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Theoretical mechanics
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§ 8–1 Concept of compositive motion of a particle
§ 8–2 Theorem of composition of velocities of a
particle
§ 8–3 Theorem of composition of accelerations
when the convected motion is translation
§ 8–4 Theorem of composition of accelerations
when the convected motion is rotation
Lesson for problem solving
Chapter 8,Compositive motion of a particle
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§ 8–1 点的合成运动的概念
§ 8–2 点的速度合成定理
§ 8–3 牵连运动为平动时点的加速度合成定理
§ 8–4 牵连运动为转动时点的加速度合成定理
习题课
第八章 点的合成运动
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§ 8-1 concept of compositive motion of a particle
1,Coordinate system,
1) Static coordinate system,A coordinate system fixed to the earth ground
is called Static coordinate system (SCS),
2) Moving coordinate system,A coordinate system fixed to a moving object
relative to the earth ground is called moving coordinate system (MCS),
For example,a running car,
In the previous two chapters,we have considered the motion of a particle
or a rigid body with respect to a system of fixed coordinate axes,In practice,
however,we often observe the motion with respect to a moving body,For
example,we observe the motion of an flying airplane from a moving car,or
observe the motion of raindrops from a traveling train,
Why do obtain different results when we observe the motion a object on
different coordinate systems or reference objects? It is known that there may
exist relationship between two different objects,Next,we are going to study
the relationship between the motions of an object and the reference object,
First of all,we begin with the introduction of some concepts,
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§ 8-1 点的合成运动的概念
一．坐标系,
1.静坐标系,把固结于地面上的坐标系称为静坐标系,简称静系。
2.动坐标系,把固结于相对于地面运动物体上的坐标系,
称为动坐标系，简称动系。例如在行驶的汽车。
前两章中我们研究点和刚体的运动，一般都是以地面为参考
体的。然而在实际问题中，还常常要在相对于地面运动着的参
考系上观察和研究物体的运动。例如，从行驶的汽车上观看飞
机的运动等，坐在行驶的火车内看下雨的雨点是向后斜落的等。
为什么在不同的坐标系或参考体上观察物体的运动会有不
同的结果呢？ 我们说事物都是相互联系着的。下面我们就将研
究参考体与观察物体运动之间的联系。为了便于研究，下面先
来介绍有关的概念。
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3,Three kinds of motion and their velocities and accelerations
1) Absolute motion,motion of the moving point relative to the SCS,
2) Relative motion,motion of the moving point relative to the MCS,
For example,motions of a person in a running bus
3) Convected motion,motion of the MCS relative to the SCS,
For example,motion of a running care relative to the earth ground,
The velocity and acceleration of the moving point in its absolute motion are called absolute velocity and absolute acceleration
The velocity and acceleration of the moving point in its relative motion are
called relative velocity and relative acceleration
The velocity and acceleration of the convected point in its absolute motion
are called convected velocity and convected acceleration
aa
ev ea
rv ra
av
convected point,the point in the MCS coinciding with the moving point at any
instant of time,Image that the moving point were fixed to the MCS and moved
together with the MCS,Such a point in the MCS is called convected point,
Motion of
a particle
Motion of a
rigid body
2,Moving point,a moving point under studying
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三．三种运动及三种速度与三种加速度。
１．绝对运动,动点对静系的运动。
２．相对运动,动点对动系的运动。
例如：人在行驶的汽车里走动。
３．牵连运动,动系相对于静系的运动
例如：行驶的汽车相对于地面的运动。
绝对运动中,动点的速度与加速度称为 绝对速度 与 绝对加速度
相对运动中,动点的速度和加速度称为 相对速度 与 相对加速度
牵连运动中,牵连点的速度和加速度称为 牵连速度 与 牵连加速度
aa
ev ea
rv ra
av
牵连点,在任意瞬时，动坐标系中与动点相重合的点，也就是
设想将该动点固结在动坐标系上，而随着动坐标系一起运动时
该点叫牵连点。
点的运动
刚体的运动
二．动点,所研究的点（运动着的点）。
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Explanation of the concepts by examples,
4,Guideline for selection of moving point,
In general,the contact point between the drive and driven objects is usually
selected as moving point,which moves relative to both SCS and MCS,
5,Guideline for MCS selection,
A MCS should be a coordinate system (object) on which the trajectory of the
moving point is known or clear,
Moving
point,
MCS,
SCS,
Point A on rod AB
Fixed to cam Ｏ '
Fixed to the ground
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下面举例说明以上各概念,
四．动点的选择原则,
一般选择主动件与从动件的连接点，它是对两个坐标系都有
运动的点。
五．动系的选择原则,
动点对动系有相对运动，且相对运动的轨迹是已知的，
或者能直接看出的。
动点,
动系,
静系,
AB杆上 A点
固结于凸轮 Ｏ '上
固结在地面上
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Relative motion
convected motion,
Curvilinear motion (arc )
Rectilinear translation
Absolute motion,Rectilinear motion
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相对运动,
牵连运动,
曲线（圆弧）
直线平动
绝对运动, 直线
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evrvav
Absolute velocity,
relative velocity,convected velocity,
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evrvav
绝对速度,
相对速度, 牵连速度,
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Absolute acceleration,
Relative acceleration,
convected acceleration,
aa
ea
ra
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绝对加速度,
相对加速度,
牵连加速度,
aa
ea
ra
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Moving point,A（ on disk)
MCS,rod O'A
SCS,support
Absolute motion,curvilinear (circle)
Relative motion,rectilinear motion
convected motion,rotation
Moving point,A1 (on rod O'A1)
MCS,circular disk
SCS,support
Absolute motion,curvilinear (arc)
Relative motion,curvilinear motion
convected motion,rotation
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动点,A（在圆盘上 )
动系,O'A摆杆
静系：机架
绝对运动：曲线（圆周）
相对运动：直线
牵连运动：定轴转动
动点,A1（在 O'A1 摆杆上 )
动系：圆盘
静系：机架
绝对运动：曲线（圆弧）
相对运动：曲线
牵连运动：定轴转动
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If the moving point is on the CAM
Moving point,A (on rod AB) A（ on the cam）
MCS,cam rod AB
SCS,earth ground earth ground
Absolute motion,rectilinear Circle ( red dashed line)
Relative motion,curvilinear (circle) curve（ path unknown
convected motion,rotation Translation
[Note] Clear state that to which
object the moving point is attached,
The moving point can not be
attached to the MCS,otherwise its
relative motion is rest on the MCS,
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若动点 A在偏心轮上时
动点,A(在 AB杆上 ) A（在偏心轮上）
动系：偏心轮 AB杆
静系：地面 地面
绝对运动：直线 圆周（红色虚线）
相对运动：圆周（曲线） 曲线（未知）
牵连运动：定轴转动 平动
[注 ] 要指明动点应在哪个
物体上,但不能选在
动系上。
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§ 8-2 Theorem of composition of velocities of a particle
The relationship among the absolute,relative and convected velocity will be
established and stated in the theorem of composition of velocities of a particle,
1MM
＝ 'MM ＋ '
1MM
When t ? t+△ t AB?A'B'
M?M'
Equivalent to M ? M１ ? M′
MM ' is the absolute path
MM ' is the absolute displacement
M1M ' is the relative path
M1M ' is the relative displacement
t
MM
t
MM
t
MM
ttt ??? ???
????
???
1
0
1
00
limlimlim
t?Divided the above equation by,
0??t,gives When
1,Proof
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§ ８－２点的速度合成定理
速度合成定理将建立动点的绝对速度，相对速度和牵连速度
之间的关系。
1MM
＝ 'MM ＋ '
1MM
当 t ? t+△ t AB?A'B'
M?M'
也可看成 M ? M１ ? M′
MM ' 为绝对轨迹
MM ' 为绝对位移
M1M ' 为相对轨迹
M1M ' 为相对位移
t
MM
t
MM
t
MM
ttt ??? ???
????
???
1
0
1
00
limlimlim
t?将上式两边同除以 后,
0??t 时的极限，得 取
一．证明
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Explanations,?va— absolute velocity of the moving point;
?vr— relative velocity of the moving point;
?ve— convected velocity of the moving point,
which is the velocity of a point
(convected
point) on the MCS;
I) When MCS is in translation,all points in it have the
same velocity,
II) When MCS rotates,ve must be the velocity of the point
in the MCS coinciding with the moving point,
Conclusion,At any instant of time,the absolute velocity of a moving point
equals to the geometric sum of its relative velocity and convected velocity,
This is the theorem of composition of velocities of a particle,
rea vvv ???
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说明,?va— 动点的绝对速度；
?vr— 动点的相对速度；
?ve— 动点的牵连速度，是动系上一点 (牵连点 )的速度
I) 动系作平动时，动系上各点速度都相等。
II) 动系作转动时,ve必须是该瞬时动系上与
动点相重合点的速度。
即在任一瞬时动点的绝对速度等于其牵连速度与相对速度的
矢量和，这就是点的速度合成定理。
rea vvv ???
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?The equation given in the theorem of composition of velocities
is a vector equation at an instant,which includes six components
such as magnitudes and directions,If any four components are
given,the other two unknowns can be solved from this equation,
2,Examples of theorem application
[Example1] The small
vehicle in a bridge cane
moves along its horizontal
track with a uniform velocity
vh,and weight A moves
vertically at a velocity vv
relative to the small vehicle,
Find the absolute velocity of
weight A,
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?点的速度合成定理是瞬时矢量式，共包括大小 ?方向 六个元
素，已知任意四个元素，就能求出其他两个。
二．应用举例
[例 1] 桥式吊车 已知：小
车水平运行，速度为 v平,
物块 A相对小车垂直上升
的速度为 v?。求物块 A的运
行速度。
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Draw the parallelogram of velocities shown in the figure,
the magnitude and direction of the velocity of the weight A can be obtained as
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?????? vvvvvv reaA
2
平
平v
v ??? 1tg?
Solution,moving point,weight A
MCS,small vehicle
SCS,earth ground
Relative motion,rectilinear;
Relative velocity vr =vv,direction?
convected motion,translation;
convected velocity ve=vh,direction
?
Absolute motion, curvilinear;
Absolute velocity va,magnitude and
direction are unknown
Employing the theorem of composition of velocities,yields
rea vvv ??
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作出速度平四边形 如图示，则物块Ａ的速度大小和方向为
222
?????? vvvvvv reaA
2
平
平v
v ??? 1tg?
解,选取 动点, 物块 A
动系, 小车
静系, 地面
相对运动, 直线 ;
相对速度 vr =v? 方向 ?
牵连运动, 平动 ;
牵连速度 ve=v平 方向 ?
绝对运动, 曲线 ;
绝对速度 va 的大小,方向待
求 由速度合成定理,rea vvv ??
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Solution,Choose point A on rod OA杆 as the moving
point,attach the MCS to rod O1B,and use the earth
ground as the SCS,
absolute velocity va = r ? direction? OA
relative velocity vr =? direction//O1B
convected velocity ve =? direction ?O1B
2
2
2
2
2
1
111
22
2
22
2
22
1
,m o r e o v e r,
s i n,s i n
lr
r
lr
r
lrAO
v
AOv
lr
r
vv
lr
r
e
e
ae
?
?
?
?
?
?????
?
???
?
?
??
??
?
??
?
?
[Example 2] In the mechanism shown in the figure,OA=
r,?,OO1=l,At this moment,OA?OO1
Find,angular velocity ?1 of rod O1B,
According to va= vr+ ve,draw the parallelogram of velocities as shown in the
figure,
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解, 取 OA杆上 A点为动点，摆杆 O1B为动系,
基座为静系。
绝对速度 va = r ? 方向 ? OA
相对速度 vr =? 方向 //O1B
牵连速度 ve =? 方向 ?O1B
2
2
2
2
2
1
111
22
2
22
2
22
1
,
s i n,s i n
lr
r
lr
r
lrAO
v
AOv
lr
r
vv
lr
r
e
e
ae
?
?
?
?
?
?????
?
???
?
?
??
??
?
??
?
?
又 （ ）
[例 2] 曲柄摆杆机构
已知, OA= r,?,OO1=l 图示瞬时 OA?OO1
求,摆杆 O1B角速度 ?1
由速度合成定理 va= vr+ ve 作出 速度平行四边形 如图示。
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According to the equation va= vr+ ve,
Draw the parallelogram of velocities as shown in the figure,
Solution,Select point A on rod AB as the moving point,
fix the MCS to the disk and the SCS to the support,
absolute velocity va =? unknown,directin//AB
relative velocity vr =? unknown,direction?CA
convected velocity ve =OA??=2e?,direction? OA
（ Look at the animation
on the next page）
)(3 32 3 3230 0 ?????? ?? evetgvv ABea
[Example 3] In the cam-rod mechanism,OC＝
e,,?（ constant）,and the moment,
OC?CA and point O,A and B are collinear,
Find,the velocity of the driven rod AB,
eR 3?
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由速度合成定理 va= vr+ ve,
作出速度平行四边形 如图示。
解,动点 取直杆上 A点,动系 固结于圆盘,
静系 固结于基座。
绝对速度 va =? 待求,方向 //AB
相对速度 vr =? 未知,方向 ?CA
牵连速度 ve =OA??=2e?,方向 ? OA
（翻页请看动画）
)(3 32 3 3230 0 ?????? ?? evetgvv ABea
[例 3] 圆盘凸轮机构
已知,OC＝ e,,?（匀角速度）
图示瞬时,OC?CA 且 O,A,B三点共线。
求,从动杆 AB的速度。
eR 3?
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As evidenced from the examples,the general steps should be followed
in solving problems about the composition of velocities,
? Select moving point,MCS and SCS;
? Analysis of the related three kinds of motions;
? Analysis of three kinds of velocities;
?According to the theorem of composition of velocities,
draw the parallelogram of velocities;
? Find unknowns from the parallelogram of velocities;
Proper selection of the moving point,MCS and SCS is the key point in
solving such problems,
,rea vvv ??
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由上述例题可看出，求解合成运动的速度问题的 一般步骤 为,
? 选取动点，动系和静系。
? 三种运动的分析。
? 三种速度的分析。
? 根据速度合成定理 作出速度平行四边形。
? 根据速度平行四边形，求出未知量。
恰当地选择动点、动系和静系是求解合成运动问题的关键。
,rea vvv ??
