1
Theoretical mechanics
2
3
§ 7–1 Translational motion of a rigid body
§ 7–2 Rotation of a rigid body about a fixed axis
§ 7–3 Velocity and acceleration of a point in a rigid
body rotating about a fixed axis
§ 7–4 Transmission between two rotating rigid bodies
§ 7–5 Vector representation of angular velocity and
angular acceleration
Lesson for problem solving
Chapter 7,Basic motions of a rigid body
4
§ 7–1 刚体的平行移动
§ 7–2 刚体的定轴转动
§ 7–3 定轴转动刚体内各点的速度与加速度
§ 7–4 绕定轴转动刚体的传动问题
§ 7–5 角速度与角速度的矢量表示
点的速度与加速度的矢积表示
习题课
第七章 刚体的基本运动
5
Chapter 7,Basic motions of a rigid body
The shape and dimensions of a rigid body should be
considered investigating its motion,However,due to
its unchanged shape,,
§ 7-1 Translational motion of a rigid body
[Examples]
translation and
rotation
Basic motions
Rectilinear translation curvilinear translation Rotation about a
fixed axis
6
第七章 刚体的基本运动
[例 ]
由于研究对象是刚体,所以运动中要考虑其本身形状和尺
寸大小,又由于刚体是几何形状不变体,所以研究它在空间的
是指刚体的平行
移动和转动
§ 7-1刚体的平行移动 (平动 )
基本运动
7
OB rotates about a fixed axis
The motion of CD is a translation
The motions of AB and of
the cam are translations
It is not necessary to determine its position by finding the
positions of all its points,Instead,its position is determined
completely by describing the position of a line or a plane in it,
8
OB作定轴转动
CD作平动 AB,凸轮均作平动
位置就不必一个点一个点地确定,只要根据刚体的各种运动形
式,确定刚体内某一个有代表性的直线或平面的位置即可。
9
)0()( ??????
dt
rdv
dt
rdrr
dt
d
dt
rdv AB
A
A
ABA
B
B ?
1,Definition of translatory motion of a rigid body,
The direction of the line linking arbitrary two points in the rigid body
never changes during its motion,
Note the equations of motion of the points A and B,i.e,
and,we have
( t )rr( t ),rr BBAA ??
ABAB rrr ??
A
A
ABA
B
B adt
rdrr
dt
d
dt
rdaS i m i l a r l y ?????
2
2
2
2
2
2
)(,
The length and direction of AB
do not change at all during its
motion,
Its trajectory may be
a straight line
a curve ???
[Example]
10
)0()( ??????? dtrdvdt rdrrdtddt rdv ABAAABABB ?
一,刚体平动的定义,
刚体在运动中,其上任意两点的连线始终保持方向不变 。
由 A,B 两点的运动方程式, 而 ( t )rr( t ),rr
BBAA ?? ABAB rrr ??
A
A
ABA
B
B adt
rdrr
dt
d
dt
rda ?????
2
2
2
2
2
2
)(:同理
[例 ] AB在运动中方向和大小始
终不变
它的轨迹
可以是直线
可以是曲线 ???
11
Conclusions,
2,Features of translational motion,
All the particles of a rigid body with translational motion
have the same motion,i,e,their trajectories,velocities and
accelerations are identical,
The translation of a rigid body can be simplified to the
motion of a particle in it,
12
得出结论,即
二,刚体平动的特点,
平动刚体在任一瞬时各点的运动轨迹形状,速度,加速度都一
样。
即,平动刚体的运动可以简化为一个点的运动 。
13
§ 7-2 Rotation of a rigid body about a fixed axis
1,Feature of rotation about a fixed axis and its simplification
Feature,There is a fixed line called axis,and every point of the
rigid body which is not on this axis moves along a circular path in a
plane perpendicular to this fixed axis,
2,Angle of rotation and equation of
rotation
? --- angle of rotation,whose unit is rad,
?=f(t)--- equation of rotation
Sign definition,follows right-hand rule,
counter-clockwise clockwise
14
§ 7-2 刚体的定轴转动
一,刚体定轴转动的特征及其简化
特点,有一条不变的线称为转轴,其余各点都在垂直于转轴的平
面上做圆周运动。
二,转角和转动方程
? ---转角,单位弧度 (rad)
?=f(t)---为转动方程
方向规定, 从 z 轴正向看去,
逆时针为正 顺时针为负
15
Unit used in engineering,
n = rpm (round per minute)
The relation between n
and w is
)nnn ( r a d / s1030602 ??? ???
,)( t h en tf ??? The unit is rad/s,
If the equation of the rotation is given as f( t )??
3,Angular velocity and angular
acceleration
(1),Definition of the angular velocity,
q u a n t i t y ) a l g e b r a i c(
Δ
Δlim
0Δ
???? ????
? dt
d
tt
16
三,定轴转动的角速度和角加速度
1.角速度,
工程中常用单位,
n = 转 /分 (r / min)
则 n与 ?的关系为,
)nnn ( r a d / s1030602 ??? ???
)(,tf ???则 单位 rad/s
若已知转动方程 f(t)??
)( ΔΔlim,
0Δ
代数量定义 ???? ????
? dt
d
tt
17
A rigid body has accelerated rotation when ? and ? have the
same signs,decelerated rotation when ? and ? have opposite
signs,
3,Uniform rotation and rotation with uniform angular
acceleration
A body has uniform rotation when ? =const,and rotation with
constant angular acceleration when ? =const,
?
?
?
?
?
?
?
??
??
??
????
???
???
2
2
1
2
0
2
2
0
0
tt
t
Frequently
used
formulae
Similar to the motion
of a particle,
2,Angular acceleration,
Denote the angular velocity at t as ? and at t +△ t as ?+△ ?,
the angular acceleration can be defined as
)(lim 2
2
0
tf
dt
d
dt
d
tt
??????
?
??
??
????? ??
unit:rad/s2 (scalar)
18
2.角加速度,
设当 t 时刻为 ?,t +△ t 时刻为 ?+△ ?
?与 ?方向一致为加速转动,?与 ? 方向相反为减速转动
3.匀速转动和匀变速转动
当 ? =常数,为 匀速转动 ；当 ? =常数,为 匀变速转动 。
?
?
?
?
?
?
?
??
??
??
????
???
???
2
2
1
2
0
2
2
0
0
tt
t
常用公式 与点的运动相类似。
)(lim,2
2
0
tfdtddtdt
t
????????
?
???? ???
?
??角加速度
单位,rad/s2 (代数量 )
19
?,? are defined for the whole rigid body
(including all points in it)；
v,a are defined for a given point in the body
(they are different for different point),
t
S
dt
dSv
t ?
???
?? 0
lim?
Rt Rv
t
????
?
???
? 0
lim
Rv ???
§ 7–3 Velocity and acceleration of a point in
a rigid body rotating about a fixed axis
1,Relationship between linear velocity and angular velocity ?
20
?,? 对整个刚体而言 (各点都一样 )；
v,a 对刚体中某个点而言 (各点不一样 )。
t
S
dt
dSv
t ?
?
? 0
lim
?
???
Rt Rv
t
????
?
???
? 0
lim
Rv ???
(即角量与线量的关系 )
§ 7-3 转动刚体内各点的速度和加速度
一,线速度 V和角速度 ?之间的关系
21
,)( RRdtdRdtddtdva ?????? ?????
