1
Theoretical Mechanics
2
3
On the basis of D‘Alembert’s Principle and of the Theorem of
Virtual Displacements in this chapter the general equation of
dynamics and the Lagrange's equations of the second kind
(abbreviated as Lagrange's equations) is deduced,The general
equation of dynamics and the Lagrange's equations are
effective means to study dynamical problems,They provide
very simple,direct and standard ways to solve dynamical
problems of unfree systems of particles,
4
本章在达朗伯原理和虚位移原理的基础上，进一步导
出动力学普遍方程和拉格朗日第二类方程（简称拉格朗日
方程）。动力学普遍方程和拉格朗日方程是研究动力学问
题的有力手段，在解决非自由质点系的动力学问题时，显
得十分简捷、规范。
5
§ 17–1 General equation of dynamics
§ 17–2 Lagrange's equations of the second kind
§ 17–3 Integrals of the Lagrange's equations of
the second kind
Chapter 17,Lagrange's equations
6
§ 17–1 动力学普遍方程
§ 17–2 拉格朗日第二类方程
§ 17–3 拉格朗日第二类方程的积分
第十七章 拉格朗日方程
7
iiiiiii amQaNFm ?? ;,,,
Suppose that there are n particles in a system,particle i being
described by, Then we have
0 i ??? QNF ii
If the system of particle is under the action of the ideal constraints,the
can be treated as positive forces and we get
iQ
0)( ???? iii rQF ?
The explicit form is
0])()()[( ??????? iiiiiiiiiiii zzmZyymYxxmX ??? ??????
§ 17-1 General equation of dynamics
This is the general equation of dynamics,
8
iiiiiiii amQaNFmM ?? ;,,,:
设质点系有 n个质点,第 i个质点
0 i ??? QNF ii
若质点系受有理想约束, 将 作为主动力处理，则,
iQ
0)( ???? iii rQF ?
解析式,
0])()()[( ??????? iiiiiiiiiiii zzmZyymYxxmX ??? ??????
§ 17-1 动力学普遍方程
动力学普遍方程。
9
[Example 1] A triangular prism B is slipping along the smooth
inclined plane of the triangular prism A,The triangular prism A is
placed on the smooth horizontal plane,The weights of A and B are M
and m,the angle of inclination is ?,Determine the acceleration of A,
Investigate the system consisting
of both triangular prisms,This
system is under the action of ideal
constraints,It has two degree of
freedom,
.,rrBeB
r
B
e
BB
A
maQmaQ
QQQ
MaQ
??
??
?
Under the action of the ideal constraints the sum of the virtual
works of the positive forces and of the inertial forces,acting on
every particle of the system,along an arbitrary virtual
displacement is zero at any moment,
Solution
10
例 1 三棱柱 B沿三棱柱 A的光滑斜面滑动，三棱柱 A置于光
滑水平面上,A和 B的质量分别为 M和 m，斜面倾角为 ? 。试求三
棱柱 A的加速度。
解,研究两三棱柱组
成的系统。该系统受理想
约束，具有两个自由度。
r
r
B
e
B
r
B
e
BB
A
maQmaQ
QQQ
MaQ
??
??
?
,
在理想约束的条件下，质点系的各质点在任一瞬时受到的
主动力与惯性力在任意虚位移上所作的虚功之和为零。
11
From the general equation of dynamics we have
.0)s i nc o s()c o s( ??????? BrBeBArBeBA sQQQxQQQ ?????
The system has two degree of freedom,we can choose
as independent virtual displacement,Moreover,
, So we get
BA sx ?? a n d
mgQ?
0s i nc o s
0c o s
???
???
r
r
mamgma
mamaMa
??
?
Solving them we get
.)s in(2 2s in 2 gmM ma ????
12
由动力学普遍方程,
0)s i nc o s()c o s( ??????? BrBeBArBeBA sQQQxQQQ ?????
系统为二自由度，取互不相关的 为独立虚位移，
且,所以
BA sx ??,
mgQ?
0s i nc o s
0c o s
???
???
r
r
mamgma
mamaMa
??
?
解得,
gmMma )s in(2 2s in 2 ????
13
§ 17-2 Lagrange's equations of the second kind
Suppose there are n particles in the system and that there are s
constraints which all are ideal ones,The number of degrees of
freedom is k=3n- s,
Here we will deduce the general equations of dynamics using
generalized coordinates,
kqqq ?,,21
Particle,
If we choose a set of generalized coordinate of the system,,
then we have
iii rmM,,
)( )2,1(
)( ),2,1( ),,,(
1
21
bni
t
r
q
q
r
dt
rd
v
anitqqqrr
k
j
i
j
j
ii
i
kii
??
??
?
?
?
?
?
??
?
?
??
??
are called the generalized velocities,
dt
dqq j
j ??
14
§ 17-2 拉格朗日第二类方程
设质点系有 n个质点，受 s个完整约束且系统所受的约束是
理想约束，自由度 k=3n- s 。
下面推导以广义坐标表示的动力学普遍方程的形式。
质点 。若取系统的一组广义坐标为,则
iii rmM,,kqqq ?,,21
)( )2,1(
)( ),2,1( ),,,(
1
21
bni
t
r
q
q
r
dt
rd
v
anitqqqrr
k
j
i
j
j
ii
i
kii
??
??
?
?
?
?
?
??
?
?
??
??
称 为 广义速度 。
dt
dqq j
j ??
15
?
?
?????
k
j
j
j
i
i cniqq
rr
1
)( ),2,1( ???
Substituting equation (c) into the general equation of dynamics
of the system of particles,we get
? ??
? ??
????? n
i
n
i iiiii
n
i iiii
dramrFramF
1 11
)( 0)( ???
?
? ?
? ????
?
? ?
? ????
??
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
k
j
jj
k
j
j
j
i
i
n
i j
i
i
j
i
i
k
j
n
i
j
j
i
i
k
j
j
j
i
n
i
i
n
i
ii
qQ
q
q
z
Z
q
y
Y
q
x
X
q
q
r
Fq
q
r
FrF
1
1 1
1 1111
)]([
)()(
?
?
???
16
?
?
?????
k
j
j
j
i
i cniqq
rr
1
)( ),2,1( ???
代入质点系动力学普遍方程，得,
? ??
? ??
????? n
i
n
i iiiii
n
i iiii
dramrFramF
1 11
)( 0)( ???
?
? ?
? ????
?
? ?
? ????
??
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
k
j
jj
k
j
j
j
i
i
n
i j
i
i
j
i
i
k
j
n
i
j
j
i
i
k
j
j
j
i
n
i
i
n
i
ii
qQ
q
q
z
Z
q
y
Y
q
x
X
q
q
r
Fq
q
r
FrF
1
1 1
1 1111
)]([
)()(
?
?
???
17
The Qj are called generalized forces,
)( )(
1
eqzZqyYqxXQ
j
i
i
j
i
i
j
i
i
n
i
j ?
??
?
??
?
?? ?
?
0)(
)()(T h e n
1 1
1 111
?
?
?
???
?
?
?
????
? ?
? ???
? ?
? ???
j
j
i
k
j
n
i
i
ij
n
i
j
k
j j
i
ii
k
j
jj
n
i
iiii
q
q
r
dt
vd
mQ
q
q
r
amqQramF
?
???
)(, ),2,1( 0
1
fkjqrdt vdmQ
j
i
n
i
i
ij ????
?? ?
?
generalized inertial forces
18
称 为 广义力
)( )(
1
eqzZqyYqxXQ
j
i
i
j
i
i
j
i
i
n
i
j ?
??
?
??
?
?? ?
?
0)(
)()(
1 1
1 111
?
?
?
???
?
?
?
????
? ?
? ???
? ?
? ???
j
j
i
k
j
n
i
i
ij
n
i
j
k
j j
i
ii
k
j
jj
n
i
iiii
q
q
r
dt
vd
mQ
q
q
r
amqQramF
?