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Guidelines for the selection of moving point,MCS and SCS
?The moving point,MCS and SCS must be in three different rigid bodies,
Otherwise,one motion out of the absolute,relative and convected motion will
be absent (always resting indeed),and consequently it will not suitable for
the application of the composition of motions,
? The relative path of the moving point relative to the MCS must be easy to
determine,(except to find the convected motion when the absolute and relative
motions are given,)
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动点、动系和静系的选择原则
? 动点、动系和静系必须分别属于三个不同的物体,
否则绝对、相对和牵连运动中就缺少一种运动，不
能成为合成运动
? 动点相对动系的相对运动轨迹易于直观判断 （已
知绝对运动和牵连运动求解相对运动的问题除外）。
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Analysis,Since the contact point between these two objects
changes with time,their contact point is not suitable to be a
moving point,otherwise the relative motion is very complicated
In this case,we need to select another point instead of the contact
point as the moving point which should satisfy the guideline,
?? 30,?v[Example 4] The radius of the cam is r,and as shown
in the diagram,Rod OA is leaning against the cam,Determine the
angular velocity of rod OA,
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分析,相接触的两个物体的接触点位置都随时间而变化，
因此两物体的接触点都不宜选为动点，否则相对运动的分析
就会很困难。这种情况下，需选择满足上述两条原则的非接
触点为动点。
[例４ ] 已知, 凸轮半径 r,图示时 杆 OA靠在凸轮上。
求：杆 OA的角速度。;30,???v
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Solution,Select point C as the
moving point,attach the MCS to
rod OA and the SCS to the base,
Absolute motion,rectilinear motion,
Absolute velocity,
Relative motion,rectilinear motion,
Relative velocity,
convected,rotation,convected velocity,
?? d ir e c t io n,vv a
OCOCv e ??? d i r e c t i o n u n k n o w n,,??
According to,
rea vvv ??
draw the parallelogram of velocities as shown in the figure,
r
vv
rr
v e
6
3
3
3
2
1
2 ????? ?
vvv ae 3 3tg ??? ?
（ ）
,2s i n ???? rrOCv e ?????
OAv r //d i r e ct i o n i t s u n k n o w n,is
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解, 取凸轮上 C点为 动点,
动系 固结于 OA杆上,
静系 固结于基座。
绝对运动, 直线运动,
绝对速度,
相对运动, 直线运动,
相对速度,
牵连运动, 定轴转动,牵连速度,
??,方向vv a
??
OCOCv e ??? 方向待求未知,,??
方向未知,rv OA
如图示。 根据速度合成定理,
rea vvv ??
做出速度平行四边形
r
vv
rr
v e
6
3
3
3
2
1
2 ????? ?
vvv ae 3 3tg ??? ?
（ ）
,2s i n ???? rrOCv e ?????又
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§ 8-3 Theorem of composition of accelerations when the
convected motion is translation
rea vvv ??
Noting the convected motion is
translation,we have
Employing theorem of composition of
velocities
'',OeOe aavv ??
''''''v a n d r kdtdzjdtdyidtdx ???
kdtdzjdtdyidtdxvv Oa ???????? ? '''
Differentiating it with respect to t,
'''''' 2
2
2
2
2
2
kdt zdjdt ydidt xddtvddtvda Oaa ????? ?
A point moves along a curved path AB attached to the MCS O'x'y'z',at the
same time,the path AB also has translatory motion together with the MCS
O'x'y'z' relative to the SCS Oxyz,
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§ 8-3 牵连运动为平动时点的加速度合成定理
rea vvv ??
由于 牵连运动为平动,故
由速度合成定理
'',OeOe aavv ??
''''''v r kdtdzjdtdyidtdx ???而
kdtdzjdtdyidtdxvv Oa ???????? ? '''
对 t求导,''''''
2
2
2
2
2
2
kdt zdjdt ydidt xddtvddtvda Oaa ????? ?
设有一动点 M按一定规律沿着固连于动系 O'x'y'z' 的曲线 AB
运动,而曲线 AB同时又随同动系 O'x'y'z' 相对静系 Oxyz平动。
47
0',0',0' ??? dt zddt yddt id
(where are the unit vectors of the MCS,and their direction do not
change due to the translation of the MCS,Hence ),
moreover
',',' kji
rea aaa ???
— Theorem of composition of accelerations when the
convected motion is translation i,e,when the convected motion is translation,
the absolute acceleration of a moving point equals to the geometric sum of its
relative and convected accelerations,
'''''',2
2
2
2
2
2
'
' k
dt
zdj
dt
ydi
dt
xdaaa
dt
vd
reO
O ??????
naaa ?? ? ?
n
rrneenaa aaaaaa ?????
???
∴ Its general form is,
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0',0',0' ??? dt zddt yddt id
(其中 为动系坐标的单位矢量，因为动系为平动，故它
们的方向不变，是常矢量，所以 ）
',',' kji
rea aaa ???
— 牵连运动为平动时点的加速度合成定理
即当牵连运动为平动时，动点的绝对加速度等于牵连加速度
与相对加速度的矢量和。
'''''',2
2
2
2
2
2
'
' k
dt
zdj
dt
ydi
dt
xdaaa
dt
vd
reO
O ??????又
naaa ?? ? ?
n
rrneenaa aaaaaa ?????
???
∴ 一般式可写为,
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Solution,Choose point A on rod AB
as the moving point,and attach the
MCS onto the cam,
[Example 1] The radius of the cam is
are also given,
Find the velocity of rod AB when ? =60o,oo avR a n d,
Look at the animation
50
解,取杆上的 A点为 动点,
动系 与凸轮固连。
[例 1] 已知：凸轮半径
求,? =60o时,顶杆 AB的加速度。
oo avR,,
请看动画
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Absolute velocity va =?,direction??AB ； absolute acceleration aa=?,
direction??AB,to be solved,
Relative velocity vr =?,direction?CA; relative acceleration ar? =?
direction?CA,direction along CA and pointing to C,
convected velocity ve=v0,direction→ ; convected acceleration ae=a0,direction
→,
Appling
,rea vvv ??
draw the parallelogram of velocities
shown in the figure above 00 3260s i ns i n vvvv oer ??? ?
? ?
Rva rnr /2?
52
绝对速度 va =?,方向 ??AB ；绝对加速度 aa=?,方向 ??AB,待求。
相对速度 vr =?,方向 ?CA; 相对加速度 ar? =? 方向 ?CA
,方向沿 CA指向 C
牵连速度 ve=v0,方向 → ; 牵连加速度 ae=a0,方向 →
由速度合成定理
,rea vvv ??
做出速度平行四边形,如图示。
0
0
3
2
60s i ns i n v
vvv
o
e
r ??? ?
? ?
Rva rnr /2?
53
Noting the convected motion is translation,
we have
nrea aaaa
r ???
?
R
vRvRvaw
r
n
r 3
4/)
3
2(/ h e r e 202
0
2 ???
Drawing vector diagram of accelerations show
in the figure,and projecting them to the normal
n,yields
nrea aaa ?? ?? c o ss i n
?? 60s i n/)
3
460c o s(s i n/)c o s( 20
0 R
vaaaa n
rea ???? ??
After arrangement,finally we have )
3
8(
3
3 20
0 R
vaaa
aAB ???
?
n
54
因 牵连运动为平动,故有
nrea aaaa
r ???
?
R
vRvRva
r
n
r 3
4/)
3
2(/ 202
0
2 ???其中
作 加速度矢量图 如图示,
将上式投影到法线上，得
nrea aaa ?? ?? c o ss i n
?? 60s i n/)
3
460c o s(s i n/)c o s( 20
0 R
vaaaa n
rea ???? ??
整理得 )
3
8(
3
3 20
0 R
vaaa
aAB ???
?
n
55
§ 8-4 Theorem of composition of accelerations
when the convected motion is rotation
In last section,we have proved the theorem of composition of
accelerations when the convected motion is translation,Is this
theorem still valid when the convected motion is rotation? Before
answering this question,let’s analyze a typical example,
A disk is rotating about a fixed axis
Ｏ with a uniform angular velocity ?,A
point M moves along a circular path
slotted in the disk with a constant vr,
what’s the absolute acceleration of the
point M?
56
§ 8-4 牵连运动为转动时点的加速度合成定理
上一节我们证明了牵连运动为平动时的点的加速度合成定
理,那么当牵连运动为转动时，上述的加速度合成定理是否还
适用呢？下面我们来分析一特例。
设一圆盘以匀角速度 ? 绕定轴 Ｏ 顺
时针转动，盘上圆槽内有一点 M以大
小不变的速度 vr 沿槽作圆周运动，那
么 M点相对于静系的绝对加速度应是
多少呢？
57
R
vav r
rr
2
,c o n s t, h a v e,we ??
Since the relative motion is a uniform circular
motion,
(direction is shown in the figure)
Applying the theorem of composition of velocities gives
c o n s t,????? rrea vRvvv ?
Choose M as the moving point,and the
MCS is attached to the disk,
In this case,the convected motion of the point
M is a uniform rotation,
RaRv ee 2,?? ??
(direction is shown in the
figure)
It is shown that the absolute motion of point M is also a uniform circular motion,
and therefore
It points to the centerＯ,
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
58
R
vav r
rr
2
,?? 常数有
相对运动 为匀速圆周运动,
（方向如图）
由速度合成定理可得出
常数????? rrea vRvvv ?
选点 M为动点，动系固结与圆盘上,
则 M点的 牵连运动 为匀速转动
RaRv ee 2,?? ?? （方向如图）
即 绝对运动 也为匀速圆周运动，所以
方向指向圆心 Ｏ 点
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
59
it can found from the above equation that in addition to
an extra term 2? vr appears,
It is clear that the absolute acceleration of a moving point is
no longer equal to the geometric sum of its relative acceleration
and convected acceleration when the convected motion is
rotation, What is the correct relationship among these
accelerations？ What is the additional term 2? vr why does it
appear? Next,we are going to seek the answers to these questions
and derive the theorem of composition of accelerations when the
convected motion is rotation,
ea
ra
aa
,a n d / 22 ?RaRva err ??
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
60
分析上式,还多出一项 2? vr 。
可见,当牵连运动为转动时，动点的绝对加速度 并不
等于牵连加速度 和相对加速度 的矢量和。 那么他们
之间的关系是什么呢？ 2? vr 又是怎样出现的呢？它是什么
呢？ 下面我们就来讨论这些问题，推证牵连运动为转动时点
的加速度合成定理。
ea ra
aa
,,/ 22 ?RaRva err ??
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
61
Analysis of three velocities
convected velocity
Relative velocity
Absolute velocity
At instant t,the system is in position Ｉ and at t+?t,it is in position II
ev
rv
rea vvv ?? ''' rea vvv ??
'ev
'rv
It can be observed that the magnitudes and directions of both
convected and relative velocities vary after an interval ?t,
A rod OA rotates about axis O with a
constant angular velocity ?,and the sleeve
M moves along rod OA with a varying
velocity,Choose sleeve M as the moving
point,and attach the MCS to rod OA and
the SCS to the static support,
62
三种速度分析
牵连速度
相对速度
绝对速度
t 瞬时在位置Ｉ t+?t 瞬时在位置 II
ev
rv
rea vvv ?? ''' rea vvv ??
'ev
'rv
可以看出，经过 ?t 时间间隔，牵连速度和相对速度的大
小和方向都变化了。
设有已知杆 OA在图示平面内以匀
? 绕轴 O转动，套筒 M（可视为点
M）沿直杆作变速运动。 取套筒 M
为动点，动系固结于杆 OA上，静
系固结于机架。
63
where -- the magnitude change of the relative velocity in ?t,which is
independent on the convected motion,
-- 在 direction change of the relative velocity caused by the convected
motion in ?t,which is dependent on the convected angular velocity ?,
Analysis of the velocity change in an interval ?t
?relative velocity,draw the triangle composed of
After cutting a length of in the vector,can be decomposed
into and,
rrr vvv ? a n d ',
'rvrv rv?
'rv? ''rv?
'''.i, e rrr vvv ?????
'rv?
''rv?
?convected velocity
draw the triangle composed of
After cutting a length of in the vector,
can be decomposed into and,
eee vvv ? a n d ',
'evev
ev? 'ev? ''ev?
''' i, e, eee vvv ?????
64
其中 -- 在 ?t内相对速度大小的改变量，它与牵连转动无关。
-- 在 ?t内由于牵连转动而引起的相对速度方向的改变
量，与牵连转动的 ? 的大小有关 。
?t 时间间隔内的速度变化分析
?相对速度,由 作速度矢量三角形,
在 矢量上截取 长度后,分解为 和
rrr vvv ?,',
'rv rv rv? 'rv? ''rv?
''' rrr vvv ?????即
'rv?
''rv?
?牵连速度,
由 作速度矢量三角形,
在 矢量上截取等于 长后,
将 分解为 和,
eee vvv ?,',
'ev ev
ev? 'ev? ''ev?
''' eee vvv ?????即
65
where
— represents the direction change of the convected velocity caused by
the convected rotation in an interval ?t,which is independent on the relative motion,
— represents the magnitude change of the convected velocity caused by
the relative motion in an interval ?t,which is dependent on the relative velocity,
'ev?
''ev?
rv
Acceleration analysis
According to the
definition,t vvvvt vva reretaata ? ??????? ???? )()''(lim'lim 00
t
v
t
v
t
vvvv r
t
e
t
rree
t ?
??
?
??
?
????
?????? 000
limlim)'()'(lim
Physical interpretations,
The first term,magnitude
eet
e
t
aOMtvtv ?????????
????
2
00
lim'lim ??
t
v
t
v
t
v
t
v r
t
r
t
e
t
e
t ?
??
?
??
?
??
?
??
????????
''lim'lim''lim'lim
0000
66
其中,
— 表示 ?t内由于牵连转动而引起的牵连速度方向的改
变量，与相对运动无关。
— 表示 ?t内动点的牵连速度，由于相对运动而引起的
大小改变量，与相对速度 有关。
'ev?
''ev?
rv
加速度分析
根据加速度定义
t
vvvv
t
vva rere
t
aa
ta ?
????
?
??
????
)()''(lim'lim
00
t
v
t
v
t
vvvv r
t
e
t
rree
t ?
??
?
??
?
????
?????? 000
limlim)'()'(lim
上式中 各项的物理意义 如下,
第一项大小,
eet
e
t
aOMtvtv ?????????
????
2
00
lim'lim ??
t
v
t
v
t
v
t
v r
t
r
t
e
t
e
t ?
??
?
??
?
??
?
??
????????
