1
Theoretical Mechanics
2
3
§ 14-1 Work done by a force
§ 14-2 Kinetic energy
§ 14-3 Theorem of kinetic energy
§ 14-4 Power,power equation
§ 14-5 Conservative forces field,potential energy,the
law of conservation of mechanical energy
§ 14-6 General theorems of dynamics and its
applications
Chapter 14:Theorem of
the change in the kinetic energy
4
§ 14–1 力的功
§ 14–2 动能
§ 14–3 动能定理
§ 14–4 功率 · 功率方程
§ 14–5 势力场 · 势能 · 机械能守恒定理
§ 14–6 动力学普遍定理及综合应用
第十四章 动能定理
5
In order to obtain the theorem of the change in the kinetic energy,we make
use of the energy method to investigate dynamical problems,In different to the
cases of the theorems of the changes in the linear and the angular moment we
make use of the vector method,The method not only has important
applications in the research of the mechanical motion,but it is also the bridge
connecting mechanical motion with other forms of motion.The theorem of the
change in the kinetic energy establishes the dependence between the physical
quantities describing motion---kinetic energy and describing the acting force---
work,It is a law describing changes between different forms of energy,
Work is a measure of the accumulated effect of the action of a force
on a body during a given displacement,
1,Work done by a constant force
SFFSW ??? c o s?
The work done by a force is a scalar quantity,
For,the work is positive,for,the work
is zero,for,the work is negative,The unit of work in the SI
system is the joule (J),
2??? 2???
2
???
m1N1J1 ??
§ 14-1 Work done by a force
6
与动量定理和动量矩定理用矢量法研究不同，动能定理用
能量法研究动力学问题。能量法不仅在机械运动的研究中有重
要的应用，而且是沟通机械运动和其它形式运动的桥梁。动能
定理建立了与运动有关的物理量 —动能和作用力的物理量 —功
之间的联系，这是一种能量传递的规律。
§ 14-1 力的功
力的功是力沿路程累积效应的度量。
SF
FSW
??
?
c o s?
力的功是代数量。 时,正功； 时,功为零； 时,负功。
单位：焦耳（Ｊ）；
2??? 2??? 2???
m1N1J1 ??
一．常力的功
7
2,Work done by a variable force,
dsFW ?? c o s?
dsF?? rdF ??
Z d zY d yX d x ???
kdzjdyidxrdkZjYiXF ??????,(
)Z d zY d yX d xrdF ????
Elementary work
The total work done by a force during a finite curvilinear
displacement is,
F
21MM
????
2
1
2
1
c o s
M
M
M
M
dsFdsFW ?? (expression in the natural form)
? ??
2
1
M
M
rdF
(vector expression)
? ???
2
1
M
M
Z d zY d yX d x
(expression in terms of rectangular
coordinates),
8
二．变力的功
dsF??
rdF ??
Z d zY d yX d x ???
kdzjdyidxrdkZjYiXF ??????,(
)Z d zY d yX d xrdF ????
力 在曲线路程 中作功为 F
21MM
????
2
1
2
1
c o s
M
M
M
M
dsFdsFW ?? （自然形式表达式）
? ??
2
1
M
M
rdF （矢量式）
? ???
2
1
M
M
Z d zY d yX d x
（直角坐标表达式）
dsFW ?? c o s?元功,
9
3,Work done by a resultant force,
If a particle is subjected to the action of n forces,the
resultant force is, The work done by the resultant force is
i.e,
The work done by the resultant force during a finite displacement is
the arithmetical sum of the work done by all the component forces
acting on the particle,
nFFF,,,21 ???
R
rdFFFrdRW n
M
M
M
M
?????????? ?? )(
2
1
2
1
21
rdFrdFrdF
M
M
n
M
M
M
M
?????????? ???
2
1
2
1
2
1
21 n
WWW ??????? 21
?? iWW
?? iFR
10
三．合力的功
质点 M 受 n个力 作用合力为 则合力
的功
nFFF,,,21 ??? ?? iFR R
rdFFFrdRW n
M
M
M
M
?????????? ?? )(
2
1
2
1
21
rdFrdFrdF
M
M
n
M
M
M
M
?????????? ???
2
1
2
1
2
1
21 n
WWW ??????? 21
即
在任一路程上，合力的功等于各分力功的代数和。
?? iWW
11
4,Work done by special forces,
1) Work done by gravity,
For a particle the projections of
the gravity force acting on it on the
three coordinate axes are
Hence,
For a system of particles,
The work done by the gravity on a system is equal to the product
of the total weight of the system times the difference between the
initial and final height of the center of gravity of the system,It does
not depend on the path of each particle,
mgZYX ????,0,0
? ????
2
1
)( 21
z
z
zzmgm g d zW
)()( 2121 CCiiii zzMgzzgmWW ????? ? ?
12
四．常见力的功
1．重力的功
? ????
2
1
)( 21
z
z
zzmgm g d zW
质点系,)()(
2121 CCiiii zzMgzzgmWW ????? ? ?
质点系重力的功，等于质点系的重量与其在始末位置重
心的高度差的乘积，而与各质点的路径无关。
mgZYX ????,0,0
质点：重力在三轴上的投影,
13
2) Work done by an elastic force
Assuming the length of the inextensible spring to be,the force in
the elastic limit is,where, The factor K is
called the stiffness of the spring,It is the force required to extend the
spring by a unit length,and its dimension is [K]= N/M,or N/cm,
0l
00 )( rlrkF ??? rrr /0?
?? ?????? 2
1
2
1
00 )(
m
M
M
M
rdrlrkrdFW
drrdrrrdrrdrrrdr ??????? )(21)(21 20
2
00 )( 2 )(h e n c e
2
1
2
1
lrdkdrlrkW
r
r
r
r
?????? ? ?
g e t we,,W i th,])()[(2 022011202201 lrlrlrlrk ???????? ??
)(2 2221 ?? ?? kW
The work done by an elastic force depends
only on the initial and final deformation and
does not depend on the path,
14
2．弹性力的功
弹簧原长,在弹性极限内
k—弹簧的刚度系数，表示使弹簧发生单位
变形时所需的力。 N/m,N/cm。 。
0l 00 )( rlrkF ???
rrr /0?
?? ?????? 2
1
2
1
00 )(
m
M
M
M
rdrlrkrdFW
drrdrrrdrrdrrrdr ??????? )(21)(21 20
2
00 )( 2 )(
2
1
2
1
lrdkdrlrkW
r
r
r
r
??????? ? ?
022011202201,])()[(2 lrlrlrlrk ???????? ??令
)( 22212 ?? ?? kW即
弹性力的功只与弹簧的起始变形和终了
变形有关，而与质点运动的路径无关。
15
3) Work done by the gravitational force,,
The work done by the gravitational force depends only on the initial
and final positions r1 and r2 and does not depend on the path,
4)Work done by forces applied to a rotating body,
Let a force F act on the point M of a rigid body rotating about an
axis Z,Determine the work done by the force F during a turn by a
finite angle,,
)11( 120 rrG m mW ??
At the locus of the point M the force is
bn FFFF ??? ?
??? ?? dFmrdFdsFW z )(???? )( 12 ??? ?? ?? 2
1
)( h e n c e
?
?
?dFmW z
?
The work done by a force applied to a rotating
body is equal to the work done by the torque,
If on a body a force couple acts lying in a plane
normal to the axis about which the body rotates
we get
??
2
1
?
?
?mdW
If m=const,then )(
12 ?? ?? mW
Note the definition of the sign of the work,
16
??? ?? dFmrdFdsFW z )(????
)( 12 ??? ??
???
2
1
)(
?
?
?dFmW z
作用于转动刚体上力的功等于力矩的功。
??
2
1
?
?
?mdW
若 m = 常量,则 )( 12 ?? ?? mW
注意：功的符号的确定。
3．万有引力的功
)11( 120 rrG m mW ??
万有引力所作的功只与质点的始末位置有关，与路径无关。
如果作用力偶,m,且力
偶的作用面垂直转轴
4．作用于转动刚体上的力的功，力偶的功
设在绕 z 轴转动的刚体上 M点作用有力,计算刚体转过
一角度 ? 时力 所作的功。 M点轨迹已知。
F
F bn FFFF ??? ?
17
5) Work done by friction,
(1) Work done by kinetic friction,
?? ???? 2121 'MMMM N d sfdsFW ?
If N=const,W= –f′N S,Thus the work depends on the actual path
of the particle,
(2)Work done by kinetic friction acting on a disc rolling without
slipping on a fixed surface,
The normal reaction force N and the frictional force F act on the
instantaneous center of velocity c,
The elementary displacement of the
instantaneous center is
0?? dtvrd C 0????? dtvFrdFW C?
(3) Work done by a couple with a moment m
resisting rolling
If m=const,then,
R
smmW ???? ?
,so
18
0?? dtvrd C 0????? dtvFrdFW
C?
正压力,摩擦力 作用于瞬心 C处，而瞬心的元位移 N F
(2) 圆轮沿固定面作纯滚动时，滑动摩擦力的功
(3) 滚动摩擦阻力偶 m的功
5．摩擦力的功
(1) 动滑动摩擦力的功
?? ???? 2121 'MMMM N d sfdsFW ?
N=常量时,W= –f′N S,与质点的路径有关。
R
smmW ???? ?若 m = 常量则
19
5,Work done by internal forces,
If the distance between two points A and B does not
change,the sum of the elementary work done by the internal
forces F and F’ is zero,
The sum of the work done by all the internal forces of a non-
deformable system is zero,Special cases of such systems are a
rigid body and an inextensible string,
BA rdFrdFW ???? '? BA rdFrdF ????
)( BA rrdF ??? )( BAdF ??
20
五．质点系内力的功
只要 A,B两点间距离保持不变,内力的元功和就等于零 。
不变质点系的内力功之和等于零。刚体的内力功之和等于零。
不可伸长的绳索内力功之和等于零 。
BA rdFrdFW ???? '? BA rdFrdF ????
)( BA rrdF ??? )( BAdF ??
21
1) Fixed smooth surface,
)( 0)( rdNrdNW N ?????2) Mobile pin-joint,fixed pin-joint and centripetal bearing,
3) Rigid body rolling without slipping along a fixed surface,
4) Smooth pin-joint joining mobile rigid bodies,
5) Flexible cable (inextensible string)
When the string is strained the sum of the
elementary works done by the internal forces is zero,
? ???? rdNrdNW N ')(?
0????? rdNrdN
6,Work done by reaction forces of constraints,
If the elementary work of the reaction force of a constraint is zero or
if the sum of the elementary work done by all the reaction forces of
a system is zero,the constraint are called ideal constraints,
22
六．理想约束反力的功
约束反力元功为零或元功之和为零的约束称为理想约束 。
1．光滑固定面约束
2．活动铰支座、固定铰支座和向心轴承
3．刚体沿固定面作纯滚动
4．联接刚体的光滑铰链（中间铰）
5．柔索约束（不可伸长的绳索）
拉紧时，内部拉力的元功之和恒等于零。
)( 0)( rdNrdNW N ?????
