1
Theoretical mechanics
2
3
§ 9–1 Introduction to plane motion of a rigid body
§ 9–2 Plane motion can be decomposed into
translation and rotation · Equations of
plane motion
§ 9–3 Velocity of a point in a plane figure
§ 9–4 Acceleration of a point in a plane figure ·
Instantaneous center of zero acceleration
Lesson for problem solving
Chapter 9,Plane motion of a rigid body
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§ 9–1 刚体平面运动的概述
§ 9–2 平面运动分解为平动和转动 ·
刚体的平面运动方程
§ 9–3 平面图形内各点的速度
§ 9–4 平面图形内各点的加速度 ·
加速度瞬心的概念
习题课
第九章 刚体的平面运动
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In engineering,we often encounter the plane motion of a rigid
body,which is more complicated,Investigation into this kind of
motion is conducted on the basis of translation and rotation of a rigid
body using composition and decomposition of motions.Decomposing
a plane motion into translation and rotation,and employing the
theory of composition of motions,the formulae for finding the
velocity and acceleration of point in the rigid body can be derived,
§ 9-1 Introduction to plane motion of a rigid body
1,Definition of plane motion of a rigid body
The distance between any point in a rigid body and a fixed plane
always keeps unchanged during its motion,In other words,any
point in the rigid body moves in a plane parallel to the fixed plane,
The motion described above is called plane motion of a rigid body,
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刚体的平面运动是工程上常见的一种运动，这是一种较为
复杂的运动．对它的研究可以在研究刚体的平动和定轴转动的
基础上，通过运动合成和分解的方法，将平面运动分解为上述
两种基本运动．然后应用合成运动的理论，推导出平面运动刚
体上一点的速度和加速度的计算公式,
§ 9-1 刚体平面运动的概述
一．平面运动的定义
在运动过程中，刚体上任一点到某一固定平面的距离始终保
持不变．也就是说，刚体上任一点都在与该固定平面平行的某一
平面内运动．具有这种特点的运动称为刚体的平面运动,
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For example, Examine the motion of the
rod in a crank-rod mechanism,Since point
A moves in a circular path,and pint B
move along a straight line,the motion of
the rod AB is neither translation nor
rotation about a fixed axis,but plane
motion,
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例如, 曲柄连杆机构中连杆 AB的运动,
A点作圆周运动,B点作直线运动,因此，
AB 杆的运动既不是平动也不是定轴转动，
而是平面运动,
9 Please look at the animation
10 请看动画
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2,Simplification of a plane motion
A plane motion of a rigid body can
simplified to a motion of a plane figure
in the plane itself,As indicated in the
figure,when studying a plane motion of a
rigid body,we do not need to consider its
geometrical shape and size are not needed，
and instead,considering the motion of a
plane figure is enough to determine the
velocity and acceleration of any point in the
rigid body,
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二．平面运动的简化
刚体的平面运动可以
简化为平面图形 S在其自
身平面内的运动,即在研
究平面运动时，不需考虑
刚体的形状和尺寸，只需
研究平面图形的运动，确
定平面图形上各点的速度
和加速度,
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§ 9–2 Plane motion can be decomposed into
translation and rotation·Equations of plane motion
1,Equations of plane motion
To determine the position of a plane figure,which represents
the plane motion of a rigid body,only the position of a line segment in
this figure is needed to be determined,
The position of a line segment AB
can be determined by the
coordinates of point A and the angle
between AB and the axis x,
Therefore,the position of the figure
S can be determined by these three
independent variables,
Hence,we have
?,,AA yx
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§ 9-2 平面运动分解为平动和转动 ·
刚体的平面运动方程
一．平面运动方程
为了确定代表平面运动刚体的平面图形的位置，我们只需确定
平面图形内任意一条线段的位置,
任意线段 AB的位置可
用 A点的坐标和 AB与 x轴夹
角表示．因此图形 S 的位
置决定于 三个
独立的参变量．所以
?,,AA yx
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2,Plane motion can be decomposed into translation and rotation
When point A in the figure S keeps static,the rigid body
rotates about a fixed axis
When the angle ? in the figure S keeps unchanged,the rigid
body has translatory motion,
Therefore,a plane motion can be viewed as the composition of a
translation and rotation,
?,,AA yx
Equations of plane motion )(1 tfx A ?
)(2 tfy A ?
)(3 tf??
The can be found from this equation at any instant t,and
thus the location of the plane figure S can be determined,
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二．平面运动分解为平动和转动
当图形 Ｓ 上 Ａ 点不动时，则刚体作定轴转动
当图形 Ｓ 上 ? 角不变时，则刚体作平动,
故刚体平面运动可以看成是平动和转动的合成运动,
?,,AA yx
平面运动方程
)(1 tfx A ?
)(2 tfy A ?
)(3 tf??
对于每一瞬时 t,都可以求出对应的, 图形 S
在该瞬时的位置也就确定了。
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Example Motion of a wheel,
The plane motion of the
wheel can be viewed as the
composition of the translation
with the vehicle body and the
rotation relative to the vehicle,
Plane motion of the wheel relative
to the static reference system (absolute motion)
Translation of the vehicle body
(moving system Ax? y? ) relative
to static system (convected motion)
Rotation of the wheel relative to
the vehicle (moving system Ax? y?) (relative motion)
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例如 车轮的运动,
车轮的平面运动可以看成
是车轮随同车厢的平动和相对
车厢的转动的合成,
车轮对于静系的平面运动 （绝对运动）
车厢（动系 Ax? y? ) 相对静系的平动 （牵连运动）
车轮相对车厢（动系 Ax? y?）的转动 （相对运动）
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The arbitrarily selected point A is called a pole,Thus,
Plane motion of a wheel
Translation with the pole A Rotation about the pole A
A plane motion of a rigid
body can be decomposed
into a translation with a pole
and a rotation about the
pole,
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我们称动系上的原点 Ａ 为 基点,于是
车轮的平面运动
随基点 A的平动 绕基点 A'的转动
刚体的平面运动可以
分解为随基点的平动
和绕基点的转动,
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Another example,plane figureＳ moves from position I to position
II in a interval ?t,
?Select point A as a pole,Translate the figure together with A to
the position A'B'',and then rotate it around A through an angle
to the final position A'B‘,
?Select point A as a pole,Translate it together with B to the
position A''B',and then rotate it around B through an angle
to the final position A'B'
Clearly,AB?? A'B'' ?? A''B', we have
21 ?? ???
1??
2??
21
21
21
2
0
1
0
,;,limlim ????????????
??
????
?? dt
d
dt
d
tt tt
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再例如, 平面图形 Ｓ 在 ?ｔ 时间内从位置 I运动到位置 II
?以 A为基点, 随基点 A平动到 A'B''后,绕基点转 角到 A'B'
?以 B为基点, 随基点 B平动到 A''B'后,绕基点转 角到 A'B'
图中看出,AB?? A'B'' ?? A''B', 于是有 21 ?? ???
1??
2??
21
21
21
2
0
1
0
,;,limlim ????????????
??
????
?? dt
d
dt
d
tt tt
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In conclusion,the translation in a plane motion
depends on the selection of the pole,however,the
rotation about the selected pole DOES NOT depend on
the choice of a pole,(In other words,the instant ?,? of
any rotations about ANY poles are the same） The
selection of the pole is arbitrary.(We usually select a
point with known motion as the pole,)
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所以,平面运动随基点平动的运动规律与基
点的选择有关，而绕基点转动的规律与基点选取
无关, (即在同一瞬间，图形绕任一基点转动的
?,?都是相同的） 基点的选取是任意的 。 (通常
选取运动情况已知的点作为基点 )
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Crank-rod mechanism
Rod AB has plane motion
Decomposition of the plane
motion
（ please look at the animation）
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曲柄连杆机构
AB杆作平面运动
平面运动的分解
（请看动画）
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§ 9-3 velocity of any point in a plane figure
Employ the theorem of velocity composition,
rea vvv ??the velocity at point B can be expressed as
BAAB vvv ??
,vv;vv;vv BArAeBa ???
The velocity of point A in figure S
and the rotational velocity ? of the
figure are given,find,
Select A as the pole,and fix the
moving reference system to point A,
The motion of the moving system
is translation,Consider the moving point B,its motion can be
viewed as the composition of the translation,the convected motion,
and the rotation,the relative motion
point to the rotation direction,
Av
Bv
1,pole-based method (composition method)
,d i r ect i o n,m a g n i t u d e ABAB ???
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§ 9-3 平面图形内各点的速度
根据速度合成定理
,rea vvv ??
则Ｂ点速度为,
BAAB vvv ??
一．基点法（合成法）
取 B为动点,则 B点的运动可视为牵连运动为平动和相对运动
为圆周运动的合成
,,ABAB ????? 方向大小 ?,vv;vv;vv BArAeBa
已知：图形 S内一点 A的速度,
图形角速度 ? 求,
指向与 ? 转向一致,
取 A为基点,将动系固结于 A点,
动系作平动。
Av
Bv
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Since point A and B are arbitrarily selected,the equation
gives the relationship between the velocities of any two point in the
figure,Noting that,projecting this equation to AB,gives
BAAB vvv ??
ABv BA ?
? ? ? ? ABAABB vv ? ——theorem of velocity projection I,e.,the velocity projections of any two point in a figure on the
line linking these two points are identical,This method for
find the velocity of a point is called velocity projection
method,
That is,the velocity of any point in the figure is obtained as
the geometric sum of the velocity of the pole and the relative
rotational velocity with respect to the pole,Such a method of
finding the velocity is called pole-based method,or composition
method,which is a basic method to find the velocity of a point in a
figure,
2,Velocity projection method
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由于 A,B点是任意的，因此 表示了图形上任
意两点速度间的关系．由于恒有,因此将上式在 AB
上投影，有
BAAB vvv ??
ABv BA ?
? ? ? ? ABAABB vv ? — 速度投影定理
即 平面图形上任意两点的速度在该两点连线上的投影彼此相
等, 这种求解速度的方法称为 速度投影法,
即 平面图形上任一点的速度等于基点的速度与该点随图形绕
基点转动的速度的矢量和, 这种求解速度的方法称为 基点法,
也称为 合成法,它是求解平面图形内一点速度的基本方法,
二．速度投影法
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3,Instantaneous velocity center method
(1) Background
If a point whose velocity is zero is selected as the pole,the process
of finding the velocity of any point will be greatly simplified,Hence,
it is natural to ask if such a point exists in any instant,If it does exist,
how to find such a point?
T h e r e f o r e
, t oo p p o s i t e a n d,d i r e c t i o n t h e,AAPA vPAvAPv ???? ?
0?Pv
(2)Concept of Instantaneous velocity center
Consider a plane figure S,The velocity of point A
is,and the angular velocity of the figure is ?,
Take a line AL along the direction of,and then
turn it 90o in the direction of ? to AL‘ and find the
point P in AL‘ by letting,we have
?/AvAP ?
Av
Av
PAAP vvv ??
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三．瞬时速度中心法（速度瞬心法）
1,问题的提出
若选取速度为零的点作为基点，求解速度问题的计算会大大
简化．于是，自然会提出，在某一瞬时图形是否有一点速度等
于零？如果存在的话，该点如何确定？
所以反向恰与方向,,,AAPA vPAvAPv ???? ?
0?Pv
２．速度瞬心的概念
平面图形 S，某瞬时其上一点 A速度,
图形角速度 ?，沿 方向取半直线 AL,然后
顺 ? 的转向转 90o至 AL'的位置,在 AL'上取长
度 则,?/
AvAP ?
Av
Av
PAAP vvv ??
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At any instant,there must exist a sole point whose velocity
is zero,which is called the instantaneous velocity center of
this figure at this instant,
3,Methods to determine the Instantaneous velocity center
① When the velocity of a point and the angular
velocity ? of the figure are known,the instantaneous
velocity center (point P) can be determined,
and point P is in the direction of
the line formed by rotating the through 90o in the
direction of ? around point A,
,,AA vAPvAP ?? ?
Av
Av
② When a plane figure rolls along a fixed surface
without slipping,the contact point P between the
figure and the fixed surface will be the
instantaneous velocity center,
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即在某一瞬时必唯一存在一点速度等于零，该点称为平
面图形在该瞬时的瞬时速度中心，简称速度瞬心,
３．几种确定速度瞬心位置的方法
① 已知图形上一点的速度 和图形角速度 ?,
可以确定速度瞬心的位置,（ P点）
且 Ｐ 在 顺 ?转向绕 A点
转 90o的 方向一侧,
,,AA vAPvAP ?? ?
Av
Av
② 已知一平面图形在固定面上作无滑动的滚
动,则图形与固定面的接触点 P为速度瞬
心,
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AB
vv
vva
BA
BA
?
?? d i r e c t i o n,s a m e
t h ep o i n t t o a n d )(
AB
vv
vvb
BA
BA
?
