We often hear physical terms such as Newtonian mechanics, relativistic mechanics, and quantum mechanics, and occasionally we hear of "Theoretical Mechanics." In fact, the term "Theoretical Mechanics" is somewhat inappropriate because it can easily lead people to mistakenly believe it is a new system of mechanics. In reality, Theoretical Mechanics is not a system of mechanics like Newtonian mechanics, but rather a method of studying mechanics, and the objects of study in most cases remain classical mechanics (this is easily discovered by opening any "Theoretical Mechanics" textbook). Simply put, it transforms the method of studying mechanics from the calculus used in Newton’s era into "variations," and changes "ordinary differential equations" into "partial differential equations." This might seem like "complicating the simple," but in fact, such a transformation brought about a giant leap in the study of mechanics.
Speaking of this, some readers might feel intimidated: this surely involves various profound and mysterious advanced mathematical methods, and we can only stay away. Indeed, the methods in theoretical mechanics are very deep; even an excellent undergraduate student in mathematics or physics might spend more than a year to finish a textbook on "Theoretical Mechanics." However, studying physics through the Principle of Least Action is so tempting. Is it impossible for beginners like us to personally experience the great convenience and unique perspective brought by the methods of theoretical mechanics?
Not necessarily. As long as the reader possesses basic knowledge of calculus (ordinary derivatives, partial derivatives, integrals, etc.) and sufficient patience, we are capable of understanding how theoretical mechanics (or the Principle of Least Action) works. Of course, the reader also needs some foundation in theoretical mechanics; at least, you should know what the action and the Principle of Least Action (within the framework of classical mechanics) are. Below, we will explore the "simple pendulum motion without a fixed point."
As shown in the figure, a ring of mass M is placed on a smooth horizontal axis. A thin string of length l is attached to the ring, and a weight of mass m is attached to the other end of the string. Analyze the motion of this pendulum. We have chosen an appropriate coordinate system such that t=0 is the moment of release (i.e., all velocities are 0).
We need to complete the following steps:
1. Use several coordinates to "position" the system under study;
2. Write down the kinetic energy T and potential energy U of each object in the system, and then write the Lagrangian L = T - U;
3. Write down the constraints between the coordinates of the objects in the system;
4. Use the constraints to simplify L (i.e., remove non-independent coordinates);
5. Substitute L into the Lagrange equations to obtain the equations of motion for the system.
First, we use Cartesian coordinates (x, y) to position m, and (z, 0) to position M. It is easy to write the total kinetic energy of both as T = \frac{1}{2} m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2} M \dot{z}^2, and the potential energy as U = mg(l - y) + Mgl (selecting the horizontal plane at distance l below the smooth axis as the zero potential surface).
We notice that x, y, z are not independent; they are subject to the constraint: (x - z)^2 + y^2 = l^2. In other words, there are only two independent coordinates. For convenience, we use a change of variables to simplify it, letting y = l \cos\theta, then x = z + l \sin\theta. Thus, we use \theta and z as our independent coordinates. Consequently, the kinetic energy becomes T = \frac{1}{2} [(M+m)\dot{z}^2 + ml^2 \dot{\theta}^2 + 2ml \cos\theta \dot{\theta} \dot{z}], and the potential energy U = mgl(1 - \cos\theta) + Mgl.
The Lagrangian is: L = \frac{1}{2} [(M+m)\dot{z}^2 + ml^2 \dot{\theta}^2 + 2ml \cos\theta \dot{\theta} \dot{z}] - mgl(1 - \cos\theta) - Mgl \tag{1} Substituting into the Lagrange equations, we have: \frac{d}{dt} \frac{\partial L}{\partial \dot{z}} = \frac{d}{dt} [(M+m)\dot{z} + ml \cos\theta \dot{\theta}] = \frac{\partial L}{\partial z} = 0 \tag{1} We immediately obtain: (M+m)\dot{z} + ml \cos\theta \dot{\theta} = C_1 \tag{2} Integrating once more: (M+m)z + ml \sin\theta = C_1 t + C_2 \tag{3} According to the initial conditions, we should have C_1 = 0 and C_2 = ml \sin\theta_0. Equation (3) becomes: (M+m)z + ml \sin\theta = ml \sin\theta_0 \tag{4} From this, we know that the motion of point M is constrained within the region \left(\frac{ml(\sin\theta_0 - 1)}{M+m}, \frac{ml(\sin\theta_0 + 1)}{M+m}\right).