''lim'lim''lim'lim
0000
67
Direction,as ?t ?0 and ?? ?0,it will gradually point to point A along the rod,
Therefore,this term is the convected acceleration at instant t,
ea
The third term,magnitude,represents the change
rate of the magnitude of, rrrt adtvdtv ?????? 'lim 0
rv
Direction,along the rod,
so,this term is the relative acceleration of the moving point at instant t,
ra
The second term,magnitude,
t
OMOM
t
vv
t
v
t
ee
t
e
t ?
????
?
??
?
?
??????
??'lim'lim''lim
000
rrt vvt
MM ????
???? ?? d i r e c t i o n,
'lim 1
0
This term is the acceleration resulting from the magnitude change of the convected
velocity due to the existence of relative motion,
The forth term：
magnitude,d i r e c t i o n,lim''lim
00 rrrt
r
t
vv
t
v
t
v ???
?
????
?
?
????
This term represents the acceleration resulting from the direction change of
the relative velocity caused by the convected rotation,
68
方向,?t ?0时,?? ?0,其方向沿着直杆指向 A点。
因此,第一项正是 t 瞬时动点的牵连加速度 。
ea
第三项大小,为对应于 大小改变
rrrt adt
vd
t
v ??
?
?
??
'lim
0
rv
方向：总是沿直杆。
因此,该项恰是 ｔ 瞬时动点的相对加速度 。 ra
第二项大小,
t
OMOM
t
vv
t
v
t
ee
t
e
t ?
????
?
??
?
?
??????
??'lim'lim''lim
000
rrt vvt
MM ???
??? ?? 方向,
'lim 1
0 ??
该项为由于相对运动的存在而引起牵连速度的大小改变的加速度。
第四项大小,
。方向,lim''lim
00 rrrt
r
t
vvtvtv ????
??
?? ????
??
这一项表明由于牵连转动而引起相对速度方向改变的加速度。
69
Finally,we obtain the theorem of composition of accelerations with
a rotating convected motion as follow
krea aaaa ???
When the convected motion is rotation,the absolute acceleration of a moving
point equals to the geometric sum of its convected,relative and Coriolis
accelerations,
More general form,
knrrneenaa aaaaaaa ?????? ???
The Coriolis acceleration can also be expressed as the following vector
form,
ka
rk va ?? ?2
w i t h s i d e s a m e t h e t op o i n t i n g,d i r e c t i o n,2 ???? rrk vva
Since the second and the fourth term have the same magnitude and direction,
they can be written into one term and denoted by,which is called Coriolis
acceleration,k
a
70
所以，当牵连运动为转动时，加速度合成定理为
krea aaaa ???
当牵连运动为转动时，动点的绝对加速度等于它的牵连加速
度，相对加速度和科氏加速度三者的矢量和。
一般式
knrrneenaa aaaaaaa ?????? ???
一般情况下 科氏加速度 的计算可以用矢积表示 ) ( 不垂直时与
rv? ka
rk va ?? ?2
转动的一边指向顺方向,,2 ?? rrk vva ??
由于第二项和第四项所表示的加速度分量的大小，方向都
相同，可以合并为一项，用 表示，称为科里奥利加速度，简
称科氏加速度。
ka
71
Solution,moving point,point A on rod AB；
MCS,cam ; SCS,earth ground,
absolute motion,rectilinear motion;
absolute velocity,va=? to be solved,direction//AB;
relative motion,curvilinear;
relative velocity,vr=? direction?n;
convected motion,rotation;
convected velocity,ve= ? r, direction?OA,? 。
),s i n (2:m a g n i t u d e rrk v va ???
direction,determined according to the right-hand rule,
0),/ / (,180 o r 0W h e n ???? ?? kr av
rkr vav ?????? ? 2),(,90W h e n
[Example 2] The cam rotates uniformly around axis O at an
angular velocity of ?, At the moment shown in the figure,
OA= r, the radius of curvature at point A is ?,and ? is
also given.Find the velocity and acceleration of the rod AB
at this moment,
72
解, 动点, 顶杆上 A点；
动系, 凸轮 ; 静系, 地面。
绝对运动, 直线 ;
绝对速度, va=? 待求,方向 //AB;
相对运动, 曲线 ; 相对速度, vr=? 方向 ?n;
牵连运动, 定轴转动 ;
牵连速度, ve= ? r, 方向 ?OA,? 。
),s i n (2,rrk v va ???大小
方向：按右手法则确定。
0),/ / ( 1 8 0 0 ?? ?? kr av?? 时或当
rkr vav ??? 2),( 90 ??? ? 时当
[例 2] 已知：凸轮机构以匀 ? 绕 O轴转动,
图示瞬时 OA= r, A点曲率半径 ?,? 已知。
求：该瞬时顶杆 AB的速度和加速度。
73
n
rva rnr
as s a m e,d i r e c t i o n
,c o s//:ona c c e l e r a t i r e l a t i v e 2222 ??? ??
ABa a //d i r e c t i o n,?:ona c c e l e r a t i a b s o l u t e ?
na r ??? d i r e c t i o n?; t op o i n t i n g d i r e c t i o n
,,0,ona c c e l e r a t i e m b r o i l i n g 2
O
raaa neeτe ????
,t oo p p o s i t ed i r e c t i o n
,c o s/22:ona c c e l e r a t i C o r i o l i s 2
n
rva rk ?????
)(tg tg ?????? ??? rvvv eaAB
??? c o s/ c o s/ rvv er ??
Employing
rea vvv ??
and drawing the parallelogram of velocities,we have
74
nθ rva rnr 方向同相对加速度,c o s//,2222 ??? ??
ABa a //,?,方向绝对加速度 ?
na r ?? 方向??;,,0, 2 Oraaa neeτe 方向指向轴心牵连加速度 ????
相反。指向与方向
科氏加速度
,//
,c o s/22,2
nn
rva rk ??? ??
)(tg tg ?????? ??? rvvv eaAB
??? c o s/ c o s/ rvv er ??
根据速度合成定理
rea vvv ??
做出速度平行四边形
75
Applying the theorem of composition of
accelerations with a rotating convected motion
knea aaaaa rr ???? ?
drawing the vector diagram of accelerations
as shown in the figure
,and projecting them to axis n,yields
k
n
rea aaaa ???? ?? c o sc o s
???????? c o s/)s e c2/s e cc o s( 22222 rrraa aAB ??????
)s e c2/s e c1( 232 ???? ???? rr
76
由 牵连运动为转动时的加速度合成定
理
knea aaaaa rr ???? ?
作出 加速度矢量图 如图示
向 n 轴投影,
k
n
rea aaaa ???? ?? c o sc o s
???????? c o s/)s e c2/s e cc o s( 22222 rrraa aAB ??????
)s e c2/s e c1( 232 ???? ???? rr
77
D A
B C Solution,The Coriolis acceleration of point M1 is
and it points inside the screen ?,
?? s i n2 11 va k ?
)//( 0 22 va k ??
[Example 3] A rectangular plate ABCD rotates
about axis z at an constant angular velocity of ?, Point
M1 and M2 move along the diagonal BD and side CD
respectively,and their velocities relative to the plate
are and respectively at this moment,Calculate
the Coriolis accelerations at point M1 and M2 and
determine their directions,
1v 2v
The Coriolis acceleration of point M2 is
78
D A
B C 解,点 M1的科氏加速度 垂直板面向里 ?。
?? s i n2 11 va k ?
)//( 0 22 va k ??
[例 3] 矩形板 ABCD以匀角速度 ? 绕固定
轴 z 转动，点 M1和点 M2分别沿板的对角线
BD和边线 CD运动，在图示位置时相对于
板的速度分别为 和,计算点 M1, M2
的科氏加速度大小,并图示方向。
1v 2v
点 M2 的科氏加速度
79
Solution,
rk va ?? 22 ?
rkr vav 22 2 ?? ????
rea vvv ??
Applying
And drawing the parallelogram of velocities gives,
)c o s (s i n),s i n (c o s 11 ???????? ??????? rvvrvv arae
11
2
2 c o s
s i n)s i n (
c o s
s i n)s i n ( ?
?
???
?
????? ??????
r
r
AO
v e
rva rk 212 c o s )22s i n (2 ?? ??? ????
direction,same as
ev
[Example 4] In the mechanism shown in the
figure,O1A＝ r,?,? and ?1 are given,Taking
point A on rod O1A as the moving point,and
attaching the MCS on the rod O2B,find the
Coriolis acceleration of point A,
80
解,
rk va ?? 22 ?
rkr vav 22 2 ?? ????
rea vvv ??
根据 做出速度平行四边形
)c o s (s i n),s i n (c o s 11 ???????? ??????? rvvrvv arae
11
2
2 c o s
s i n)s i n (
c o s
s i n)s i n ( ?
?
???
?
????? ??????
r
r
AO
v e
rva rk 212 c o s )22s i n (2 ?? ??? ????
方向：与 相同。
ev
[例 4] 曲柄摆杆机构
已知,O1A＝ r,?,?,?1;
取 O1A杆上 A点为动点，动系固结 O2B上,
试计算动点 A的科氏加速度。
81
rea vvv ??
rea aaa ??
Chapter 8 Lesson for problem solving
1,Concepts and formulae
1),One point,two systems and three motions
The absolute motion of a point is the composition its
relative and convected motions
Composition of motions
2),Theorem of velocity composition
3,Theorem of acceleration composition
when the convected motion is translation
when the convected motion is rotation
)2( rkkrea vaaaaa ????? ?
82
rea vvv ??
rea aaa ??
第八章 点的合成运动习题课
一．概念及公式
1,一点、二系、三运动
点的绝对运动为点的相对运动与牵连
运动的合成,
2,速度合成定理
3,加速度合成定理
牵连运动为平动时
牵连运动为转动时 )2(
rkkrea vaaaaa ????? ?
83
2,Steps for problem solving
1),Choose moving point,MCS and SCS,
2),Analyze three kind of motions,absolute,relative and convected motions,
3),Based on velocity analysis,draw the parallelogram of velocities and solve
The relevant unknowns such as velocities and angular velocities,
4),Based on acceleration analysis,draw the vector diagram of accelerations
and solve the relevant unknowns such as accelerations and angular accelerations,
84
二．解题步骤
1,选择动点、动系、静系。
2,分析三种运动：绝对运动、相对运动和牵连运动。
3,作速度分析,画出速度平行四边形,求出有关未知量 (速度,
角速度）。
4,作加速度分析，画出加速度矢量图，求出有关的加速度,
角加速度未知量。
85
2,Some skills for problem solving
1),Properly select moving point,MCS and SCS according to the guideline,
which includes,
? To find the relative velocity between two points,
select one of them as the moving point,and attach a translating MCS to the
other one,
? When a point travels on a rigid body which itself also moves,In this case this
point has a complicated motion,To solve such a problem,
select this point as the moving point and attach the MCS to the rigid body,
? In the transmission two rigid bodies where there is a point on one body always
keeps contact with but moves relative to the other one,
For Guiding rod-block mechanisms,a typical way is to attach the MCS to the
guiding rod and take the sliding block as the moving point;
For cam-rod mechanism,a typical way is to attach the MCS to the cam and
select the contacting point on the rod as the moving point,
86
二．解题技巧
1,恰当地选择动点,动系和静系,应满足选择原则,，具体地有,
? 两个不相关的动点，求二者的相对速度。
根据题意,选择其中之一为动点,动系为固结于另一点的平动
坐标系。
? 运动刚体上有一动点，点作复杂运动。
该点取为动点，动系固结于运动刚体上。
? 机构传动,传动特点是在一个刚体上存在一个不变的接触点,
相对于另一个刚体运动。
导杆滑块机构：典型方法是动系固结于导杆，取滑块为动点。
凸轮挺杆机构：典型方法是动系固结与凸轮，取挺杆上与凸轮
接触点为动点。
87
? As a class of special problems where the contact point between two
rigid bodies varies with time relative to both of them,the contact point is not
suitable as a moving point,In this case,a suitable point other than the
contact point should be carefully selected,as illustrated in the example
given above,
2) Velocity-related problems,It is simple to find the unknown
velocity by geometric method,i.e,drawing the parallelogram of
velocities；
Acceleration-related problems,Since more than three vectors are
involved,analytical (projection) method should be applied to find the
unknowns and the axes to be projected on should be chosen to make
most of the projections be easily calculated,
88
? 特殊问题,特点是相接触两个物体的接触点位置都随时间而
变化, 此时,这两个物体的接触点都不宜选为动点，应选择满
足前述的选择原则的非接触点为动点。
2,速度问题,一般采用几何法求解简便,即作出速度平行四边形；
加速度问题,往往超过三个矢量,一般采用解析（投影）法求
解，投影轴的选取依解题简便的要求而定。
89
4,cautions
1) convected velocity and acceleration are those of the convected
point,
2) When convected motion is rotation,make sure that the Coriolis
acceleration is included and analyzed and calculated correctly,
3) Put the projections of vectors at left-hand side in a vector
equation of accelerations in the left-hand side,and those at right-hand
side still in the right-hand side of the equation,Beware that an
equation of acceleration has different form from that of equilibrium,
4) For a circular motion,
For a non-circular curvilinear motion,
( is the radius of curvature)
ka
RRva n 22 / ???
??? 22 / ?? va n
ka
?
90
四．注意问题
1,牵连速度及加速度是牵连点的速度及加速度。
2,牵连转动时作加速度分析不要丢掉,正确分析和计算 。
3,加速度矢量方程的投影是等式两端的投影，与静平衡方程
的投影式不同。
4,圆周运动时,
非圆周运动时,( 为曲率半径 )
ka
RRva n 22 / ???
??? 22 / ?? va n
ka
?
91
?
Solution,moving point,point A on rod OA;
MCS,attached to the T-rod ;
SCS,attached to the support,
Absolute motion,circular motion,
Relative motion,rectilinear,
convected motion,translation；
)i r e c t i o nd( OAlv a ?? ?
) t op o i n t i n g a l o n g( ),i r e c t i o nd( 2 OAOlaOAla naa ??? ???
er t i ca lv ?? rr av
s o l v e d be t o,h o r i z o n t a l ?? ee av
[Example 1] In the mechanism shown in
the figure,OA＝ l,?and ? are known when
= 45o, Find the velocity and
acceleration of the small vehicle,
Look at the animation
92
已知, OA＝ l,= 45o 时,?,? ;
求,小车的速度与加速度,
?
