1
Theoretical Mechanics
2
3
§ 13–1 Moment of momentum
§ 13–2 moment of momentum theorem
§ 13–3 Differential equations for the rotation of a rigid body
around a fixed-axis
§ 13–4 Moment of inertia of a rigid body with respect to an axis
§ 13–5 moment of momentum theorem for a system with respect
to its center of mass,differential equations of plane motion
of a rigid body
Exercises
Chapter 13,Moment of Momentum Theorem
4
§ 13–1 动量矩
§ 13–2 动量矩定理
§ 13–3 刚体定轴转动微分方程
§ 13–4 刚体对轴的转动惯量
§ 13–5 质点系相对于质心的动量矩定理,
刚体平面运动微分方程
习题课
第十三章 动量矩定理
5
Theorem of momentum,
The change of the momentum of a particle (or of a system of
particles) is the result of external forces (the principal vector of an
external force system),
Theorem of motion of the center of mass,
The motion of the center of mass is the result of external forces
(the principal vector of an external force system),
When the center of mass coincides with a certain point of a fixed
axis,then the momentum of a rotating body is always zero because
Vc=0,But in this case the system is still subjected to the action of
external forces,The moment of momentum theorem establishes
the dependence between the change of the moment of momentum
of a particle or a system with respect to a center (or a fixed axis)
and the torque given by all external forces acting on the particle or
the system with respect to the same center (or axis),
6
质点
质点系 动量定理,动量的改变 — ?外力（外力系主矢）
若当质心为固定轴上一点时,vC=0，则其动量恒等于零,
质心无运动，可是质点系确受外力的作用。 动量矩定理建立了
质点和质点系相对于某固定点（固定轴）的动量矩的改变与外
力对同一点（轴）之矩两者之间的关系。
质心运动定理,质心的运动 — ?外力（外力系主矢）
7
§ 13-1 Moment of momentum
Moment of momentum of a particle with respect to a center O is
,it is a vector,
Moment of momentum of a particle with respect to an axis Z is
,it is an algebraic quantity,
vmrvmm O ??)(
)()( xyOz vmmvmm ?
O A Bvmm O ?2)( ?
''2)( BOAvmm z ???The definition of the sign of the
moment of momentum with respect
to an axis is the same as that of the
moment of a force with respect to an
axis,Looking from the positive end
of the axis,it’s positive if it is
counterclockwise,and it is negative if
it is clockwise,
8
§ 13-1 动量矩
一．质点的动量矩
质点对点 O的动量矩,矢量
质点对轴 z 的动量矩,代数量
vmrvmm O ??)(
)()( xyOz vmmvmm ?
O A Bvmm O ?2)( ?
''2)( BOAvmm z ???
正负号规定与力对轴矩的规定相同
对着轴看：顺时针为负
逆时针为正
9
The relation between the moment of momentum of a particle with
respect to an axis and a center is
2,Moment of momentum of a system of particles,
The moment of momentum of a system with respect to a center O is
The moment of momentum of a system with respect to an axis Z is iiiiiOO vmrvmmL ??? ?? )(
? ? zOiizz LvmmL )( ?? ?
kg·ｍ 2/s,
Moment of momentum measures the intensity of rotation of a
body rotating around a fixed center or an axis at any instant,
? ? )( )( vmmvmm zzO ?
Calculation of moment of momentum of a rigid body,
1) Rigid body in translational motion
CCCOO vmrvmmL ??? )(
)( CCCiiiii vmrvrmvmr ????? ?? )( Czz vmmL ?
The moment of momentum of a rigid body in translational motion
with respect to a fixed center (axis) is equal to the moment of
momentum of the center of mass of the rigid body with respect to
that center (axis),
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质点对点 O的动量矩与对轴 z 的动量矩之间的关系,
二．质点系的动量矩
质系对点 O动量矩,
质系对轴 z 动量矩,
iiiiiOO vmrvmmL ??? ?? )(
? ? zOiizz LvmmL )( ?? ?
kg·ｍ 2/s。 动量矩度量物体在任一瞬时绕固定点 (轴 )转动的强弱
? ? )( )( vmmvmm zzO ?
