1
Theoretical mechanics
2
3
§ 10–1 Composition of rotations about
two parallel axes
Chapter 10,General motion of a rigid body
4
§ 10–1 刚体绕平行轴转动的合成
第十章 刚体的一般运动
5
Chapter 10,General motion of a rigid body
§ 10-1 Composition of rotations about two parallel axes
Composition problem of rotations about parallel axes is
often encountered in mechanical engineering,For instance,the
planet gear wheel in a planet gear mechanism has plane motion,
In previous study,we dealt with a plane motion of a planet gear
by decomposing it into a translation and a rotation,However,it
is more convenient to view the plane motion of a planet gear as
a composition of two rotations,
6
第十章 刚体的一般运动
§ 10-1 刚体绕平行轴转动的合成
刚体绕平行轴转动的合成问题在机械中经常遇到。
例如，行星圆柱齿轮机构，行星轮作平面运动。前面
所研究的平面运动是把它看成为平动和转动的合成运
动，但是在分析行星轮系的传动问题时，将行星轮的
平面运动看成为转动与转动的合成运动则比较方便。
7
（ Look for the animation on the next page）
Take a planet gear mechanism as an example,
Static system is O1xy
Moving system is O1x'y'
Relative motion,rotation about axis O2,
?r is the relative
angular velocity,
Embroiling motion,rotation about axis O1,
?e is the embroiling
angular velocity,
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静系,O1xy
动系,O1x'y'
相对运动, 绕 O2轴转动,
?r为相对角速度。
牵连运动, 绕 O1轴转动,
?e为牵连角速度。
（翻页请看动画）
例如在行星轮系中
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10
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dt
d
dt
d
dt
d rea ??? ??
rea ??? ??? In conclusion,The absolute angular velocity ?
a of the plane
figure (the planet gear) equals to the algebraic sum of the
embroiling angular velocity ?e and the relative one ?r,
When ?e and ?r have same direction,the direction of ?a is
the same as them,
When ?e and ?r have opposite directions,the direction of
?a is the same as the bigger one,
rea ??? ??
rea ??? ??
At moment t,it is in position O2A;
At moment t + ? t,it is in position O2'A‘,
rea ??? ??
The figure shows that
Differentiating it with respect to t gives,
14
由图看出
对 t 求导,
rea ??? ??
dt
d
dt
d
dt
d rea ??? ??
rea ??? ???
即,平面图形（这里指行星轮） 的绝对角速度 ?a等于牵连角速
度 ?e 与相对角速度 ?r的代表和,
当 ?e 与 ?r 转向相同时 转向与两者相同,
当 ?e与 ?r 转向相异时 转向与大者的相同,
rea ??? ??
rea ??? ??
ｔ 时刻,O2A 位置；
t + ? t 时刻,O2'A'位置
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Next,we determine the position of the instantaneous
velocity center for the plane figure S,
re POPO ?? ??? 21
P is the instantaneous center of velocity of the figure,The axis
across point P and perpendicular to axes O1 and O2 is called
instantaneous axis,every point of which has zero velocity,
Since ve=vr,and they
have opposite
directions,vp=0,P is
the instantaneous
center of velocity,
Hence,
r
e
PO
PO
?
???
1
2
?e and ?r have the
same direction
?e and ?r have the
opposite directions
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下面来确定图形 S的瞬心的位置
re POPO ?? ??? 21
P点为图形的速度瞬心,通过点 P且与轴 O1,O2平行
的轴称为 瞬时轴,该轴上各点的速度都等于零 。
由于 ve=vr,且方向相
反,因此 vp=0,P为速
度瞬心。此时
r
e
PO
PO
?
???
1
2?
e 与 ?r同转向 ?e 与 ?r同 反 向
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Conclusion,Two rotations of a rigid body about two parallel
axes can be composed into a rotation about a instantaneous axis,
The instantaneous axis is parallel to the original axes,and the
ratio of its distance to one axis and that to the other is inversely
proportional to the ratio of their angular velocities,When these
rotations turn to the same direction,the instantaneous axis lies
between the two axes,,Otherwise,it is in the
extension of the line segment between the two axes and lies in
the side of the axis with larger angular velocity,
rea ??? ??
rea ??? ??
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即,刚体绕两平行轴的转动可合成为绕瞬轴的转动，
瞬轴与原两轴共面且平行，到两轴的距离与两角速
度大小成反比。同向转动时，瞬轴在两轴之间,
,转向与两者相同；反向转动时,瞬轴
在两轴之外,在角速度值大的一侧,，转
向与大者的相同。
rea ??? ??
rea ??? ??