解, 动点,OA杆上 A点 ;
动系：固结在滑杆上 ;
静系：固结在机架上。
绝对运动：圆周运动,
相对运动：直线运动,
牵连运动：平动；
)( OAlv a ?? 方向?
)( ),( 2 OAOlaOAla naa 指向沿方向 ??? ???
铅直方向 ?? rr av
., 待求量水平方向?? ee av
[例 1] 曲柄滑杆机构
请看动画
93
Velocity of the small vehicle,
evv ?
According to draw the parallelogram of velocities shown in
the figure,rea
vvv ??
)(c o sc o s ???? ??? llvv ae 2 245 ?
Projecting to axis x,yields
e
n
aa aaa ?? ??
? s i nc o s
?? 4545 2 s i nc o s ?? lla e ???
,direction is shown in the figure
l)(2 2 2?? ??
acceleration of the small vehicle,
eaa ?
Employing the theorem of composition of
Accelerations when the convected motion is
translation
renaa aaaa ????
The vector diagram is shown in the figure,
94
小车的速度,
evv ?
根据速度合成定理 做出速度平行四边形,如图示
rea vvv ??
)(c o sc o s ???? ??? llvv ae 2 245 ?
投至 x轴,
e
n
aa aaa ?? ??
? s i nc o s
?? 4545 2 s i nc o s ?? lla e ???
，方向如图示
l)(2 2 2?? ??小车的加速度,
eaa ?
根据牵连平动的加速度合成定理
renaa aaaa ????
做出速度矢量图如图示 。
95
[Example 2]
Solution,moving point,pin D (on rod BC); MCS,OA； SCS,support
absolute motion,rectilinear motion,
relative motion,rectilinear motion,along OA
convected motion,rotation,
aavv aa ??,
,?? rr av
OODaOAODa nee a t p o i n t?;?,2 ??????? ???
OAODv e ?????,?
???? s i ns i n,c o sc o s vvvvvv arae ????
hv
hvODv
e
?
???
2c o s )
c o s/(c o s/ ???
（ ）
avh,,,,?
are given,Find ? and ? of rod OA,
According to Draw the parallelogram of velocities,as shown in
the figure,rea vvv ??
Se
e an
im
ati
on
96
[例 2] 摇杆滑道机构
解, 动点,销子 D (BC上 ); 动系, 固结于 OA；静系, 固结于机架。
绝对运动：直线运动,
相对运动：直线运动，,沿 OA 线
牵连运动：定轴转动,
aavv aa ??,
,?? rr av
OODaOAODa nee 指向?;?,2 ??????? ???
OAODv e ?????,?
???? s i ns i n,c o sc o s vvvvvv arae ????
hv
hvODv
e
?
???
2c o s )
c o s/(c o s/ ???
（ ）
avh,,,,?
已知 求, OA杆的 ?,? 。
根据 速度合成定理 做出速度平行四边形,如图示。
rea vvv ??
请
看
动
画
97
Projecting to axis?,
kea aaa ?? ??c o s
????? co ss i nco s2co s 22 ahvaaa ake ????
????
?
22
2
2
c o s2s i nc o s hahvODa e ??? （ ）
Employing the theorem of composition of
Accelerations when the convected motion is rotation,
we have
krneea aaaaa ???? ?
???
??
?
s i nc o s22
,c o s)c o s(
c o s
2
32
2
2
v
h
vva
h
v
h
vha
rk
n
e
???
???
98
投至 ? 轴,kea aaa ?? ??c o s
????? co ss i nco s2co s 22 ahvaaa ake ????
????
?
22
2
2
c o s2s i nc o s hahvODa e ???
（ ）
根据 牵连转动的加速度合成定理
krneea aaaaa ???? ?
???
??
?
s i nc o s22
,c o s)c o s(
c o s
2
32
2
2
v
h
vva
h
v
h
vha
rk
n
e
???
???
99
See the animation
[Example 3]
Solution,moving point,point A on O1A; MCS,attached to BCD,SCS,
attached to the support,
absolute motion,circular motion;
relative motion,rectilinear motion;
convected motion, translation;, horizontal
AOrv a 11,?? ?
BCv r / /?,?
?ev
and h;are
given,At the moment,
find the angular velocity ?2 of
rod O2E,
EOAO 21 //
??,,11 rAO ?
100
请看动画
[例 3] 曲柄滑块机构
解, 动点,O1A上 A点 ; 动系,固结于 BCD上,静系固结于机架上。
绝对运动：圆周运动 ;
相对运动：直线运动 ;
牵连运动：平动 ;,水平方向
AOrv a 11,?? ?
BCv r / /?,?
?ev
已知,h;
图示瞬时 ;
求, 该瞬时 杆的 ?2 。
EOAO 21 //
EO 2
,,,11 ??rAO ?
101
According to draw
the parallelogram of velocities rea vvv ??
Select moving point,point F on BCD
MCS,attached to O2E,
SCS,attached to the support
absolute motion,rectilinear motion;
relative motion,rectilinear motion;
convected motion, rotation;
)(s i n1 ?? ??rv Fa
)( / /?,2 EOv Fr ?
)(?,2 EOv Fe ??
??? s i ns i n 1rvv ae ??
According to draw the parallelogram of
velocities,FrFeFa vvv ??
?????? 211 s i ns i ns i ns i n rrvv FaFe ????
?? s i n/,222 hFOFOv eF ????又
?????? 3121
2
2 s i n
s i ns i n
h
r
h
r
FO
v eF ?????
） （
102
根据
做出速度平行四边形 rea vvv ??
再选动点,BCD上 F点
动系：固结于 O2E上,
静系固结于机架上
绝对运动：直线运动,
相对运动：直线运动,
牵连运动：定轴转动,
)(s i n1 ?? ??rv Fa
)( / /?,2 EOv Fr ?
)(?,2 EOv Fe ??
??? s i ns i n 1rvv ae ??
根据 做出速度平行四边形
FrFeFa vvv ??
?????? 211 s i ns i ns i ns i n rrvv FaFe ????
?? s i n/,222 hFOFOv eF ????又
?????? 3121
2
2 s i n
s i ns i n
h
r
h
r
FO
v eF ?????
） （
103
solution,Take C on the cam as the moving
point,and attach MCS to rod OA and SCS
to the earth ground,
absolute motion,rectilinear motion;
relative motion,rectilinear motion;
convected motion, rotation; OAav rr //i r e c t i o n d,??
OCv e ?? d i r e c t i o n?,
The radius of cam C is R,point O and C
in the same line at this moment,are
known,Find the angular velocity and acceleration
of rod OA at this instant,
av a n d,?
Analysis,Since the contact point between these
two rigid bodies varies with time relative to both
of them,the contact point is not suitable as a
moving point,
[Example 4]; t op o i n t i n g?2 OOCa ne ??? ??,??? ?? OCa e
direction OC?
See the animation
aavv aa ??,
104
解, 取凸轮上 C点为动点,
动系固结于 OA杆上,
静系固结于地面上,
绝对运动, 直线运动,
相对运动, 直线运动,
牵连运动, 定轴转动,
aavv aa ??,
OAav rr //,方向??
OCv e ?? 方向?,
已知,凸轮半径为 R，图示瞬时 O,C
在一条铅直线上 ; 已知 ;
求, 该瞬时 OA杆的角速度和角加速度。
av,,?
分析, 由于接触点在两个物体上的位
置均是变化的，因此不宜选接触点为
动点。
[例 4] 凸轮机构;?2 OOCa ne 指向??? ??,??? ?? OCa e
方向 OC?
请看动画
105
??? s i ns i n/ ;,0 RvR vOC vvvvv eaer ??????
） (
drawing the parallelogram of velocities,we have
According to
rea vvv ??
According to
kr
n
eea aaaaa ????
?
Drawing the vectorial diagram of accelerations,gives
02,s i n)s i n(s i n
2
2 ?????
rk
n
e vaR
v
R
vRa ???
?
Projecting them to axis ?,we have,
??? ? c o ss i nc o s enea aaa ?? ?? tgneae aaa ??
2
222 s i ns i n
s i n/
/s i n
R
v
R
a
R
Rva
OC
a e ??
?
?? ? ?????
Rotating direction depends on its sign.,if >0, and if <0
106
??? s i ns i n/ ;,0 RvR vOC vvvvv eaer ??????
） (
做出速度平行四边形,知
根据
rea vvv ??
根据
kr
n
eea aaaaa ????
?
做出加速度矢量图
02,s i n)s i n(s i n
2
2 ?????
rk
n
e vaR
v
R
vRa ???
?
投至 ? 轴,??? ? c o ss i nc o s
e
n
ea aaa ??
?? tgneae aaa ??
2
222 s i ns i n
s i n/
/s i n
R
v
R
a
R
Rva
OC
a e ??
?
?? ? ?????
转向由上式符号决定,>0则, <0 则
107
(see the animation)
[Example 5]
The rotating speed of the drive wheel is n=30 r/min
and OA=150mm, At this moment,OA?OO1,
Find ? 1 and ?1 of rod O1D and the of
block B,BB av,
108
（请看动画）
[例 5] 刨床机构
已知, 主动轮 O转速 n=30 r/min
OA=150mm,图示瞬时,OA?OO1
求, O1D 杆的 ? 1,?1
和滑块 B的 。
BB av,
109
where
m / s 15.03015.0 ??? ????? nOAv a
r a d / s
5515.0
503.0
m / s 503.0s i n
1
1
??
?
??
???
???
AO
v
vv
e
ae
） (
Solution,moving point,point A on wheel
O,MCs,O1D,SCS,support,
According to
draw the parallelogram of velocities
rea vvv ??
m / s 506.0c o s
)
5
5
s i n,
5
52
( c o s
??
??
??
??
ar vv
110
其中
m / s 15.03015.0 ??? ????? nOAv a
r a d / s
5515.0
503.0
m / s 503.0s i n
1
1
??
?
??
???
???
AO
v
vv
e
ae
） (
解,动点：轮 O上 A点
动系,O1D,静系：机架
根据 做出速度平行四边形 。
rea vvv ??
m / s 506.0c o s
)
5
5
s i n,
5
52
( c o s
??
??
??
??
ar vv
111
According to
kr
n
eea aaaaa ????
?
Draw the vector diagram of accelerations
rka vaa 12 2 15.0 ?? ??
Projecting them to gives,
ka ?? eka aaa ??c o s
222 m / s
5
518.0506.0
525
5215.0 ????? ??????
ea
22211 r a d / s
25
6
515.0
1
5
518.0/ ??? ? ???? AOa
e
） (
Again,select block B as the moving point; MCS,O1D; SCS,support,
112
根据
kr
n
eea aaaaa ????
?
做出加速度矢量图
rka vaa 12 2 15.0 ?? ??
投至 方向,
ka ?? eka aaa ??c o s
222 m / s
5
518.0506.0
525
5215.0 ????? ??????
ea
22211 r a d / s
25
6
515.0
1
5
518.0/ ??? ? ???? AOa
e
） (
再选动点,滑块 B; 动系, O1D; 静系, 机架。
113
According to
BrBeBa vvv ??
draw the vector
diagram of velocities,
,m / s 506.02 ??? eeB vv
m / s 503.0tg
m / s 15.0c o s/
??
??
???
????
eBrB
eBaBB
vv
vvv
Projecting them to axis x yields,
kBeBaB aaa ??
??c o s
2222 m / s 15.0
5
52/)
5
506.0
5
536.0( ??? ?????
aBB aa
According to
kBrB
n
eBBeBa aaaaa ????
?
draw the vector diagram of accelerations,
2 2 m / s5 536.02 ?? ?? eeB aa
where
22
1 m / s 5
506.0 503.0
522 ??
?? ?????
rBkB va
114
根据
BrBeBa vvv ??
做出速度矢量图 。
,m / s 506.02 ??? eeB vv
m / s 503.0tg
m / s 15.0c o s/
??
??
???
????
eBrB
eBaBB
vv
vvv
投至 x 轴,
kBeBaB aaa ??
??c o s
2222 m / s 15.0
5
52/)
5
506.0
5
536.0( ??? ?????
aBB aa
根据
kBrB
n
eBBeBa aaaaa ????
?
做出加速度矢量图
2 2 m / s5 536.02 ?? ?? eeB aa
其中
22
1 m / s 5
506.0 503.0
522 ??
?? ?????
rBkB va
115
[Example 6]
In the mechanism shown in the figure,and h
are given,Find ? and ? of sleeve O.。
av,,?
Solution,scheme 1,
Point A has rectilinear motion with
?tg?? hx A
hxx AA /)2s i nc o s( 2 ?????? ??? ????
hxhx AA /c o s s e c 22 ???? ???? ??? 即
Substituting the given quantities,we have
????? 2222 c o s)2s i n(,c o s hvhahv ??? （ ） （ ）
请看动画
116
[例 6] 套筒滑道机构
图示瞬时,h已知,
求,套筒 O的 ?, ? 。
av,,?
解,方法 1,
A点作直线运动
?tg?? hx A
hxx AA /)2s i nc o s( 2 ?????? ??? ????
hxhx AA /c o s s e c 22 ???? ???? ??? 即
代入图示瞬时的已知量，得
????? 2222 c o s)2s i n(,c o s hvhahv ??? （ ） （ ）
请看动画
117
Try to compare these two
schemes,
（ ） ??? ? 2
2
2
c o s)2s in(
h
v
h
a
OA
a e
???
Projecting to direction gives,
??
eka aaa ??c o s
ka
c o s2s i nc o s
2
????
h
vaa
e ??
h
vOAv
e
?? 2c o s/ ?? （ ）
Scheme 2,moving point,point A on CD,
MCS,sleeve O,SCS,support
rea vvv ??
??? s i n,c o sc o s vvvvv rae ????
kr
n
eea aaaaa ????
?
??? c o s2s i n2,
2
???? hvvaaa rka
vv a ??
118
对比两种方法 （ ） ??? ? 2
2
2
c o s)2s in(
h
v
h
a
OA
a e
???
投至 方向,
??
eka aaa ??c o ska
c o s2s i nc o s
2
????
h
vaa
e ??
h
vOAv
e
?? 2c o s/ ?? （ ）
方法 2,动点, CD上 A点,
动系, 套筒 O，静系, 机架
rea vvv ??
??? s i n,c o sc o s vvvvv rae ????
kr
n
eea aaaaa ????
?