? ???? rdNrdNW N ')(?
0????? rdNrdN
23
The kinetic energy of a material body is the energy due to the
motion,the kinetic energy is a measure of the mechanical motion,
1,Kinetic energy of a particle,,
221 mvT ?
Kinetic energy is an instantaneous quantity,It is a positive scalar
quantity not depending on the direction of the velocity,Its dimension
is the same as that of work,Joule (J),
§ 14–2 Kinetic energy
2,Kinetic energy of a system of particles,
, 221 ii vmT ??
??? 22 '2121 iiC vmMvT
For a system of particles where '
iv
is the velocity of the i-th particle
relative to the center of mass of the system we get
,
24
§ 14-2 动能
物体的动能是由于物体运动而具有的能量，是机械运动强弱
的又一种度量。
一．质点的动能
二．质点系的动能
221 mvT ?
瞬时量,与速度方向无关的正标量,具有与功相同的量纲,单位也是 J。
221 ii vmT ??
对于任一质点系：（ 为第 i个质点相对质心的速度） '
iv
??? 22 '2121 iiC vmMvT 柯尼希定理
25
3,Kinetic energy of a rigid body
1) Translational motion
2222 2121)(2121 Ciii MvMvvmvmT ???? ??
2) Rotational motion about an fixed axis
2222 21)(2121 ?? ziiii IrmvmT ?????
3) Plane motion
221 ?PIT ?
(P is the instantaneous center of the velocity)
2MdII CP ?? 22222
2
1
2
1)(
2
1
2
1 ???
CCC IvMdMI ????
26
221 ?PIT ?
（ P为速度瞬心） 2MdII
CP ??
22222 21 21)(2121 ??? CCC IvMdMI ????
2222 2121)(2121 Ciii MvMvvmvmT ???? ??
2222 21)(2121 ?? ziiii IrmvmT ?????
1．平动刚体
2．定轴转动刚体
3．平面运动刚体
三．刚体的动能
27
§ 14-3 Theorem of kinetic energy
1.Theorem of kinetic energy of a particle,
If both sides of the equation are multiplied by
we get dtvrd ?? ? ? rdFdtvvmdtd ???
Wmvd ??)21( 2
).21()(2)(B u t 2mvdvvdmdtvvmdtd ????
Therefore
Integrating this expression along the path 21MM
we obtain Wmvmv ?? 2
122 2121
FvmdtdFam ??? )(
This is the differential form of
the theorem of kinetic energy,
This is the integral form of the
theorem of kinetic energy,
28
§ 14-3 动能定理
1．质点的动能定理,
)21()(2)( 2mvdvvdmdtvvmdtd ????而
Wmvd ??)21( 2
因此 动能定理的微分形式
将上式沿路径 积分，可得
21MM
Wmvmv ?? 2122 2121 动能定理的积分形式
两边点乘以,有 dtvrd ?? ? ? rdFdtvvmdtd ???
FvmdtdFam ??? )(
29
,
This equation is called the theorem of the change in the kinetic
energy of a system,
In the case of ideal constraints,the theorem of the change in the
kinetic energy of a system of particles can be expressed as
.o r )(12)( ?? ??? FF WTTWdT ?
2,Theorem of kinetic energy of a system of particles,
For any particle in a system we have
iM iii Wvmd ??)21( 2
Then for the whole system we have
,???? ???
iiiiii WvmdWvmd ?? )21( )21( 22
i.e., This is the differential form of the theorem of
kinetic energy of a system of particles,?? iWdT ?
Integrating this expression along the path,we obtain
21MM
??? WTT 12 This is the integral form of the theorem of kinetic
energy of a system of particles,
30
对质点系中的一质点,
iM
iii Wvmd ??)21( 2
即 质点系动能定理的微分形式 ??
iWdT ?
21MM
??? WTT 12 质点系动能定理的积分形式
在理想约束的条件下，质点系的动能定理可写成以下的形式
?? ??? )(12)( ; FF WTTWdT ?
???? ??? iiiiii WvmdWvmd ?? )21( )21( 22
对整个质点系，有
2．质点系的动能定理
将上式沿路径 积分，可得
31
[Example 1] In the system shown in the diagram the homogeneous
disks A and B each are of weight P and radius R,the line through the
centers of the two discs is a horizontal one,The disc A is subjected to
a constant couple with moment m and a load weight Q,Determine the
velocity and the acceleration of the load after it has fallen through a
distance h,Neglect the mass of the string,The string is inextensible,
the disc B rolls without slipping,At the initial instant the whole
system is at rest,
32
[例 1] 图示系统中,均质圆盘 A,B各重 P，半径均为 R,两盘中心
线为水平线,盘 A上作用矩为 M(常量 )的一力偶；重物 D重 Q。问
下落距离 h时重物的速度与加速度。 (绳重不计，绳不可伸长，
盘 B作纯滚动，初始时系统静止 )
33
Solution,Take the whole system as the
object to be investigated,
),/( )( RhQhmW F ???? ??,01 ?T
222
2 2
1
2
1
2
1
BCAO Ivg
QIT ??
????
)78(
16
2
3
2
1
2
1
22
1
2
22222
PQ
g
v
R
g
Pv
g
Q
R
g
P
BA
??
????? ??
g e t w e F r o m )(12 ??? FWTT
PQ
hgQRMvhQ
R
MPQ
g
v
78
)/(4 )(0)78(
16
2
?
???????
),(,)(216 78 dtdhvdtdhQRMdtdvvg PQ ?????
Differentiation of both sides of the equation with respect to time
results in
.78 )/(8 PQ gQRMa ? ??
34
解,取系统为研究对象
)/( )( RhQhmW F ???? ?? 01 ?T
222
2 2
1
2
1
2
1
BCAO Ivg
QIT ??
????
)78(
16
2
3
2
1
2
1
22
1
2
22222
PQ
g
v
R
g
Pv
g
Q
R
g
P
BA
??
????? ??
??? )(12 FWTT由
PQ
hgQRMvhQ
R
MPQ
g
v
78
)/(4 )(0)78(
16
2
?
???????
上式求导得,
)( )(216 78 dtdhvdtdhQRMdtdvvg PQ ?????
PQ
gQRMa
78
)/(8
?
??
35
Exercises concerning the application of the theorem of the
change in kinetic energy
1,The homogeneous rod OA is of mass =30kg,the string is
inextensible as long as the rod is in the vertical position,The stiffness
of the spring is k=30KN/m,Determine the minimum angular velocity
of the rod which allows it to rotate to the horizontal position,
Solution,Consider the rod OA
)(212.1 2221)( ?? ????? kPW F ])22.14.2(0[3 0 0 0212.18.930 22 ????????
),J(4.388??
,8.284.2303121 202021 ?? ?????T
.02 ?T From g e t w e)(12 ??? FWTT
r a d /s,67.3
4.3888.280
0
2
0
??
???
?
?
36
动能定理的应用练习题
1．图示的均质杆 OA的质量为 30kg，杆在铅垂位置时弹簧处
于自然状态。设弹簧常数 k =3kN/m，为使杆能由铅直位置 OA
转到水平位置 OA'，在铅直位置时的角速度至少应为多大？
解,研究 OA杆
)(212.1 2221)( ?? ????? kPW F ])22.14.2(0[3 0 0 0212.18.930 22 ????????
)J(4.388??
,8.284.2303121 202021 ?? ?????T
02?T 由 ??? )(12 FWTT
r a d /s67.3
4.3 8 88.280
0
2
0
?
???
?
?
37
2,A planetary gearing is placed in a horizontal plane,the movable
gear of radius r and weight p may be regarded as a homogeneous
disc,The crank of weight Q and length L is subjected to a couple
with moment M=constant,giving rise to a rotation starting from rest,
Determine the angular velocity and the angular
acceleration of the crank,
Solution,Choose the whole system to be
investigated ? ? ?MW F )(
01?T 21221222 2 2121321 ?? grPvgPgQlT ????
??? rlrvlv ??? 111,22222222 12 92)( 4 )(26 ???? lg PQrlgrPlgPgQlT ?????
According to the theorem of the change in kinetic energy
We obtain
?? Mlg PQ ??? 012 92 22 ①,PQgMl 9232 ?? ??
2)92(
6
lPQ
gM
???
Differentiation of the expression (1)
with respect to time t,gives,
38
2．行星齿轮传动机构,放在水平面内。 动齿轮半径 r,重 P,视为
均质圆盘；曲柄重 Q,长 l,作用一力偶,矩为 M(常量 ),曲柄由静止
开始转动； 求曲柄的角速度 (以转角 ? 的函数表示 ) 和角加速度。
解,取整个系统为研究对象
? ? ?MW F )(
01?T 21221222 2 2121321 ?? grPvgPgQlT ????
??? rlrvlv ??? 111,
22222222 12 92)( 4 )(26 ???? lg PQrlgrPlgPgQlT ?????
根据动能定理，得
?? Mlg PQ ??? 012 92 22 ? PQ
gM
l 92
32
??
??
将 ?式对 t 求导数，得
2)92(
6
lPQ
gM
???
39
3,A mechanism consisting of two straight homogeneous rods and
their sizes are shown in the diagram,The mass of the rod OA is twice
as large as that of rod OB,Neglecting friction determine the velocity
at the point B of the rod AB when the rod OA swings into a horizontal
configuration after the mechanism
in the position shown in the figure is
released from rest,
Solution,Take the whole system as the
object to be investigated,
mgmgmgW F 35.1)15.06.0(2 9.02)( ?????
01 ?T 2222 219.023121 mvmT ?????? ?v??9.0
o b t a in w e
e q u a t io n in t o gse x p r e s s io n f o u n d t h engS u b s t i t u t i,
6
5T h e r e f o r e
)(
12
2
2
???
?
FWTT
mvT
m / s,98.3 35.1065 2 ???? vmgmv
40
3．两根均质直杆组成的机构及尺寸如图示； OA杆质量是
AB杆质量的两倍，各处摩擦不计，如机构在图示位置从静止
释放，求当 OA杆转到铅垂位置时,AB杆 B 端的速度。
mgmgmgW F 35.1)15.06.0(2 9.02)( ?????
01 ?T
2222 219.023121 mvmT ?????? ?
v??9.0
????? 得代入到 )(1222 65 FWTTmvT
m / s98.3 35.1065 2 ??? vmgmv
解,取整个系统为研究对象
41
§ 14-4 Power,power equation
1,Power
Power is the work done by a force in a unit of time,It is an
important index that measures the working capacity of a machine,
Power is a scalar quantity and has a instantaneous characteristic,
Power of an acting force,
Power of the moment of a force,
The unit of power is Watt (W) or Kilowatt (KW),where1W=1 J/s,
dtWN ??
vFvFdt rdFdtWN ?? ??????