2
22 )(
??? RRRva n ???
4222
r e s u l t a n t |||| ???? ??????? Raaaaa nn
22 tg ?
?
?
?? ? ???
R
R
a
a
n
2,Relation between angular
acceleration ? and an,a?
22
,)( RRdtdRdtddtdva ?????? ?????
2
22 )(
??? RRRva n ???
4222|||| ???? ??????? Raaaaa nn
全
22 tg ?
?
?
?? ? ???
R
R
a
a
n
二,角加速度 ? 与 an,a? 的关系
23
Conclusions,① v is proportional to R and its direction is
perpendicular to R,It is defined as positive when its direction
is the same as that of ?,
② The angle ? between the resultant acceleration of any
point and the radius through it is identical and less than 90o,
The distribution diagrams of velocities and accelerations of a
rotating body are shown below,
Distribution of the velocities of points
Distribution of the
accelerations of points
24
结论, ① v方向与 ? 相同时为正,?R,与 R 成正比。
②各点的全加速度方向与各点转动半径夹角 ? 都一致,且
小于 90o,在同一瞬间的速度和加速度的分布图为,
各点速度分布图 各点加速度分布图
25
In engineering practice,several gears meshed with each other are
usually used to change velocities of rotation and transmit power,
which can be explained as follows,
§ 7–4 Transmission between two rotating rigid bodies
I.Transmission of gear wheels
It is assumed that the gear rolls
without slipping,hence,
The transmission ratio of gear
E to gear F can be defined as
E
F
E
F
F
E
EF Z
Z
r
ri ???
?
?
EF vv ?? EF vv ??
EEFF rr ?? ?
1.Inner meshing
26
我们常见到在工程中,用一系列互相啮合的齿轮来实现变速,
它们变速的基本原理是什么呢?
§ 7-4 绕定轴转动刚体的传动问题
一,齿轮传动
因为是做纯滚动 (即没有相对滑动 )
定义 齿轮传动比
E
F
E
F
F
E
EF Z
Z
r
ri ???
?
?
EF vv ?? EF vv ??
EEFF rr ?? ?
1.内啮合
27
DDCCDCDC rrvvvv ?? ???????
C
D
C
D
D
C
CD Z
Z
r
ri ???????
?
?
t
rZ ?2?
E
F
E
F
E
F
Z
Z
tr
tr
r
r ???
/ 2
/ 2
?
?
2,External meshing
Number
of teeth,
28
DDCCDCDC rrvvvv ?? ???????
C
D
C
D
D
C
CD Z
Z
r
ri ???????
?
?
t
rZ ?2?齿数
E
F
E
F
E
F
Z
Z
tr
tr
r
r ???
/ 2
/ 2
?
?
2.外啮合
29
Since
2
1
2
1
60
2
n
nn ???
?
???
g e a rd r i v e n
g e a r d r i v i n g i, e,
1
2
1
2
2
1
2
1
2,1 ?????????? z
z
r
r
n
ni
?
?
Obviously,when,,transmission for increasing
the rotating velocity of the driven gear；
when,,transmission for decreasing
the rotating velocity of the driven gear,
1|| 2,1 ?i 12 ?? ?
1|| 2,1 ?i 12 ?? ?
30
由于转速 n与 ? 有如下关系,
成正比
2
1
2
1
60
2
n
nn ???
?
???
从动轮
主动轮即 ??????????
1
2
1
2
2
1
2
1
2,1,z
z
r
r
n
ni
?
?
显然当, 时,,为升速转动；
时,,为降速转动。 1|| 2,1 ?i 12
?? ?
1|| 2,1 ?i 12 ?? ?
31
III,Pulley system,For a system consisting of pulleys
A,B,C,D,E,F,G,H,the transmission ration of the pulleys A to
H is given by,
,HGGFFEEDDCCBBA
m
H
G
G
F
F
E
E
D
D
C
C
B
B
Am
H
A
HA
iiiiiii
i
,,,,,,,
,
)1()1(
???????
??????????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
where m represents the number of the pulleys,and the negative
sign indicates that the rotation direction of the last pulley is
opposite to that of the first one,
II.Transmission system
consisting of belt and pulley
BA vv ? ?
(but not due to the difference
between their directions ) BA vv ?
??? BBAA rr ?? Transmission ratio,
A
B
B
A
AB r
ri ??
?
?
32
三,链轮系, 设有,A,B,C,D,E,F,G,H 轮系,则总传动比为,
HGGFFEEDDCCBBA
m
H
G
G
F
F
E
E
D
D
C
C
B
B
Am
H
A
HA
iiiiiii
i
,,,,,,,
,
)1()1(
???????
??????????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
其中 m代表外啮合的个数；负号表示最后一个轮转向与
第一个轮转向相反。
二,皮带轮系传动
BA vv ? ? (而不是 方向不同 ) ? BA vv ?
??? BBAA rr ?? 皮带传动
A
B
B
A
AB r
ri ??
?
?
33
§ 7–5 Vector representation of
angular velocity and angular acceleration
1,Vector representation of angular velocity and angular
acceleration,
|||| dtd ?? ?
k?? ?,f i g u r e i n t h es h o w n isd i r e c t i o n T h e kkdtddtd ???? ???
The directions of
and are determined
according to the right-
hand rule,
?
?
Magnitude,
34
§ 7-5 角速度和角加速度的矢量表示
点的速度和加速度的矢量表示
一, 角速度和角加速度的矢量表示
|||:| dtd ?? ?大小
k?? ?方向如图
kkdtddtd ???? ???
按右手定则规定
, 的方向。 ? ?
35
2,Vectorial representation of the linear velocity and
acceleration of any point in a rigid body
??????? ??????????? RrrrRv s i n||s i n?
vr ????
dt
rdr
dt
d
dt
rd
dt
vda ??????? ??? )(?
vra ????? ??
Rvva
Rrra
o
n
290s i n||||
s i n||
???
?????
????
????????
ra ??? ?? va n ??? ?
ra
rv
??
??
?
?
?
va n ?? ?
36
二 刚体内任一点的线速度和线加速度的矢积表示
??????? ??????????? RrrrRv s i n||s i n?
vr ????
dt
rdr
dt
d
dt
rd
dt
vda ??????? ??? )(?
vra ????? ??
Rvva
Rrra
o
n
290s i n||||
s i n||
???
?????
????
????????
ra ??? ?? va n ??? ?
ra
rv
??
??
?
?
?
va n ?? ?
37
1,Basic concepts,features of motions and equations
1) Basic concepts,rectilinear motion,curvilinear motion (of a
particle);
translation,rotation about a fixed axis ( of a rigid body),
2) Basic features of motions and basic equations,
Chapter 6,Kinematics of a particle
Chapter 7,Basic motions of a rigid body
Lesson for
problem
solving
38
一,基本概念和基本运动规律及基本公式
1,基本概念：直线运动,曲线运动 (点 ) ;
平动,定轴转动 (刚体 )。
2,基本运动规律与公式,
第六章 点的运动学,
第七章 刚体的基本运动
习题课
39
dt
dv
a ?
?
点 的 运 动
加速度
a
?
a
n
a
v s
匀
速
0 0 0
匀
变
a
?
