Theoretical Mechanics
3
An important theorem of dynamics,D’Alembert’s principle,will be
introduced in this chapter,By this principle dynamical problems can
be transformed formally formally into those of statics,Then they can
be solved by the theorem of equilibrium,Therefore this method to
solve the dynamical problems of dynamics is called the dynamic-
static method,
4
本章介绍动力学的一个重要原理 —— 达朗伯原理 。应用
这一原理，就将动力学问题从形式上转化为静力学问题，从
而根据关于平衡的理论来求解。这种解答动力学问题的方法，
因而也称 动静法 。
§ 15–1 The concept of the inertial force and the
D’Alembert’s principle for one particle
§ 15–2 D’Alembert’s principle for a system of
particles
§ 15–3 The simplification of a system of
inertial forces of a rigid body
§ 15–4 Dynamical reaction of the bearing of a
rigid body under fixed-axis
rotation and the concepts of static
and dynamic equilibrium
The application of D’Alembert’s principle
Chapter 15,D'Alembert's principle
§ 15–1 惯性力的概念 · 质点的达朗伯原理
§ 15–2 质点系的达朗伯原理
§ 15–3 刚体惯性力系的简化
§ 15–4 定轴转动刚体的轴承动反力 ?
静平衡与动平衡的概念
达朗伯原理的应用
第十五章 达朗伯原理
7
§ 15-1 The concept of the inertial force
and D’Alembert’s principle for a particle
If a man is pushing the cart by hands,then
amFQ ????The force comes from the inertia of the cart,
keep the force diagram as the original motion
state,the reaction force to the object (hands)
applying the force is called the inertial force of
the cart,
Q
Concept of the inertial force of a particle,
If a particle is moving accelerated,the sum of the inertial
reactions of the particle to the objects which force it to
produce the accelerated motion is called the inertial force of
the particle,
amQ ??
1,The concept of the inertial force
8
§ 15-1 惯性力的概念 · 质点的达朗伯原理
人用手推车 amFF ????'
力 是由于小车具有惯性，力图保持原来
的运动状态，对于施力物体 (人手 )产生的
反抗力。称为小车的 惯性力 。
'F
定义：质点惯性力
加速运动的质点，对迫使其产生加速运动的物体的惯
性反抗的总和。
amQ ??
一,惯性力的概念
9
。
2
2
2
2
2
2
,
,
dt
zd
mmaQ
dt
yd
mmaQ
dt
xd
mmaQ
zz
yy
xx
????
????
????
。0
,
,
2
2
2
???
????
????
bb
nn
maQ
v
mmaQ
dt
sd
mmaQ
?
??
[notice]
The inertial force is not the real force acting on the particle,it
is the resultant force of reaction of the particle to the object
applying the force,
10
2
2
2
2
2
2
dt
zd
mmaQ
dt
yd
mmaQ
dt
xd
mmaQ
zz
yy
xx
????
????
????
0
2
2
2
???
????
????
bb
nn
maQ
v
mmaQ
dt
sd
mmaQ
?
??
[注 ] 质点惯性力不是作用在质点上的真实力，它是质点对施
力体反作用力的合力。
11
The mass of an unfree particle M is m,It is
subjected to a positive force,The reaction force
is, The resultant force is,
F
N amNFR ???
0??? amNF
0??? QNF
This is D’Alembert’s
principle for a
particle,
2,D’Alembert’s principle for a particle
12
非自由质点 M，质量 m，受主动力,
约束反力,合力
F
N amNFR ???
0??? amNF
0??? QNF
质点的达朗伯原理
二、质点的达朗伯原理
13
This equation is an equilibrium form of a dynamical problem,it
does not change the essential of the problems,The most
outstanding advantage of the dynamic-static method is that the
methods of statics can be applied to solve dynamical problems,
14
该方程对动力学问题来说只是形式上的平衡，并没有
改变动力学问题的实质。采用动静法解决动力学问题的最
大优点，可以利用静力学提供的解题方法，给动力学问题
一种统一的解题格式。
15
[Example 1] A train is running along a horizontal railway,and a
single pendulum is hanging in the carriage,When the carriage
moves to the right with an uniform acceleration,the single
pendulum will turn to the left by an angle ?,and does not move
relative to the carriage,Determine the acceleration of the
carriage, a
16
[例 1] 列车在水平轨道上行驶，车厢内悬挂一单摆，当车厢向
右作匀加速运动时，单摆左偏角度 ?,相对于车厢静止。求车
厢的加速度 。 a
17
Investigate the single pendulum,
add the virtual inertial forces,
.) ( maQamQ ???
.0c o ss i n,0 ????? ?? QmgX
.tg ??? ga
The angle ? changes with the acceleration,when does not
change,the angle ? does not change,too,If we know the angle ?
the acceleration of the train can be calculated, This is the
theory of a pendulum accelerometer,
a a
a
Solution
According to the dynamic-static method
we have
Solving it we get
18
选单摆的摆锤为研究对象
虚加惯性力
) ( maQamQ ???
0co ss i n,0 ????? ?? QmgX
?tg?? ga
?角随着加速度 的变化而变化，当 不变时,? 角也
不变。只要测出 ? 角，就能知道列车的加速度 。摆式加速
计的原理。
a a
a
解,
由动静法,有
解得
19
§ 15-2 D’Alembert’s principle for a system of particles
A system of particles consists of n particles,For every particle
we have
),1,2,.,,,,, ( 0 niQNF iii ????
For the whole system,the positive force system,the system of the
reaction forces of constraints and the inertial force system form
together of an equivalent force system,This is D’Alembert’s
principle for a system of particles,It can be expressed by as
follows,
.0)()()(
,0
???
???
???
???
iOiOiO
iii
QmNmFm
QNF
Please,note that, If we classify the forces by
internal and external forces then we have ?? ?? 0)(,0
)()( iiOii FmF
。? ?
? ?
??
??
0)()(
,0
)(
)(
iO
e
iO
i
e
i
QmFm
QF
20
§ 15-2 质点系的达朗伯原理
对整个质点系，主动力系、约束反力系、惯性力系形式上
构成平衡力系。这就是 质点系的达朗伯原理 。可用方程表示为,
0)()()(
0
???
???
???
???
iOiOiO
iii
QmNmFm
QNF
设有一质点系由 n个质点组成，对每一个质点，有
),1,2,.,,,,, ( 0 niQNF iii ????
注意到,将质点系受力按内力、外力
划分,则 ?? ?? 0)(,0
)()( iiOii FmF
? ?
? ?
??
??
0)()(
0
)(
)(
iO
e
iO
i
e
i
QmFm
QF
21
The equations show that,for the whole system,the equilibrium
equations given by the dynamic-static method just express the
equilibrium between the inertial force system and the external
force system,the internal forces are not involved,
,)( ??? ???????? dt KdvmdtdaMamQ iiCiii
。dtLdvmmdtdammQm OiiOiiOiO ?????? ??? )()()(
22
表明：对整个质点系来说，动静法给出的平衡方程，只
是质点系的惯性力系与其外力的平衡，而与内力无关。
??? ???????? dt KdvmdtdaMamQ iiCiii )(
dt
Ldvmm
dt
dammQm O
iiOiiOiO ?????? ??? )()()(
23
you have for an arbitrary
force system in a plane
。? ?
? ?
? ?
??
??
??
0)()(
,0
,0
)(
)(
)(
iO
e
iO
iy
e
i
ix
e
i
QmFm
QY
QX
and you for an arbitrary
force system in space
。0)()(,0
,0)()(,0
,0)()(,0
)()(
)()(
)()(
????
????
????
??? ?
??? ?
??? ?
iz
e
iziz
e
i
iy
e
iyiy
e
i
ix
e
ixix
e
i
QmFmQZ
QmFmQY
QmFmQX
In practice,we can select the object to e investigate a arbitrarily as in
the case al statics,Write dour the equilibrium equations and then solve
them,
When you solve dynamical problems of dynamics by the dynamic-static
method,
24
对平面任意力系,
? ?
? ?
? ?
??
??
??
0)()(
0
0
)(
)(
)(
iO
e
iO
iy
e
i
ix
e
i
QmFm
QY
QX
对于空间任意力系,
0)()(,0
0)()(,0
0)()(,0
)()(
)()(
)()(
????
????
????
??? ?
??? ?
??? ?
iz
e
iziz
e
i
iy
e
iyiy
e
i
ix
e
ixix
e
i
QmFmQZ
QmFmQY
QmFmQX
实际应用时,同静力学一样任意选取研究对象,列平衡方
程求解。
用动静法求解动力学问题时,
25
§ 15-3 The simplification of a system of inertial forces of a rigid body
The method of simplification is the same as in the case of the
theorem of simplification of a force systems in statics,We should
treat the virtual inertial force system as an inertial force and an
inertial force couple,which are the results of the simplification
of the force system to a point O,
QR
QOM
r e d u c t i o n, ofc e n t e r o n t h e d e p e n d sw h ic h,)(
r e d u c t i o n, ofc e n t e r o n t h e in d e p e n d e d is w h ic h,
?
??
?
?????
QmM
aMamQR
OQO
CQ
No matter what motion the rigid body is doing,the principle vector
of the inertial force system is equal to the product of the weight of
the rigid body and the acceleration of the center of mass,the
direction is opposite to the direction of the acceleration of the
center of mass,
26
§ 15-3 刚体惯性力系的简化
简化方法就是采用静力学中的力系简化的理论。将虚拟的
惯性力系视作力系向任一点 O简化而得到一个惯性力 和一个
惯性力偶 。
QR
QOM
)(
与简化中心有关
与简化中心无关
?
??
?
?????
QmM
aMamQR
OQO
CQ
无论刚体作什么运动，惯性力系主矢都等于刚体质量与质
心加速度的乘积，方向与质心加速度方向相反。
27
1,Transactional motion al a rigid body,
Reducing to the center of mass C we get,
CQ aMR ??
.0)()(? ?? ???????? CiiCiiiCQC armamrQmM
cQ aMR ??
If the rigid body performs a translation,the inertial force
system composes a resultant inertial force passing
through the center of mass,
Look for
the
animation
on the
next page,
28
一、刚体作平动
向质心 C简化,
CQ aMR ??
? ?? ???????? 0)()( CiiCiiiCQC armamrQmM
cQ aMR ??
刚体平动时惯性力系合成为一过质心的合惯性力。
翻
页
请
看
动
画
29
30
31
The inertial force system in space can be simplified
to the inertial force system in plane,Point O is the
intersect point at which axis z intersects the mass
symmetry Plane.Reducing to point O we get
.iii amQ ??
the principle vector as and
the principle moment as CQ aMR ??
2,Fixed-axis rotation of a rigid body,
Firstly,we discuss the simple situation in which
the system has a mass symmetry plane being
perpendicular to the axis of rotation,
For a line i while passes point Mi we have
O
)()( ?? ?? niOiOQO QmQmM ?
,0 2 ??? Oiiiii Irmrmr ???????? ??
(The negative sign expresses that it is opposite to the direction of
?.)
32
空间惯性力系 — >平面惯性力系（质量对称面）
O为转轴 z与质量对称平面的交点，向 O点简化,
iii amQ ??
主矢,
主矩,
CQ aMR ??
)(
0
)()(
2
反向负号表示与 ?
??
?
?
Oii
iii
n
iOiOQO
Irm
rmr
QmQmM
????
????
??
?
?
??
二、定轴转动刚体
先讨论具有垂直于转轴的质量对称平面
的简单情况。
O 直线 i, 平动,过 M
i点,
33
Reducing to point O,
,CQ aMR ??
.?OQO IM ??
Reducing to the center of mass C,
,CQ aMR ??
.???? CQC IM
They are acting at point C,
They are acting at point O,
34
向 O点简化,
CQ aMR ??
?OQO IM ??
向质点 C点简化,
CQ aMR ??
???? CQC IM
作用在 C点
作用在 O点
35
Discussion
① The rigid body is performing an uniform rotation,the axis of
rotation does not pass through the center of mass C,2?meR Q ?
36
讨论,
① 刚体作匀速转动，转轴不通过质点 C 。 2?meR
Q ?
37
② The axis of rotation passes the center of mass C,but ??0,the
inertial force couple is (the direction is opposite to ?),?
CQ IM ??
38
讨论,
② 转轴过质点 C，但 ??0，惯性力偶 （与 ?反向） ?
CQ IM ??
39
③ The rigid body is performing an uniform rotation,moreover,the
axis of rotation passes the center of mass,Then we have
(the principle vector and the principle moment are both zero),0,0 ?? QCQ MR
40
讨论,
③ 刚体作匀速转动，且转轴过质心，则 0,0 ??
