0 引言

完整系统的第二类Lagrange方程是受约束质点系且不包含约束力的运动微分方程^[1].故为研究有多余坐标完整约束系统，需用第一类Lagrange方程或有多余坐标带乘子形式的Lagrange方程^[2-4].对于完整力学系统通常采用第二类Lagrange方程来组建系统的运动微分方程，其中的坐标是彼此独立的.但是，有多余坐标的完整系统力学不仅在运动学描述上有重要意义，而且在诸多动力学问题中，如四连杆机构的运动学描述^[5-7]，工程中振动仪器^[7]，多体系统动力学^[8-10]等有重要意义.在多体系统的动力学中更多地采用微分---代数方程，即有多余坐标完整系统的方程，以便更好地实施计算^[11-13].故有时选多余坐标反而会带来方便.

有多余坐标完整系统的自由运动是指系统的约束力为零的运动. 关于约束系统的自由运动问题取得了一些成果.文献^[14]给出了非完整系统的自由运动，研究了Chaplygin雪橇的自由运动，并讨论了有外力时实现非完整系统自由运动的可能性，具有一定的实际意义.2005年，Yushkov利用Maggi方程实现了Chaplygin雪橇问题的自由运动，并研究了在有主动力时非完整系统自由运动的可能性^[15-16]. 在此基础之上，推广并研究了非完整力学的可控运动^[17].关于自由运动问题，一些学者从不同方面进行了讨论研究^[18-19].本文主要提出并研究有多余坐标完整系统的一类特殊运动，即自由运动，给出实现自由运动的条件.

1 系统的运动微分方程

设系统的位形由$r$个广义坐标$q_d \left( {d = 1,2,\cdots,r} \right)$来确定. 由于某些要求，需选$g$个多余坐标$q_{r + 1}$,$q_{r + 2}$,$\cdots,q_{r + g} \left( {g = n - r} \right)$，并有$g$个双面理想完整约束

$ f_\beta \left( {q_s ,t} \right) = 0 \quad \left( {\beta = 1,2,\cdots ,g ; s = 1,2,\cdots,n} \right)$

(1)

d'Alembert-Lagrange原理有形式

$ \left( {\dfrac {d }{d }\dfrac {\partial T}{\partial \dot {q}_s } - \dfrac {\partial T}{\partial q_s } - Q_s } \right)\delta q_s = 0 \quad \left( {s = 1,2,\cdots ,n} \right) $

(2)

这里及以后,同一项中相同指标表示求和. 式(1)加在虚位移$\delta q_s $上的限制为

$ \dfrac {\partial f_\beta }{\partial q_s }\delta q_s = 0 $

(3)

由式(2)和式(3)，利用Lagrange乘子法，得到方程

$ \dfrac {d }{d }\dfrac {\partial T}{\partial \dot {q}_s } - \dfrac {\partial T}{\partial q_s } = Q_s + \lambda _\beta \dfrac {\partial f_\beta }{\partial q_s } \\ \left( {\beta = 1,2,\cdots ,g ; \ s = 1,2,\cdots,n} \right) $

(4)

式(4)右端带乘子的项$\lambda _\beta $代表约束力. 由式(4)和式(1)可求系统的运动，并且可求约束力.

2 约束力的确定

将系统动能写成

$ T = T_2 + T_1 + T_0 $

(5)

其中

$ T_2 = \dfrac {1}{2}A_{sk} \dot {q}_s \dot {q}_k ,T_1 = B_s \dot {q}_s $

(6)

则有

$ \dfrac {d }{d t }\dfrac {\partial T_2 }{\partial \dot {q}_s } - \dfrac {\partial T_2 }{\partial q_s } = \\ A_{sk} \ddot {q}_k + \left( {\dfrac {\partial A_{sk} }{\partial q_m } - \dfrac {1}{2}\dfrac {\partial A_{km} }{\partial q_s }} \right)\dot {q}_k \dot {q}_m + \dfrac {\partial A_{ks} }{\partial t}\dot {q}_k =\\ A_{sk} \ddot {q}_k + \left[{k,m;s} \right] \dot {q}_k \dot {q}_m + \dfrac {\partial A_{ks} }{\partial t}\dot {q}_k $

(7)

其中

$ \left[{k,m;s} \right] = \dfrac {1}{2}\left( {\dfrac {\partial A_{ks} }{\partial q_m } + \dfrac {\partial A_{ms} }{\partial q_k } - \dfrac {\partial A_{km} }{\partial q_s }} \right) $

(8)

为系数矩阵$\left( {A_{sk} } \right)$的第一类Christoffel记号. 又

$\dfrac {d }{d t }\dfrac {\partial T_1 }{\partial \dot {q}_s } - \dfrac {\partial T_1 }{\partial q_s } = \dfrac {\partial B_s }{\partial t} - \left( {\dfrac {\partial B_k }{\partial q_s } - \dfrac {\partial B_s }{\partial q_k }} \right)\dot {q}_k $