=C 0 a = C
直
线
运
动
变
速
0
匀
速
0
匀
变
a
?
=C
曲
线
运
动
变
速
dt
dv
a ?
Cv ?
vttfs ?? )(
atvv ??
0
2
0
2
1
attvs ??
?
??
t
a d tvv
0
0
?
?
t
v d ts
0
dt
dv
a ?
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a ?
22
n
aa
a
?
?
?
22
n
aa
a
?
?
?
Cv ?
tavv
?
??
0
?
??
t
dtavv
0
0 ?
vts ?
2
0
2
1
tatvs
?
??
?
?
t
v d ts
0
Motion of a particle
acceleration
Rectili-
near
motion
Uniform
velocity
Uniform
acceler-
ation
General
case
Uniform
velocity
Uniform
acceler-
ation
General
case
Curvili-
near
motion
40
dt
dv
a ?
?
点 的 运 动
加速度
a
?
a
n
a
v s
匀
速
0 0 0
匀
变
a
?
=C 0 a = C
直
线
运
动
变
速
0
匀
速
0
匀
变
a
?
=C
曲
线
运
动
变
速
dt
dv
a ?
Cv ?
vttfs ?? )(
atvv ??
0
2
0
2
1
attvs ??
?
??
t
a d tvv
0
0
?
?
t
v d ts
0
dt
dv
a ?
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a ?
22
n
aa
a
?
?
?
22
n
aa
a
?
?
?
Cv ?
tavv
?
??
0
?
??
t
dtavv
0
0 ?
vts ?
2
0
2
1
tatvs
?
??
?
?
t
v d ts
0
41
Rotation of a body about a fixed axis
Equation of rotation,
Angular velocity,
)( tf??
dt
d?? ?
2
2
dt
d
dt
d ??? ??Angular acceleration,
Uniform rotation,
Rotation with uniformly varying velocity,
t??? ?? 0
t??? ?? 02
00 2
1 tt ???? ??? ???? 2202 ??
42
刚体定轴转动
转动方程,
角速度,
)( tf??
dt
d?? ?
2
2
dt
d
dt
d ??? ??角加速度,
匀速转动,
匀变速运动,
t??? ?? 0
t??? ?? 0
2
00 2
1 tt ???? ??? ???? 2
2
02 ??
43
2.Steps for solving problems and things to consider carefully
1) Steps for solving problems,
① Analyze the problem carefully and make sure what the known
and what the unknowns are,
② Select a proper coordinate system,namely a rectangular
coordinate system or a natural coordinate system which can
describe the equations of motion most conveniently,
③ Perform the differential or integral calculation according to the
known variables,
④ Finally determine the related constants using initial conditions,
2) Pay special attention to,
① Geometric relations and directions of motion,
② Eliminating the parameter,t” while obtaining the
equation for the trajectories,
③ Selection of the appropriate coordinate systems,
44
二,解题步骤及注意问题
1.解题步骤,
① 弄清题意,明确已知条件和所求的问题。
②选好坐标系：直角坐标法，自然法。
③ 根据已知条件进行微分，或积分运算。
④ 用初始条件定积分常数。
对常见的特殊运动,
可直接应用公式计算 。
２．注意问题,
①几何关系和运动方向。
②求轨迹方程时要消去参数, t”。
③ 坐标系（参考系）的选择。
45
3,Examples
[Example 1] A locomotive moves along a curved track of radius
R=300m with uniformly varying velocity,The length of the
curved track is l=200m,The velocity of the locomotive when it
enters the track is v0=30km/h,and that when it leaves it is v1＝
48km/h,Find its accelerations when it enters and leaves the
curved track,respectively,
46
三,例题
[例 1]列车在 R=300m的曲线上匀变速行驶。轨道上曲线部分长
l=200m，当列车开始走上曲线时的速度 v0=30km/h，而将要离开
曲线轨道时的速度是 v1＝ 48km/h。
求列车走上曲线与将要离开曲线时的加速度？
47
??a
Solution,Since it move with uniformly varying velocity,const,
Employing and noting we have
savv ?2202 ??,m2 0 0?? ls
m / s
3
40
3 6 0 0
1 0 0 048
,m / s
3
25
3 6 0 0
1 0 0 030
m / s27.0
2 0 029
6 2 51 6 0 0
2
10
2
2
0
2
1
?
?
??
?
?
?
??
?
?
?
?
vv
s
vv
a
?
When it enters the curve,
Resultant acceleration
When it leaves the curve,
Resultant acceleration
2
22
0
0
2 m / s23.0
300
)3/25(,m / s27.0 ????
R
vaa
n?
'
0
1
0
22
0
2
0 2949tg,m / s356.0
?????? ?
n
n a
aaaa ?
? ?
2
22
1
1
2 m / s593.0
300
)3/40(,m / s27.0 ????
R
vaa
n?
'
1
1
1
22
1
2
1 3424tg,m / s652.0
?????? ?
n
n a
aaaa ?
? ?
48
??a
解,由于是匀变速运动，则 常量。
由公式 而由已知
savv ?2202 ??,m2 0 0?? ls
m / s
3
40
3600
100048
,m / s
3
25
3600
100030
m / s27.0
20029
6251600
2
10
2
2
0
2
1
?
?
??
?
?
?
??
?
?
?
??
vv
s
vv
a ?
列车走上曲线时,
全加速度
列车将要离开曲线时,
全加速度
2
22
0
0
2 m / s23.0
300
)3/25(,m / s27.0 ????
R
vaa
n?
'
0
1
0
22
0
2
0 2949tg,m / s356.0
?????? ?
n
n a
aaaa ?
? ?
2
22
1
1
2 m / s593.0
300
)3/40(,m / s27.0 ????
R
vaa
n?
'
1
1
1
22
1
2
1 3424tg,m / s652.0
?????? ?
n
n a
aaaa ?
? ?
49
〔 Example 2〕 Find the initial velocity and acceleration of a
projectile shown in the figure which has its hit point at A,
Analysis,Only when the projectile reaches its highest point A,
then vy＝ 0,
gtvdtdyv y ???? ?s i n0
Solution,the displacement can be
expressed as
2
0
0
2
1
s i n
c o s
gttvy
tvx
??
?
?
?
?
?
At point A vy=0,The time it takes from the origin to point A is,
g
vt ?s in0?
50
〔 例 2〕 已知如图，求 时正好射到 A点且用力最小。
,0 ?? v?
分析：只有在 A点,vy＝ 0且为最大高度时，用力才最小。
gtvdtdyv y ???? ?s i n0
解,由
2
0
0
2
1s i n
c o s
gttvy
tvx
??
?
?
?
?
?
由于在 A点时,vy=0，所以上升到最大高度 A点时所用
时间为,
g
vt ?s in0?
51
Substituting the above
equation into ① and ②,
yields
g
v
y
g
v
x AA
2
s i n
,
2
2s i n 22020 ??
??
?
?
?
?
?
?
tg
2
1
2s i n
s i n
5
5.1 2
2
2s i n
2
s i n
2
0
22
0
????
g
v
g
v
A
A
x
y
?????? 31,6.05 5.12tg ??