??,d i r e c t i o n s
o p p o s i t e h a v e a n d )(
④ The magnitudes of the velocities
of two points A and B at any instant are
given,and,
BA vv,
ABvABv BA ??,
(b) (a)
③ When the directions of the velocities
at two points A and B in a figure are known,and
,draw lines from A and B
perpendicular to respectively,and the
cross point P of these two lines will be the
instantaneous velocity center,
BA vv,
BA vv t op a r al l e ln o t is
BA vv a n d
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AB
vvvva BA
BA
???,)( 同向与
AB
vvvvb BA
BA
???,)( 反向与
④ 已知某瞬时图形上 A,B两点速度
大小,且
BA vv,
ABvABv BA ??,
(b) (a)
③ 已知某瞬间平面图形上 A,B两点速度
的方向，且 。
过 A,B两点分别作速度 的垂线,交点
P即为该瞬间的速度瞬心。
BA vv,
BA vv 不平行
BA vv,
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In addition,if vA＝ vB in case ④,
the motion is instantaneous translation
too,
⑤ The velocities of two points A and B point to the same direction
at any instant,but they are not perpendicular to line AB,
In this case,the instantaneous velocity center is indefinitely far
away,and the angular velocity ? =0,i,e,all point in the figure have
the same velocity at this instant of time,Such a motion is called
instantaneous translation,(but their accelerations are not identical),
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另：对 ?种 (a)的情况，若 vA＝ vB,
则是瞬时平动,
⑤ 已知某瞬时图形上 A,B两点的速度方向相同，且不与 AB连线
垂直,
此时,图形的瞬心在无穷远处,图形的角速度 ? =0,图形上
各点速度相等,这种情况称为 瞬时平动, (此时各点的加速度不
相等 )
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Example,At the instant of time shown in the figure,the rod
BC in the crank-rod mechanism has instantaneous translation,
The angular velocity of the rod BC,The velocities of all
points in BC are identical,but their accelerations are not,
If ? is uniform, then
)(2 ???? ?ABaa nBB
But the direction of is along AC,
Instantaneous translation is different from translation,
ca cB aa ?
0?BC?
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例如, 曲柄连杆机构在图示位置时，连杆 BC作瞬时平动,
此时连杆 BC的图形角速度,
BC杆上各点的速度都相等, 但各点的加速度并不相等,
设匀 ?，则
)(2 ???? ?ABaa nBB
而 的方向沿 AC的,瞬时平动与平动不同
ca cB aa ?
0?BC?
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４, Instantaneous velocity center method
The method for finding the velocity of a point in a figure using
instantaneous velocity center is called instantaneous velocity center
method,
At any instant of time,the motion of a figure can be viewed as the
rotation around the instantaneous velocity center,
If P is the instantaneous velocity center,the velocity of any point A
and its direction ?AP,pointing the same direction with ?, ??? APv
A
５, Cautions
?The position of the the instantaneous velocity center is not fixed
at all time,it changes instantly with time,and exists uniquely at any
instant of time,
?Only the velocity at the instantaneous velocity center is zero,
but its acceleration is not certainly zero,This differs from the rotation
about a fixed axis,
?When a rigid body is in instantaneous translation,although the
velocities at all points in it are identical,but their accelerations are not
necessarily identical,This is different from the translatory motion,
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４, 速度瞬心法
利用速度瞬心求解平面图形上点的速度的方法,称为速度瞬心法,
平面图形在任一瞬时的运动可以视为绕速度瞬心的瞬时转
动，速度瞬心又称为平面图形的瞬时转动中心。
若 P点为速度瞬心，则任意一点 A的速度
方向 ?AP，指向与 ? 一致。 ??? APv A
５, 注意的问题
?速度瞬心在平面图形上的位置不是固定的，而是随时间不
断变化的。在任一瞬时是唯一存在的。
?速度瞬心处的速度为零,加速度不一定为零。不同于定轴转动
?刚体作瞬时平动时，虽然各点的速度相同，但各点的加速
度是不一定相同的。不同于刚体作平动。
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Solution,In this mechanism,OA rotates
about a fixed axis,AB has plane motion,
and piston B has translation,
?Pole-based method（ composition method）
Consider AB and take A as the pole, Its direction is
shown in the figures,?lv A ?
???
???
??
?
????
????
???
?
llABv
llvv
ll
vv
BAAB
ABA
AB
//
45tgtg
)(245c o s/
c o s/
?
?
（ ）
[Example 1] In a crank-rod mechanism,
OA=AB=l,and the crank OA rotates with
a uniform ?,Find,when ? =45o,the
velocity of piston B and the angular
velocity of rod AB,
,BAAB vvv ??Since Draw the parallelogram of
velocities at point B,as shown in the figure,
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解：机构中,OA作定轴转动,AB作平面运
动,滑块 B作平动。 ?
基点法（合成法）
研究 AB，以 A为基点，且 方向如图示。
,?lv A ?
???
???
??
?
????
????
???
?
llABv
llvv
ll
vv
BAAB
ABA
AB
//
45tgtg
)(245c o s/
c o s/
?
?
（ ）
[例 1] 已知：曲柄连杆机构 OA=AB=l，
取柄 OA以匀 ? 转动。 求：当 ? =45o时,
滑块 B的速度及 AB杆的角速度,
根据,
BAAB vvv ??
在 Ｂ 点做 速度平行四边形，如图示。
45
)(2
//
,
????
????
??
??
???
?
lBPv
llAPv
lAPlv
ABB
AAB
A? （ ）
Try to compare these three methods,
? ? ? ? ABAABB vv ?
Employing theorem of velocity projection
,We have ?c o s
BA vv ?
)(245c o s/ c o s/ ????? ??? llvv AB ?
We can not find, AB?
?Velocity projection method,
Consider AB.,and its direction?OA,
is along BO,
?lv A ?
Bv
?Instantaneous velocity center method
Consider AB,Since the directions of
are known,we can determine the instantaneous
center of velocity at point P,
BA vv an d
46
)(2
//
,
????
????
??
??
???
?
lBPv
llAPv
lAPlv
ABB
AAB
A? （ ）
试比较上述三种方法的特点。
? ? ? ? ABAABB vv ?
根据速度投影定理
?c o sBA vv ?
)(245c o s/
c o s/
???
??
??
?
ll
vv AB
?不能求出 AB?
?速度投影法 研究 AB,,
方向 ?OA,方向沿 BO直线
?lv A ?
Bv
?速度瞬心法
研究 AB，已知 的方向，因此
可确定出 P点为速度瞬心
BA vv,
47
§ 9-4 Acceleration of a point in a plane
figure·Instantaneous center of acceleration
Take A as the pole,and fix the moving
reference system onto A,Take B as the
moving point,the motion of point B can be
decomposed into a relative motion (circular
motion) and a convected motion
(translation) with the pole,
nBABABArAeBa aaaaaaaa ????? ? ; ;
Hence,employing the theorem of acceleration composition
,the following formula can be obtained,
rea aaa ??
n
BABAAB aaaa ???
?
1,Pole-based method (composition method)
At an instant of time,the acceleration of a
point A in a figure S,? and ? are given,Find
the acceleration of any point B in the figure,
Aa
48
§ 9-4 平面图形内各点的加速度
加速度瞬心的概念
取 A为基点，将平动坐标系固结于 A点
取 B动点，则 B点的运动分解为相对运动
为圆周运动和牵连运动为平动,
nBABABArAeBa aaaaaaaa ????? ? ; ;
于是,由牵连平动时加速度合成定理 可得如下公式,
rea aaa ??
n
BABAAB aaaa ???
?
一, 基点法 (合成法 ) 已知：图形 S 内一点 A 的加速度 和图形
的 ?,?（某一瞬时）。
求,该瞬时图形上任一点 B的加速度。
Aa
49
where,direction?AB,pointing to the same direction
with ? ; which is along AB and points to A,
?? ?? ABa BA
2??? ABa nBA
The acceleration of any point in a figure equals to the geometric
sum of the acceleration of the pole,tangential and normal
accelerations of this point rotating about the pole together with the
figure,This method to find the acceleration of a point is call pole-
based method,or composition method,which is the basic method
to finding the acceleration of a point in a plane figure,
The formula given above is an equation in terms of plane vectors,
Therefore,we can solve two unknowns from it provided that the
other variables are given,Since the directions of are
always known,to solve the unknowns,only four other variables
are needed,
nBABA aa,?
n
BABAAB aaaa ???
?
50
其中：,方向 ?AB，指向与 ? 一致；
,方向沿 AB，指向 A点。
?? ?? ABa BA
2??? ABa nBA
即 平面图形内任一点的加速度等于基点的加速度与该点随图形绕
基点转动的切向加速度和法向加速度的矢量和 。这种求解加速度
的方法称为 基点法,也称为 合成法 。是求解平面图形内一点加速
度的基本方法。
上述公式是一平面矢量方程。需知其中六个要素，方能求
出其余两个。由于 方位总是已知，所以在使用该公式
中，只要再知道四个要素，即可解出问题的待求量。
nBABA aa,?
n
BABAAB aaaa ???
?
51
n
BABA aa,
?
2,Instantaneous center of acceleration
Due to the dependent of on the point B,we can always
find a point Q in the figure, at which the relative acceleration
has the same magnitude but opposite direction with the acceleration
in the pole so that the absolute acceleration, Such a
point Q is called the instantaneous center of acceleration,
QAa
Aa 0?
Qa
52
n
BABA aa,
?
二．加速度瞬心,
由于 的大小和方向随 B点的不同而不同,所以总可以
在图形内找到一点 Q，在此瞬时，相对加速度 大小恰与基点
A的加速度 等值反向，其绝对加速度,Q点就称为图形
在该瞬时的 加速度瞬心,
QAa
Aa 0?Qa
53
[Note] ?In general,the instantaneous center of acceleration does
not coincide with the instantaneous center of velocity,
? In general, there is no similar relationship between the
accelerations at two points given in the theorem of velocity
projection,That is to say,the following relationship does not hold in
general,
? ? ? ? ABBABA aa ?
Only in the special case where ? =0,the figure is in instantaneous
translation,holds,
In other words,if the angular velocity of a plane figure is zero at an
instant of time,the acceleration projections of any two point in a
figure on the line linking these two points are identical,
? ? ? ? ABBABA aa ?
54
[注 ] ?一般情况下,加速度瞬心与速度瞬心不是同一个点,
?一般情况下，对于加速度没有类似于速度投影定理的关
系式, 即一般情况下,图形上任意两点 A,B的加速度
? ? ? ? ABBABA aa ?
若某瞬时图形 ? =0,即瞬时平动,则有
即 若平面图形在运动过程中某瞬时的角速度等于零，则该瞬时
图形上任意两点的加速度在这两点连线上的投影相等,
? ? ? ? ABBABA aa ?
55
? ?Since it is not easy to find the instantaneous center of
acceleration and there does not exists a relationship similar to the
theorem of velocity projection,instantaneous center of acceleration is
not widely used to find accelerations of a given point,Instead,the pole
method is often employed,
Analysis,
magnitude？ √ Ｒ ? Ｒ ? 2
direction？ √ √ √
Therefore,? and ? should be solved first,
n
POPOOP aaaa ???
?
Rv O /?? ? （ ）
[Example 1] A wheel with radius R rolls on a plane surface without
slipping,The velocity and acceleration at the its center O are
given,Find the acceleration at the contact point P,O
v Oa
Solution,The wheel has plane motion and P is the
instantaneous velocity center,
56
? ?由于加速度瞬心的位置不象速度瞬心那样容易确定，且一
般
情况下又不存在类似于速度投影定理的关系式，故常采用基
点法求图形上各点的加速度或图形角加速度,
分析,
大小？ √ Ｒ ? Ｒ ? 2
方向？ √ √ √
故应先求出 ?? ?,
n
POPOOP aaaa ???
?
Rv O /?? ? （ ）
[例 1] 半径为 R的车轮沿直线作纯滚动,已知轮心 O点的速度
及加速度,求车轮与轨道接触点 P的加速度,O
v
Oa
解：轮 O作平面运动,P为速度瞬心,
57
Since the equation given above is valid at any instant,and point O
moves along a straight line,we obtain
R
a
dt
dv
Rdt
d OO ??? 1??
（ ）
It is evident that the acceleration at the instantaneous center of
velocity is not zero,which means that point P is not the instantaneous
center of acceleration,In this case,the acceleration of at the
instantaneous center of velocity P points to the center of the wheel,
Taking O as the pole,yields
As shown in the vector diagram of the
accelerations,（ and are equal in magnitude but
opposite in direction）
i,e,
nPOPOOP aaaa ??? ?
R
v
R
vRRaaRa OOn
POOPO
2
22 )(,??????? ???
n
POP aa ? Oa
?
POa
)(/2 ?? Rva OP
58
由于此式在任何瞬时都成立，且 O点作直线运动，故而
R
a
dt
dv
Rdt
d OO ??? 1?? （ ）
由此看出，速度瞬心 P的加速度并不等于零，即它不是加速度
瞬心．当车轮沿固定的直线轨道作纯滚动时，其速度瞬心 P的加
速度指向轮心,
以 O为基点，有 其中,
做出加速度矢量图，由图中看出,
（ 与 等值反向）
即
nPOPOOP aaaa ??? ?
R
v
R
vRRaaRa OOn
POOPO
2
22 )(,??????? ???
n
POP aa ? Oa ?POa
)(/2 ?? Rva OP
59
Solution,
(a) AB has translation,),(,n
B
n
ABABABA aaaaaavv ?????
??
BOAO
BOaAOa
BOvAOv
BA
BA
21
2211
2211
a n d;/,/;/,/ a d d t i o n,I n
?
??
??
??
??
???
.; 2121 ???? ???