刚体动量矩计算,
1．平动刚体 CCCOO vmrvmmL ??? )(
)( CCCiiiii vmrvrmvmr ????? ??
)( Czz vmmL ?
平动刚体对固定点（轴）的动量矩等于刚体质心的动量
对该点（轴）的动量矩。
11
3) Rigid body in plane motion,
The moment of momentum of a rigid body in plane motion with
respect to an axis perpendicular to the symmetrical plane of mass is
equal to the sum of the moment of momentum of the center of mass
of the rigid body in translational motion together with the center of
mass about that axis and the moment of momentum of the rigid
body when rotating around the center of mass with respect to a
parallel axis through the center of mass of the body,
?? ?????? ziiiizz IrmvmmL 2)(
???? CCzz IvmmL )(
2) Rigid body in fixed-axis rotation,
The moment of momentum of a rigid body in fixed-axis rotation
with respect to the axis of rotation is equal to the product of the
moment of inertia of the rigid body with respect to that axis and its
angular velocity,
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3．平面运动刚体
平面运动刚体对垂直于质量对称平面的固定轴的动量矩，等于
刚体随同质心作平动时质心的动量对该轴的动量矩与绕质心轴
作转动时的动量矩之和。
?? ?????? ziiiizz IrmvmmL 2)(
???? CCzz IvmmL )(
2．定轴转动刚体
定轴转动刚体对转轴的动量矩等于刚体对该轴转动惯量与角速
度的乘积。
13
112223 2
1 ?? RRvv ???
.)( 32322
2
2
2
2
1 vRmm
R
I
R
IL
O ????
OCOBOAO LLLL ???
2332222211 )( RvmRvmII ???? ??
Solution,
[Example 1] Pulley A,m1,R1,R1=2R2,I1
Pulley B,m2,R2,I2 ；
Material body C,m3
Determine the moment of momentum of the
system with respect to the axis O,
14
112223 2
1 ?? RRvv ???
32322
2
2
2
2
1 )( vRmm
R
I
R
IL
O ????
OCOBOAO LLLL ???
2332222211 )( RvmRvmII ???? ??
解,
[例 1] 滑轮 A,m1,R1,R1=2R2,I1
滑轮 B,m2,R2,I2 ；物体 C,m3
求 系统对 O轴的动量矩。
15
§ 13-2 Moment of momentum theorem
1,Moment of momentum theorem of a particle,
vmdt rdvmrdtddt vmdr ????? )()(
,)( a n d 0 B u t,FmFrvmvvmdt rd O??????
)()]([,)( FmvmmdtdFrvmrdtd OO ????
The time-derivative of the moment of momentum of a particle
with respect to any fixed center with respect to time is equal to the
moment of the force acting on the particle with respect to the same
center,This is the moment of momentum theorem of a particle,
Therefore,
Fdt vmd ?)(
Frdt vmdr ??? )(
r
, Now multiplying both sides of
this equation by the radius vector according
to vector multiplication,we get,
The left-hand side can be expressed in the form
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Fdt vmd ?)(
§ 13-2 动量矩定理
一．质点的动量矩定理
两边叉乘矢径,有 Fr
dt
vmdr ??? )(r
左边可写成
vmdt rdvmrdtddt vmdr ????? )()(
,)(,0 FmFrvmvvmdt rd O??????而
)()]([,)( FmvmmdtdFrvmrdtd OO ????
质点对任一固定点的动量矩对时间的导数，等于作用在质
点上的力对同一点之矩。这就是 质点对固定点的动量矩定理。
故,
17
Projecting the equation given above on the axes of a Cartesian
coordinate system through a fixed center O we obtain
)()( ),()( ),()( FmvmmdtdFmvmmdtdFmvmmdtd zzyyxx ???
These equations are the moment of momentum theorem of a
particle with respect to a fixed axis,They are also called the
projection forms of the principle of moments with respect to a fixed
axis,The derivative of the moment of momentum of a particle with
respect to any fixed axis is equal to the moment of the force acting
on the particle with respect to the same axis,
If
These results express the conservation law of the moment of
momentum of a particle,
)0)(( 0)( ?? FmFm zO then ?)( vmm O const,c o n s t ))(( ?vmm z
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将上式在通过固定点 O的三个直角坐标轴上投影，得
)()( ),()( ),()( FmvmmdtdFmvmmdtdFmvmmdtd zzyyxx ???