Substituting ?31?? into ③ we have
1 1 162s i n 58.922s i n220 ?????? ?Agxv m / s5.10
0 ?? v
52
将上式代入 ①和②，得,
g
vy
g
vx
AA 2
s i n,
2
2s i n 22020 ?? ?? ?
?
?
?
?
?
tg
2
1
2s i n
s i n
5
5.1 2
2
2s i n
2
s i n
2
0
22
0
????
g
v
g
v
A
A
x
y
?????? 31,6.05 5.12tg ??
将 ?31?? 代入③,得
1 1 162s i n 58.922s i n220 ?????? ?Agxv
m / s5.100 ?? v
53
① The number of revolution of the pulley in 3
seconds；
② The displacement of weight B in 3 seconds；
③ The velocity of weight B at ｔ ＝ 3s；
④ The acceleration of point C on the rim of
the pulley at the initial time；
⑤ The acceleration of point C at t＝ 3s,
,
Find,
,m5.0?R
〔 Example 3〕
The acceleration of weight A in the figure is 2m /s1?
Aa
and its initial velocity is m/s5.1
0 ?v
m.30,r ?
（ constant）
54
〔 例 3〕 已知：重物 A的 2m/s1?
Aa
（常数）初瞬时速度 m/s5.1
0 ?v
方向如图示。 求：,m5.0?R ? m30,r ?
① 滑轮 3s内的转数；
②重物 B在 3s内的行程；
③重物 B在 ｔ ＝ 3s时的速度；
④滑轮边上 C点在初瞬时的加速度；
⑤滑轮边上 C点在 ｔ ＝ 3s时的加速度。
55
2r a d / s 2
5.0
1 ???
R
a C ??
r a d / s35.0 5.1,m / s5.1 0 ????? Rvvv CAC ?
） constant （
（ ）
,m / s1 2?? AC aa ?
Solution,
① If the rope is not extensible we have
m4.5183.0 ???? ?rs②
? ?????? m / s7.293.0?rv B
）,
③ r a d / s 9323 0 ?????? t???
（
86.2
2
,r a d1832
2
1
33
2
1 22
0
??
?????????
?
?
???
n
tt
(revolutions)
56
2r a d / s 2
5.0
1 ???
R
a C ??
r a d / s35.0 5.1,m / s5.1 0 ????? Rvvv CAC ?
）常数 （
（ ）
,m / s1 2?? AC aa ?
解, ① 因为绳子不可以伸长，所以有
转）(86.22,r a d1832213321 220 ??????????? ????? ntt
m4.5183.0 ???? ?rs②
? ?????? m / s7.293.0?rv B
）,
③ r a d / s 9323 0 ?????? t???
（
57
④ At t = 0,
222
0
2
m / s5.435.0
,m / s1
?????
??
?
?
Ra
aa
n
C
AC
?????
??????
5.12,222.0
5.4
1
tg
m / s 61.45.41)()( 22222
??
?
?
n
C
C
n
CCC
a
a
aaa
⑤ At t=3s,
222 m / s5.4095.0,m / s1 2 ??????? ?? Raaa nCAC
???????? 41.1,0247.051.40 1 t g,m / s51.405.401 222 ??Ca
58
④ t = 0 时,
222
0
2
m / s5.435.0
,m / s1
?????
??
?
?
Ra
aa
n
C
AC
?????
??????
5.12,222.0
5.4
1
tg
m / s 61.45.41)()( 22222
??
?
?
n
C
C
n
CCC
a
a
aaa
⑤ t=3s 时,
222 m / s5.4095.0,m / s1 2 ??????? ?? Raaa nCAC
???????? 41.1,0247.051.40 1 t g,m / s51.405.401 222 ??Ca
59
[Example 4] A wheel rotates starting at rest with constant angular
acceleration,OM＝ 0.4m,At a certain time,the acceleration of point
M is measured to be, ??? 30,m / s 40
2 ?Ma
Find, The equation of rotation;
The velocity and the centripetal
acceleration of point M at t＝ 5s,
??? s in??? aRa
2r a d / s 50
4.0
30s i n40s i n ???????
R
a
R
a ?? ?
Solution,?
22200 25502121,0 tttt ???????? ?????
225:r o t a t i o n ofE q u a t i o n t??
M
60
[例 4] 已知：圆轮 O由静止开始作等加速转动,OM＝ 0.4m,
在 某瞬时测得 ??? 30,m / s 40
2 ?Ma
求,?转动方程 ;
? t＝ 5s时,Ｍ 点的速度和
向心加速度的大小。
??? s in??? aRa
2r a d / s 50
4.0
30s i n40s i n ???????
R
a
R
a ?? ?
解,?
22200 25502121,0 tttt ???????? ?????
225 t??转动方程
M
61
?
ttRvtt M 20504.0,50 0 ??????? ????
2
22
m / s2 5 0 0 04.01 0 0 ??? Rva MnM
?atｔ ＝ 5s,m / s1 0 0520 ???
Mv
M
62
?
ttRvtt M 20504.0,50 0 ??????? ????
2
22
m / s2 5 0 0 04.01 0 0 ??? Rva MnM
?当 ｔ ＝ 5s时,m / s1 0 0520 ???
Mv
M
63
〔 Example 5〕 Draw the velocity and acceleration vectors of the
pointsＭ and Ｎ on the rigid bodies shown in the following figures,
),( 2121 ABOOBOAO ??
64
〔 例 5〕 试画出图中刚体上 Ｍ ?Ｎ 两点在图示位置时的速度和
加速度。 ),(
2121 ABOOBOAO ??
65
〔 Example6〕 Referring to the figure,a projectile is fired from the
point O with an angle of elevation, Show that the angle under which
the projectile reaches the line L in the shortest time is, 2?
For y = h,the relation between t and
is
0?
? 2
00 2
1s i n gttvh ?? ?
200
2
1s i n gttvy ?? ? ?
Proof,choosing the coordinate
system,we get xoy
O
0?
66
〔 例 6〕 已知如图，从 O点以
任一角度抛出一质点，试证明
质点最早到达直线Ｌ的抛角为 。
2
?
（与上升的最大高度无关，只要
求时间对抛射角度的变化率）
到达高度为 h 时,t 与 的
关系有下式确定
0?
? 2
00 2
1s i n gttvh ?? ?
200
2
1s i n gttvy ?? ? ?
解,选 坐标系，则 xoy
O
67
00
000 22
1)s i nc o s(0
???? d
dttg
d
dttv ?????
To minimize the time you have to solve
0
0
??d dt
?
The derivative of ? with respect to
gives 0?
00
0000 s i nc o s0 ???? d
dttg
d
dtvtv ?????
??
Inserting into ? yields
0
0
??ddt ;0c o s 00 ??tv
note 0c o s,0
00 ??? ?tv 2
?? ??
which gives the angle for the shortest time
which the projectile takes from O to the line L,
68
00
000 22
1)s i nc o s(0
???? d
dttg
d
dttv ?????
欲使最早到达，必须满足
0
0
??d dt
?
将 ?对 求导数 0?
00
0000 s i nc o s0 ???? d
dttg
d
dtvtv ?????
??
将 （最早到达的条件） 代入 ?，得
0
0
??ddt ;0c o s 00 ??tv
又 0c o s,0
00 ??? ?tv 2?? ??