[Example 2] At the instant shown in the figure,O1A=O2B and
O1A?/O2B,Are ?1 and ? 2 or ?1 and ?2 identical in case (a) and (b),
(a) (b)
60
解,(a) AB作平动,),(,n
B
n
ABABABA aaaaaavv ?????
??
BOAO
BOaAOa
BOvAOv
BA
BA
21
2211
2211
;/,/;/,/
?
??
??
而
又
??
??
???,;
2121 ???? ???
[例 2] 已知 O1A=O2B,图示瞬时 O1A?/O2B
试问 (a),(b)两种情况下 ?1和 ? 2,?1和 ?2是否相等？
(a) (b)
61
(b) AB has plane motion,and it is in instantaneous translation at
this instant of time,Therefore,
BAAB vv ??,0?
21221121,/,/,???? ????? BOvAOvBOAO BA?
? ? ? ? ? ? ? ? ? ? ? ? ABnBABBABnAABAABBABA aaaaaa ???? ??,即
???????? c o ss i nc o ss i n 2222221111 ??????? BOBOAOAO
,a t c l e a r l y t hs e e n bec a n I t, i, e, c t g2 212112 BA aa ????? ??????
62
(b) AB作平面运动,图示瞬时作瞬时平动,此时 BAAB vv ??,0?
21221121,/,/,???? ????? BOvAOvBOAO BA?
? ? ? ? ? ? ? ? ? ? ? ? ABnBABBABnAABAABBABA aaaaaa ???? ??,即
???????? c o ss i nc o ss i n 2222221111 ??????? BOBOAOAO
BA aaAB ????? 作瞬时平动时并由此看出即,c t g2 212112 ??????
63
[Example 3] In a crank-rod mechanism,the radius of
the wheel is R=15cm,n=60 rpm
Find ?Ｂ and ?Ｂ of the wheel when ? =60o and OA?AB,
Look for the animation on the next page
?
64
[例 3] 曲柄滚轮机构 滚子半径 R=15cm,n=60 rpm
求：当 ? =60o时 (OA?AB),滚轮的 ?Ｂ, ?Ｂ,
翻页请看动画
?
65
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66
请
看
动
画
67
?
Solution,OA rotates about a fixed axis,
AB and wheel B have plane motions,
Consider rod AB,
r a d / s 32153/30/ 1 ??? ????? APv AAB
（ ）
c m / s 30215
r a d / s 230/6030/
???
????
?????
???
OAv
n
A
P１ is the instantaneous velocity center
of AB
)(cm / s 3203215321 ??????? ??? ABB BPv
Analysis,To find ?Ｂ,?Ｂ of the wheel,vB and aB
should be worked out first,
P2
P1
vB P2 is the instantaneous velocity center
of the wheel
68
?
解, OA定轴转动,AB杆和轮 B作平面运动
研究 AB,
r a d / s 32153/30/ 1 ??? ????? APv AAB
（ ）
c m / s 30215
r a d / s 230/6030/
???
????
?????
???
OAv
n
A
P１ 为其速度瞬心
)(cm / s 3203215321 ??????? ??? ABB BPv
分析, 要想求出滚轮的 ?Ｂ,?Ｂ 先要求出 vB,aB
P2
P1
vB P2为轮速度瞬心
69
Take A as the pole,2222 cm / s60)2(15 ??? ????? OAa
A
Pointing to O
n
BABAAB aaaa ???
?
) a l o n g,
3
320)
3
2(153( 222 BAABa
AB
n
BA ?
?? ??????
Magnitude？ √？ √
Direction √ √ √ √
Draw the vector diagram for accelerations,
Projecting the above equation onto BA,
gives
n
BAB aa ??? 0030c o s ?
)(c m / s5.1313402 3/3 32030c o s/ 222 ????? ???nBAB aa
r a d / s25.715/320/ 2 ??? ?? BPv BB
22 r a d / s77.815/5.131/ ??? BPa BB?
） (
） (
Consider wheel B,its instantaneous velocity center is P2
70
取 A为基点,2222 cm / s60)2(15 ??? ????? OAa A 指向 O点
n
BABAAB aaaa ???
? ),
3
320)
3
2(153( 222 BAABa
AB
n
BA 沿?
?? ??????
大小？ √？ √
方向 √ √ √ √
作加速度矢量图，将上式向 BA线上投影
n
BAB aa ??? 0030c o s ?
)(c m / s5.1313402 3/3 32030c o s/ 222 ????? ???nBAB aa
r a d / s25.715/320/ 2 ??? ?? BPv BB
22 r a d / s77.815/5.131/ ??? BPa BB?
） (
） (
研究轮 B,P2为其速度瞬心
71
Lesson of problem solving for Chapter 9
1,Concepts and content
(1) Definition of plane motion
The distance between any point in a rigid body and a fixed
plane always keeps unchanged during its motion,
(2) Simplification of plane motion
The motion of a plane figure in the rigid body parallel with the
fixed plane S in its own plane can be used to represent the plane
motion of the entire rigid body,
(3) Decomposition of plane motion,
It can be decomposed into,
(4) Pole
Theoretically speaking,any a point in the figure can be
selected as the pole,We usually chose a point with known motion as
the pole,
Translation together with the pole（ it depends on the choice of pole）
Rotation around the pole（ it does not depend on the selection of pole）
72
第九章 刚体平面运动习题课
一．概念与内容
1,刚体平面运动的定义
刚体运动时，其上任一点到某固定平面的距离保持不变,
2,刚体平面运动的简化
可以用刚体上一个与固定平面平行的平面图形 S在自身平
面内的运动代替刚体的整体运动,
3,刚体平面运动的分解
分解为
4,基点
可以选择平面图形内任意一点,通常是运动状态已知的点,
随基点的平动（平动规律与基点的选择有关）
绕基点的转动（转动规律与基点的选择无关）
73
(5) Instantaneous center of velocity
?At any instant of time,there exists a sole point in the figure or its
extension whose velocity is zero,
?The location of the instantaneous center of velocity changes with
time,
?At any instant,the motion of a plane figure can be viewed as the
instant rotation around the instantaneous center of velocity,However,
this instant rotation is different from the rotation about a fixed axis,
? ? =0,instantaneous center of velocity is indefinitely far away,
In this case,all points have the same velocity and the rigid body is in
instantaneous translation,
Note that instantaneous translation is different from translation,
74
5,瞬心（速度瞬心）
?任一瞬时,平面图形或扩大部分都唯一存在一个速度为零的点
?瞬心位置随时间改变,
?每一瞬时平面图形的运动可视为绕该瞬时瞬心的转动．这
种瞬时绕瞬心的转动与定轴转动不同,
? ? =0,瞬心位于无穷远处,各点速度相同,刚体作瞬时平动,
瞬时平动与平动不同,
75
(6) Rotation and translation are two special cases of plane motion,
(7) Methods to find velocity of any point in a figure
?pole method,
?velocity projection method,
? instantaneous velocity center method,
In these methods,the pole methods is the basic method,and the
instantaneous velocity center method is a special case of the pole
method,
p o l e t h eis,Avvv BAAB ??
? ? ? ? ABAABB vv ?
v e l o c i t yofc e n t e r o u si n s t a n t a n e
t h esi,,PBPvBPv BB ??? ?
76
6,刚体定轴转动和平面平动是刚体平面运动的特例,
7,求平面图形上任一点速度的方法
?基点法,
?速度投影法,
?速度瞬心法,
其中，基点法是最基本的公式，瞬心法是基点法的特
例,
为基点Avvv BAAB,??
? ? ? ? ABAABB vv ?
为瞬心一致与 PBPvBPv BB,,,?? ???
77
(8) Methods to find acceleration of a point in a figure
pole method：,A is the pole,This is the most
useful method,In addition,when ? =0,i.e,instantaneous translation,
we can also employ, It is the special case of the pole
method when ? =0,
n
BABAAB aaaa ???
?
? ? ? ? ABAABB aa ?
(9) Conditions for the applications of the methods used in plane
motion and composition of motions
? methods used in plane motion are applied to find the relationship
between the velocities/accelerations of any two points or determine
the relationship among the velocity,acceleration of a point and the
angular velocity and angular acceleration of ONE rigid body having
plane motion,
?The methods used in the composition of motions are usually
applied to determine the transfer of motion at the contact point where
TWO rigid bodies are in contact and relative slipping exists,
78
8,求平面图形上一点加速度的方法
基点法：, A为基点,是最常用的方法
此外，当 ? =0,瞬时平动时也可采用方法
它是基点法在 ? =0时的特例。
n
BABAAB aaaa ???
?
? ? ? ? ABAABB aa ?
9,平面运动方法与合成运动方法的应用条件
?平面运动方法用于研究 一个平面运动刚体 上任意两点的速
度、加速度之间的关系及任意一点的速度、加速度与图形
角速度、角加速度之间的关系,
?合成运动方法常用来确定 两个相接触的物体 在接触点处有
相对滑动时的运动关系的传递,
79
2,Steps in problem solving and some key points
(1) According to the information given in the problem and the
definitions of the motions of a rigid body,determine the type of
motions of the rigid bodies appeared in the problem,Note that only
one rigid body should be considered each time,
(2) For a rigid body with plane motion,,properly choose a suitable
method to find the velocity (or angular velocity) of a point based on
the known and unknown conditions,To find accelerations,pole
method is recommended,
(3) Do the thorough analysis for velocities and accelerations,and finally
solve the unknowns,
(Pole method,properly select a pole and then draw the parallelogram
for velocities and the vector diagram for accelerations；
Velocity projection method,its limitation is that ? can not be found;
Instantaneous velocity center method,find the Instantaneous velocity
center is the key step.)
80
二．解题步骤和要点
1,根据题意和刚体各种运动的定义，判断机构中各刚体的运动
形式．注意每一次的研究对象只是一个刚体,
2,对作平面运动的刚体，根据已知条件和待求量，选择求解速
度 (图形角速度 )问题的方法,用基点法求加速度 (图形角加速
度 )
3,作速度分析和加速度分析，求出待求量,
(基点法, 恰当选取基点，作速度平行四边形，加速度矢量图；
速度投影法, 不能求出图形 ? ;
速度瞬心法：确定瞬心的位置是关键．）
81
[Example 1] Consider a mechanism shown in the figure,
OA=0.15m,n=300 rpm,AB=0.76m,BC=BD=0.53m,At
the instant shown in the figure,rod AB in a horizontal plane,
Find, and at this instant of time,
ABBD ??
Dv
Look for the animation
on the next page
82
[例 1] 曲柄肘杆压床机构
已知,OA=0.15m,n=300 rpm,AB=0.76m,
BC=BD=0.53m,图示位置时,AB水平
求该位置时的, 及
ABBD ?? Dv
翻页请看动画
83
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84
请
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动
画
85
Solution,OA and BC rotate around
their fixed axes,AB,BD have plane
motions,
Consider AB,P１ is its instantaneous
velocity center,
r ad / s10303 0 030 ???? ??? n
m / s 5.11015.0 ??? ????? OAv A
（ ）
r a d / s 16.7376.0 25.160s i n5.1
1
?? ????? ??? ?ABAPv AAB
m / s 72.216.75.076.016.760co s1 ???????? ?ABBPv ABB ?
Next,consider BD,P2is its instantaneous velocity center,
Note ?BDP2 is an equilateral triangleDP2=BP2=BD
r a d / s 13.553.0 73.2
2
???? BPv BBD?
)(m / s 72.213.553.02 ?????? BDD DPv ?
（ ）
86
解,OA,BC作定轴转动,
AB,BD均作平面运动
根据题意,
研究 AB,P１ 为其速度瞬心
r ad / s10303 0 030 ???? ??? n
m / s 5.11015.0 ??? ????? OAv A
（ ） r a d / s 16.7376.0 25.160s i n5.1
1
?? ????? ??? ?ABAPv AAB
m / s 72.216.75.076.016.760co s1 ???????? ?ABBPv ABB ?
研究 BD,P2为其速度瞬心,?BDP2为等边三角形 DP2=BP2=BD
r a d / s 13.553.0 73.2
2
???? BPv BBD?
)(m / s 72.213.553.02 ?????? BDD DPv ?
（ ）
87
[Example 2] the mechanism of planet gears L
oo
k a
t th
e a
nim
ati
on
88
[例 2] 行星齿轮机构
请
看
动
画
89
Solution,OA rotate around O; Wheel A has
plane motion and P is the instantaneous
velocity center,
,)(2211 ooM rRr rRrPMv ??? ???????
ooA r
rRrrRv ???? ?????? )( ） (
f i g u r e, i n t h es h o w n a r e d i r e c t i o n s
,)(2222 ooM rR
r
rR
rPMv ??? ??
?
????
[Example 2] a mechanism of planet gears
R,r and ?o are given and wheel A rolls
without slipping,Find,
21 a n d MM vv
90
解,OA定轴转动 ; 轮 A作平面运动,瞬心 P点
,)(2211 ooM rRr rRrPMv ??? ???????
ooA r
rRrrRv ???? ?????? )( ） (
[例 2] 行星齿轮机构
已知, R,r,?o 轮 A作纯滚动，求
21,MM vv
方向均如图示,)(2222 ooM rRr rRrPMv ??? ???????