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[Example 1] Gear ? and ??? have the same radius R,and the
radius of gear ?? is r,They are meshed one by one,and gear ? is
fixed,Gear ?? and ??? is pined to the crank O1O3 respectively，
and they can rotate around axes O2 and O3,Crank O1O3 rotate
with ? 0, Fine the angular velocity ?3 r of gear III relative to
the crank,its absolute angular velocity ?3,and the velocities of
point A and B respectively at the instant shown in the figure,
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[例 1] 齿轮 ?,???半径均为 R,齿轮 ??半径为 r,依次互啮合,轮 ? 固
定不动，轮 ?? 和轮 ??? 装在曲柄 O1O3上，可分别绕 O2,O3轴转
动。设曲柄 O1O3以 ? 0顺时针转动．试求齿轮 III相对于曲柄转
动的角速度 ?3 r 和齿轮 ???的绝对角速度 ?3 以及图示瞬时 A,
B 两点的速度。
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Solution,Take O1O3 as the moving coordinate system,and
denote ?1 r,?2 r and ?３ r as the angular velocities of gear ?,?? and
??? relative to the connecting rod,The transmission ratio can be
expressed as
rr
r
r
R
r
r
R
31
3
1 1 ??
?
? ?????
Employing the theory developed above,we have
orre ?????? 1011 ?????
00033013 ????????? ???????? rerr ;
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解,取系杆 O1O3 为动系,?1 r, ?2 r, ?３ r 分别是 ?, ??, ???
轮相对于系杆的角速度,根据传动比公式,可得
rr
r
r
R
r
r
R
31
3
1 1 ??
?
? ?????
由平行轴转动的合成理论，得
orre ?????? 1011 ?????
00033013 ????????? ???????? rerr ;
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It is clear that gear ??? is in translation,Noting that all points in a
translating rigid body have the same velocity,we obtain
)(20313 rROOvvv OBA ?????? ?
Direction,perpendicular to rod O1O3,and pointing downward,
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由此可知，齿轮 ???作平动,平动刚体上各点的速度相同,故
)(20313 rROOvvv OBA ?????? ?
方向：垂直于 O1O3杆，指向朝下,
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Theoretical mechanics
2
3
§ 10–1 Composition of rotations about
two parallel axes
Chapter 10,General motion of a rigid body
4
§ 10–1 刚体绕平行轴转动的合成
第十章 刚体的一般运动
5
Chapter 10,General motion of a rigid body
§ 10-1 Composition of rotations about two parallel axes
Composition problem of rotations about parallel axes is
often encountered in mechanical engineering,For instance,the
planet gear wheel in a planet gear mechanism has plane motion,
In previous study,we dealt with a plane motion of a planet gear
by decomposing it into a translation and a rotation,However,it
is more convenient to view the plane motion of a planet gear as
a composition of two rotations,
6
第十章 刚体的一般运动
§ 10-1 刚体绕平行轴转动的合成
刚体绕平行轴转动的合成问题在机械中经常遇到。
例如，行星圆柱齿轮机构，行星轮作平面运动。前面
所研究的平面运动是把它看成为平动和转动的合成运
动，但是在分析行星轮系的传动问题时，将行星轮的
平面运动看成为转动与转动的合成运动则比较方便。
7
（ Look for the animation on the next page）
Take a planet gear mechanism as an example,
Static system is O1xy
Moving system is O1x'y'
Relative motion,rotation about axis O2,
?r is the relative
angular velocity,
Embroiling motion,rotation about axis O1,
?e is the embroiling
angular velocity,
8
静系,O1xy
动系,O1x'y'
相对运动, 绕 O2轴转动,
?r为相对角速度。
牵连运动, 绕 O1轴转动,
?e为牵连角速度。
（翻页请看动画）
例如在行星轮系中
9
10
11
12
13
dt
d
dt
d
dt
d rea ??? ??
rea ??? ??? In conclusion,The absolute angular velocity ?
a of the plane
figure (the planet gear) equals to the algebraic sum of the
embroiling angular velocity ?e and the relative one ?r,
When ?e and ?r have same direction,the direction of ?a is
the same as them,
When ?e and ?r have opposite directions,the direction of
?a is the same as the bigger one,
rea ??? ??
rea ??? ??
At moment t,it is in position O2A;
At moment t + ? t,it is in position O2'A‘,
rea ??? ??