证毕。
表示出在某一角度下时间会最短。（极值）
69
70
Theoretical mechanics
2
3
§ 7–1 Translational motion of a rigid body
§ 7–2 Rotation of a rigid body about a fixed axis
§ 7–3 Velocity and acceleration of a point in a rigid
body rotating about a fixed axis
§ 7–4 Transmission between two rotating rigid bodies
§ 7–5 Vector representation of angular velocity and
angular acceleration
Lesson for problem solving
Chapter 7,Basic motions of a rigid body
4
§ 7–1 刚体的平行移动
§ 7–2 刚体的定轴转动
§ 7–3 定轴转动刚体内各点的速度与加速度
§ 7–4 绕定轴转动刚体的传动问题
§ 7–5 角速度与角速度的矢量表示
点的速度与加速度的矢积表示
习题课
第七章 刚体的基本运动
5
Chapter 7,Basic motions of a rigid body
The shape and dimensions of a rigid body should be
considered investigating its motion,However,due to
its unchanged shape,,
§ 7-1 Translational motion of a rigid body
[Examples]
translation and
rotation
Basic motions
Rectilinear translation curvilinear translation Rotation about a
fixed axis
6
第七章 刚体的基本运动
[例 ]
由于研究对象是刚体,所以运动中要考虑其本身形状和尺
寸大小,又由于刚体是几何形状不变体,所以研究它在空间的
是指刚体的平行
移动和转动
§ 7-1刚体的平行移动 (平动 )
基本运动
7
OB rotates about a fixed axis
The motion of CD is a translation
The motions of AB and of
the cam are translations
It is not necessary to determine its position by finding the
positions of all its points,Instead,its position is determined
completely by describing the position of a line or a plane in it,
8
OB作定轴转动
CD作平动 AB,凸轮均作平动
位置就不必一个点一个点地确定,只要根据刚体的各种运动形
式,确定刚体内某一个有代表性的直线或平面的位置即可。
9
)0()( ??????
dt
rdv
dt
rdrr
dt
d
dt
rdv AB
A
A
ABA
B
B ?
1,Definition of translatory motion of a rigid body,
The direction of the line linking arbitrary two points in the rigid body
never changes during its motion,
Note the equations of motion of the points A and B,i.e,
and,we have
( t )rr( t ),rr BBAA ??
ABAB rrr ??
A
A
ABA
B
B adt
rdrr
dt
d
dt
rdaS i m i l a r l y ?????
2
2
2
2
2
2
)(,
The length and direction of AB
do not change at all during its
motion,
Its trajectory may be
a straight line
a curve ???
[Example]
10
)0()( ??????? dtrdvdt rdrrdtddt rdv ABAAABABB ?
一,刚体平动的定义,
刚体在运动中,其上任意两点的连线始终保持方向不变 。
由 A,B 两点的运动方程式, 而 ( t )rr( t ),rr
BBAA ?? ABAB rrr ??
A
A
ABA
B
B adt
rdrr
dt
d
dt
rda ?????
2
2
2
2
2
2
)(:同理
[例 ] AB在运动中方向和大小始
终不变
它的轨迹
可以是直线
可以是曲线 ???
11
Conclusions,
2,Features of translational motion,
All the particles of a rigid body with translational motion
have the same motion,i,e,their trajectories,velocities and
accelerations are identical,
The translation of a rigid body can be simplified to the
motion of a particle in it,
12
得出结论,即
二,刚体平动的特点,
平动刚体在任一瞬时各点的运动轨迹形状,速度,加速度都一
样。
即,平动刚体的运动可以简化为一个点的运动 。
13
§ 7-2 Rotation of a rigid body about a fixed axis
1,Feature of rotation about a fixed axis and its simplification
Feature,There is a fixed line called axis,and every point of the
rigid body which is not on this axis moves along a circular path in a
plane perpendicular to this fixed axis,
2,Angle of rotation and equation of
rotation
? --- angle of rotation,whose unit is rad,
?=f(t)--- equation of rotation
Sign definition,follows right-hand rule,
counter-clockwise clockwise
14
§ 7-2 刚体的定轴转动
一,刚体定轴转动的特征及其简化
特点,有一条不变的线称为转轴,其余各点都在垂直于转轴的平
面上做圆周运动。
二,转角和转动方程
? ---转角,单位弧度 (rad)
?=f(t)---为转动方程
方向规定, 从 z 轴正向看去,
逆时针为正 顺时针为负
15
Unit used in engineering,
n = rpm (round per minute)
The relation between n
and w is
)nnn ( r a d / s1030602 ??? ???
,)( t h en tf ??? The unit is rad/s,
If the equation of the rotation is given as f( t )??
3,Angular velocity and angular
acceleration
(1),Definition of the angular velocity,
q u a n t i t y ) a l g e b r a i c(
Δ
Δlim
0Δ
???? ????
? dt
d
tt
16
三,定轴转动的角速度和角加速度
1.角速度,
工程中常用单位,
n = 转 /分 (r / min)
则 n与 ?的关系为,
)nnn ( r a d / s1030602 ??? ???
)(,tf ???则 单位 rad/s
若已知转动方程 f(t)??
)( ΔΔlim,
0Δ
代数量定义 ???? ????
? dt
d
tt
17
A rigid body has accelerated rotation when ? and ? have the
same signs,decelerated rotation when ? and ? have opposite
signs,
3,Uniform rotation and rotation with uniform angular
acceleration
A body has uniform rotation when ? =const,and rotation with
constant angular acceleration when ? =const,
?
?
?
?
?
?
?
??
??
??
????
???
???
2
2
1
2
0
2
2
0
0
tt
t
Frequently
used
formulae
Similar to the motion
of a particle,
2,Angular acceleration,
Denote the angular velocity at t as ? and at t +△ t as ?+△ ?,
the angular acceleration can be defined as
)(lim 2
2
0
tf
dt
d
dt
d
tt
??????
?
??
??
????? ??
unit:rad/s2 (scalar)
18
2.角加速度,
设当 t 时刻为 ?,t +△ t 时刻为 ?+△ ?
?与 ?方向一致为加速转动,?与 ? 方向相反为减速转动
3.匀速转动和匀变速转动
当 ? =常数,为 匀速转动 ；当 ? =常数,为 匀变速转动 。
?
?
?
?
?
?
?
??
??
??
????
???
???
2
2
1
2
0
2
2
0
0
tt
t
常用公式 与点的运动相类似。
)(lim,2
2
0
tfdtddtdt
t
????????
?
???? ???
?
??角加速度
单位,rad/s2 (代数量 )
19
?,? are defined for the whole rigid body
(including all points in it)；
v,a are defined for a given point in the body
(they are different for different point),
t
S
dt
dSv
t ?
???
?? 0
lim?
Rt Rv
t
????
?
???
? 0
lim
Rv ???
§ 7–3 Velocity and acceleration of a point in
a rigid body rotating about a fixed axis
1,Relationship between linear velocity and angular velocity ?
20
?,? 对整个刚体而言 (各点都一样 )；
v,a 对刚体中某个点而言 (各点不一样 )。
t
S
dt
dSv
t ?
?
? 0
lim
?
???
Rt Rv
t
????
?
???
? 0
lim
Rv ???
(即角量与线量的关系 )
§ 7-3 转动刚体内各点的速度和加速度
一,线速度 V和角速度 ?之间的关系
21
,)( RRdtdRdtddtdva ?????? ?????