QCQ MR
（主矢、主矩均为零）
41
Suppose that the rigid body has a mass symmetry plane,moreover,
it is moving in this plane,Here,the inertial force system of the rigid
body can be simplified to a coplanar force system in the mass
symmetry plane,
The movement of the rigid body in a plane
can be decomposed into two parts,
One is the translation along the base point (the
center of mass C),,
The other is the rotation around the axis which
passes the center of mass,,
So are acting on the center of
mass,
.CQ aMR ??
?CQC IM ??
CQ aMR ??
?CQC IM ??
3,Planar motion of a rigid body,
42
假设刚体具有质量对称平面，并且平行于该平面作平面运
动。此时，刚体的惯性力系可先简化为对称平面内的平面力系。
刚体平面运动可分解为
随基点（质点 C）的平动,
绕通过质心轴的转动,
? 作用于质心
CQ aMR ??
?CQC IM ??
CQ aMR ??
?CQC IM ??
三、刚体作平面运动
43
44
45
For a rigid body moving in a plane,three equations can be written
down by the dynamic-static method as follows,
.0)(,0)(
,0,0
,0,0
)(
)(
)(
???
???
???
??
??
??
QC
e
CC
Qy
e
Qx
e
MFmFm
RYY
RXX
In fact,
,)(,,)(2
2
)(
2
2
)(
2
2 ???
??? eCCeCeC FmdtdIYdt ydMXdt xdM ?
46
对于平面运动刚体：由动静法可列出如下三个方程,
0)(,0)(
0,0
0,0
)(
)(
)(
???
???
???
??
??
??
QC
e
CC
Qy
e
Qx
e
MFmFm
RYY
RXX
实质上,
)(,,)(22)(22)(22 ??? ??? eCCeCeC FmdtdIYdt ydMXdt xdM ?
47
[Example 1] The length of an isotropic rod is l,its weight is m,It is
joined the horizontal plane and it drops from the position where the
angle between the rod and the plane is ?0,Determine the angular
acceleration of the rod AB and the reaction force at the support A at
the beginning of the drop,
Investigate the rod AB,add the virtual
inertial force system as follows,
,2?? mlR Q ?
.3,0
2 ?
? mlIMmaR AQAnnQ ????
Solution
According to the dynamic-static method we obtain
48
[例 1] 均质杆长 l,质量 m,与水平面铰接,杆由与平面成 ?0角位
置静止落下。求开始落下时杆 AB的角加速度及 A点支座反力。
选杆 AB为研究对象
虚加惯性力系,
2
?? mlR
Q ?
3,0
2 ?? mlIMmaR
AQAn
n
Q ????
解,
根据动静法，有
49
( 3 ),02/c o s,0)(
( 2 ),0s i n,0
( 1 ),0c o s,0
0
0
0
????
????
????
?
?
?
QAA
n
Q
n
An
QA
MlmgFm
RmgRF
RmgRF
?
?
?
??
?
.c o s
4
g e t ( 1 ) w ee q u a t io n in t o ngS u b s t it u t i
,c o s
2
3
h a v e w e)3(e q u a t io n b y
,s i n h a v e w e)2(e q u a t io n B y
0
0
0
?
??
?
? mg
R
l
g
mgR
A
n
A
??
?
?
50
( 3 ) 02/c o s,0)(
( 2 ) 0s in,0
( 1 ) 0c o s,0
0
0
0
????
????
????
?
?
?
QAA
n
Q
n
An
QA
MlmgFm
RmgRF
RmgRF
?
?
?
??
?
。得代入
得由
得由
c o s
4
,( 1 ); c o s
2
3
,)3(; s in,)2(
0
0
0
?
??
?
? mg
R
l
g
mgR
A
n
A
??
?
?
51
.c o s
2
3
3
1
c o s
2
2
?
?
?
l
g
ml
l
mg
??
0 a n d c o s23g,,0A t 00 ???? ????? lt
This problem can be solved by the insular momentum theorem of
moment of momentum and the motion theorem of the motion center
of mass,too,
2c o s lmgI A ?? ??
Solution,
Study the rod AB,From we get
By the motion theorem of the
motion al the center of mass,
.s in0
,c o s
4
3
2
,c o s
0
00
n
An
A
Rmgma
g
ε
l
amgRma
???
????
?
?? ???
.c o s4,s i n T h e r e f o r e 00 ?? ? mgRmgR AnA ???
52
?
?
? c o s23
3
1
c o s2
2 l
g
ml
lmg
??
0,co s23g,,???? ????? 此时时 000 lt
用动量矩定理 +质心运动定理再求解此题,
解,选 AB为研究对象
2c o s lmgI A ?? ??
由 得,
由质心运动定理,
n
An
A
Rmgma
gεlamgRma
???
????
0
00
s in0
c o s
4
3
2
c o s
?
?? ???
00 co s4,s i n ??
? mgRmgR
A
n
A ????
53
[Example 2] The mass of the positive wheel of a tractor is m,the
radius is R,The wheel is rolling along a horizontal railway,The
force acting on this wheel can be decomposed into two forces
acting on the center of mass and a driving force couple M,An
axis passes the center of mass C and is perpendicular to the wheel,
The radius of gyration of the wheel with respect to this axis is ?,
the coefficient of friction between the wheel and railway is f,
Determine the condition under which the wheel is rolling but not
sliding,the maximum value of the driving force couple M,
TS,
Investigate the wheel,adding the
virtual inertial force system,
.
,
2 ???
?
mIM
mRmaR
CQC
CQ
??
??
Solution,
According to the dynamic-static
method we get
O
54
[例 2] 牵引车的主动轮质量为 m，半径为 R，沿水平直线轨道
滚动，设车轮所受的主动力可简化为作用于质心的两个力
及驱动力偶矩 M，车轮对于通过质心 C并垂直于轮盘的轴的回
转半径为 ?，轮与轨道间摩擦系数为 f,试求在车轮滚动而不滑
动的条件下，驱动力偶矩 M 之最大值。
TS,
取轮为研究对象
虚加惯性力系,
???
?
2mIM
mRmaR
CQC
CQ
??
??
解,
由动静法，得,
O
55
( 3 ), 0,0)(
( 2 ),0,0
( 1 ),0,0
?????
????
????
?
?
?
QCC
Q
MFRMFm
SPNY
RTFX
By equation (1) we have, TFmRR
Q ??? ?
g e t ( 3 ) w ee q u a t io n i n t oo n s u b s t it u t i,B y H e u c e,mR TF ???
.2 mR TFmFRMFRM QC ????? ?
From equation (2) we have N= P +S,In order to
keep the wheel not sliding F<f N =f (P+S) (5)
must be satisfied,
.))((
22
RTRRSPfM
?? ????
It is obvious that
the larger f is,it
is more difficult
for the wheel to
slide,Mmax is the
value of the
right- hand side
of the east
equation,
Substituting equation (5) into (4)
we get
O
( 4 ),)()( T h e r e f o r e
222
RTRRFTFRFRM
??? ??????
56
( 3 ) 0,0)(
( 2 ) 0,0
( 1 ) 0,0
?????
????
????
?
?
?
QCC
Q
MFRMFm
SPNY
RTFX
由 (1)得 TFmRR
Q ??? ?
得代入所以 ( 3 ) mR TF ???
( 4 ) )()(
222
2
R
TR
R
FTF
R
FRM
mR
TFmFRMFRM
QC
???
?
???????
?????
由 (2)得 N= P +S，要保证车轮不滑动,
必须 F<f N =f (P+S) (5)
RTRRSPfM
22 ))(( ?? ????
可见,f 越
大越不易滑动。
Mmax的值
为上式右端的
值。 把 (5)代入 (4)得,
O
57
§ 15-4 Dynamical reaction of the bearing
of a rigid body under fixed-axis rotation
and the concepts of static and dynamic equilibrium
'R
'QR
OM
QOM
,)()()(
)(,
???
??
???
???
??????
kQmjQmiQm
kMjMiM
QmQrMaMR
iziyix
QzQyQx
iOiiQOCQ
1,Dynamical reaction of the bearing of a rigid body
The angle velocity of the rigid body is ?,the
angular acceleration is ? (counterclockwise)
Simplifying the positive force system to point O,
we get the principle vector and the
principle torque,
Simplifying the inertial force system to point O,
we get the principle vector and the
principle torque,
58
§ 15-4 定轴转动刚体的轴承动反力 ?
静平衡与动平衡的概念
一、刚体的轴承动反力
刚体的角速度 ?,角加速度 ?（ 逆时针）
主动力系向 O点简化, 主矢,主矩
惯性力系向 O点简化, 主矢,主矩
'R
'QR
OM
QOM
)()()(
)(
???
??
???
???
???
???
kQmjQmiQm
kMjMiM
QmQrM
aMR
iziyix
QzQyQx
iOiiQO
CQ
59
.c o ss i n
c o ss i n
)()()(
2 ??
??
? ??
???
???
???
iiiiiiii
iiiii
n
iii
ix
n
ixixQx
RzmRzm
amzamz
QmQmQmM
????
??
?
?
.)(
,g e t r e a s o n w e s a m e B y t h e
2
2
??
??
??
ziiiiiizQz
yzzxQy
IRmRamQmM
IIM
???????
??
???
.)()( H e u c e,
./co /s inB u t
2 ?? ??
??
iiiiiiQx
iiiiii
zymxzmM
RxsRy
??
??
2 c a u w r i le we
,
,w r t u
?? yzzxQx
iiiyz
iiizx
IIM
zymI
xzmI
??
?
?
?
?
60
??
??
? ??
???
???
???
iiiiiiii
iiiii
n
iii
ix
n
ixixQx
RzmRzm
amzamz
QmQmQmM
????
?? ?
?
c o ss in
c o ss in
)()()(
2
??
??
??
ziiiiiizQz
yzzxQy
IRmRamQmM
IIM
???????
??
??? 2
2
)(
同理可得
?? ??
??
)()(
/co /s in
2
iiiiiiQx
iiiiii
zymxzmM
RxsRy
??
?? 故而
2
,
?? yzzxQx
iiiyziiizx
IIM
zymIxzmI
???
?? ? ? 惯性积令
61
According to the dynamic-static method
,0
,0
,0
,0 '
,0 '
,0 '
'
'
??
??????
??????
??
????
????
Qzz
BAQyy
ABQxx
zB
Qy
yBA
Qx
xBA
MM
OBXOAXMM
OAYOBYMM
RZ
RRYY
RRXX
Five of the above equations are connected to with the reaction
forces of constraints,Supposing AB=l,OA=l and OB=l2 we obtain
62
根据动静法,
,0
,0
,0
,0 '
,0 '
,0 '
'
'
??
??????
??????
??
????
????
Qzz
BAQyy
ABQxx
zB
Qy
yBA
Qx
xBA
MM
OBXOAXMM
OAYOBYMM
RZ
RRYY
RRXX
其中有五个式子与约束反力有关。设 AB=l,OA=l1,OB=l2 可得
63
'.
,/)]'()'[(
,/)]'()'[(
,/)]'()'[(
,/)]'()'[(
11
11
22
22
xB
QxQyxyB
QyQxyxB
QyQxyxA
QxQyxyA
RZ
llRMlRMX
llRMlRMY
llRMlRMY
llRMlRMX
??
??????
???????
??????
???????
They consist of two parts,one is the result of the positive forces,It
cannot be avoided and is called the static reaction force,The other
part is the result of an unbalance of the inertial force system.It is
called the additional dynamic reaction force and can be avoided
by adjustment,
If you want to make the
additional dynamic reaction
force to be zero you have to
proceed as follows,
Static reaction force
Additional dynamic
reaction force
Dynamic
reaction
force
64
'
/)]'()'[(
/)]'()'[(
/)]'()'[(
/)]'()'[(
11
11
22
22
xB
QxQyxyB
QyQxyxB
QyQxyxA
QxQyxyA
RZ
llRMlRMX
llRMlRMY
llRMlRMY
llRMlRMX
??
??????
???????
??????
???????
由两部分组成，一部分由主动力引起的，不能消除，称为
静反力 ；一部分是由于惯性力系的不平衡引起的，称为 附加动
反力,它可以通过调整加以消除。
使附加动反力为零，须有 静反力 附加动反力 动反力
65
If the axis of rotation of the rigid body is the principle axis of
inertia,the additional dynamic reaction force of the bearing is
zero,
0?? QyQx MM 0'' ?? QyQx RR
0
0
2
2
??
??
??
??
yzzx
yzzx
II
II
)0(
0
42
2
2
???
?
???
??
??
??
yzzxxz III
0
0
?
?