???则
)( ),2,1( 0
1
fkjqrdt vdmQ
j
in
i
i
ij ????
?? ?
?
广义惯性力
19
The generalized inertial force can be obtained from the kinetic
energy of the system of particle in the following way,
).()(
111 j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i q
r
dt
dvm
q
rv
dt
dm
q
r
dt
vdm
?
???
?
???
?
?? ???
???
For the following calculation we need the two expression,
j
i
j
i
j
i
j
i
q
v
q
r
dt
d
q
v
q
r
?
??
?
?
?
??
?
? a n d
?
The first expression is obtained by differentiation of both sides
of the equation (b) by,
jq?
20
广义惯性力可改变为用质点系的动能表示,因此
)()(
111 j
in
i
ii
j
i
i
n
i
i
j
in
i
i
i q
r
dt
dvm
q
rv
dt
dm
q
r
dt
vdm
?
???
?
???
?
?? ???
???
为简化计算,需要用到以下两个关系式,
j
i
j
i
j
i
j
i
q
v
q
r
dt
d
q
v
q
r
?
??
?
?
?
??
?
? ;
?
下面来推导这两个关系式,
第一式只须将 (b)式两边对 求偏导数即可得到。
jq?
21
The second expression is obtained by comparing two results,one is
given by the differentiation of equation (a) twice by ql and t,the other
is given by the differentiation of equation (b) by ql,
We get,
h a v e w e),( i n t o s u b s t i t u t e
)
2
1
()]
2
1
([
)(
1
2
1
2
111
f
q
T
q
T
dt
d
vm
q
vm
qdt
d
q
v
vm
q
r
v
dt
d
m
q
r
dt
vd
m
jj
n
i
ii
j
n
i
i
j
j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
??
???
??
???
?
?
?
),1,2,( kjQqTqTdtd j
jj
?? ???????
These are the Lagrange's equations of the second kind,
abbreviated as Lagrange's equations,
22
第二式可比较 (a)式先对 ql求偏导数 再对 t求导数与 (b)式对
ql求偏导数的结论得出。
:,)(
)
2
1
()]
2
1
([
)(
1
2
1
2
111
得式代入 f
q
T
q
T
dt
d
vm
q
vm
qdt
d
q
v
vm
q
r
v
dt
d
m
q
r
dt
vd
m
jj
n
i
ii
j
n
i
i
j
j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
??
???
??
???
?
?
?
),1,2,( kjQqTqTdtd j
jj
?? ???????
拉格朗日第二类动力学方程，简称拉格朗日方程。
23
If the forces acting on the system of particles all are potential forces,the generalized forces can be expressed by the potential energy,
jQ
).,1,2,(
),(
),,1,2,( )(
1
1
kj
q
U
Q
q
z
z
U
q
y
y
U
q
x
x
U
kj
q
z
Z
q
y
Y
q
x
XQ
j
j
j
i
ij
i
ij
i
i
n
i
j
i
i
j
i
i
j
i
i
n
i
j
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
According to the Lagrange's equations
),1,2,( kjqUqTqTdtd
jjj
?? ??????????
Introducing the Lagrange's function L=T-U we get
),1,2,( 0 )( kjqLqLdtd
jj
?? ???????
These are the Lagrange's equations of a potential system,
24
如果作用于质点系的力是有势力，则广义力 可用质点系的势
能来表达。
jQ
),1,2,(
)(
),1,2,( )(
1
1
kj
q
U
Q
q
z
z
U
q
y
y
U
q
x
x
U
kj
q
z
Z
q
y
Y
q
x
XQ
j
j
j
i
ij
i
ij
i
i
n
i
j
i
i
j
i
i
j
i
i
n
i
j
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
而拉氏方程为,
),1,2,( kjqUqTqTdtd
jjj
?? ??????????
引入拉格朗日函数,L=T-U 则,
),1,2,( 0 )( kjqLqLdtd
jj
?? ???????
保守系统的拉格朗日方程。
25
The steps of solving problems,Using the Lagrange's equations are,1,Determine the number of degrees of freedom of the system k and
choose appropriate generalized coordinates,Be careful,do not neglect
independent coordinates or add redundant (dependent) coordinates,
2,Calculate the kinetic energy of the system T which can be
expressed as a function of the generalized velocities and the
generalized coordinates,
3,Calculate the generalized forces,the formulae
are ),1,2,( kjQ j ??
)(
1 j
i
i
j
i
i
j
i
i
n
i
j q
zZ
q
yY
q
xXQ
?
??
?
??
?
?? ?
?
or
j
j
j q
WQ
?
? )(?
If the positive forces are potential forces,the potential energy U
should be expressed as a function of the generalized coordinates,
4,Writing down the Lagrange's equations and collate,we obtain
k second order ordinary differential equations,
5,Solve this set of differential equations,
26
应用拉氏方程解题的步骤,
1,判定质点系的自由度 k，选取适宜的广义坐标。必须注意：
不能遗漏独立的坐标，也不能有多余的（不独立）坐标。
2,计算质点系的动能 T，表示为广义速度和广义坐标的函数。
3,计算广义力,计算公式为,),1,2,( kjQ
j ??
)(
1 j
i
i
j
i
i
j
i
i
n
i
j q
zZ
q
yY
q
xXQ
?
??
?
??
?
?? ?
?
或
j
j
j q
WQ
?
? )(?
若主动力为有势力，须将势能 U表示为广义坐标的函数。
4,建立拉氏方程并加以整理，得出 k个二阶常微分方程。
5,求出上述一组微分方程的积分。
27
[Example 1] A planet gear moves in the horizontal plane,The weight of the rod OA is P,it can rotate about point O,The weight of
the smaller gear is Q,its radius is r,It rotates around the bigger gear
which is fixed,The radius of the bigger gear is R,At the beginning,
the system was at rest,the rod OA is placed at the position OA0 as
shown in the fig,A constant force couple M is acting on the rod OA,
Determine the equation of motion of the rod OA,
,
)(
??
?
?
?
r
rR
r
v
rRv
A
A
A
???
??
Solution
The mechanism shown in the fig,only
has only one degree of freedom,The
constraints are ideal and steady,
the rotation angle ? of the rod OA can
be chosen as the generalized coordinate,
28
[例 1] 水平面内运动的行星齿轮机构。均质杆 OA：重 P，
可绕 O点转动；均质小齿轮：重 Q，半径 r,沿半径为 R的固
定大齿轮滚动。系统初始静止，系杆 OA位于图示 OA0位置。
系杆 OA受大小不变力偶 M作用后，求系杆 OA的运动方程。
所受约束皆为完整、理想、定常的，
可取 OA杆转角 ? 为广义坐标。
??
?
?
?
r
rR
r
v
rRv
A
A
A
???
?? )(
解,图示机构只有一个自由度
29
22
2
2
2
22222
222
)(
92
12
1
)(
2
1
2
1
)(
2
1
)(
3
1
2
1
2
1
2
1
2
1
?
???
??
?
???
?
rR
g
QP
r
rR
r
g
Q
rR
g
Q
rR
g
P
Iv
g
Q
IT
AAAO
?
?
??
?
??????
???
,0 ; )(
92
6
1; )(
92
6
1
2
2
)(
)(
?
?
?
?
?
??
?
?
?
?
??
?
?
??
?
?
?
?
?
?
??
?
???
?
?
?
T
rR
g
QPT
dt
d
rR
g
QPT
M
W
Q
MW
??
?
?
?
30
22
2
2
2
22222
222
)(
92
12
1
)(
2
1
2
1
)(
2
1
)(
3
1
2
1
2
1
2
1
2
1
?
???
??
?
???
?
rR
g
QP
r
rR
r
g
Q
rR
g
Q
rR
g
P
Iv
g
Q
IT
AAAO
?
