1
Theoretical Mechanics
2
3
Summary of the General Theorems of Dynamics
Problem for one particle dynamics Establish the differential
equations of motion for a particle and solve them,
Problem for the dynamics of a system of particles,In principle,we
can write down 3n differential equations for a system of n particles
and then solve them,
Practical problems are,
1.Combining and solving differential equations (performing the
integral operation) is very difficult,
2.In a great number of problems we only need to investigate the
motion of the system of particles as a whole without the necessity
to know the motion of every particle in the system,
Starting with the present chapter we introduce other methods of
solving dynamical problems,The general theorems of dynamics
(including the theorem of momentum,the theorem of kinetic energy,
the theorem of moment of momentum and some others theorems
derived from them) will be introduced,
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实际上的问题是,1、联立求解微分方程 (尤其是积分问题 )非
常困难。
2、大量的问题中，不需要了解每一个质
点的运 动,仅需要研究质点系整体的运
动情况。
动力学普遍定理概述
对 质点 动力学问题,建立质点运动微分方程求解。
对 质点系 动力学问题,理论上讲,n个质点列出 3n个微分方
程,联立求解它们即可。
从本章起,将要讲述解答动力学问题的其它方法,而首先要讨论
的是 动力学普遍定理 (包括动量定理, 动量矩定理, 动能定理及由此
推导出来的其它一些定理 )。
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They show the dependence between two kinds of quantities in concise
mathematical forms,One kind are the quantities related to the
characteristics of motion (momentum,moment of momentum,kinetics
energy etc),The second kind are the quantities related to the forces
(impulse,moment of a force,work,etc.)
In this chapter we will investigate the theorem of momentum of a
particle or a system of particles and establish the relation between
the change of momentum and the impulse of a force,In addition we
will study another important form of the theorem of momentum,the
theorem of motion of the center of mass,
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它们以简明的数学形式,表明两种量 —— 一种是同运动
特征相关的量 (动量、动量矩、动能等 )，一种是同力相关的量
(冲量、力 矩、功等 ) —— 之间的关系，从不同侧面对物体的
机械运动进行深入的研究。在一定条件下，用这些定理来解答
动力学问题非常方便简捷 。
本章中研究 质点和质点系的动量定理,建立了 动量的改变
与力的冲量之间的关系,并研究质点系动量定理的另一重要形
式 —— 质心运动定理 。
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§ 12–1 The center of mass of a system
of particles,external forces and
internal forces
§ 12–2 Momentum and impulse
§ 12–3 Theorem of momentum
§ 12–4 Theorem of motion of the center
of mass
Chapter 12,Theorem of momentum
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§ 12–1 质点系的质心,内力与外力
§ 12–2 动量与冲量
§ 12–3 动量定理
§ 12–4 质心运动定理
第十二章 动量定理
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1,The center of mass,
The center of mass of a system of particles is called center of
mass,It is an important concept representing the distribution
of mass in any system of particles,
§ 12-1 The center of mass of a system
of particles,external forces and internal forces
)( ?? imM
?? ?? iiCiiC rmrMM rmr o r
g o t we,F r o m kzjyixr cccc ???
M
zmz
M
ymy
M
xmx ii
CiiCiiC
??? ???,,
The position of the
center of mass c is
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一,质点系的质心
质点系的质量中心称为质心 。是表征质点系质量分布情况的
一个重要概念。
§ 12-1 质点系的质心，内力与外力
)( ?? imM
?? ?? iiCiiC rmrMM rmr 或
则设,kzjyixr cccc ???
M
zmz
M
ymy
M
xmx ii
CiiCiiC
??? ???,,
质心 C 点的位置,
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In a homogeneous gravitational field the centers of mass and the
center gravity coincide.Using any method of determination of the
center of gravity determines the position of the center of mass of a
system,But they are not identical,The concept of the center of mass
has more mechanical meaning than that of the center of gravity,
External forces are the forces exerted on the members of a system by
particles or bodies not belonging to the given system
Internal forces are the forces of interaction between the members of
the same system,
As far as the whole system of particles is concerned,the geometrical
sum (the principal vector) of all the internal forces of a system is zero,
The sum of the moments (the principal moment )of all the internal
forces of a system with respect to any center of axis is zero,too,
??? ??? 。 0)( or 0)( ;0 )()()( iixiiOii FmFmF
2,External forces and internal forces of a system of particles,
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在均匀重力场中，质点系的质心与重心的位置重合 。可 采
用静力学中确定重心的各种方法来确定质心的位置 。但是，质
心与重心是两个不同的概念，质心比重心具有更加广泛的力学
意义。
内力,所考察的质点系内各质点之间相互作用的力。
对整个质点系来讲，内力系的主矢恒等于零，内力系对任一
点（或轴）的主矩恒等于零。即,
??? ??? 。或 0)( 0)( ;0 )()()( iixiiOii FmFmF
二、质点系的内力与外力
外力,所考察的质点系以外的物体作用于该质点系中各质点的力。
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§ 12-2 Momentum and impulse
1,Momentum
1) Momentum of a particle.The product of the mass of a particle
and its velocity is called the momentum of a particle,It is a
time-dependent vector with the same direction as the
velocity,the unit of which is kg?m/s,
Momentum is a physical quantity measuring the intensity of the
mechanical motion of a material body,For example,the velocity
of a bullet is big but its mass is small,In the case of a boat it is
just opposite,
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§ 12-2 动量与冲量
一、动量
1.质点的动量：质点的质量与速度的乘积 mv 称为
质点的动量。 是瞬时矢量，方向与 v 相同。单位是
kg?m/s。
动量是度量物体机械运动强弱程度的一个物理量。
例,枪弹：速度大，质量小； 船：速度小，质量大。
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2) The momentum of a system of particles is defined as the vector
equal to the geometric sum of the momenta of all the particles of the
system,
Cii vMvmK ?? ?
) t i m e,r e s p e c t t o w i t h e q u a t i o n t h ei a t e( D i f f e r e n t Cii rMrm ??
CCzzCCyyCCxx zMMvKyMMvKxMMvK ??? ??????,,
3)Momentum of a system of rigid bodies,Assume that the mss
and the velocity of the center of mass of the i-th rigid body
are, For the whole system we get then
cii vm,
?? Cii vmK
??
??
??
??
??
??
CiiC i ziz
CiiC i yiy
CiiC i xix
zmvmK
ymvmK
xmvmK
?
?
?
The momentum of a system is equal to the product of the mass of
the whole system and the velocity of its center of mass,In terms
of projections on cartesian axes we have
,
,
,
,
,
,
,
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2.质点系的动量：质点系中所有各质点的动量的矢量和 。
Cii vMvmK ?? ? ) ( 求导Cii rMrm ??
质点系的质量与其质心速度的乘积就等于质点系的动量。则,
CCzzCCyyCCxx zMMvKyMMvKxMMvK ??? ??????,,
3.刚体系统的动量,设第 i个刚体 则整个系统,
cii vm,
?? Cii vmK
??
??
??
??
??
??
CiiC i ziz
CiiC i yiy
CiiC i xix
zmvmK
ymvmK
xmvmK
?
?
?
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?? ?? ABlPC,252
ivvvm
vmvmvmK
CCC
CCC
)c o ss in[( 321
321
????
???
?? ])s i nc o s( 21 jvv CC ?? ??