2
22 )(
??? RRRva n ???
4222
r e s u l t a n t |||| ???? ??????? Raaaaa nn
22 tg ?
?
?
?? ? ???
R
R
a
a
n
2,Relation between angular
acceleration ? and an,a?
22
,)( RRdtdRdtddtdva ?????? ?????
2
22 )(
??? RRRva n ???
4222|||| ???? ??????? Raaaaa nn
全
22 tg ?
?
?
?? ? ???
R
R
a
a
n
二,角加速度 ? 与 an,a? 的关系
23
Conclusions,① v is proportional to R and its direction is
perpendicular to R,It is defined as positive when its direction
is the same as that of ?,
② The angle ? between the resultant acceleration of any
point and the radius through it is identical and less than 90o,
The distribution diagrams of velocities and accelerations of a
rotating body are shown below,
Distribution of the velocities of points
Distribution of the
accelerations of points
24
结论, ① v方向与 ? 相同时为正,?R,与 R 成正比。
②各点的全加速度方向与各点转动半径夹角 ? 都一致,且
小于 90o,在同一瞬间的速度和加速度的分布图为,
各点速度分布图 各点加速度分布图
25
In engineering practice,several gears meshed with each other are
usually used to change velocities of rotation and transmit power,
which can be explained as follows,
§ 7–4 Transmission between two rotating rigid bodies
I.Transmission of gear wheels
It is assumed that the gear rolls
without slipping,hence,
The transmission ratio of gear
E to gear F can be defined as
E
F
E
F
F
E
EF Z
Z
r
ri ???
?
?
EF vv ?? EF vv ??
EEFF rr ?? ?
1.Inner meshing
26
我们常见到在工程中,用一系列互相啮合的齿轮来实现变速,
它们变速的基本原理是什么呢?
§ 7-4 绕定轴转动刚体的传动问题
一,齿轮传动
因为是做纯滚动 (即没有相对滑动 )
定义 齿轮传动比
E
F
E
F
F
E
EF Z
Z
r
ri ???
?
?
EF vv ?? EF vv ??
EEFF rr ?? ?
1.内啮合
27
DDCCDCDC rrvvvv ?? ???????
C
D
C
D
D
C
CD Z
Z
r
ri ???????
?
?
t
rZ ?2?
E
F
E
F
E
F
Z
Z
tr
tr
r
r ???
/ 2
/ 2
?
?
2,External meshing
Number
of teeth,
28
DDCCDCDC rrvvvv ?? ???????
C
D
C
D
D
C
CD Z
Z
r
ri ???????
?
?
t
rZ ?2?齿数
E
F
E
F
E
F
Z
Z
tr
tr
r
r ???
/ 2
/ 2
?
?
2.外啮合
29
Since
2
1
2
1
60
2
n
nn ???
?
???
g e a rd r i v e n
g e a r d r i v i n g i, e,
1
2
1
2
2
1
2
1
2,1 ?????????? z
z
r
r
n
ni
?
?
Obviously,when,,transmission for increasing
the rotating velocity of the driven gear；
when,,transmission for decreasing
the rotating velocity of the driven gear,
1|| 2,1 ?i 12 ?? ?
1|| 2,1 ?i 12 ?? ?
30
由于转速 n与 ? 有如下关系,
成正比
2
1
2
1
60
2
n
nn ???
?
???
从动轮
主动轮即 ??????????
1
2
1
2
2
1
2
1
2,1,z
z
r
r
n
ni
?
?
显然当, 时,,为升速转动；
时,,为降速转动。 1|| 2,1 ?i 12
?? ?
1|| 2,1 ?i 12 ?? ?
31
III,Pulley system,For a system consisting of pulleys
A,B,C,D,E,F,G,H,the transmission ration of the pulleys A to
H is given by,
,HGGFFEEDDCCBBA
m
H
G
G
F
F
E
E
D
D
C
C
B
B
Am
H
A
HA
iiiiiii
i
,,,,,,,
,
)1()1(
???????
??????????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
where m represents the number of the pulleys,and the negative
sign indicates that the rotation direction of the last pulley is
opposite to that of the first one,
II.Transmission system
consisting of belt and pulley
BA vv ? ?
(but not due to the difference
between their directions ) BA vv ?
??? BBAA rr ?? Transmission ratio,
A
B
B
A
AB r
ri ??
?
?
32
三,链轮系, 设有,A,B,C,D,E,F,G,H 轮系,则总传动比为,
HGGFFEEDDCCBBA
m
H
G
G
F
F
E
E
D
D
C
C
B
B
Am
H
A
HA
iiiiiii
i
,,,,,,,
,
)1()1(
???????
??????????
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
?
其中 m代表外啮合的个数；负号表示最后一个轮转向与
第一个轮转向相反。
二,皮带轮系传动
BA vv ? ? (而不是 方向不同 ) ? BA vv ?
??? BBAA rr ?? 皮带传动
A
B
B
A
AB r
ri ??
?
?
33
§ 7–5 Vector representation of
angular velocity and angular acceleration
1,Vector representation of angular velocity and angular
acceleration,
|||| dtd ?? ?
k?? ?,f i g u r e i n t h es h o w n isd i r e c t i o n T h e kkdtddtd ???? ???
The directions of
and are determined
according to the right-
hand rule,
?
?
Magnitude,
34
§ 7-5 角速度和角加速度的矢量表示
点的速度和加速度的矢量表示
一, 角速度和角加速度的矢量表示
|||:| dtd ?? ?大小
k?? ?方向如图
kkdtddtd ???? ???
按右手定则规定
, 的方向。 ? ?
35
2,Vectorial representation of the linear velocity and
acceleration of any point in a rigid body
??????? ??????????? RrrrRv s i n||s i n?
vr ????
dt
rdr
dt
d
dt
rd
dt
vda ??????? ??? )(?
vra ????? ??
Rvva
Rrra
o
n
290s i n||||
s i n||
???
?????
????
????????
ra ??? ?? va n ??? ?
ra
rv
??
??
?
?
?
va n ?? ?
36
二 刚体内任一点的线速度和线加速度的矢积表示
??????? ??????????? RrrrRv s i n||s i n?
vr ????
dt
rdr
dt
d
dt
rd
dt
vda ??????? ??? )(?
vra ????? ??
Rvva
Rrra
o
n
290s i n||||
s i n||
???
?????
????
????????
ra ??? ?? va n ??? ?
ra
rv
??
??
?
?
?
va n ?? ?
37
1,Basic concepts,features of motions and equations
1) Basic concepts,rectilinear motion,curvilinear motion (of a
particle);
translation,rotation about a fixed axis ( of a rigid body),
2) Basic features of motions and basic equations,
Chapter 6,Kinematics of a particle
Chapter 7,Basic motions of a rigid body
Lesson for
problem
solving
38
一,基本概念和基本运动规律及基本公式
1,基本概念：直线运动,曲线运动 (点 ) ;
平动,定轴转动 (刚体 )。
2,基本运动规律与公式,
第六章 点的运动学,
第七章 刚体的基本运动
习题课
39
dt
dv
a ?
?
点 的 运 动
加速度
a
?
a
n
a
v s
匀
速
0 0 0
匀
变
a
?