?
??
?
??????
???
0 ; )(
92
6
1; )(
92
6
1
2
2
)(
)(
?
?
?
?
?
??
?
?
?
?
??
?
?
??
?
?
?
?
?
?
??
?
???
?
?
?
T
rR
g
QPT
dt
d
rR
g
QPT
M
W
Q
MW
??
?
?
?
31
Substitution into the Lagrange's equations,results in
g
))(92(
6
0 )(
92
6
1
2
2
rRQP
M
M rR
g
QP
??
?
???
?
?
?
?
??
??
Integration gives
2122))(92(
3 CtCgt
rRQP
M ??
????
.))(92( 3 22 gtrRQP M ????Therefore,
Using as the initial conditions
t =0,we obtain, 0 0,0
2100 ?????? C C???? ??
32
代入拉氏方程,
g
))(92(
6
0 )(
92
6
1
2
2
rRQP
M
M rR
g
QP
??
?
???
?
?
?
?
??
??
积分，得,
2122))(92(
3 CtCgt
rRQP
M ??
????
2
2))(92(
3 gt
rRQP
M
????
故,
代入初始条件,t =0 时,得 0 0,0
2100 ?????? C C???? ??
33
[Example 2] A slider A is connected to a spring of stiffness k and
mass m1,They can slip on the smooth horizontal plane,Further,a
simple pendulum is connected to the slider A,the length is l,the
mass of the pendulum bob is m2,Determine the differential
equation of motion of this system,
Solution
This system has two degree of
freedom,All constraints are
ideal ones,The system is a
potential system,We can
choose x and ? as generalized
coordinates,The zero point of the axis x is at the place of the
common length of the spring,For an anticlockwise displacement
of B ? is chosen to be positive,
34
[例 2] 与刚度为 k 的弹簧相连的滑块 A，质量为 m1,可在光
滑水平面上滑动。滑块 A上又连一单摆，摆长 l, 摆锤质量为
m2,试列出该系统的运动微分方程。
解,将弹簧力计入主
动力，则系统成为具
有完整、理想约束的
二自由度系统。保守
系统。取 x,?为广义
坐标,x 轴 原点位于
弹簧自然长度位置,
? 逆时针转向为正。
35
??
?
??
??
c o s2
)s in(
)c o s(
222
2
22
??
??
?
??
lx
lx
l
lxv
B
???
?
??
The kinetic energy of the system is
.c o s
2
1
)(
2
1
)c o s2(
2
1
2
1
2
1
2
1
2
22
2
2
21
222
2
2
1
2
2
2
1
???
???
????
??????
lxmlmxmm
lxlxmxmvmxmT B
????
??????
36
??
?
??
??
c o s2
)s in(
)c o s(
222
2
22
??
??
?
??
lx
lx
l
lxv
B
???
?
??
系统动能,
???
???
c o s
2
1)(
2
1
)c o s2(
2
1
2
1
2
1
2
1
2
22
2
2
21
222
2
2
1
2
2
2
1
????
??????
lxmlmxmm
lxlxmxmvmxmT B
????
??????
37
The potential energy of the system is (choosing the common
length of the spring as the zero point of the potential energy)
.c o s21 22 ?glmkxU ??
The Lagrange's function is
.,c o s)(
.c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
??
????
??
?
????
38
系统势能,（以弹簧原长为弹性势能零点，滑块 A所在平面为
重力势能零点）
?c o s21 22 glmkxU ??
拉格朗日函数,
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
,c o s)(
c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
??
????
??
?
????
39
results after
simplification in
.s i nc o s)(
,s i ns i n,c o s
,s i nc o s)(
22
2
2
222
2
2
2
2221
????
?
???
?
??
?
????
??????
?
????
?
?????
?
lxmlxmlm
L
dt
d
glmlxm
L
lxmlm
L
lmlmxmm
x
L
dt
d
???
?
?
???
?
?
??
?
?
????
?
?
Substitution into
.0s i n c o s
,0s i nc o s)(
),1,2,( 0)(
2
2221
???
?????
??
?
?
?
?
?
???
????
glx
kxlmlmxmm
kj
q
L
q
L
dt
d
jj
????
?????
?
?
40
????
?
???
?
??
?
????
s i nc os)(
s i ns i n,c os
s i nc os)(
22
2
2
222
2
2
2
2221
??????
?
????
?
?????
?
lxmlxmlm
L
dt
d
glmlxm
L
lxmlm
L
lmlmxmm
x
L
dt
d
???
?
?
???
?
?
??
?
?
????
?
?
代入,
0s in c o s
0s inc o s)(
),1,2,( 0)(
2
2221
???
?????
??
?
?
?
?
?
???
????
glx
kxlmlmxmm
kj
q
L
q
L
dt
d
jj
????
?????
?
?
并适当化简得,
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
,c o s)(
c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
??
????
??
?
????
41
0s in c o s
0s inc o s)( 22221
???
?????
???
????
glx
kxlmlmxmm
????
?????
These are the differential equations of motion of the system,
0
0)( 221
???
????
??
?
glx
kxlmxmm
????
????
These are the differential equations of motion for small amplitude
around the equilibrium position (x =0,? =0),
If the system is moving with a small amplitude around the
equilibrium position,we can use ? <<1o,cos? 1and sin? ?,
Neglecting all terms of higher then first order,we get
? ?
42
0s in c o s
0s inc o s)( 22221
???
?????
???
????
glx
kxlmlmxmm
????
?????
系统的运动微分方程。
0
0)( 221
???
????
??
?
glx
kxlmxmm
????
????
上式为系统在平衡位置 (x =0,? =0)附近微幅运动的微分方程。
若系统在平衡位置附近作微幅运动，此时 ? <<1o,
cos? 1,sin? ?，略去二阶以上无穷小量，则 ??
43
§ 17-3 Integrals of the
Lagrange's equations of the second kind
For a potential system a first integration of the Lagrange's equations
can be obtained by an unified formalism,Therefore,the solution of
the dynamical problem for a potential system is simplified,
The energy integral and cyclic integral are involved in the first
integration of the Lagrange's equations for a potential system,
1,Energy integral
If all positive forces of the system all are potential forces the
Lagrange's function is L = T - U the time which t does not appear
explicitly in it,
44
§ 17-3 拉格朗日第二类方程的积分
对于保守系统，可以得到拉格朗日方程的某些统一形式
的首次积分，从而使得保守系统动力学问题的求解过程进一
步简化。
保守系统拉格朗日方程的首次积分包括：能量积分、循
环积分。
一、能量积分
设系统所受的主动力是有势力，且拉格朗日函数 L = T - U
中不显含 t,则
45
j
j
k
j j
j
k
j j
j
k
j j
j
k
j j
q
q
L
q
L
dt
d
q
q
L
dt
d
q
q
L
q
q
L
dt
dL
?
?
?
?
??
?
?
)()(
11
11
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
??
??
a n d,0)( T h e r e f o r e,
1
?????
?
LqqLdtd j
k
j j
??
( c o n s t )
1
CLqqL j
k
j j
?????
?
??
This is the generalized
energy integral,
The time t is not involved in the Lagrange's equations for a potential
system,so the result is called the generalized conservation of energy
for a potential system,It can be proven that in the case of steady
constraints,the equation above can be rewritten as
0
46
j
j
k
j j
j
k
j j
j
k
j j
j
k
j j
q
q
L
q
L
dt
d
q
q
L
dt
d
q
q
L
q
q
L
dt
dL
?
?
?
?
??
?
?
)()(
11
11
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
??
??
0)(
1
??? ???
?
LqqLdtd jk
j j
??
)(
1
常数CLqqL jk
j j
??? ??
?
??