30
???? nMM
dt
dM
dt
WN zzz ?????
42
§ 14-4 功率 · 功率方程
一．功率,力在单位时间内所作的功（它是衡量机器工作能力
的一个重要指标）。功率是代数量，并有瞬时性。
dtWN ??
作用力的功率,vFvF
dt rdFdtWN ?? ??????
力矩的功率,
30
???? nMM
dt
dM
dt
WN zzz ?????
功率的单位：瓦特（ W），千瓦（ kW），１ W=１ J/s 。
43
2,Power equation
, Dividing both sides of the equation by dt ?? WdT ?
we obtain ?? ?????
u s e l e s su s e f u li m p o r t t h a t so NNNNdt
dT
dt
W
dt
dT ?
Analysis.In a starting state,
In a braking state,
In the steady state,
0?dtdT
0?dtdT
0?dtdT
u s e l e s su s e f u li n p u t NNN ??
When a machine operates in a steady state,dT/dt=0,the mechanical
efficiency is
%1 0 0
i m p u t
u s e f u l ??
N
N?
?is one of the most important indices for judging the quality of a
?machine,Generally,? ＜ 100%,
u s e l e s su s e f u li n p u t NNN ??
u s e l e s su s e f u li n p u t NNN ??
44
二．功率方程,
由 的两边同除以 dt 得 ?? WdT ?
?? ????? 无用有用输入即 NNNNdtdTdt WdtdT ?
分析：起动阶段（加速）,即
制动阶段（减速）,即
稳定阶段（匀速）,即
0?dtdT
0?dtdT
0?dtdT
无用有用输入 NNN ??
无用有用输入 NNN ??
无用有用输入 NNN ??
机器稳定运行时,机械效率 0/ ?dtdT
%1 0 0??
输入
有用
N
N?
?是评定机器质量优劣的重要指标之一。一般情况下 ? ＜１ 。
45
§ 14-5 Conservative force field,potential energy
and the law of conservation of the mechanical energy
1,Conservative force field
1) Force field,If in a region of space any particle experiences a
force of a certain magnitude and direction depending on position,
the region of space is called to be a force field,
2) Conservative force field,If in a force field the work done by the
force acting on a moving particle depends only on the initial and the
final position of the particle and does not depend on path the force
field is called to be a conservative force field,
The fields of gravity,of elastic and of gravitational forces are
conservative field,
The forces acting in a conservative force field are called
conservative forces,such as gravity,elastic force etc,
46
§ 14-5 势力场、势能、机械能守恒定律
一．势力场
1．力场,若质点在某空间内的任何位置都受到一个大小和
方向完全由所在位置确定的力的作用，则此空间称为 力场 。
重力场、万有引力场、弹性力场都是势力场。
质点在势力场中受到的场力称为有势力 (保守力 ),如重力、弹力
等。
2．势力场, 在力场中,如果作用于质点的场力作功只决定于
质点的始末位置，与运动路径无关，这种力场称为 势力场 。
47
,dzzVdyyVdxxVdV ?????????,,zVZyVYxVX ?????????????
The potential energy of a system of particles is
.)(),,,,,,( 111 ? ? ??????
io
i
M
M
iiiiiinnn dzZdyYdxXzyxzyxV
is an uniquely defined continuous function of the
coordinates,The potential energy of any particle on the so called
equal potential energy surface is constant,
dVZ d zY d yX d x ????
),,( zyxVV ?
The potential energy at the position M0 as a basic standard position is
defined as zero,the zero point of the potential energy,The potential
energy is a relative quantity,
2,Potential energy
In a conservative force field the work done by the conservative force
when a particle move from position M to position M0 is called the
potential energy of the particle at the position M relative to position M0,
It is denoted by V and given by,
? ???? ?? 00
M
M
M
M
Zd zYd yX d xrdFV
48
二．势能
在势力场中,质点从位置 M 运动到任选位置 M0,有势力所作的
功称为质点在位置 M 相对于位置 M0的势能，用 V 表示。
? ???? ?? 00
M
M
M
M
Zd zYd yX d xrdFV
M0作为基准位置，势能为零，称为零势能点。势能具有相对性。
dVZ d zY d yX d x ????
),,( zyxVV ? 是坐标的单值连续函数。
等势面：质点位于该面上任何地方，势能都相等。
dzzVdyyVdxxVdV ?????????,,zVZyVYxVX ?????????????
质点系的势能,
? ? ?????? io
i
M
M
iiiiiinnn dzZdyYdxXzyxzyxV )(),,,,,,( 111
49
,At point
20
__
22
0
2
,WrdFvM
M
M
??? ?
21201012 VVWWW ????
M1→ M2,
The work done by a conservative force is equal to the difference of
the potential energies between the initial and final position,
3,Work done by conservative force,
At point
10
__
11
2
1
,WrdFvM
M
M
??? ?
For one particle
For a system of particles
1) Field of gravity,
2) Field of elastic force,,
Choosing the position of a spring in its natural state as the zero
point of the potential energy we get,
3) Field of gravitational force,
Choosing the infinitely distant from the center of the
gravitation position as the zero point of the potential energy
( ) we get
phzzpV ???? )( 0
phzzpV coc ???? )(
2
2
1 ?kv ?
??0r
r
mGmv 21??
50
1.重力场
质点,
质点系,
2,弹性力场,取弹簧的自然位置为零势能点
3,万有引力场,取与引力中心相距无穷远处为零势能位置
PhzzPV ???? )( 0
hPzzPV CC ???? )( 0
221 ?kV ?
)( 0 ??r
r
mGmV 21??
有势力的功等于质点系在运动的始末位置的势能之差。
三．有势力的功
在 M1位置,
101
0
1
WrdFV
M
M
??? ? 202 0
2
WrdFV M
M
??? ?M2位置,
21201012 VVWWW ????
M1→ M2,
51
[Example 1] A straight homogeneous rod of
mass m stand in a vertical position on the
smooth surface of a table without initial
velocity,Determine the velocity of the
center of mass (expressed in the inclination
of the rod and the position of the center of
mass) if the rod falls down,
?
4,The Law of Conservation of the Mechanical Energy
Mechanical energy is defined as the arithmetical sum of the kinetic
and the potential energy of a system,
Assume a system to be subjected only to the action of conservative
forces,then, 211212 VVWTT ????
Therefore co n s t,
2211 ???? VTVT
This is the law of conservative of
mechanical energy,
Such system are called conservative systems,
For a non-conservative system,assuming the work done by non-
conservative forces is,
we have
121122 )()( WVTVT ?????
'12W
52
设质点系只受到有势力 (或同时受到不作功的非有势力 ) 作用，
则 211212 VVWTT ????
—机械能守恒定律 常量?????
2211 VTVT
对非保守系统，设非保守力的功为 W12',则有
121122 )()( WVTVT ?????
四．机械能守恒定律
机械能：系统的动能与势能的代数和 。
这样的系统成为保守系统 。
[例 1] 长为 l，质量为 m的均质直杆，初瞬
时直立于光滑的桌面上。当杆无初速度
地倾倒后，求质心的速度（用杆的倾角 ?
和质心的位置表达）。
53
Substituting into the expression and simplifying
we obtain,
?? sin2l y?? ?
ygy ??2
2
s in31
s in6
???
Solution, Because the rod is not subjected
to an external force in horizontal direction,
the center of mass will fall vertically,Again,
because the reaction force does not perform
work and the active force is a conservative
one,we may use the law of conservation of mechanical energy to find
the solution,
At initial moment,,
At any instant:,
,
From the law of conservation of mechanical energy we have
mglVT ??? 2,0 11
222222 212412121 ymmlymIT C ???? ???? ?? )2(2 ylmgV ??
? ? ????? s i n2,s i n2 i, e c o s12 a n d l ylyly ???? ?????
)2(2124120 222 ylmgymmlmgl ????? ???
54
解,由于水平方向不受外力,
且初始静止，故质心 C铅垂下降。
由于约束反力不作功,主动力为有势力,
因此可用机械能守恒定律求解。
? ? ????? s i n2,s i n2 co s12 l ylyly ????? ????? 即又
由机械能守恒定律,
)2(2124120 222 ylmgymmlmgl ????? ???
将 代入上式，化简后得 ?? sin2l y?? ?
ygy ??2
2
s in31
s in6
???
mglVT ??? 2,0 11初瞬时,
222222 212412121 ymmlymIT C ???? ???? ??
)2(2 ylmgV ??
任一瞬时,
55
§ 14-6 General theorems of
dynamics and its applications
The general theorems of dynamics include the theorems of the change in the
linear,in the angular momentum and in the kinetic energy for a particle or a
system of particles,The theorem of the change in momentum and the theorem
of moments are expressed in vector forms,while the theorem of the change in
kinetic energy is expressed in scalar forms,They all may be applied to study
mechanical motion,The theorem of the change in kinetic energy can also solve
problems concerning the transformation between the kinetic energy and other
forms of energy,
In general,the theorems of dynamics offer methods to solve dynamical
problems,The applications of the general theorems include two aspects:Ⅰ
According to the initial conditions of a problem and the quantity to be
determined,choose proper theorem and solve,including the judgment of
various conservative condition and the applications of the corresponding law
of conservation; Directly obtain the required result,eliminating the irrelevant
unknown quantities,Ⅱ For more complicated problems,according to necessity,
choose two or three theorems and combine them to solve the problem,
Moreover,during solving a problem,we should correctly perform the analysis
of motion and write down the proper complementary equations of kinematics,
56
§ 14-6 动力学普遍定理及综合应用
动力学普遍定理包括质点和质点系的动量定理、动量矩定理和
动能定理。 动量定理和动量矩定理是矢量形式，动能定理是标量形
式，他们都可应用研究机械运动，而动能定理还可以研究其它形式
的运动能量转化问题。
动力学普遍定理提供了解决动力学问题的一般方法。动力学普
遍定理的综合应用，大体上包括两方面的含义,一是能根据问题的
已知条件和待求量，选择适当的定理求解，包括各种守恒情况的判
断，相应守恒定理的应用。避开那些无关的未知量，直接求得需求
的结果。二是对比较复杂的问题，能根据需要选用两、三个定理联
合求解。
求解过程中,要正确进行运动分析,提供正确的运动学补充方程。
57
Illustration of the application of the general theorems of dynamics,
[Example 1] Two homogeneous rods AC and BC each of weight p and
length l are,attached to each other by means of a joint,they are placed
on a smooth horizontal plane,Assuming the axes of the two rods
remain in vertical plane throughout,At initial instant the system is at
rest and the height of point c is h,Determine the velocity when the
joint c reaches the ground,
58
举例说明动力学普遍定理的综合应用,
[例 1] 两根均质杆 AC和 BC各重为 P，长为 l，在 C处光滑铰接，置于
光滑水平面上；设两杆轴线始终在铅垂面内，初始静止,C点高度
为 h，求铰 C到达地面时的速度。
59
Discussion
(1)Applying the law of conservation of linear momentum plus the theorem of
the change in kinetic energy solves the problem,
(2)Make use of the kinematics of planar motion when computing the kinetic
energy,
Substituting these expressions into the theorem of the change in kinetic
energy we get,
ghvPhvgP CC 3 e w h e n c031 2 ???