The figure shows that
Differentiating it with respect to t gives,
14
由图看出
对 t 求导,
rea ??? ??
dt
d
dt
d
dt
d rea ??? ??
rea ??? ???
即,平面图形（这里指行星轮） 的绝对角速度 ?a等于牵连角速
度 ?e 与相对角速度 ?r的代表和,
当 ?e 与 ?r 转向相同时 转向与两者相同,
当 ?e与 ?r 转向相异时 转向与大者的相同,
rea ??? ??
rea ??? ??
ｔ 时刻,O2A 位置；
t + ? t 时刻,O2'A'位置
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Next,we determine the position of the instantaneous
velocity center for the plane figure S,
re POPO ?? ??? 21
P is the instantaneous center of velocity of the figure,The axis
across point P and perpendicular to axes O1 and O2 is called
instantaneous axis,every point of which has zero velocity,
Since ve=vr,and they
have opposite
directions,vp=0,P is
the instantaneous
center of velocity,
Hence,
r
e
PO
PO
?
???
1
2
?e and ?r have the
same direction
?e and ?r have the
opposite directions
16
下面来确定图形 S的瞬心的位置
re POPO ?? ??? 21
P点为图形的速度瞬心,通过点 P且与轴 O1,O2平行
的轴称为 瞬时轴,该轴上各点的速度都等于零 。
由于 ve=vr,且方向相
反,因此 vp=0,P为速
度瞬心。此时
r
e
PO
PO
?
???
1
2?
e 与 ?r同转向 ?e 与 ?r同 反 向
17
Conclusion,Two rotations of a rigid body about two parallel
axes can be composed into a rotation about a instantaneous axis,
The instantaneous axis is parallel to the original axes,and the
ratio of its distance to one axis and that to the other is inversely
proportional to the ratio of their angular velocities,When these
rotations turn to the same direction,the instantaneous axis lies
between the two axes,,Otherwise,it is in the
extension of the line segment between the two axes and lies in
the side of the axis with larger angular velocity,
rea ??? ??
rea ??? ??
18
即,刚体绕两平行轴的转动可合成为绕瞬轴的转动，
瞬轴与原两轴共面且平行，到两轴的距离与两角速
度大小成反比。同向转动时，瞬轴在两轴之间,
,转向与两者相同；反向转动时,瞬轴
在两轴之外,在角速度值大的一侧,，转
向与大者的相同。
rea ??? ??
rea ??? ??
19
[Example 1] Gear ? and ??? have the same radius R,and the
radius of gear ?? is r,They are meshed one by one,and gear ? is
fixed,Gear ?? and ??? is pined to the crank O1O3 respectively，
and they can rotate around axes O2 and O3,Crank O1O3 rotate
with ? 0, Fine the angular velocity ?3 r of gear III relative to
the crank,its absolute angular velocity ?3,and the velocities of
point A and B respectively at the instant shown in the figure,
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[例 1] 齿轮 ?,???半径均为 R,齿轮 ??半径为 r,依次互啮合,轮 ? 固
定不动，轮 ?? 和轮 ??? 装在曲柄 O1O3上，可分别绕 O2,O3轴转
动。设曲柄 O1O3以 ? 0顺时针转动．试求齿轮 III相对于曲柄转
动的角速度 ?3 r 和齿轮 ???的绝对角速度 ?3 以及图示瞬时 A,
B 两点的速度。
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Solution,Take O1O3 as the moving coordinate system,and
denote ?1 r,?2 r and ?３ r as the angular velocities of gear ?,?? and
??? relative to the connecting rod,The transmission ratio can be
expressed as
rr
r
r
R
r
r
R
31
3
1 1 ??
?
? ?????
Employing the theory developed above,we have
orre ?????? 1011 ?????
00033013 ????????? ???????? rerr ;
22
解,取系杆 O1O3 为动系,?1 r, ?2 r, ?３ r 分别是 ?, ??, ???
轮相对于系杆的角速度,根据传动比公式,可得
rr
r
r
R
r
r
R
31
3
1 1 ??
?
? ?????
由平行轴转动的合成理论，得
orre ?????? 1011 ?????
00033013 ????????? ???????? rerr ;
23
It is clear that gear ??? is in translation,Noting that all points in a
translating rigid body have the same velocity,we obtain
)(20313 rROOvvv OBA ?????? ?
Direction,perpendicular to rod O1O3,and pointing downward,
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由此可知，齿轮 ???作平动,平动刚体上各点的速度相同,故
)(20313 rROOvvv OBA ?????? ?
方向：垂直于 O1O3杆，指向朝下,
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