1
Theoretical Mechanics
2
3
Introduction
1,The subject of dynamics is the relation between the mechanical
motion of a material body and the force acting on it,
2,Two models of mechanics,
1) A particle is a material body possessing mass,the shape and size of
which can be neglected in investigating its motion,
Examples,When we study a satellite’s orbit the satellite may be
regarded as a particle,I.e,satellite particle,
A rigid body in translation motion
can always be treated as a particle,
I.e,rigid body particle,
2) A system of particles is a collection of material points in which the
position and motion of each particle of the system depends on the
position and motion of all the other particles,
A rigid body is a special system of particles composed of a large
number of particles,the distance between any two points of which
remains the same during the whole motion,It is also called to be a
non-deformable system,
4
引 言
一,研究对象,
二,力学模型,
研究物体的机械运动与作用力之间的关系
2.质点系,由有限或无限个有着一定联系
的质点组成的系统。
1.质点,具有一定质量而不考虑其形状大小的物体。
刚体 是一个特殊的质点系，由无数个相互间保持距离
不变的质点组成。又称为不变质点系。
例如,研究卫星的轨道时，卫星 质点；
刚体作平动时,刚体 质点。
5
System of free particles,a system in which the motion of each
particle is not restricted by any constraints,
System of non-free particles refers to a system in which the
motion of particle is restricted by constraints,
A system of particles is the most general abstract model of
mechanics including rigid bodies,elastic bodies and fluids,
3,Classification of dynamics,
Particles dynamics Particle dynamics is the basis of the
Dynamics of a system of particles dynamics of a system of particles
4,The fundamental problems of dynamics,
There are two cases in general,One is to determine the acting forces
when the condition of motion is known,the other is to determine
the motion of the body when the acting forces are known,
There are still some additional comprehensive problems,Sometimes
part of the forces and part of the motion are known,we have to
determine the other part,In some cases,the positive forces are
known,we should determine the motion firstly,then determine the
reaction forces of constraints,
??
?
6
自由质点系,质点系中各质点的运动不受约束的限制。
非自由质点系,质点系中的质点的运动受到约束的限制。
质点系是力学中最普遍的抽象化模型；包括刚体,弹性体,流体。
三,动力学分类, 质点动力学
质点系动力学 ???
质点动力学是质点
系动力学的基础。
四,动力学的基本问题,大体上可分为两类,
第一类：已知物体的运动情况，求作用力；
第二类：已知物体的受力情况，求物体的运动。
综合性问题：已知部分力，部分运动求另一部分力、部分运动。
已知主动力，求运动，再由运动求约束反力。
7
Theoretical Mechanics
8
9
§ 11–1 The differential equations of
motion of a particle
§ 11–2 The two types of problems of
particle dynamics
Chapter 11,Differential equations of
motion of a particle
10
§ 11–1 质点运动微分方程的形式
§ 11–2 质点动力学两类问题
第十一章 质点运动微分方程
11
Chapter11,Differential equations of motion of a particle
§ 11-1 The differential equations of motion of a particle
The basic equation of dynamics represented by
equations in differential forms are called the differential
equations of motion of a particle,
)( Fam ?
1.Vector form (In this formula,
describes the equation of motion of a particle in terms of its
radius vector),
2
2
Fdt rdm ?
)( trr ?
?
?
?
?
?
?
?
?
?
?
?
?
Z
dt
yd
m
Y
dt
yd
m
X
dt
xd
m
2
2
2
2
2
2
2,Forms in rectangular coordinates,
(In these formulae,describe the motion of
a particle in terms of rectangular coordinates),
)(
)(
)(
??
?
?
?
?
?
?
tzz
tyy
txx
12
将动力学基本方程 表示为微分形式的方程，称为
质点的运动微分方程。
)( Fam ?
1.矢量形式
) )( ( 22 方程为质点矢径形式的运动式中 trrFdt rdm ??
第十一章 质点运动微分方程
§ 11-1 质点运动微分方程的形式
)
)(
)(
)(
(
2
2
2
2
2
2
运动方程为质点直角坐标形式的式中
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
tzz
tyy
txx
Z
dt
yd
m
Y
dt
yd
m
X
dt
xd
m
2.直角坐标形式
13
Besides these three forms,differential equations of the motion of a
particle may be represented in the forms of polar and other coordinates,
Applying the equations we can solve two types of problems of
particle dynamics,
3.Natural form
b
n
F
F
v
m
F
dt
sd
m
?