Cy
Cx
Ma
Ma
0?? CC yx
The product of inertia about axis z
is zero,Axis z is principal axis of
inertia of the rigid body at point O,
Pass the center of mass
xzI
66
0?? QyQx MM 0'' ?? QyQx RR
当刚体转轴为中心惯性主轴时，轴承的附加动反力为零。
0
0
2
2
??
??
??
??
yzzx
yzzx
II
II
)0(
0
42
2
2
???
?
???
??
??
??
yzzxxz III
0
0
?
?
Cy
Cx
Ma
Ma
0?? CC yx
对 z 轴惯性积为零,z 轴为
刚体在 O点的惯性主轴； 过质心
67
Static equilibrium,if the axis of rotation of the rigid body passes
the center of mass,and if the rigid body is subjected only to gravity
and to no others positive force,the system is in equilibrium
independent on its position,
Dynamic equilibrium,if the axis of rotation of the rigid body is
the principle axis of inertia,the additional dynamic reaction force
of the bearing is zero,
2,The concepts of static equilibrium and of dynamic equilibrium
68
静平衡,刚体转轴过质心，则刚体在仅受重力而不受其它
主动力时，不论位置如何，总能平衡。
动平衡,转动为中心惯性主轴时，转动时不产生附加动反
力。
二、静平衡与动平衡的概念
69
[Example 1] A rigid bearing without mass is rotating with the
uniform angular velocity ?,two spheres A and B which have the same
mass m are connected to the bearing,Determine in which of the
following cases we have static equilibrium respectively dynamic
equilibrium,
( a) is in dynamic equilibrium,but (b) and (d) are in static equilibrium
70
[例 1] 质量不计的刚轴以角速度 ?匀速转动，其上固结着两个
质量均为 m的小球 A和 B。指出在图示各种情况下，哪些是静平
衡的？哪些是动平衡的？
静平衡,(b),(d) 动平衡,( a)
71
If a rigid body is in dynamic equilibrium,it is in static
equilibrium,too,But if a rigid body is in static equilibrium,it
may not be in dynamic equilibrium,
Grr
g
G
mrGrrRMb
GrmrGrMa
QQ
Q
?????
???
2
2
2
2
2
1
2
1
2
1
,0:)( c a s eI n
2
1
,0,)( c a s eI n
??
?
21
21,
??
??
?
?
[Example 2] Two identical crown blocks,as shown in the diagram
below,start to rotate from rest,Determine whose angular acceleration
is larger,
(a) A force G acting on the rope (b) A object whose weight is G is hanging on the rope,
O
O
72
动平衡的刚体，一定是静平衡的；反过来，静平衡的刚体，
不一定是动平衡的。
Grr
g
GmrGrrRMb
GrmrGrMa
QQ
Q
?????
???
2
2
2
2
2
1
2
1
2
1,0,)(
2
1,0,)(
??
?
对
对
21
21,
??
??
?
?
[例 2] 两个相同的定滑轮如下图示，开始时都处于静止，问哪
个角速度大？
(a) 绳子上加力 G (b) 绳子上挂一重 G的物体
O
O
73
According to D’Alembert’s principle,the method of transforming
the dynamic equations into the form of equilibrium equations of
statics is called the dynamic-static method,Not only the motion,
the acceleration and angular acceleration,but the forces can be
determined as well,This method is often used to determine the
dynamic reaction forces of constraints of the system of particles
when the motion is known,
Using the dynamic-static method you can use all the equilibrium
equations of statics,For example,the center of moment can be
selected arbitrarily and the equations of two moments or the
equations of three moments are both can be used,Therefore it is
more convenient to use the dynamic-static method in order to
solve the problems with several reaction forces of constraints,
The application of D’Alembert’s principle
74
根据达朗伯原理，以静力学平衡方程的形式来建立动力学
方程的方法，称为动静法。应用动静法既可求运动，例如加
速度、角加速度；也可以求力，并且多用于已知运动，求质
点系运动时的动约束反力。
应用动静法可以利用静力学建立平衡方程的一切形式上
的便利。例如，矩心可以任意选取，二矩式，三矩式等等。
因此当问题中有多个约束反力时，应用动静法求解它们时就
方便得多。
达朗伯原理的应用
75
① Select an object to study,The principles are the same as in
statics,
② Analyze the forces,Draw all the positive forces and the external
reaction forces of constraints,
③ Analyze the motion,Determine the acceleration of the angular
of mass,the angle acceleration of the rigid body and their directions,
④ Add the virtual inertial forces,Draw all inertial forces and all
inertial force couples into a force diagrams,This step is essential for
the correct analysis of motion,The result of the simplification of the
inertial force system of the rigid body should be remembered by
heart,
The steps and key points of the application of the dynamic-static
method to solve dynamical problems are
76
① 选取研究对象 。原则与静力学相同。
②受力分析。 画出全部主动力和外约束反力。
③运动分析。 主要是刚体质心加速度，刚体角加速度，标出
方向。
应用动静法求动力学问题的步骤及要点,
④ 虚加惯性力。 在受力图上画上惯性力和惯性力偶，一定要
在 正确进行运动分析的基础上。熟记刚体惯
性力系的简化结果。
77
⑤ Write down the dynamic-static equation,Select the center of
reduction and the projection axes properly,
⑥ Write down the supplementary equations,They are the
supplementary equations of kinetics (they show the relationships
among the quantities describing the motion),
⑦ Solve the equations to determine the unknown quantities,
[Notice] Because the direction of have been drawn into the
diagram,nothing has to be done but substituting these quantities into
the equations when you write down the
equations,
QOQ MR a n d
?OQOCQ IMmaR ?? a n d
78
⑤ 列动静方程。 选取适当的矩心和投影轴。
⑥建立补充方程。 运动学补充方程（运动量之间的关系）。
⑦求解求知量。
[注 ] 的方向及转向已在受力图中标出，建立方程时，
只需按 代入即可。
QOQ MR,
?OQOCQ IMmaR ??,
79
[Example 1] The masse of two weight,are m1and m2,respective,
They are hanging on two ropes which are wrapped on two coaxial
tub wheels of radius r1and r2,The moment of inertia of the two
wheels with respect to the axis of rotation O is I,the system moves
under the action of gravity,Determine the angular acceleration of the
tub wheels,
Select the system as the object to be
investigated,
Solution,
Method 1,Solve it by D'Alembert's principle
80
[例 1] 质量为 m1和 m2的两重物，分别挂在两条绳子上，绳又分
别绕在半径为 r1和 r2并装在同一轴的两鼓轮上，已知两鼓轮对于
转轴 O的转动惯量为 I，系统在重力作用下发生运动，求鼓轮的
角加速度。
取系统为研究对象
解,方法 1 用达朗伯原理求解
81
Add the virtual inertial force and the
inertial force couple,
?? IIMamRamR OQOQQ ????,,222111
According to the dynamic-static method we
get
.0
,0
,0)(
2221112211
22112211
?????
?????
??
?Iramramgrmgrm
MrRrRgrmgrm
Fm
QOQQ
O
Writing down the supplementary equations and
substituting them into the equation above you get ?? 2211,rara ??
g
Irmrm
rmrm
??
??
2
22
2
11
2211?
82
虚加惯性力和惯性力偶,
?? IIMamRamR OQOQQ ????,,222111
由动静法,
0
0
,0)(
2221112211
22112211
?????
?????
??
?Iramramgrmgrm
MrRrRgrmgrm
Fm
QOQQ
O
列补充方程,代入
上式
得,
?? 2211,rara ??
g
Irmrm
rmrm
??
??
2
22
2
11
2211?
83
Method 2,Solve the problem by the angular momentum
theorem,
.
,)(
2211
)(
2
22
2
11
222111
grmgrmM
Irmrm
IrvmrvmL
e
O
O
??
???
???
?
?
,T h e r e f o r e 2
22
2
11
2211 g
Irmrm
rmrm
??
???
According to the angular momentum
theorem
.])[( 2211222211 grmgrmIrmrmdtd ???? ?
Select the system as the object to be investigated,
84
方法 2 用动量矩定理求解
2211
)(
2
22
2
11
222111
)(
grmgrmM
Irmrm
IrvmrvmL
e
O
O
??
???
???
?
?
g
Irmrm
rmrm
??
???
2
22
2
11
2211 ?
根据动量矩定理,
2211222211 ])[( grmgrmIrmrmdtd ???? ?
取系统为研究对象
85
,2
22
2
11
2211 g
Irmrm
rmrm
??
???
)(
2
2
1
2
1
2
1
2
22
2
11
2
22
22
2
11
Irmrm
IvmvmT
???
???
?
?
??? gdrmrmIrmrmdWdT F )()](2[ g e t w e F r o m 2211222211
2
????? ?
Select the system as the object to be studied,
The kinetic energy of the system at any
instant is
,
is w o r k e l e m e n t a r y T h e
2211
2211
2211
) g dr-mr(m
dgrmdgrm
g d smg d smW F
?
??
?
?
??
???
By dt both sides of the equation and solve for the derivative
you get
Method 3,Solve the problem by the theorem of kinetic energy,
86
g
Irmrm
rmrm
??
??
2
22
2
11
2211 ?
)(
2
2
1
2
1
2
1
2
22
2
11
2
22
22
2
11
Irmrm
IvmvmT
???
???
?
?
?? gdrmrmIrmrmdδ WdT F )()](2[ 22112222112 ????? ? 得由
取系统为研究对象，任一瞬时系统的
) g dr-mr(m
dgrmdgrm
gdsmgdsmW F
?
??
?
2211
2211
2211
?
??
???元功
两边除以 dt，并求导数，得
方法 3 用动能定理求解
87
[Example 2] In the mechanism shown below the cylinders and the
tub wheel O are all homogenous,the weight of the two cylinders are
P and Q,the radii of both are R,They are Performing pure rolling,
The rope (without weight) can not be stretched,the angle of
inclination is ?,A constant torque M is acting on the tub wheel,
Determine (1)the angle acceleration of the tub wheel,(2) the tensile
force of the rope,(3) the reaction forces at point O on the bearing,
(4)the friction force between the cylinder and the inclined plane
(neglecting the rolling friction),
88
[例 2] 在图示机构中，沿斜面向上作纯滚动的圆柱体和鼓轮 O均
为均质物体，各重为 P和 Q，半径均为 R，绳子不可伸长，其质
量不计，斜面倾角 ?，如在鼓轮上作用一常力偶矩 M,试求：
(1)鼓轮的角加速度？ (2)绳子的拉力？ (3)轴承 O处的支反力？
(4)圆柱体与斜面间的摩擦力（不计滚动摩擦）？
89
Solution
Method 1,solution by e D’Alembert’s principle,
Study the wheel O,add the virtual inertial force couple,
OOOQ Rg
QIM ?? 2
2
1??
Write down the dynamic-static equations as
( 3 ), 0 s in0
( 2 ),0c o s0
( 1 ),0,0)(
????
???
????
?
?
?
?
?
TQ,YY
T,XX
MMTRFm
O
O
QO
.21M,2QA AAQ RgPagPR ???
Study the wheel A,add the virtual inertial force and the inertial
force couple MQC as shown in the diagram,QR
90
解,方法 1 用达朗伯原理求解
取轮 O为研究对象，虚加惯性力偶
OOOQ Rg
QIM ?? 2
2
1??
列出动静方程,
( 3 ) 0 s i n0
( 2 ) 0c o s0
( 1 ) 0,0)(
????
???
????
?
?
?
?
?
TQ,YY
T,XX
MMTRFm
O
O
QO
AAQ RgPagPR ?2QA 21M,??
取轮 A为研究对象，虚加惯性力 和惯性力偶 MQC如图示。
QR
91
Writing down the dynamic-static equations we get
( 5 ), 0s i n,0
( 4 ),0's i n,0)(
?????
?????
?
?
?
?
PFRT'X
MRTRRRPFm
Q
QAQC
We know that the kinematical relationships are
.,OAOAA RRa ???? ???
Substitute MQ,RQ,MQA and the equation above into equation (1) and
(4) and then solve them,to obtain
,
)3(
)s in3(
,
)3(
)s in(2
2
RPQ
QRMP
T
g
RPQ
RPM
O
?
?
?
?
?
?
?
?
?
92
列出动静方程,
( 5 ) 0s i n,0
( 4 ) 0's i n,0)(
?????
?????
?
?
?
?
PFRT'X
MRTRRRPFm
Q
QAQC
运动学关系：,
OAOAA RRa ???? ???
将 MQ,RQ,MQA及运动学关系代入到 (1)和 (4)式并联立求解得,
。
)3(
)s in3(
,
)3(
)s in(2
2
RPQ
QRMP
T
g
RPQ
RPM
O
?
?
?
?
?
?
?
?
?