])
10
1
2
5
2
2
2
1()2
10
3
2
5
2
2
2
1[(
])s i n
2
545c o s
2
1()2c o s
2
545s i n
2
1[(
jiml
jllilllm
??????????
??????
?
??????? ??
]212[2 jiml ??? ?
?lvm C 21,1 ?
?? llvm ABC 2 52 5,2 ??
?lvm C 2,3 ?
( point P is the instantaneous
center of velocity ),
[Example 1] In the mechanism shown in the figure,
OA rotate with a constant angular velocity ?,
Assume that OA=L,OB=L,The rods OA and AB
are homogeneous and of mass m,The mass of the
slide block at B is also m,Determine the
momentum of the system when ?=45o,
Solution,For the rod OA mass =,
For the slide block mass =,
For the rod AB,mass=
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解, 曲柄 OA,
滑块 B,
连杆 AB,（ P为速度瞬心,）
?? ?? ABlPC ;252
〔 例１ 〕 曲柄连杆机构的曲柄 OA以匀
? 转动，设 OA=AB=l,曲柄 OA及连杆
AB都是匀质杆,质量各为 m,滑块 B的质
量也为 m。 求 当 ? = 45o时系统的动量。
ivvvm
vmvmvmK
CCC
CCC
)c o ss in[( 321
321
????
???
?? ])s i nc o s( 21 jvv CC ?? ??
])
10
1
2
5
2
2
2
1()2
10
3
2
5
2
2
2
1[(
])s i n
2
545c o s
2
1()2c o s
2
545s i n
2
1[(
jiml
jllilllm
??????????
??????
?
??????? ??
]212[2 jiml ??? ?
?lvm C 21,1 ?
?? llvm ABC 2 52 5,2 ??
?lvm C 2,3 ?
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F2) Force is a variable vector (Include magnitude and direction.)
The elementary impulse is
The impulse is
)( 12 ttFS ??
dtFSd ?
??
2
1
t
t
dtFS
1) Force is a constant vector,F
2,Impulse,
The product of a force and the action time of the force is
called impulse,Impulse is used to characterize the
accumulated effect on a body of a force acting during a
certain time interval,For instance,when pushed,a cart
exertedly a larger force during a shorter time interval may
obtain the same general effect as that exertedly a smaller
force during a longer time interval,
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2．力 是变矢量：（包括大小和方向的变化）
元冲量,
冲量,
F
)( 12 ttFS ??
dtFSd ?
??
2
1
t
t
dtFS
1．力 是常矢量,F
二．冲量
力与其作用时间的乘积称为力的冲量,冲量表示力在其作
用时间内对物体作用的累积效应的度量。例如，推动车子时，
较大的力作用较短的时间，与较小的力作用较长的时间，可得
到同样的总效应。
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3) The impulse of a resultant force is equal to the geometric sum
of the impulses of all component forces,
?? ?? ? ? ????? i
t
t
t
t
t
t
SdtFdtFdtRS
2
1
2
1
2
1
m / s,kg s)m / s( k g sN 2 ??????
The unit of impulse is the same as that of momentum,
? ? ????
2
1
2
1
2
1
,,
t
t
t
t
t
t
zzyyxx dtFSdtFSdtFS
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3．合力的冲量,等于各分力冲量的矢量和,
??? ???? ? ??? i
t
t
t
t
t
t
SdtFdtFdtRS 2
1
2
1
2
1
冲量的单位,m / skg sm / skg sN
2 ?????? 与动量单位同,
? ? ????
2
1
2
1
2
1
,,
t
t
t
t
t
t
zzyyxx dtFSdtFSdtFS
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§ 12-3 Theorem of momentum
1,Theorem of momentum for one particle,
FvmdtdFdt vdmam ??? )( so B e c a u s e
The derivative of the momentum of a particle with respect to time is
equal to the force acting on the particle,This is the momentum
theorem for one particle,
In a certain time interval,the change of the momentum of a particle
is equal to the impulse of the force during the same interval of time,
SdtFvmvm
t
t
??? ?2
1
12
② Integral form,
SddtFvmd ??)(
The differential of the momentum of a particle equals the elementary
impulse of the force acting on it,
① Differential form,
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§ 12-3 动量定理
一．质点的动量定理
FvmdtdFdt vdmam ???? )( ?
质点的动量对时间的导数等于作用于质点的力
（在某一时间间隔内，动量的增量等于力在该时间内的冲量）
SddtFvmd ??)(
SdtFvmvm
t
t
??? ?2
1
12
— 质点的动量定理
1.微分形式, （动量的微分等于力的元冲量）
2,积分形式,
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③ Projection form,
xx Fmvdtd ?)(
yy Fmvdtd ?)(
zz Fmvdtd ?)(
???? 2
1
12
t
t xxxx
dtFSmvmv
???? 2
1
12
t
t yyyy
dtFSmvmv
???? 2
1
12
t
t zzzz
dtFSmvmv④ The law of conservation of the linear momentum of a particle
If,then is a constant vector and the particle is an inertial motion,
If,then is a const and the motion of the particle along the axis
X is an inertial motion,
0?F
0?xF
vm
xmv
2.Theorem of momentum of a system of particles
)()()( e
i
i
iii FFvmdt
d ??
? ? g e t w ea n d 0B u t, )( )()( ???? ??? i
i
e
i
i
iii FFFvmdt
d
)( e
iFdtKd ??
This is the momentum theorem
of a system of particles,
For the whole system of particles,we have
For any particle i in the system,we have,
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3投影形式,
xx Fmvdtd ?)(
yy Fmvdtd ?)(
zz Fmvdtd ?)(
???? 2
1
12
t
t xxxx
dtFSmvmv
???? 2
1
12
t
t yyyy
dtFSmvmv
???? 2
1
12
t
t zzzz
dtFSmvmv4.质点的动量守恒
若,则 常矢量，质点作惯性运动
若,则 常量，质点沿 x 轴的运动是惯性运动
0?F
0?xF
?vm
?xmv
二、质点系的动量定理
)()()( e
i
i
iii FFvmdt
d ??
? ????? ??? )0( )( )()( i
i
eiiiii FFFvmdtd 而
)( e
iFdtKd ??
(质点系的动量定理）
对整个质点系,
对质点系内任一质点 I,
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The derivative of the linear momentum of a system of particles
with respect to time is equal to the geometric sum of all the
external forces acting on the system,
The differential of the linear momentum of a system of particles
is equal to the geometric sum of the elementary impulses of all
the external forces acting on the system,
During a certain time interval,the change in the linear
momentum of a system of particles is equal to the geometric sum
of the impulses of all the external forces acting on the system
during the same time interval,
Integral form,2)
1) Differential form,i e S dt F ? e i K ) ( ) ( d d ? ? ?,
) (
1 2
e
i S K K ? ? ?
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质点系动量对时间的导数等于作用在质点系上所有外力的矢量和。
质点系动量的微分等于作用在质点系上所有外力元冲量的矢量和。
在某一时间间隔内，质点系动量的改变量等于作用在质点系上
的所有外力在同一时间间隔内的冲量的矢量和,
积分形式 ) ( 1 2 e i S K K ? ? ?
1,e i K i e S dt F ? 微分形式 ) ( ) ( d d ? ? ?