上式称 质点对固定轴的动量矩定理,也称为质点动量矩定
理的投影形式。即质点对任一固定轴的动量矩对时间的导数，
等于作用在质点上的力对同一轴之矩。
称为 质点的动量矩守恒 。
若 )0)(( 0)( ?? FmFm
zO 则 ?)( vmm O
常矢量 ))(( 常量?vmm z
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The force analysis is shown in figure,
By the analysis of motion we obtain
?? ?? 2)( mllmlvmm O ??
OMlv ??,??
Applying the moment of momentum theorem we have
,hence,
When the pendulum swing with small amplitude we can use,
Let,then,Solving the differential equation and
taking at into account,we obtain the equation of
motion of the simple pendulum for small amplitude as
The period of the pendulum is
)()( Fmvmmdtd OO ? 0s i n,s i n)( 2 ???? ???? lgm g lmldtd ???
sin ?? ?
l
g
n ?
2? 02 ?? ??? n??
tlgc o s0?? ?
l
gT ?2?
?s i n)()()( m g lgmmTmFm OOO ????
Solution Choose the small ball m as a particle,
[Example 2] A pendulum of mass m and length l is
released from ?= ?0 at t =0,Deduce its equation of
motion,
0,,0 00 ??? ??? ?t
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运动分析,。 ?? ?? 2)( mllmlvmm
O ??OMlv ??,??
由动量矩定理
即
)()( Fmvmmdtd OO ?
0s i n,s i n)( 2 ???? ???? lgm g lmldtd ???
微幅摆动时,并令,则,sin ?? ?
l
g
n ?
2? 02 ?? ??? n??
解微分方程,并代入初始条件 则运动方程 )0,,0(
00 ??? ??? ?t
tlgc o s0?? ?,摆动周期
l
gT ?2?
?s i n)()()( m g lgmmTmFm OOO ????
解,将小球视为质点。
受力分析；受力图如图示。
[例 2] 单摆 已知 m,l,t =0时 ?= ?0，从静止
开始释放。 求 单摆的运动规律。
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Note,When we calculate a moment of momentum and a
moment of force,the definition of their signs should be
identical (here we call a moment positive if it tends to rotate
the pendulum counterclockwise),
Apply the moment of momentum theorem of
a particle when
1) the particle is subjected to a central force,
2) the particle rotates around a center (an axis),
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注：计算动量矩与力矩时，符号规定应一致（本题规定逆时
针转向为正）
质点动量矩定理的应用,
?在质点受有心力的作用时。
?质点绕某心（轴）转动的问题。
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2,Moment of momentum theorem of a system of
particles,
On the left-hand side we exchange the order of the operations
(summation and derivative) and get
hence,,
,0)( ),( )( ?? ?? iiOiiOO FmvmmL
? ?? )()( )( eOeiOO MFmdtLd
This is the moment of momentum
theorem of a system with respect to
any fixed center,
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ?????? ???