91
[Example 3] In the plane mechanism shown below,the angle in the
wedge M is ? =30o,v=12cm/s ; r = 4cm for the wheel,No slipping
is assumed between the wheel and the wedge,Find the angular
velocity ? of the wheel,velocity at point O and B,Loo
k a
t th
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on
92
[例 3] 平面机构中,楔块 M,? =30o,v=12cm/s ; 盘, r = 4cm,与 楔
块间无滑动．求圆盘的 ?及轴 O的速度和 B点速度,
请
看
动
画
93
Solution,Shaft O,rod OC and wedge M are
in translatory motions,The wheel has plane
motion and P is the instantaneous velocity
center,
,c m / s 12?? vv A
r a d / s 3230c o s4/12c o s/12/ ???? ??? rPAv A
)(m / s 343230s i n4s i n ???????? ???? rPOv o
m722142242120c o s2 2222 ???????????? ?OBPOOBPOPB
) ( m / s 3.182143272 PBPBv B ??????? ?
） (
94
解,轴 O,杆 OC,楔块 M均作平动,
圆盘作平面运动,P为速度瞬心
,c m / s 12?? vv A
r a d / s 3230c o s4/12c o s/12/ ???? ??? rPAv A
)(m / s 343230s i n4s i n ???????? ???? rPOv o
m722142242120c o s2 2222 ???????????? ?OBPOOBPOPB
) ( m / s 3.182143272 PBPBv B ??????? ?
） (
95
? It can be seen from the comparison between [Example 2] and
[Example3] that the contact point between the wheel and the
surface is not always the instantaneous velocity center,Only
when the surface is fixed,the point on the wheel in contact with
this surface is the instantaneous velocity center,
? Every rigid body has its own instantaneous velocity center
and angular velocity that may be either with the body or in its
extension,but not in other rigid bodies,For instance,in
[Example 1],the instantaneous velocity center of rod AB lies in
point P1,and that of rod BD in point P2,Note P1 is not a point
on rod CB,
96
? 比较 [例 2]和 [例 3]可以看出,不能认为圆轮只滚不滑时,接
触点就是瞬心,只有在接触面是固定面时,圆轮上接触点
才是速度瞬心
? 每个作平面运动的刚体在每一瞬时都有自己的速度瞬心和
角速度,并且瞬心在刚体或其扩大部分上,不能认为瞬心在
其他刚体上, 例如,[例 1] 中 AB的瞬心在 P1点,BD的瞬心在 P2
点,而且 P1也不是 CB杆上的点
97
[Example 4] A slotted bar-sliding block mechanism
Look at the animation
98
[例 4] 导槽滑块机构 请看动画
99
[Example 4] In the mechanism show in the figure,OA= r,the
crank rotates with uniform ?,and a block attached at the center C
of rod AB can slid along the slotted bar O1D,AB=l,At the moment
shown in the figure,point O,A and O1 in the same horizontal line,
and OA?AB,?AO1C= ?=30,
Find,the angular velocity of O1D,
Solution,OA and O1D rotate about their
fixed axes and rod AB is in plane motion,
? Consider AB:,AB is in
instantaneous translation,therefore,
?? rvvrv AcB ??? ;
?rv A ?
?Composition of motions method
Fine the velocity of the contact point in O1D with block C,
moving point is point C on rod AB (or block C ),
moving system is rod O1D,
static system is supports,
100
[例 4] 导槽滑块机构
已知, 曲柄 OA= r,匀角速度 ? 转动,连杆 AB的中点 C处连接一
滑块 C可沿导槽 O1D滑动,AB=l,图示瞬时 O,A,O1三点
在同一水平线上,OA?AB,?AO1C= ?=30。
求,该瞬时 O1D的角速度,
解, OA,O1D均作定轴转动,
AB作平面运动
? 研究 AB:,图示位置,
作 瞬时平动,所以
?? rvvrv AcB ??? ;
?rv A ?
?用合成运动方法
求 O1D杆上与滑块 C 接触的点的速度
动点, AB杆上 C (或滑块 C ),动系, O1D杆,静系, 机架
101
Absolute motion,curvilinear motion
,direction ?
Relative motion,rectilinear motion
,direction// O1D
Convected motion,rotation,,
direction? O1D
?rvv ca ??
?rv
?ev
According to,draw the
parallelogram of velocities rea vvv ??
??? rrvv Ce 2 330c o sc o s ????? ?
?
?
?
?
?
l
r
l
r
CO
v
COv
e
DO
DOe
2
3
s i n/
2
23
a n d
1
1
1
1
????
???
This is a comprehensive problem,which can be solve by applying
both composition of motions and plane motion,Pay attention to the
solving method for such kind of problems,See next example,
102
绝对运动,曲线运动,方向 ?
相对运动,直线运动，,方向 // O1D
牵连运动,定轴转动，,方向 ? O1D
?rvv ca ??
?rv
?ev
根据, 作速度平行四边形
rea vvv ??
??? rrvv Ce 2 330c o sc o s ????? ?
?
?
?
?
?
l
r
l
r
CO
v
COv
e
DO
DOe
2
3
s i n/
2
23
1
1
1
1
????
???又
） (
这是一个需要联合应用点的合成运动和刚体平面运动理论
求解的综合性问题,注意这类题的解法，再看下例,
103
[Example 5] Planar mechanism
Please look at the animation
104
[例 5] 平面机构
请
看
动
画
105
[Example 5] In the planar mechanism
shown in the figure,point O is in the
center of rod AB and ? =60o,BC?AB,In
addition,point O and C in the same
horizontal line,and AB=20cm,vA=16cm/s,
Find the angular velocities of rod AB and
BC respectively and the velocity of block C,
Solution,Wheel A,rod AB and rod BC have plane motions,Sleeve O
rotates about a fixed axis,and block C is in translation,
Take point O in the sleeve as the moving point,the moving system is
attached on AB,and static system on the fixed support,
rea vvv ??
,due to along AB,The direction of is
along AB and opposite with, In such a way,the direction of the
velocity at the point on AB in contact with sleeve O is determined,
ra vv,0? e
v
rv
Consider AB,P1 is the instantaneous center of velocity,
106
[例 5] 平面机构
图示瞬时,O点在 AB中点,? =60o,
BC?AB,已知 O,C在同一水平线上,
AB=20cm,vA=16cm/s,
试求 该瞬时 AB杆,BC杆的角速度
及滑块 C的速度,
解, 轮 A,杆 AB,杆 BC均作平面运动,套筒 O作定轴转动,滑块 C平动,
取套筒上 O点为 动点,动系 固结于 AB杆 ; 静系 固结于机架,
rea vvv ??
,由于 沿 AB,
所以 方向沿 AB并且与 反向。 从而确定了 AB杆上与 O点接
触点的速度方向。
ra vv,0?
ev rv
研究 AB,P1为速度瞬心
107
?BC and vC are simply found using
instantaneous velocity center method,
c m / s31631660
c m / s321622
60c o s
?????
?????
?
?
tgvv
v
v
v
BCB
B
B
C
cm3103 ????? OBBCBCv BCCB ??
Consider BC,and take B as the pole,
According to
draw the parallelogram of velocities CBBC
vvv ??
c m / s 1611 ?????? AABABB vAPBPv ??
r a d / s 35460s i n/10 16s i n/16
1
???? ??? OAAPv AAB
） (
（ ）
r a d / s 6.1
310
316 ????
BC
v CB
BC?
108 也可以用瞬心法求 ?BC和 vC，很简便
c m / s31631660
c m / s321622
60c o s
?????
?????
?
?
tgvv
v
v
v
BCB
B
B
C
cm3103 ????? OBBCBCv BCCB ??
研究 BC,以 B为基点,
根据
作 速度平行四边形
CBBC vvv ??
c m / s 1611 ?????? AABABB vAPBPv ??
r a d / s 35460s i n/10 16s i n/16
1
???? ??? OAAPv AAB
） (
（ ）
r a d / s 6.1
310
316 ????
BC
v CB
BC?
109
Solution,Rod OA rotates about a fixed axis,
rod AB and BC have plane motions,block B
and C have translatory motions,
?Find
cvApplying the theorem of velocity projection
to rod AB,gives
?? 30c o s60c o s
AB vv ? oAB rvv ?33 ???
Applying the theorem of velocity projection to rod BC,we have
?60s in
Bc vv ?
)( ???? ooc rrv ?? 232 33
[Example 6] In the mechanism,OA= r,rod OA rotates with uniform
?o,At the moment,? = 60o,AB?BC,AB=6 r,BC=,
Find the velocity and acceleration of block C at this moment,
r33
110
解,OA定轴转动 ; AB,BC均作平面运动,
滑块 B和 C均作平动
?求
cv
对 AB杆应用速度投影定理,
?? 30c o s60c o s
AB vv ? oAB rvv ?33 ???
对 BC杆应用速度投影定理,?60s in
Bc vv ?
)( ???? ooc rrv ?? 232 33
[例 6] 已知,配气机构中,OA= r,以等 ?o转动,在某瞬时 ? = 60o
AB?BC,AB=6 r,BC=, 求 该瞬时滑块 C的
速度和加速度,
r33
111
?Find,
caTake A as the pole to find acceleration at point B,
n
BABAAB aaaa ???
? ( a )
,,22 ABnBAoA ABara ?? ??? P1 is the instantaneous velocity center
of rod AB and, rAP 3
1 ?
22
1
3
2
3
6
33
o
on
BA
ooA
AB
rra
r
r
AP
v
?
?
??
?
???
????
)(
,
Draw the vector diagram of
accelerations,and project them to BA
222
33
4
6060
oooB
n
BAAB
r
rra
aaa
??? ?????
???? c o sc o s ??
112
?求
ca
以 A为基点 求 B点加速度,n
BABAAB aaaa ???
? ( a )
,,22 ABnBAoA ABara ?? ??? P1为 AB杆速度瞬心，而 rAP 31 ?
22
1
3
2
3
6
33
o
on
BA
ooA
AB
rra
r
r
AP
v
?
?
??
?
???
????
)(
,
作加速度矢量图,
并沿 BA方向投影
222
33
4
6060
oooB
n
BAAB
r
rra
aaa
??? ?????
???? c o sc o s ??
113
)( baaaa nCBCBBc ??? ?
Draw the vector diagram of
accelerations P2 is the
instantaneous center of velocity
for rod BC and P2C = 9 r,
Then take B as the pole to find
ca
69
1
2
3
2
o
o
c
BC rrCP
v ??? ?????
222
12
3
633 o
o
BC
n
CB rrBCa ?
?? ????? )(
Projecting Equ,(b) to BC gives
222
12
3
12
3
2
3
330 ooo
n
CBBc rr
raaa ??? ?????? ?c o s
[Note] We can assume the direction of, If the solved
acceleration has a positive sign,the assumed direction is right,
otherwise,the true direction is opposite to the assumed one,
cB aa or
30o
114
)( baaaa nCBCBBc ??? ?
作加速度矢量图,
P2 为 BC的瞬心,而 P2C = 9 r
再以 B为基点,求
ca
69
1
2
3
2
o
o
c
BC rrCP
v ??? ?????
222
12
3
633 o
o
BC
n
CB rrBCa ?
?? ????? )(
将 (b) 式在 BC方向线上投影
222
12
3
12
3
2
3
330 ooo
n
CBBc rr
raaa ??? ?????? ?c o s
[注 ] 指向可假设，结果为正说明假设与实际指向相同,
反之，结果为负，说明假设与实际指向相反, cB aa,
30o
115
[Example 7] Slotted bar-block mechanism
Please look at the animation
116
[例 7] 导槽滑块机构 请看动画
117
Solution,
? Apply the composition method of motions to determine the
relationship between the velocities at point C on rod CD and point
C‘ in contact with rod AE,Take point C on rod CD as the moving
point,the moving system is attached on rod AE,and the static system
is on the fixed support,we have
(a)
rcc vvv ?? '?Determine the relationship between the velocity at point A on
AE and that at point C' based on the analysis of the plane motion,
(b)
AcAc vvv ???'
[Example 7] In the slotted bar-sliding block
mechanism,the velocity of rod AB and that
of rod CD as well as the angle ? are given
At this moment,AC= l,Find the angular
velocity of the slotted bar AE,
u
v
118
解, ? 应用点的合成运动方法
确定 CD杆上 C点与 AE杆上接触 点 C'之间的速度关系
取 CD杆上 C为动点，动系固结于 AE，静系固结于机架；则
(a)
rcc vvv ?? '
?应用平面运动方法确定 AE上 A,C' 点之间速度关系
(b)
AcAc vvv ???'
[例 7] 导槽滑块机构
图示瞬时,杆 AB速度,杆 CD速度
及 ? 角已知，且 AC= l,
求导槽 AE的图形角速度,
u
v
119
Inserting (b) into (a) gives,Drawing vector
diagram of velocities and projecting them to the axis ?, and
noting vC＝ v,v＝ u,we have
rACAC vvvv ??? ?
??
??
s i n c o s e,i.
s i nc o s
uvv
vvv
AC
ACAC
??
???
?
?
l
uv
AC
v AC
AE
??? s i nc o s ??? ? （ ）
120
将 (b) 代入 (a) 得,作速度矢量图投至 ? 轴，
且 vC＝ v,v＝ u，有
rACAC vvvv ??? ?
??
??
s i n c o s
s i nc o s
uvv
vvv
AC
ACAC
??
???
?
?