2,
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3) Projection form,
? ? ) ( e ix x F dt dK
? ? ) ( e iy y F dt dK
? ? ) ( e iz z F dt dK
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
ix
e
x x dt F Six K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iy
e
y y dt F Siy K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iz
e
z z dt F Siz K K
4)The conservation law of the linear momentum of a system of
particles,
If then const,
If then const,
,0)( ?? eiF
,0)( ?? eixF
?? ? ii vmK
?? ? ixix vmK
Only external forces can change the total momentum of a
system of particles,while internal forces are incapable of
changing it,They only may cause that the momenta of the
particles or parts of it are exchanged between the particles,
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3.投影形式,
? ? ) ( e ix x F dt dK
? ? ) ( e iy y F dt dK
? ? ) ( e iz z F dt dK
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
ix
e
x x dt F Six K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iy
e
y y dt F Siy K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iz
e
z z dt F Siz K K
4,质点系的动量守恒
若 则 常矢量。
若 则 常量。
,0)( ?? eiF
,0)( ?? eixF
?? ? ii vmK
??? ixix vmK
只有外力才能改变质点系的动量，内力不能改变整个质点系
的动量，但可以引起系统内各质点动量的传递。
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[Example 2] A big triangular column of mass M is placed on a smooth
horizontal plane,on the slope of it lying a small triangular column of mass m,
Determine the displacement of the big triangular column when the small
column slides down the shape to the end,
0)( ??? axmvvM
Solution,Choose the system composed of these two
bodies as the object to be investigated,
Analysis of force ? ?,0)( e
xF
in horizontal direction
?xK const,
From the conservation law of the linear
momentum in horizontal direction (the system
being at rest at the initial moment) we obtain
0)()( ????? vvmvM rx
).( bamM mSmM mS rx ?????
rv
v
Analysis of motion Assume that the velocity of the big
triangular block is and the velocity of the small
triangular block with respect to the big triangular block is,
,As m mMSSm mMvv rxrx ?????
rea vvv ??
Then for the small triangular block we get,
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[例 2] 质量为 M的大三角形柱体,放于光滑水平面上,斜面上另
放一质量为 m的小三角形柱体,求小三角形柱体滑到底时,大三角
形柱体的位移。
0)( ??? axmvvM
解, 选 两物体组成的系统为 研究对象。
受力分析, ? ?,0)( e
xF ?xK
水平方向 常量。
由水平方向动量守恒及初始静止 ；则
0)()( ????? vvmvM rx
)( bamM mSmM mS rx ??????
rv
ra vvv ??
v设大三角块速度,
小三角块相对大三角块速度为,
则小三角块
运动分析,
m mMSSm mMvv rxrx ??????
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[Example 3] A fluid flows through a bend,the fluid velocities at sections A
and B being respect,Determine the dynamic force (additionally
the dynamic reaction ) exerted by the fluid on the bend.Assume the fluid is
incompressible,the flow rate Q(m3/s) is constant and density of the the fluid is
? (kg/m3),
Solution,Choose the fluid between the sections A and B as the system of
particles under consideration,The analysis of the force is shown in the figure
below,Assume that the fluid AB moves to the position ab during the time t,
)m /s( a n d 21 vv
])([])[( 12 aBAaBbaBABab KKKKKKK ???????
,
,)()(f o r
12
12
vtQvtQKKK
KK
AaBb
aBaB
?????????
?
??
From the momentum theorem of
a system of particles we obtain
,)( lim 21120 RPPWvvQtKdtKd t ????????? ?? ?
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运动分析，设经过 ?ｔ 时间后，流体 AB
运动到位置 ab,
[例 3] 流体流过弯管时,在截面 A和 B处的平均流速分别为
求流体对弯管产生的动压力 (附加动压力 )。 设流体
不可压缩，流量 Q(m3/s)为常量,密度为 ? (kg/m3）。
),m /s(,21 vv
])([])[( 12 aBAaBbaBABab KKKKKKK ???????
12
12
)()(
vtQvtQKKK
KK
AaBb
aBaB
???????
?
?????
?
解,取截面 A与 B之间的流体作为研究的质点系。 受力分析如图示。
由质点系动量定理；得
RPPWvvQtKdtKd t ??????? ? 21120 )( lim ????
35
Static reaction,dynamic reaction )('
21 PPWR ???? )('' 12 vvQR ?? ?
When calculating we often use the projection form ''R
)( '' 12 xxx vvQR ?? ?
)( '' 12 yyy vvQR ?? ?
RPPWvvQtKdtKd t ??????? ? 21120 )(lim ????
)()( 1221 vvQPPWR ?????? ?
The force contrary to the force R is just the dynamic force exerted
by the fluid to the bend,
36
静反力, 动反力 )('
21 PPWR ???? )('' 12 vvQR ?? ?
计算 时，常采用投影形式 ''R
)( '' 12 xxx vvQR ?? ?
)( '' 12 yyy vvQR ?? ?
与 相反的力就是管壁上受到的流体作用的动压力,''R
RPPWvvQtKdtKd t ??????? ? 21120 )(lim ????
)()( 1221 vvQPPWR ?????? ?
即
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§ 12-4 Theorem of motion of the center of mass
Substituting into the momentum theorem of a system of particles
we obtain,CvMK ?
.)( )(?? eiC FvMdtd
If the mass of the system does not change or
)( e
iC FaM ?? ??
)( eiC FrM ??
The formulas above are the theorem of motion of the center of mass (or the
differential equations of the motion of the center of mass).The product of the
acceleration of the center of mass of a system and the mass of the whole
system is equal to the geometric sum of all external forces acting on the
system,the principal vector of the external force system,
1,Projection forms,
① ;,,)()()( ??? ?????? e
izCCzeiyCCyeixCCx FzMMaFyMMaFxMMa ??????
②; 0,,)()(
2
)( ??? ????? e
ib
e
in
C
Cn
e
iC FF
vMMaF
dt
dvMMa
???
38
§ 12-4 质心运动定理
将 代入到质点系动量定理，得
CvMK ? ??
)()( e
iC FvMdt
d
若质点系质量不变,则 或 )( e
iC FaM ?? ?? )( eiC FrM ??
上式称为质心运动定理（或质心运动微分方程）。 质点系
的质量与加速度的乘积，等于作用于质点系上所有外力的矢量
和（外力系的主矢）。
1,投影形式,
① 。??? ??????,,)()()( e
izCCzeiyCCyeixCCx FzMMaFyMMaFxMMa ??????
②
。??? ????? 0,,)()(2)( eibeinCCneiC FFvMMaFdtdvMMa ???
39
? ? ??? )( eixCiiC i xi Fxmam ??
? ? ??? )( eiyCiiC i yi Fymam ??
? ? ??? )( eizCiiC i zi Fzmam ??
3,The theorem of motion of the center of mass is an equivalent
expression to the momentum theorem of a system,being similar in
form to the differential equations of motion for one particle,For a
system of particles in arbitrary motion,the center of mass of the
system moves as if it would be a particle of mass equal to the mass of
the whole system to which all the external forces acting on the system
are applied,
2,System of rigid bodies,
Denoting by the No.i the body of mass mi and velocity vCi we have
or )( e
iCii Fam ?? ? ?? ? )( eiCii Frm ??
)( e
iC FaM ??
?? )( eiC FrM ??
40
? ? ??? )( eixCiiC i xi Fxmam ??