For the system of particles,
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ??????For the particle Mi,,
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二．质点系的动量矩定理
左边交换求和与导数运算的顺序，而
则,0)( ),( )( ?? ?? iiOiiOO FmvmmL
? ?? )()( )( eOeiOO MFmdtLd 一 质点系对固定点的动量矩定理
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ?????? ???对质点系，有
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ??????对质点 Mi,
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The time-derivative of the total moment of momentum of a system
with respect to any fixed center is equal to the geometric sum (the
principal moment )of the moments of all the external forces acting
on the same system with respect to the same center,
??? ?????? )()()()()()( )(,)(,)( ezeizzeyeiyyexeixx MFmdtdLMFmdtdLMFmdtdL
Projecting the equation above on the axes of a Cartesian
coordinate system through a fixed center O we obtain
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质点系对任一固定点的动量矩对时间的
导数，等于作用在质点系上所有外力对同一
点之矩的矢量和（外力系的主矩）。
??? ?????? )()()()()()( )(,)(,)( ezeizzeyeiyyexeixx MFmdtdLMFmdtdLMFmdtdL
将上式在通过固定点 O的三个直角坐标轴上投影，得,
27
The formulae above are called the moment of momentum theorem of
a system of particles with respect to any fixed axis, The time-
derivative of the total moment of momentum of a system with respect
to any fixed axis is equal to the algebraic sum of the moments of all
the external forces acting on the system with respect to the same axis
(the principal moment of external force system with respect to the
same axis),
The theorem states that only external forces can change the moment
of momentum of the system,internal forces ca not do this,
The conversation law of the total moment of momentum of a system,
① If then constant vector,
② If then const,
0)( ?eOM
0)( ?ezM
?OL
?zL
28
上式称为 质点系对固定轴的动量矩定理 。即质点系对任一固
定轴的动量矩对时间的导数，等于作用在质点系上所有外力对同
一固定轴之矩的代数和（外力系对同一轴的主矩）。
质点系的动量矩守恒
?当 时,常矢量。
?当 时,常量。
0)( ?eOM
0)( ?ezM
?OL
?zL
定理说明内力不会改变质点系的动量矩，只有外力才能改
变质点系的动量矩。
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[Example3], D et er m i n e,k n o w n ar e an d,?rPPP
BA ?
Solution Take the whole system as the object to
be investigated,The force analysis is shown in
the diagram,The analysis of motion gives
v =ｒ ?,
rPPrPrPM BABAeO )()( ????
?OBAO IrvgPrvgPL ?????
)2(
2 P
PPgrL BAO ??? ?
Applying the moment of momentum theorem we obtain
rPPPPPgrdtd BABA )()]2([ 2 ?????
2/PPP
PP
r
g
dt
d
BA
BA ?? ????? ??
2
2
1 r
g
PI
O ?
Substituting into the equation we get
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解, 取整个系统为研究对象,
受力分析如图示。
运动分析,v =ｒ ?
rPPrPrPM BABAeO )()( ????
?OBAO IrvgPrvgPL ?????
)2(,21 22 PPPgrLrgPI BAOO ???? ?得代入将
由动量矩定理,
rPPPPPgrdtd BABA )()]2([ 2 ?????
2/PPP
PP
r
g
dt
d
BA
BA ?? ????? ??
[例 3] 已知, 。求。 ; ; ?rPPP
BA ?
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Solution so the moment of
momentum of the system does not change,
? ?,0)( )( eO Fm
rvvmrvm ABAA )(0 ??? 2vv A ?
Therefore,the absolute velocities of the
monkeys A and B are identical,Both
are,
2v
[Example 4] The weight of two monkeys A an B are knowing,The
monkey B climbs upwards with a velocity v relative to the rope,while
the monkey A does not move relative to the rope,Determine
(1)the motion of the monkey A move when the monkey B climbs
upward,
(2)the velocity of the monkey A (neglecting the weight of the rope),
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解, 系统的动量矩守恒。 ? ??,0)( )( e
O Fm
rvvmrvm ABAA )(0 ???
2
vv
A ?
猴 A与猴 B向上的绝对速度是一样的,
均为 。
2v
[例 4] 已知：猴子 A重 =猴子 B重，猴 B以相对绳速度
上爬，猴 A不动，问当猴 B向上爬时，猴 A将如何动？
动的速度多大？（轮重不计）
v
33
§ 13-3 Differential equations for
the rotation of a rigid body around a fixed-axis
For a rigid body in fixed-axis rotation we have ?
zz IL ?
)()( e
zz MIdt
d ??
)(
2
2)(
o r ezzezz MdtdIMI ?? ??
Substituting this equation into the formula of the moment of
momentum theorem,we get
This is the differential equation of
fixed-axis rotation for a rigid body,Two types of problems can be solved in this way,
(1) Knowing the moment of the external forces acting on a rigid
body we can determine the law of the rotational motion of the
rigid body,
(2) Knowing the law of the rotational motion of a rigid body we
can determine the external forces (or moment of force ) acting
on the body,but we can not determine the reactions at bearings,
For this problem,the theorem of motion of the center of mass
has to be applied,
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§ 13-3 刚体定轴转动微分方程
对于 一个定轴转动刚体
代入质点系动量矩定理，有
?zz IL ?