93
Substitution into equations (2),(3) and (5) yields
。
)3(
)s i n(
,s i n
)3(
)s i n3(
,c os
)3(
)s i n3(
RPQ
PRMP
F
Q
RPQ
QRMP
Y
RPQ
QRMP
X
O
O
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
94
代入 (2),(3),(5)式，得,
。
)3(
)s i n(
,s i n
)3(
)s i n3(
,c os
)3(
)s i n3(
RPQ
PRMP
F
Q
RPQ
QRMP
Y
RPQ
QRMP
X
O
O
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
95
(1) Determine the angular acceleration of the
tub wheel by the theorem of kinetic energy,
Study the whole system,
,)s i n(
s i n
??
???
PRM
PRMW F
??
???
.)s i n()3(4 g e t w e F o r m 2
2
12 ??
? PRMCRPQ
gWTT
OF ?????? ?
)( AO R ωR ωv ??
,)3(
4
22
1
2
1
22
1
( c o n s t ),
2
2
22222
2
1
RPQ
g
R
g
P
v
g
P
R
g
Q
T
CT
O
AO ?????????
?
?
??
gRPQ PRMO ???? 2)3( )s in(2 ??
Determine the time-
derivative,OOO PRMRPQg ???? )s in(2)3(4 1 2 ?????
Method 2,solution by the general theorems of dynamics
96
方法 2 用动力学普遍定理求解
(1) 用动能定理求鼓轮角加速度。
取系统为研究对象
??
???
)s i n(
s i n
PRM
PRMW F
??
???
??? )s i n()3(4,2
2
12 PRMCRPQgWTT
OF ?????? ? 得由
)( AO R ωR ωv ??
2
2
22222
2
1
)3(
4
22
1
2
1
22
1
)(
RPQ
g
R
g
Pv
g
PR
g
QT
CT
O
AO ?????????
?
?
??
常量
gRPQ PRMO ???? 2)3( )s in(2 ??
两边对 t求导数,
)s i n(2)3(4 1 2 OOO PRMRPQg ???? ?????
97
(2) Determine the tensile force of the rope by the
angular momentum theorem (differential equation
of a fixed-axis rotation),
Study the wheel O,form the angular momentum
theorem we get
,2 2 TRMRgQ O ???,)3(
)s in3(
RPQ
QRMPT
?
?? ?
(3) Determine the reaction forces on the bearing O by the motion
theorem of motion of the center of mass,
Investigate the wheel O,from the theorem of motion of the center
of mass we get
,s i n0,
,c o s0,
?
?
????
???
?
?
TQYYMa
TXXMa
OCy
OCx
,s i n)3( )s i n3(,c o s)3( )s i n3( QRPQ QRMPYRPQ QRMPX OO ????????? ????
98
(2) 用动量矩定理求绳子拉力
（定轴转动微分方程）
取轮 O为研究对象，由动量矩定理得
TRMRgQ O ???22 RPQ QRMPT )3( )s in3( ??? ?
(3) 用质心运动定理求解轴承 O处支反力
取轮 O为研究对象，根据质心运动定理,
?
?
????
???
?
?
s i n0,
c o s0,
TQYYMa
TXXMa
OCy
OCx
QRPQ QRMPYRPQ QRMPX OO ????????? s i n)3( )s i n3(,c o s)3( )s i n3( ????
99
(4) Determine the friction force by the
differential equation of plane motion,
Investigate cylinder A,from the differential
equation of plane motion we gem
)(,OAAA FRI ??? ??
.)3( )s i n()3( )s i n(221 22 RPQ PRMPgRPQ PRMRgPRRIF AA ???????? ???
Method 3 determine the angular acceleration of the tub wheel
by the theorem of kinetic energy,
We can get the reaction forces of the constraints by
D’Alembert’s principle (the tensile force of the rope,the
reaction forces at the point O on the bearing and and the
friction force ),
T
OX
F
OY
100
(4) 用刚体平面运动微分方程求摩擦力
取圆柱体 A为研究对象,
根据刚体平面运动微分方程
)( OAAA FRI ??? ??
RPQ
PRMPg
RPQ
PRMR
g
P
RR
IF AA
)3(
)s i n(
)3(
)s i n(2
2
1
2
2
?
??
?
???? ???
方法 3：用动能定理求鼓轮的角加速度
用达朗伯原理求约束反力 （绳子拉力,轴承 O处反
力 和 及摩擦力 ）。
T
OX OY
F
101
[Example 3] The weight of a homogeneous cylinder is P,its radius
is R,It is rolling down from point O along the inclined plane
without sliding,At the beginning it does not move,The angle
between the plane and the horizon is ?,At OA=S determine the
reaction force of the constraint of the board at point O,Neglect the
weight of the board,
Solution,
(1) Determine the velocity and the
acceleration by the theorem of kinetic
energy,
The motion of the cylinder is a plane
movement,At the beginning it is
at test,at the end the angle velocity is
? and vC = R ? The kinetic energy is,
P
01 ?T
2T
102
[例 3] 均质圆柱体重为 P，半径为 R，无滑动地沿倾斜平板由静
止自 O点开始滚动。平板对水平线的倾角为 ?,试求 OA=S时平
板在 O点的约束反力。板的重力略去不计。
解, (1) 用动能定理求速度，加速度
圆柱体作平面运动。在初始位置时，
处于静止状态，故 T1=0；在末位置时，
设角速度为 ?，则 vC = R ?,动能为,
P
103
.4322121 22222 CC vgPRgPvgPT ??? ?
The work of the
positive forces is ? ?? ?s i nPSW F
From the theorem of kinetic energy
we get ??? FWTT 12
?? s i n34 s i n043 22 ?????? gSvPSvgP CC
Form the time-derivative we obtain
.s i n32,s i n32 ??? Rgga C ??(2) Determine the reaction force of the constraint by D’Alembert’s
principle,
Study the whole system,add the virtual inertial force and
the inertial force couple MQC, Q
R
P
104
2222
2 4
3
22
1
2
1
CC vg
PR
g
Pv
g
PT ??? ?
主动力的功,? ?? ?s i nPSW F
由动能定理 得 ??? FWTT 12
?? s i n34 s i n043 22 ?????? gSvPSvgP CC
对 t 求导数，则,??? s i n
3
2,s i n
3
2
R
gga
C ??
(2) 用达朗伯原理求约束反力
取系统为研究对象，虚加惯性力 和惯性力偶 MQC
QR
P
105
.s in
3
s in
3
2
2
1
,s in
3
2
2 ??
?
PR
R
g
R
g
P
M
P
a
g
P
R
QC
CQ
??
??
,0s inc o ss in
3
2
s in
3
,0)(
0s ins in
3
2
,0
0c o ss in
3
2
,0
????????
?????
????
?
?
?
RPPSR
P
R
P
MFm
,αα
P
P YY
,
P
XX
OO
O
O
????
??Write down the dynamic-static equations as
.c o s SP M O ?? ?
,2s in3 ?P X O ?
),s in321 2 ??? P( Y O
106
??
?
s i n
3
s i n
3
2
2
1
,s i n
3
2
2 PR
R
g
R
g
PM
Pa
g
PR
QC
CQ
??
??
0s inc o ss in
3
2
s in
3
,0)(
0s ins in
3
2
,0
0c o ss in
3
2
,0
????????
?????
????
?
?
?
RPPSR
P
R
P
MFm
,αα
P
P YY
,
P
XX
OO
O
O
????
??
列出动静方程,
SP M O ?? ?c o s
?2s in3P X O ?
)s in321 2 ??? P( Y O
107
[Example 4] The weight of a bobbin winder pulley is P,the radii
are R and r,the moment of inertia about the center of mass O is IO,
Under the action of a constant force T which makes an angle ?
with the horizon it is purely rolling,The rolling resistance can be
neglected,Determine (1)the acceleration of the center of the wheel,
(2)the conditions under which the wheel is purely rolling,
Solution,
Solution by D'Alembert's principle,The bobbin
winder pulley is Performing plane motion (pure
rolling)
) (,?RaaRIMagPR OOOQOOQ ???By D’Alembert’s principle we have
.0c o s,0)( ?????? RTTrRRMFm QQOC ?
Substituting RQ and MQO into the equation
above we obtain,)c o s(
2R
g
PI
rRTRa
O
O
?
?? ?
108
[例 4] 绕线轮重 P，半径为 R及 r,对质心 O转动惯量为 IO，在
与水平成 ? 角的常力 T 作用下纯滚动，不计滚阻，求,(1)轮心
的加速度； (2)分析纯滚动的条件。
解,用达朗伯原理求解
绕线轮作平面运动 （纯滚动）
) (,?RaaRIMagPR OOOQOOQ ???
由达朗伯原理，得
0c o s,0)( ?????? RTTrRRMFm QQOC ?
将 RQ, MQO代入上式，可得 2
)c os(
RgPI
rRTRa
O
O
?
?? ?
109
,0c o s,0 ????? QRFTX ?
,
)c o s(
)c o s(
c o s
c o s
22
R
g
P
I
Rr
g
P
IT
R
g
P
I
rRTR
g
P
T
RTF
O
O
O
Q
?
?
?
?
?
???
??
?
?
?
?
,s in
,0s in,0
?
?
TPN
TPNY
??
?????
The condition of pure rolling
is that F ≤f N,
),s i n(
)c o s(
2
?
?
TPf
R
g
P
I
Rr
g
P
IT
O
O
??
?
?
.
))(s in(
)c os(
2R
g
P
ITP
Rr
g
P
IT
f
O
O
??
?
?
?
?
110
0c o s,0 ????? QRFTX ?
22
)c o s(
)c o s(
c o s
c o s
R
g
P
I
Rr
g
P
IT
R
g
P
I
rRTR
g
P
T
RTF
O
O
O
Q
?
?
?
?
?
???
??
?
?
?
?
?
?
s in
0s in,0
TPN
TPNY
??
?????
纯滚动的条件,F ≤f N
)s in(
)c o s(
2
?
?
TPf
R
g
PI
Rr
g
PIT
O
O
??
?
?
))(s in(
)c o s(
2R
g
PITP
Rr
g
PIT
f
O
O
??
?
?
?
?
111
1,The system consists of two blocks A and
B,their weights both are m,they are placed
on a smooth horizontal plane,A horizontal
force F is acting on A,Please explain
whether the force with which block A acts
on B is equal to F,
Questions to think about,
Solution,
.
,
,0
' FNN
FmaFN
NRF Q
??
???
???
112
1,物体系统由质量均为 m的两物块
A和 B组成，放在光滑水平面上,
物体 A上作用一水平力 F，试用动静
法说明 A物体对 B物体作用力大小是
否等于 F？
思考题,
解,
FNN
FmaFN
NRF Q
??
???
???
'
0
113
?c o s2 212221 aaaamR Q ??? ??? c o ss intg 21 21 aa a?? ?
Solution
2,A triangular prism A,whose weight is M,is moving towards right
with acceleration,a sliding block B,whose weight is m,is sliding
down the inclined plane of A with the acceleration, Determine the
magnitude and direction of the inertial force of block B,
1a
2a
114
?c o s2 212221 aaaamR Q ??? ??? c o ss intg 21 21 aa a?? ?
解,
2,质量为 M的三棱柱体 A 以加速度 向右移动，质量为 m的滑
块 B以加速度 相对三棱柱体的斜面滑动，试问滑块 B的惯性
力的大小和方向如何？
1a
2a
115
3,The weight of a homogeneous wheel is P,its radius is r,It is
purely rolling on the horizontal plane,At a given time the angle
velocity is ?,the angle acceleration is ?,Determine the moment of
inertia of the wheel with respect to the center of mass C,the
momentum and the kinetic energy of the wheel,the angular
momentum with respect to C,the principle vector and the principle
moment of the inertial force system with respect to the center of
mass,
Solution,
???
????
?
g
rP
IMr
g
P
a
g
P
R
g
rP
IL
g
rP
IT
r
g
P
v
g
P
Kr
g
P
I
CQCCQ
CCC
CC
2
,
2
42
1
)(
2
2
2
2
2
2
2
?
????
?
??
?
??
????
116
3,匀质轮重为 P，半径为 r,在水平面上作纯滚动。某瞬时角
速度 ?,角加速度为 ?,求轮对质心 C 的转动惯量，轮的动量、
动能，对质心的动量矩，向质心简化的惯性力系主矢与主矩。
解,
???
??
??
?
g
rP
IMr
g
P
a
g
P
R
g
rP
IL
g
rP
IT
r
g
P
v
g
P
K
r
g
P
I
CQCCQ
CC
C
C
C
2
,
2
42
1
)(
2
2
2
2
2
2
2
?
????
?
??
?
??
???
?