? ? ??? )( eiyCiiC i yi Fymam ??
? ? ??? )( eizCiiC i zi Fzmam ??
3,质心运动定理是动量定理的另一种表现形式，与质点运动微
分方程形式相似。 对于任意一个质点系,无论它作什么形式的
运动,质点系质心的运动可以看成为一个质点的运动,并设想
把整个质点系的质量都集中在质心这个点上,所有外力也集中
作用在质心这个点上 。
2,刚体系统,设第 i 个刚体 mi,vCi，则有
或 )( e
iCii Fam ?? ? ?? ? )( eiCii Frm ?? )( eiC FaM ??
?? )( eiC FrM ??
41
4,The conservation law of the motion of the center of mass,
? ? 0)( eiF ?? CC voa an d
00?Cv
?Cr
? ? 0)( eixF ?? CxCx va an d 0
00?Cxv
?Cx
5,The theorem of motion of the center of mass may solve two
types of dynamical problems,
1)Knowing the motion of the center of mass of a system,
determine the external forces acting on the system (including
the reactions of constraints),
2) Knowing the external forces acting on the system,determine
the law of motion of the center of mass,
Only external forces can change the motion of the center of mass of
system,while internal forces are incapable of changing it,They change
only the motion of the individual particles of the system,
1)If then constant,The center of mass
moves uniformly and rectilinearly,If the center of mass was
initially at rest,,then const,the position of the
center of mass will remain constant,
2)If then constant,The projection of the
velocity of the center of mass of the system on the axis x is a
constant quantity,If at the initial moment then const.,
the coordinate x remains constant,
42
4,质心运动守恒定律
?若,则 常矢量，质心作匀速直线运动；
若开始时系统静止，即 则 常矢量,质心位置守恒。
?若 则 常量，质心沿 x方向速度不变；
若存在 则 常量，质心在 x 轴的位置坐标保持不变。
? ? 0)( eiF ?? CC voa,
00?Cv ?Cr
,)(? ?0eixF ?? CxCx va,0
00?Cxv ?Cx
5．质心运动定理可求解两类动力学问题,
?已知质点系质心的运动,求作用 于 质点系的外力 (包括约束反力 )。
?已知作用于质点系的外力，求质心的运动规律。
只有外力才能改变质点系质心的运动,内力不能改变质心
的运动，但可以改变系统内各质点的运动。
43
Solution choose the motor as the system of
particles to be investigated the analysis of forces is
shown in the figure Analysis of motion;The
acceleration a1 at the center of mass of the stator
O1 is zero,
The acceleration a2 at the center of mass of the
rotor O2 is e?2 (directed to O1),
[Example 4] A motor body is mounted on a horizontal foundation,The mass
of the rotor is m2 and the mass of stator is m1,The shaft of the rotor passes
through the center of mass O1 of stator,but there is a distance e from the
center of mass O2 of the rotor to O1 due to an inaccuracy in the production,
Determine the reaction forces of the constraints exerted on the mount of the
motor by the foundation when the rotor is rotating with the angular velocity ?,
44
解, 取整个电动机作为质点系研究,
分析受力,受力图如图示
运动分析：定子质心加速度 a1=0,
转子质心 O2的加速度 a2=e?2,
方向指向 O1。
[例 4] 电动机的外壳固定在水平基础上，定子的质量为 m1,转子质
量为 m2,转子的轴通过定子的质心 O1,但由于制造误差,转子的质
心 O2到 O1的距离为 e 。 求转子以角速度 ? 作匀速转动时，基础作
用在电动机底座上的约束反力。
45
According to the theorem of motion of
the center of mass we have
? ? ???? xxeixC i xi NtemamFam ?? c o s,2222)(
? ? ?????? gmgmNtemamFam yyeiyC i yi 212222)( s i n,??
temgmgmNtemN yx ???? s i n,c o sT h e r e f o r e 222122 ?????
The dynamic reaction caused by the deviation from the
center is a periodic function of time,
,s i n,c o s 22 22 teatea yx ???? ????
a1=0, a2=e?2,
46
teatea yx s i n,c o s 22 22 ???? ????
根据质心运动定理，有
? ? ???? xxeixC i xi NtemamFam ?? c o s,2222)(
? ? ?????? gmgmNtemamFam yyeiyC i yi 212222)( s i n,??
temgmgmNtemN yx ???? s i n,c o s 222122 ??????
可见，由于偏心引起的动反力是随时间而变化的周期函数。
a1=0,a2=e?2
47
321
332211
321
332211 '''
mmm
xmxmxm
mmm
xmxmxm
??
???
??
??
Solution,Choose the system consisting of the hoisting boat with the jib and
the load as the object under study,
0 ??? ii xP ?
[Example 5] There is a floating hoisting boat of weight P1=200kN with the
jib of weight P2=10kN and length l=8m,The weight of the lifted load
P3=20kN,Assume that at the initial moment the whole system at rest and the
angle between the jib OA and the vertical was ?1=60o,Neglecting the
resistance of the water determine the displacement of the boat at the instant
when the jib OA makes an angle ?2 =30o to the vertical,
The analysis of the force is
shown in the figure,Because
and the whole system was initially at rest
the coordinate of the center of mass of the
whole system will remain at rest,
? ? 0)( exF
? ??? 0ii xm ?
48
321
332211
321
332211 '''
mmm
xmxmxm
mmm
xmxmxm
??
???
??
??
解,取起重船，起重杆和重物组成的质点系为研究对象。
0 ??? ii xP ?
[例 5] 浮动起重船,船的重量为 P1=200kN,起重杆的重量为
P2=10kN,长 l=8m，起吊物体的重量为 P3=20kN 。 设开始起吊
时整个系统处于静止，起重杆 OA与铅直位置的夹角为 ?1=60o,
水的阻力不计,求起重杆 OA与铅直位置成角 ?2 =30o时船的位移。
受力分析如图示，,且初始
时系统静止，所以系统质心的位置坐标
XC保持不变。
? ? 0)( exF
? ??? 0ii xm ?
49
The displacement of the boat is ?x１,
The displacement of the jib is
.2/)s i n( s i n 2112 lxx ?? ?????
The displacement of the weight is
.)s i n( s i n 2113 lxx ?? ?????
0]/)s i n( s i n[]2/)s i n( s i n[ T h e r e f o r e 2113211211 ???????????? lxPlxPxP ????
)s in( s in)(2 2 21
321
321 ??? ?????? lPPP PPx
)30s in60( s i n8)2010200(2 20210 ?????? ????
m,318.0??
The negative sign
shows that the boat
moves to the left,
0 ??? ii xP ?? ??? 0B e c a u s e ii xm
50
船的位移 ?x１,杆的位移
,2/)s i n( s i n 2112 lxx ???? ???
重物的位移
lxx )s i n( s i n 2113 ???? ???
0]/)s i n( s i n[]2/)s i n( s i n[ 2113211211 ?????????? lxPlxPxP ???????
)s in( s in)(2 2 21
321
321 ??? ?????? lPPP PPx
)30s in60( s i n8)2010200(2 20210 ?????? ????
m 318.0??
计算结果为负值，表明
船的位移水平向左。
0 ??? ii xP ?? ??? 0ii xm ?