)()( e
zz MIdt
d ??
)(
2
2)( e
zzezz Mdt
dIMI ??? ?? 或— 刚体定轴转动微分方程
解决两类问题,
?已知作用在刚体的外力矩，求刚体的转动规律。
?已知刚体的转动规律，求作用于刚体的外力（矩）。
但不能求出轴承处的约束反力，需用质心运动定理求解。
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Special cases are,
? If,then const,the rigid body will
rotate uniformly or remain at rest,
? If const,then ? =const,the rigid body will be in uniform
rotation,Comparing with,we see that the moment
of inertia is a measure of a body’s inertia concerning its rotational
motion,
? ?? 0)( )()( ezez FmM ???,0
?)(ezM
)( ezz MI ?? Fam ?
zI
36
特殊情况,
? 若,则 恒量，刚体作匀速转动或
保持静止。
? 若 常量，则 ? =常量，刚体作匀变速转动。
将 与 比较，刚体的转动惯量 是刚体
转动惯性的度量。
? ?? 0)( )()( ezez FmM ?? ??,0
?)(ezM
)( ezz MI ?? Fam ?
zI
37
§ 13-4 Moment of inertia of
a rigid body with respect to an axis
1,Definition,
If the mass in a rigid body is
distributed continuously,then
?? 2iiz rmI
?? dmrI mz 2
The moment of inertia is a measure of a rigid body’s inertia
concerning its rotational motion around a fixed axis,Its
magnitude shows whether it is difficult or easy for the rotational
state of a rigid body to be changed,The moment of inertia is
always positive,The unit for it in the SI system is kg·m2,
38
§ 13-4 刚体对轴的转动惯量
一．定义,
若刚体的质量是连续分布，则
?? 2iiz rmI
?? dmrI mz 2
刚体的转动惯量是刚体对某轴转动惯性大小的度量，它的
大小表现了刚体转动状态改变的难易程度。
转动惯量恒为正值，国际单位制中单位 kg·m2 。
39
1)Method of integration
This method may be applied to any homogenous rigid body with
regular geometric shape,
[Example 1] There is a thin homogeneous rod
of length and mass m,Determine
(1) its moment of inertia IZ with respect to the axis Z and
(2) its moment of inertia IZ with respect to the axis Z,
2,Calculation of the moment of Inertia,
222
2 12
1 mldx
l
mxI l
lz ? ??
2
0
2' 31 mldxlmxI lz ? ???
Solution,
40
１．积分法 （具有规则几何形状的均匀刚体可采用）
[例 1] 匀质细直杆长为 l,质量为 m 。
求, ?对 z轴的转动惯量 ；
?对 z' 轴的转动惯量 。
zI
'zI
二．转动惯量的计算
222
2 12
1 mldx
l
mxI l
lz ? ??
2
0
2' 31 mldxlmxI lz ? ???
解,
41
2) Radius of gyration
The radius of gyration of a rigid body with respect to an axis Z is
defined by the equation,
and,
m
I z??
z?
2zz mI ??
For a homogeneous rigid body,the radius of gyration is given
only by its geometric shape,but not by its density,Two
homogenous rigid bodies with identical shapes,but made of
different materials (different densities),have the same radius of
gyration,
In the designing handbooks of mechanical engineering the moments
of inertia and the radius of gyration for parts with simple geometric
shapes can be found,The and of some common homogenous
rigid body are given in these books,too,
z?
zI z?
42
2,回转半径
由 所定义的长度 称为刚体对 z 轴的回转半径。
m
I z?? z?
2zz mI ??
对于均质刚体,仅与几何形状有关，与密度无关。对
于几何形状相同而材料不同（密度不同）的均质刚体，其回
转半径是相同的。
z?
在机械工程设计手册中，可以查阅到简单几何形状或已
标准化的零件的转动惯量和回转半径。书中列出几种常见均质
刚体的,以供参考。
zzI ?和
43
3) The parallel axis theorem,
The moments of inertia of a body with respect to different axes
are generally different,
2' mdII zCz ??