PhhPW F ??? ?? 22)( 01?T
22222 3123121 ?? lgPlgPT ????
22 31 CC vgPTlv ??? ?
Because the system initially is at rest the
position of the center of mass in horizontal
direction remains at rest,
,0)(? ?exFAnalysis of forces,
Solution Because we do not have to determine the
internal force of the system the two rods do not
have to be treated separately,The object under study
is the whole system,
60
讨论 ? 动量守恒定理＋动能定理求解。
? 计算动能时，利用平面运动的运动学关系。
解,由于不求系统的内力，可以不拆开。
研究对象：整体
分析受力：,
且初始静止，所以水平方向质心位置守恒。
? ? 0)( exF
PhhPW F ??? ?? 22)( 01?T
22222 3123121 ?? lgPlgPT ????
代入动能定理,
ghvPhvgP CC 3 031 2 ????
22 31 CC vgPTlv ??? ?
61
[Example2] The circular homogeneous
disc A is of mass m and radius r,the slide
block of mass m,and the rod AB is of
negligible mass and parallel to the angle of
inclination,The coefficient of friction
is f,Determine the acceleration of the slide
block when the circular disk roll without
slipping,The system initially is at rest,
Solution Choose the whole system as the object under study
? ???? )co ss i n2( co s s i n 2)( ???? fSmgm g SfSmgW F
222221 21212121 0 ?mrmvmvTT ?????
Kinematics,
?rv ? 22 45 mvT ?From the theorem of the change in kinetic energy
we get )co ss i n2(0
4
5 2 ?? fm g Smv ???
Differentiation of the expression
with respect to time t,results in,
gfa )c o s52s i n54( ?? ??
?
62
[例 2] 均质圆盘 A,m,r；滑块 B：
m；杆 AB：质量不 计，平行于斜
面。斜面倾角 ?，摩擦系数 f，圆盘
作纯滚动，系统初始静止。求：
滑块的加速度。
解：选系统为研究对象
? ???? )co ss i n2( co s s i n 2)( ???? fSmgm g SfSmgW F
222221 21212121 0 ?mrmvmvTT ?????
运动学关系,?rv?
22 45 mvT ??
由动能定理,
)co ss i n2(045 2 ?? fm g Smv ???
对 ｔ 求导，得
gfa )c o s52s i n54( ?? ??
63
[Example 3]A circular homogeneous disc of weight 150N and a
homogeneous rod of weight 60 N and length 24cm are joined at the
point B,The system is released without initial velocity from the
position shown in the diagram,
Determine the velocity of the point B and the
reaction force of the constraint at mount A when the
system passes through its lowest position,
Solution,
(1) Take the disc as the object under study,; 0)(? ?Fm B 0 0 ??? BBBI ??
00 ?? ?? B,the disc is in
translational motion,
64
[例 3] 重 150N的均质圆盘与重 60N、长 24cm的均质杆 AB在 B处用
铰链连接。 系统由图示位置无初速地释放。 求 系统经过最低位
置 B'点时的速度及支座 A的约束反力。
解,（ 1）取圆盘为研究对象; 0)(? ?Fm B 0 0 ??? BBBI ??
00 ?? ?? B,圆盘平动。
65
(2)Applying the theorem of the change in kinetic energy determine
the velocity of point B,
Considerate the whole system,
At the initial moment,
At the lowest position
,
01 ?T
2222 2121 BA vgGIT ??? ?
2212221 6 3213121 BBB vg GGvgGvgG ??? ?????
)30s i n)(2()30s i n()30s i n22( 2121)( ??????????? llGGllGllGW F
??? )(12 FWTT )30s in)(
2(06
3
2
1221 ??????
? llG
Gv
g
GG
B
Substituting the known values into this expression we get
, m /s 58.1
' ?Bv
66
（ 2）用动能定理求速度 。
取系统研究。初始时 T1=0,
最低位置时,
2222 2121 BA vgGIT ??? ?
2212221 6 3213121 BBB vg GGvgGvgG ??? ?????
)30s i n)(2()30s i n()30s i n22( 2121)( ??????????? llGGllGllGW F
??? )(12 FWTT
)30s in)(2(06 3 21221 ?????? ? llGGvg GG B
代入数据，得 m /s 58.1' ?Bv
67
Substituting the known values
into this expression we get, N4 0 1,0 ??
AA YX; 0'21? ???? ABcixi XagGagGam ?? ? ????? 212221 2 GGYlgGlgGam Aiyi ??
(4) Calculate the reaction force at mount A from the theorem of the
motion of the center of mass,
Investigate the whole system.,
(3) Applying the theorem of the change in angular momentum
determine the angular acceleration, ?
?? )31(31 2221221 lgGlgGvlgGlgGL A ???? ? ?? 0)( )( eAA FmdtdL
?＝ 0,
?
?
)0( 2 2 ???? ?? CnCC alaa
The acceleration of the center of mass of the rod is
The acceleration of the center of the disc is
ca
'Ba
)0( 2' ???? ?? ?? BnBB alaa
r ad / s 58.624.0 58.1' ??? lv B?
68
（ 3）用动量矩定理求杆的角加速度 ? 。
?? )31(31 2221221 lgGlgGvlgGlgGL A ????
由于 ? ?? 0)( )( e
AA FmdtdL
所以 ?＝ 0 。
杆质心 C的加速度,
盘质心加速度,
)0( 2 2 ???? ?? CnCC alaa
)0( 2' ???? ?? ?? BnBB alaa
r ad / s 58.624.0 58.1' ??? lv B?
（ 4）由质心运动定理求支座反力。 研究整个系统。; 0'21? ???? ABcixi XagGagGam ??
代入数据，得 N4 0 1,0 ??
AA YX
? ????? 212221 2 GGYlgGlgGam Aiyi ??
69
(1) Using the theorem of the conservation of the total angular
momentum with respect to the center of mass of a system plus
the theorem of the change in kinetic energy plus the theorem
of the change in angular momentum plus the theorem of
motion of center of mass,
(2) we may calculate by differentiation of the expression of the
theorem of the change in kinetic energy in integral form with
respect to time t,but we can not analyze the motion of the rod
AB at an arbitrary position,
?
70
(1) 相对质心动量矩守恒定理 +动能定理 +动量矩定理 +质心
运动定理。
(2) 可用对积分形式的动能定理求导计算 ?，但要注意需取
杆 AB在一般位置进行分析 。
71
?mLmvK C 61??
?? ])6(121[ 22 LmmLIL OO ???
?291 mL?
222 18121 ?? mLIT O ??
?223 mRL O ?
2243 ?mRT ?
mvK ?
?221 mRL C ?
222 4121 ?mRmvT ??
?mRK ?
[Example 4] Calculate the basic quantities (linear momentum,
angular momentum and kinetic energy) for the following systems,
72
?mLmvK C 61??
?? ])6(121[ 22 LmmLIL OO ???
?291 mL?
222 18121 ?? mLIT O ??
?223 mRL O ?
2243 ?mRT ?
?mRK ? mvK ?
?221 mRL C ?
222 4121 ?mRmvT ??
[例 4] 基本量计算 (动量,动量矩,动能 )
73
[Example 5] A rod of mass m is supported on two cylindrical
rollers of radius r and mass m/2 each,The rollers are placed on a
horizontal plane, Determine the acceleration of the rod when it
is subjected to a horizontal force F taking into account friction of
contact but no relative slipping,
Solution,
(1) Solve this problem applying the theorem of the change in
kinetic energy,
Choose the system as the object under study,The rod is in
translational motion and the cylindrical rollers in plane motion,
Assuming that the instantaneous velocity of the rod is v,the
velocity of the centers of mass of the rollers is v/2,and the
angular velocity of the rollers is, rv2??
74
[例５ ] 质量为 m 的杆置于两个半径为 r,质量为 的实心
圆柱上，圆柱放在水平面上，求当杆上加水平力 时，
杆的加速度。设接触处都有摩擦，而无相对滑动。
2m
P
解, (1)用动能定理求解。
取系统为研究对象，杆作平动，圆柱体作平面运动。
设任一瞬时，杆的速度为 v，则圆柱体质心速度为 v/2，
角速度, rv2??
75
From the theorem of the change in kinetic energy in differential
form we have
dSPmvd ??)1611( 2?? )( FWdT ?
Dividing both side by d t,
and differentiating we get
vPavm ???? 21611,118 mPa ??
22222 1611])2)(221(21)2(221[221 mvrvrmvmmvT ?????
The sum of the elementary work done by the active force is
,)(? ?? dSPW F?
Then the kinetic energy of the system is
,
76
系统的动能 22222
1611])2)(221(21)2(221[221 mvrvrmvmmvT ?????
主动力的元功之和,? ?? dSPW F )(?
由动能定理的微分形式,
?? )( FWdT ? dSPmvd ??)1611( 2
两边除以,并求导数，得 dt
vPavm ???? 21611 mPa 118 ??
77
(2) Applying the theorem of the change in angular
momentum solve this problem,
Choose the whole system as the object under study,
m v rrvrmrvmrmvL O 411)222122(22 2 ??????? ? ?? rPFm eO 2)( )(
According the principle of the moments we get
?? )( FmdtdL OO
and hence,
rPm v rdtd 2)411( ??
m
P
a
rPm r a
11
8
t h e r e f or e,
a n d,2
4
11
f ol low sI t
?
??
78
(2) 用动量矩定理求解
取系统为研究对象
m v rrvrmrvmrmvL O 411)222122(22 2 ??????? ? ?? rPFm eO 2)( )(
根据动量矩定理：,得 ?? )( Fm
dtdL OO
rPm v rdtd 2)411( ??
mParPm r a 118 2411 ????
79
[Example 6] A homogeneous rod OA has weight p and length l,
Determine the angular acceleration of the rod and the reaction force at
the position o at a moment at which the string suddenly breaks,
Solution,Choose the rod as the object
under study,According to the theorem
of moments we have
231 2 lPlgP ???? lg 2/3 H e n c e ??
According to the theorem of motion of the center of mass
we get,
OCx Xag
P ?? 0
PYPYlgPagP OOCy 41 w h e n c e2 ????? ?
80
解,取杆为研究对象
231 2 lPlgP ???? lg 2/3??
由质心运动定理,
OCx Xag
P ?? 0
PYPYlgPagP OOCy 41 2 ?????? ?