?
?
0
2
2
2
?
? In these formulae,s=s(t) describes the motion
of a particle in terms of the arc coordinate
of which F?,Fn and Fb are respectively the
projections of force F on axes ?,n and b,
14
3.自然形式
),
,,
)((
轴上的投影轴和轴自然轴系
在分别为力运动方程。
为质点的弧坐标形式的式中
bn
FFFF
tss
bn
?
?
?
质点运动微分方程除以上三种基本形式外,还可有极坐标形式,
柱坐标形式等等。
应用质点运动微分方程，可以求解质点动力学的两类问题。
b
n
F
F
v
m
F
dt
sd
m
?
?
?
0
2
2
2
?
?
15
1,The first is,knowing the motion of a particle determine the
force acting on it,The steps and the important points of
solving the problem are,
§ 11-2 The two types of problems of particle dynamics
1)Choose the object under study correctly (generally choose
the particle which relates the known quantity with the
quantity to be determined),
2)Conduct the analysis of forces correctly,drawing all the
forces acting on the particle ( we should analyze the moving
particle in an arbitrary position),
3)Conduct the analysis of motion correctly,analyzing the
characteristic quantities of motion of the particle,
4)Choose and write down the proper differential equations of
motion of the particle (establish a coordinate system),
5)Solve for the unknown quantities,
16
1.第一类, 已知质点的运动，求作用在质点上的力（微分问题）
§ 11-2 质点动力学两类问题
解题步骤和要点,
①正确选择研究对象 （一般选择联系已知量和待求量的质点）。
②正确进行受力分析,画出受力图 (应在一般位置上进行分析 )。
③ 正确进行运动分析 （分析质点运动的特征量）。
④ 选择并列出适当形式的质点运动微分方程 （建立坐标系）。
⑤求解未知量 。
17
0v
[Example 1] A bridge crane suspending a load whose weight is G
moves uniformly,Its velocity is,The distance from the center of
the load to the point of suspension is l,If the crane suddenly
breaks the load will swing forward due to inertia,Determine the
maximum tension in the steel cable,
Solution,
① Choose the load as
the object to be investigated,
② The analysis of force is shown in
the diagram,
③ Analysis of motion,the load
swings along a circular are with
center O and radius l,
18
0v
[例 1] 桥式起重机跑车吊挂一重为 G的重物，沿水平横梁作匀速
运动，速度为,重物中心至悬挂点距离为 L。突然刹车，重物
因惯性绕悬挂点 O向前摆动,求 钢丝绳的最大拉力。
解, ① 选重物 (抽象为质点 )为研究对象
② 受力分析如图所示
③ 运动分析，沿以 O为圆心,
L为半径的圆弧摆动。
19
????? ? 1 s i n,??? GdtdvgGFma
????? ? 2 c o s,2 ?GTlvgGFma nn
④ Write the differential equations of
motion in the natural form,
[Notes]
① The methods to decrease the tension in the rope are decreasing
the velocity of the crane or increasing the length of the rope,
② The tension Tmax consists of two parts,One,equals the weight of
the load,is called static tension,the other which is the cause by the
acceleration is called complementary dynamic tension,Tmax is
called dynamic tension,
According to equation<1> the load is in retarded motion,Therefore at
,
)1( 20m a x glvGT ??
),( c o so b t a i n we2F r o m
2
gl
vGT ???? ?
0??
⑤ Solve the unknown quantities,
.q u a n t i t i e s v a r i a b l ea r e a n d W h e r e v?
ma x TT ?
20
????? ? 1 s i n,??? GdtdvgGFma
????? ? 2 c o s,2 ?GTlvgGFma nn
④ 列出自然形式的质点运动微方程
.,
,)( c o s 2
2
为变量其中
式得由
v
gl
vGT
?
? ????
m a x,0,1 TT ???? 时因此重物作减速运动式知由 ?
)1( 20m a x glvGT ??