??? c o s2s i n2,
2
???? hvvaaa rka
vva ?其中
119
120
Theoretical mechanics
2
3
§ 8–1 Concept of compositive motion of a particle
§ 8–2 Theorem of composition of velocities of a
particle
§ 8–3 Theorem of composition of accelerations
when the convected motion is translation
§ 8–4 Theorem of composition of accelerations
when the convected motion is rotation
Lesson for problem solving
Chapter 8,Compositive motion of a particle
4
§ 8–1 点的合成运动的概念
§ 8–2 点的速度合成定理
§ 8–3 牵连运动为平动时点的加速度合成定理
§ 8–4 牵连运动为转动时点的加速度合成定理
习题课
第八章 点的合成运动
5
§ 8-1 concept of compositive motion of a particle
1,Coordinate system,
1) Static coordinate system,A coordinate system fixed to the earth ground
is called Static coordinate system (SCS),
2) Moving coordinate system,A coordinate system fixed to a moving object
relative to the earth ground is called moving coordinate system (MCS),
For example,a running car,
In the previous two chapters,we have considered the motion of a particle
or a rigid body with respect to a system of fixed coordinate axes,In practice,
however,we often observe the motion with respect to a moving body,For
example,we observe the motion of an flying airplane from a moving car,or
observe the motion of raindrops from a traveling train,
Why do obtain different results when we observe the motion a object on
different coordinate systems or reference objects? It is known that there may
exist relationship between two different objects,Next,we are going to study
the relationship between the motions of an object and the reference object,
First of all,we begin with the introduction of some concepts,
6
§ 8-1 点的合成运动的概念
一．坐标系,
1.静坐标系,把固结于地面上的坐标系称为静坐标系,简称静系。
2.动坐标系,把固结于相对于地面运动物体上的坐标系,
称为动坐标系，简称动系。例如在行驶的汽车。
前两章中我们研究点和刚体的运动，一般都是以地面为参考
体的。然而在实际问题中，还常常要在相对于地面运动着的参
考系上观察和研究物体的运动。例如，从行驶的汽车上观看飞
机的运动等，坐在行驶的火车内看下雨的雨点是向后斜落的等。
为什么在不同的坐标系或参考体上观察物体的运动会有不
同的结果呢？ 我们说事物都是相互联系着的。下面我们就将研
究参考体与观察物体运动之间的联系。为了便于研究，下面先
来介绍有关的概念。
7
3,Three kinds of motion and their velocities and accelerations
1) Absolute motion,motion of the moving point relative to the SCS,
2) Relative motion,motion of the moving point relative to the MCS,
For example,motions of a person in a running bus
3) Convected motion,motion of the MCS relative to the SCS,
For example,motion of a running care relative to the earth ground,
The velocity and acceleration of the moving point in its absolute motion are called absolute velocity and absolute acceleration
The velocity and acceleration of the moving point in its relative motion are
called relative velocity and relative acceleration
The velocity and acceleration of the convected point in its absolute motion
are called convected velocity and convected acceleration
aa
ev ea
rv ra
av
convected point,the point in the MCS coinciding with the moving point at any
instant of time,Image that the moving point were fixed to the MCS and moved
together with the MCS,Such a point in the MCS is called convected point,
Motion of
a particle
Motion of a
rigid body
2,Moving point,a moving point under studying
8
三．三种运动及三种速度与三种加速度。
１．绝对运动,动点对静系的运动。
２．相对运动,动点对动系的运动。
例如：人在行驶的汽车里走动。
３．牵连运动,动系相对于静系的运动
例如：行驶的汽车相对于地面的运动。
绝对运动中,动点的速度与加速度称为 绝对速度 与 绝对加速度
相对运动中,动点的速度和加速度称为 相对速度 与 相对加速度
牵连运动中,牵连点的速度和加速度称为 牵连速度 与 牵连加速度
aa
ev ea
rv ra
av
牵连点,在任意瞬时，动坐标系中与动点相重合的点，也就是
设想将该动点固结在动坐标系上，而随着动坐标系一起运动时
该点叫牵连点。
点的运动
刚体的运动
二．动点,所研究的点（运动着的点）。
9
Explanation of the concepts by examples,
4,Guideline for selection of moving point,
In general,the contact point between the drive and driven objects is usually
selected as moving point,which moves relative to both SCS and MCS,
5,Guideline for MCS selection,
A MCS should be a coordinate system (object) on which the trajectory of the
moving point is known or clear,
Moving
point,
MCS,
SCS,
Point A on rod AB
Fixed to cam Ｏ '
Fixed to the ground
10
下面举例说明以上各概念,
四．动点的选择原则,
一般选择主动件与从动件的连接点，它是对两个坐标系都有
运动的点。
五．动系的选择原则,
动点对动系有相对运动，且相对运动的轨迹是已知的，
或者能直接看出的。
动点,
动系,
静系,
AB杆上 A点
固结于凸轮 Ｏ '上
固结在地面上
11
Relative motion
convected motion,
Curvilinear motion (arc )
Rectilinear translation
Absolute motion,Rectilinear motion
12
相对运动,
牵连运动,
曲线（圆弧）
直线平动
绝对运动, 直线
13
evrvav
Absolute velocity,
relative velocity,convected velocity,
14
evrvav
绝对速度,
相对速度, 牵连速度,
15
Absolute acceleration,
Relative acceleration,
convected acceleration,
aa
ea
ra
16
绝对加速度,
相对加速度,
牵连加速度,
aa
ea
ra
17
Moving point,A（ on disk)
MCS,rod O'A
SCS,support
Absolute motion,curvilinear (circle)
Relative motion,rectilinear motion
convected motion,rotation
Moving point,A1 (on rod O'A1)
MCS,circular disk
SCS,support
Absolute motion,curvilinear (arc)
Relative motion,curvilinear motion
convected motion,rotation
18
动点,A（在圆盘上 )
动系,O'A摆杆
静系：机架
绝对运动：曲线（圆周）
相对运动：直线
牵连运动：定轴转动
动点,A1（在 O'A1 摆杆上 )
动系：圆盘
静系：机架
绝对运动：曲线（圆弧）
相对运动：曲线
牵连运动：定轴转动
19
If the moving point is on the CAM
Moving point,A (on rod AB) A（ on the cam）
MCS,cam rod AB
SCS,earth ground earth ground
Absolute motion,rectilinear Circle ( red dashed line)
Relative motion,curvilinear (circle) curve（ path unknown
convected motion,rotation Translation
[Note] Clear state that to which
object the moving point is attached,
The moving point can not be
attached to the MCS,otherwise its
relative motion is rest on the MCS,
20
若动点 A在偏心轮上时
动点,A(在 AB杆上 ) A（在偏心轮上）
动系：偏心轮 AB杆
静系：地面 地面
绝对运动：直线 圆周（红色虚线）
相对运动：圆周（曲线） 曲线（未知）
牵连运动：定轴转动 平动
[注 ] 要指明动点应在哪个
物体上,但不能选在
动系上。
21
§ 8-2 Theorem of composition of velocities of a particle
The relationship among the absolute,relative and convected velocity will be
established and stated in the theorem of composition of velocities of a particle,
1MM
＝ 'MM ＋ '
1MM
When t ? t+△ t AB?A'B'
M?M'
Equivalent to M ? M１ ? M′
MM ' is the absolute path
MM ' is the absolute displacement
M1M ' is the relative path
M1M ' is the relative displacement
t
MM
t
MM
t
MM
ttt ??? ???
????
???
1
0
1
00
limlimlim
t?Divided the above equation by,
0??t,gives When
1,Proof
22
§ ８－２点的速度合成定理
速度合成定理将建立动点的绝对速度，相对速度和牵连速度
之间的关系。
1MM
＝ 'MM ＋ '
1MM
当 t ? t+△ t AB?A'B'
M?M'
也可看成 M ? M１ ? M′
MM ' 为绝对轨迹
MM ' 为绝对位移
M1M ' 为相对轨迹
M1M ' 为相对位移
t
MM
t
MM
t
MM
ttt ??? ???
????
???
1
0
1
00
limlimlim
t?将上式两边同除以 后,
0??t 时的极限，得 取
一．证明
23
24
25
Explanations,?va— absolute velocity of the moving point;
?vr— relative velocity of the moving point;
?ve— convected velocity of the moving point,
which is the velocity of a point
(convected
point) on the MCS;
I) When MCS is in translation,all points in it have the
same velocity,
II) When MCS rotates,ve must be the velocity of the point
in the MCS coinciding with the moving point,
Conclusion,At any instant of time,the absolute velocity of a moving point
equals to the geometric sum of its relative velocity and convected velocity,
This is the theorem of composition of velocities of a particle,
rea vvv ???
26
说明,?va— 动点的绝对速度；
?vr— 动点的相对速度；
?ve— 动点的牵连速度，是动系上一点 (牵连点 )的速度
I) 动系作平动时，动系上各点速度都相等。
II) 动系作转动时,ve必须是该瞬时动系上与
动点相重合点的速度。
即在任一瞬时动点的绝对速度等于其牵连速度与相对速度的
矢量和，这就是点的速度合成定理。
rea vvv ???
27
?The equation given in the theorem of composition of velocities
is a vector equation at an instant,which includes six components
such as magnitudes and directions,If any four components are
given,the other two unknowns can be solved from this equation,
2,Examples of theorem application
[Example1] The small
vehicle in a bridge cane
moves along its horizontal
track with a uniform velocity
vh,and weight A moves
vertically at a velocity vv
relative to the small vehicle,
Find the absolute velocity of
weight A,
28
?点的速度合成定理是瞬时矢量式，共包括大小 ?方向 六个元
素，已知任意四个元素，就能求出其他两个。
二．应用举例
[例 1] 桥式吊车 已知：小
车水平运行，速度为 v平,
物块 A相对小车垂直上升
的速度为 v?。求物块 A的运
行速度。
29
Draw the parallelogram of velocities shown in the figure,
the magnitude and direction of the velocity of the weight A can be obtained as
222
?????? vvvvvv reaA
2
平
平v
v ??? 1tg?
Solution,moving point,weight A
MCS,small vehicle
SCS,earth ground
Relative motion,rectilinear;
Relative velocity vr =vv,direction?
convected motion,translation;
convected velocity ve=vh,direction
?
Absolute motion, curvilinear;
Absolute velocity va,magnitude and
direction are unknown
Employing the theorem of composition of velocities,yields
rea vvv ??
30
作出速度平四边形 如图示，则物块Ａ的速度大小和方向为
222
?????? vvvvvv reaA
2
平
平v
v ??? 1tg?
解,选取 动点, 物块 A
动系, 小车
静系, 地面
相对运动, 直线 ;
相对速度 vr =v? 方向 ?
牵连运动, 平动 ;
牵连速度 ve=v平 方向 ?
绝对运动, 曲线 ;
绝对速度 va 的大小,方向待
求 由速度合成定理,rea vvv ??
31
Solution,Choose point A on rod OA杆 as the moving
point,attach the MCS to rod O1B,and use the earth
ground as the SCS,
absolute velocity va = r ? direction? OA
relative velocity vr =? direction//O1B
convected velocity ve =? direction ?O1B
2
2
2
2
2
1
111
22
2
22
2
22
1
,m o r e o v e r,
s i n,s i n
lr
r
lr
r
lrAO
v
AOv
lr
r
vv
lr
r
e
e
ae
?
?
?
?
?
?????
?
???
?
?
??
??
?
??
?
?
[Example 2] In the mechanism shown in the figure,OA=
r,?,OO1=l,At this moment,OA?OO1
Find,angular velocity ?1 of rod O1B,
According to va= vr+ ve,draw the parallelogram of velocities as shown in the
figure,
32
解, 取 OA杆上 A点为动点，摆杆 O1B为动系,
基座为静系。
绝对速度 va = r ? 方向 ? OA
相对速度 vr =? 方向 //O1B
牵连速度 ve =? 方向 ?O1B
2
2
2
2
2
1
111
22
2
22
2
22
1
,
s i n,s i n
lr
r
lr
r
lrAO
v
AOv
lr
r
vv
lr
r
e
e
ae
?
?
?
?
?
?????
?
???
?
?
??
??
?
??
?
?
又 （ ）
[例 2] 曲柄摆杆机构
已知, OA= r,?,OO1=l 图示瞬时 OA?OO1
求,摆杆 O1B角速度 ?1
由速度合成定理 va= vr+ ve 作出 速度平行四边形 如图示。
33
According to the equation va= vr+ ve,
Draw the parallelogram of velocities as shown in the figure,
Solution,Select point A on rod AB as the moving point,
fix the MCS to the disk and the SCS to the support,
absolute velocity va =? unknown,directin//AB
relative velocity vr =? unknown,direction?CA
convected velocity ve =OA??=2e?,direction? OA
（ Look at the animation
on the next page）
)(3 32 3 3230 0 ?????? ?? evetgvv ABea
[Example 3] In the cam-rod mechanism,OC＝
e,,?（ constant）,and the moment,
OC?CA and point O,A and B are collinear,
Find,the velocity of the driven rod AB,
eR 3?
34
由速度合成定理 va= vr+ ve,
作出速度平行四边形 如图示。
解,动点 取直杆上 A点,动系 固结于圆盘,
静系 固结于基座。
绝对速度 va =? 待求,方向 //AB
相对速度 vr =? 未知,方向 ?CA
牵连速度 ve =OA??=2e?,方向 ? OA
（翻页请看动画）
)(3 32 3 3230 0 ?????? ?? evetgvv ABea
[例 3] 圆盘凸轮机构
已知,OC＝ e,,?（匀角速度）
图示瞬时,OC?CA 且 O,A,B三点共线。
求,从动杆 AB的速度。
eR 3?
35
36
37
As evidenced from the examples,the general steps should be followed
in solving problems about the composition of velocities,
? Select moving point,MCS and SCS;
? Analysis of the related three kinds of motions;
? Analysis of three kinds of velocities;
?According to the theorem of composition of velocities,
draw the parallelogram of velocities;
? Find unknowns from the parallelogram of velocities;
Proper selection of the moving point,MCS and SCS is the key point in
solving such problems,
,rea vvv ??
38
由上述例题可看出，求解合成运动的速度问题的 一般步骤 为,
? 选取动点，动系和静系。
? 三种运动的分析。
? 三种速度的分析。
? 根据速度合成定理 作出速度平行四边形。
? 根据速度平行四边形，求出未知量。
恰当地选择动点、动系和静系是求解合成运动问题的关键。
,rea vvv ??