即
l
uv
AC
v AC
AE
??? s i nc o s ??? ? （ ）
121
122
Theoretical mechanics
2
3
§ 9–1 Introduction to plane motion of a rigid body
§ 9–2 Plane motion can be decomposed into
translation and rotation · Equations of
plane motion
§ 9–3 Velocity of a point in a plane figure
§ 9–4 Acceleration of a point in a plane figure ·
Instantaneous center of zero acceleration
Lesson for problem solving
Chapter 9,Plane motion of a rigid body
4
§ 9–1 刚体平面运动的概述
§ 9–2 平面运动分解为平动和转动 ·
刚体的平面运动方程
§ 9–3 平面图形内各点的速度
§ 9–4 平面图形内各点的加速度 ·
加速度瞬心的概念
习题课
第九章 刚体的平面运动
5
In engineering,we often encounter the plane motion of a rigid
body,which is more complicated,Investigation into this kind of
motion is conducted on the basis of translation and rotation of a rigid
body using composition and decomposition of motions.Decomposing
a plane motion into translation and rotation,and employing the
theory of composition of motions,the formulae for finding the
velocity and acceleration of point in the rigid body can be derived,
§ 9-1 Introduction to plane motion of a rigid body
1,Definition of plane motion of a rigid body
The distance between any point in a rigid body and a fixed plane
always keeps unchanged during its motion,In other words,any
point in the rigid body moves in a plane parallel to the fixed plane,
The motion described above is called plane motion of a rigid body,
6
刚体的平面运动是工程上常见的一种运动，这是一种较为
复杂的运动．对它的研究可以在研究刚体的平动和定轴转动的
基础上，通过运动合成和分解的方法，将平面运动分解为上述
两种基本运动．然后应用合成运动的理论，推导出平面运动刚
体上一点的速度和加速度的计算公式,
§ 9-1 刚体平面运动的概述
一．平面运动的定义
在运动过程中，刚体上任一点到某一固定平面的距离始终保
持不变．也就是说，刚体上任一点都在与该固定平面平行的某一
平面内运动．具有这种特点的运动称为刚体的平面运动,
7
For example, Examine the motion of the
rod in a crank-rod mechanism,Since point
A moves in a circular path,and pint B
move along a straight line,the motion of
the rod AB is neither translation nor
rotation about a fixed axis,but plane
motion,
8
例如, 曲柄连杆机构中连杆 AB的运动,
A点作圆周运动,B点作直线运动,因此，
AB 杆的运动既不是平动也不是定轴转动，
而是平面运动,
9 Please look at the animation
10 请看动画
11
2,Simplification of a plane motion
A plane motion of a rigid body can
simplified to a motion of a plane figure
in the plane itself,As indicated in the
figure,when studying a plane motion of a
rigid body,we do not need to consider its
geometrical shape and size are not needed，
and instead,considering the motion of a
plane figure is enough to determine the
velocity and acceleration of any point in the
rigid body,
12
二．平面运动的简化
刚体的平面运动可以
简化为平面图形 S在其自
身平面内的运动,即在研
究平面运动时，不需考虑
刚体的形状和尺寸，只需
研究平面图形的运动，确
定平面图形上各点的速度
和加速度,
13
§ 9–2 Plane motion can be decomposed into
translation and rotation·Equations of plane motion
1,Equations of plane motion
To determine the position of a plane figure,which represents
the plane motion of a rigid body,only the position of a line segment in
this figure is needed to be determined,
The position of a line segment AB
can be determined by the
coordinates of point A and the angle
between AB and the axis x,
Therefore,the position of the figure
S can be determined by these three
independent variables,
Hence,we have
?,,AA yx
14
§ 9-2 平面运动分解为平动和转动 ·
刚体的平面运动方程
一．平面运动方程
为了确定代表平面运动刚体的平面图形的位置，我们只需确定
平面图形内任意一条线段的位置,
任意线段 AB的位置可
用 A点的坐标和 AB与 x轴夹
角表示．因此图形 S 的位
置决定于 三个
独立的参变量．所以
?,,AA yx
15
2,Plane motion can be decomposed into translation and rotation
When point A in the figure S keeps static,the rigid body
rotates about a fixed axis
When the angle ? in the figure S keeps unchanged,the rigid
body has translatory motion,
Therefore,a plane motion can be viewed as the composition of a
translation and rotation,
?,,AA yx
Equations of plane motion )(1 tfx A ?
)(2 tfy A ?
)(3 tf??
The can be found from this equation at any instant t,and
thus the location of the plane figure S can be determined,
16
二．平面运动分解为平动和转动
当图形 Ｓ 上 Ａ 点不动时，则刚体作定轴转动
当图形 Ｓ 上 ? 角不变时，则刚体作平动,
故刚体平面运动可以看成是平动和转动的合成运动,
?,,AA yx
平面运动方程
)(1 tfx A ?
)(2 tfy A ?
)(3 tf??
对于每一瞬时 t,都可以求出对应的, 图形 S
在该瞬时的位置也就确定了。
17
Example Motion of a wheel,
The plane motion of the
wheel can be viewed as the
composition of the translation
with the vehicle body and the
rotation relative to the vehicle,
Plane motion of the wheel relative
to the static reference system (absolute motion)
Translation of the vehicle body
(moving system Ax? y? ) relative
to static system (convected motion)
Rotation of the wheel relative to
the vehicle (moving system Ax? y?) (relative motion)
18
例如 车轮的运动,
车轮的平面运动可以看成
是车轮随同车厢的平动和相对
车厢的转动的合成,
车轮对于静系的平面运动 （绝对运动）
车厢（动系 Ax? y? ) 相对静系的平动 （牵连运动）
车轮相对车厢（动系 Ax? y?）的转动 （相对运动）
19
The arbitrarily selected point A is called a pole,Thus,
Plane motion of a wheel
Translation with the pole A Rotation about the pole A
A plane motion of a rigid
body can be decomposed
into a translation with a pole
and a rotation about the
pole,
20
我们称动系上的原点 Ａ 为 基点,于是
车轮的平面运动
随基点 A的平动 绕基点 A'的转动
刚体的平面运动可以
分解为随基点的平动
和绕基点的转动,
21
Another example,plane figureＳ moves from position I to position
II in a interval ?t,
?Select point A as a pole,Translate the figure together with A to
the position A'B'',and then rotate it around A through an angle
to the final position A'B‘,
?Select point A as a pole,Translate it together with B to the
position A''B',and then rotate it around B through an angle
to the final position A'B'
Clearly,AB?? A'B'' ?? A''B', we have
21 ?? ???
1??
2??
21
21
21
2
0
1
0
,;,limlim ????????????
??
????
?? dt
d
dt
d
tt tt
22
再例如, 平面图形 Ｓ 在 ?ｔ 时间内从位置 I运动到位置 II
?以 A为基点, 随基点 A平动到 A'B''后,绕基点转 角到 A'B'
?以 B为基点, 随基点 B平动到 A''B'后,绕基点转 角到 A'B'
图中看出,AB?? A'B'' ?? A''B', 于是有 21 ?? ???
1??
2??
21
21
21
2
0
1
0
,;,limlim ????????????
??
????
?? dt
d
dt
d
tt tt
23
In conclusion,the translation in a plane motion
depends on the selection of the pole,however,the
rotation about the selected pole DOES NOT depend on
the choice of a pole,(In other words,the instant ?,? of
any rotations about ANY poles are the same） The
selection of the pole is arbitrary.(We usually select a
point with known motion as the pole,)
24
所以,平面运动随基点平动的运动规律与基
点的选择有关，而绕基点转动的规律与基点选取
无关, (即在同一瞬间，图形绕任一基点转动的
?,?都是相同的） 基点的选取是任意的 。 (通常
选取运动情况已知的点作为基点 )
25
Crank-rod mechanism
Rod AB has plane motion
Decomposition of the plane
motion
（ please look at the animation）
26
曲柄连杆机构
AB杆作平面运动
平面运动的分解
（请看动画）
27
§ 9-3 velocity of any point in a plane figure
Employ the theorem of velocity composition,
rea vvv ??the velocity at point B can be expressed as
BAAB vvv ??
,vv;vv;vv BArAeBa ???
The velocity of point A in figure S
and the rotational velocity ? of the
figure are given,find,
Select A as the pole,and fix the
moving reference system to point A,
The motion of the moving system
is translation,Consider the moving point B,its motion can be
viewed as the composition of the translation,the convected motion,
and the rotation,the relative motion
point to the rotation direction,
Av
Bv
1,pole-based method (composition method)
,d i r ect i o n,m a g n i t u d e ABAB ???
28
§ 9-3 平面图形内各点的速度
根据速度合成定理
,rea vvv ??
则Ｂ点速度为,
BAAB vvv ??
一．基点法（合成法）
取 B为动点,则 B点的运动可视为牵连运动为平动和相对运动
为圆周运动的合成
,,ABAB ????? 方向大小 ?,vv;vv;vv BArAeBa
已知：图形 S内一点 A的速度,
图形角速度 ? 求,
指向与 ? 转向一致,
取 A为基点,将动系固结于 A点,
动系作平动。
Av
Bv
29
Since point A and B are arbitrarily selected,the equation
gives the relationship between the velocities of any two point in the
figure,Noting that,projecting this equation to AB,gives
BAAB vvv ??
ABv BA ?
? ? ? ? ABAABB vv ? ——theorem of velocity projection I,e.,the velocity projections of any two point in a figure on the
line linking these two points are identical,This method for
find the velocity of a point is called velocity projection
method,
That is,the velocity of any point in the figure is obtained as
the geometric sum of the velocity of the pole and the relative
rotational velocity with respect to the pole,Such a method of
finding the velocity is called pole-based method,or composition
method,which is a basic method to find the velocity of a point in a
figure,
2,Velocity projection method
30
由于 A,B点是任意的，因此 表示了图形上任
意两点速度间的关系．由于恒有,因此将上式在 AB
上投影，有
BAAB vvv ??
ABv BA ?
? ? ? ? ABAABB vv ? — 速度投影定理
即 平面图形上任意两点的速度在该两点连线上的投影彼此相
等, 这种求解速度的方法称为 速度投影法,
即 平面图形上任一点的速度等于基点的速度与该点随图形绕
基点转动的速度的矢量和, 这种求解速度的方法称为 基点法,
也称为 合成法,它是求解平面图形内一点速度的基本方法,
二．速度投影法
31
3,Instantaneous velocity center method
(1) Background
If a point whose velocity is zero is selected as the pole,the process
of finding the velocity of any point will be greatly simplified,Hence,
it is natural to ask if such a point exists in any instant,If it does exist,
how to find such a point?
T h e r e f o r e
, t oo p p o s i t e a n d,d i r e c t i o n t h e,AAPA vPAvAPv ???? ?
0?Pv
(2)Concept of Instantaneous velocity center
Consider a plane figure S,The velocity of point A
is,and the angular velocity of the figure is ?,
Take a line AL along the direction of,and then
turn it 90o in the direction of ? to AL‘ and find the
point P in AL‘ by letting,we have
?/AvAP ?
Av
Av
PAAP vvv ??
32
三．瞬时速度中心法（速度瞬心法）
1,问题的提出
若选取速度为零的点作为基点，求解速度问题的计算会大大
简化．于是，自然会提出，在某一瞬时图形是否有一点速度等
于零？如果存在的话，该点如何确定？
所以反向恰与方向,,,AAPA vPAvAPv ???? ?
0?Pv
２．速度瞬心的概念
平面图形 S，某瞬时其上一点 A速度,
图形角速度 ?，沿 方向取半直线 AL,然后
顺 ? 的转向转 90o至 AL'的位置,在 AL'上取长
度 则,?/
AvAP ?
Av
Av
PAAP vvv ??
33
At any instant,there must exist a sole point whose velocity
is zero,which is called the instantaneous velocity center of
this figure at this instant,
3,Methods to determine the Instantaneous velocity center
① When the velocity of a point and the angular
velocity ? of the figure are known,the instantaneous
velocity center (point P) can be determined,
and point P is in the direction of
the line formed by rotating the through 90o in the
direction of ? around point A,
,,AA vAPvAP ?? ?
Av
Av
② When a plane figure rolls along a fixed surface
without slipping,the contact point P between the
figure and the fixed surface will be the
instantaneous velocity center,
34
即在某一瞬时必唯一存在一点速度等于零，该点称为平
面图形在该瞬时的瞬时速度中心，简称速度瞬心,
３．几种确定速度瞬心位置的方法
① 已知图形上一点的速度 和图形角速度 ?,
可以确定速度瞬心的位置,（ P点）
且 Ｐ 在 顺 ?转向绕 A点
转 90o的 方向一侧,
,,AA vAPvAP ?? ?
Av
Av
② 已知一平面图形在固定面上作无滑动的滚
动,则图形与固定面的接触点 P为速度瞬
心,
35
AB
vv
vva
BA
BA
?
?? d i r e c t i o n,s a m e
t h ep o i n t t o a n d )(
AB
vv
vvb
BA
BA
?