[例 6] 均质杆 OA，重 P，长 l，绳子突然剪断。 求 该瞬时，角加
速度及 O处反力。
由动量矩定理,
81
82
Theoretical Mechanics
2
3
§ 14-1 Work done by a force
§ 14-2 Kinetic energy
§ 14-3 Theorem of kinetic energy
§ 14-4 Power,power equation
§ 14-5 Conservative forces field,potential energy,the
law of conservation of mechanical energy
§ 14-6 General theorems of dynamics and its
applications
Chapter 14:Theorem of
the change in the kinetic energy
4
§ 14–1 力的功
§ 14–2 动能
§ 14–3 动能定理
§ 14–4 功率 · 功率方程
§ 14–5 势力场 · 势能 · 机械能守恒定理
§ 14–6 动力学普遍定理及综合应用
第十四章 动能定理
5
In order to obtain the theorem of the change in the kinetic energy,we make
use of the energy method to investigate dynamical problems,In different to the
cases of the theorems of the changes in the linear and the angular moment we
make use of the vector method,The method not only has important
applications in the research of the mechanical motion,but it is also the bridge
connecting mechanical motion with other forms of motion.The theorem of the
change in the kinetic energy establishes the dependence between the physical
quantities describing motion---kinetic energy and describing the acting force---
work,It is a law describing changes between different forms of energy,
Work is a measure of the accumulated effect of the action of a force
on a body during a given displacement,
1,Work done by a constant force
SFFSW ??? c o s?
The work done by a force is a scalar quantity,
For,the work is positive,for,the work
is zero,for,the work is negative,The unit of work in the SI
system is the joule (J),
2??? 2???
2
???
m1N1J1 ??
§ 14-1 Work done by a force
6
与动量定理和动量矩定理用矢量法研究不同，动能定理用
能量法研究动力学问题。能量法不仅在机械运动的研究中有重
要的应用，而且是沟通机械运动和其它形式运动的桥梁。动能
定理建立了与运动有关的物理量 —动能和作用力的物理量 —功
之间的联系，这是一种能量传递的规律。
§ 14-1 力的功
力的功是力沿路程累积效应的度量。
SF
FSW
??
?
c o s?
力的功是代数量。 时,正功； 时,功为零； 时,负功。
单位：焦耳（Ｊ）；
2??? 2??? 2???
m1N1J1 ??
一．常力的功
7
2,Work done by a variable force,
dsFW ?? c o s?
dsF?? rdF ??
Z d zY d yX d x ???
kdzjdyidxrdkZjYiXF ??????,(
)Z d zY d yX d xrdF ????
Elementary work
The total work done by a force during a finite curvilinear
displacement is,
F
21MM
????
2
1
2
1
c o s
M
M
M
M
dsFdsFW ?? (expression in the natural form)
? ??
2
1
M
M
rdF
(vector expression)
? ???
2
1
M
M
Z d zY d yX d x
(expression in terms of rectangular
coordinates),
8
二．变力的功
dsF??
rdF ??
Z d zY d yX d x ???
kdzjdyidxrdkZjYiXF ??????,(
)Z d zY d yX d xrdF ????
力 在曲线路程 中作功为 F
21MM
????
2
1
2
1
c o s
M
M
M
M
dsFdsFW ?? （自然形式表达式）
? ??
2
1
M
M
rdF （矢量式）
? ???
2
1
M
M
Z d zY d yX d x
（直角坐标表达式）
dsFW ?? c o s?元功,
9
3,Work done by a resultant force,
If a particle is subjected to the action of n forces,the
resultant force is, The work done by the resultant force is
i.e,
The work done by the resultant force during a finite displacement is
the arithmetical sum of the work done by all the component forces
acting on the particle,
nFFF,,,21 ???
R
rdFFFrdRW n
M
M
M
M
?????????? ?? )(
2
1
2
1
21
rdFrdFrdF
M
M
n
M
M
M
M
?????????? ???
2
1
2
1
2
1
21 n
WWW ??????? 21
?? iWW
?? iFR
10
三．合力的功
质点 M 受 n个力 作用合力为 则合力
的功
nFFF,,,21 ??? ?? iFR R
rdFFFrdRW n
M
M
M
M
?????????? ?? )(
2
1
2
1
21
rdFrdFrdF
M
M
n
M
M
M
M
?????????? ???
2
1
2
1
2
1
21 n
WWW ??????? 21
即
在任一路程上，合力的功等于各分力功的代数和。
?? iWW
11
4,Work done by special forces,
1) Work done by gravity,
For a particle the projections of
the gravity force acting on it on the
three coordinate axes are
Hence,
For a system of particles,
The work done by the gravity on a system is equal to the product
of the total weight of the system times the difference between the
initial and final height of the center of gravity of the system,It does
not depend on the path of each particle,
mgZYX ????,0,0
? ????
2
1
)( 21
z
z
zzmgm g d zW
)()( 2121 CCiiii zzMgzzgmWW ????? ? ?
12
四．常见力的功
1．重力的功
? ????
2
1
)( 21
z
z
zzmgm g d zW
质点系,)()(
2121 CCiiii zzMgzzgmWW ????? ? ?
质点系重力的功，等于质点系的重量与其在始末位置重
心的高度差的乘积，而与各质点的路径无关。
mgZYX ????,0,0
质点：重力在三轴上的投影,
13
2) Work done by an elastic force
Assuming the length of the inextensible spring to be,the force in
the elastic limit is,where, The factor K is
called the stiffness of the spring,It is the force required to extend the
spring by a unit length,and its dimension is [K]= N/M,or N/cm,
0l
00 )( rlrkF ??? rrr /0?
?? ?????? 2
1
2
1
00 )(
m
M
M
M
rdrlrkrdFW
drrdrrrdrrdrrrdr ??????? )(21)(21 20
2
00 )( 2 )(h e n c e
2
1
2
1
lrdkdrlrkW
r
r
r
r
?????? ? ?
g e t we,,W i th,])()[(2 022011202201 lrlrlrlrk ???????? ??
)(2 2221 ?? ?? kW
The work done by an elastic force depends
only on the initial and final deformation and
does not depend on the path,
14
2．弹性力的功
弹簧原长,在弹性极限内
k—弹簧的刚度系数，表示使弹簧发生单位
变形时所需的力。 N/m,N/cm。 。
0l 00 )( rlrkF ???
rrr /0?
?? ?????? 2
1
2
1
00 )(
m
M
M
M
rdrlrkrdFW
drrdrrrdrrdrrrdr ??????? )(21)(21 20
2
00 )( 2 )(
2
1
2
1
lrdkdrlrkW
r
r
r
r
??????? ? ?
022011202201,])()[(2 lrlrlrlrk ???????? ??令
)( 22212 ?? ?? kW即
弹性力的功只与弹簧的起始变形和终了
变形有关，而与质点运动的路径无关。
15
3) Work done by the gravitational force,,
The work done by the gravitational force depends only on the initial
and final positions r1 and r2 and does not depend on the path,
4)Work done by forces applied to a rotating body,
Let a force F act on the point M of a rigid body rotating about an
axis Z,Determine the work done by the force F during a turn by a
finite angle,,
)11( 120 rrG m mW ??
At the locus of the point M the force is
bn FFFF ??? ?
??? ?? dFmrdFdsFW z )(???? )( 12 ??? ?? ?? 2
1
)( h e n c e
?
?
?dFmW z
?
The work done by a force applied to a rotating
body is equal to the work done by the torque,
If on a body a force couple acts lying in a plane
normal to the axis about which the body rotates
we get
??
2
1
?
?
?mdW
If m=const,then )(
12 ?? ?? mW
Note the definition of the sign of the work,
16
??? ?? dFmrdFdsFW z )(????
)( 12 ??? ??
???
2
1
)(
?
?
?dFmW z
作用于转动刚体上力的功等于力矩的功。
??
2
1
?
?
?mdW
若 m = 常量,则 )( 12 ?? ?? mW
注意：功的符号的确定。
3．万有引力的功
)11( 120 rrG m mW ??
万有引力所作的功只与质点的始末位置有关，与路径无关。
如果作用力偶,m,且力
偶的作用面垂直转轴
4．作用于转动刚体上的力的功，力偶的功
设在绕 z 轴转动的刚体上 M点作用有力,计算刚体转过
一角度 ? 时力 所作的功。 M点轨迹已知。
F
F bn FFFF ??? ?
17
5) Work done by friction,
(1) Work done by kinetic friction,
?? ???? 2121 'MMMM N d sfdsFW ?
If N=const,W= –f′N S,Thus the work depends on the actual path
of the particle,
(2)Work done by kinetic friction acting on a disc rolling without
slipping on a fixed surface,
The normal reaction force N and the frictional force F act on the
instantaneous center of velocity c,
The elementary displacement of the
instantaneous center is
0?? dtvrd C 0????? dtvFrdFW C?
(3) Work done by a couple with a moment m
resisting rolling
If m=const,then,
R
smmW ???? ?
,so
18
0?? dtvrd C 0????? dtvFrdFW
C?
正压力,摩擦力 作用于瞬心 C处，而瞬心的元位移 N F
(2) 圆轮沿固定面作纯滚动时，滑动摩擦力的功
(3) 滚动摩擦阻力偶 m的功
5．摩擦力的功
(1) 动滑动摩擦力的功
?? ???? 2121 'MMMM N d sfdsFW ?
N=常量时,W= –f′N S,与质点的路径有关。
R
smmW ???? ?若 m = 常量则
19
5,Work done by internal forces,
If the distance between two points A and B does not
change,the sum of the elementary work done by the internal
forces F and F’ is zero,
The sum of the work done by all the internal forces of a non-
deformable system is zero,Special cases of such systems are a
rigid body and an inextensible string,
BA rdFrdFW ???? '? BA rdFrdF ????
)( BA rrdF ??? )( BAdF ??
20
五．质点系内力的功
只要 A,B两点间距离保持不变,内力的元功和就等于零 。
不变质点系的内力功之和等于零。刚体的内力功之和等于零。
不可伸长的绳索内力功之和等于零 。
BA rdFrdFW ???? '? BA rdFrdF ????
)( BA rrdF ??? )( BAdF ??
21
1) Fixed smooth surface,
)( 0)( rdNrdNW N ?????2) Mobile pin-joint,fixed pin-joint and centripetal bearing,
3) Rigid body rolling without slipping along a fixed surface,
4) Smooth pin-joint joining mobile rigid bodies,
5) Flexible cable (inextensible string)
When the string is strained the sum of the
elementary works done by the internal forces is zero,
? ???? rdNrdNW N ')(?
0????? rdNrdN
6,Work done by reaction forces of constraints,
If the elementary work of the reaction force of a constraint is zero or
if the sum of the elementary work done by all the reaction forces of
a system is zero,the constraint are called ideal constraints,
22
六．理想约束反力的功
约束反力元功为零或元功之和为零的约束称为理想约束 。
1．光滑固定面约束
2．活动铰支座、固定铰支座和向心轴承
3．刚体沿固定面作纯滚动
4．联接刚体的光滑铰链（中间铰）
5．柔索约束（不可伸长的绳索）
拉紧时，内部拉力的元功之和恒等于零。
)( 0)( rdNrdNW N ?????