⑤ 求解未知量
[注 ]① 减小绳子拉力途径：减小跑车速度或者增加绳子长度。
② 拉力 Tmax由两部分组成,一部分等于物体重量,称为静拉力
一部分由加速度引起，称为附加动拉力。全部拉力称为动拉力。
21
2.The second type is,knowing the forces acting on a particle,
determine the motion of the particle (problem of integration),
The known acting forces are possibly constant forces,If they are
not constant they may depend only on time or only on position or only
on velocity or on all these quantities,
The steps of problem solution are the following,
① Correctly choose the object to be investigated,
② Correctly perform the analysis of forces and draw the force
diagram including all applied forces judging the natures of the
Forces,(Analyze the applied forces on the moving material point
in arbitrary position and establish the expressions of variable
forces),
③ Analyze the motion Correctly (We should confirm the initial
conditions of motion as well as analyzing the features of motion
of the particle),
22
2.第二类：已知作用在质点上的力，求质点的运动（积分问题）
已知的作用力可能是常力,也可能是变力 。 变力可能是时间,
位置, 速度或者同时是上述几种变量的函数。
解题步骤如下,
①正确选择研究对象 。
②正确进行受力分析，画出受力图 。判断力是什么性质的力
（应放在一般位置上进行分析，对变力建立力的表达式）。
③正确进行运动分析。 (除应分析质点的运动特征外，还要确定
出其运动初始条件） 。
23
④ Choose and write down the appropriate differential equations
of motion of the particle,
⑤ Solve for the unknown quantities,
The integration is carried out according to the form of the force
function,Employing the initial conditions determine the motion of
the particle,
dt
dv
dt
dv
When the forces are constant quantities or depend on time and
velocity only the equations of motion can be integrated by the
method of separating,
If the force depends on the position it is necessary to transform the
variable into, After that we may carry out the integration be
the method of separating the variables,ds
dvv
24
④ 选择并列出适当的质点运动微分方程。
如力是常量或是时间及速度函数时,
可直接分离变量 。 积分
dt
dv
⑤ 求解未知量。 应根据力的函数形式决定如何积分，并利用运
动的初始条件，求出质点的运动。
再分离变量积分。,dsdvvdtdv ?
如力是位置的函数，需进行变量置换
25
Write the differential equations of motion of a particle in terms of Cartesian
rectangular coordinates and perform the integration
differential equations ? the first integration ? the second integration
??
???
????
??
?
42
2
31
2
1 ctcgty
ctcx
?
?
?
?
?
???
?
?
?
?
?
??
?
?
??
?
2
10
cgtdtdy
cdtdx
mgdt
dv
m
dt
dvm
y
x
[Example 2.]A stuffing mechanism is employed
to stuff the material in a coal mine,The stuffing
material has to be thrown to the position A of
the roof board at distances S=5 meter and
H=1.5 meter as shown in the figure,Determine
1.the initial velocity of the stuffing material,
2.the angle a0 that the initial velocity makes
with the horizon,0v
0v
,s i n,c o s;0,0,0 00000000 ?? vvvvyxt yx ?????
yx vvHySxAM,,,,???
Solution:This problem belongs to the second type of problems,the known force
is a constant quantity,Choose the stuffing material M to be investigated,The
applied force is shown in the diagram M being the projectile,At
where and is to be determined,At the time t, 0? 0v
26
待求0000000000,,s i n,c o s;0,0,0 ??? vvvvvyxt yx ?????
yx vvHySxAMt,,,,,???瞬时
列直角坐标形式的质点运动微分方程并对其积分运算
微分方程 积分一次 再积分一次
解, 属于已知力为常量的第二类问题。
选择填充材料 M为研究对象，受力如图所示,M作斜抛运动。
[例 2] 煤矿用填充机进行填充,为保证充
填材料抛到距离为 S=5米,H=1.5米的顶
板 A处。 求 (1)充填材料需有多大的初速
度 v0? (2)初速 与水平的夹角 a0？
0v
??
?
?
?
????
??
?
42
2
31
2
1
ctcgty
ctcx
?
?
?
?
?
???
?
?
?
?
?
??
?
?
??
?