??,d i r e c t i o n s
o p p o s i t e h a v e a n d )(
④ The magnitudes of the velocities
of two points A and B at any instant are
given,and,
BA vv,
ABvABv BA ??,
(b) (a)
③ When the directions of the velocities
at two points A and B in a figure are known,and
,draw lines from A and B
perpendicular to respectively,and the
cross point P of these two lines will be the
instantaneous velocity center,
BA vv,
BA vv t op a r al l e ln o t is
BA vv a n d
36
AB
vvvva BA
BA
???,)( 同向与
AB
vvvvb BA
BA
???,)( 反向与
④ 已知某瞬时图形上 A,B两点速度
大小,且
BA vv,
ABvABv BA ??,
(b) (a)
③ 已知某瞬间平面图形上 A,B两点速度
的方向，且 。
过 A,B两点分别作速度 的垂线,交点
P即为该瞬间的速度瞬心。
BA vv,
BA vv 不平行
BA vv,
37
In addition,if vA＝ vB in case ④,
the motion is instantaneous translation
too,
⑤ The velocities of two points A and B point to the same direction
at any instant,but they are not perpendicular to line AB,
In this case,the instantaneous velocity center is indefinitely far
away,and the angular velocity ? =0,i,e,all point in the figure have
the same velocity at this instant of time,Such a motion is called
instantaneous translation,(but their accelerations are not identical),
38
另：对 ?种 (a)的情况，若 vA＝ vB,
则是瞬时平动,
⑤ 已知某瞬时图形上 A,B两点的速度方向相同，且不与 AB连线
垂直,
此时,图形的瞬心在无穷远处,图形的角速度 ? =0,图形上
各点速度相等,这种情况称为 瞬时平动, (此时各点的加速度不
相等 )
39
Example,At the instant of time shown in the figure,the rod
BC in the crank-rod mechanism has instantaneous translation,
The angular velocity of the rod BC,The velocities of all
points in BC are identical,but their accelerations are not,
If ? is uniform, then
)(2 ???? ?ABaa nBB
But the direction of is along AC,
Instantaneous translation is different from translation,
ca cB aa ?
0?BC?
40
例如, 曲柄连杆机构在图示位置时，连杆 BC作瞬时平动,
此时连杆 BC的图形角速度,
BC杆上各点的速度都相等, 但各点的加速度并不相等,
设匀 ?，则
)(2 ???? ?ABaa nBB
而 的方向沿 AC的,瞬时平动与平动不同
ca cB aa ?
0?BC?
41
４, Instantaneous velocity center method
The method for finding the velocity of a point in a figure using
instantaneous velocity center is called instantaneous velocity center
method,
At any instant of time,the motion of a figure can be viewed as the
rotation around the instantaneous velocity center,
If P is the instantaneous velocity center,the velocity of any point A
and its direction ?AP,pointing the same direction with ?, ??? APv
A
５, Cautions
?The position of the the instantaneous velocity center is not fixed
at all time,it changes instantly with time,and exists uniquely at any
instant of time,
?Only the velocity at the instantaneous velocity center is zero,
but its acceleration is not certainly zero,This differs from the rotation
about a fixed axis,
?When a rigid body is in instantaneous translation,although the
velocities at all points in it are identical,but their accelerations are not
necessarily identical,This is different from the translatory motion,
42
４, 速度瞬心法
利用速度瞬心求解平面图形上点的速度的方法,称为速度瞬心法,
平面图形在任一瞬时的运动可以视为绕速度瞬心的瞬时转
动，速度瞬心又称为平面图形的瞬时转动中心。
若 P点为速度瞬心，则任意一点 A的速度
方向 ?AP，指向与 ? 一致。 ??? APv A
５, 注意的问题
?速度瞬心在平面图形上的位置不是固定的，而是随时间不
断变化的。在任一瞬时是唯一存在的。
?速度瞬心处的速度为零,加速度不一定为零。不同于定轴转动
?刚体作瞬时平动时，虽然各点的速度相同，但各点的加速
度是不一定相同的。不同于刚体作平动。
43
Solution,In this mechanism,OA rotates
about a fixed axis,AB has plane motion,
and piston B has translation,
?Pole-based method（ composition method）
Consider AB and take A as the pole, Its direction is
shown in the figures,?lv A ?
???
???
??
?
????
????
???
?
llABv
llvv
ll
vv
BAAB
ABA
AB
//
45tgtg
)(245c o s/
c o s/
?
?
（ ）
[Example 1] In a crank-rod mechanism,
OA=AB=l,and the crank OA rotates with
a uniform ?,Find,when ? =45o,the
velocity of piston B and the angular
velocity of rod AB,
,BAAB vvv ??Since Draw the parallelogram of
velocities at point B,as shown in the figure,
44
解：机构中,OA作定轴转动,AB作平面运
动,滑块 B作平动。 ?
基点法（合成法）
研究 AB，以 A为基点，且 方向如图示。
,?lv A ?
???
???
??
?
????
????
???
?
llABv
llvv
ll
vv
BAAB
ABA
AB
//
45tgtg
)(245c o s/
c o s/
?
?
（ ）
[例 1] 已知：曲柄连杆机构 OA=AB=l，
取柄 OA以匀 ? 转动。 求：当 ? =45o时,
滑块 B的速度及 AB杆的角速度,
根据,
BAAB vvv ??
在 Ｂ 点做 速度平行四边形，如图示。
45
)(2
//
,
????
????
??
??
???
?
lBPv
llAPv
lAPlv
ABB
AAB
A? （ ）
Try to compare these three methods,
? ? ? ? ABAABB vv ?
Employing theorem of velocity projection
,We have ?c o s
BA vv ?
)(245c o s/ c o s/ ????? ??? llvv AB ?
We can not find, AB?
?Velocity projection method,
Consider AB.,and its direction?OA,
is along BO,
?lv A ?
Bv
?Instantaneous velocity center method
Consider AB,Since the directions of
are known,we can determine the instantaneous
center of velocity at point P,
BA vv an d
46
)(2
//
,
????
????
??
??
???
?
lBPv
llAPv
lAPlv
ABB
AAB
A? （ ）
试比较上述三种方法的特点。
? ? ? ? ABAABB vv ?
根据速度投影定理
?c o sBA vv ?
)(245c o s/
c o s/
???
??
??
?
ll
vv AB
?不能求出 AB?
?速度投影法 研究 AB,,
方向 ?OA,方向沿 BO直线
?lv A ?
Bv
?速度瞬心法
研究 AB，已知 的方向，因此
可确定出 P点为速度瞬心
BA vv,
47
§ 9-4 Acceleration of a point in a plane
figure·Instantaneous center of acceleration
Take A as the pole,and fix the moving
reference system onto A,Take B as the
moving point,the motion of point B can be
decomposed into a relative motion (circular
motion) and a convected motion
(translation) with the pole,
nBABABArAeBa aaaaaaaa ????? ? ; ;
Hence,employing the theorem of acceleration composition
,the following formula can be obtained,
rea aaa ??
n
BABAAB aaaa ???
?
1,Pole-based method (composition method)
At an instant of time,the acceleration of a
point A in a figure S,? and ? are given,Find
the acceleration of any point B in the figure,
Aa
48
§ 9-4 平面图形内各点的加速度
加速度瞬心的概念
取 A为基点，将平动坐标系固结于 A点
取 B动点，则 B点的运动分解为相对运动
为圆周运动和牵连运动为平动,
nBABABArAeBa aaaaaaaa ????? ? ; ;
于是,由牵连平动时加速度合成定理 可得如下公式,
rea aaa ??
n
BABAAB aaaa ???
?
一, 基点法 (合成法 ) 已知：图形 S 内一点 A 的加速度 和图形
的 ?,?（某一瞬时）。
求,该瞬时图形上任一点 B的加速度。
Aa
49
where,direction?AB,pointing to the same direction
with ? ; which is along AB and points to A,
?? ?? ABa BA
2??? ABa nBA
The acceleration of any point in a figure equals to the geometric
sum of the acceleration of the pole,tangential and normal
accelerations of this point rotating about the pole together with the
figure,This method to find the acceleration of a point is call pole-
based method,or composition method,which is the basic method
to finding the acceleration of a point in a plane figure,
The formula given above is an equation in terms of plane vectors,
Therefore,we can solve two unknowns from it provided that the
other variables are given,Since the directions of are
always known,to solve the unknowns,only four other variables
are needed,
nBABA aa,?
n
BABAAB aaaa ???
?
50
其中：,方向 ?AB，指向与 ? 一致；
,方向沿 AB，指向 A点。
?? ?? ABa BA
2??? ABa nBA
即 平面图形内任一点的加速度等于基点的加速度与该点随图形绕
基点转动的切向加速度和法向加速度的矢量和 。这种求解加速度
的方法称为 基点法,也称为 合成法 。是求解平面图形内一点加速
度的基本方法。
上述公式是一平面矢量方程。需知其中六个要素，方能求
出其余两个。由于 方位总是已知，所以在使用该公式
中，只要再知道四个要素，即可解出问题的待求量。
nBABA aa,?
n
BABAAB aaaa ???
?
51
n
BABA aa,
?
2,Instantaneous center of acceleration
Due to the dependent of on the point B,we can always
find a point Q in the figure, at which the relative acceleration
has the same magnitude but opposite direction with the acceleration
in the pole so that the absolute acceleration, Such a
point Q is called the instantaneous center of acceleration,
QAa
Aa 0?
Qa
52
n
BABA aa,
?
二．加速度瞬心,
由于 的大小和方向随 B点的不同而不同,所以总可以
在图形内找到一点 Q，在此瞬时，相对加速度 大小恰与基点
A的加速度 等值反向，其绝对加速度,Q点就称为图形
在该瞬时的 加速度瞬心,
QAa
Aa 0?Qa
53
[Note] ?In general,the instantaneous center of acceleration does
not coincide with the instantaneous center of velocity,
? In general, there is no similar relationship between the
accelerations at two points given in the theorem of velocity
projection,That is to say,the following relationship does not hold in
general,
? ? ? ? ABBABA aa ?
Only in the special case where ? =0,the figure is in instantaneous
translation,holds,
In other words,if the angular velocity of a plane figure is zero at an
instant of time,the acceleration projections of any two point in a
figure on the line linking these two points are identical,
? ? ? ? ABBABA aa ?
54
[注 ] ?一般情况下,加速度瞬心与速度瞬心不是同一个点,
?一般情况下，对于加速度没有类似于速度投影定理的关
系式, 即一般情况下,图形上任意两点 A,B的加速度
? ? ? ? ABBABA aa ?
若某瞬时图形 ? =0,即瞬时平动,则有
即 若平面图形在运动过程中某瞬时的角速度等于零，则该瞬时
图形上任意两点的加速度在这两点连线上的投影相等,
? ? ? ? ABBABA aa ?
55
? ?Since it is not easy to find the instantaneous center of
acceleration and there does not exists a relationship similar to the
theorem of velocity projection,instantaneous center of acceleration is
not widely used to find accelerations of a given point,Instead,the pole
method is often employed,
Analysis,
magnitude？ √ Ｒ ? Ｒ ? 2
direction？ √ √ √
Therefore,? and ? should be solved first,
n
POPOOP aaaa ???
?
Rv O /?? ? （ ）
[Example 1] A wheel with radius R rolls on a plane surface without
slipping,The velocity and acceleration at the its center O are
given,Find the acceleration at the contact point P,O
v Oa
Solution,The wheel has plane motion and P is the
instantaneous velocity center,
56
? ?由于加速度瞬心的位置不象速度瞬心那样容易确定，且一
般
情况下又不存在类似于速度投影定理的关系式，故常采用基
点法求图形上各点的加速度或图形角加速度,
分析,
大小？ √ Ｒ ? Ｒ ? 2
方向？ √ √ √
故应先求出 ?? ?,
n
POPOOP aaaa ???
?
Rv O /?? ? （ ）
[例 1] 半径为 R的车轮沿直线作纯滚动,已知轮心 O点的速度
及加速度,求车轮与轨道接触点 P的加速度,O
v
Oa
解：轮 O作平面运动,P为速度瞬心,
57
Since the equation given above is valid at any instant,and point O
moves along a straight line,we obtain
R
a
dt
dv
Rdt
d OO ??? 1??
（ ）
It is evident that the acceleration at the instantaneous center of
velocity is not zero,which means that point P is not the instantaneous
center of acceleration,In this case,the acceleration of at the
instantaneous center of velocity P points to the center of the wheel,
Taking O as the pole,yields
As shown in the vector diagram of the
accelerations,（ and are equal in magnitude but
opposite in direction）
i,e,
nPOPOOP aaaa ??? ?
R
v
R
vRRaaRa OOn
POOPO
2
22 )(,??????? ???
n
POP aa ? Oa
?
POa
)(/2 ?? Rva OP
58
由于此式在任何瞬时都成立，且 O点作直线运动，故而
R
a
dt
dv
Rdt
d OO ??? 1?? （ ）
由此看出，速度瞬心 P的加速度并不等于零，即它不是加速度
瞬心．当车轮沿固定的直线轨道作纯滚动时，其速度瞬心 P的加
速度指向轮心,
以 O为基点，有 其中,
做出加速度矢量图，由图中看出,
（ 与 等值反向）
即
nPOPOOP aaaa ??? ?
R
v
R
vRRaaRa OOn
POOPO
2
22 )(,??????? ???
n
POP aa ? Oa ?POa
)(/2 ?? Rva OP
59
Solution,
(a) AB has translation,),(,n
B
n
ABABABA aaaaaavv ?????
??
BOAO
BOaAOa
BOvAOv
BA
BA
21
2211
2211
a n d;/,/;/,/ a d d t i o n,I n
?
??
??
??
??
???
.; 2121 ???? ???