(9)

$\dfrac {d }{d t }\dfrac {\partial T_0 }{\partial \dot {q}_s } - \dfrac {\partial T_0 }{\partial q_s } = - \dfrac {\partial T_0 }{\partial q_s } $

(10)

将式(7)、 式(9)和式(10)代入式(4)，得到

$ A_{sk} \ddot {q}_k + \left[{k,m; s} \right] \dot {q}_k \dot {q}_m = \\ \left( {\dfrac {\partial B_k }{\partial q_s } - \dfrac {\partial B_s }{\partial q_k }} \right)\dot {q}_k + Q_s - \dfrac {\partial B_s }{\partial t} + \dfrac {\partial T_0 }{\partial q_s } -\\ \dfrac {\partial A_{sk} }{\partial t}\dot {q}_k + \lambda _\beta \dfrac {\partial f_\beta }{\partial q_s } \quad \left( {s = 1,2,\cdots ,n} \right) $

(11)

因${\rm{det}}\left( {{A_{sk}}} \right) \ne 0$，故可由式(11)求出所有广义加速度

$ \ddot {q}_l = A^{ls}\left[{ - \left[{k,m; s} \right] \dot {q}_k \dot {q}_m + \left( {\dfrac {\partial B_k }{\partial q_s } - \dfrac {\partial B_s }{\partial q_k }} \right)\dot {q}_k + Q_s } \right. -$

$ \left. { \dfrac {\partial B_s }{\partial t} + \dfrac {\partial T_0 }{\partial q_s } - \dfrac {\partial A_{sk} }{\partial t}\dot {q}_k + \lambda _\beta \dfrac {\partial f_\beta }{\partial q_s }} \right] $

(12)

其中

$ A^{ls}A_{sk} = \delta _k^l $

(13)

为求得约束力，将式(1)对$t$求两次导数，得

$ \dfrac {\partial ^2f_\beta }{\partial t^2} + 2\dfrac {\partial ^2f_\beta }{\partial t\partial q_s }\dot {q}_s + \dfrac {\partial ^2f_\beta }{\partial q_s \partial q_k }\dot {q}_s \dot {q}_k + \dfrac {\partial f_\beta }{\partial q_s }\ddot {q}_s = 0 $

(14)

将式(12)代入式(14)，消去广义加速度，得到

$\eqalign{ & {{{\partial ^2}{f_\beta }} \over {\partial {t^2}}} + 2{{{\partial ^2}{f_\beta }} \over {\partial t\partial {q_s}}}{{\dot q}_s} + {{{\partial ^2}{f_\beta }} \over {\partial {q_s}\partial {q_k}}}{{\dot q}_s}{{\dot q}_k} + {{\partial {f_\beta }} \over {\partial {q_l}}}{A^{ls}} \cr & \left[ { - \left[ {k,m;s} \right]{{\dot q}_k}{{\dot q}_m} + \left( {{{\partial {B_k}} \over {\partial {q_s}}} - {{\partial {B_s}} \over {\partial {q_k}}}} \right){{\dot q}_k} + {Q_s}} \right. - \left. {{{\partial {B_s}} \over {\partial t}} + {{\partial {T_0}} \over {\partial {q_s}}} - {{\partial {A_{sk}}} \over {\partial t}}{{\dot q}_k} + {\lambda _\gamma }{{\partial {f_\gamma }} \over {\partial {q_s}}}} \right] \cr & = 0\left( {\beta = 1,2, \cdots ,g} \right) \cr} $

(15)

当

${\rm{det}}\left( {{{\partial {f_\beta }} \over {\partial {q_l}}}{A^{ls}}{{\partial {f_\gamma }} \over {\partial {q_s}}}} \right) \ne 0$

(16)

时，可由式(15)求出所有$\lambda _\gamma \left( {\gamma = 1,2,\cdots ,g}\right)$，进而可求出约束力

$ \varLambda _s = \lambda _\gamma \dfrac {\partial f_\gamma }{\partial q_s } \quad \left( {\gamma = 1,2,\cdots ,g;s = 1,2,\cdots ,n} \right) $

(17)

3 自由运动的条件

系统的自由运动是指约束力为零的运动，将$\lambda _\gamma = 0 \left( {\gamma =1,2,\cdots ,g} \right)$代入式(15)，得到