117
118
3
An important theorem of dynamics,D’Alembert’s principle,will be
introduced in this chapter,By this principle dynamical problems can
be transformed formally formally into those of statics,Then they can
be solved by the theorem of equilibrium,Therefore this method to
solve the dynamical problems of dynamics is called the dynamic-
static method,
4
本章介绍动力学的一个重要原理 —— 达朗伯原理 。应用
这一原理，就将动力学问题从形式上转化为静力学问题，从
而根据关于平衡的理论来求解。这种解答动力学问题的方法，
因而也称 动静法 。
§ 15–1 The concept of the inertial force and the
D’Alembert’s principle for one particle
§ 15–2 D’Alembert’s principle for a system of
particles
§ 15–3 The simplification of a system of
inertial forces of a rigid body
§ 15–4 Dynamical reaction of the bearing of a
rigid body under fixed-axis
rotation and the concepts of static
and dynamic equilibrium
The application of D’Alembert’s principle
Chapter 15,D'Alembert's principle
§ 15–1 惯性力的概念 · 质点的达朗伯原理
§ 15–2 质点系的达朗伯原理
§ 15–3 刚体惯性力系的简化
§ 15–4 定轴转动刚体的轴承动反力 ?
静平衡与动平衡的概念
达朗伯原理的应用
第十五章 达朗伯原理
7
§ 15-1 The concept of the inertial force
and D’Alembert’s principle for a particle
If a man is pushing the cart by hands,then
amFQ ????The force comes from the inertia of the cart,
keep the force diagram as the original motion
state,the reaction force to the object (hands)
applying the force is called the inertial force of
the cart,
Q
Concept of the inertial force of a particle,
If a particle is moving accelerated,the sum of the inertial
reactions of the particle to the objects which force it to
produce the accelerated motion is called the inertial force of
the particle,
amQ ??
1,The concept of the inertial force
8
§ 15-1 惯性力的概念 · 质点的达朗伯原理
人用手推车 amFF ????'
力 是由于小车具有惯性，力图保持原来
的运动状态，对于施力物体 (人手 )产生的
反抗力。称为小车的 惯性力 。
'F
定义：质点惯性力
加速运动的质点，对迫使其产生加速运动的物体的惯
性反抗的总和。
amQ ??
一,惯性力的概念
9
。
2
2
2
2
2
2
,
,
dt
zd
mmaQ
dt
yd
mmaQ
dt
xd
mmaQ
zz
yy
xx
????
????
????
。0
,
,
2
2
2
???
????
????
bb
nn
maQ
v
mmaQ
dt
sd
mmaQ
?
??
[notice]
The inertial force is not the real force acting on the particle,it
is the resultant force of reaction of the particle to the object
applying the force,
10
2
2
2
2
2
2
dt
zd
mmaQ
dt
yd
mmaQ
dt
xd
mmaQ
zz
yy
xx
????
????
????
0
2
2
2
???
????
????
bb
nn
maQ
v
mmaQ
dt
sd
mmaQ
?
??
[注 ] 质点惯性力不是作用在质点上的真实力，它是质点对施
力体反作用力的合力。
11
The mass of an unfree particle M is m,It is
subjected to a positive force,The reaction force
is, The resultant force is,
F
N amNFR ???
0??? amNF
0??? QNF
This is D’Alembert’s
principle for a
particle,
2,D’Alembert’s principle for a particle
12
非自由质点 M，质量 m，受主动力,
约束反力,合力
F
N amNFR ???
0??? amNF
0??? QNF
质点的达朗伯原理
二、质点的达朗伯原理
13
This equation is an equilibrium form of a dynamical problem,it
does not change the essential of the problems,The most
outstanding advantage of the dynamic-static method is that the
methods of statics can be applied to solve dynamical problems,
14
该方程对动力学问题来说只是形式上的平衡，并没有
改变动力学问题的实质。采用动静法解决动力学问题的最
大优点，可以利用静力学提供的解题方法，给动力学问题
一种统一的解题格式。
15
[Example 1] A train is running along a horizontal railway,and a
single pendulum is hanging in the carriage,When the carriage
moves to the right with an uniform acceleration,the single
pendulum will turn to the left by an angle ?,and does not move
relative to the carriage,Determine the acceleration of the
carriage, a
16
[例 1] 列车在水平轨道上行驶，车厢内悬挂一单摆，当车厢向
右作匀加速运动时，单摆左偏角度 ?,相对于车厢静止。求车
厢的加速度 。 a
17
Investigate the single pendulum,
add the virtual inertial forces,
.) ( maQamQ ???
.0c o ss i n,0 ????? ?? QmgX
.tg ??? ga
The angle ? changes with the acceleration,when does not
change,the angle ? does not change,too,If we know the angle ?
the acceleration of the train can be calculated, This is the
theory of a pendulum accelerometer,
a a
a
Solution
According to the dynamic-static method
we have
Solving it we get
18
选单摆的摆锤为研究对象
虚加惯性力
) ( maQamQ ???
0co ss i n,0 ????? ?? QmgX
?tg?? ga
?角随着加速度 的变化而变化，当 不变时,? 角也
不变。只要测出 ? 角，就能知道列车的加速度 。摆式加速
计的原理。
a a
a
解,
由动静法,有
解得
19
§ 15-2 D’Alembert’s principle for a system of particles
A system of particles consists of n particles,For every particle
we have
),1,2,.,,,,, ( 0 niQNF iii ????
For the whole system,the positive force system,the system of the
reaction forces of constraints and the inertial force system form
together of an equivalent force system,This is D’Alembert’s
principle for a system of particles,It can be expressed by as
follows,
.0)()()(
,0
???
???
???
???
iOiOiO
iii
QmNmFm
QNF
Please,note that, If we classify the forces by
internal and external forces then we have ?? ?? 0)(,0
)()( iiOii FmF
。? ?
? ?
??
??
0)()(
,0
)(
)(
iO
e
iO
i
e
i
QmFm
QF
20
§ 15-2 质点系的达朗伯原理
对整个质点系，主动力系、约束反力系、惯性力系形式上
构成平衡力系。这就是 质点系的达朗伯原理 。可用方程表示为,
0)()()(
0
???
???
???
???
iOiOiO
iii
QmNmFm
QNF
设有一质点系由 n个质点组成，对每一个质点，有
),1,2,.,,,,, ( 0 niQNF iii ????
注意到,将质点系受力按内力、外力
划分,则 ?? ?? 0)(,0
)()( iiOii FmF
? ?
? ?
??
??
0)()(
0
)(
)(
iO
e
iO
i
e
i
QmFm
QF
21
The equations show that,for the whole system,the equilibrium
equations given by the dynamic-static method just express the
equilibrium between the inertial force system and the external
force system,the internal forces are not involved,
,)( ??? ???????? dt KdvmdtdaMamQ iiCiii
。dtLdvmmdtdammQm OiiOiiOiO ?????? ??? )()()(
22
表明：对整个质点系来说，动静法给出的平衡方程，只
是质点系的惯性力系与其外力的平衡，而与内力无关。
??? ???????? dt KdvmdtdaMamQ iiCiii )(
dt
Ldvmm
dt
dammQm O
iiOiiOiO ?????? ??? )()()(
23
you have for an arbitrary
force system in a plane
。? ?
? ?
? ?
??
??
??
0)()(
,0
,0
)(
)(
)(
iO
e
iO
iy
e
i
ix
e
i
QmFm
QY
QX
and you for an arbitrary
force system in space
。0)()(,0
,0)()(,0
,0)()(,0
)()(
)()(
)()(
????
????
????
??? ?
??? ?
??? ?
iz
e
iziz
e
i
iy
e
iyiy
e
i
ix
e
ixix
e
i
QmFmQZ
QmFmQY
QmFmQX
In practice,we can select the object to e investigate a arbitrarily as in
the case al statics,Write dour the equilibrium equations and then solve
them,
When you solve dynamical problems of dynamics by the dynamic-static
method,
24
对平面任意力系,
? ?
? ?
? ?
??
??
??
0)()(
0
0
)(
)(
)(
iO
e
iO
iy
e
i
ix
e
i
QmFm
QY
QX
对于空间任意力系,
0)()(,0
0)()(,0
0)()(,0
)()(
)()(
)()(
????
????
????
??? ?
??? ?
??? ?
iz
e
iziz
e
i
iy
e
iyiy
e
i
ix
e
ixix
e
i
QmFmQZ
QmFmQY
QmFmQX
实际应用时,同静力学一样任意选取研究对象,列平衡方
程求解。
用动静法求解动力学问题时,
25
§ 15-3 The simplification of a system of inertial forces of a rigid body
The method of simplification is the same as in the case of the
theorem of simplification of a force systems in statics,We should
treat the virtual inertial force system as an inertial force and an
inertial force couple,which are the results of the simplification
of the force system to a point O,
QR
QOM
r e d u c t i o n, ofc e n t e r o n t h e d e p e n d sw h ic h,)(
r e d u c t i o n, ofc e n t e r o n t h e in d e p e n d e d is w h ic h,
?
??
?
?????
QmM
aMamQR
OQO
CQ
No matter what motion the rigid body is doing,the principle vector
of the inertial force system is equal to the product of the weight of
the rigid body and the acceleration of the center of mass,the
direction is opposite to the direction of the acceleration of the
center of mass,
26
§ 15-3 刚体惯性力系的简化
简化方法就是采用静力学中的力系简化的理论。将虚拟的
惯性力系视作力系向任一点 O简化而得到一个惯性力 和一个
惯性力偶 。
QR
QOM
)(
与简化中心有关
与简化中心无关
?
??
?
?????
QmM
aMamQR
OQO
CQ
无论刚体作什么运动，惯性力系主矢都等于刚体质量与质
心加速度的乘积，方向与质心加速度方向相反。
27
1,Transactional motion al a rigid body,
Reducing to the center of mass C we get,
CQ aMR ??
.0)()(? ?? ???????? CiiCiiiCQC armamrQmM
cQ aMR ??
If the rigid body performs a translation,the inertial force
system composes a resultant inertial force passing
through the center of mass,
Look for
the
animation
on the
next page,
28
一、刚体作平动
向质心 C简化,
CQ aMR ??
? ?? ???????? 0)()( CiiCiiiCQC armamrQmM
cQ aMR ??
刚体平动时惯性力系合成为一过质心的合惯性力。
翻
页
请
看
动
画
29
30
31
The inertial force system in space can be simplified
to the inertial force system in plane,Point O is the
intersect point at which axis z intersects the mass
symmetry Plane.Reducing to point O we get
.iii amQ ??
the principle vector as and
the principle moment as CQ aMR ??
2,Fixed-axis rotation of a rigid body,
Firstly,we discuss the simple situation in which
the system has a mass symmetry plane being
perpendicular to the axis of rotation,
For a line i while passes point Mi we have
O
)()( ?? ?? niOiOQO QmQmM ?
,0 2 ??? Oiiiii Irmrmr ???????? ??
(The negative sign expresses that it is opposite to the direction of
?.)
32
空间惯性力系 — >平面惯性力系（质量对称面）
O为转轴 z与质量对称平面的交点，向 O点简化,
iii amQ ??
主矢,
主矩,
CQ aMR ??
)(
0
)()(
2
反向负号表示与 ?
??
?
?
Oii
iii
n
iOiOQO
Irm
rmr
QmQmM
????
????
??
?
?
??
二、定轴转动刚体
先讨论具有垂直于转轴的质量对称平面
的简单情况。
O 直线 i, 平动,过 M
i点,
33
Reducing to point O,
,CQ aMR ??
.?OQO IM ??
Reducing to the center of mass C,
,CQ aMR ??
.???? CQC IM
They are acting at point C,
They are acting at point O,
34
向 O点简化,
CQ aMR ??
?OQO IM ??
向质点 C点简化,
CQ aMR ??
???? CQC IM
作用在 C点
作用在 O点
35
Discussion
① The rigid body is performing an uniform rotation,the axis of
rotation does not pass through the center of mass C,2?meR Q ?
36
讨论,
① 刚体作匀速转动，转轴不通过质点 C 。 2?meR
Q ?
37
② The axis of rotation passes the center of mass C,but ??0,the
inertial force couple is (the direction is opposite to ?),?
CQ IM ??
38
讨论,
② 转轴过质点 C，但 ??0，惯性力偶 （与 ?反向） ?
CQ IM ??
39
③ The rigid body is performing an uniform rotation,moreover,the
axis of rotation passes the center of mass,Then we have
(the principle vector and the principle moment are both zero),0,0 ?? QCQ MR
40
讨论,
③ 刚体作匀速转动，且转轴过质心，则 0,0 ??