2
10
cgt
dt
dy
c
dt
dx
mg
dt
dv
m
dt
dv
m
y
x
27
Because the projection of the acceleration at the highest point on the axis
y is zero we obtain,
Substituting the time of flight from 0 to A,x=s and y=H into the
equations of motion we obtain,
,0s i n 00 ??? gtvdtdy ? gvt 00 sin ??
gH
sgv
2c o s 00 ??
gHv 2s i n 00 ??
Hence the initial velocity v0 and the initial angle are
,
,
0?
m / s 5.1022)s i n()c o s( 222002000 ????? gHgH sgvvv ??
???? ?? 31 2tgc o ss intg 1
00
0010 sHvv ???
0,s i n,co s 43002001 ???? ccvcvc ??
Substituting the initial conditions into these expression we obtain
2
0000 2
1s in,c o s gttvytvx ??? ??
0
22
0
2
0
0 c o s2
1tg
?? v
xgxy ??
The equation of motion are,
The path is given by
28
0,s i n,c o s, 43002001 ???? ccvcvc ??代入初始条件得
20000 21s i n,co s, gttvytvx ??? ??则运动方程为
0
22
0
2
0
0 c o s2
1tg,
?? v
xgxy ??则轨迹方程为
代入最高点 A处值，得,即
将到达 A点时的时间 t,x=S,y=H 代入运动方程，得
,0s i n 00 ??? gtvdtdy ? gvt 00 sin ??
gH
sgv
2c o s 00 ??
gHv 2s i n 00 ??
发射初速度大小与初发射角 为 0?
m / s 5.1022)s i n()c o s( 222002000 ????? gHgH sgvvv ??
???? ?? 31 2tgc o ss intg 1
00
0010 sHvv ???
29
[Example 3] Projecting a rocket determine the
minimum velocity for the rocket to escape the earth,
Solution,This problem belongs to the second type
of problems,the known force depends on position,
Choose the rocket as the object to be investigated
and the coordinate axis as shown in the figure,The
rocket is subjected to the action of the earth’s
gravitation at arbitrary position X,
hence,
Write down the differential equation of
motion of the particle
2
2
22 x
m g RF
R
mMfmg
x
mMfF ????? ?
2
2
2
2
xm g Rdtdxm ??
Then the velocity at an arbitrary position is
),,0w h e n ( 02
2
0
vvRxtdxxm g RRxdvmvvv xxx ?????? ??
Thus )(
2
2
2
2
dx
dvv
dtdxdx
dv
dt
dv
dt xdx
m g R
dx
dvmv xxxxx
x ??????
30
[例 3] 发射火箭,求 脱离地球引力的最小速度。
解, 属于已知力是位置的函数的第二类问题。
取火箭 (质点 )为研究对象,建立坐标如图示。
火箭在任意位置 x 处受地球引力 F 的作用。
2
2
22 x
m g RF
R
mMfmg
x
mMfF ????? ?
2
2
2
2
x
m g R
dtdxm ??建立质点运动微分方程
则在任意位置时的速度
),0( 02 2
0
vvRxtdxxm g RRxdvmvvv xxx ?????? ?? 时
即,)(
2
2
2
2
dx
dvv
dtdxdx
dv
dt
dv
dt xdx
m g R
dx
dvmv xxxxx
x ??????
31
It is obviously that v decreases with x,If the velocity
becomes zero at a certain position x=R+H and the rocket falls back
to earth,if,no matter how large (even infinite) the x is,the
rocket is not back,The minimum velocity for the rocket to escape
the earth’s gravitational force and to proceed infinitely away from
the earth ( ) is,
(the second escape velocity)
gRv 220 ?
gRv 220 ?
??x
k m / s )( 2.116 3 7 0108.922 30 ?????? ?gRv
x
gRgRvv 2
02
2)2( ???
32
可见,v 随着 x 的增加而减小。若 则在某一位置
x=R+H 时速度将减小到零，火箭回落。若 时，无论 x
多大（甚至为 ∞）,火箭也不会回落。因此脱离地球引力而一
去不返 时（ ）的最小初速度
gRv 220 ?
gRv 220 ?
??x
k m / s )( 2.116 3 7 0108.922 30 ?????? ?gRv （第二宇宙速度）
x
gRgRvv 2
02
2)2( ???