The moment of inertia of a rigid body with respect to any axis is
equal to the moment of inertia of the body with respect to a
parallel axis through the center of mass of the body plus the
product of the mass of the body with the square of the distance d
between the two axes,
44
3,平行移轴定理
同一个刚体对不同轴的转动惯量一般是不相同的。
2' mdII zCz ??
刚体对某轴的转动惯量等于刚体对通过质心且与该轴平行的
轴的转动惯量，加上刚体的质量与两轴间距离的平方之乘积。
45
Proof,Suppose that the center of mass of a rigid body of mass M is
C and, CzzO //''
In example 1 e.g.,the moment of inertia of the thin homogeneous rod
with respect to the axis Z′ is
2222 3141121)2(' mlmlmllmII zz ?????
Consequently,the moment of inertia is the smallest with respect to the
axis through the center of mass of a rigid body,
? ? ??? )( 222 iiiiizC yxmrmI
? ? ??? )''(' 222' iiiiiz yxmrmI
? ????
???
])([
','
22
' dyxmI
dyyxx
iiiz
iiii?
?
? ?
?
???
ii
iiii
ymd
dmyxm
2
)()( 222
? ? ?????? 2' 0,mdIImyymmm zCzCiii?
46
? ? ??? )( 222 iiiiizC yxmrmI
? ? ??? )''(' 222' iiiiiz yxmrmI
? ????
???
])([
','
22
' dyxmI
dyyxx
iiiz
iiii?
?
? ?
?
???
ii
iiii
ymd
dmyxm
2
)()( 222
证明,设质量为 m的刚体，质心为 C,CzzO //''
? ? ?????? 2' 0,mdIImyymmm zCzCiii?
例如,对于例 1中均质细杆 z' 轴的转动惯量为
2222 3141121)2(' mlmlmllmII zz ?????
刚体对通过质心轴的转动惯量具有最小值 。
47
When a material body consists of few bodies of regular geometric
shape,computing the moments of inertia of every part (every
material body)and then adding them,we can obtain the moment of
inertia of the whole body,If a body has a hollow part,we should
treat the moment of inertia of that part as a negative quantity,
4) Combination method of the calculation of a moment of inertia,
d i s cr o d OOO III ?? 222221 )(2131 RlmRmlm ????
)423(2131 22221 lRlRmlm ????
Solution,
[Example 2] A clock pendulum is formed by
a homogeneous rod of mass m1and length l
and a circular homogeneous disc of mass
m2 and radius R,Determine IO,
48
当物体由几个规则几何形状的物体组成时，可先计算每一
部分 (物体 )的转动惯量,然后再加起来就是整个物体的转动惯量。
若物体有空心部分,要把此部分的转动惯量视为负值来处理。
4．计算转动惯量的组合法
盘杆 OOO III ?? 222221 )(2131 RlmRmlm ????
)423(2131 22221 lRlRmlm ????
解,
[例 2] 钟摆,均质直杆 m1,l ；
均质圆盘,m2,R 。 求 IO 。
49
[Example 3] In the crane the weights of the wheels A and B,
considered as circular homogeneous discs,are P1,P2 and their radii
are r1,r2,A constant moment of force M1 acts on the wheel,
Determine the acceleration when the load moves upward,
Considering the wheel B together with the load C
we obtain,
( 2 ) ')21( 232232222 rPrTvrgPrgPdtd ?????
Complementary conditions of kinematics give
112222,??? rarvr ???
Simplifying equation (1) we have,
Simplifying equation (2) we get
332 '2
2 PTa
g
PP ???
TrMagP ??
1
11
2
gPPP PrMa 22/
321
311 ?
??
??
( 1 ) 21 111211 TrMrgP ????
Solution,Taking the wheel A as the object
under consideration,we have
Therefore
50
[例 3] 提升装置中，轮 A,B的重量分别为 P1, P2,半径分别为
r1, r2,可视为均质圆盘 ; 物体 C 的重
量为 P3 ; 轮 A上作用常力矩 M1 。
求 物体 C上升的加速度。
取轮 B连同物体 C为研究对象
( 2 ) ')21( 232232222 rPrTvrgPrgPdtd ?????
补充运动学条件 112222,??? rarvr ???
化简 (1) 得,
化简 (2) 得,
332 '2
2 PTa
g
PP ???