=C 0 a = C
直
线
运
动
变
速
0
匀
速
0
匀
变
a
?
=C
曲
线
运
动
变
速
dt
dv
a ?
Cv ?
vttfs ?? )(
atvv ??
0
2
0
2
1
attvs ??
?
??
t
a d tvv
0
0
?
?
t
v d ts
0
dt
dv
a ?
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a ?
22
n
aa
a
?
?
?
22
n
aa
a
?
?
?
Cv ?
tavv
?
??
0
?
??
t
dtavv
0
0 ?
vts ?
2
0
2
1
tatvs
?
??
?
?
t
v d ts
0
Motion of a particle
acceleration
Rectili-
near
motion
Uniform
velocity
Uniform
acceler-
ation
General
case
Uniform
velocity
Uniform
acceler-
ation
General
case
Curvili-
near
motion
40
dt
dv
a ?
?
点 的 运 动
加速度
a
?
a
n
a
v s
匀
速
0 0 0
匀
变
a
?
=C 0 a = C
直
线
运
动
变
速
0
匀
速
0
匀
变
a
?
=C
曲
线
运
动
变
速
dt
dv
a ?
Cv ?
vttfs ?? )(
atvv ??
0
2
0
2
1
attvs ??
?
??
t
a d tvv
0
0
?
?
t
v d ts
0
dt
dv
a ?
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a
n
?
?
2
v
a ?
22
n
aa
a
?
?
?
22
n
aa
a
?
?
?
Cv ?
tavv
?
??
0
?
??
t
dtavv
0
0 ?
vts ?
2
0
2
1
tatvs
?
??
?
?
t
v d ts
0
41
Rotation of a body about a fixed axis
Equation of rotation,
Angular velocity,
)( tf??
dt
d?? ?
2
2
dt
d
dt
d ??? ??Angular acceleration,
Uniform rotation,
Rotation with uniformly varying velocity,
t??? ?? 0
t??? ?? 02
00 2
1 tt ???? ??? ???? 2202 ??
42
刚体定轴转动
转动方程,
角速度,
)( tf??
dt
d?? ?
2
2
dt
d
dt
d ??? ??角加速度,
匀速转动,
匀变速运动,
t??? ?? 0
t??? ?? 0
2
00 2
1 tt ???? ??? ???? 2
2
02 ??
43
2.Steps for solving problems and things to consider carefully
1) Steps for solving problems,
① Analyze the problem carefully and make sure what the known
and what the unknowns are,
② Select a proper coordinate system,namely a rectangular
coordinate system or a natural coordinate system which can
describe the equations of motion most conveniently,
③ Perform the differential or integral calculation according to the
known variables,
④ Finally determine the related constants using initial conditions,
2) Pay special attention to,
① Geometric relations and directions of motion,
② Eliminating the parameter,t” while obtaining the
equation for the trajectories,
③ Selection of the appropriate coordinate systems,
44
二,解题步骤及注意问题
1.解题步骤,
① 弄清题意,明确已知条件和所求的问题。
②选好坐标系：直角坐标法，自然法。
③ 根据已知条件进行微分，或积分运算。
④ 用初始条件定积分常数。
对常见的特殊运动,
可直接应用公式计算 。
２．注意问题,
①几何关系和运动方向。
②求轨迹方程时要消去参数, t”。
③ 坐标系（参考系）的选择。
45
3,Examples
[Example 1] A locomotive moves along a curved track of radius
R=300m with uniformly varying velocity,The length of the
curved track is l=200m,The velocity of the locomotive when it
enters the track is v0=30km/h,and that when it leaves it is v1＝
48km/h,Find its accelerations when it enters and leaves the
curved track,respectively,
46
三,例题
[例 1]列车在 R=300m的曲线上匀变速行驶。轨道上曲线部分长
l=200m，当列车开始走上曲线时的速度 v0=30km/h，而将要离开
曲线轨道时的速度是 v1＝ 48km/h。
求列车走上曲线与将要离开曲线时的加速度？
47
??a
Solution,Since it move with uniformly varying velocity,const,
Employing and noting we have
savv ?2202 ??,m2 0 0?? ls
m / s
3
40
3 6 0 0
1 0 0 048
,m / s
3
25
3 6 0 0
1 0 0 030
m / s27.0
2 0 029
6 2 51 6 0 0
2
10
2
2
0
2
1
?
?
??
?
?
?
??
?
?
?
?
vv
s
vv
a
?
When it enters the curve,
Resultant acceleration
When it leaves the curve,
Resultant acceleration
2
22
0
0
2 m / s23.0
300
)3/25(,m / s27.0 ????
R
vaa
n?
'
0
1
0
22
0
2
0 2949tg,m / s356.0
?????? ?
n
n a
aaaa ?
? ?
2
22
1
1
2 m / s593.0
300
)3/40(,m / s27.0 ????
R
vaa
n?
'
1
1
1
22
1
2
1 3424tg,m / s652.0
?????? ?
n
n a
aaaa ?
? ?
48
??a
解,由于是匀变速运动，则 常量。
由公式 而由已知
savv ?2202 ??,m2 0 0?? ls
m / s
3
40
3600
100048
,m / s
3
25
3600
100030
m / s27.0
20029
6251600
2
10
2
2
0
2
1
?
?
??
?
?
?
??
?
?
?
??
vv
s
vv
a ?
列车走上曲线时,
全加速度
列车将要离开曲线时,
全加速度
2
22
0
0
2 m / s23.0
300
)3/25(,m / s27.0 ????
R
vaa
n?
'
0
1
0
22
0
2
0 2949tg,m / s356.0
?????? ?
n
n a
aaaa ?
? ?
2
22
1
1
2 m / s593.0
300
)3/40(,m / s27.0 ????
R
vaa
n?
'
1
1
1
22
1
2
1 3424tg,m / s652.0
?????? ?
n
n a
aaaa ?
? ?
49
〔 Example 2〕 Find the initial velocity and acceleration of a
projectile shown in the figure which has its hit point at A,
Analysis,Only when the projectile reaches its highest point A,
then vy＝ 0,
gtvdtdyv y ???? ?s i n0
Solution,the displacement can be
expressed as
2
0
0
2
1
s i n
c o s
gttvy
tvx
??
?
?
?
?
?
At point A vy=0,The time it takes from the origin to point A is,
g
vt ?s in0?
50
〔 例 2〕 已知如图，求 时正好射到 A点且用力最小。
,0 ?? v?
分析：只有在 A点,vy＝ 0且为最大高度时，用力才最小。
gtvdtdyv y ???? ?s i n0
解,由
2
0
0
2
1s i n
c o s
gttvy
tvx
??
?
?
?
?
?
由于在 A点时,vy=0，所以上升到最大高度 A点时所用
时间为,
g
vt ?s in0?
51
Substituting the above
equation into ① and ②,
yields
g
v
y
g
v
x AA
2
s i n
,
2
2s i n 22020 ??
??
?
?
?
?
?
?
tg
2
1
2s i n
s i n
5
5.1 2
2
2s i n
2
s i n
2
0
22
0
????
g
v
g
v
A
A
x
y
?????? 31,6.05 5.12tg ??