广义能量积分。
保守系统的拉格朗日函数不显含时间 t 时，保守系统的 广
义能量守恒 。可以证明，当系统约束为定常时，上式为
0
47
The equation of the generalized energy integral of the system is just
the equation of conservation of the mechanical energy of system,
( c o n s t ) )(2
1
CUTUTTLqqL j
k
j j
??????????
?
??
2,Cyclic integrals
If a certain generalized coordinate qr does not appear in the
Lagrange's function L this coordinate is called a cyclic coordinate
or an ignorable coordinate of the potential system,
If is a cyclic coordinate of the system we have )( krq r ?
0???
rq
L
Therefore,the corresponding Lagrange's equation yields
0)( ??????
rr q
LqLdtd ?
48
系统的广义能量积分式就是系统的机械能守恒方程式。
)( )(2
1
常数CUTUTTLqqL jk
j j
???????? ??
?
??
二、循环积分
如果拉格朗日函数 L中不显含某一广义坐标 qr,则该坐标
称为保守系统的 循环坐标或可遗坐标 。
当 为系统的循环坐标时，必有 )( krq r ?
0???
rq
L
于是拉氏方程成为
0)( ??????
rr q
LqLdtd ?
49
)( )c o n s t( krCqL
r
???? ?
After integration
we get This is a cyclic integral,
Because L = T – U,and is not involved in U explicitly,we can
rewrite the equation above as
)c o n s t( )( CPqTUTqqL r
rrr
??????? ???? ???
rq?
Pr is called generalized momentum,so the cyclic integral can be also
called the integral of the generalized momentum of the system,A
generalized momentum of a potential system corresponding to a
cyclic coordinate is conserved,
The energy integral and cyclic integrals are obtained by integrating
the Lagrange's equations of a potential system once,They are
differential equations which are one order lower than the Lagrange's
equations,
There is only one energy integral of a potential system,However,
there may be several cyclic coordinates and (the same number of)
cyclic integrals,
50
)( )( krCqL
r
???? 常数?
积分得,循环积分
因 L = T - U，而 U中不显含,故上式可写成
)( )( 常数CPqTUTqqL r
rrr
??????? ???? ???
rq?
Pr称为广义动量，因此循环积分也可称为系统的广义动量积分。
保守系统对应于循环坐标的广义动量守恒。
一个系统的能量积分只可能有一个；而循环积分可能不止
一个，有几个循环坐标，便有几个相应的循环积分。
能量积分和循环积分都是由保守系统拉格朗日方程积分一
次得到的，它们都是比拉格朗日方程低一阶的微分方程。
51
[Example 3] The weight of the wedge is P,the inclination angle of the inclined plane is ?,the horizontal surface is smooth,The
weight of the cylinder is Q,its radius is r,it can not slide but can
roll only,At the beginning the system is at rest,the cylinder is at
the top of the inclined plane,Determine (1)the differential equation
of motion of the system,(2) the acceleration of the wedge,(3) the
energy integral and cyclic integral of the system,
Solution
Investigate the system consisting
of the wedge and the cylinder,The
constraints are ideal,and steady,
The system has two degrees of
freedom,We choose x and s as the
generalized coordinates,the origin
of both coordinates are at the
initial position,
52
[例 3] 楔形体重 P，斜面倾角 ?，置于光滑水平面上。均
质圆柱体重 Q，半径为 r,在楔形体的斜面上只滚不滑。初始
系统静止，且圆柱体位于斜面最高点。试求,(1)系统的运动
微分方程； (2)楔形体的加速度； (3)系统的能量积分与循环积
分。
解,研究楔形体与圆柱体组成
的系统。系统受理想、完整、
定常约束，具有两个自由度。
取广义坐标为 x,s ；各坐标原
点均在初始位置。
53
The kinetic energy of the system is
)( c o s
4
3
2
1
)(
2
1
2
1)c o s2(
2
1
2
1
22
22222
asx
g
Q
s
g
Q
x
g
QP
r
sr
g
Q
sxsx
g
Q
x
g
PT
?
?
????
??????
??
?
??
??????
The potential energy of the system is
(choose the horizontal plane as the
zero point of the gravitational
potential energy.)
)( )c o ss i n(31 brshQPhU ?? ?????
The Lagrange's function is
)( )c o ss i n31 c o s 4321 22 crsQ ( hPhsxgQsgQxg QP
UTL
??? ?????????
??
????
54
系统的动能,
)( c o s
4
3
2
1
)(
2
1
2
1)c o s2(
2
1
2
1
22
22222
asx
g
Q
s
g
Q
x
g
QP
r
sr
g
Q
sxsx
g
Q
x
g
PT
?
?
????
??????
??
?
??
??????
系统的势能,
取水平面为重力势能零点。
)( )c o ss i n(31 brshQPhU ?? ?????
拉格朗日函数,
)( )c o ss i n31 c o s 4321 22 crsQ ( hPhsxgQsgQxg QP
UTL
??? ?????????
??
????
55
Substitution into the Lagrange's equations for a potential system gives after appropriate simplification the differential equations of
motion of the system,
??
?
s in2c o s23
0c o s)(
gxs
sQxQP
??
????
????
???? （ d）
Eliminating from (d) we obtain the acceleration of
the wedge is
gQQP Qx ???? ?? 2s in23 2s in??
The system has an energy integral because t does not appear in
the Lagrange's function L explicitly,
s?
56
代入保守系统拉氏方程，并适当化简，得到系统的运动微分
方程。
??
?
s in2c o s23
0c o s)(
gxs
sQxQP
??
????
????
???? （ d）
解得楔形体的加速度为
gQQP Qx ???? ?? 2s in23 2s in??
拉格朗日函数 L中不显含 t,故系统存在能量积分。
57
1
22 )c o ss in(
3
1
c o s
4
3
2
1
CrshQPhsx
g
Q
s
g
Q
x
g
QP
Lq
q
L
j
j
????????
?
?
?
?
?
?
???????
?
?
At t =0,and x = s = 0,Substitution into the equation above
yields 0?? sx ??
).c o s(311 ?rhQPhC ???
)(, 0s i nc o s4321 22 f sQ sxgQs gQxg QP ???????? ??????
58
1
22 )c o ss in(
3
1
c o s
4
3
2
1
CrshQPhsx
g
Q
s
g
Q
x
g
QP
Lq
q
L
j
j
????????
?
?
?
?
?
?
???????
?
?
当 t =0时，, x = s = 0,代入上式中，得 0?? sx ??
)c o s(311 ?rhQPhC ???
)( 0s i nc o s4321 22 f sQ sxgQs gQxg QP ????????? ??????
59
The generalized coordinate x is a cyclic coordinate of the system,it
does not appear in the Lagrange's function L explicitly,Therefore,
the cyclic integral is
.c o s 2CsgQxg QPxTxLP x ?????????? ?????
At t = 0,,hence C2 = 0,We obtain 0?? sx ??
)(, 0c o s)( gsQxQP ??? ???
Equation ( f ) and ( g ) are the energy integral and the cyclic
integral of the system,In fact,equation ( f ) is the equation of
conservation of the mechanical energy,equation ( g ) is the
conservation of momentum in the direction of x,
60
由于拉格朗日函数 L中不显含广义坐标 x，故 x 为系统循环坐
标，故有循环积分,
2c o s Csg
Qx
g
QP
x
T
x
LP
x ??
??
?
??
?
?? ???
??
t = 0时,故上式中 C2 = 0,可得 0?? sx ??
)( 0c o s)( gsQxQP ??? ???