51
52
Theoretical Mechanics
2
3
Summary of the General Theorems of Dynamics
Problem for one particle dynamics Establish the differential
equations of motion for a particle and solve them,
Problem for the dynamics of a system of particles,In principle,we
can write down 3n differential equations for a system of n particles
and then solve them,
Practical problems are,
1.Combining and solving differential equations (performing the
integral operation) is very difficult,
2.In a great number of problems we only need to investigate the
motion of the system of particles as a whole without the necessity
to know the motion of every particle in the system,
Starting with the present chapter we introduce other methods of
solving dynamical problems,The general theorems of dynamics
(including the theorem of momentum,the theorem of kinetic energy,
the theorem of moment of momentum and some others theorems
derived from them) will be introduced,
4
实际上的问题是,1、联立求解微分方程 (尤其是积分问题 )非
常困难。
2、大量的问题中，不需要了解每一个质
点的运 动,仅需要研究质点系整体的运
动情况。
动力学普遍定理概述
对 质点 动力学问题,建立质点运动微分方程求解。
对 质点系 动力学问题,理论上讲,n个质点列出 3n个微分方
程,联立求解它们即可。
从本章起,将要讲述解答动力学问题的其它方法,而首先要讨论
的是 动力学普遍定理 (包括动量定理, 动量矩定理, 动能定理及由此
推导出来的其它一些定理 )。
5
They show the dependence between two kinds of quantities in concise
mathematical forms,One kind are the quantities related to the
characteristics of motion (momentum,moment of momentum,kinetics
energy etc),The second kind are the quantities related to the forces
(impulse,moment of a force,work,etc.)
In this chapter we will investigate the theorem of momentum of a
particle or a system of particles and establish the relation between
the change of momentum and the impulse of a force,In addition we
will study another important form of the theorem of momentum,the
theorem of motion of the center of mass,
6
它们以简明的数学形式,表明两种量 —— 一种是同运动
特征相关的量 (动量、动量矩、动能等 )，一种是同力相关的量
(冲量、力 矩、功等 ) —— 之间的关系，从不同侧面对物体的
机械运动进行深入的研究。在一定条件下，用这些定理来解答
动力学问题非常方便简捷 。
本章中研究 质点和质点系的动量定理,建立了 动量的改变
与力的冲量之间的关系,并研究质点系动量定理的另一重要形
式 —— 质心运动定理 。
7
§ 12–1 The center of mass of a system
of particles,external forces and
internal forces
§ 12–2 Momentum and impulse
§ 12–3 Theorem of momentum
§ 12–4 Theorem of motion of the center
of mass
Chapter 12,Theorem of momentum
8
§ 12–1 质点系的质心,内力与外力
§ 12–2 动量与冲量
§ 12–3 动量定理
§ 12–4 质心运动定理
第十二章 动量定理
9
1,The center of mass,
The center of mass of a system of particles is called center of
mass,It is an important concept representing the distribution
of mass in any system of particles,
§ 12-1 The center of mass of a system
of particles,external forces and internal forces
)( ?? imM
?? ?? iiCiiC rmrMM rmr o r
g o t we,F r o m kzjyixr cccc ???
M
zmz
M
ymy
M
xmx ii
CiiCiiC
??? ???,,
The position of the
center of mass c is
10
一,质点系的质心
质点系的质量中心称为质心 。是表征质点系质量分布情况的
一个重要概念。
§ 12-1 质点系的质心，内力与外力
)( ?? imM
?? ?? iiCiiC rmrMM rmr 或
则设,kzjyixr cccc ???
M
zmz
M
ymy
M
xmx ii
CiiCiiC
??? ???,,
质心 C 点的位置,
11
In a homogeneous gravitational field the centers of mass and the
center gravity coincide.Using any method of determination of the
center of gravity determines the position of the center of mass of a
system,But they are not identical,The concept of the center of mass
has more mechanical meaning than that of the center of gravity,
External forces are the forces exerted on the members of a system by
particles or bodies not belonging to the given system
Internal forces are the forces of interaction between the members of
the same system,
As far as the whole system of particles is concerned,the geometrical
sum (the principal vector) of all the internal forces of a system is zero,
The sum of the moments (the principal moment )of all the internal
forces of a system with respect to any center of axis is zero,too,
??? ??? 。 0)( or 0)( ;0 )()()( iixiiOii FmFmF
2,External forces and internal forces of a system of particles,
12
在均匀重力场中，质点系的质心与重心的位置重合 。可 采
用静力学中确定重心的各种方法来确定质心的位置 。但是，质
心与重心是两个不同的概念，质心比重心具有更加广泛的力学
意义。
内力,所考察的质点系内各质点之间相互作用的力。
对整个质点系来讲，内力系的主矢恒等于零，内力系对任一
点（或轴）的主矩恒等于零。即,
??? ??? 。或 0)( 0)( ;0 )()()( iixiiOii FmFmF
二、质点系的内力与外力
外力,所考察的质点系以外的物体作用于该质点系中各质点的力。
13
§ 12-2 Momentum and impulse
1,Momentum
1) Momentum of a particle.The product of the mass of a particle
and its velocity is called the momentum of a particle,It is a
time-dependent vector with the same direction as the
velocity,the unit of which is kg?m/s,
Momentum is a physical quantity measuring the intensity of the
mechanical motion of a material body,For example,the velocity
of a bullet is big but its mass is small,In the case of a boat it is
just opposite,
14
§ 12-2 动量与冲量
一、动量
1.质点的动量：质点的质量与速度的乘积 mv 称为
质点的动量。 是瞬时矢量，方向与 v 相同。单位是
kg?m/s。
动量是度量物体机械运动强弱程度的一个物理量。
例,枪弹：速度大，质量小； 船：速度小，质量大。
15
2) The momentum of a system of particles is defined as the vector
equal to the geometric sum of the momenta of all the particles of the
system,
Cii vMvmK ?? ?
) t i m e,r e s p e c t t o w i t h e q u a t i o n t h ei a t e( D i f f e r e n t Cii rMrm ??
CCzzCCyyCCxx zMMvKyMMvKxMMvK ??? ??????,,
3)Momentum of a system of rigid bodies,Assume that the mss
and the velocity of the center of mass of the i-th rigid body
are, For the whole system we get then
cii vm,
?? Cii vmK
??
??
??
??
??
??
CiiC i ziz
CiiC i yiy
CiiC i xix
zmvmK
ymvmK
xmvmK
?
?
?
The momentum of a system is equal to the product of the mass of
the whole system and the velocity of its center of mass,In terms
of projections on cartesian axes we have
,
,
,
,
,
,
,
16
2.质点系的动量：质点系中所有各质点的动量的矢量和 。
Cii vMvmK ?? ? ) ( 求导Cii rMrm ??
质点系的质量与其质心速度的乘积就等于质点系的动量。则,
CCzzCCyyCCxx zMMvKyMMvKxMMvK ??? ??????,,
3.刚体系统的动量,设第 i个刚体 则整个系统,
cii vm,
?? Cii vmK
??
??
??
??
??
??
CiiC i ziz
CiiC i yiy
CiiC i xix
zmvmK
ymvmK
xmvmK
?
?
?
17
?? ?? ABlPC,252
ivvvm
vmvmvmK
CCC
CCC
)c o ss in[( 321
321
????
???
?? ])s i nc o s( 21 jvv CC ?? ??