33
34
Theoretical Mechanics
2
3
Introduction
1,The subject of dynamics is the relation between the mechanical
motion of a material body and the force acting on it,
2,Two models of mechanics,
1) A particle is a material body possessing mass,the shape and size of
which can be neglected in investigating its motion,
Examples,When we study a satellite’s orbit the satellite may be
regarded as a particle,I.e,satellite particle,
A rigid body in translation motion
can always be treated as a particle,
I.e,rigid body particle,
2) A system of particles is a collection of material points in which the
position and motion of each particle of the system depends on the
position and motion of all the other particles,
A rigid body is a special system of particles composed of a large
number of particles,the distance between any two points of which
remains the same during the whole motion,It is also called to be a
non-deformable system,
4
引 言
一,研究对象,
二,力学模型,
研究物体的机械运动与作用力之间的关系
2.质点系,由有限或无限个有着一定联系
的质点组成的系统。
1.质点,具有一定质量而不考虑其形状大小的物体。
刚体 是一个特殊的质点系，由无数个相互间保持距离
不变的质点组成。又称为不变质点系。
例如,研究卫星的轨道时，卫星 质点；
刚体作平动时,刚体 质点。
5
System of free particles,a system in which the motion of each
particle is not restricted by any constraints,
System of non-free particles refers to a system in which the
motion of particle is restricted by constraints,
A system of particles is the most general abstract model of
mechanics including rigid bodies,elastic bodies and fluids,
3,Classification of dynamics,
Particles dynamics Particle dynamics is the basis of the
Dynamics of a system of particles dynamics of a system of particles
4,The fundamental problems of dynamics,
There are two cases in general,One is to determine the acting forces
when the condition of motion is known,the other is to determine
the motion of the body when the acting forces are known,
There are still some additional comprehensive problems,Sometimes
part of the forces and part of the motion are known,we have to
determine the other part,In some cases,the positive forces are
known,we should determine the motion firstly,then determine the
reaction forces of constraints,
??
?
6
自由质点系,质点系中各质点的运动不受约束的限制。
非自由质点系,质点系中的质点的运动受到约束的限制。
质点系是力学中最普遍的抽象化模型；包括刚体,弹性体,流体。
三,动力学分类, 质点动力学
质点系动力学 ???
质点动力学是质点
系动力学的基础。
四,动力学的基本问题,大体上可分为两类,
第一类：已知物体的运动情况，求作用力；
第二类：已知物体的受力情况，求物体的运动。
综合性问题：已知部分力，部分运动求另一部分力、部分运动。
已知主动力，求运动，再由运动求约束反力。
7
Theoretical Mechanics
8
9
§ 11–1 The differential equations of
motion of a particle
§ 11–2 The two types of problems of
particle dynamics
Chapter 11,Differential equations of
motion of a particle
10
§ 11–1 质点运动微分方程的形式
§ 11–2 质点动力学两类问题
第十一章 质点运动微分方程
11
Chapter11,Differential equations of motion of a particle
§ 11-1 The differential equations of motion of a particle
The basic equation of dynamics represented by
equations in differential forms are called the differential
equations of motion of a particle,
)( Fam ?
1.Vector form (In this formula,
describes the equation of motion of a particle in terms of its
radius vector),
2
2
Fdt rdm ?
)( trr ?
?
?
?
?
?
?
?
?
?
?
?
?
Z
dt
yd
m
Y
dt
yd
m
X
dt
xd
m
2
2
2
2
2
2
2,Forms in rectangular coordinates,
(In these formulae,describe the motion of
a particle in terms of rectangular coordinates),
)(
)(
)(
??
?
?
?
?
?
?
tzz
tyy
txx
12
将动力学基本方程 表示为微分形式的方程，称为
质点的运动微分方程。
)( Fam ?
1.矢量形式
) )( ( 22 方程为质点矢径形式的运动式中 trrFdt rdm ??
第十一章 质点运动微分方程
§ 11-1 质点运动微分方程的形式
)
)(
)(
)(
(
2
2
2
2
2
2
运动方程为质点直角坐标形式的式中
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
tzz
tyy
txx
Z
dt
yd
m
Y
dt
yd
m
X
dt
xd
m
2.直角坐标形式
13
Besides these three forms,differential equations of the motion of a
particle may be represented in the forms of polar and other coordinates,
Applying the equations we can solve two types of problems of
particle dynamics,
3.Natural form
b
n
F
F
v
m
F
dt
sd
m
?