39
Guidelines for the selection of moving point,MCS and SCS
?The moving point,MCS and SCS must be in three different rigid bodies,
Otherwise,one motion out of the absolute,relative and convected motion will
be absent (always resting indeed),and consequently it will not suitable for
the application of the composition of motions,
? The relative path of the moving point relative to the MCS must be easy to
determine,(except to find the convected motion when the absolute and relative
motions are given,)
40
动点、动系和静系的选择原则
? 动点、动系和静系必须分别属于三个不同的物体,
否则绝对、相对和牵连运动中就缺少一种运动，不
能成为合成运动
? 动点相对动系的相对运动轨迹易于直观判断 （已
知绝对运动和牵连运动求解相对运动的问题除外）。
41
Analysis,Since the contact point between these two objects
changes with time,their contact point is not suitable to be a
moving point,otherwise the relative motion is very complicated
In this case,we need to select another point instead of the contact
point as the moving point which should satisfy the guideline,
?? 30,?v[Example 4] The radius of the cam is r,and as shown
in the diagram,Rod OA is leaning against the cam,Determine the
angular velocity of rod OA,
42
分析,相接触的两个物体的接触点位置都随时间而变化，
因此两物体的接触点都不宜选为动点，否则相对运动的分析
就会很困难。这种情况下，需选择满足上述两条原则的非接
触点为动点。
[例４ ] 已知, 凸轮半径 r,图示时 杆 OA靠在凸轮上。
求：杆 OA的角速度。;30,???v
43
Solution,Select point C as the
moving point,attach the MCS to
rod OA and the SCS to the base,
Absolute motion,rectilinear motion,
Absolute velocity,
Relative motion,rectilinear motion,
Relative velocity,
convected,rotation,convected velocity,
?? d ir e c t io n,vv a
OCOCv e ??? d i r e c t i o n u n k n o w n,,??
According to,
rea vvv ??
draw the parallelogram of velocities as shown in the figure,
r
vv
rr
v e
6
3
3
3
2
1
2 ????? ?
vvv ae 3 3tg ??? ?
（ ）
,2s i n ???? rrOCv e ?????
OAv r //d i r e ct i o n i t s u n k n o w n,is
44
解, 取凸轮上 C点为 动点,
动系 固结于 OA杆上,
静系 固结于基座。
绝对运动, 直线运动,
绝对速度,
相对运动, 直线运动,
相对速度,
牵连运动, 定轴转动,牵连速度,
??,方向vv a
??
OCOCv e ??? 方向待求未知,,??
方向未知,rv OA
如图示。 根据速度合成定理,
rea vvv ??
做出速度平行四边形
r
vv
rr
v e
6
3
3
3
2
1
2 ????? ?
vvv ae 3 3tg ??? ?
（ ）
,2s i n ???? rrOCv e ?????又
45
§ 8-3 Theorem of composition of accelerations when the
convected motion is translation
rea vvv ??
Noting the convected motion is
translation,we have
Employing theorem of composition of
velocities
'',OeOe aavv ??
''''''v a n d r kdtdzjdtdyidtdx ???
kdtdzjdtdyidtdxvv Oa ???????? ? '''
Differentiating it with respect to t,
'''''' 2
2
2
2
2
2
kdt zdjdt ydidt xddtvddtvda Oaa ????? ?
A point moves along a curved path AB attached to the MCS O'x'y'z',at the
same time,the path AB also has translatory motion together with the MCS
O'x'y'z' relative to the SCS Oxyz,
46
§ 8-3 牵连运动为平动时点的加速度合成定理
rea vvv ??
由于 牵连运动为平动,故
由速度合成定理
'',OeOe aavv ??
''''''v r kdtdzjdtdyidtdx ???而
kdtdzjdtdyidtdxvv Oa ???????? ? '''
对 t求导,''''''
2
2
2
2
2
2
kdt zdjdt ydidt xddtvddtvda Oaa ????? ?
设有一动点 M按一定规律沿着固连于动系 O'x'y'z' 的曲线 AB
运动,而曲线 AB同时又随同动系 O'x'y'z' 相对静系 Oxyz平动。
47
0',0',0' ??? dt zddt yddt id
(where are the unit vectors of the MCS,and their direction do not
change due to the translation of the MCS,Hence ),
moreover
',',' kji
rea aaa ???
— Theorem of composition of accelerations when the
convected motion is translation i,e,when the convected motion is translation,
the absolute acceleration of a moving point equals to the geometric sum of its
relative and convected accelerations,
'''''',2
2
2
2
2
2
'
' k
dt
zdj
dt
ydi
dt
xdaaa
dt
vd
reO
O ??????
naaa ?? ? ?
n
rrneenaa aaaaaa ?????
???
∴ Its general form is,
48
0',0',0' ??? dt zddt yddt id
(其中 为动系坐标的单位矢量，因为动系为平动，故它
们的方向不变，是常矢量，所以 ）
',',' kji
rea aaa ???
— 牵连运动为平动时点的加速度合成定理
即当牵连运动为平动时，动点的绝对加速度等于牵连加速度
与相对加速度的矢量和。
'''''',2
2
2
2
2
2
'
' k
dt
zdj
dt
ydi
dt
xdaaa
dt
vd
reO
O ??????又
naaa ?? ? ?
n
rrneenaa aaaaaa ?????
???
∴ 一般式可写为,
49
Solution,Choose point A on rod AB
as the moving point,and attach the
MCS onto the cam,
[Example 1] The radius of the cam is
are also given,
Find the velocity of rod AB when ? =60o,oo avR a n d,
Look at the animation
50
解,取杆上的 A点为 动点,
动系 与凸轮固连。
[例 1] 已知：凸轮半径
求,? =60o时,顶杆 AB的加速度。
oo avR,,
请看动画
51
Absolute velocity va =?,direction??AB ； absolute acceleration aa=?,
direction??AB,to be solved,
Relative velocity vr =?,direction?CA; relative acceleration ar? =?
direction?CA,direction along CA and pointing to C,
convected velocity ve=v0,direction→ ; convected acceleration ae=a0,direction
→,
Appling
,rea vvv ??
draw the parallelogram of velocities
shown in the figure above 00 3260s i ns i n vvvv oer ??? ?
? ?
Rva rnr /2?
52
绝对速度 va =?,方向 ??AB ；绝对加速度 aa=?,方向 ??AB,待求。
相对速度 vr =?,方向 ?CA; 相对加速度 ar? =? 方向 ?CA
,方向沿 CA指向 C
牵连速度 ve=v0,方向 → ; 牵连加速度 ae=a0,方向 →
由速度合成定理
,rea vvv ??
做出速度平行四边形,如图示。
0
0
3
2
60s i ns i n v
vvv
o
e
r ??? ?
? ?
Rva rnr /2?
53
Noting the convected motion is translation,
we have
nrea aaaa
r ???
?
R
vRvRvaw
r
n
r 3
4/)
3
2(/ h e r e 202
0
2 ???
Drawing vector diagram of accelerations show
in the figure,and projecting them to the normal
n,yields
nrea aaa ?? ?? c o ss i n
?? 60s i n/)
3
460c o s(s i n/)c o s( 20
0 R
vaaaa n
rea ???? ??
After arrangement,finally we have )
3
8(
3
3 20
0 R
vaaa
aAB ???
?
n
54
因 牵连运动为平动,故有
nrea aaaa
r ???
?
R
vRvRva
r
n
r 3
4/)
3
2(/ 202
0
2 ???其中
作 加速度矢量图 如图示,
将上式投影到法线上，得
nrea aaa ?? ?? c o ss i n
?? 60s i n/)
3
460c o s(s i n/)c o s( 20
0 R
vaaaa n
rea ???? ??
整理得 )
3
8(
3
3 20
0 R
vaaa
aAB ???
?
n
55
§ 8-4 Theorem of composition of accelerations
when the convected motion is rotation
In last section,we have proved the theorem of composition of
accelerations when the convected motion is translation,Is this
theorem still valid when the convected motion is rotation? Before
answering this question,let’s analyze a typical example,
A disk is rotating about a fixed axis
Ｏ with a uniform angular velocity ?,A
point M moves along a circular path
slotted in the disk with a constant vr,
what’s the absolute acceleration of the
point M?
56
§ 8-4 牵连运动为转动时点的加速度合成定理
上一节我们证明了牵连运动为平动时的点的加速度合成定
理,那么当牵连运动为转动时，上述的加速度合成定理是否还
适用呢？下面我们来分析一特例。
设一圆盘以匀角速度 ? 绕定轴 Ｏ 顺
时针转动，盘上圆槽内有一点 M以大
小不变的速度 vr 沿槽作圆周运动，那
么 M点相对于静系的绝对加速度应是
多少呢？
57
R
vav r
rr
2
,c o n s t, h a v e,we ??
Since the relative motion is a uniform circular
motion,
(direction is shown in the figure)
Applying the theorem of composition of velocities gives
c o n s t,????? rrea vRvvv ?
Choose M as the moving point,and the
MCS is attached to the disk,
In this case,the convected motion of the point
M is a uniform rotation,
RaRv ee 2,?? ??
(direction is shown in the
figure)
It is shown that the absolute motion of point M is also a uniform circular motion,
and therefore
It points to the centerＯ,
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
58
R
vav r
rr
2
,?? 常数有
相对运动 为匀速圆周运动,
（方向如图）
由速度合成定理可得出
常数????? rrea vRvvv ?
选点 M为动点，动系固结与圆盘上,
则 M点的 牵连运动 为匀速转动
RaRv ee 2,?? ?? （方向如图）
即 绝对运动 也为匀速圆周运动，所以
方向指向圆心 Ｏ 点
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
59
it can found from the above equation that in addition to
an extra term 2? vr appears,
It is clear that the absolute acceleration of a moving point is
no longer equal to the geometric sum of its relative acceleration
and convected acceleration when the convected motion is
rotation, What is the correct relationship among these
accelerations？ What is the additional term 2? vr why does it
appear? Next,we are going to seek the answers to these questions
and derive the theorem of composition of accelerations when the
convected motion is rotation,
ea
ra
aa
,a n d / 22 ?RaRva err ??
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
60
分析上式,还多出一项 2? vr 。
可见,当牵连运动为转动时，动点的绝对加速度 并不
等于牵连加速度 和相对加速度 的矢量和。 那么他们
之间的关系是什么呢？ 2? vr 又是怎样出现的呢？它是什么
呢？ 下面我们就来讨论这些问题，推证牵连运动为转动时点
的加速度合成定理。
ea ra
aa
,,/ 22 ?RaRva err ??
r
rra
a vR
vR
R
vR
R
va ??? 2)( 2222 ??????
61
Analysis of three velocities
convected velocity
Relative velocity
Absolute velocity
At instant t,the system is in position Ｉ and at t+?t,it is in position II
ev
rv
rea vvv ?? ''' rea vvv ??
'ev
'rv
It can be observed that the magnitudes and directions of both
convected and relative velocities vary after an interval ?t,
A rod OA rotates about axis O with a
constant angular velocity ?,and the sleeve
M moves along rod OA with a varying
velocity,Choose sleeve M as the moving
point,and attach the MCS to rod OA and
the SCS to the static support,
62
三种速度分析
牵连速度
相对速度
绝对速度
t 瞬时在位置Ｉ t+?t 瞬时在位置 II
ev
rv
rea vvv ?? ''' rea vvv ??
'ev
'rv
可以看出，经过 ?t 时间间隔，牵连速度和相对速度的大
小和方向都变化了。
设有已知杆 OA在图示平面内以匀
? 绕轴 O转动，套筒 M（可视为点
M）沿直杆作变速运动。 取套筒 M
为动点，动系固结于杆 OA上，静
系固结于机架。
63
where -- the magnitude change of the relative velocity in ?t,which is
independent on the convected motion,
-- 在 direction change of the relative velocity caused by the convected
motion in ?t,which is dependent on the convected angular velocity ?,
Analysis of the velocity change in an interval ?t
?relative velocity,draw the triangle composed of
After cutting a length of in the vector,can be decomposed
into and,
rrr vvv ? a n d ',
'rvrv rv?
'rv? ''rv?
'''.i, e rrr vvv ?????
'rv?
''rv?
?convected velocity
draw the triangle composed of
After cutting a length of in the vector,
can be decomposed into and,
eee vvv ? a n d ',
'evev
ev? 'ev? ''ev?
''' i, e, eee vvv ?????
64
其中 -- 在 ?t内相对速度大小的改变量，它与牵连转动无关。
-- 在 ?t内由于牵连转动而引起的相对速度方向的改变
量，与牵连转动的 ? 的大小有关 。
?t 时间间隔内的速度变化分析
?相对速度,由 作速度矢量三角形,
在 矢量上截取 长度后,分解为 和
rrr vvv ?,',
'rv rv rv? 'rv? ''rv?
''' rrr vvv ?????即
'rv?
''rv?
?牵连速度,
由 作速度矢量三角形,
在 矢量上截取等于 长后,
将 分解为 和,
eee vvv ?,',
'ev ev
ev? 'ev? ''ev?
''' eee vvv ?????即
65
where
— represents the direction change of the convected velocity caused by
the convected rotation in an interval ?t,which is independent on the relative motion,
— represents the magnitude change of the convected velocity caused by
the relative motion in an interval ?t,which is dependent on the relative velocity,
'ev?
''ev?
rv
Acceleration analysis
According to the
definition,t vvvvt vva reretaata ? ??????? ???? )()''(lim'lim 00
t
v
t
v
t
vvvv r
t
e
t
rree
t ?
??
?
??
?
????
?????? 000
limlim)'()'(lim
Physical interpretations,
The first term,magnitude
eet
e
t
aOMtvtv ?????????
????
2
00
lim'lim ??
t
v
t
v
t
v
t
v r
t
r
t
e
t
e
t ?
??
?
??
?
??
?
??
????????
''lim'lim''lim'lim
0000
66
其中,
— 表示 ?t内由于牵连转动而引起的牵连速度方向的改
变量，与相对运动无关。
— 表示 ?t内动点的牵连速度，由于相对运动而引起的
大小改变量，与相对速度 有关。
'ev?
''ev?
rv
加速度分析
根据加速度定义
t
vvvv
t
vva rere
t
aa
ta ?
????
?
??
????
)()''(lim'lim
00
t
v
t
v
t
vvvv r
t
e
t
rree
t ?
??
?
??
?
????
?????? 000
limlim)'()'(lim
上式中 各项的物理意义 如下,
第一项大小,
eet
e
t
aOMtvtv ?????????
????
2
00
lim'lim ??
t
v
t
v
t
v
t
v r
t
r
t
e
t
e
t ?