QCQ MR
（主矢、主矩均为零）
41
Suppose that the rigid body has a mass symmetry plane,moreover,
it is moving in this plane,Here,the inertial force system of the rigid
body can be simplified to a coplanar force system in the mass
symmetry plane,
The movement of the rigid body in a plane
can be decomposed into two parts,
One is the translation along the base point (the
center of mass C),,
The other is the rotation around the axis which
passes the center of mass,,
So are acting on the center of
mass,
.CQ aMR ??
?CQC IM ??
CQ aMR ??
?CQC IM ??
3,Planar motion of a rigid body,
42
假设刚体具有质量对称平面，并且平行于该平面作平面运
动。此时，刚体的惯性力系可先简化为对称平面内的平面力系。
刚体平面运动可分解为
随基点（质点 C）的平动,
绕通过质心轴的转动,
? 作用于质心
CQ aMR ??
?CQC IM ??
CQ aMR ??
?CQC IM ??
三、刚体作平面运动
43
44
45
For a rigid body moving in a plane,three equations can be written
down by the dynamic-static method as follows,
.0)(,0)(
,0,0
,0,0
)(
)(
)(
???
???
???
??
??
??
QC
e
CC
Qy
e
Qx
e
MFmFm
RYY
RXX
In fact,
,)(,,)(2
2
)(
2
2
)(
2
2 ???
??? eCCeCeC FmdtdIYdt ydMXdt xdM ?
46
对于平面运动刚体：由动静法可列出如下三个方程,
0)(,0)(
0,0
0,0
)(
)(
)(
???
???
???
??
??
??
QC
e
CC
Qy
e
Qx
e
MFmFm
RYY
RXX
实质上,
)(,,)(22)(22)(22 ??? ??? eCCeCeC FmdtdIYdt ydMXdt xdM ?
47
[Example 1] The length of an isotropic rod is l,its weight is m,It is
joined the horizontal plane and it drops from the position where the
angle between the rod and the plane is ?0,Determine the angular
acceleration of the rod AB and the reaction force at the support A at
the beginning of the drop,
Investigate the rod AB,add the virtual
inertial force system as follows,
,2?? mlR Q ?
.3,0
2 ?
? mlIMmaR AQAnnQ ????
Solution
According to the dynamic-static method we obtain
48
[例 1] 均质杆长 l,质量 m,与水平面铰接,杆由与平面成 ?0角位
置静止落下。求开始落下时杆 AB的角加速度及 A点支座反力。
选杆 AB为研究对象
虚加惯性力系,
2
?? mlR
Q ?
3,0
2 ?? mlIMmaR
AQAn
n
Q ????
解,
根据动静法，有
49
( 3 ),02/c o s,0)(
( 2 ),0s i n,0
( 1 ),0c o s,0
0
0
0
????
????
????
?
?
?
QAA
n
Q
n
An
QA
MlmgFm
RmgRF
RmgRF
?
?
?
??
?
.c o s
4
g e t ( 1 ) w ee q u a t io n in t o ngS u b s t it u t i
,c o s
2
3
h a v e w e)3(e q u a t io n b y
,s i n h a v e w e)2(e q u a t io n B y
0
0
0
?
??
?
? mg
R
l
g
mgR
A
n
A
??
?
?
50
( 3 ) 02/c o s,0)(
( 2 ) 0s in,0
( 1 ) 0c o s,0
0
0
0
????
????
????
?
?
?
QAA
n
Q
n
An
QA
MlmgFm
RmgRF
RmgRF
?
?
?
??
?
。得代入
得由
得由
c o s
4
,( 1 ); c o s
2
3
,)3(; s in,)2(
0
0
0
?
??
?
? mg
R
l
g
mgR
A
n
A
??
?
?
51
.c o s
2
3
3
1
c o s
2
2
?
?
?
l
g
ml
l
mg
??
0 a n d c o s23g,,0A t 00 ???? ????? lt
This problem can be solved by the insular momentum theorem of
moment of momentum and the motion theorem of the motion center
of mass,too,
2c o s lmgI A ?? ??
Solution,
Study the rod AB,From we get
By the motion theorem of the
motion al the center of mass,
.s in0
,c o s
4
3
2
,c o s
0
00
n
An
A
Rmgma
g
ε
l
amgRma
???
????
?
?? ???
.c o s4,s i n T h e r e f o r e 00 ?? ? mgRmgR AnA ???
52
?
?
? c o s23
3
1
c o s2
2 l
g
ml
lmg
??
0,co s23g,,???? ????? 此时时 000 lt
用动量矩定理 +质心运动定理再求解此题,
解,选 AB为研究对象
2c o s lmgI A ?? ??
由 得,
由质心运动定理,
n
An
A
Rmgma
gεlamgRma
???
????
0
00
s in0
c o s
4
3
2
c o s
?
?? ???
00 co s4,s i n ??
? mgRmgR
A
n
A ????
53
[Example 2] The mass of the positive wheel of a tractor is m,the
radius is R,The wheel is rolling along a horizontal railway,The
force acting on this wheel can be decomposed into two forces
acting on the center of mass and a driving force couple M,An
axis passes the center of mass C and is perpendicular to the wheel,
The radius of gyration of the wheel with respect to this axis is ?,
the coefficient of friction between the wheel and railway is f,
Determine the condition under which the wheel is rolling but not
sliding,the maximum value of the driving force couple M,
TS,
Investigate the wheel,adding the
virtual inertial force system,
.
,
2 ???
?
mIM
mRmaR
CQC
CQ
??
??
Solution,
According to the dynamic-static
method we get
O
54
[例 2] 牵引车的主动轮质量为 m，半径为 R，沿水平直线轨道
滚动，设车轮所受的主动力可简化为作用于质心的两个力
及驱动力偶矩 M，车轮对于通过质心 C并垂直于轮盘的轴的回
转半径为 ?，轮与轨道间摩擦系数为 f,试求在车轮滚动而不滑
动的条件下，驱动力偶矩 M 之最大值。
TS,
取轮为研究对象
虚加惯性力系,
???
?
2mIM
mRmaR
CQC
CQ
??
??
解,
由动静法，得,
O
55
( 3 ), 0,0)(
( 2 ),0,0
( 1 ),0,0
?????
????
????
?
?
?
QCC
Q
MFRMFm
SPNY
RTFX
By equation (1) we have, TFmRR
Q ??? ?
g e t ( 3 ) w ee q u a t io n i n t oo n s u b s t it u t i,B y H e u c e,mR TF ???
.2 mR TFmFRMFRM QC ????? ?
From equation (2) we have N= P +S,In order to
keep the wheel not sliding F<f N =f (P+S) (5)
must be satisfied,
.))((
22
RTRRSPfM
?? ????
It is obvious that
the larger f is,it
is more difficult
for the wheel to
slide,Mmax is the
value of the
right- hand side
of the east
equation,
Substituting equation (5) into (4)
we get
O
( 4 ),)()( T h e r e f o r e
222
RTRRFTFRFRM
??? ??????
56
( 3 ) 0,0)(
( 2 ) 0,0
( 1 ) 0,0
?????
????
????
?
?
?
QCC
Q
MFRMFm
SPNY
RTFX
由 (1)得 TFmRR
Q ??? ?
得代入所以 ( 3 ) mR TF ???
( 4 ) )()(
222
2
R
TR
R
FTF
R
FRM
mR
TFmFRMFRM
QC
???
?
???????
?????
由 (2)得 N= P +S，要保证车轮不滑动,
必须 F<f N =f (P+S) (5)
RTRRSPfM
22 ))(( ?? ????
可见,f 越
大越不易滑动。
Mmax的值
为上式右端的
值。 把 (5)代入 (4)得,
O
57
§ 15-4 Dynamical reaction of the bearing
of a rigid body under fixed-axis rotation
and the concepts of static and dynamic equilibrium
'R
'QR
OM
QOM
,)()()(
)(,
???
??
???
???
??????
kQmjQmiQm
kMjMiM
QmQrMaMR
iziyix
QzQyQx
iOiiQOCQ
1,Dynamical reaction of the bearing of a rigid body
The angle velocity of the rigid body is ?,the
angular acceleration is ? (counterclockwise)
Simplifying the positive force system to point O,
we get the principle vector and the
principle torque,
Simplifying the inertial force system to point O,
we get the principle vector and the
principle torque,
58
§ 15-4 定轴转动刚体的轴承动反力 ?
静平衡与动平衡的概念
一、刚体的轴承动反力
刚体的角速度 ?,角加速度 ?（ 逆时针）
主动力系向 O点简化, 主矢,主矩
惯性力系向 O点简化, 主矢,主矩
'R
'QR
OM
QOM
)()()(
)(
???
??
???
???
???
???
kQmjQmiQm
kMjMiM
QmQrM
aMR
iziyix
QzQyQx
iOiiQO
CQ
59
.c o ss i n
c o ss i n
)()()(
2 ??
??
? ??
???
???
???
iiiiiiii
iiiii
n
iii
ix
n
ixixQx
RzmRzm
amzamz
QmQmQmM
????
??
?
?
.)(
,g e t r e a s o n w e s a m e B y t h e
2
2
??
??
??
ziiiiiizQz
yzzxQy
IRmRamQmM
IIM
???????
??
???
.)()( H e u c e,
./co /s inB u t
2 ?? ??
??
iiiiiiQx
iiiiii
zymxzmM
RxsRy
??
??
2 c a u w r i le we
,
,w r t u
?? yzzxQx
iiiyz
iiizx
IIM
zymI
xzmI
??
?
?
?
?
60
??
??
? ??
???
???
???
iiiiiiii
iiiii
n
iii
ix
n
ixixQx
RzmRzm
amzamz
QmQmQmM
????
?? ?
?
c o ss in
c o ss in
)()()(
2
??
??
??
ziiiiiizQz
yzzxQy
IRmRamQmM
IIM
???????
??
??? 2
2
)(
同理可得
?? ??
??
)()(
/co /s in
2
iiiiiiQx
iiiiii
zymxzmM
RxsRy
??
?? 故而
2
,
?? yzzxQx
iiiyziiizx
IIM
zymIxzmI
???
?? ? ? 惯性积令
61
According to the dynamic-static method
,0
,0
,0
,0 '
,0 '
,0 '
'
'
??
??????
??????
??
????
????
Qzz
BAQyy
ABQxx
zB
Qy
yBA
Qx
xBA
MM
OBXOAXMM
OAYOBYMM
RZ
RRYY
RRXX
Five of the above equations are connected to with the reaction
forces of constraints,Supposing AB=l,OA=l and OB=l2 we obtain
62
根据动静法,
,0
,0
,0
,0 '
,0 '
,0 '
'
'
??
??????
??????
??
????
????
Qzz
BAQyy
ABQxx
zB
Qy
yBA
Qx
xBA
MM
OBXOAXMM
OAYOBYMM
RZ
RRYY
RRXX
其中有五个式子与约束反力有关。设 AB=l,OA=l1,OB=l2 可得
63
'.
,/)]'()'[(
,/)]'()'[(
,/)]'()'[(
,/)]'()'[(
11
11
22
22
xB
QxQyxyB
QyQxyxB
QyQxyxA
QxQyxyA
RZ
llRMlRMX
llRMlRMY
llRMlRMY
llRMlRMX
??
??????
???????
??????
???????
They consist of two parts,one is the result of the positive forces,It
cannot be avoided and is called the static reaction force,The other
part is the result of an unbalance of the inertial force system.It is
called the additional dynamic reaction force and can be avoided
by adjustment,
If you want to make the
additional dynamic reaction
force to be zero you have to
proceed as follows,
Static reaction force
Additional dynamic
reaction force
Dynamic
reaction
force
64
'
/)]'()'[(
/)]'()'[(
/)]'()'[(
/)]'()'[(
11
11
22
22
xB
QxQyxyB
QyQxyxB
QyQxyxA
QxQyxyA
RZ
llRMlRMX
llRMlRMY
llRMlRMY
llRMlRMX
??
??????
???????
??????
???????
由两部分组成，一部分由主动力引起的，不能消除，称为
静反力 ；一部分是由于惯性力系的不平衡引起的，称为 附加动
反力,它可以通过调整加以消除。
使附加动反力为零，须有 静反力 附加动反力 动反力
65
If the axis of rotation of the rigid body is the principle axis of
inertia,the additional dynamic reaction force of the bearing is
zero,
0?? QyQx MM 0'' ?? QyQx RR
0
0
2
2
??
??
??
??
yzzx
yzzx
II
II
)0(
0
42
2
2
???
?
???
??
??
??
yzzxxz III
0
0
?
?