?
?
0
2
2
2
?
? In these formulae,s=s(t) describes the motion
of a particle in terms of the arc coordinate
of which F?,Fn and Fb are respectively the
projections of force F on axes ?,n and b,
14
3.自然形式
),
,,
)((
轴上的投影轴和轴自然轴系
在分别为力运动方程。
为质点的弧坐标形式的式中
bn
FFFF
tss
bn
?
?
?
质点运动微分方程除以上三种基本形式外,还可有极坐标形式,
柱坐标形式等等。
应用质点运动微分方程，可以求解质点动力学的两类问题。
b
n
F
F
v
m
F
dt
sd
m
?
?
?
0
2
2
2
?
?
15
1,The first is,knowing the motion of a particle determine the
force acting on it,The steps and the important points of
solving the problem are,
§ 11-2 The two types of problems of particle dynamics
1)Choose the object under study correctly (generally choose
the particle which relates the known quantity with the
quantity to be determined),
2)Conduct the analysis of forces correctly,drawing all the
forces acting on the particle ( we should analyze the moving
particle in an arbitrary position),
3)Conduct the analysis of motion correctly,analyzing the
characteristic quantities of motion of the particle,
4)Choose and write down the proper differential equations of
motion of the particle (establish a coordinate system),
5)Solve for the unknown quantities,
16
1.第一类, 已知质点的运动，求作用在质点上的力（微分问题）
§ 11-2 质点动力学两类问题
解题步骤和要点,
①正确选择研究对象 （一般选择联系已知量和待求量的质点）。
②正确进行受力分析,画出受力图 (应在一般位置上进行分析 )。
③ 正确进行运动分析 （分析质点运动的特征量）。
④ 选择并列出适当形式的质点运动微分方程 （建立坐标系）。
⑤求解未知量 。
17
0v
[Example 1] A bridge crane suspending a load whose weight is G
moves uniformly,Its velocity is,The distance from the center of
the load to the point of suspension is l,If the crane suddenly
breaks the load will swing forward due to inertia,Determine the
maximum tension in the steel cable,
Solution,
① Choose the load as
the object to be investigated,
② The analysis of force is shown in
the diagram,
③ Analysis of motion,the load
swings along a circular are with
center O and radius l,
18
0v
[例 1] 桥式起重机跑车吊挂一重为 G的重物，沿水平横梁作匀速
运动，速度为,重物中心至悬挂点距离为 L。突然刹车，重物
因惯性绕悬挂点 O向前摆动,求 钢丝绳的最大拉力。
解, ① 选重物 (抽象为质点 )为研究对象
② 受力分析如图所示
③ 运动分析，沿以 O为圆心,
L为半径的圆弧摆动。
19
????? ? 1 s i n,??? GdtdvgGFma
????? ? 2 c o s,2 ?GTlvgGFma nn
④ Write the differential equations of
motion in the natural form,
[Notes]
① The methods to decrease the tension in the rope are decreasing
the velocity of the crane or increasing the length of the rope,
② The tension Tmax consists of two parts,One,equals the weight of
the load,is called static tension,the other which is the cause by the
acceleration is called complementary dynamic tension,Tmax is
called dynamic tension,
According to equation<1> the load is in retarded motion,Therefore at
,
)1( 20m a x glvGT ??
),( c o so b t a i n we2F r o m
2
gl
vGT ???? ?
0??
⑤ Solve the unknown quantities,
.q u a n t i t i e s v a r i a b l ea r e a n d W h e r e v?
ma x TT ?
20
????? ? 1 s i n,??? GdtdvgGFma
????? ? 2 c o s,2 ?GTlvgGFma nn
④ 列出自然形式的质点运动微方程
.,
,)( c o s 2
2
为变量其中
式得由
v
gl
vGT
?
? ????
m a x,0,1 TT ???? 时因此重物作减速运动式知由 ?
)1( 20m a x glvGT ??