? ???? rdNrdNW N ')(?
0????? rdNrdN
23
The kinetic energy of a material body is the energy due to the
motion,the kinetic energy is a measure of the mechanical motion,
1,Kinetic energy of a particle,,
221 mvT ?
Kinetic energy is an instantaneous quantity,It is a positive scalar
quantity not depending on the direction of the velocity,Its dimension
is the same as that of work,Joule (J),
§ 14–2 Kinetic energy
2,Kinetic energy of a system of particles,
, 221 ii vmT ??
??? 22 '2121 iiC vmMvT
For a system of particles where '
iv
is the velocity of the i-th particle
relative to the center of mass of the system we get
,
24
§ 14-2 动能
物体的动能是由于物体运动而具有的能量，是机械运动强弱
的又一种度量。
一．质点的动能
二．质点系的动能
221 mvT ?
瞬时量,与速度方向无关的正标量,具有与功相同的量纲,单位也是 J。
221 ii vmT ??
对于任一质点系：（ 为第 i个质点相对质心的速度） '
iv
??? 22 '2121 iiC vmMvT 柯尼希定理
25
3,Kinetic energy of a rigid body
1) Translational motion
2222 2121)(2121 Ciii MvMvvmvmT ???? ??
2) Rotational motion about an fixed axis
2222 21)(2121 ?? ziiii IrmvmT ?????
3) Plane motion
221 ?PIT ?
(P is the instantaneous center of the velocity)
2MdII CP ?? 22222
2
1
2
1)(
2
1
2
1 ???
CCC IvMdMI ????
26
221 ?PIT ?
（ P为速度瞬心） 2MdII
CP ??
22222 21 21)(2121 ??? CCC IvMdMI ????
2222 2121)(2121 Ciii MvMvvmvmT ???? ??
2222 21)(2121 ?? ziiii IrmvmT ?????
1．平动刚体
2．定轴转动刚体
3．平面运动刚体
三．刚体的动能
27
§ 14-3 Theorem of kinetic energy
1.Theorem of kinetic energy of a particle,
If both sides of the equation are multiplied by
we get dtvrd ?? ? ? rdFdtvvmdtd ???
Wmvd ??)21( 2
).21()(2)(B u t 2mvdvvdmdtvvmdtd ????
Therefore
Integrating this expression along the path 21MM
we obtain Wmvmv ?? 2
122 2121
FvmdtdFam ??? )(
This is the differential form of
the theorem of kinetic energy,
This is the integral form of the
theorem of kinetic energy,
28
§ 14-3 动能定理
1．质点的动能定理,
)21()(2)( 2mvdvvdmdtvvmdtd ????而
Wmvd ??)21( 2
因此 动能定理的微分形式
将上式沿路径 积分，可得
21MM
Wmvmv ?? 2122 2121 动能定理的积分形式
两边点乘以,有 dtvrd ?? ? ? rdFdtvvmdtd ???
FvmdtdFam ??? )(
29
,
This equation is called the theorem of the change in the kinetic
energy of a system,
In the case of ideal constraints,the theorem of the change in the
kinetic energy of a system of particles can be expressed as
.o r )(12)( ?? ??? FF WTTWdT ?
2,Theorem of kinetic energy of a system of particles,
For any particle in a system we have
iM iii Wvmd ??)21( 2
Then for the whole system we have
,???? ???
iiiiii WvmdWvmd ?? )21( )21( 22
i.e., This is the differential form of the theorem of
kinetic energy of a system of particles,?? iWdT ?
Integrating this expression along the path,we obtain
21MM
??? WTT 12 This is the integral form of the theorem of kinetic
energy of a system of particles,
30
对质点系中的一质点,
iM
iii Wvmd ??)21( 2
即 质点系动能定理的微分形式 ??
iWdT ?
21MM
??? WTT 12 质点系动能定理的积分形式
在理想约束的条件下，质点系的动能定理可写成以下的形式
?? ??? )(12)( ; FF WTTWdT ?
???? ??? iiiiii WvmdWvmd ?? )21( )21( 22
对整个质点系，有
2．质点系的动能定理
将上式沿路径 积分，可得
31
[Example 1] In the system shown in the diagram the homogeneous
disks A and B each are of weight P and radius R,the line through the
centers of the two discs is a horizontal one,The disc A is subjected to
a constant couple with moment m and a load weight Q,Determine the
velocity and the acceleration of the load after it has fallen through a
distance h,Neglect the mass of the string,The string is inextensible,
the disc B rolls without slipping,At the initial instant the whole
system is at rest,
32
[例 1] 图示系统中,均质圆盘 A,B各重 P，半径均为 R,两盘中心
线为水平线,盘 A上作用矩为 M(常量 )的一力偶；重物 D重 Q。问
下落距离 h时重物的速度与加速度。 (绳重不计，绳不可伸长，
盘 B作纯滚动，初始时系统静止 )
33
Solution,Take the whole system as the
object to be investigated,
),/( )( RhQhmW F ???? ??,01 ?T
222
2 2
1
2
1
2
1
BCAO Ivg
QIT ??
????
)78(
16
2
3
2
1
2
1
22
1
2
22222
PQ
g
v
R
g
Pv
g
Q
R
g
P
BA
??
????? ??
g e t w e F r o m )(12 ??? FWTT
PQ
hgQRMvhQ
R
MPQ
g
v
78
)/(4 )(0)78(
16
2
?
???????
),(,)(216 78 dtdhvdtdhQRMdtdvvg PQ ?????
Differentiation of both sides of the equation with respect to time
results in
.78 )/(8 PQ gQRMa ? ??
34
解,取系统为研究对象
)/( )( RhQhmW F ???? ?? 01 ?T
222
2 2
1
2
1
2
1
BCAO Ivg
QIT ??
????
)78(
16
2
3
2
1
2
1
22
1
2
22222
PQ
g
v
R
g
Pv
g
Q
R
g
P
BA
??
????? ??
??? )(12 FWTT由
PQ
hgQRMvhQ
R
MPQ
g
v
78
)/(4 )(0)78(
16
2
?
???????
上式求导得,
)( )(216 78 dtdhvdtdhQRMdtdvvg PQ ?????
PQ
gQRMa
78
)/(8
?
??
35
Exercises concerning the application of the theorem of the
change in kinetic energy
1,The homogeneous rod OA is of mass =30kg,the string is
inextensible as long as the rod is in the vertical position,The stiffness
of the spring is k=30KN/m,Determine the minimum angular velocity
of the rod which allows it to rotate to the horizontal position,
Solution,Consider the rod OA
)(212.1 2221)( ?? ????? kPW F ])22.14.2(0[3 0 0 0212.18.930 22 ????????
),J(4.388??
,8.284.2303121 202021 ?? ?????T
.02 ?T From g e t w e)(12 ??? FWTT
r a d /s,67.3
4.3888.280
0
2
0
??
???
?
?
36
动能定理的应用练习题
1．图示的均质杆 OA的质量为 30kg，杆在铅垂位置时弹簧处
于自然状态。设弹簧常数 k =3kN/m，为使杆能由铅直位置 OA
转到水平位置 OA'，在铅直位置时的角速度至少应为多大？
解,研究 OA杆
)(212.1 2221)( ?? ????? kPW F ])22.14.2(0[3 0 0 0212.18.930 22 ????????
)J(4.388??
,8.284.2303121 202021 ?? ?????T
02?T 由 ??? )(12 FWTT
r a d /s67.3
4.3 8 88.280
0
2
0
?
???
?
?
37
2,A planetary gearing is placed in a horizontal plane,the movable
gear of radius r and weight p may be regarded as a homogeneous
disc,The crank of weight Q and length L is subjected to a couple
with moment M=constant,giving rise to a rotation starting from rest,
Determine the angular velocity and the angular
acceleration of the crank,
Solution,Choose the whole system to be
investigated ? ? ?MW F )(
01?T 21221222 2 2121321 ?? grPvgPgQlT ????
??? rlrvlv ??? 111,22222222 12 92)( 4 )(26 ???? lg PQrlgrPlgPgQlT ?????
According to the theorem of the change in kinetic energy
We obtain
?? Mlg PQ ??? 012 92 22 ①,PQgMl 9232 ?? ??
2)92(
6
lPQ
gM
???
Differentiation of the expression (1)
with respect to time t,gives,
38
2．行星齿轮传动机构,放在水平面内。 动齿轮半径 r,重 P,视为
均质圆盘；曲柄重 Q,长 l,作用一力偶,矩为 M(常量 ),曲柄由静止
开始转动； 求曲柄的角速度 (以转角 ? 的函数表示 ) 和角加速度。
解,取整个系统为研究对象
? ? ?MW F )(
01?T 21221222 2 2121321 ?? grPvgPgQlT ????
??? rlrvlv ??? 111,
22222222 12 92)( 4 )(26 ???? lg PQrlgrPlgPgQlT ?????
根据动能定理，得
?? Mlg PQ ??? 012 92 22 ? PQ
gM
l 92
32
??
??
将 ?式对 t 求导数，得
2)92(
6
lPQ
gM
???
39
3,A mechanism consisting of two straight homogeneous rods and
their sizes are shown in the diagram,The mass of the rod OA is twice
as large as that of rod OB,Neglecting friction determine the velocity
at the point B of the rod AB when the rod OA swings into a horizontal
configuration after the mechanism
in the position shown in the figure is
released from rest,
Solution,Take the whole system as the
object to be investigated,
mgmgmgW F 35.1)15.06.0(2 9.02)( ?????
01 ?T 2222 219.023121 mvmT ?????? ?v??9.0
o b t a in w e
e q u a t io n in t o gse x p r e s s io n f o u n d t h engS u b s t i t u t i,
6
5T h e r e f o r e
)(
12
2
2
???
?
FWTT
mvT
m / s,98.3 35.1065 2 ???? vmgmv
40
3．两根均质直杆组成的机构及尺寸如图示； OA杆质量是
AB杆质量的两倍，各处摩擦不计，如机构在图示位置从静止
释放，求当 OA杆转到铅垂位置时,AB杆 B 端的速度。
mgmgmgW F 35.1)15.06.0(2 9.02)( ?????
01 ?T
2222 219.023121 mvmT ?????? ?
v??9.0
????? 得代入到 )(1222 65 FWTTmvT
m / s98.3 35.1065 2 ??? vmgmv
解,取整个系统为研究对象
41
§ 14-4 Power,power equation
1,Power
Power is the work done by a force in a unit of time,It is an
important index that measures the working capacity of a machine,
Power is a scalar quantity and has a instantaneous characteristic,
Power of an acting force,
Power of the moment of a force,
The unit of power is Watt (W) or Kilowatt (KW),where1W=1 J/s,
dtWN ??
vFvFdt rdFdtWN ?? ??????