Cy
Cx
Ma
Ma
0?? CC yx
The product of inertia about axis z
is zero,Axis z is principal axis of
inertia of the rigid body at point O,
Pass the center of mass
xzI
66
0?? QyQx MM 0'' ?? QyQx RR
当刚体转轴为中心惯性主轴时，轴承的附加动反力为零。
0
0
2
2
??
??
??
??
yzzx
yzzx
II
II
)0(
0
42
2
2
???
?
???
??
??
??
yzzxxz III
0
0
?
?
Cy
Cx
Ma
Ma
0?? CC yx
对 z 轴惯性积为零,z 轴为
刚体在 O点的惯性主轴； 过质心
67
Static equilibrium,if the axis of rotation of the rigid body passes
the center of mass,and if the rigid body is subjected only to gravity
and to no others positive force,the system is in equilibrium
independent on its position,
Dynamic equilibrium,if the axis of rotation of the rigid body is
the principle axis of inertia,the additional dynamic reaction force
of the bearing is zero,
2,The concepts of static equilibrium and of dynamic equilibrium
68
静平衡,刚体转轴过质心，则刚体在仅受重力而不受其它
主动力时，不论位置如何，总能平衡。
动平衡,转动为中心惯性主轴时，转动时不产生附加动反
力。
二、静平衡与动平衡的概念
69
[Example 1] A rigid bearing without mass is rotating with the
uniform angular velocity ?,two spheres A and B which have the same
mass m are connected to the bearing,Determine in which of the
following cases we have static equilibrium respectively dynamic
equilibrium,
( a) is in dynamic equilibrium,but (b) and (d) are in static equilibrium
70
[例 1] 质量不计的刚轴以角速度 ?匀速转动，其上固结着两个
质量均为 m的小球 A和 B。指出在图示各种情况下，哪些是静平
衡的？哪些是动平衡的？
静平衡,(b),(d) 动平衡,( a)
71
If a rigid body is in dynamic equilibrium,it is in static
equilibrium,too,But if a rigid body is in static equilibrium,it
may not be in dynamic equilibrium,
Grr
g
G
mrGrrRMb
GrmrGrMa
Q
?????
???
2
2
2
2
2
1
2
1
2
1
,0:)( c a s eI n
2
1
,0,)( c a s eI n
??
?
21
21,
??
??
?
?
[Example 2] Two identical crown blocks,as shown in the diagram
below,start to rotate from rest,Determine whose angular acceleration
is larger,
(a) A force G acting on the rope (b) A object whose weight is G is hanging on the rope,
O
O
72
动平衡的刚体，一定是静平衡的；反过来，静平衡的刚体，
不一定是动平衡的。
Grr
g
GmrGrrRMb
GrmrGrMa
Q
?????
???
2
2
2
2
2
1
2
1
2
1,0,)(
2
1,0,)(
??
?
对
对
21
21,
??
??
?
?
[例 2] 两个相同的定滑轮如下图示，开始时都处于静止，问哪
个角速度大？
(a) 绳子上加力 G (b) 绳子上挂一重 G的物体
O
O
73
According to D’Alembert’s principle,the method of transforming
the dynamic equations into the form of equilibrium equations of
statics is called the dynamic-static method,Not only the motion,
the acceleration and angular acceleration,but the forces can be
determined as well,This method is often used to determine the
dynamic reaction forces of constraints of the system of particles
when the motion is known,
Using the dynamic-static method you can use all the equilibrium
equations of statics,For example,the center of moment can be
selected arbitrarily and the equations of two moments or the
equations of three moments are both can be used,Therefore it is
more convenient to use the dynamic-static method in order to
solve the problems with several reaction forces of constraints,
The application of D’Alembert’s principle
74
根据达朗伯原理，以静力学平衡方程的形式来建立动力学
方程的方法，称为动静法。应用动静法既可求运动，例如加
速度、角加速度；也可以求力，并且多用于已知运动，求质
点系运动时的动约束反力。
应用动静法可以利用静力学建立平衡方程的一切形式上
的便利。例如，矩心可以任意选取，二矩式，三矩式等等。
因此当问题中有多个约束反力时，应用动静法求解它们时就
方便得多。
达朗伯原理的应用
75
① Select an object to study,The principles are the same as in
statics,
② Analyze the forces,Draw all the positive forces and the external
reaction forces of constraints,
③ Analyze the motion,Determine the acceleration of the angular
of mass,the angle acceleration of the rigid body and their directions,
④ Add the virtual inertial forces,Draw all inertial forces and all
inertial force couples into a force diagrams,This step is essential for
the correct analysis of motion,The result of the simplification of the
inertial force system of the rigid body should be remembered by
heart,
The steps and key points of the application of the dynamic-static
method to solve dynamical problems are
76
① 选取研究对象 。原则与静力学相同。
②受力分析。 画出全部主动力和外约束反力。
③运动分析。 主要是刚体质心加速度，刚体角加速度，标出
方向。
应用动静法求动力学问题的步骤及要点,
④ 虚加惯性力。 在受力图上画上惯性力和惯性力偶，一定要
在 正确进行运动分析的基础上。熟记刚体惯
性力系的简化结果。
77
⑤ Write down the dynamic-static equation,Select the center of
reduction and the projection axes properly,
⑥ Write down the supplementary equations,They are the
supplementary equations of kinetics (they show the relationships
among the quantities describing the motion),
⑦ Solve the equations to determine the unknown quantities,
[Notice] Because the direction of have been drawn into the
diagram,nothing has to be done but substituting these quantities into
the equations when you write down the
equations,
QOQ MR a n d
?OQOCQ IMmaR ?? a n d
78
⑤ 列动静方程。 选取适当的矩心和投影轴。
⑥建立补充方程。 运动学补充方程（运动量之间的关系）。
⑦求解求知量。
[注 ] 的方向及转向已在受力图中标出，建立方程时，
只需按 代入即可。
QOQ MR,
?OQOCQ IMmaR ??,
79
[Example 1] The masse of two weight,are m1and m2,respective,
They are hanging on two ropes which are wrapped on two coaxial
tub wheels of radius r1and r2,The moment of inertia of the two
wheels with respect to the axis of rotation O is I,the system moves
under the action of gravity,Determine the angular acceleration of the
tub wheels,
Select the system as the object to be
investigated,
Solution,
Method 1,Solve it by D'Alembert's principle
80
[例 1] 质量为 m1和 m2的两重物，分别挂在两条绳子上，绳又分
别绕在半径为 r1和 r2并装在同一轴的两鼓轮上，已知两鼓轮对于
转轴 O的转动惯量为 I，系统在重力作用下发生运动，求鼓轮的
角加速度。
取系统为研究对象
解,方法 1 用达朗伯原理求解
81
Add the virtual inertial force and the
inertial force couple,
?? IIMamRamR OQOQQ ????,,222111
According to the dynamic-static method we
get
.0
,0
,0)(
2221112211
22112211
?????
?????
??
?Iramramgrmgrm
MrRrRgrmgrm
Fm
QOQQ
O
Writing down the supplementary equations and
substituting them into the equation above you get ?? 2211,rara ??
g
Irmrm
rmrm
??
??
2
22
2
11
2211?
82
虚加惯性力和惯性力偶,
?? IIMamRamR OQOQQ ????,,222111
由动静法,
0
0
,0)(
2221112211
22112211
?????
?????
??
?Iramramgrmgrm
MrRrRgrmgrm
Fm
QOQQ
O
列补充方程,代入
上式
得,
?? 2211,rara ??
g
Irmrm
rmrm
??
??
2
22
2
11
2211?
83
Method 2,Solve the problem by the angular momentum
theorem,
.
,)(
2211
)(
2
22
2
11
222111
grmgrmM
Irmrm
IrvmrvmL
e
O
O
??
???
???
?
?
,T h e r e f o r e 2
22
2
11
2211 g
Irmrm
rmrm
??
???
According to the angular momentum
theorem
.])[( 2211222211 grmgrmIrmrmdtd ???? ?
Select the system as the object to be investigated,
84
方法 2 用动量矩定理求解
2211
)(
2
22
2
11
222111
)(
grmgrmM
Irmrm
IrvmrvmL
e
O
O
??
???
???
?
?
g
Irmrm
rmrm
??
???
2
22
2
11
2211 ?
根据动量矩定理,
2211222211 ])[( grmgrmIrmrmdtd ???? ?
取系统为研究对象
85
,2
22
2
11
2211 g
Irmrm
rmrm
??
???
)(
2
2
1
2
1
2
1
2
22
2
11
2
22
22
2
11
Irmrm
IvmvmT
???
???
?
?
??? gdrmrmIrmrmdWdT F )()](2[ g e t w e F r o m 2211222211
2
????? ?
Select the system as the object to be studied,
The kinetic energy of the system at any
instant is
,
is w o r k e l e m e n t a r y T h e
2211
2211
2211
) g dr-mr(m
dgrmdgrm
g d smg d smW F
?
??
?
?
??
???
By dt both sides of the equation and solve for the derivative
you get
Method 3,Solve the problem by the theorem of kinetic energy,
86
g
Irmrm
rmrm
??
??
2
22
2
11
2211 ?
)(
2
2
1
2
1
2
1
2
22
2
11
2
22
22
2
11
Irmrm
IvmvmT
???
???
?
?
?? gdrmrmIrmrmdδ WdT F )()](2[ 22112222112 ????? ? 得由
取系统为研究对象，任一瞬时系统的
) g dr-mr(m
dgrmdgrm
gdsmgdsmW F
?
??
?
2211
2211
2211
?
??
???元功
两边除以 dt，并求导数，得
方法 3 用动能定理求解
87
[Example 2] In the mechanism shown below the cylinders and the
tub wheel O are all homogenous,the weight of the two cylinders are
P and Q,the radii of both are R,They are Performing pure rolling,
The rope (without weight) can not be stretched,the angle of
inclination is ?,A constant torque M is acting on the tub wheel,
Determine (1)the angle acceleration of the tub wheel,(2) the tensile
force of the rope,(3) the reaction forces at point O on the bearing,
(4)the friction force between the cylinder and the inclined plane
(neglecting the rolling friction),
88
[例 2] 在图示机构中，沿斜面向上作纯滚动的圆柱体和鼓轮 O均
为均质物体，各重为 P和 Q，半径均为 R，绳子不可伸长，其质
量不计，斜面倾角 ?，如在鼓轮上作用一常力偶矩 M,试求：
(1)鼓轮的角加速度？ (2)绳子的拉力？ (3)轴承 O处的支反力？
(4)圆柱体与斜面间的摩擦力（不计滚动摩擦）？
89
Solution
Method 1,solution by e D’Alembert’s principle,
Study the wheel O,add the virtual inertial force couple,
OOOQ Rg
QIM ?? 2
2
1??
Write down the dynamic-static equations as
( 3 ), 0 s in0
( 2 ),0c o s0
( 1 ),0,0)(
????
???
????
?
?
?
?
?
TQ,YY
T,XX
MMTRFm
O
O
QO
.21M,2QA AAQ RgPagPR ???
Study the wheel A,add the virtual inertial force and the inertial
force couple MQC as shown in the diagram,QR
90
解,方法 1 用达朗伯原理求解
取轮 O为研究对象，虚加惯性力偶
OOOQ Rg
QIM ?? 2
2
1??
列出动静方程,
( 3 ) 0 s i n0
( 2 ) 0c o s0
( 1 ) 0,0)(
????
???
????
?
?
?
?
?
TQ,YY
T,XX
MMTRFm
O
O
QO
AAQ RgPagPR ?2QA 21M,??
取轮 A为研究对象，虚加惯性力 和惯性力偶 MQC如图示。
QR
91
Writing down the dynamic-static equations we get
( 5 ), 0s i n,0
( 4 ),0's i n,0)(
?????
?????
?
?
?
?
PFRT'X
MRTRRRPFm
Q
QAQC
We know that the kinematical relationships are
.,OAOAA RRa ???? ???
Substitute MQ,RQ,MQA and the equation above into equation (1) and
(4) and then solve them,to obtain
,
)3(
)s in3(
,
)3(
)s in(2
2
RPQ
QRMP
T
g
RPQ
RPM
O
?
?
?
?
?
?
?
?
?
92
列出动静方程,
( 5 ) 0s i n,0
( 4 ) 0's i n,0)(
?????
?????
?
?
?
?
PFRT'X
MRTRRRPFm
Q
QAQC
运动学关系：,
OAOAA RRa ???? ???
将 MQ,RQ,MQA及运动学关系代入到 (1)和 (4)式并联立求解得,
。
)3(
)s in3(
,
)3(
)s in(2
2
RPQ
QRMP
T
g
RPQ
RPM
O
?
?
?
?
?
?
?
?
?