$ \dfrac {\partial ^2f_\beta }{\partial t^2} + 2\dfrac {\partial ^2f_\beta }{\partial t\partial q_s }\dot {q}_s + \dfrac {\partial ^2f_\beta }{\partial q_k \partial q_s }\dot {q}_k \dot {q}_s +\\ \dfrac {\partial f_\beta }{\partial q_l }A^{ls}\Bigg [- \left[{k,m;s} \right] \dot {q}_k \dot {q}_m + Q_s - \\ \dfrac {\partial B_s }{\partial t} + \dfrac {\partial T_0 }{\partial q_s } - \dfrac {\partial A_{sk} }{\partial t}\dot {q}_k \Bigg] = 0 \\ \left( {\beta = 1,2,\cdots ,g} \right) $

(18)

这就是系统发生自由运动的条件. 特别地，如果

$ \left.\!\!\begin{array}{l} B_s = 0 \left( {s = 1,2,\cdots ,n} \right) \\ T_0 = 0 \\ \dfrac {\partial A_{ks} }{\partial t} = 0 \left( {k,s = 1,2,\cdots ,n} \right) \end{array} \right\} $

(19)

则式(18)成为

$ \dfrac {\partial ^2f_\beta }{\partial t^2} + 2\dfrac {\partial ^2f_\beta }{\partial t\partial q_s }\dot {q}_s + \dfrac {\partial ^2f_\beta }{\partial q_s \partial q_k }\dot {q}_s \dot {q}_k + \\ \dfrac {\partial f_\beta }{\partial q_l }A^{ls}\left( { - \left[{k,m; s} \right] \dot {q}_k \dot {q}_m + Q_s } \right) = 0 $

(20)

进而，如果$A_{sk} = {\rm const.} \left( {s,k = 1,2,\cdots ,n}\right)$，则式(20)成为

$ \dfrac {\partial ^2f_\beta }{\partial t^2} + 2\dfrac {\partial ^2f_\beta }{\partial t\partial q_s }\dot {q}_s + \dfrac {\partial ^2f_\beta }{\partial q_s \partial q_k }\dot {q}_s \dot {q}_k + \\ \dfrac {\partial f_\beta }{\partial q_l }A^{ls}Q_s = 0 \quad \left( {\beta = 1,2,\cdots ,g} \right) $

(21)

当系统发生自由运动时，式(4)成为

$ \dfrac {d }{d }\dfrac {\partial T}{\partial \dot {q}_s } - \dfrac {\partial T}{\partial q_s } = Q_s \quad \left( {s = 1,2,\cdots ,n} \right) $

(22)

4 算例

例1 系统的动能和约束分别为

$ T = \dfrac {1}{2}\left( {\dot {q}_1^2 + \dot {q}_2^2 + \dot {q}_3^2 } \right) ,f = tq_1 + 2q_2 - 3q_3 = 0 $

(23)

其中的量已无量纲化，试求实现自由运动所应施加的主动力之间的关系.

解：式(21)给出

$ 2\dot {q}_1 + tQ_1 + 2Q_2 - 3Q_3 = 0 $

(24)

如不施加主动力，即$Q_1 = Q_2 = Q_3 = 0$，则不能实现自由运动. 式(24)给出主动力应满足的条件，例如，取$Q_1 = 0$,$Q_2 = - \dot {q}_1$,$Q_3 = 0$，则可实现自由运动，此时约束力为零. 当然，还有其他选择.

例2 系统动能和约束分别为

$\left. {\matrix{ {T = {1 \over 2}\left[ {\dot q_1^2\left( {2 + \sin t} \right) + \dot q_2^2 + \dot q_3^2} \right]} \cr {f = {t^2}{q_1} - {q_2} - {q_3} = 0} \cr } } \right\}$

(25)

其中量已无量纲化，试求系统实现自由运动的条件.

解：由$T$表达式知

$ A_{11} = 2 + \sin t ,A_{22} = 1 ,A_{33} = 1 $

于是有

$ A_{11} = \dfrac {1}{2 + \sin t} ,A_{22} = 1 ,A_{33} = 1 $

以及

$ \left[{k,m;s} \right] = 0 \quad \left( {k,m,s = 1,2} \right)$

式(18)给出

$ 2q_1 + 4t\dot {q}_1 + \dfrac {t^2}{2 + \sin t}\left( {Q_1 - \dot {q}_1 \cos t} \right) - \\ Q_2 - Q_3 = 0 $

(26)

这就是系统实现自由运动对主动力$Q_1 ,Q_2 ,Q_3 $ 的限制. 显然，不施加主动力是不能实现自由运动的.

5 结论

本文研究了有多余坐标完整力学系统的自由运动，得到实现自由运动的条件. 如果需要自由运动，必须满足这些条件；如果不需要自由运动，必须避开这些条件. 同时，正如文献^[15]所指出的，如果系统的运动对自由运动有小的偏离，那么就有小的约束力，可将其作为干扰来研究以实现规划运动.