[Example 2] At the instant shown in the figure,O1A=O2B and
O1A?/O2B,Are ?1 and ? 2 or ?1 and ?2 identical in case (a) and (b),
(a) (b)
60
解,(a) AB作平动,),(,n
B
n
ABABABA aaaaaavv ?????
??
BOAO
BOaAOa
BOvAOv
BA
BA
21
2211
2211
;/,/;/,/
?
??
??
而
又
??
??
???,;
2121 ???? ???
[例 2] 已知 O1A=O2B,图示瞬时 O1A?/O2B
试问 (a),(b)两种情况下 ?1和 ? 2,?1和 ?2是否相等？
(a) (b)
61
(b) AB has plane motion,and it is in instantaneous translation at
this instant of time,Therefore,
BAAB vv ??,0?
21221121,/,/,???? ????? BOvAOvBOAO BA?
? ? ? ? ? ? ? ? ? ? ? ? ABnBABBABnAABAABBABA aaaaaa ???? ??,即
???????? c o ss i nc o ss i n 2222221111 ??????? BOBOAOAO
,a t c l e a r l y t hs e e n bec a n I t, i, e, c t g2 212112 BA aa ????? ??????
62
(b) AB作平面运动,图示瞬时作瞬时平动,此时 BAAB vv ??,0?
21221121,/,/,???? ????? BOvAOvBOAO BA?
? ? ? ? ? ? ? ? ? ? ? ? ABnBABBABnAABAABBABA aaaaaa ???? ??,即
???????? c o ss i nc o ss i n 2222221111 ??????? BOBOAOAO
BA aaAB ????? 作瞬时平动时并由此看出即,c t g2 212112 ??????
63
[Example 3] In a crank-rod mechanism,the radius of
the wheel is R=15cm,n=60 rpm
Find ?Ｂ and ?Ｂ of the wheel when ? =60o and OA?AB,
Look for the animation on the next page
?
64
[例 3] 曲柄滚轮机构 滚子半径 R=15cm,n=60 rpm
求：当 ? =60o时 (OA?AB),滚轮的 ?Ｂ, ?Ｂ,
翻页请看动画
?
65
Lo
ok
at
th
e a
nim
ati
on
66
请
看
动
画
67
?
Solution,OA rotates about a fixed axis,
AB and wheel B have plane motions,
Consider rod AB,
r a d / s 32153/30/ 1 ??? ????? APv AAB
（ ）
c m / s 30215
r a d / s 230/6030/
???
????
?????
???
OAv
n
A
P１ is the instantaneous velocity center
of AB
)(cm / s 3203215321 ??????? ??? ABB BPv
Analysis,To find ?Ｂ,?Ｂ of the wheel,vB and aB
should be worked out first,
P2
P1
vB P2 is the instantaneous velocity center
of the wheel
68
?
解, OA定轴转动,AB杆和轮 B作平面运动
研究 AB,
r a d / s 32153/30/ 1 ??? ????? APv AAB
（ ）
c m / s 30215
r a d / s 230/6030/
???
????
?????
???
OAv
n
A
P１ 为其速度瞬心
)(cm / s 3203215321 ??????? ??? ABB BPv
分析, 要想求出滚轮的 ?Ｂ,?Ｂ 先要求出 vB,aB
P2
P1
vB P2为轮速度瞬心
69
Take A as the pole,2222 cm / s60)2(15 ??? ????? OAa
A
Pointing to O
n
BABAAB aaaa ???
?
) a l o n g,
3
320)
3
2(153( 222 BAABa
AB
n
BA ?
?? ??????
Magnitude？ √？ √
Direction √ √ √ √
Draw the vector diagram for accelerations,
Projecting the above equation onto BA,
gives
n
BAB aa ??? 0030c o s ?
)(c m / s5.1313402 3/3 32030c o s/ 222 ????? ???nBAB aa
r a d / s25.715/320/ 2 ??? ?? BPv BB
22 r a d / s77.815/5.131/ ??? BPa BB?
） (
） (
Consider wheel B,its instantaneous velocity center is P2
70
取 A为基点,2222 cm / s60)2(15 ??? ????? OAa A 指向 O点
n
BABAAB aaaa ???
? ),
3
320)
3
2(153( 222 BAABa
AB
n
BA 沿?
?? ??????
大小？ √？ √
方向 √ √ √ √
作加速度矢量图，将上式向 BA线上投影
n
BAB aa ??? 0030c o s ?
)(c m / s5.1313402 3/3 32030c o s/ 222 ????? ???nBAB aa
r a d / s25.715/320/ 2 ??? ?? BPv BB
22 r a d / s77.815/5.131/ ??? BPa BB?
） (
） (
研究轮 B,P2为其速度瞬心
71
Lesson of problem solving for Chapter 9
1,Concepts and content
(1) Definition of plane motion
The distance between any point in a rigid body and a fixed
plane always keeps unchanged during its motion,
(2) Simplification of plane motion
The motion of a plane figure in the rigid body parallel with the
fixed plane S in its own plane can be used to represent the plane
motion of the entire rigid body,
(3) Decomposition of plane motion,
It can be decomposed into,
(4) Pole
Theoretically speaking,any a point in the figure can be
selected as the pole,We usually chose a point with known motion as
the pole,
Translation together with the pole（ it depends on the choice of pole）
Rotation around the pole（ it does not depend on the selection of pole）
72
第九章 刚体平面运动习题课
一．概念与内容
1,刚体平面运动的定义
刚体运动时，其上任一点到某固定平面的距离保持不变,
2,刚体平面运动的简化
可以用刚体上一个与固定平面平行的平面图形 S在自身平
面内的运动代替刚体的整体运动,
3,刚体平面运动的分解
分解为
4,基点
可以选择平面图形内任意一点,通常是运动状态已知的点,
随基点的平动（平动规律与基点的选择有关）
绕基点的转动（转动规律与基点的选择无关）
73
(5) Instantaneous center of velocity
?At any instant of time,there exists a sole point in the figure or its
extension whose velocity is zero,
?The location of the instantaneous center of velocity changes with
time,
?At any instant,the motion of a plane figure can be viewed as the
instant rotation around the instantaneous center of velocity,However,
this instant rotation is different from the rotation about a fixed axis,
? ? =0,instantaneous center of velocity is indefinitely far away,
In this case,all points have the same velocity and the rigid body is in
instantaneous translation,
Note that instantaneous translation is different from translation,
74
5,瞬心（速度瞬心）
?任一瞬时,平面图形或扩大部分都唯一存在一个速度为零的点
?瞬心位置随时间改变,
?每一瞬时平面图形的运动可视为绕该瞬时瞬心的转动．这
种瞬时绕瞬心的转动与定轴转动不同,
? ? =0,瞬心位于无穷远处,各点速度相同,刚体作瞬时平动,
瞬时平动与平动不同,
75
(6) Rotation and translation are two special cases of plane motion,
(7) Methods to find velocity of any point in a figure
?pole method,
?velocity projection method,
? instantaneous velocity center method,
In these methods,the pole methods is the basic method,and the
instantaneous velocity center method is a special case of the pole
method,
p o l e t h eis,Avvv BAAB ??
? ? ? ? ABAABB vv ?
v e l o c i t yofc e n t e r o u si n s t a n t a n e
t h esi,,PBPvBPv BB ??? ?
76
6,刚体定轴转动和平面平动是刚体平面运动的特例,
7,求平面图形上任一点速度的方法
?基点法,
?速度投影法,
?速度瞬心法,
其中，基点法是最基本的公式，瞬心法是基点法的特
例,
为基点Avvv BAAB,??
? ? ? ? ABAABB vv ?
为瞬心一致与 PBPvBPv BB,,,?? ???
77
(8) Methods to find acceleration of a point in a figure
pole method：,A is the pole,This is the most
useful method,In addition,when ? =0,i.e,instantaneous translation,
we can also employ, It is the special case of the pole
method when ? =0,
n
BABAAB aaaa ???
?
? ? ? ? ABAABB aa ?
(9) Conditions for the applications of the methods used in plane
motion and composition of motions
? methods used in plane motion are applied to find the relationship
between the velocities/accelerations of any two points or determine
the relationship among the velocity,acceleration of a point and the
angular velocity and angular acceleration of ONE rigid body having
plane motion,
?The methods used in the composition of motions are usually
applied to determine the transfer of motion at the contact point where
TWO rigid bodies are in contact and relative slipping exists,
78
8,求平面图形上一点加速度的方法
基点法：, A为基点,是最常用的方法
此外，当 ? =0,瞬时平动时也可采用方法
它是基点法在 ? =0时的特例。
n
BABAAB aaaa ???
?
? ? ? ? ABAABB aa ?
9,平面运动方法与合成运动方法的应用条件
?平面运动方法用于研究 一个平面运动刚体 上任意两点的速
度、加速度之间的关系及任意一点的速度、加速度与图形
角速度、角加速度之间的关系,
?合成运动方法常用来确定 两个相接触的物体 在接触点处有
相对滑动时的运动关系的传递,
79
2,Steps in problem solving and some key points
(1) According to the information given in the problem and the
definitions of the motions of a rigid body,determine the type of
motions of the rigid bodies appeared in the problem,Note that only
one rigid body should be considered each time,
(2) For a rigid body with plane motion,,properly choose a suitable
method to find the velocity (or angular velocity) of a point based on
the known and unknown conditions,To find accelerations,pole
method is recommended,
(3) Do the thorough analysis for velocities and accelerations,and finally
solve the unknowns,
(Pole method,properly select a pole and then draw the parallelogram
for velocities and the vector diagram for accelerations；
Velocity projection method,its limitation is that ? can not be found;
Instantaneous velocity center method,find the Instantaneous velocity
center is the key step.)
80
二．解题步骤和要点
1,根据题意和刚体各种运动的定义，判断机构中各刚体的运动
形式．注意每一次的研究对象只是一个刚体,
2,对作平面运动的刚体，根据已知条件和待求量，选择求解速
度 (图形角速度 )问题的方法,用基点法求加速度 (图形角加速
度 )
3,作速度分析和加速度分析，求出待求量,
(基点法, 恰当选取基点，作速度平行四边形，加速度矢量图；
速度投影法, 不能求出图形 ? ;
速度瞬心法：确定瞬心的位置是关键．）
81
[Example 1] Consider a mechanism shown in the figure,
OA=0.15m,n=300 rpm,AB=0.76m,BC=BD=0.53m,At
the instant shown in the figure,rod AB in a horizontal plane,
Find, and at this instant of time,
ABBD ??
Dv
Look for the animation
on the next page
82
[例 1] 曲柄肘杆压床机构
已知,OA=0.15m,n=300 rpm,AB=0.76m,
BC=BD=0.53m,图示位置时,AB水平
求该位置时的, 及
ABBD ?? Dv
翻页请看动画
83
Lo
ok
at
th
e a
nim
ati
on
84
请
看
动
画
85
Solution,OA and BC rotate around
their fixed axes,AB,BD have plane
motions,
Consider AB,P１ is its instantaneous
velocity center,
r ad / s10303 0 030 ???? ??? n
m / s 5.11015.0 ??? ????? OAv A
（ ）
r a d / s 16.7376.0 25.160s i n5.1
1
?? ????? ??? ?ABAPv AAB
m / s 72.216.75.076.016.760co s1 ???????? ?ABBPv ABB ?
Next,consider BD,P2is its instantaneous velocity center,
Note ?BDP2 is an equilateral triangleDP2=BP2=BD
r a d / s 13.553.0 73.2
2
???? BPv BBD?
)(m / s 72.213.553.02 ?????? BDD DPv ?
（ ）
86
解,OA,BC作定轴转动,
AB,BD均作平面运动
根据题意,
研究 AB,P１ 为其速度瞬心
r ad / s10303 0 030 ???? ??? n
m / s 5.11015.0 ??? ????? OAv A
（ ） r a d / s 16.7376.0 25.160s i n5.1
1
?? ????? ??? ?ABAPv AAB
m / s 72.216.75.076.016.760co s1 ???????? ?ABBPv ABB ?
研究 BD,P2为其速度瞬心,?BDP2为等边三角形 DP2=BP2=BD
r a d / s 13.553.0 73.2
2
???? BPv BBD?
)(m / s 72.213.553.02 ?????? BDD DPv ?
（ ）
87
[Example 2] the mechanism of planet gears L
oo
k a
t th
e a
nim
ati
on
88
[例 2] 行星齿轮机构
请
看
动
画
89
Solution,OA rotate around O; Wheel A has
plane motion and P is the instantaneous
velocity center,
,)(2211 ooM rRr rRrPMv ??? ???????
ooA r
rRrrRv ???? ?????? )( ） (
f i g u r e, i n t h es h o w n a r e d i r e c t i o n s
,)(2222 ooM rR
r
rR
rPMv ??? ??
?
????
[Example 2] a mechanism of planet gears
R,r and ?o are given and wheel A rolls
without slipping,Find,
21 a n d MM vv
90
解,OA定轴转动 ; 轮 A作平面运动,瞬心 P点
,)(2211 ooM rRr rRrPMv ??? ???????
ooA r
rRrrRv ???? ?????? )( ） (
[例 2] 行星齿轮机构
已知, R,r,?o 轮 A作纯滚动，求
21,MM vv
方向均如图示,)(2222 ooM rRr rRrPMv ??? ???????