30
???? nMM
dt
dM
dt
WN zzz ?????
42
§ 14-4 功率 · 功率方程
一．功率,力在单位时间内所作的功（它是衡量机器工作能力
的一个重要指标）。功率是代数量，并有瞬时性。
dtWN ??
作用力的功率,vFvF
dt rdFdtWN ?? ??????
力矩的功率,
30
???? nMM
dt
dM
dt
WN zzz ?????
功率的单位：瓦特（ W），千瓦（ kW），１ W=１ J/s 。
43
2,Power equation
, Dividing both sides of the equation by dt ?? WdT ?
we obtain ?? ?????
u s e l e s su s e f u li m p o r t t h a t so NNNNdt
dT
dt
W
dt
dT ?
Analysis.In a starting state,
In a braking state,
In the steady state,
0?dtdT
0?dtdT
0?dtdT
u s e l e s su s e f u li n p u t NNN ??
When a machine operates in a steady state,dT/dt=0,the mechanical
efficiency is
%1 0 0
i m p u t
u s e f u l ??
N
N?
?is one of the most important indices for judging the quality of a
?machine,Generally,? ＜ 100%,
u s e l e s su s e f u li n p u t NNN ??
u s e l e s su s e f u li n p u t NNN ??
44
二．功率方程,
由 的两边同除以 dt 得 ?? WdT ?
?? ????? 无用有用输入即 NNNNdtdTdt WdtdT ?
分析：起动阶段（加速）,即
制动阶段（减速）,即
稳定阶段（匀速）,即
0?dtdT
0?dtdT
0?dtdT
无用有用输入 NNN ??
无用有用输入 NNN ??
无用有用输入 NNN ??
机器稳定运行时,机械效率 0/ ?dtdT
%1 0 0??
输入
有用
N
N?
?是评定机器质量优劣的重要指标之一。一般情况下 ? ＜１ 。
45
§ 14-5 Conservative force field,potential energy
and the law of conservation of the mechanical energy
1,Conservative force field
1) Force field,If in a region of space any particle experiences a
force of a certain magnitude and direction depending on position,
the region of space is called to be a force field,
2) Conservative force field,If in a force field the work done by the
force acting on a moving particle depends only on the initial and the
final position of the particle and does not depend on path the force
field is called to be a conservative force field,
The fields of gravity,of elastic and of gravitational forces are
conservative field,
The forces acting in a conservative force field are called
conservative forces,such as gravity,elastic force etc,
46
§ 14-5 势力场、势能、机械能守恒定律
一．势力场
1．力场,若质点在某空间内的任何位置都受到一个大小和
方向完全由所在位置确定的力的作用，则此空间称为 力场 。
重力场、万有引力场、弹性力场都是势力场。
质点在势力场中受到的场力称为有势力 (保守力 ),如重力、弹力
等。
2．势力场, 在力场中,如果作用于质点的场力作功只决定于
质点的始末位置，与运动路径无关，这种力场称为 势力场 。
47
,dzzVdyyVdxxVdV ?????????,,zVZyVYxVX ?????????????
The potential energy of a system of particles is
.)(),,,,,,( 111 ? ? ??????
io
i
M
M
iiiiiinnn dzZdyYdxXzyxzyxV
is an uniquely defined continuous function of the
coordinates,The potential energy of any particle on the so called
equal potential energy surface is constant,
dVZ d zY d yX d x ????
),,( zyxVV ?
The potential energy at the position M0 as a basic standard position is
defined as zero,the zero point of the potential energy,The potential
energy is a relative quantity,
2,Potential energy
In a conservative force field the work done by the conservative force
when a particle move from position M to position M0 is called the
potential energy of the particle at the position M relative to position M0,
It is denoted by V and given by,
? ???? ?? 00
M
M
M
M
Zd zYd yX d xrdFV
48
二．势能
在势力场中,质点从位置 M 运动到任选位置 M0,有势力所作的
功称为质点在位置 M 相对于位置 M0的势能，用 V 表示。
? ???? ?? 00
M
M
M
M
Zd zYd yX d xrdFV
M0作为基准位置，势能为零，称为零势能点。势能具有相对性。
dVZ d zY d yX d x ????
),,( zyxVV ? 是坐标的单值连续函数。
等势面：质点位于该面上任何地方，势能都相等。
dzzVdyyVdxxVdV ?????????,,zVZyVYxVX ?????????????
质点系的势能,
? ? ?????? io
i
M
M
iiiiiinnn dzZdyYdxXzyxzyxV )(),,,,,,( 111
49
,At point
20
__
22
0
2
,WrdFvM
M
M
??? ?
21201012 VVWWW ????
M1→ M2,
The work done by a conservative force is equal to the difference of
the potential energies between the initial and final position,
3,Work done by conservative force,
At point
10
__
11
2
1
,WrdFvM
M
M
??? ?
For one particle
For a system of particles
1) Field of gravity,
2) Field of elastic force,,
Choosing the position of a spring in its natural state as the zero
point of the potential energy we get,
3) Field of gravitational force,
Choosing the infinitely distant from the center of the
gravitation position as the zero point of the potential energy
( ) we get
phzzpV ???? )( 0
phzzpV coc ???? )(
2
2
1 ?kv ?
??0r
r
mGmv 21??
50
1.重力场
质点,
质点系,
2,弹性力场,取弹簧的自然位置为零势能点
3,万有引力场,取与引力中心相距无穷远处为零势能位置
PhzzPV ???? )( 0
hPzzPV CC ???? )( 0
221 ?kV ?
)( 0 ??r
r
mGmV 21??
有势力的功等于质点系在运动的始末位置的势能之差。
三．有势力的功
在 M1位置,
101
0
1
WrdFV
M
M
??? ? 202 0
2
WrdFV M
M
??? ?M2位置,
21201012 VVWWW ????
M1→ M2,
51
[Example 1] A straight homogeneous rod of
mass m stand in a vertical position on the
smooth surface of a table without initial
velocity,Determine the velocity of the
center of mass (expressed in the inclination
of the rod and the position of the center of
mass) if the rod falls down,
?
4,The Law of Conservation of the Mechanical Energy
Mechanical energy is defined as the arithmetical sum of the kinetic
and the potential energy of a system,
Assume a system to be subjected only to the action of conservative
forces,then, 211212 VVWTT ????
Therefore co n s t,
2211 ???? VTVT
This is the law of conservative of
mechanical energy,
Such system are called conservative systems,
For a non-conservative system,assuming the work done by non-
conservative forces is,
we have
121122 )()( WVTVT ?????
'12W
52
设质点系只受到有势力 (或同时受到不作功的非有势力 ) 作用，
则 211212 VVWTT ????
—机械能守恒定律 常量?????
2211 VTVT
对非保守系统，设非保守力的功为 W12',则有
121122 )()( WVTVT ?????
四．机械能守恒定律
机械能：系统的动能与势能的代数和 。
这样的系统成为保守系统 。
[例 1] 长为 l，质量为 m的均质直杆，初瞬
时直立于光滑的桌面上。当杆无初速度
地倾倒后，求质心的速度（用杆的倾角 ?
和质心的位置表达）。
53
Substituting into the expression and simplifying
we obtain,
?? sin2l y?? ?
ygy ??2
2
s in31
s in6
???
Solution, Because the rod is not subjected
to an external force in horizontal direction,
the center of mass will fall vertically,Again,
because the reaction force does not perform
work and the active force is a conservative
one,we may use the law of conservation of mechanical energy to find
the solution,
At initial moment,,
At any instant:,
,
From the law of conservation of mechanical energy we have
mglVT ??? 2,0 11
222222 212412121 ymmlymIT C ???? ???? ?? )2(2 ylmgV ??
? ? ????? s i n2,s i n2 i, e c o s12 a n d l ylyly ???? ?????
)2(2124120 222 ylmgymmlmgl ????? ???
54
解,由于水平方向不受外力,
且初始静止，故质心 C铅垂下降。
由于约束反力不作功,主动力为有势力,
因此可用机械能守恒定律求解。
? ? ????? s i n2,s i n2 co s12 l ylyly ????? ????? 即又
由机械能守恒定律,
)2(2124120 222 ylmgymmlmgl ????? ???
将 代入上式，化简后得 ?? sin2l y?? ?
ygy ??2
2
s in31
s in6
???
mglVT ??? 2,0 11初瞬时,
222222 212412121 ymmlymIT C ???? ???? ??
)2(2 ylmgV ??
任一瞬时,
55
§ 14-6 General theorems of
dynamics and its applications
The general theorems of dynamics include the theorems of the change in the
linear,in the angular momentum and in the kinetic energy for a particle or a
system of particles,The theorem of the change in momentum and the theorem
of moments are expressed in vector forms,while the theorem of the change in
kinetic energy is expressed in scalar forms,They all may be applied to study
mechanical motion,The theorem of the change in kinetic energy can also solve
problems concerning the transformation between the kinetic energy and other
forms of energy,
In general,the theorems of dynamics offer methods to solve dynamical
problems,The applications of the general theorems include two aspects:Ⅰ
According to the initial conditions of a problem and the quantity to be
determined,choose proper theorem and solve,including the judgment of
various conservative condition and the applications of the corresponding law
of conservation; Directly obtain the required result,eliminating the irrelevant
unknown quantities,Ⅱ For more complicated problems,according to necessity,
choose two or three theorems and combine them to solve the problem,
Moreover,during solving a problem,we should correctly perform the analysis
of motion and write down the proper complementary equations of kinematics,
56
§ 14-6 动力学普遍定理及综合应用
动力学普遍定理包括质点和质点系的动量定理、动量矩定理和
动能定理。 动量定理和动量矩定理是矢量形式，动能定理是标量形
式，他们都可应用研究机械运动，而动能定理还可以研究其它形式
的运动能量转化问题。
动力学普遍定理提供了解决动力学问题的一般方法。动力学普
遍定理的综合应用，大体上包括两方面的含义,一是能根据问题的
已知条件和待求量，选择适当的定理求解，包括各种守恒情况的判
断，相应守恒定理的应用。避开那些无关的未知量，直接求得需求
的结果。二是对比较复杂的问题，能根据需要选用两、三个定理联
合求解。
求解过程中,要正确进行运动分析,提供正确的运动学补充方程。
57
Illustration of the application of the general theorems of dynamics,
[Example 1] Two homogeneous rods AC and BC each of weight p and
length l are,attached to each other by means of a joint,they are placed
on a smooth horizontal plane,Assuming the axes of the two rods
remain in vertical plane throughout,At initial instant the system is at
rest and the height of point c is h,Determine the velocity when the
joint c reaches the ground,
58
举例说明动力学普遍定理的综合应用,
[例 1] 两根均质杆 AC和 BC各重为 P，长为 l，在 C处光滑铰接，置于
光滑水平面上；设两杆轴线始终在铅垂面内，初始静止,C点高度
为 h，求铰 C到达地面时的速度。
59
Discussion
(1)Applying the law of conservation of linear momentum plus the theorem of
the change in kinetic energy solves the problem,
(2)Make use of the kinematics of planar motion when computing the kinetic
energy,
Substituting these expressions into the theorem of the change in kinetic
energy we get,
ghvPhvgP CC 3 e w h e n c031 2 ???