⑤ 求解未知量
[注 ]① 减小绳子拉力途径：减小跑车速度或者增加绳子长度。
② 拉力 Tmax由两部分组成,一部分等于物体重量,称为静拉力
一部分由加速度引起，称为附加动拉力。全部拉力称为动拉力。
21
2.The second type is,knowing the forces acting on a particle,
determine the motion of the particle (problem of integration),
The known acting forces are possibly constant forces,If they are
not constant they may depend only on time or only on position or only
on velocity or on all these quantities,
The steps of problem solution are the following,
① Correctly choose the object to be investigated,
② Correctly perform the analysis of forces and draw the force
diagram including all applied forces judging the natures of the
Forces,(Analyze the applied forces on the moving material point
in arbitrary position and establish the expressions of variable
forces),
③ Analyze the motion Correctly (We should confirm the initial
conditions of motion as well as analyzing the features of motion
of the particle),
22
2.第二类：已知作用在质点上的力，求质点的运动（积分问题）
已知的作用力可能是常力,也可能是变力 。 变力可能是时间,
位置, 速度或者同时是上述几种变量的函数。
解题步骤如下,
①正确选择研究对象 。
②正确进行受力分析，画出受力图 。判断力是什么性质的力
（应放在一般位置上进行分析，对变力建立力的表达式）。
③正确进行运动分析。 (除应分析质点的运动特征外，还要确定
出其运动初始条件） 。
23
④ Choose and write down the appropriate differential equations
of motion of the particle,
⑤ Solve for the unknown quantities,
The integration is carried out according to the form of the force
function,Employing the initial conditions determine the motion of
the particle,
dt
dv
dt
dv
When the forces are constant quantities or depend on time and
velocity only the equations of motion can be integrated by the
method of separating,
If the force depends on the position it is necessary to transform the
variable into, After that we may carry out the integration be
the method of separating the variables,ds
dvv
24
④ 选择并列出适当的质点运动微分方程。
如力是常量或是时间及速度函数时,
可直接分离变量 。 积分
dt
dv
⑤ 求解未知量。 应根据力的函数形式决定如何积分，并利用运
动的初始条件，求出质点的运动。
再分离变量积分。,dsdvvdtdv ?
如力是位置的函数，需进行变量置换
25
Write the differential equations of motion of a particle in terms of Cartesian
rectangular coordinates and perform the integration
differential equations ? the first integration ? the second integration
??
???
????
??
?
42
2
31
2
1 ctcgty
ctcx
?
?
?
?
?
???
?
?
?
?
?
??
?
?
??
?
2
10
cgtdtdy
cdtdx
mgdt
dv
m
dt
dvm
y
x
[Example 2.]A stuffing mechanism is employed
to stuff the material in a coal mine,The stuffing
material has to be thrown to the position A of
the roof board at distances S=5 meter and
H=1.5 meter as shown in the figure,Determine
1.the initial velocity of the stuffing material,
2.the angle a0 that the initial velocity makes
with the horizon,0v
0v
,s i n,c o s;0,0,0 00000000 ?? vvvvyxt yx ?????
yx vvHySxAM,,,,???
Solution:This problem belongs to the second type of problems,the known force
is a constant quantity,Choose the stuffing material M to be investigated,The
applied force is shown in the diagram M being the projectile,At
where and is to be determined,At the time t, 0? 0v
26
待求0000000000,,s i n,c o s;0,0,0 ??? vvvvvyxt yx ?????
yx vvHySxAMt,,,,,???瞬时
列直角坐标形式的质点运动微分方程并对其积分运算
微分方程 积分一次 再积分一次
解, 属于已知力为常量的第二类问题。
选择填充材料 M为研究对象，受力如图所示,M作斜抛运动。
[例 2] 煤矿用填充机进行填充,为保证充
填材料抛到距离为 S=5米,H=1.5米的顶
板 A处。 求 (1)充填材料需有多大的初速
度 v0? (2)初速 与水平的夹角 a0？
0v
??
?
?
?
????
??
?
42
2
31
2
1
ctcgty
ctcx
?
?
?
?
?
???
?
?
?
?
?
??
?
?
??
?