（ f ),( g ) 式即为系统的能量积分和循环积分。 ( f ) 式
实际上是系统的机械能守恒方程。 ( g )式实质上是系统的动
量在 x方向守恒。
61
62
Theoretical Mechanics
2
3
On the basis of D‘Alembert’s Principle and of the Theorem of
Virtual Displacements in this chapter the general equation of
dynamics and the Lagrange's equations of the second kind
(abbreviated as Lagrange's equations) is deduced,The general
equation of dynamics and the Lagrange's equations are
effective means to study dynamical problems,They provide
very simple,direct and standard ways to solve dynamical
problems of unfree systems of particles,
4
本章在达朗伯原理和虚位移原理的基础上，进一步导
出动力学普遍方程和拉格朗日第二类方程（简称拉格朗日
方程）。动力学普遍方程和拉格朗日方程是研究动力学问
题的有力手段，在解决非自由质点系的动力学问题时，显
得十分简捷、规范。
5
§ 17–1 General equation of dynamics
§ 17–2 Lagrange's equations of the second kind
§ 17–3 Integrals of the Lagrange's equations of
the second kind
Chapter 17,Lagrange's equations
6
§ 17–1 动力学普遍方程
§ 17–2 拉格朗日第二类方程
§ 17–3 拉格朗日第二类方程的积分
第十七章 拉格朗日方程
7
iiiiiii amQaNFm ?? ;,,,
Suppose that there are n particles in a system,particle i being
described by, Then we have
0 i ??? QNF ii
If the system of particle is under the action of the ideal constraints,the
can be treated as positive forces and we get
iQ
0)( ???? iii rQF ?
The explicit form is
0])()()[( ??????? iiiiiiiiiiii zzmZyymYxxmX ??? ??????
§ 17-1 General equation of dynamics
This is the general equation of dynamics,
8
iiiiiiii amQaNFmM ?? ;,,,:
设质点系有 n个质点,第 i个质点
0 i ??? QNF ii
若质点系受有理想约束, 将 作为主动力处理，则,
iQ
0)( ???? iii rQF ?
解析式,
0])()()[( ??????? iiiiiiiiiiii zzmZyymYxxmX ??? ??????
§ 17-1 动力学普遍方程
动力学普遍方程。
9
[Example 1] A triangular prism B is slipping along the smooth
inclined plane of the triangular prism A,The triangular prism A is
placed on the smooth horizontal plane,The weights of A and B are M
and m,the angle of inclination is ?,Determine the acceleration of A,
Investigate the system consisting
of both triangular prisms,This
system is under the action of ideal
constraints,It has two degree of
freedom,
.,rrBeB
r
B
e
BB
A
maQmaQ
QQQ
MaQ
??
??
?
Under the action of the ideal constraints the sum of the virtual
works of the positive forces and of the inertial forces,acting on
every particle of the system,along an arbitrary virtual
displacement is zero at any moment,
Solution
10
例 1 三棱柱 B沿三棱柱 A的光滑斜面滑动，三棱柱 A置于光
滑水平面上,A和 B的质量分别为 M和 m，斜面倾角为 ? 。试求三
棱柱 A的加速度。
解,研究两三棱柱组
成的系统。该系统受理想
约束，具有两个自由度。
r
r
B
e
B
r
B
e
BB
A
maQmaQ
QQQ
MaQ
??
??
?
,
在理想约束的条件下，质点系的各质点在任一瞬时受到的
主动力与惯性力在任意虚位移上所作的虚功之和为零。
11
From the general equation of dynamics we have
.0)s i nc o s()c o s( ??????? BrBeBArBeBA sQQQxQQQ ?????
The system has two degree of freedom,we can choose
as independent virtual displacement,Moreover,
, So we get
BA sx ?? a n d
mgQ?
0s i nc o s
0c o s
???
???
r
r
mamgma
mamaMa
??
?
Solving them we get
.)s in(2 2s in 2 gmM ma ????
12
由动力学普遍方程,
0)s i nc o s()c o s( ??????? BrBeBArBeBA sQQQxQQQ ?????
系统为二自由度，取互不相关的 为独立虚位移，
且,所以
BA sx ??,
mgQ?
0s i nc o s
0c o s
???
???
r
r
mamgma
mamaMa
??
?
解得,
gmMma )s in(2 2s in 2 ????
13
§ 17-2 Lagrange's equations of the second kind
Suppose there are n particles in the system and that there are s
constraints which all are ideal ones,The number of degrees of
freedom is k=3n- s,
Here we will deduce the general equations of dynamics using
generalized coordinates,
kqqq ?,,21
Particle,
If we choose a set of generalized coordinate of the system,,
then we have
iii rmM,,
)( )2,1(
)( ),2,1( ),,,(
1
21
bni
t
r
q
q
r
dt
rd
v
anitqqqrr
k
j
i
j
j
ii
i
kii
??
??
?
?
?
?
?
??
?
?
??
??
are called the generalized velocities,
dt
dqq j
j ??
14
§ 17-2 拉格朗日第二类方程
设质点系有 n个质点，受 s个完整约束且系统所受的约束是
理想约束，自由度 k=3n- s 。
下面推导以广义坐标表示的动力学普遍方程的形式。
质点 。若取系统的一组广义坐标为,则
iii rmM,,kqqq ?,,21
)( )2,1(
)( ),2,1( ),,,(
1
21
bni
t
r
q
q
r
dt
rd
v
anitqqqrr
k
j
i
j
j
ii
i
kii
??
??
?
?
?
?
?
??
?
?
??
??
称 为 广义速度 。
dt
dqq j
j ??
15
?
?
?????
k
j
j
j
i
i cniqq
rr
1
)( ),2,1( ???
Substituting equation (c) into the general equation of dynamics
of the system of particles,we get
? ??
? ??
????? n
i
n
i iiiii
n
i iiii
dramrFramF
1 11
)( 0)( ???
?
? ?
? ????
?
? ?
? ????
??
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
k
j
jj
k
j
j
j
i
i
n
i j
i
i
j
i
i
k
j
n
i
j
j
i
i
k
j
j
j
i
n
i
i
n
i
ii
q
q
z
Z
q
y
Y
q
x
X
q
q
r
Fq
q
r
FrF
1
1 1
1 1111
)]([
)()(
?
?
???
16
?
?
?????
k
j
j
j
i
i cniqq
rr
1
)( ),2,1( ???
代入质点系动力学普遍方程，得,
? ??
? ??
????? n
i
n
i iiiii
n
i iiii
dramrFramF
1 11
)( 0)( ???
?
? ?
? ????
?
? ?
? ????
??
?
?
?
?
?
?
?
?
?
?
?
?
???
?
?
???
k
j
jj
k
j
j
j
i
i
n
i j
i
i
j
i
i
k
j
n
i
j
j
i
i
k
j
j
j
i
n
i
i
n
i
ii
q
q
z
Z
q
y
Y
q
x
X
q
q
r
Fq
q
r
FrF
1
1 1
1 1111
)]([
)()(
?
?
???
17
The Qj are called generalized forces,
)( )(
1
eqzZqyYqxXQ
j
i
i
j
i
i
j
i
i
n
i
j ?
??
?
??
?
?? ?
?
0)(
)()(T h e n
1 1
1 111
?
?
?
???
?
?
?
????
? ?
? ???
? ?
? ???
j
j
i
k
j
n
i
i
ij
n
i
j
k
j j
i
ii
k
j
jj
n
i
iiii
q
q
r
dt
vd
mQ
q
q
r
amqQramF
?
???
)(, ),2,1( 0
1
fkjqrdt vdmQ
j
i
n
i
i
ij ????
?? ?
?
generalized inertial forces
18
称 为 广义力
)( )(
1
eqzZqyYqxXQ
j
i
i
j
i
i
j
i
i
n
i
j ?
??
?
??
?
?? ?
?
0)(
)()(
1 1
1 111
?
?
?
???
?
?
?
????
? ?
? ???
? ?
? ???
j
j
i
k
j
n
i
i
ij
n
i
j
k
j j
i
ii
k
j
jj
n
i
iiii
q
q
r
dt
vd
mQ
q
q
r
amqQramF
?