])
10
1
2
5
2
2
2
1()2
10
3
2
5
2
2
2
1[(
])s i n
2
545c o s
2
1()2c o s
2
545s i n
2
1[(
jiml
jllilllm
??????????
??????
?
??????? ??
]212[2 jiml ??? ?
?lvm C 21,1 ?
?? llvm ABC 2 52 5,2 ??
?lvm C 2,3 ?
( point P is the instantaneous
center of velocity ),
[Example 1] In the mechanism shown in the figure,
OA rotate with a constant angular velocity ?,
Assume that OA=L,OB=L,The rods OA and AB
are homogeneous and of mass m,The mass of the
slide block at B is also m,Determine the
momentum of the system when ?=45o,
Solution,For the rod OA mass =,
For the slide block mass =,
For the rod AB,mass=
18
解, 曲柄 OA,
滑块 B,
连杆 AB,（ P为速度瞬心,）
?? ?? ABlPC ;252
〔 例１ 〕 曲柄连杆机构的曲柄 OA以匀
? 转动，设 OA=AB=l,曲柄 OA及连杆
AB都是匀质杆,质量各为 m,滑块 B的质
量也为 m。 求 当 ? = 45o时系统的动量。
ivvvm
vmvmvmK
CCC
CCC
)c o ss in[( 321
321
????
???
?? ])s i nc o s( 21 jvv CC ?? ??
])
10
1
2
5
2
2
2
1()2
10
3
2
5
2
2
2
1[(
])s i n
2
545c o s
2
1()2c o s
2
545s i n
2
1[(
jiml
jllilllm
??????????
??????
?
??????? ??
]212[2 jiml ??? ?
?lvm C 21,1 ?
?? llvm ABC 2 52 5,2 ??
?lvm C 2,3 ?
19
F2) Force is a variable vector (Include magnitude and direction.)
The elementary impulse is
The impulse is
)( 12 ttFS ??
dtFSd ?
??
2
1
t
t
dtFS
1) Force is a constant vector,F
2,Impulse,
The product of a force and the action time of the force is
called impulse,Impulse is used to characterize the
accumulated effect on a body of a force acting during a
certain time interval,For instance,when pushed,a cart
exertedly a larger force during a shorter time interval may
obtain the same general effect as that exertedly a smaller
force during a longer time interval,
20
2．力 是变矢量：（包括大小和方向的变化）
元冲量,
冲量,
F
)( 12 ttFS ??
dtFSd ?
??
2
1
t
t
dtFS
1．力 是常矢量,F
二．冲量
力与其作用时间的乘积称为力的冲量,冲量表示力在其作
用时间内对物体作用的累积效应的度量。例如，推动车子时，
较大的力作用较短的时间，与较小的力作用较长的时间，可得
到同样的总效应。
21
3) The impulse of a resultant force is equal to the geometric sum
of the impulses of all component forces,
?? ?? ? ? ????? i
t
t
t
t
t
t
SdtFdtFdtRS
2
1
2
1
2
1
m / s,kg s)m / s( k g sN 2 ??????
The unit of impulse is the same as that of momentum,
? ? ????
2
1
2
1
2
1
,,
t
t
t
t
t
t
zzyyxx dtFSdtFSdtFS
22
3．合力的冲量,等于各分力冲量的矢量和,
??? ???? ? ??? i
t
t
t
t
t
t
SdtFdtFdtRS 2
1
2
1
2
1
冲量的单位,m / skg sm / skg sN
2 ?????? 与动量单位同,
? ? ????
2
1
2
1
2
1
,,
t
t
t
t
t
t
zzyyxx dtFSdtFSdtFS
23
§ 12-3 Theorem of momentum
1,Theorem of momentum for one particle,
FvmdtdFdt vdmam ??? )( so B e c a u s e
The derivative of the momentum of a particle with respect to time is
equal to the force acting on the particle,This is the momentum
theorem for one particle,
In a certain time interval,the change of the momentum of a particle
is equal to the impulse of the force during the same interval of time,
SdtFvmvm
t
t
??? ?2
1
12
② Integral form,
SddtFvmd ??)(
The differential of the momentum of a particle equals the elementary
impulse of the force acting on it,
① Differential form,
24
§ 12-3 动量定理
一．质点的动量定理
FvmdtdFdt vdmam ???? )( ?
质点的动量对时间的导数等于作用于质点的力
（在某一时间间隔内，动量的增量等于力在该时间内的冲量）
SddtFvmd ??)(
SdtFvmvm
t
t
??? ?2
1
12
— 质点的动量定理
1.微分形式, （动量的微分等于力的元冲量）
2,积分形式,
25
③ Projection form,
xx Fmvdtd ?)(
yy Fmvdtd ?)(
zz Fmvdtd ?)(
???? 2
1
12
t
t xxxx
dtFSmvmv
???? 2
1
12
t
t yyyy
dtFSmvmv
???? 2
1
12
t
t zzzz
dtFSmvmv④ The law of conservation of the linear momentum of a particle
If,then is a constant vector and the particle is an inertial motion,
If,then is a const and the motion of the particle along the axis
X is an inertial motion,
0?F
0?xF
vm
xmv
2.Theorem of momentum of a system of particles
)()()( e
i
i
iii FFvmdt
d ??
? ? g e t w ea n d 0B u t, )( )()( ???? ??? i
i
e
i
i
iii FFFvmdt
d
)( e
iFdtKd ??
This is the momentum theorem
of a system of particles,
For the whole system of particles,we have
For any particle i in the system,we have,
26
3投影形式,
xx Fmvdtd ?)(
yy Fmvdtd ?)(
zz Fmvdtd ?)(
???? 2
1
12
t
t xxxx
dtFSmvmv
???? 2
1
12
t
t yyyy
dtFSmvmv
???? 2
1
12
t
t zzzz
dtFSmvmv4.质点的动量守恒
若,则 常矢量，质点作惯性运动
若,则 常量，质点沿 x 轴的运动是惯性运动
0?F
0?xF
?vm
?xmv
二、质点系的动量定理
)()()( e
i
i
iii FFvmdt
d ??
? ????? ??? )0( )( )()( i
i
eiiiii FFFvmdtd 而
)( e
iFdtKd ??
(质点系的动量定理）
对整个质点系,
对质点系内任一质点 I,
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The derivative of the linear momentum of a system of particles
with respect to time is equal to the geometric sum of all the
external forces acting on the system,
The differential of the linear momentum of a system of particles
is equal to the geometric sum of the elementary impulses of all
the external forces acting on the system,
During a certain time interval,the change in the linear
momentum of a system of particles is equal to the geometric sum
of the impulses of all the external forces acting on the system
during the same time interval,
Integral form,2)
1) Differential form,i e S dt F ? e i K ) ( ) ( d d ? ? ?,
) (
1 2
e
i S K K ? ? ?
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质点系动量对时间的导数等于作用在质点系上所有外力的矢量和。
质点系动量的微分等于作用在质点系上所有外力元冲量的矢量和。
在某一时间间隔内，质点系动量的改变量等于作用在质点系上
的所有外力在同一时间间隔内的冲量的矢量和,
积分形式 ) ( 1 2 e i S K K ? ? ?
1,e i K i e S dt F ? 微分形式 ) ( ) ( d d ? ? ?
