双摆和三摆

贡献者： addis; FFjet

预备知识　拉格朗日方程

1. 双摆

图 1：双摆

图 2：数值模拟结果，动画见这里，代码见 “双摆的数值计算（Matlab）”。

未完成：文字说明

N

摆的动能可以参考这里。

\begin{matrix} (1) & \begin{aligned} T & = \frac{1}{2} m_{1} (l_{1} ω_{1})^{2} + \frac{1}{2} m_{2} [(l_{1} ω_{1} \cos θ_{1} + l_{2} ω_{2} \cos θ_{2})^{2} + (l_{1} ω_{1} \sin θ_{1} + l_{2} ω_{2} \sin θ_{2})^{2}], \end{aligned} \end{matrix}

\begin{matrix} (2) & V = - m_{1} g l_{1} \cos θ_{1} - m_{2} g (l_{1} \cos θ_{1} + l_{2} \cos θ_{2}), \end{matrix}

\begin{matrix} (3) & L (θ_{1}, θ_{2}, ω_{1}, ω_{2}) = T (ω_{1}, ω_{2}) - V (θ_{1}, θ_{2}), \end{matrix}

\begin{matrix} (4) & \frac{d}{d t} \frac{\partial L}{\partial ω_{i}} = \frac{\partial L}{\partial θ_{i}} (i = 1, 2) . \end{matrix}

　　令 $α = θ_{2} - θ_{1}$ ，得

\begin{matrix} (5) & \frac{\partial T}{\partial ω_{1}} = (m_{1} + m_{2}) l_{1}^{2} ω_{1} + m_{2} l_{1} l_{2} \cos α \cdot ω_{2}, \end{matrix}

\begin{matrix} (6) & \frac{\partial T}{\partial ω_{2}} = m_{2} l_{1} l_{2} \cos α \cdot ω_{1} + m_{2} l_{2}^{2} ω_{2}, \end{matrix}

\begin{matrix} (7) & \frac{d}{d t} \frac{\partial T}{\partial ω_{1}} = (m_{1} + m_{2}) l_{1}^{2} {\dot{ω}}_{1} + m_{2} l_{1} l_{2} \cos α \cdot {\dot{ω}}_{2} - m_{2} l_{1} l_{2} \sin α (ω_{2} - ω_{1}), \end{matrix}

\begin{matrix} (8) & \frac{d}{d t} \frac{\partial T}{\partial ω_{2}} = m_{2} l_{1} l_{2} \cos α \cdot {\dot{ω}}_{1} + m_{2} l_{2}^{2} {\dot{ω}}_{2} - m_{2} l_{1} l_{2} \sin α (ω_{2} - ω_{1}) ω_{1}, \end{matrix}

\begin{matrix} (9) & \frac{\partial T}{\partial θ_{1}} = m_{2} l_{1} l_{2} \sin α \cdot ω_{1} ω_{2}, \end{matrix}

\begin{matrix} (10) & \frac{\partial T}{\partial θ_{2}} = - m_{2} l_{1} l_{2} \sin α \cdot ω_{1} ω_{2}, \end{matrix}

\begin{matrix} (11) & \frac{\partial V}{\partial θ_{1}} = m_{1} g l_{1} \sin θ_{1} + m_{2} g l_{1} \sin θ_{1}, \end{matrix}

\begin{matrix} (12) & \frac{\partial V}{\partial θ_{2}} = m_{2} g l_{2} \sin θ_{2} . \end{matrix}

代入式 4 得

\begin{matrix} (13) & (m_{1} + m_{2}) l_{1} {\dot{ω}}_{1} + m_{2} l_{2} \cos α \cdot {\dot{ω}}_{2} = m_{2} l_{2} \sin α \cdot ω_{2}^{2} - (m_{1} + m_{2}) g \sin θ_{1}, \end{matrix}

\begin{matrix} (14) & l_{1} \cos α \cdot {\dot{ω}}_{1} + l_{2} {\dot{ω}}_{2} = l_{1} \sin (θ_{1} - θ_{2}) ω_{1}^{2} - g \sin θ_{2}, \end{matrix}

解得

\begin{matrix} (15) & {\dot{ω}}_{1} = \frac{m_{2} l_{2} \sin α \cdot ω_{2}^{2} - (m_{1} + m_{2}) g \sin θ_{1} + m_{2} [l_{1} \sin α \cdot ω_{1}^{2} + g \sin θ_{2}] \cos α}{m_{1} l_{1} + m_{2} l_{1} \sin^{2} α}, \end{matrix}

\begin{matrix} (16) & {\dot{ω}}_{2} = \frac{[(m_{1} + m_{2}) g \sin θ_{1} - m_{2} l_{2} \sin α \cdot ω_{2}^{2}] \cos α - (m_{1} + m_{2}) [l_{1} \sin α \cdot ω_{1}^{2} + g \sin θ_{2}]}{m_{1} l_{2} + m_{2} l_{2} \sin^{2} α} . \end{matrix}

