After discussing the analysis of an inverted pendulum, Brother Hu Xiong (a good friend of the author) raised a question: For a uniform rod, where the mass is naturally not negligible, if a single force (for simplicity, assume it is a constant force) acts on one point (for simplicity, assume it is an endpoint), how does the rod move?
In fact, although I have studied a fair amount of classical mechanics and analyzed many problems, I remain somewhat vague about concepts like torque and angular momentum. For me, most classical mechanics problems are simply a matter of "action + variation," and this problem is no exception. To make the experimental significance of the problem clearer, we might as well rephrase it as follows:
A neutral uniform rod has a point charge at one of its endpoints. What is its motion in a (uniform) electric field?
Here, we simplify further and only consider the planar problem. The rod is a rigid body. To describe its motion, we need to describe the motion of a point on the rod and the rotation of the rod about that point. That is to say, even considering only the planar case, the system has three degrees of freedom. Let the coordinates of the charged endpoint of the rod be (x, y). To describe the rotation, we establish a polar coordinate system centered at this endpoint and let the polar angle of the rod be \theta. Let the potential function be U(x, y). Since only one point is charged (subject to force), the potential energy is simple.
Equations of Motion
However, since both rotation and translation exist, the kinetic energy is slightly more complex. Let the linear density of the rod be \rho, the length be R, and the mass be m = \rho R. Then the coordinates of a point on the rod at a distance r from the endpoint (relative to an inertial frame) are: (X, Y) = (x, y) + (r \cos \theta, r \sin \theta) Its velocity is: (\dot{X}, \dot{Y}) = (\dot{x} - r \sin \theta \dot{\theta}, \dot{y} + r \cos \theta \dot{\theta}) The kinetic energy is the integral: \begin{aligned} E_k &= \int_0^R \frac{1}{2} \left(\dot{X}^2 + \dot{Y}^2\right) \rho dr \\ &= \int_0^R \frac{1}{2} \left(\dot{x}^2 + \dot{y}^2 + r^2 \dot{\theta}^2 + 2r \dot{\theta} \dot{y} \cos \theta - 2r \dot{\theta} \dot{x} \sin \theta\right) \rho dr \\ &= \frac{1}{2} \rho R \left(\dot{x}^2 + \dot{y}^2 + \frac{1}{3} R^2 \dot{\theta}^2 + R \dot{\theta} \dot{y} \cos \theta - R \dot{\theta} \dot{x} \sin \theta\right) \\ &= \frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2 + \frac{1}{3} R^2 \dot{\theta}^2 + R \dot{\theta} \dot{y} \cos \theta - R \dot{\theta} \dot{x} \sin \theta\right) \end{aligned} Although it is somewhat complex, we can now write the action: S = \int \left[\frac{1}{2} m \left(\dot{x}^2 + \dot{y}^2 + \frac{1}{3} R^2 \dot{\theta}^2 + R \dot{\theta} \dot{y} \cos \theta - R \dot{\theta} \dot{x} \sin \theta\right) - U(x, y)\right] dt Varying it, i.e., substituting into the Euler-Lagrange equations, we obtain respectively: \begin{aligned} m \frac{d}{dt} \left(\dot{x} - \frac{1}{2} R \dot{\theta} \sin \theta\right) &= -\frac{\partial U}{\partial x} \\ m \frac{d}{dt} \left(\dot{y} + \frac{1}{2} R \dot{\theta} \cos \theta\right) &= -\frac{\partial U}{\partial y} \\ m \frac{d}{dt} \left(\frac{1}{3} R^2 \dot{\theta} + \frac{1}{2} R \dot{y} \cos \theta - \frac{1}{2} R \dot{x} \sin \theta\right) &= -\frac{1}{2} m R \dot{\theta} \dot{y} \sin \theta - \frac{1}{2} m R \dot{\theta} \dot{x} \cos \theta \end{aligned} Even without the principle of least action, it is easy to list the first two equations. They