2
10
cgt
dt
dy
c
dt
dx
mg
dt
dv
m
dt
dv
m
y
x
27
Because the projection of the acceleration at the highest point on the axis
y is zero we obtain,
Substituting the time of flight from 0 to A,x=s and y=H into the
equations of motion we obtain,
,0s i n 00 ??? gtvdtdy ? gvt 00 sin ??
gH
sgv
2c o s 00 ??
gHv 2s i n 00 ??
Hence the initial velocity v0 and the initial angle are
,
,
0?
m / s 5.1022)s i n()c o s( 222002000 ????? gHgH sgvvv ??
???? ?? 31 2tgc o ss intg 1
00
0010 sHvv ???
0,s i n,co s 43002001 ???? ccvcvc ??
Substituting the initial conditions into these expression we obtain
2
0000 2
1s in,c o s gttvytvx ??? ??
0
22
0
2
0
0 c o s2
1tg
?? v
xgxy ??
The equation of motion are,
The path is given by
28
0,s i n,c o s, 43002001 ???? ccvcvc ??代入初始条件得
20000 21s i n,co s, gttvytvx ??? ??则运动方程为
0
22
0
2
0
0 c o s2
1tg,
?? v
xgxy ??则轨迹方程为
代入最高点 A处值，得,即
将到达 A点时的时间 t,x=S,y=H 代入运动方程，得
,0s i n 00 ??? gtvdtdy ? gvt 00 sin ??
gH
sgv
2c o s 00 ??
gHv 2s i n 00 ??
发射初速度大小与初发射角 为 0?
m / s 5.1022)s i n()c o s( 222002000 ????? gHgH sgvvv ??
???? ?? 31 2tgc o ss intg 1
00
0010 sHvv ???
29
[Example 3] Projecting a rocket determine the
minimum velocity for the rocket to escape the earth,
Solution,This problem belongs to the second type
of problems,the known force depends on position,
Choose the rocket as the object to be investigated
and the coordinate axis as shown in the figure,The
rocket is subjected to the action of the earth’s
gravitation at arbitrary position X,
hence,
Write down the differential equation of
motion of the particle
2
2
22 x
m g RF
R
mMfmg
x
mMfF ????? ?
2
2
2
2
xm g Rdtdxm ??
Then the velocity at an arbitrary position is
),,0w h e n ( 02
2
0
vvRxtdxxm g RRxdvmvvv xxx ?????? ??
Thus )(
2
2
2
2
dx
dvv
dtdxdx
dv
dt
dv
dt xdx
m g R
dx
dvmv xxxxx
x ??????
30
[例 3] 发射火箭,求 脱离地球引力的最小速度。
解, 属于已知力是位置的函数的第二类问题。
取火箭 (质点 )为研究对象,建立坐标如图示。
火箭在任意位置 x 处受地球引力 F 的作用。
2
2
22 x
m g RF
R
mMfmg
x
mMfF ????? ?
2
2
2
2
x
m g R
dtdxm ??建立质点运动微分方程
则在任意位置时的速度
),0( 02 2
0
vvRxtdxxm g RRxdvmvvv xxx ?????? ?? 时
即,)(
2
2
2
2
dx
dvv
dtdxdx
dv
dt
dv
dt xdx
m g R
dx
dvmv xxxxx
x ??????
31
It is obviously that v decreases with x,If the velocity
becomes zero at a certain position x=R+H and the rocket falls back
to earth,if,no matter how large (even infinite) the x is,the
rocket is not back,The minimum velocity for the rocket to escape
the earth’s gravitational force and to proceed infinitely away from
the earth ( ) is,
(the second escape velocity)
gRv 220 ?
gRv 220 ?
??x
k m / s )( 2.116 3 7 0108.922 30 ?????? ?gRv
x
gRgRvv 2
02
2)2( ???
32
可见,v 随着 x 的增加而减小。若 则在某一位置
x=R+H 时速度将减小到零，火箭回落。若 时，无论 x
多大（甚至为 ∞）,火箭也不会回落。因此脱离地球引力而一
去不返 时（ ）的最小初速度
gRv 220 ?
gRv 220 ?
??x
k m / s )( 2.116 3 7 0108.922 30 ?????? ?gRv （第二宇宙速度）
x
gRgRvv 2
02
2)2( ???
33
34