93
Substitution into equations (2),(3) and (5) yields
。
)3(
)s i n(
,s i n
)3(
)s i n3(
,c os
)3(
)s i n3(
RPQ
PRMP
F
Q
RPQ
QRMP
Y
RPQ
QRMP
X
O
O
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
94
代入 (2),(3),(5)式，得,
。
)3(
)s i n(
,s i n
)3(
)s i n3(
,c os
)3(
)s i n3(
RPQ
PRMP
F
Q
RPQ
QRMP
Y
RPQ
QRMP
X
O
O
?
?
?
??
?
?
?
?
?
?
?
?
?
?
?
?
95
(1) Determine the angular acceleration of the
tub wheel by the theorem of kinetic energy,
Study the whole system,
,)s i n(
s i n
??
???
PRM
PRMW F
??
???
.)s i n()3(4 g e t w e F o r m 2
2
12 ??
? PRMCRPQ
gWTT
OF ?????? ?
)( AO R ωR ωv ??
,)3(
4
22
1
2
1
22
1
( c o n s t ),
2
2
22222
2
1
RPQ
g
R
g
P
v
g
P
R
g
Q
T
CT
O
AO ?????????
?
?
??
gRPQ PRMO ???? 2)3( )s in(2 ??
Determine the time-
derivative,OOO PRMRPQg ???? )s in(2)3(4 1 2 ?????
Method 2,solution by the general theorems of dynamics
96
方法 2 用动力学普遍定理求解
(1) 用动能定理求鼓轮角加速度。
取系统为研究对象
??
???
)s i n(
s i n
PRM
PRMW F
??
???
??? )s i n()3(4,2
2
12 PRMCRPQgWTT
OF ?????? ? 得由
)( AO R ωR ωv ??
2
2
22222
2
1
)3(
4
22
1
2
1
22
1
)(
RPQ
g
R
g
Pv
g
PR
g
QT
CT
O
AO ?????????
?
?
??
常量
gRPQ PRMO ???? 2)3( )s in(2 ??
两边对 t求导数,
)s i n(2)3(4 1 2 OOO PRMRPQg ???? ?????
97
(2) Determine the tensile force of the rope by the
angular momentum theorem (differential equation
of a fixed-axis rotation),
Study the wheel O,form the angular momentum
theorem we get
,2 2 TRMRgQ O ???,)3(
)s in3(
RPQ
QRMPT
?
?? ?
(3) Determine the reaction forces on the bearing O by the motion
theorem of motion of the center of mass,
Investigate the wheel O,from the theorem of motion of the center
of mass we get
,s i n0,
,c o s0,
?
?
????
???
?
?
TQYYMa
TXXMa
OCy
OCx
,s i n)3( )s i n3(,c o s)3( )s i n3( QRPQ QRMPYRPQ QRMPX OO ????????? ????
98
(2) 用动量矩定理求绳子拉力
（定轴转动微分方程）
取轮 O为研究对象，由动量矩定理得
TRMRgQ O ???22 RPQ QRMPT )3( )s in3( ??? ?
(3) 用质心运动定理求解轴承 O处支反力
取轮 O为研究对象，根据质心运动定理,
?
?
????
???
?
?
s i n0,
c o s0,
TQYYMa
TXXMa
OCy
OCx
QRPQ QRMPYRPQ QRMPX OO ????????? s i n)3( )s i n3(,c o s)3( )s i n3( ????
99
(4) Determine the friction force by the
differential equation of plane motion,
Investigate cylinder A,from the differential
equation of plane motion we gem
)(,OAAA FRI ??? ??
.)3( )s i n()3( )s i n(221 22 RPQ PRMPgRPQ PRMRgPRRIF AA ???????? ???
Method 3 determine the angular acceleration of the tub wheel
by the theorem of kinetic energy,
We can get the reaction forces of the constraints by
D’Alembert’s principle (the tensile force of the rope,the
reaction forces at the point O on the bearing and and the
friction force ),
T
OX
F
OY
100
(4) 用刚体平面运动微分方程求摩擦力
取圆柱体 A为研究对象,
根据刚体平面运动微分方程
)( OAAA FRI ??? ??
RPQ
PRMPg
RPQ
PRMR
g
P
RR
IF AA
)3(
)s i n(
)3(
)s i n(2
2
1
2
2
?
??
?
???? ???
方法 3：用动能定理求鼓轮的角加速度
用达朗伯原理求约束反力 （绳子拉力,轴承 O处反
力 和 及摩擦力 ）。
T
OX OY
F
101
[Example 3] The weight of a homogeneous cylinder is P,its radius
is R,It is rolling down from point O along the inclined plane
without sliding,At the beginning it does not move,The angle
between the plane and the horizon is ?,At OA=S determine the
reaction force of the constraint of the board at point O,Neglect the
weight of the board,
Solution,
(1) Determine the velocity and the
acceleration by the theorem of kinetic
energy,
The motion of the cylinder is a plane
movement,At the beginning it is
at test,at the end the angle velocity is
? and vC = R ? The kinetic energy is,
P
01 ?T
2T
102
[例 3] 均质圆柱体重为 P，半径为 R，无滑动地沿倾斜平板由静
止自 O点开始滚动。平板对水平线的倾角为 ?,试求 OA=S时平
板在 O点的约束反力。板的重力略去不计。
解, (1) 用动能定理求速度，加速度
圆柱体作平面运动。在初始位置时，
处于静止状态，故 T1=0；在末位置时，
设角速度为 ?，则 vC = R ?,动能为,
P
103
.4322121 22222 CC vgPRgPvgPT ??? ?
The work of the
positive forces is ? ?? ?s i nPSW F
From the theorem of kinetic energy
we get ??? FWTT 12
?? s i n34 s i n043 22 ?????? gSvPSvgP CC
Form the time-derivative we obtain
.s i n32,s i n32 ??? Rgga C ??(2) Determine the reaction force of the constraint by D’Alembert’s
principle,
Study the whole system,add the virtual inertial force and
the inertial force couple MQC, Q
R
P
104
2222
2 4
3
22
1
2
1
CC vg
PR
g
Pv
g
PT ??? ?
主动力的功,? ?? ?s i nPSW F
由动能定理 得 ??? FWTT 12
?? s i n34 s i n043 22 ?????? gSvPSvgP CC
对 t 求导数，则,??? s i n
3
2,s i n
3
2
R
gga
C ??
(2) 用达朗伯原理求约束反力
取系统为研究对象，虚加惯性力 和惯性力偶 MQC
QR
P
105
.s in
3
s in
3
2
2
1
,s in
3
2
2 ??
?
PR
R
g
R
g
P
M
P
a
g
P
R
QC
CQ
??
??
,0s inc o ss in
3
2
s in
3
,0)(
0s ins in
3
2
,0
0c o ss in
3
2
,0
????????
?????
????
?
?
?
RPPSR
P
R
P
MFm
,αα
P
P YY
,
P
XX
OO
O
O
????
??Write down the dynamic-static equations as
.c o s SP M O ?? ?
,2s in3 ?P X O ?
),s in321 2 ??? P( Y O
106
??
?
s i n
3
s i n
3
2
2
1
,s i n
3
2
2 PR
R
g
R
g
PM
Pa
g
PR
QC
CQ
??
??
0s inc o ss in
3
2
s in
3
,0)(
0s ins in
3
2
,0
0c o ss in
3
2
,0
????????
?????
????
?
?
?
RPPSR
P
R
P
MFm
,αα
P
P YY
,
P
XX
OO
O
O
????
??
列出动静方程,
SP M O ?? ?c o s
?2s in3P X O ?
)s in321 2 ??? P( Y O
107
[Example 4] The weight of a bobbin winder pulley is P,the radii
are R and r,the moment of inertia about the center of mass O is IO,
Under the action of a constant force T which makes an angle ?
with the horizon it is purely rolling,The rolling resistance can be
neglected,Determine (1)the acceleration of the center of the wheel,
(2)the conditions under which the wheel is purely rolling,
Solution,
Solution by D'Alembert's principle,The bobbin
winder pulley is Performing plane motion (pure
rolling)
) (,?RaaRIMagPR OOOQOOQ ???By D’Alembert’s principle we have
.0c o s,0)( ?????? RTTrRRMFm QQOC ?
Substituting RQ and MQO into the equation
above we obtain,)c o s(
2R
g
PI
rRTRa
O
O
?
?? ?
108
[例 4] 绕线轮重 P，半径为 R及 r,对质心 O转动惯量为 IO，在
与水平成 ? 角的常力 T 作用下纯滚动，不计滚阻，求,(1)轮心
的加速度； (2)分析纯滚动的条件。
解,用达朗伯原理求解
绕线轮作平面运动 （纯滚动）
) (,?RaaRIMagPR OOOQOOQ ???
由达朗伯原理，得
0c o s,0)( ?????? RTTrRRMFm QQOC ?
将 RQ, MQO代入上式，可得 2
)c os(
RgPI
rRTRa
O
O
?
?? ?
109
,0c o s,0 ????? QRFTX ?
,
)c o s(
)c o s(
c o s
c o s
22
R
g
P
I
Rr
g
P
IT
R
g
P
I
rRTR
g
P
T
RTF
O
O
O
Q
?
?
?
?
?
???
??
?
?
?
?
,s in
,0s in,0
?
?
TPN
TPNY
??
?????
The condition of pure rolling
is that F ≤f N,
),s i n(
)c o s(
2
?
?
TPf
R
g
P
I
Rr
g
P
IT
O
O
??
?
?
.
))(s in(
)c os(
2R
g
P
ITP
Rr
g
P
IT
f
O
O
??
?
?
?
?
110
0c o s,0 ????? QRFTX ?
22
)c o s(
)c o s(
c o s
c o s
R
g
P
I
Rr
g
P
IT
R
g
P
I
rRTR
g
P
T
RTF
O
O
O
Q
?
?
?
?
?
???
??
?
?
?
?
?
?
s in
0s in,0
TPN
TPNY
??
?????
纯滚动的条件,F ≤f N
)s in(
)c o s(
2
?
?
TPf
R
g
PI
Rr
g
PIT
O
O
??
?
?
))(s in(
)c o s(
2R
g
PITP
Rr
g
PIT
f
O
O
??
?
?
?
?
111
1,The system consists of two blocks A and
B,their weights both are m,they are placed
on a smooth horizontal plane,A horizontal
force F is acting on A,Please explain
whether the force with which block A acts
on B is equal to F,
Questions to think about,
Solution,
.
,
,0
' FNN
FmaFN
NRF Q
??
???
???
112
1,物体系统由质量均为 m的两物块
A和 B组成，放在光滑水平面上,
物体 A上作用一水平力 F，试用动静
法说明 A物体对 B物体作用力大小是
否等于 F？
思考题,
解,
FNN
FmaFN
NRF Q
??
???
???
'
0
113
?c o s2 212221 aaaamR Q ??? ??? c o ss intg 21 21 aa a?? ?
Solution
2,A triangular prism A,whose weight is M,is moving towards right
with acceleration,a sliding block B,whose weight is m,is sliding
down the inclined plane of A with the acceleration, Determine the
magnitude and direction of the inertial force of block B,
1a
2a
114
?c o s2 212221 aaaamR Q ??? ??? c o ss intg 21 21 aa a?? ?
解,
2,质量为 M的三棱柱体 A 以加速度 向右移动，质量为 m的滑
块 B以加速度 相对三棱柱体的斜面滑动，试问滑块 B的惯性
力的大小和方向如何？
1a
2a
115
3,The weight of a homogeneous wheel is P,its radius is r,It is
purely rolling on the horizontal plane,At a given time the angle
velocity is ?,the angle acceleration is ?,Determine the moment of
inertia of the wheel with respect to the center of mass C,the
momentum and the kinetic energy of the wheel,the angular
momentum with respect to C,the principle vector and the principle
moment of the inertial force system with respect to the center of
mass,
Solution,
???
????
?
g
rP
IMr
g
P
a
g
P
R
g
rP
IL
g
rP
IT
r
g
P
v
g
P
Kr
g
P
I
CQCCQ
CCC
CC
2
,
2
42
1
)(
2
2
2
2
2
2
2
?
????
?
??
?
??
????
116
3,匀质轮重为 P，半径为 r,在水平面上作纯滚动。某瞬时角
速度 ?,角加速度为 ?,求轮对质心 C 的转动惯量，轮的动量、
动能，对质心的动量矩，向质心简化的惯性力系主矢与主矩。
解,
???
??
??
?
g
rP
IMr
g
P
a
g
P
R
g
rP
IL
g
rP
IT
r
g
P
v
g
P
K
r
g
P
I
CQCCQ
CC
C
C
C
2
,
2
42
1
)(
2
2
2
2
2
2
2
?
????
?
??
?
??
???
?
117
118