??
?
??
?
??
?
??
????????
''lim'lim''lim'lim
0000
67
Direction,as ?t ?0 and ?? ?0,it will gradually point to point A along the rod,
Therefore,this term is the convected acceleration at instant t,
ea
The third term,magnitude,represents the change
rate of the magnitude of, rrrt adtvdtv ?????? 'lim 0
rv
Direction,along the rod,
so,this term is the relative acceleration of the moving point at instant t,
ra
The second term,magnitude,
t
OMOM
t
vv
t
v
t
ee
t
e
t ?
????
?
??
?
?
??????
??'lim'lim''lim
000
rrt vvt
MM ????
???? ?? d i r e c t i o n,
'lim 1
0
This term is the acceleration resulting from the magnitude change of the convected
velocity due to the existence of relative motion,
The forth term：
magnitude,d i r e c t i o n,lim''lim
00 rrrt
r
t
vv
t
v
t
v ???
?
????
?
?
????
This term represents the acceleration resulting from the direction change of
the relative velocity caused by the convected rotation,
68
方向,?t ?0时,?? ?0,其方向沿着直杆指向 A点。
因此,第一项正是 t 瞬时动点的牵连加速度 。
ea
第三项大小,为对应于 大小改变
rrrt adt
vd
t
v ??
?
?
??
'lim
0
rv
方向：总是沿直杆。
因此,该项恰是 ｔ 瞬时动点的相对加速度 。 ra
第二项大小,
t
OMOM
t
vv
t
v
t
ee
t
e
t ?
????
?
??
?
?
??????
??'lim'lim''lim
000
rrt vvt
MM ???
??? ?? 方向,
'lim 1
0 ??
该项为由于相对运动的存在而引起牵连速度的大小改变的加速度。
第四项大小,
。方向,lim''lim
00 rrrt
r
t
vvtvtv ????
??
?? ????
??
这一项表明由于牵连转动而引起相对速度方向改变的加速度。
69
Finally,we obtain the theorem of composition of accelerations with
a rotating convected motion as follow
krea aaaa ???
When the convected motion is rotation,the absolute acceleration of a moving
point equals to the geometric sum of its convected,relative and Coriolis
accelerations,
More general form,
knrrneenaa aaaaaaa ?????? ???
The Coriolis acceleration can also be expressed as the following vector
form,
ka
rk va ?? ?2
w i t h s i d e s a m e t h e t op o i n t i n g,d i r e c t i o n,2 ???? rrk vva
Since the second and the fourth term have the same magnitude and direction,
they can be written into one term and denoted by,which is called Coriolis
acceleration,k
a
70
所以，当牵连运动为转动时，加速度合成定理为
krea aaaa ???
当牵连运动为转动时，动点的绝对加速度等于它的牵连加速
度，相对加速度和科氏加速度三者的矢量和。
一般式
knrrneenaa aaaaaaa ?????? ???
一般情况下 科氏加速度 的计算可以用矢积表示 ) ( 不垂直时与
rv? ka
rk va ?? ?2
转动的一边指向顺方向,,2 ?? rrk vva ??
由于第二项和第四项所表示的加速度分量的大小，方向都
相同，可以合并为一项，用 表示，称为科里奥利加速度，简
称科氏加速度。
ka
71
Solution,moving point,point A on rod AB；
MCS,cam ; SCS,earth ground,
absolute motion,rectilinear motion;
absolute velocity,va=? to be solved,direction//AB;
relative motion,curvilinear;
relative velocity,vr=? direction?n;
convected motion,rotation;
convected velocity,ve= ? r, direction?OA,? 。
),s i n (2:m a g n i t u d e rrk v va ???
direction,determined according to the right-hand rule,
0),/ / (,180 o r 0W h e n ???? ?? kr av
rkr vav ?????? ? 2),(,90W h e n
[Example 2] The cam rotates uniformly around axis O at an
angular velocity of ?, At the moment shown in the figure,
OA= r, the radius of curvature at point A is ?,and ? is
also given.Find the velocity and acceleration of the rod AB
at this moment,
72
解, 动点, 顶杆上 A点；
动系, 凸轮 ; 静系, 地面。
绝对运动, 直线 ;
绝对速度, va=? 待求,方向 //AB;
相对运动, 曲线 ; 相对速度, vr=? 方向 ?n;
牵连运动, 定轴转动 ;
牵连速度, ve= ? r, 方向 ?OA,? 。
),s i n (2,rrk v va ???大小
方向：按右手法则确定。
0),/ / ( 1 8 0 0 ?? ?? kr av?? 时或当
rkr vav ??? 2),( 90 ??? ? 时当
[例 2] 已知：凸轮机构以匀 ? 绕 O轴转动,
图示瞬时 OA= r, A点曲率半径 ?,? 已知。
求：该瞬时顶杆 AB的速度和加速度。
73
n
rva rnr
as s a m e,d i r e c t i o n
,c o s//:ona c c e l e r a t i r e l a t i v e 2222 ??? ??
ABa a //d i r e c t i o n,?:ona c c e l e r a t i a b s o l u t e ?
na r ??? d i r e c t i o n?; t op o i n t i n g d i r e c t i o n
,,0,ona c c e l e r a t i e m b r o i l i n g 2
O
raaa neeτe ????
,t oo p p o s i t ed i r e c t i o n
,c o s/22:ona c c e l e r a t i C o r i o l i s 2
n
rva rk ?????
)(tg tg ?????? ??? rvvv eaAB
??? c o s/ c o s/ rvv er ??
Employing
rea vvv ??
and drawing the parallelogram of velocities,we have
74
nθ rva rnr 方向同相对加速度,c o s//,2222 ??? ??
ABa a //,?,方向绝对加速度 ?
na r ?? 方向??;,,0, 2 Oraaa neeτe 方向指向轴心牵连加速度 ????
相反。指向与方向
科氏加速度
,//
,c o s/22,2
nn
rva rk ??? ??
)(tg tg ?????? ??? rvvv eaAB
??? c o s/ c o s/ rvv er ??
根据速度合成定理
rea vvv ??
做出速度平行四边形
75
Applying the theorem of composition of
accelerations with a rotating convected motion
knea aaaaa rr ???? ?
drawing the vector diagram of accelerations
as shown in the figure
,and projecting them to axis n,yields
k
n
rea aaaa ???? ?? c o sc o s
???????? c o s/)s e c2/s e cc o s( 22222 rrraa aAB ??????
)s e c2/s e c1( 232 ???? ???? rr
76
由 牵连运动为转动时的加速度合成定
理
knea aaaaa rr ???? ?
作出 加速度矢量图 如图示
向 n 轴投影,
k
n
rea aaaa ???? ?? c o sc o s
???????? c o s/)s e c2/s e cc o s( 22222 rrraa aAB ??????
)s e c2/s e c1( 232 ???? ???? rr
77
D A
B C Solution,The Coriolis acceleration of point M1 is
and it points inside the screen ?,
?? s i n2 11 va k ?
)//( 0 22 va k ??
[Example 3] A rectangular plate ABCD rotates
about axis z at an constant angular velocity of ?, Point
M1 and M2 move along the diagonal BD and side CD
respectively,and their velocities relative to the plate
are and respectively at this moment,Calculate
the Coriolis accelerations at point M1 and M2 and
determine their directions,
1v 2v
The Coriolis acceleration of point M2 is
78
D A
B C 解,点 M1的科氏加速度 垂直板面向里 ?。
?? s i n2 11 va k ?
)//( 0 22 va k ??
[例 3] 矩形板 ABCD以匀角速度 ? 绕固定
轴 z 转动，点 M1和点 M2分别沿板的对角线
BD和边线 CD运动，在图示位置时相对于
板的速度分别为 和,计算点 M1, M2
的科氏加速度大小,并图示方向。
1v 2v
点 M2 的科氏加速度
79
Solution,
rk va ?? 22 ?
rkr vav 22 2 ?? ????
rea vvv ??
Applying
And drawing the parallelogram of velocities gives,
)c o s (s i n),s i n (c o s 11 ???????? ??????? rvvrvv arae
11
2
2 c o s
s i n)s i n (
c o s
s i n)s i n ( ?
?
???
?
????? ??????
r
r
AO
v e
rva rk 212 c o s )22s i n (2 ?? ??? ????
direction,same as
ev
[Example 4] In the mechanism shown in the
figure,O1A＝ r,?,? and ?1 are given,Taking
point A on rod O1A as the moving point,and
attaching the MCS on the rod O2B,find the
Coriolis acceleration of point A,
80
解,
rk va ?? 22 ?
rkr vav 22 2 ?? ????
rea vvv ??
根据 做出速度平行四边形
)c o s (s i n),s i n (c o s 11 ???????? ??????? rvvrvv arae
11
2
2 c o s
s i n)s i n (
c o s
s i n)s i n ( ?
?
???
?
????? ??????
r
r
AO
v e
rva rk 212 c o s )22s i n (2 ?? ??? ????
方向：与 相同。
ev
[例 4] 曲柄摆杆机构
已知,O1A＝ r,?,?,?1;
取 O1A杆上 A点为动点，动系固结 O2B上,
试计算动点 A的科氏加速度。
81
rea vvv ??
rea aaa ??
Chapter 8 Lesson for problem solving
1,Concepts and formulae
1),One point,two systems and three motions
The absolute motion of a point is the composition its
relative and convected motions
Composition of motions
2),Theorem of velocity composition
3,Theorem of acceleration composition
when the convected motion is translation
when the convected motion is rotation
)2( rkkrea vaaaaa ????? ?
82
rea vvv ??
rea aaa ??
第八章 点的合成运动习题课
一．概念及公式
1,一点、二系、三运动
点的绝对运动为点的相对运动与牵连
运动的合成,
2,速度合成定理
3,加速度合成定理
牵连运动为平动时
牵连运动为转动时 )2(
rkkrea vaaaaa ????? ?
83
2,Steps for problem solving
1),Choose moving point,MCS and SCS,
2),Analyze three kind of motions,absolute,relative and convected motions,
3),Based on velocity analysis,draw the parallelogram of velocities and solve
The relevant unknowns such as velocities and angular velocities,
4),Based on acceleration analysis,draw the vector diagram of accelerations
and solve the relevant unknowns such as accelerations and angular accelerations,
84
二．解题步骤
1,选择动点、动系、静系。
2,分析三种运动：绝对运动、相对运动和牵连运动。
3,作速度分析,画出速度平行四边形,求出有关未知量 (速度,
角速度）。
4,作加速度分析，画出加速度矢量图，求出有关的加速度,
角加速度未知量。
85
2,Some skills for problem solving
1),Properly select moving point,MCS and SCS according to the guideline,
which includes,
? To find the relative velocity between two points,
select one of them as the moving point,and attach a translating MCS to the
other one,
? When a point travels on a rigid body which itself also moves,In this case this
point has a complicated motion,To solve such a problem,
select this point as the moving point and attach the MCS to the rigid body,
? In the transmission two rigid bodies where there is a point on one body always
keeps contact with but moves relative to the other one,
For Guiding rod-block mechanisms,a typical way is to attach the MCS to the
guiding rod and take the sliding block as the moving point;
For cam-rod mechanism,a typical way is to attach the MCS to the cam and
select the contacting point on the rod as the moving point,
86
二．解题技巧
1,恰当地选择动点,动系和静系,应满足选择原则,，具体地有,
? 两个不相关的动点，求二者的相对速度。
根据题意,选择其中之一为动点,动系为固结于另一点的平动
坐标系。
? 运动刚体上有一动点，点作复杂运动。
该点取为动点，动系固结于运动刚体上。
? 机构传动,传动特点是在一个刚体上存在一个不变的接触点,
相对于另一个刚体运动。
导杆滑块机构：典型方法是动系固结于导杆，取滑块为动点。
凸轮挺杆机构：典型方法是动系固结与凸轮，取挺杆上与凸轮
接触点为动点。
87
? As a class of special problems where the contact point between two
rigid bodies varies with time relative to both of them,the contact point is not
suitable as a moving point,In this case,a suitable point other than the
contact point should be carefully selected,as illustrated in the example
given above,
2) Velocity-related problems,It is simple to find the unknown
velocity by geometric method,i.e,drawing the parallelogram of
velocities；
Acceleration-related problems,Since more than three vectors are
involved,analytical (projection) method should be applied to find the
unknowns and the axes to be projected on should be chosen to make
most of the projections be easily calculated,
88
? 特殊问题,特点是相接触两个物体的接触点位置都随时间而
变化, 此时,这两个物体的接触点都不宜选为动点，应选择满
足前述的选择原则的非接触点为动点。
2,速度问题,一般采用几何法求解简便,即作出速度平行四边形；
加速度问题,往往超过三个矢量,一般采用解析（投影）法求
解，投影轴的选取依解题简便的要求而定。
89
4,cautions
1) convected velocity and acceleration are those of the convected
point,
2) When convected motion is rotation,make sure that the Coriolis
acceleration is included and analyzed and calculated correctly,
3) Put the projections of vectors at left-hand side in a vector
equation of accelerations in the left-hand side,and those at right-hand
side still in the right-hand side of the equation,Beware that an
equation of acceleration has different form from that of equilibrium,
4) For a circular motion,
For a non-circular curvilinear motion,
( is the radius of curvature)
ka
RRva n 22 / ???
??? 22 / ?? va n
ka
?
90
四．注意问题
1,牵连速度及加速度是牵连点的速度及加速度。
2,牵连转动时作加速度分析不要丢掉,正确分析和计算 。
3,加速度矢量方程的投影是等式两端的投影，与静平衡方程
的投影式不同。
4,圆周运动时,
非圆周运动时,( 为曲率半径 )
ka
RRva n 22 / ???
??? 22 / ?? va n
ka
?
91
?
Solution,moving point,point A on rod OA;
MCS,attached to the T-rod ;
SCS,attached to the support,
Absolute motion,circular motion,
Relative motion,rectilinear,
convected motion,translation；
)i r e c t i o nd( OAlv a ?? ?
) t op o i n t i n g a l o n g( ),i r e c t i o nd( 2 OAOlaOAla naa ??? ???
er t i ca lv ?? rr av
s o l v e d be t o,h o r i z o n t a l ?? ee av
[Example 1] In the mechanism shown in
the figure,OA＝ l,?and ? are known when
= 45o, Find the velocity and
acceleration of the small vehicle,
Look at the animation
92
已知, OA＝ l,= 45o 时,?,? ;
求,小车的速度与加速度,
?