图 3：刚体双摆的模拟结果，动画见这里，代码见 “双摆的数值计算（Matlab）”。

　　若第一个质量为 $m_{1}$ 刚体的质心用 $l_{1}, θ_{1}$ 描述，关于质心的转动惯量为 $I_{1}$ ，第二个刚体质量为 $m_{2}$ ，转轴与第一个刚体的转轴距离 $r_{1}$ ，与第一个刚体质心的夹角为 $α$ ，那么

\begin{matrix} (17) & \begin{aligned} T & = \frac{1}{2} m_{1} l_{1}^{2} ω_{1}^{2} + \frac{1}{2} I_{1} ω_{1}^{2} + \frac{1}{2} m_{2} [(r_{1} ω_{1} \cos (θ_{1} + α) + l_{2} ω_{2} \cos θ_{2})^{2} \\ + (r_{1} ω_{1} \sin (θ_{1} + α_{1}) + l_{2} ω_{2} \sin θ_{2})^{2}] + \frac{1}{2} I_{2} ω_{2}^{2} \\ = \frac{1}{2} (m_{1} l_{1}^{2} + m_{2} r_{1}^{2} + I_{1}) ω_{1}^{2} + \frac{1}{2} (m_{2} l_{2}^{2} + I_{2}) ω_{2}^{2} + m_{2} r_{1} l_{2} \cos (θ_{1} + α - θ_{2}) ω_{1} ω_{2}, \end{aligned} \end{matrix}

令

T = a ω_{1}^{2} + b ω_{2}^{2} + c ω_{1} ω_{2}

，注意

a, b

是常数，而

c

是时间的函数。

\begin{matrix} (18) & V = - m_{1} g l_{1} \cos θ_{1} - m_{2} g (r_{1} \cos (θ_{1} + α) + l_{2} \cos θ_{2}) . \end{matrix}

这样，

\begin{matrix} (19) & \frac{d}{d t} \frac{\partial T}{\partial ω_{1}} = 2 a {\dot{ω}}_{1} + c {\dot{ω}}_{2} + \frac{\partial c}{\partial θ_{1}} ω_{1} ω_{2} + \frac{\partial c}{\partial θ_{2}} ω_{2}^{2}, \end{matrix}

\begin{matrix} (20) & \frac{d}{d t} \frac{\partial T}{\partial ω_{2}} = 2 b {\dot{ω}}_{2} + c {\dot{ω}}_{1} + \frac{\partial c}{\partial θ_{1}} ω_{1}^{2} + \frac{\partial c}{\partial θ_{2}} ω_{1} ω_{2}, \end{matrix}

\begin{matrix} (21) & \frac{\partial T}{\partial θ_{1}} = \frac{\partial c}{\partial θ_{1}} ω_{1} ω_{2}, \frac{\partial T}{\partial θ_{2}} = \frac{\partial c}{\partial θ_{2}} ω_{1} ω_{2}, \end{matrix}

\begin{matrix} (22) & \frac{\partial V}{\partial θ_{1}} = m_{1} g l_{1} \sin θ_{1} + m_{2} g r_{1} \sin (θ_{1} + α), \frac{\partial V}{\partial θ_{2}} = m_{2} g l_{2} \sin θ_{2} . \end{matrix}

代入拉格朗日方程式 4 得

\begin{matrix} (23) & 2 a {\dot{ω}}_{1} + c {\dot{ω}}_{2} = - \frac{\partial c}{\partial θ_{2}} ω_{2}^{2} - \frac{\partial V}{\partial θ_{1}}, \end{matrix}

\begin{matrix} (24) & c {\dot{ω}}_{1} + 2 b {\dot{ω}}_{2} = - \frac{\partial c}{\partial θ_{1}} ω_{1}^{2} - \frac{\partial V}{\partial θ_{2}}, \end{matrix}

这里不具体写出

{\dot{ω}}_{1}, {\dot{ω}}_{2}

的解，可以直接使用克拉默法则。

图 4：三摆

　　如图 4 所示，三根质量不计的杆长度分别为 $l_{1}, l_{2}, l_{3}$ ，三个质点的质量分别为 $m_{1}, m_{2}, m_{3}$ 。我们把这个模型叫做三摆。三摆常用于演示物理学中的混沌现象。

未完成：文字说明

\begin{matrix} (25) & \begin{aligned} T & = \frac{1}{2} m_{1} (l_{1} ω_{1})^{2} + \frac{1}{2} m_{2} [(l_{1} ω_{1} \cos θ_{1} + l_{2} ω_{2} \cos θ_{2})^{2} + \\ (l_{1} ω_{1} \sin θ_{1} + l_{2} ω_{2} \sin θ_{2})^{2}] \\ + \frac{1}{2} m_{3} [(l_{1} ω_{1} \cos θ_{1} + l_{2} ω_{2} \cos θ_{2} + l_{3} ω_{3} \cos θ_{3})^{2} + \\ (l_{1} ω_{1} \sin θ_{1} + l_{2} ω_{2} \sin θ_{2} + l_{3} ω_{3} \sin θ_{3})^{2}] . \end{aligned} \end{matrix}

\begin{matrix} (26) & \begin{aligned} V & = - m_{1} g l_{1} \cos θ_{1} - m_{2} g (l_{1} \cos θ_{1} + l_{2} \cos θ_{2}) \\ - m_{3} g (l_{1} \cos θ_{1} + l_{2} \cos θ_{2} + l_{3} \cos θ_{3}) . \end{aligned} \end{matrix}

\begin{matrix} (27) & L (θ_{1}, θ_{2}, θ_{3}, ω_{1}, ω_{2}, ω_{3}) = T (ω_{1}, ω_{2}, ω_{3}) - V (θ_{1}, θ_{2}, θ_{3}) . \end{matrix}

\begin{matrix} (28) & \frac{d}{d t} \frac{\partial L}{\partial ω_{i}} = \frac{\partial L}{\partial θ_{i}} (i = 1, 2, 3) . \end{matrix}

　　这是一个二阶微分方程组。