Substituting ?31?? into ③ we have
1 1 162s i n 58.922s i n220 ?????? ?Agxv m / s5.10
0 ?? v
52
将上式代入 ①和②，得,
g
vy
g
vx
AA 2
s i n,
2
2s i n 22020 ?? ?? ?
?
?
?
?
?
tg
2
1
2s i n
s i n
5
5.1 2
2
2s i n
2
s i n
2
0
22
0
????
g
v
g
v
A
A
x
y
?????? 31,6.05 5.12tg ??
将 ?31?? 代入③,得
1 1 162s i n 58.922s i n220 ?????? ?Agxv
m / s5.100 ?? v
53
① The number of revolution of the pulley in 3
seconds；
② The displacement of weight B in 3 seconds；
③ The velocity of weight B at ｔ ＝ 3s；
④ The acceleration of point C on the rim of
the pulley at the initial time；
⑤ The acceleration of point C at t＝ 3s,
,
Find,
,m5.0?R
〔 Example 3〕
The acceleration of weight A in the figure is 2m /s1?
Aa
and its initial velocity is m/s5.1
0 ?v
m.30,r ?
（ constant）
54
〔 例 3〕 已知：重物 A的 2m/s1?
Aa
（常数）初瞬时速度 m/s5.1
0 ?v
方向如图示。 求：,m5.0?R ? m30,r ?
① 滑轮 3s内的转数；
②重物 B在 3s内的行程；
③重物 B在 ｔ ＝ 3s时的速度；
④滑轮边上 C点在初瞬时的加速度；
⑤滑轮边上 C点在 ｔ ＝ 3s时的加速度。
55
2r a d / s 2
5.0
1 ???
R
a C ??
r a d / s35.0 5.1,m / s5.1 0 ????? Rvvv CAC ?
） constant （
（ ）
,m / s1 2?? AC aa ?
Solution,
① If the rope is not extensible we have
m4.5183.0 ???? ?rs②
? ?????? m / s7.293.0?rv B
）,
③ r a d / s 9323 0 ?????? t???
（
86.2
2
,r a d1832
2
1
33
2
1 22
0
??
?????????
?
?
???
n
tt
(revolutions)
56
2r a d / s 2
5.0
1 ???
R
a C ??
r a d / s35.0 5.1,m / s5.1 0 ????? Rvvv CAC ?
）常数 （
（ ）
,m / s1 2?? AC aa ?
解, ① 因为绳子不可以伸长，所以有
转）(86.22,r a d1832213321 220 ??????????? ????? ntt
m4.5183.0 ???? ?rs②
? ?????? m / s7.293.0?rv B
）,
③ r a d / s 9323 0 ?????? t???
（
57
④ At t = 0,
222
0
2
m / s5.435.0
,m / s1
?????
??
?
?
Ra
aa
n
C
AC
?????
??????
5.12,222.0
5.4
1
tg
m / s 61.45.41)()( 22222
??
?
?
n
C
C
n
CCC
a
a
aaa
⑤ At t=3s,
222 m / s5.4095.0,m / s1 2 ??????? ?? Raaa nCAC
???????? 41.1,0247.051.40 1 t g,m / s51.405.401 222 ??Ca
58
④ t = 0 时,
222
0
2
m / s5.435.0
,m / s1
?????
??
?
?
Ra
aa
n
C
AC
?????
??????
5.12,222.0
5.4
1
tg
m / s 61.45.41)()( 22222
??
?
?
n
C
C
n
CCC
a
a
aaa
⑤ t=3s 时,
222 m / s5.4095.0,m / s1 2 ??????? ?? Raaa nCAC
???????? 41.1,0247.051.40 1 t g,m / s51.405.401 222 ??Ca
59
[Example 4] A wheel rotates starting at rest with constant angular
acceleration,OM＝ 0.4m,At a certain time,the acceleration of point
M is measured to be, ??? 30,m / s 40
2 ?Ma
Find, The equation of rotation;
The velocity and the centripetal
acceleration of point M at t＝ 5s,
??? s in??? aRa
2r a d / s 50
4.0
30s i n40s i n ???????
R
a
R
a ?? ?
Solution,?
22200 25502121,0 tttt ???????? ?????
225:r o t a t i o n ofE q u a t i o n t??
M
60
[例 4] 已知：圆轮 O由静止开始作等加速转动,OM＝ 0.4m,
在 某瞬时测得 ??? 30,m / s 40
2 ?Ma
求,?转动方程 ;
? t＝ 5s时,Ｍ 点的速度和
向心加速度的大小。
??? s in??? aRa
2r a d / s 50
4.0
30s i n40s i n ???????
R
a
R
a ?? ?
解,?
22200 25502121,0 tttt ???????? ?????
225 t??转动方程
M
61
?
ttRvtt M 20504.0,50 0 ??????? ????
2
22
m / s2 5 0 0 04.01 0 0 ??? Rva MnM
?atｔ ＝ 5s,m / s1 0 0520 ???
Mv
M
62
?
ttRvtt M 20504.0,50 0 ??????? ????
2
22
m / s2 5 0 0 04.01 0 0 ??? Rva MnM
?当 ｔ ＝ 5s时,m / s1 0 0520 ???
Mv
M
63
〔 Example 5〕 Draw the velocity and acceleration vectors of the
pointsＭ and Ｎ on the rigid bodies shown in the following figures,
),( 2121 ABOOBOAO ??
64
〔 例 5〕 试画出图中刚体上 Ｍ ?Ｎ 两点在图示位置时的速度和
加速度。 ),(
2121 ABOOBOAO ??
65
〔 Example6〕 Referring to the figure,a projectile is fired from the
point O with an angle of elevation, Show that the angle under which
the projectile reaches the line L in the shortest time is, 2?
For y = h,the relation between t and
is
0?
? 2
00 2
1s i n gttvh ?? ?
200
2
1s i n gttvy ?? ? ?
Proof,choosing the coordinate
system,we get xoy
O
0?
66
〔 例 6〕 已知如图，从 O点以
任一角度抛出一质点，试证明
质点最早到达直线Ｌ的抛角为 。
2
?
（与上升的最大高度无关，只要
求时间对抛射角度的变化率）
到达高度为 h 时,t 与 的
关系有下式确定
0?
? 2
00 2
1s i n gttvh ?? ?
200
2
1s i n gttvy ?? ? ?
解,选 坐标系，则 xoy
O
67
00
000 22
1)s i nc o s(0
???? d
dttg
d
dttv ?????
To minimize the time you have to solve
0
0
??d dt
?
The derivative of ? with respect to
gives 0?
00
0000 s i nc o s0 ???? d
dttg
d
dtvtv ?????
??
Inserting into ? yields
0
0
??ddt ;0c o s 00 ??tv
note 0c o s,0
00 ??? ?tv 2
?? ??
which gives the angle for the shortest time
which the projectile takes from O to the line L,
68
00
000 22
1)s i nc o s(0
???? d
dttg
d
dttv ?????
欲使最早到达，必须满足
0
0
??d dt
?
将 ?对 求导数 0?
00
0000 s i nc o s0 ???? d
dttg
d
dtvtv ?????
??
将 （最早到达的条件） 代入 ?，得
0
0
??ddt ;0c o s 00 ??tv
又 0c o s,0
00 ??? ?tv 2?? ??
证毕。
表示出在某一角度下时间会最短。（极值）
69
70