2,
29
3) Projection form,
? ? ) ( e ix x F dt dK
? ? ) ( e iy y F dt dK
? ? ) ( e iz z F dt dK
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
ix
e
x x dt F Six K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iy
e
y y dt F Siy K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iz
e
z z dt F Siz K K
4)The conservation law of the linear momentum of a system of
particles,
If then const,
If then const,
,0)( ?? eiF
,0)( ?? eixF
?? ? ii vmK
?? ? ixix vmK
Only external forces can change the total momentum of a
system of particles,while internal forces are incapable of
changing it,They only may cause that the momenta of the
particles or parts of it are exchanged between the particles,
30
3.投影形式,
? ? ) ( e ix x F dt dK
? ? ) ( e iy y F dt dK
? ? ) ( e iz z F dt dK
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
ix
e
x x dt F Six K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iy
e
y y dt F Siy K K
? ? ? ? ? ? 2
1
) ( ) (
1 2
t
t
e
iz
e
z z dt F Siz K K
4,质点系的动量守恒
若 则 常矢量。
若 则 常量。
,0)( ?? eiF
,0)( ?? eixF
?? ? ii vmK
??? ixix vmK
只有外力才能改变质点系的动量，内力不能改变整个质点系
的动量，但可以引起系统内各质点动量的传递。
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[Example 2] A big triangular column of mass M is placed on a smooth
horizontal plane,on the slope of it lying a small triangular column of mass m,
Determine the displacement of the big triangular column when the small
column slides down the shape to the end,
0)( ??? axmvvM
Solution,Choose the system composed of these two
bodies as the object to be investigated,
Analysis of force ? ?,0)( e
xF
in horizontal direction
?xK const,
From the conservation law of the linear
momentum in horizontal direction (the system
being at rest at the initial moment) we obtain
0)()( ????? vvmvM rx
).( bamM mSmM mS rx ?????
rv
v
Analysis of motion Assume that the velocity of the big
triangular block is and the velocity of the small
triangular block with respect to the big triangular block is,
,As m mMSSm mMvv rxrx ?????
rea vvv ??
Then for the small triangular block we get,
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[例 2] 质量为 M的大三角形柱体,放于光滑水平面上,斜面上另
放一质量为 m的小三角形柱体,求小三角形柱体滑到底时,大三角
形柱体的位移。
0)( ??? axmvvM
解, 选 两物体组成的系统为 研究对象。
受力分析, ? ?,0)( e
xF ?xK
水平方向 常量。
由水平方向动量守恒及初始静止 ；则
0)()( ????? vvmvM rx
)( bamM mSmM mS rx ??????
rv
ra vvv ??
v设大三角块速度,
小三角块相对大三角块速度为,
则小三角块
运动分析,
m mMSSm mMvv rxrx ??????
33
[Example 3] A fluid flows through a bend,the fluid velocities at sections A
and B being respect,Determine the dynamic force (additionally
the dynamic reaction ) exerted by the fluid on the bend.Assume the fluid is
incompressible,the flow rate Q(m3/s) is constant and density of the the fluid is
? (kg/m3),
Solution,Choose the fluid between the sections A and B as the system of
particles under consideration,The analysis of the force is shown in the figure
below,Assume that the fluid AB moves to the position ab during the time t,
)m /s( a n d 21 vv
])([])[( 12 aBAaBbaBABab KKKKKKK ???????
,
,)()(f o r
12
12
vtQvtQKKK
KK
AaBb
aBaB
?????????
?
??
From the momentum theorem of
a system of particles we obtain
,)( lim 21120 RPPWvvQtKdtKd t ????????? ?? ?
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运动分析，设经过 ?ｔ 时间后，流体 AB
运动到位置 ab,
[例 3] 流体流过弯管时,在截面 A和 B处的平均流速分别为
求流体对弯管产生的动压力 (附加动压力 )。 设流体
不可压缩，流量 Q(m3/s)为常量,密度为 ? (kg/m3）。
),m /s(,21 vv
])([])[( 12 aBAaBbaBABab KKKKKKK ???????
12
12
)()(
vtQvtQKKK
KK
AaBb
aBaB
???????
?
?????
?
解,取截面 A与 B之间的流体作为研究的质点系。 受力分析如图示。
由质点系动量定理；得
RPPWvvQtKdtKd t ??????? ? 21120 )( lim ????
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Static reaction,dynamic reaction )('
21 PPWR ???? )('' 12 vvQR ?? ?
When calculating we often use the projection form ''R
)( '' 12 xxx vvQR ?? ?
)( '' 12 yyy vvQR ?? ?
RPPWvvQtKdtKd t ??????? ? 21120 )(lim ????
)()( 1221 vvQPPWR ?????? ?
The force contrary to the force R is just the dynamic force exerted
by the fluid to the bend,
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静反力, 动反力 )('
21 PPWR ???? )('' 12 vvQR ?? ?
计算 时，常采用投影形式 ''R
)( '' 12 xxx vvQR ?? ?
)( '' 12 yyy vvQR ?? ?
与 相反的力就是管壁上受到的流体作用的动压力,''R
RPPWvvQtKdtKd t ??????? ? 21120 )(lim ????
)()( 1221 vvQPPWR ?????? ?
即
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§ 12-4 Theorem of motion of the center of mass
Substituting into the momentum theorem of a system of particles
we obtain,CvMK ?
.)( )(?? eiC FvMdtd
If the mass of the system does not change or
)( e
iC FaM ?? ??
)( eiC FrM ??
The formulas above are the theorem of motion of the center of mass (or the
differential equations of the motion of the center of mass).The product of the
acceleration of the center of mass of a system and the mass of the whole
system is equal to the geometric sum of all external forces acting on the
system,the principal vector of the external force system,
1,Projection forms,
① ;,,)()()( ??? ?????? e
izCCzeiyCCyeixCCx FzMMaFyMMaFxMMa ??????
②; 0,,)()(
2
)( ??? ????? e
ib
e
in
C
Cn
e
iC FF
vMMaF
dt
dvMMa
???
38
§ 12-4 质心运动定理
将 代入到质点系动量定理，得
CvMK ? ??
)()( e
iC FvMdt
d
若质点系质量不变,则 或 )( e
iC FaM ?? ?? )( eiC FrM ??
上式称为质心运动定理（或质心运动微分方程）。 质点系
的质量与加速度的乘积，等于作用于质点系上所有外力的矢量
和（外力系的主矢）。
1,投影形式,
① 。??? ??????,,)()()( e
izCCzeiyCCyeixCCx FzMMaFyMMaFxMMa ??????
②
。??? ????? 0,,)()(2)( eibeinCCneiC FFvMMaFdtdvMMa ???
39
? ? ??? )( eixCiiC i xi Fxmam ??
? ? ??? )( eiyCiiC i yi Fymam ??
? ? ??? )( eizCiiC i zi Fzmam ??
3,The theorem of motion of the center of mass is an equivalent
expression to the momentum theorem of a system,being similar in
form to the differential equations of motion for one particle,For a
system of particles in arbitrary motion,the center of mass of the
system moves as if it would be a particle of mass equal to the mass of
the whole system to which all the external forces acting on the system
are applied,
2,System of rigid bodies,
Denoting by the No.i the body of mass mi and velocity vCi we have
or )( e
iCii Fam ?? ? ?? ? )( eiCii Frm ??
)( e
iC FaM ??