Additionally, we have energy conservation: T + U = \text{constant}, namely: E = \frac{1}{2} [(M+m)\dot{z}^2 + ml^2 \dot{\theta}^2 + 2ml \cos\theta \dot{\theta} \dot{z}] + mgl(1 - \cos\theta) + Mgl = C_3 \tag{5} Combining (4) and (5) allows us to describe the entire motion process. Note that we obtained the equations of motion and some primary integrals without performing any force analysis; furthermore, our operations on the energy function felt very "smooth" without any sense of being "stuck," which allows us to feel the convenience of theoretical mechanics research methods. If we used Newton’s second law, we would have to carefully identify all relevant acting forces. This is a task that is both difficult and prone to error. Next, from (3) we can obtain: \dot{z} = -\frac{ml \cos\theta \dot{\theta}}{M+m}
Substituting this into (5) yields: E = \frac{ml^2 \dot{\theta}^2}{2(M+m)}(M + m \sin^2 \theta) + mgl(1 - \cos\theta) + Mgl = C_3 \tag{6} From (6) and the initial conditions, we get C_3 = mgl(1 - \cos\theta_0) + Mgl. Therefore: \dot{\theta}^2 = \frac{2g(\cos\theta - \cos\theta_0)(M+m)}{Ml + ml \sin^2\theta} \tag{7} If we could solve (7), we could completely describe the entire motion. Unfortunately, (7) is difficult to solve. We now consider several limiting cases:
1. When \theta is very small (small amplitude)
We replace \cos\theta with 1 - \frac{\theta^2}{2} and replace \sin^2 \theta with 0 (terms in the denominator have a "lower status" and are more easily ignored by us), then (7) can be simplified to: \dot{\theta}^2 = \frac{g(M+m)}{Ml}(\theta_0^2 - \theta^2) \tag{8} Solving this gives: \theta = \theta_0 \cos\left(\sqrt{\frac{g(M+m)}{Ml}} t\right) \tag{9} From this, it can be seen that the period for small amplitudes is T = 2\pi \sqrt{\frac{Ml}{g(M+m)}}. At the same time, equation (7) indicates that in this case M cannot be too small, otherwise it will cause the denominator to diverge, making the approximation meaningless.
2. When M tends to infinity
In this case, (7) becomes: \dot{\theta}^2 = 2g(\cos\theta - \cos\theta_0) \tag{10} This is consistent with the equation of motion for a simple pendulum with a fixed point, which can be referenced here. We will not elaborate further.
3. When M = 0
This special case is enough to give us a glimpse of the weirdness of simple pendulum motion without a fixed point.
In this case, (7) becomes: \dot{\theta}^2
= \frac{2g(\cos\theta - \cos\theta_0)}{l \sin^2\theta} \tag{11}
That is, t = -\sqrt{\frac{l}{2g}} \int
\sin\theta \sqrt{\frac{1}{\cos\theta - \cos\theta_0}}
d\theta
(Think about why a negative sign is taken
here?)
This can be integrated exactly quite simply: t = \sqrt{\frac{l}{2g}} \int \sqrt{\frac{1}{\cos\theta - \cos\theta_0}} d(\cos\theta) = \sqrt{\frac{2l}{g}(\cos\theta - \cos\theta_0)} + C_4
It is easy to see that C_4 = 0, so we get: \cos\theta = \cos\theta_0 + \frac{gt^2}{2l} \tag{12} What does (12) imply? It implies that the motion time is only t = \sqrt{\frac{2l}{g}(1 - \cos\theta_0)}.
This clearly contradicts our intuition. Moreover, according to the conservation of energy, it is impossible for it to simply stop. So, what happens at this point?
The situation then becomes very chaotic... (to be continued)
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