???则
)( ),2,1( 0
1
fkjqrdt vdmQ
j
in
i
i
ij ????
?? ?
?
广义惯性力
19
The generalized inertial force can be obtained from the kinetic
energy of the system of particle in the following way,
).()(
111 j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i q
r
dt
dvm
q
rv
dt
dm
q
r
dt
vdm
?
???
?
???
?
?? ???
???
For the following calculation we need the two expression,
j
i
j
i
j
i
j
i
q
v
q
r
dt
d
q
v
q
r
?
??
?
?
?
??
?
? a n d
?
The first expression is obtained by differentiation of both sides
of the equation (b) by,
jq?
20
广义惯性力可改变为用质点系的动能表示,因此
)()(
111 j
in
i
ii
j
i
i
n
i
i
j
in
i
i
i q
r
dt
dvm
q
rv
dt
dm
q
r
dt
vdm
?
???
?
???
?
?? ???
???
为简化计算,需要用到以下两个关系式,
j
i
j
i
j
i
j
i
q
v
q
r
dt
d
q
v
q
r
?
??
?
?
?
??
?
? ;
?
下面来推导这两个关系式,
第一式只须将 (b)式两边对 求偏导数即可得到。
jq?
21
The second expression is obtained by comparing two results,one is
given by the differentiation of equation (a) twice by ql and t,the other
is given by the differentiation of equation (b) by ql,
We get,
h a v e w e),( i n t o s u b s t i t u t e
)
2
1
()]
2
1
([
)(
1
2
1
2
111
f
q
T
q
T
dt
d
vm
q
vm
qdt
d
q
v
vm
q
r
v
dt
d
m
q
r
dt
vd
m
jj
n
i
ii
j
n
i
i
j
j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
??
???
??
???
?
?
?
),1,2,( kjQqTqTdtd j
jj
?? ???????
These are the Lagrange's equations of the second kind,
abbreviated as Lagrange's equations,
22
第二式可比较 (a)式先对 ql求偏导数 再对 t求导数与 (b)式对
ql求偏导数的结论得出。
:,)(
)
2
1
()]
2
1
([
)(
1
2
1
2
111
得式代入 f
q
T
q
T
dt
d
vm
q
vm
qdt
d
q
v
vm
q
r
v
dt
d
m
q
r
dt
vd
m
jj
n
i
ii
j
n
i
i
j
j
i
n
i
ii
j
i
i
n
i
i
j
i
n
i
i
i
i
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
??
?
?
?
??
???
??
???
?
?
?
),1,2,( kjQqTqTdtd j
jj
?? ???????
拉格朗日第二类动力学方程，简称拉格朗日方程。
23
If the forces acting on the system of particles all are potential forces,the generalized forces can be expressed by the potential energy,
jQ
).,1,2,(
),(
),,1,2,( )(
1
1
kj
q
U
Q
q
z
z
U
q
y
y
U
q
x
x
U
kj
q
z
Z
q
y
Y
q
x
XQ
j
j
j
i
ij
i
ij
i
i
n
i
j
i
i
j
i
i
j
i
i
n
i
j
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
According to the Lagrange's equations
),1,2,( kjqUqTqTdtd
jjj
?? ??????????
Introducing the Lagrange's function L=T-U we get
),1,2,( 0 )( kjqLqLdtd
jj
?? ???????
These are the Lagrange's equations of a potential system,
24
如果作用于质点系的力是有势力，则广义力 可用质点系的势
能来表达。
jQ
),1,2,(
)(
),1,2,( )(
1
1
kj
q
U
Q
q
z
z
U
q
y
y
U
q
x
x
U
kj
q
z
Z
q
y
Y
q
x
XQ
j
j
j
i
ij
i
ij
i
i
n
i
j
i
i
j
i
i
j
i
i
n
i
j
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
?
?
而拉氏方程为,
),1,2,( kjqUqTqTdtd
jjj
?? ??????????
引入拉格朗日函数,L=T-U 则,
),1,2,( 0 )( kjqLqLdtd
jj
?? ???????
保守系统的拉格朗日方程。
25
The steps of solving problems,Using the Lagrange's equations are,1,Determine the number of degrees of freedom of the system k and
choose appropriate generalized coordinates,Be careful,do not neglect
independent coordinates or add redundant (dependent) coordinates,
2,Calculate the kinetic energy of the system T which can be
expressed as a function of the generalized velocities and the
generalized coordinates,
3,Calculate the generalized forces,the formulae
are ),1,2,( kjQ j ??
)(
1 j
i
i
j
i
i
j
i
i
n
i
j q
zZ
q
yY
q
xXQ
?
??
?
??
?
?? ?
?
or
j
j
j q
WQ
?
? )(?
If the positive forces are potential forces,the potential energy U
should be expressed as a function of the generalized coordinates,
4,Writing down the Lagrange's equations and collate,we obtain
k second order ordinary differential equations,
5,Solve this set of differential equations,
26
应用拉氏方程解题的步骤,
1,判定质点系的自由度 k，选取适宜的广义坐标。必须注意：
不能遗漏独立的坐标，也不能有多余的（不独立）坐标。
2,计算质点系的动能 T，表示为广义速度和广义坐标的函数。
3,计算广义力,计算公式为,),1,2,( kjQ
j ??
)(
1 j
i
i
j
i
i
j
i
i
n
i
j q
zZ
q
yY
q
xXQ
?
??
?
??
?
?? ?
?
或
j
j
j q
WQ
?
? )(?
若主动力为有势力，须将势能 U表示为广义坐标的函数。
4,建立拉氏方程并加以整理，得出 k个二阶常微分方程。
5,求出上述一组微分方程的积分。
27
[Example 1] A planet gear moves in the horizontal plane,The weight of the rod OA is P,it can rotate about point O,The weight of
the smaller gear is Q,its radius is r,It rotates around the bigger gear
which is fixed,The radius of the bigger gear is R,At the beginning,
the system was at rest,the rod OA is placed at the position OA0 as
shown in the fig,A constant force couple M is acting on the rod OA,
Determine the equation of motion of the rod OA,
,
)(
??
?
?
?
r
rR
r
v
rRv
A
A
A
???
??
Solution
The mechanism shown in the fig,only
has only one degree of freedom,The
constraints are ideal and steady,
the rotation angle ? of the rod OA can
be chosen as the generalized coordinate,
28
[例 1] 水平面内运动的行星齿轮机构。均质杆 OA：重 P，
可绕 O点转动；均质小齿轮：重 Q，半径 r,沿半径为 R的固
定大齿轮滚动。系统初始静止，系杆 OA位于图示 OA0位置。
系杆 OA受大小不变力偶 M作用后，求系杆 OA的运动方程。
所受约束皆为完整、理想、定常的，
可取 OA杆转角 ? 为广义坐标。
??
?
?
?
r
rR
r
v
rRv
A
A
A
???
?? )(
解,图示机构只有一个自由度
29
22
2
2
2
22222
222
)(
92
12
1
)(
2
1
2
1
)(
2
1
)(
3
1
2
1
2
1
2
1
2
1
?
???
??
?
???
?
rR
g
QP
r
rR
r
g
Q
rR
g
Q
rR
g
P
Iv
g
Q
IT
AAAO
?
?
??
?
??????
???
,0 ; )(
92
6
1; )(
92
6
1
2
2
)(
)(
?
?
?
?
?
??
?
?
?
?
??
?
?
??
?
?
?
?
?
?
??
?
???
?
?
?
T
rR
g
QPT
dt
d
rR
g
QPT
M
W
Q
MW
??
?
?
?
30
22
2
2
2
22222
222
)(
92
12
1
)(
2
1
2
1
)(
2
1
)(
3
1
2
1
2
1
2
1
2
1
?
???
??
?
???
?
rR
g
QP
r
rR
r
g
Q
rR
g
Q
rR
g
P
Iv
g
Q
IT
AAAO
?