?? )( eiC FrM ??
40
? ? ??? )( eixCiiC i xi Fxmam ??
? ? ??? )( eiyCiiC i yi Fymam ??
? ? ??? )( eizCiiC i zi Fzmam ??
3,质心运动定理是动量定理的另一种表现形式，与质点运动微
分方程形式相似。 对于任意一个质点系,无论它作什么形式的
运动,质点系质心的运动可以看成为一个质点的运动,并设想
把整个质点系的质量都集中在质心这个点上,所有外力也集中
作用在质心这个点上 。
2,刚体系统,设第 i 个刚体 mi,vCi，则有
或 )( e
iCii Fam ?? ? ?? ? )( eiCii Frm ?? )( eiC FaM ??
?? )( eiC FrM ??
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4,The conservation law of the motion of the center of mass,
? ? 0)( eiF ?? CC voa an d
00?Cv
?Cr
? ? 0)( eixF ?? CxCx va an d 0
00?Cxv
?Cx
5,The theorem of motion of the center of mass may solve two
types of dynamical problems,
1)Knowing the motion of the center of mass of a system,
determine the external forces acting on the system (including
the reactions of constraints),
2) Knowing the external forces acting on the system,determine
the law of motion of the center of mass,
Only external forces can change the motion of the center of mass of
system,while internal forces are incapable of changing it,They change
only the motion of the individual particles of the system,
1)If then constant,The center of mass
moves uniformly and rectilinearly,If the center of mass was
initially at rest,,then const,the position of the
center of mass will remain constant,
2)If then constant,The projection of the
velocity of the center of mass of the system on the axis x is a
constant quantity,If at the initial moment then const.,
the coordinate x remains constant,
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4,质心运动守恒定律
?若,则 常矢量，质心作匀速直线运动；
若开始时系统静止，即 则 常矢量,质心位置守恒。
?若 则 常量，质心沿 x方向速度不变；
若存在 则 常量，质心在 x 轴的位置坐标保持不变。
? ? 0)( eiF ?? CC voa,
00?Cv ?Cr
,)(? ?0eixF ?? CxCx va,0
00?Cxv ?Cx
5．质心运动定理可求解两类动力学问题,
?已知质点系质心的运动,求作用 于 质点系的外力 (包括约束反力 )。
?已知作用于质点系的外力，求质心的运动规律。
只有外力才能改变质点系质心的运动,内力不能改变质心
的运动，但可以改变系统内各质点的运动。
43
Solution choose the motor as the system of
particles to be investigated the analysis of forces is
shown in the figure Analysis of motion;The
acceleration a1 at the center of mass of the stator
O1 is zero,
The acceleration a2 at the center of mass of the
rotor O2 is e?2 (directed to O1),
[Example 4] A motor body is mounted on a horizontal foundation,The mass
of the rotor is m2 and the mass of stator is m1,The shaft of the rotor passes
through the center of mass O1 of stator,but there is a distance e from the
center of mass O2 of the rotor to O1 due to an inaccuracy in the production,
Determine the reaction forces of the constraints exerted on the mount of the
motor by the foundation when the rotor is rotating with the angular velocity ?,
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解, 取整个电动机作为质点系研究,
分析受力,受力图如图示
运动分析：定子质心加速度 a1=0,
转子质心 O2的加速度 a2=e?2,
方向指向 O1。
[例 4] 电动机的外壳固定在水平基础上，定子的质量为 m1,转子质
量为 m2,转子的轴通过定子的质心 O1,但由于制造误差,转子的质
心 O2到 O1的距离为 e 。 求转子以角速度 ? 作匀速转动时，基础作
用在电动机底座上的约束反力。
45
According to the theorem of motion of
the center of mass we have
? ? ???? xxeixC i xi NtemamFam ?? c o s,2222)(
? ? ?????? gmgmNtemamFam yyeiyC i yi 212222)( s i n,??
temgmgmNtemN yx ???? s i n,c o sT h e r e f o r e 222122 ?????
The dynamic reaction caused by the deviation from the
center is a periodic function of time,
,s i n,c o s 22 22 teatea yx ???? ????
a1=0, a2=e?2,
46
teatea yx s i n,c o s 22 22 ???? ????
根据质心运动定理，有
? ? ???? xxeixC i xi NtemamFam ?? c o s,2222)(
? ? ?????? gmgmNtemamFam yyeiyC i yi 212222)( s i n,??
temgmgmNtemN yx ???? s i n,c o s 222122 ??????
可见，由于偏心引起的动反力是随时间而变化的周期函数。
a1=0,a2=e?2
47
321
332211
321
332211 '''
mmm
xmxmxm
mmm
xmxmxm
??
???
??
??
Solution,Choose the system consisting of the hoisting boat with the jib and
the load as the object under study,
0 ??? ii xP ?
[Example 5] There is a floating hoisting boat of weight P1=200kN with the
jib of weight P2=10kN and length l=8m,The weight of the lifted load
P3=20kN,Assume that at the initial moment the whole system at rest and the
angle between the jib OA and the vertical was ?1=60o,Neglecting the
resistance of the water determine the displacement of the boat at the instant
when the jib OA makes an angle ?2 =30o to the vertical,
The analysis of the force is
shown in the figure,Because
and the whole system was initially at rest
the coordinate of the center of mass of the
whole system will remain at rest,
? ? 0)( exF
? ??? 0ii xm ?
48
321
332211
321
332211 '''
mmm
xmxmxm
mmm
xmxmxm
??
???
??
??
解,取起重船，起重杆和重物组成的质点系为研究对象。
0 ??? ii xP ?
[例 5] 浮动起重船,船的重量为 P1=200kN,起重杆的重量为
P2=10kN,长 l=8m，起吊物体的重量为 P3=20kN 。 设开始起吊
时整个系统处于静止，起重杆 OA与铅直位置的夹角为 ?1=60o,
水的阻力不计,求起重杆 OA与铅直位置成角 ?2 =30o时船的位移。
受力分析如图示，,且初始
时系统静止，所以系统质心的位置坐标
XC保持不变。
? ? 0)( exF
? ??? 0ii xm ?
49
The displacement of the boat is ?x１,
The displacement of the jib is
.2/)s i n( s i n 2112 lxx ?? ?????
The displacement of the weight is
.)s i n( s i n 2113 lxx ?? ?????
0]/)s i n( s i n[]2/)s i n( s i n[ T h e r e f o r e 2113211211 ???????????? lxPlxPxP ????
)s in( s in)(2 2 21
321
321 ??? ?????? lPPP PPx
)30s in60( s i n8)2010200(2 20210 ?????? ????
m,318.0??
The negative sign
shows that the boat
moves to the left,
0 ??? ii xP ?? ??? 0B e c a u s e ii xm
50
船的位移 ?x１,杆的位移
,2/)s i n( s i n 2112 lxx ???? ???
重物的位移
lxx )s i n( s i n 2113 ???? ???
0]/)s i n( s i n[]2/)s i n( s i n[ 2113211211 ?????????? lxPlxPxP ???????
)s in( s in)(2 2 21
321
321 ??? ?????? lPPP PPx
)30s in60( s i n8)2010200(2 20210 ?????? ????
m 318.0??
计算结果为负值，表明
船的位移水平向左。
0 ??? ii xP ?? ??? 0ii xm ?
51
52