simply state that the net external force acting on the system equals the total mass of the system multiplied by the acceleration of its center of mass. In plain terms, when discussing the motion of the center of mass, the system is equivalent to a single particle. Therefore, considering that the force might not be conservative, the most general equations should be written as: \begin{aligned} &m \frac{d}{dt} \left(\dot{x} - \frac{1}{2} R \dot{\theta} \sin \theta\right) = F_x \\ &m \frac{d}{dt} \left(\dot{y} + \frac{1}{2} R \dot{\theta} \cos \theta\right) = F_y \end{aligned}
The equation regarding \dot{\theta} is more difficult to derive directly. I would be very grateful if any friends could provide a mechanical analysis or explanation for it. If we substitute the expressions for \ddot{x} and \ddot{y} into the expression for \dot{\theta}, we obtain a result with a relatively more obvious physical meaning: \begin{aligned} \frac{1}{3} m R^2 \ddot{\theta} + \frac{1}{2} m R \left(-\frac{1}{2} R \ddot{\theta} \cos \theta + \frac{1}{2} R \dot{\theta}^2 \sin \theta + \frac{F_y}{m}\right) \cos \theta \\ - \frac{1}{2} m R \left(\frac{1}{2} R \ddot{\theta} \sin \theta + \frac{1}{2} R \dot{\theta}^2 \cos \theta + \frac{F_x}{m}\right) \sin \theta = 0 \end{aligned} i.e., \frac{1}{6} m R \ddot{\theta} + F_y \cos \theta - F_x \sin \theta = 0 The physical meaning seems obvious, but I still cannot find a precise physical explanation. I look forward to your guidance. At this point, we have completed the first step, which is to list the equations of motion.
Constant Force
If the force F = (F_x, F_y) is a constant force, one can always choose an appropriate coordinate system such that one of the components is 0. Let us assume F_y = 0, then: \begin{aligned} m \frac{d}{dt} \left(\dot{x} - \frac{1}{2} R \dot{\theta} \sin \theta\right) &= F_x \\ m \frac{d}{dt} \left(\dot{y} + \frac{1}{2} R \dot{\theta} \cos \theta\right) &= 0 \\ \frac{1}{6} m R \ddot{\theta} - F_x \sin \theta &= 0 \end{aligned} The first two equations are easy to integrate: \begin{aligned} x + \frac{1}{2} R \cos \theta &= \frac{F_x}{2m} t^2 + C_1 t + C_2 \\ y + \frac{1}{2} R \sin \theta &= C_3 t + C_4 \end{aligned} This indicates that the motion of the center of mass is similar to projectile motion.
The third equation, \frac{1}{6} m R \ddot{\theta} - F_x \sin \theta = 0, is similar to the equation of a simple pendulum. When F_x is positive, it corresponds to an inverted pendulum; when F_x is negative, it corresponds to an ordinary pendulum. For a simple demonstration, let us assume the initial \theta is small and F_x < 0, yielding the approximate solution: \theta = C_5 \cos \omega t + C_6 \sin \omega t, \quad \omega = \sqrt{\frac{-6F_x}{mR}}
Taking a specific case, the approximate animation of its motion is as follows:
[View GIF Animation: Uniform Rod with a Point Charge]
![[View GIF Animation: Uniform Rod with a Point Charge]](https://kexue.fm/usr/uploads/2014/02/4133040644.gif){kind=link}
(The point of force application is the left endpoint, the force is constant and directed horizontally to the left, and the initial velocity is 0.)
[View GIF Animation: Uniform Rod with a Point Charge 2]
![[View GIF Animation: Uniform Rod with a Point Charge 2]](https://kexue.fm/usr/uploads/2014/02/2789803481.gif){kind=link}
(The point of force application is the left endpoint, the force is constant and directed horizontally to the left, with a vertical downward initial velocity.)
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