?
??
?
??????
???
0 ; )(
92
6
1; )(
92
6
1
2
2
)(
)(
?
?
?
?
?
??
?
?
?
?
??
?
?
??
?
?
?
?
?
?
??
?
???
?
?
?
T
rR
g
QPT
dt
d
rR
g
QPT
M
W
Q
MW
??
?
?
?
31
Substitution into the Lagrange's equations,results in
g
))(92(
6
0 )(
92
6
1
2
2
rRQP
M
M rR
g
QP
??
?
???
?
?
?
?
??
??
Integration gives
2122))(92(
3 CtCgt
rRQP
M ??
????
.))(92( 3 22 gtrRQP M ????Therefore,
Using as the initial conditions
t =0,we obtain, 0 0,0
2100 ?????? C C???? ??
32
代入拉氏方程,
g
))(92(
6
0 )(
92
6
1
2
2
rRQP
M
M rR
g
QP
??
?
???
?
?
?
?
??
??
积分，得,
2122))(92(
3 CtCgt
rRQP
M ??
????
2
2))(92(
3 gt
rRQP
M
????
故,
代入初始条件,t =0 时,得 0 0,0
2100 ?????? C C???? ??
33
[Example 2] A slider A is connected to a spring of stiffness k and
mass m1,They can slip on the smooth horizontal plane,Further,a
simple pendulum is connected to the slider A,the length is l,the
mass of the pendulum bob is m2,Determine the differential
equation of motion of this system,
Solution
This system has two degree of
freedom,All constraints are
ideal ones,The system is a
potential system,We can
choose x and ? as generalized
coordinates,The zero point of the axis x is at the place of the
common length of the spring,For an anticlockwise displacement
of B ? is chosen to be positive,
34
[例 2] 与刚度为 k 的弹簧相连的滑块 A，质量为 m1,可在光
滑水平面上滑动。滑块 A上又连一单摆，摆长 l, 摆锤质量为
m2,试列出该系统的运动微分方程。
解,将弹簧力计入主
动力，则系统成为具
有完整、理想约束的
二自由度系统。保守
系统。取 x,?为广义
坐标,x 轴 原点位于
弹簧自然长度位置,
? 逆时针转向为正。
35
??
?
??
??
c o s2
)s in(
)c o s(
222
2
22
??
??
?
??
lx
lx
l
lxv
B
???
?
??
The kinetic energy of the system is
.c o s
2
1
)(
2
1
)c o s2(
2
1
2
1
2
1
2
1
2
22
2
2
21
222
2
2
1
2
2
2
1
???
???
????
??????
lxmlmxmm
lxlxmxmvmxmT B
????
??????
36
??
?
??
??
c o s2
)s in(
)c o s(
222
2
22
??
??
?
??
lx
lx
l
lxv
B
???
?
??
系统动能,
???
???
c o s
2
1)(
2
1
)c o s2(
2
1
2
1
2
1
2
1
2
22
2
2
21
222
2
2
1
2
2
2
1
????
??????
lxmlmxmm
lxlxmxmvmxmT B
????
??????
37
The potential energy of the system is (choosing the common
length of the spring as the zero point of the potential energy)
.c o s21 22 ?glmkxU ??
The Lagrange's function is
.,c o s)(
.c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
??
????
??
?
????
38
系统势能,（以弹簧原长为弹性势能零点，滑块 A所在平面为
重力势能零点）
?c o s21 22 glmkxU ??
拉格朗日函数,
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
,c o s)(
c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
??
????
??
?
????
39
results after
simplification in
.s i nc o s)(
,s i ns i n,c o s
,s i nc o s)(
22
2
2
222
2
2
2
2221
????
?
???
?
??
?
????
??????
?
????
?
?????
?
lxmlxmlm
L
dt
d
glmlxm
L
lxmlm
L
lmlmxmm
x
L
dt
d
???
?
?
???
?
?
??
?
?
????
?
?
Substitution into
.0s i n c o s
,0s i nc o s)(
),1,2,( 0)(
2
2221
???
?????
??
?
?
?
?
?
???
????
glx
kxlmlmxmm
kj
q
L
q
L
dt
d
jj
????
?????
?
?
40
????
?
???
?
??
?
????
s i nc os)(
s i ns i n,c os
s i nc os)(
22
2
2
222
2
2
2
2221
??????
?
????
?
?????
?
lxmlxmlm
L
dt
d
glmlxm
L
lxmlm
L
lmlmxmm
x
L
dt
d
???
?
?
???
?
?
??
?
?
????
?
?
代入,
0s in c o s
0s inc o s)(
),1,2,( 0)(
2
2221
???
?????
??
?
?
?
?
?
???
????
glx
kxlmlmxmm
kj
q
L
q
L
dt
d
jj
????
?????
?
?
并适当化简得,
kx
x
L
lmxmm
x
L
glmkxlxmlmxmm
UTL
??
?
?
???
?
?
??????
??
,c o s)(
c o s
2
1
c o s
2
1
)(
2
1
221
2
2
2
22
2
2
21
??
????
??
?
????
41
0s in c o s
0s inc o s)( 22221
???
?????
???
????
glx
kxlmlmxmm
????
?????
These are the differential equations of motion of the system,
0
0)( 221
???
????
??
?
glx
kxlmxmm
????
????
These are the differential equations of motion for small amplitude
around the equilibrium position (x =0,? =0),
If the system is moving with a small amplitude around the
equilibrium position,we can use ? <<1o,cos? 1and sin? ?,
Neglecting all terms of higher then first order,we get
? ?
42
0s in c o s
0s inc o s)( 22221
???
?????
???
????
glx
kxlmlmxmm
????
?????
系统的运动微分方程。
0
0)( 221
???
????
??
?
glx
kxlmxmm
????
????
上式为系统在平衡位置 (x =0,? =0)附近微幅运动的微分方程。
若系统在平衡位置附近作微幅运动，此时 ? <<1o,
cos? 1,sin? ?，略去二阶以上无穷小量，则 ??
43
§ 17-3 Integrals of the
Lagrange's equations of the second kind
For a potential system a first integration of the Lagrange's equations
can be obtained by an unified formalism,Therefore,the solution of
the dynamical problem for a potential system is simplified,
The energy integral and cyclic integral are involved in the first
integration of the Lagrange's equations for a potential system,
1,Energy integral
If all positive forces of the system all are potential forces the
Lagrange's function is L = T - U the time which t does not appear
explicitly in it,
44
§ 17-3 拉格朗日第二类方程的积分
对于保守系统，可以得到拉格朗日方程的某些统一形式
的首次积分，从而使得保守系统动力学问题的求解过程进一
步简化。
保守系统拉格朗日方程的首次积分包括：能量积分、循
环积分。
一、能量积分
设系统所受的主动力是有势力，且拉格朗日函数 L = T - U
中不显含 t,则
45
j
j
k
j j
j
k
j j
j
k
j j
j
k
j j
q
q
L
q
L
dt
d
q
q
L
dt
d
q
q
L
q
q
L
dt
dL
?
?
?
?
??
?
?
)()(
11
11
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
??
??
a n d,0)( T h e r e f o r e,
1
?????
?
LqqLdtd j
k
j j
??
( c o n s t )
1
CLqqL j
k
j j
?????
?
??
This is the generalized
energy integral,
The time t is not involved in the Lagrange's equations for a potential
system,so the result is called the generalized conservation of energy
for a potential system,It can be proven that in the case of steady
constraints,the equation above can be rewritten as
0
46
j
j
k
j j
j
k
j j
j
k
j j
j
k
j j
q
q
L
q
L
dt
d
q
q
L
dt
d
q
q
L
q
q
L
dt
dL
?
?
?
?
??
?
?
)()(
11
11
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
??
??
??
??
0)(
1
??? ???
?