解, 动点,OA杆上 A点 ;
动系：固结在滑杆上 ;
静系：固结在机架上。
绝对运动：圆周运动,
相对运动：直线运动,
牵连运动：平动；
)( OAlv a ?? 方向?
)( ),( 2 OAOlaOAla naa 指向沿方向 ??? ???
铅直方向 ?? rr av
., 待求量水平方向?? ee av
[例 1] 曲柄滑杆机构
请看动画
93
Velocity of the small vehicle,
evv ?
According to draw the parallelogram of velocities shown in
the figure,rea
vvv ??
)(c o sc o s ???? ??? llvv ae 2 245 ?
Projecting to axis x,yields
e
n
aa aaa ?? ??
? s i nc o s
?? 4545 2 s i nc o s ?? lla e ???
,direction is shown in the figure
l)(2 2 2?? ??
acceleration of the small vehicle,
eaa ?
Employing the theorem of composition of
Accelerations when the convected motion is
translation
renaa aaaa ????
The vector diagram is shown in the figure,
94
小车的速度,
evv ?
根据速度合成定理 做出速度平行四边形,如图示
rea vvv ??
)(c o sc o s ???? ??? llvv ae 2 245 ?
投至 x轴,
e
n
aa aaa ?? ??
? s i nc o s
?? 4545 2 s i nc o s ?? lla e ???
，方向如图示
l)(2 2 2?? ??小车的加速度,
eaa ?
根据牵连平动的加速度合成定理
renaa aaaa ????
做出速度矢量图如图示 。
95
[Example 2]
Solution,moving point,pin D (on rod BC); MCS,OA； SCS,support
absolute motion,rectilinear motion,
relative motion,rectilinear motion,along OA
convected motion,rotation,
aavv aa ??,
,?? rr av
OODaOAODa nee a t p o i n t?;?,2 ??????? ???
OAODv e ?????,?
???? s i ns i n,c o sc o s vvvvvv arae ????
hv
hvODv
e
?
???
2c o s )
c o s/(c o s/ ???
（ ）
avh,,,,?
are given,Find ? and ? of rod OA,
According to Draw the parallelogram of velocities,as shown in
the figure,rea vvv ??
Se
e an
im
ati
on
96
[例 2] 摇杆滑道机构
解, 动点,销子 D (BC上 ); 动系, 固结于 OA；静系, 固结于机架。
绝对运动：直线运动,
相对运动：直线运动，,沿 OA 线
牵连运动：定轴转动,
aavv aa ??,
,?? rr av
OODaOAODa nee 指向?;?,2 ??????? ???
OAODv e ?????,?
???? s i ns i n,c o sc o s vvvvvv arae ????
hv
hvODv
e
?
???
2c o s )
c o s/(c o s/ ???
（ ）
avh,,,,?
已知 求, OA杆的 ?,? 。
根据 速度合成定理 做出速度平行四边形,如图示。
rea vvv ??
请
看
动
画
97
Projecting to axis?,
kea aaa ?? ??c o s
????? co ss i nco s2co s 22 ahvaaa ake ????
????
?
22
2
2
c o s2s i nc o s hahvODa e ??? （ ）
Employing the theorem of composition of
Accelerations when the convected motion is rotation,
we have
krneea aaaaa ???? ?
???
??
?
s i nc o s22
,c o s)c o s(
c o s
2
32
2
2
v
h
vva
h
v
h
vha
rk
n
e
???
???
98
投至 ? 轴,kea aaa ?? ??c o s
????? co ss i nco s2co s 22 ahvaaa ake ????
????
?
22
2
2
c o s2s i nc o s hahvODa e ???
（ ）
根据 牵连转动的加速度合成定理
krneea aaaaa ???? ?
???
??
?
s i nc o s22
,c o s)c o s(
c o s
2
32
2
2
v
h
vva
h
v
h
vha
rk
n
e
???
???
99
See the animation
[Example 3]
Solution,moving point,point A on O1A; MCS,attached to BCD,SCS,
attached to the support,
absolute motion,circular motion;
relative motion,rectilinear motion;
convected motion, translation;, horizontal
AOrv a 11,?? ?
BCv r / /?,?
?ev
and h;are
given,At the moment,
find the angular velocity ?2 of
rod O2E,
EOAO 21 //
??,,11 rAO ?
100
请看动画
[例 3] 曲柄滑块机构
解, 动点,O1A上 A点 ; 动系,固结于 BCD上,静系固结于机架上。
绝对运动：圆周运动 ;
相对运动：直线运动 ;
牵连运动：平动 ;,水平方向
AOrv a 11,?? ?
BCv r / /?,?
?ev
已知,h;
图示瞬时 ;
求, 该瞬时 杆的 ?2 。
EOAO 21 //
EO 2
,,,11 ??rAO ?
101
According to draw
the parallelogram of velocities rea vvv ??
Select moving point,point F on BCD
MCS,attached to O2E,
SCS,attached to the support
absolute motion,rectilinear motion;
relative motion,rectilinear motion;
convected motion, rotation;
)(s i n1 ?? ??rv Fa
)( / /?,2 EOv Fr ?
)(?,2 EOv Fe ??
??? s i ns i n 1rvv ae ??
According to draw the parallelogram of
velocities,FrFeFa vvv ??
?????? 211 s i ns i ns i ns i n rrvv FaFe ????
?? s i n/,222 hFOFOv eF ????又
?????? 3121
2
2 s i n
s i ns i n
h
r
h
r
FO
v eF ?????
） （
102
根据
做出速度平行四边形 rea vvv ??
再选动点,BCD上 F点
动系：固结于 O2E上,
静系固结于机架上
绝对运动：直线运动,
相对运动：直线运动,
牵连运动：定轴转动,
)(s i n1 ?? ??rv Fa
)( / /?,2 EOv Fr ?
)(?,2 EOv Fe ??
??? s i ns i n 1rvv ae ??
根据 做出速度平行四边形
FrFeFa vvv ??
?????? 211 s i ns i ns i ns i n rrvv FaFe ????
?? s i n/,222 hFOFOv eF ????又
?????? 3121
2
2 s i n
s i ns i n
h
r
h
r
FO
v eF ?????
） （
103
solution,Take C on the cam as the moving
point,and attach MCS to rod OA and SCS
to the earth ground,
absolute motion,rectilinear motion;
relative motion,rectilinear motion;
convected motion, rotation; OAav rr //i r e c t i o n d,??
OCv e ?? d i r e c t i o n?,
The radius of cam C is R,point O and C
in the same line at this moment,are
known,Find the angular velocity and acceleration
of rod OA at this instant,
av a n d,?
Analysis,Since the contact point between these
two rigid bodies varies with time relative to both
of them,the contact point is not suitable as a
moving point,
[Example 4]; t op o i n t i n g?2 OOCa ne ??? ??,??? ?? OCa e
direction OC?
See the animation
aavv aa ??,
104
解, 取凸轮上 C点为动点,
动系固结于 OA杆上,
静系固结于地面上,
绝对运动, 直线运动,
相对运动, 直线运动,
牵连运动, 定轴转动,
aavv aa ??,
OAav rr //,方向??
OCv e ?? 方向?,
已知,凸轮半径为 R，图示瞬时 O,C
在一条铅直线上 ; 已知 ;
求, 该瞬时 OA杆的角速度和角加速度。
av,,?
分析, 由于接触点在两个物体上的位
置均是变化的，因此不宜选接触点为
动点。
[例 4] 凸轮机构;?2 OOCa ne 指向??? ??,??? ?? OCa e
方向 OC?
请看动画
105
??? s i ns i n/ ;,0 RvR vOC vvvvv eaer ??????
） (
drawing the parallelogram of velocities,we have
According to
rea vvv ??
According to
kr
n
eea aaaaa ????
?
Drawing the vectorial diagram of accelerations,gives
02,s i n)s i n(s i n
2
2 ?????
rk
n
e vaR
v
R
vRa ???
?
Projecting them to axis ?,we have,
??? ? c o ss i nc o s enea aaa ?? ?? tgneae aaa ??
2
222 s i ns i n
s i n/
/s i n
R
v
R
a
R
Rva
OC
a e ??
?
?? ? ?????
Rotating direction depends on its sign.,if >0, and if <0
106
??? s i ns i n/ ;,0 RvR vOC vvvvv eaer ??????
） (
做出速度平行四边形,知
根据
rea vvv ??
根据
kr
n
eea aaaaa ????
?
做出加速度矢量图
02,s i n)s i n(s i n
2
2 ?????
rk
n
e vaR
v
R
vRa ???
?
投至 ? 轴,??? ? c o ss i nc o s
e
n
ea aaa ??
?? tgneae aaa ??
2
222 s i ns i n
s i n/
/s i n
R
v
R
a
R
Rva
OC
a e ??
?
?? ? ?????
转向由上式符号决定,>0则, <0 则
107
(see the animation)
[Example 5]
The rotating speed of the drive wheel is n=30 r/min
and OA=150mm, At this moment,OA?OO1,
Find ? 1 and ?1 of rod O1D and the of
block B,BB av,
108
（请看动画）
[例 5] 刨床机构
已知, 主动轮 O转速 n=30 r/min
OA=150mm,图示瞬时,OA?OO1
求, O1D 杆的 ? 1,?1
和滑块 B的 。
BB av,
109
where
m / s 15.03015.0 ??? ????? nOAv a
r a d / s
5515.0
503.0
m / s 503.0s i n
1
1
??
?
??
???
???
AO
v
vv
e
ae
） (
Solution,moving point,point A on wheel
O,MCs,O1D,SCS,support,
According to
draw the parallelogram of velocities
rea vvv ??
m / s 506.0c o s
)
5
5
s i n,
5
52
( c o s
??
??
??
??
ar vv
110
其中
m / s 15.03015.0 ??? ????? nOAv a
r a d / s
5515.0
503.0
m / s 503.0s i n
1
1
??
?
??
???
???
AO
v
vv
e
ae
） (
解,动点：轮 O上 A点
动系,O1D,静系：机架
根据 做出速度平行四边形 。
rea vvv ??
m / s 506.0c o s
)
5
5
s i n,
5
52
( c o s
??
??
??
??
ar vv
111
According to
kr
n
eea aaaaa ????
?
Draw the vector diagram of accelerations
rka vaa 12 2 15.0 ?? ??
Projecting them to gives,
ka ?? eka aaa ??c o s
222 m / s
5
518.0506.0
525
5215.0 ????? ??????
ea
22211 r a d / s
25
6
515.0
1
5
518.0/ ??? ? ???? AOa
e
） (
Again,select block B as the moving point; MCS,O1D; SCS,support,
112
根据
kr
n
eea aaaaa ????
?
做出加速度矢量图
rka vaa 12 2 15.0 ?? ??
投至 方向,
ka ?? eka aaa ??c o s
222 m / s
5
518.0506.0
525
5215.0 ????? ??????
ea
22211 r a d / s
25
6
515.0
1
5
518.0/ ??? ? ???? AOa
e
） (
再选动点,滑块 B; 动系, O1D; 静系, 机架。
113
According to
BrBeBa vvv ??
draw the vector
diagram of velocities,
,m / s 506.02 ??? eeB vv
m / s 503.0tg
m / s 15.0c o s/
??
??
???
????
eBrB
eBaBB
vv
vvv
Projecting them to axis x yields,
kBeBaB aaa ??
??c o s
2222 m / s 15.0
5
52/)
5
506.0
5
536.0( ??? ?????
aBB aa
According to
kBrB
n
eBBeBa aaaaa ????
?
draw the vector diagram of accelerations,
2 2 m / s5 536.02 ?? ?? eeB aa
where
22
1 m / s 5
506.0 503.0
522 ??
?? ?????
rBkB va
114
根据
BrBeBa vvv ??
做出速度矢量图 。
,m / s 506.02 ??? eeB vv
m / s 503.0tg
m / s 15.0c o s/
??
??
???
????
eBrB
eBaBB
vv
vvv
投至 x 轴,
kBeBaB aaa ??
??c o s
2222 m / s 15.0
5
52/)
5
506.0
5
536.0( ??? ?????
aBB aa
根据
kBrB
n
eBBeBa aaaaa ????
?
做出加速度矢量图
2 2 m / s5 536.02 ?? ?? eeB aa
其中
22
1 m / s 5
506.0 503.0
522 ??
?? ?????
rBkB va
115
[Example 6]
In the mechanism shown in the figure,and h
are given,Find ? and ? of sleeve O.。
av,,?
Solution,scheme 1,
Point A has rectilinear motion with
?tg?? hx A
hxx AA /)2s i nc o s( 2 ?????? ??? ????
hxhx AA /c o s s e c 22 ???? ???? ??? 即
Substituting the given quantities,we have
????? 2222 c o s)2s i n(,c o s hvhahv ??? （ ） （ ）
请看动画
116
[例 6] 套筒滑道机构
图示瞬时,h已知,
求,套筒 O的 ?, ? 。
av,,?
解,方法 1,
A点作直线运动
?tg?? hx A
hxx AA /)2s i nc o s( 2 ?????? ??? ????
hxhx AA /c o s s e c 22 ???? ???? ??? 即
代入图示瞬时的已知量，得
????? 2222 c o s)2s i n(,c o s hvhahv ??? （ ） （ ）
请看动画
117
Try to compare these two
schemes,
（ ） ??? ? 2
2
2
c o s)2s in(
h
v
h
a
OA
a e
???
Projecting to direction gives,
??
eka aaa ??c o s
ka
c o s2s i nc o s
2
????
h
vaa
e ??
h
vOAv
e
?? 2c o s/ ?? （ ）
Scheme 2,moving point,point A on CD,
MCS,sleeve O,SCS,support
rea vvv ??
??? s i n,c o sc o s vvvvv rae ????
kr
n
eea aaaaa ????
?
??? c o s2s i n2,
2
???? hvvaaa rka
vv a ??
118
对比两种方法 （ ） ??? ? 2
2
2
c o s)2s in(
h
v
h
a
OA
a e
???
投至 方向,
??
eka aaa ??c o ska
c o s2s i nc o s
2
????
h
vaa
e ??
h
vOAv
e
?? 2c o s/ ?? （ ）
方法 2,动点, CD上 A点,
动系, 套筒 O，静系, 机架
rea vvv ??
??? s i n,c o sc o s vvvvv rae ????
kr
n
eea aaaaa ????
?
??? c o s2s i n2,
2
???? hvvaaa rka
vva ?其中
119
120