PhhPW F ??? ?? 22)( 01?T
22222 3123121 ?? lgPlgPT ????
22 31 CC vgPTlv ??? ?
Because the system initially is at rest the
position of the center of mass in horizontal
direction remains at rest,
,0)(? ?exFAnalysis of forces,
Solution Because we do not have to determine the
internal force of the system the two rods do not
have to be treated separately,The object under study
is the whole system,
60
讨论 ? 动量守恒定理＋动能定理求解。
? 计算动能时，利用平面运动的运动学关系。
解,由于不求系统的内力，可以不拆开。
研究对象：整体
分析受力：,
且初始静止，所以水平方向质心位置守恒。
? ? 0)( exF
PhhPW F ??? ?? 22)( 01?T
22222 3123121 ?? lgPlgPT ????
代入动能定理,
ghvPhvgP CC 3 031 2 ????
22 31 CC vgPTlv ??? ?
61
[Example2] The circular homogeneous
disc A is of mass m and radius r,the slide
block of mass m,and the rod AB is of
negligible mass and parallel to the angle of
inclination,The coefficient of friction
is f,Determine the acceleration of the slide
block when the circular disk roll without
slipping,The system initially is at rest,
Solution Choose the whole system as the object under study
? ???? )co ss i n2( co s s i n 2)( ???? fSmgm g SfSmgW F
222221 21212121 0 ?mrmvmvTT ?????
Kinematics,
?rv ? 22 45 mvT ?From the theorem of the change in kinetic energy
we get )co ss i n2(0
4
5 2 ?? fm g Smv ???
Differentiation of the expression
with respect to time t,results in,
gfa )c o s52s i n54( ?? ??
?
62
[例 2] 均质圆盘 A,m,r；滑块 B：
m；杆 AB：质量不 计，平行于斜
面。斜面倾角 ?，摩擦系数 f，圆盘
作纯滚动，系统初始静止。求：
滑块的加速度。
解：选系统为研究对象
? ???? )co ss i n2( co s s i n 2)( ???? fSmgm g SfSmgW F
222221 21212121 0 ?mrmvmvTT ?????
运动学关系,?rv?
22 45 mvT ??
由动能定理,
)co ss i n2(045 2 ?? fm g Smv ???
对 ｔ 求导，得
gfa )c o s52s i n54( ?? ??
63
[Example 3]A circular homogeneous disc of weight 150N and a
homogeneous rod of weight 60 N and length 24cm are joined at the
point B,The system is released without initial velocity from the
position shown in the diagram,
Determine the velocity of the point B and the
reaction force of the constraint at mount A when the
system passes through its lowest position,
Solution,
(1) Take the disc as the object under study,; 0)(? ?Fm B 0 0 ??? BBBI ??
00 ?? ?? B,the disc is in
translational motion,
64
[例 3] 重 150N的均质圆盘与重 60N、长 24cm的均质杆 AB在 B处用
铰链连接。 系统由图示位置无初速地释放。 求 系统经过最低位
置 B'点时的速度及支座 A的约束反力。
解,（ 1）取圆盘为研究对象; 0)(? ?Fm B 0 0 ??? BBBI ??
00 ?? ?? B,圆盘平动。
65
(2)Applying the theorem of the change in kinetic energy determine
the velocity of point B,
Considerate the whole system,
At the initial moment,
At the lowest position
,
01 ?T
2222 2121 BA vgGIT ??? ?
2212221 6 3213121 BBB vg GGvgGvgG ??? ?????
)30s i n)(2()30s i n()30s i n22( 2121)( ??????????? llGGllGllGW F
??? )(12 FWTT )30s in)(
2(06
3
2
1221 ??????
? llG
Gv
g
GG
B
Substituting the known values into this expression we get
, m /s 58.1
' ?Bv
66
（ 2）用动能定理求速度 。
取系统研究。初始时 T1=0,
最低位置时,
2222 2121 BA vgGIT ??? ?
2212221 6 3213121 BBB vg GGvgGvgG ??? ?????
)30s i n)(2()30s i n()30s i n22( 2121)( ??????????? llGGllGllGW F
??? )(12 FWTT
)30s in)(2(06 3 21221 ?????? ? llGGvg GG B
代入数据，得 m /s 58.1' ?Bv
67
Substituting the known values
into this expression we get, N4 0 1,0 ??
AA YX; 0'21? ???? ABcixi XagGagGam ?? ? ????? 212221 2 GGYlgGlgGam Aiyi ??
(4) Calculate the reaction force at mount A from the theorem of the
motion of the center of mass,
Investigate the whole system.,
(3) Applying the theorem of the change in angular momentum
determine the angular acceleration, ?
?? )31(31 2221221 lgGlgGvlgGlgGL A ???? ? ?? 0)( )( eAA FmdtdL
?＝ 0,
?
?
)0( 2 2 ???? ?? CnCC alaa
The acceleration of the center of mass of the rod is
The acceleration of the center of the disc is
ca
'Ba
)0( 2' ???? ?? ?? BnBB alaa
r ad / s 58.624.0 58.1' ??? lv B?
68
（ 3）用动量矩定理求杆的角加速度 ? 。
?? )31(31 2221221 lgGlgGvlgGlgGL A ????
由于 ? ?? 0)( )( e
AA FmdtdL
所以 ?＝ 0 。
杆质心 C的加速度,
盘质心加速度,
)0( 2 2 ???? ?? CnCC alaa
)0( 2' ???? ?? ?? BnBB alaa
r ad / s 58.624.0 58.1' ??? lv B?
（ 4）由质心运动定理求支座反力。 研究整个系统。; 0'21? ???? ABcixi XagGagGam ??
代入数据，得 N4 0 1,0 ??
AA YX
? ????? 212221 2 GGYlgGlgGam Aiyi ??
69
(1) Using the theorem of the conservation of the total angular
momentum with respect to the center of mass of a system plus
the theorem of the change in kinetic energy plus the theorem
of the change in angular momentum plus the theorem of
motion of center of mass,
(2) we may calculate by differentiation of the expression of the
theorem of the change in kinetic energy in integral form with
respect to time t,but we can not analyze the motion of the rod
AB at an arbitrary position,
?
70
(1) 相对质心动量矩守恒定理 +动能定理 +动量矩定理 +质心
运动定理。
(2) 可用对积分形式的动能定理求导计算 ?，但要注意需取
杆 AB在一般位置进行分析 。
71
?mLmvK C 61??
?? ])6(121[ 22 LmmLIL OO ???
?291 mL?
222 18121 ?? mLIT O ??
?223 mRL O ?
2243 ?mRT ?
mvK ?
?221 mRL C ?
222 4121 ?mRmvT ??
?mRK ?
[Example 4] Calculate the basic quantities (linear momentum,
angular momentum and kinetic energy) for the following systems,
72
?mLmvK C 61??
?? ])6(121[ 22 LmmLIL OO ???
?291 mL?
222 18121 ?? mLIT O ??
?223 mRL O ?
2243 ?mRT ?
?mRK ? mvK ?
?221 mRL C ?
222 4121 ?mRmvT ??
[例 4] 基本量计算 (动量,动量矩,动能 )
73
[Example 5] A rod of mass m is supported on two cylindrical
rollers of radius r and mass m/2 each,The rollers are placed on a
horizontal plane, Determine the acceleration of the rod when it
is subjected to a horizontal force F taking into account friction of
contact but no relative slipping,
Solution,
(1) Solve this problem applying the theorem of the change in
kinetic energy,
Choose the system as the object under study,The rod is in
translational motion and the cylindrical rollers in plane motion,
Assuming that the instantaneous velocity of the rod is v,the
velocity of the centers of mass of the rollers is v/2,and the
angular velocity of the rollers is, rv2??
74
[例５ ] 质量为 m 的杆置于两个半径为 r,质量为 的实心
圆柱上，圆柱放在水平面上，求当杆上加水平力 时，
杆的加速度。设接触处都有摩擦，而无相对滑动。
2m
P
解, (1)用动能定理求解。
取系统为研究对象，杆作平动，圆柱体作平面运动。
设任一瞬时，杆的速度为 v，则圆柱体质心速度为 v/2，
角速度, rv2??
75
From the theorem of the change in kinetic energy in differential
form we have
dSPmvd ??)1611( 2?? )( FWdT ?
Dividing both side by d t,
and differentiating we get
vPavm ???? 21611,118 mPa ??
22222 1611])2)(221(21)2(221[221 mvrvrmvmmvT ?????
The sum of the elementary work done by the active force is
,)(? ?? dSPW F?
Then the kinetic energy of the system is
,
76
系统的动能 22222
1611])2)(221(21)2(221[221 mvrvrmvmmvT ?????
主动力的元功之和,? ?? dSPW F )(?
由动能定理的微分形式,
?? )( FWdT ? dSPmvd ??)1611( 2
两边除以,并求导数，得 dt
vPavm ???? 21611 mPa 118 ??
77
(2) Applying the theorem of the change in angular
momentum solve this problem,
Choose the whole system as the object under study,
m v rrvrmrvmrmvL O 411)222122(22 2 ??????? ? ?? rPFm eO 2)( )(
According the principle of the moments we get
?? )( FmdtdL OO
and hence,
rPm v rdtd 2)411( ??
m
P
a
rPm r a
11
8
t h e r e f or e,
a n d,2
4
11
f ol low sI t
?
??
78
(2) 用动量矩定理求解
取系统为研究对象
m v rrvrmrvmrmvL O 411)222122(22 2 ??????? ? ?? rPFm eO 2)( )(
根据动量矩定理：,得 ?? )( Fm
dtdL OO
rPm v rdtd 2)411( ??
mParPm r a 118 2411 ????
79
[Example 6] A homogeneous rod OA has weight p and length l,
Determine the angular acceleration of the rod and the reaction force at
the position o at a moment at which the string suddenly breaks,
Solution,Choose the rod as the object
under study,According to the theorem
of moments we have
231 2 lPlgP ???? lg 2/3 H e n c e ??
According to the theorem of motion of the center of mass
we get,
OCx Xag
P ?? 0
PYPYlgPagP OOCy 41 w h e n c e2 ????? ?
80
解,取杆为研究对象
231 2 lPlgP ???? lg 2/3??
由质心运动定理,
OCx Xag
P ?? 0
PYPYlgPagP OOCy 41 2 ?????? ?
[例 6] 均质杆 OA，重 P，长 l，绳子突然剪断。 求 该瞬时，角加
速度及 O处反力。
由动量矩定理,
81
82