LqqLdtd jk
j j
??
)(
1
常数CLqqL jk
j j
??? ??
?
??
广义能量积分。
保守系统的拉格朗日函数不显含时间 t 时，保守系统的 广
义能量守恒 。可以证明，当系统约束为定常时，上式为
0
47
The equation of the generalized energy integral of the system is just
the equation of conservation of the mechanical energy of system,
( c o n s t ) )(2
1
CUTUTTLqqL j
k
j j
??????????
?
??
2,Cyclic integrals
If a certain generalized coordinate qr does not appear in the
Lagrange's function L this coordinate is called a cyclic coordinate
or an ignorable coordinate of the potential system,
If is a cyclic coordinate of the system we have )( krq r ?
0???
rq
L
Therefore,the corresponding Lagrange's equation yields
0)( ??????
rr q
LqLdtd ?
48
系统的广义能量积分式就是系统的机械能守恒方程式。
)( )(2
1
常数CUTUTTLqqL jk
j j
???????? ??
?
??
二、循环积分
如果拉格朗日函数 L中不显含某一广义坐标 qr,则该坐标
称为保守系统的 循环坐标或可遗坐标 。
当 为系统的循环坐标时，必有 )( krq r ?
0???
rq
L
于是拉氏方程成为
0)( ??????
rr q
LqLdtd ?
49
)( )c o n s t( krCqL
r
???? ?
After integration
we get This is a cyclic integral,
Because L = T – U,and is not involved in U explicitly,we can
rewrite the equation above as
)c o n s t( )( CPqTUTqqL r
rrr
??????? ???? ???
rq?
Pr is called generalized momentum,so the cyclic integral can be also
called the integral of the generalized momentum of the system,A
generalized momentum of a potential system corresponding to a
cyclic coordinate is conserved,
The energy integral and cyclic integrals are obtained by integrating
the Lagrange's equations of a potential system once,They are
differential equations which are one order lower than the Lagrange's
equations,
There is only one energy integral of a potential system,However,
there may be several cyclic coordinates and (the same number of)
cyclic integrals,
50
)( )( krCqL
r
???? 常数?
积分得,循环积分
因 L = T - U，而 U中不显含,故上式可写成
)( )( 常数CPqTUTqqL r
rrr
??????? ???? ???
rq?
Pr称为广义动量，因此循环积分也可称为系统的广义动量积分。
保守系统对应于循环坐标的广义动量守恒。
一个系统的能量积分只可能有一个；而循环积分可能不止
一个，有几个循环坐标，便有几个相应的循环积分。
能量积分和循环积分都是由保守系统拉格朗日方程积分一
次得到的，它们都是比拉格朗日方程低一阶的微分方程。
51
[Example 3] The weight of the wedge is P,the inclination angle of the inclined plane is ?,the horizontal surface is smooth,The
weight of the cylinder is Q,its radius is r,it can not slide but can
roll only,At the beginning the system is at rest,the cylinder is at
the top of the inclined plane,Determine (1)the differential equation
of motion of the system,(2) the acceleration of the wedge,(3) the
energy integral and cyclic integral of the system,
Solution
Investigate the system consisting
of the wedge and the cylinder,The
constraints are ideal,and steady,
The system has two degrees of
freedom,We choose x and s as the
generalized coordinates,the origin
of both coordinates are at the
initial position,
52
[例 3] 楔形体重 P，斜面倾角 ?，置于光滑水平面上。均
质圆柱体重 Q，半径为 r,在楔形体的斜面上只滚不滑。初始
系统静止，且圆柱体位于斜面最高点。试求,(1)系统的运动
微分方程； (2)楔形体的加速度； (3)系统的能量积分与循环积
分。
解,研究楔形体与圆柱体组成
的系统。系统受理想、完整、
定常约束，具有两个自由度。
取广义坐标为 x,s ；各坐标原
点均在初始位置。
53
The kinetic energy of the system is
)( c o s
4
3
2
1
)(
2
1
2
1)c o s2(
2
1
2
1
22
22222
asx
g
Q
s
g
Q
x
g
QP
r
sr
g
Q
sxsx
g
Q
x
g
PT
?
?
????
??????
??
?
??
??????
The potential energy of the system is
(choose the horizontal plane as the
zero point of the gravitational
potential energy.)
)( )c o ss i n(31 brshQPhU ?? ?????
The Lagrange's function is
)( )c o ss i n31 c o s 4321 22 crsQ ( hPhsxgQsgQxg QP
UTL
??? ?????????
??
????
54
系统的动能,
)( c o s
4
3
2
1
)(
2
1
2
1)c o s2(
2
1
2
1
22
22222
asx
g
Q
s
g
Q
x
g
QP
r
sr
g
Q
sxsx
g
Q
x
g
PT
?
?
????
??????
??
?
??
??????
系统的势能,
取水平面为重力势能零点。
)( )c o ss i n(31 brshQPhU ?? ?????
拉格朗日函数,
)( )c o ss i n31 c o s 4321 22 crsQ ( hPhsxgQsgQxg QP
UTL
??? ?????????
??
????
55
Substitution into the Lagrange's equations for a potential system gives after appropriate simplification the differential equations of
motion of the system,
??
?
s in2c o s23
0c o s)(
gxs
sQxQP
??
????
????
???? （ d）
Eliminating from (d) we obtain the acceleration of
the wedge is
gQQP Qx ???? ?? 2s in23 2s in??
The system has an energy integral because t does not appear in
the Lagrange's function L explicitly,
s?
56
代入保守系统拉氏方程，并适当化简，得到系统的运动微分
方程。
??
?
s in2c o s23
0c o s)(
gxs
sQxQP
??
????
????
???? （ d）
解得楔形体的加速度为
gQQP Qx ???? ?? 2s in23 2s in??
拉格朗日函数 L中不显含 t,故系统存在能量积分。
57
1
22 )c o ss in(
3
1
c o s
4
3
2
1
CrshQPhsx
g
Q
s
g
Q
x
g
QP
Lq
q
L
j
j
????????
?
?
?
?
?
?
???????
?
?
At t =0,and x = s = 0,Substitution into the equation above
yields 0?? sx ??
).c o s(311 ?rhQPhC ???
)(, 0s i nc o s4321 22 f sQ sxgQs gQxg QP ???????? ??????
58
1
22 )c o ss in(
3
1
c o s
4
3
2
1
CrshQPhsx
g
Q
s
g
Q
x
g
QP
Lq
q
L
j
j
????????
?
?
?
?
?
?
???????
?
?
当 t =0时，, x = s = 0,代入上式中，得 0?? sx ??
)c o s(311 ?rhQPhC ???
)( 0s i nc o s4321 22 f sQ sxgQs gQxg QP ????????? ??????
59
The generalized coordinate x is a cyclic coordinate of the system,it
does not appear in the Lagrange's function L explicitly,Therefore,
the cyclic integral is
.c o s 2CsgQxg QPxTxLP x ?????????? ?????
At t = 0,,hence C2 = 0,We obtain 0?? sx ??
)(, 0c o s)( gsQxQP ??? ???
Equation ( f ) and ( g ) are the energy integral and the cyclic
integral of the system,In fact,equation ( f ) is the equation of
conservation of the mechanical energy,equation ( g ) is the
conservation of momentum in the direction of x,
60
由于拉格朗日函数 L中不显含广义坐标 x，故 x 为系统循环坐
标，故有循环积分,
2c o s Csg
Qx
g
QP
x
T
x
LP
x ??
??
?
??
?
?? ???
??
t = 0时,故上式中 C2 = 0,可得 0?? sx ??
)( 0c o s)( gsQxQP ??? ???
（ f ),( g ) 式即为系统的能量积分和循环积分。 ( f ) 式
实际上是系统的机械能守恒方程。 ( g )式实质上是系统的动
量在 x方向守恒。
61
62
