can use knowledge of calculus to derive the formula for the curvature radius. However, I prefer a physical approach. By combining physics with vector knowledge to derive the curvature radius formula, I find it to have a "unique flavor."
Derivation 1:
First, let us consider the two-dimensional case. We know that the centripetal acceleration formula is a_c = \frac{v^2}{R}. If we can determine the centripetal acceleration and the velocity, we can find R (the curvature radius). For any given equation of motion \vec{r} = \vec{r}(t), the velocity is already known as \vec{v} = \dot{\vec{r}}. The remaining unknown is the centripetal acceleration.
What does "centripetal" mean? It is actually the projection of the acceleration onto the radius of the circle at that point. We know that the direction of velocity in circular motion is always perpendicular to the radial direction, so the "centripetal" direction is actually the direction perpendicular to the velocity (this is also called the "normal" direction, which is why centripetal acceleration is also called "normal acceleration").
\vec{a} = \ddot{\vec{r}}, \quad |a_c| = |\vec{a}| \cdot \sin\theta = |\dot{\vec{r}} \times \ddot{\vec{r}}| \div |\dot{\vec{r}}|
Therefore, R = \frac{v^2}{a_c} = \frac{|\dot{\vec{r}}|^3}{|\dot{\vec{r}} \times \ddot{\vec{r}}|}
Let \vec{r} = (x, y, 0), then we have: \begin{aligned} \dot{\vec{r}} &= (\dot{x}, \dot{y}, 0), \quad \ddot{\vec{r}} = (\ddot{x}, \ddot{y}, 0) \\ \dot{\vec{r}} \times \ddot{\vec{r}} &= (0, 0, \dot{x}\ddot{y} - \dot{y}\ddot{x}) \end{aligned}
Substituting these into the formula, we get: R = \frac{(\dot{x}^2 + \dot{y}^2)^{3/2}}{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}
If we use three-dimensional coordinates, we can obtain the curvature radius of a curve in 3D space: R = \frac{(x'^2 + y'^2 + z'^2)^{3/2}}{\sqrt{(z''y' - y''z')^2 + (x''z' - z''x')^2 + (y''x' - x''y')^2}}
Derivation 2:
In this article, we obtained two equations regarding circular motion: \vec{R} \cdot \dot{\vec{r}} = 0 \dot{\vec{r}}^2 + \vec{R} \cdot \ddot{\vec{r}} = 0
Similarly, let \vec{r} = (x, y) and \vec{R} = (a, b). Substituting these gives: \dot{x}^2 + \dot{y}^2 + a\ddot{x} + b\ddot{y} = 0; \quad a\dot{x} + b\dot{y} = 0
Solving these equations, we get: \begin{aligned} a &= -\frac{\dot{x}^2\dot{y} + \dot{y}^3}{\ddot{x}\dot{y} - \dot{x}\ddot{y}} \\ b &= \frac{\dot{y}^2\dot{x} + \dot{x}^3}{\ddot{x}\dot{y} - \dot{x}\ddot{y}} \end{aligned}
Thus, R = \sqrt{a^2 + b^2} = \frac{(\dot{x}^2 + \dot{y}^2)^{3/2}}{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}
When reprinting, please include the original address: https://kexue.fm/archives/714
For more details on reprinting, please refer to: Scientific Space FAQ