91
[Example 3] In the plane mechanism shown below,the angle in the
wedge M is ? =30o,v=12cm/s ; r = 4cm for the wheel,No slipping
is assumed between the wheel and the wedge,Find the angular
velocity ? of the wheel,velocity at point O and B,Loo
k a
t th
e a
nim
ati
on
92
[例 3] 平面机构中,楔块 M,? =30o,v=12cm/s ; 盘, r = 4cm,与 楔
块间无滑动．求圆盘的 ?及轴 O的速度和 B点速度,
请
看
动
画
93
Solution,Shaft O,rod OC and wedge M are
in translatory motions,The wheel has plane
motion and P is the instantaneous velocity
center,
,c m / s 12?? vv A
r a d / s 3230c o s4/12c o s/12/ ???? ??? rPAv A
)(m / s 343230s i n4s i n ???????? ???? rPOv o
m722142242120c o s2 2222 ???????????? ?OBPOOBPOPB
) ( m / s 3.182143272 PBPBv B ??????? ?
） (
94
解,轴 O,杆 OC,楔块 M均作平动,
圆盘作平面运动,P为速度瞬心
,c m / s 12?? vv A
r a d / s 3230c o s4/12c o s/12/ ???? ??? rPAv A
)(m / s 343230s i n4s i n ???????? ???? rPOv o
m722142242120c o s2 2222 ???????????? ?OBPOOBPOPB
) ( m / s 3.182143272 PBPBv B ??????? ?
） (
95
? It can be seen from the comparison between [Example 2] and
[Example3] that the contact point between the wheel and the
surface is not always the instantaneous velocity center,Only
when the surface is fixed,the point on the wheel in contact with
this surface is the instantaneous velocity center,
? Every rigid body has its own instantaneous velocity center
and angular velocity that may be either with the body or in its
extension,but not in other rigid bodies,For instance,in
[Example 1],the instantaneous velocity center of rod AB lies in
point P1,and that of rod BD in point P2,Note P1 is not a point
on rod CB,
96
? 比较 [例 2]和 [例 3]可以看出,不能认为圆轮只滚不滑时,接
触点就是瞬心,只有在接触面是固定面时,圆轮上接触点
才是速度瞬心
? 每个作平面运动的刚体在每一瞬时都有自己的速度瞬心和
角速度,并且瞬心在刚体或其扩大部分上,不能认为瞬心在
其他刚体上, 例如,[例 1] 中 AB的瞬心在 P1点,BD的瞬心在 P2
点,而且 P1也不是 CB杆上的点
97
[Example 4] A slotted bar-sliding block mechanism
Look at the animation
98
[例 4] 导槽滑块机构 请看动画
99
[Example 4] In the mechanism show in the figure,OA= r,the
crank rotates with uniform ?,and a block attached at the center C
of rod AB can slid along the slotted bar O1D,AB=l,At the moment
shown in the figure,point O,A and O1 in the same horizontal line,
and OA?AB,?AO1C= ?=30,
Find,the angular velocity of O1D,
Solution,OA and O1D rotate about their
fixed axes and rod AB is in plane motion,
? Consider AB:,AB is in
instantaneous translation,therefore,
?? rvvrv AcB ??? ;
?rv A ?
?Composition of motions method
Fine the velocity of the contact point in O1D with block C,
moving point is point C on rod AB (or block C ),
moving system is rod O1D,
static system is supports,
100
[例 4] 导槽滑块机构
已知, 曲柄 OA= r,匀角速度 ? 转动,连杆 AB的中点 C处连接一
滑块 C可沿导槽 O1D滑动,AB=l,图示瞬时 O,A,O1三点
在同一水平线上,OA?AB,?AO1C= ?=30。
求,该瞬时 O1D的角速度,
解, OA,O1D均作定轴转动,
AB作平面运动
? 研究 AB:,图示位置,
作 瞬时平动,所以
?? rvvrv AcB ??? ;
?rv A ?
?用合成运动方法
求 O1D杆上与滑块 C 接触的点的速度
动点, AB杆上 C (或滑块 C ),动系, O1D杆,静系, 机架
101
Absolute motion,curvilinear motion
,direction ?
Relative motion,rectilinear motion
,direction// O1D
Convected motion,rotation,,
direction? O1D
?rvv ca ??
?rv
?ev
According to,draw the
parallelogram of velocities rea vvv ??
??? rrvv Ce 2 330c o sc o s ????? ?
?
?
?
?
?
l
r
l
r
CO
v
COv
e
DO
DOe
2
3
s i n/
2
23
a n d
1
1
1
1
????
???
This is a comprehensive problem,which can be solve by applying
both composition of motions and plane motion,Pay attention to the
solving method for such kind of problems,See next example,
102
绝对运动,曲线运动,方向 ?
相对运动,直线运动，,方向 // O1D
牵连运动,定轴转动，,方向 ? O1D
?rvv ca ??
?rv
?ev
根据, 作速度平行四边形
rea vvv ??
??? rrvv Ce 2 330c o sc o s ????? ?
?
?
?
?
?
l
r
l
r
CO
v
COv
e
DO
DOe
2
3
s i n/
2
23
1
1
1
1
????
???又
） (
这是一个需要联合应用点的合成运动和刚体平面运动理论
求解的综合性问题,注意这类题的解法，再看下例,
103
[Example 5] Planar mechanism
Please look at the animation
104
[例 5] 平面机构
请
看
动
画
105
[Example 5] In the planar mechanism
shown in the figure,point O is in the
center of rod AB and ? =60o,BC?AB,In
addition,point O and C in the same
horizontal line,and AB=20cm,vA=16cm/s,
Find the angular velocities of rod AB and
BC respectively and the velocity of block C,
Solution,Wheel A,rod AB and rod BC have plane motions,Sleeve O
rotates about a fixed axis,and block C is in translation,
Take point O in the sleeve as the moving point,the moving system is
attached on AB,and static system on the fixed support,
rea vvv ??
,due to along AB,The direction of is
along AB and opposite with, In such a way,the direction of the
velocity at the point on AB in contact with sleeve O is determined,
ra vv,0? e
v
rv
Consider AB,P1 is the instantaneous center of velocity,
106
[例 5] 平面机构
图示瞬时,O点在 AB中点,? =60o,
BC?AB,已知 O,C在同一水平线上,
AB=20cm,vA=16cm/s,
试求 该瞬时 AB杆,BC杆的角速度
及滑块 C的速度,
解, 轮 A,杆 AB,杆 BC均作平面运动,套筒 O作定轴转动,滑块 C平动,
取套筒上 O点为 动点,动系 固结于 AB杆 ; 静系 固结于机架,
rea vvv ??
,由于 沿 AB,
所以 方向沿 AB并且与 反向。 从而确定了 AB杆上与 O点接
触点的速度方向。
ra vv,0?
ev rv
研究 AB,P1为速度瞬心
107
?BC and vC are simply found using
instantaneous velocity center method,
c m / s31631660
c m / s321622
60c o s
?????
?????
?
?
tgvv
v
v
v
BCB
B
B
C
cm3103 ????? OBBCBCv BCCB ??
Consider BC,and take B as the pole,
According to
draw the parallelogram of velocities CBBC
vvv ??
c m / s 1611 ?????? AABABB vAPBPv ??
r a d / s 35460s i n/10 16s i n/16
1
???? ??? OAAPv AAB
） (
（ ）
r a d / s 6.1
310
316 ????
BC
v CB
BC?
108 也可以用瞬心法求 ?BC和 vC，很简便
c m / s31631660
c m / s321622
60c o s
?????
?????
?
?
tgvv
v
v
v
BCB
B
B
C
cm3103 ????? OBBCBCv BCCB ??
研究 BC,以 B为基点,
根据
作 速度平行四边形
CBBC vvv ??
c m / s 1611 ?????? AABABB vAPBPv ??
r a d / s 35460s i n/10 16s i n/16
1
???? ??? OAAPv AAB
） (
（ ）
r a d / s 6.1
310
316 ????
BC
v CB
BC?
109
Solution,Rod OA rotates about a fixed axis,
rod AB and BC have plane motions,block B
and C have translatory motions,
?Find
cvApplying the theorem of velocity projection
to rod AB,gives
?? 30c o s60c o s
AB vv ? oAB rvv ?33 ???
Applying the theorem of velocity projection to rod BC,we have
?60s in
Bc vv ?
)( ???? ooc rrv ?? 232 33
[Example 6] In the mechanism,OA= r,rod OA rotates with uniform
?o,At the moment,? = 60o,AB?BC,AB=6 r,BC=,
Find the velocity and acceleration of block C at this moment,
r33
110
解,OA定轴转动 ; AB,BC均作平面运动,
滑块 B和 C均作平动
?求
cv
对 AB杆应用速度投影定理,
?? 30c o s60c o s
AB vv ? oAB rvv ?33 ???
对 BC杆应用速度投影定理,?60s in
Bc vv ?
)( ???? ooc rrv ?? 232 33
[例 6] 已知,配气机构中,OA= r,以等 ?o转动,在某瞬时 ? = 60o
AB?BC,AB=6 r,BC=, 求 该瞬时滑块 C的
速度和加速度,
r33
111
?Find,
caTake A as the pole to find acceleration at point B,
n
BABAAB aaaa ???
? ( a )
,,22 ABnBAoA ABara ?? ??? P1 is the instantaneous velocity center
of rod AB and, rAP 3
1 ?
22
1
3
2
3
6
33
o
on
BA
ooA
AB
rra
r
r
AP
v
?
?
??
?
???
????
)(
,
Draw the vector diagram of
accelerations,and project them to BA
222
33
4
6060
oooB
n
BAAB
r
rra
aaa
??? ?????
???? c o sc o s ??
112
?求
ca
以 A为基点 求 B点加速度,n
BABAAB aaaa ???
? ( a )
,,22 ABnBAoA ABara ?? ??? P1为 AB杆速度瞬心，而 rAP 31 ?
22
1
3
2
3
6
33
o
on
BA
ooA
AB
rra
r
r
AP
v
?
?
??
?
???
????
)(
,
作加速度矢量图,
并沿 BA方向投影
222
33
4
6060
oooB
n
BAAB
r
rra
aaa
??? ?????
???? c o sc o s ??
113
)( baaaa nCBCBBc ??? ?
Draw the vector diagram of
accelerations P2 is the
instantaneous center of velocity
for rod BC and P2C = 9 r,
Then take B as the pole to find
ca
69
1
2
3
2
o
o
c
BC rrCP
v ??? ?????
222
12
3
633 o
o
BC
n
CB rrBCa ?
?? ????? )(
Projecting Equ,(b) to BC gives
222
12
3
12
3
2
3
330 ooo
n
CBBc rr
raaa ??? ?????? ?c o s
[Note] We can assume the direction of, If the solved
acceleration has a positive sign,the assumed direction is right,
otherwise,the true direction is opposite to the assumed one,
cB aa or
30o
114
)( baaaa nCBCBBc ??? ?
作加速度矢量图,
P2 为 BC的瞬心,而 P2C = 9 r
再以 B为基点,求
ca
69
1
2
3
2
o
o
c
BC rrCP
v ??? ?????
222
12
3
633 o
o
BC
n
CB rrBCa ?
?? ????? )(
将 (b) 式在 BC方向线上投影
222
12
3
12
3
2
3
330 ooo
n
CBBc rr
raaa ??? ?????? ?c o s
[注 ] 指向可假设，结果为正说明假设与实际指向相同,
反之，结果为负，说明假设与实际指向相反, cB aa,
30o
115
[Example 7] Slotted bar-block mechanism
Please look at the animation
116
[例 7] 导槽滑块机构 请看动画
117
Solution,
? Apply the composition method of motions to determine the
relationship between the velocities at point C on rod CD and point
C‘ in contact with rod AE,Take point C on rod CD as the moving
point,the moving system is attached on rod AE,and the static system
is on the fixed support,we have
(a)
rcc vvv ?? '?Determine the relationship between the velocity at point A on
AE and that at point C' based on the analysis of the plane motion,
(b)
AcAc vvv ???'
[Example 7] In the slotted bar-sliding block
mechanism,the velocity of rod AB and that
of rod CD as well as the angle ? are given
At this moment,AC= l,Find the angular
velocity of the slotted bar AE,
u
v
118
解, ? 应用点的合成运动方法
确定 CD杆上 C点与 AE杆上接触 点 C'之间的速度关系
取 CD杆上 C为动点，动系固结于 AE，静系固结于机架；则
(a)
rcc vvv ?? '
?应用平面运动方法确定 AE上 A,C' 点之间速度关系
(b)
AcAc vvv ???'
[例 7] 导槽滑块机构
图示瞬时,杆 AB速度,杆 CD速度
及 ? 角已知，且 AC= l,
求导槽 AE的图形角速度,
u
v
119
Inserting (b) into (a) gives,Drawing vector
diagram of velocities and projecting them to the axis ?, and
noting vC＝ v,v＝ u,we have
rACAC vvvv ??? ?
??
??
s i n c o s e,i.
s i nc o s
uvv
vvv
AC
ACAC
??
???
?
?
l
uv
AC
v AC
AE
??? s i nc o s ??? ? （ ）
120
将 (b) 代入 (a) 得,作速度矢量图投至 ? 轴，
且 vC＝ v,v＝ u，有
rACAC vvvv ??? ?
??
??
s i n c o s
s i nc o s
uvv
vvv
AC
ACAC
??
???
?
?
即
l
uv
AC
v AC
AE
??? s i nc o s ??? ? （ ）
121
122