TrMagP ??
1
11
2
gPPP PrMa 22/
321
311 ??? ???
( 1 ) 21 111211 TrMrgP ????解, 取轮 A为研究对象
51
§ 13-5 Moment of momentum theorem
for a system with respect to its center of mass,
differential equations of plane motion of a rigid body
1,The moment of momentum of a system of particles is
)( rCCrCCCO LLLvmrL ????
? ?? )()( )( eCeiCrC MFmdtLd
2,The moment of momentum theorem of a system of particles
with respect to its
center of mass is
The change of the moment of momentum of a system of particles with
respect to the center of mass only involves the external forces acting
on the system,not the internal forces,
The moment of momentum theorem of a system with respect to the
center of mass has the same mathematical form as that with respect to
a fixed center,for the moment of momentum to other moving points
this simple relation is not valid,in general,
52
§ 13-5 质点系相对于质心的
动量矩定理刚体平面运动微分方程
一．质点系动量矩 )(
rCCrCCCO LLLvmrL ????
质点系相对于质心和固定点的动量矩定理，具有完全相似
的数学形式，而对于质心以外的其它动点，一般并不存在这种
简单的关系 。
? ?? )()( )( eCeiCrC MFmdtLd二．质点系相对质心的动量矩定理
质点系相对于质心的动量矩的改变，只与作用在质点系上
的外力有关，而与内力无关。
53
3,Differential equations of plane motion of a rigid body,
Assuming that a rigid body in plane motion has a plane of symmetry
of its mass,then the force system may be simplified into a
force system in that plane,Denoting the plane of symmetry of the
mass by S,the center of mass is in the plane S,
nFFF,,,21 ???
Choosing the center of mass to coincide with the origin of the moving
coordinate system,the plane motion of the rigid body can be
decomposed into,
① a translation (all the points of the plane move
in the same way as the center of mass (xC,yC)),
② a rotation about the center of mass (?),
Employing the theorem of motion of the center
of mass and the moment of momentum theorem
with respect to the center of mass we have
??? ??CCrCCrC IIdtdLIL,???
?? ?? )(,t h e r e f o r e )( eCCC FmIFam ?
54
三．刚体平面运动微分方程
设有一平面运动刚体具有质量对称平面，力系
可以简化为该平面内的一个力系。取质量对称平面为平面图形 S，
质心一定位于 S内。
nFFF,,,21 ???
取质心 C为动系原点，则此平面运动可分解为
? 随质心 C的平动 (xC,yC)
? 绕质心 C的平动 （ ?）
可通过质心运动定理和相对质心的动量
矩定理来确定。
??? ??? CCrCCrC IIdtdLIL,???
?? ??? )(,)( eCCC FmIFam ?
55
The projected equations are
?? ? ??? )(,,)( eCCyCxC FmIYmaXma ?
or
?? ? ??? )(,,)( eCCCC FmIYymXxm ? ??????
The equations above are called the differential equations of
plane motion,
56
写成 投影形式
?? ? ??? )(,,)( eCCyCxC FmIYmaXma ?
或
?? ? ??? )(,,)( eCCCC FmIYymXxm ? ??????
上式称为 平面运动微分方程 。
57
[Example 4] A circular homogeneous wheel of mass m and radius R is
placed on a plane of inclination angle a,Under the action of gravity it start
to move,Assuming that the coefficients of the static and kinetic friction are
f and f’ respectively,Neglecting the rolling resistance analyze the motion of
the wheel,
Solution,Investigate the wheel,The force
analysis is shown in the figure,Analysis of
motion,Choosing the cartesian coordinate
system Oxy,shown in figure,we get aC y =0,
aC x =aC,Because the wheel is in plane motion
from the differential equations of plane
motion we obtain
From equation ② we get
,
The two equations ①
and ③ contain three
unknown quantities aC,
F,?,Additional
conditions have to be
complemented,
Fmgma C ?? as i n
Nmg ??? ac o s 0
FRI C ??
aco smgN ?
①
②
③
④
58
[例 4] 质量为 m半径为 R的均质圆轮置放于倾角为 a 的斜面上，
在重力作用下由静止开始运动。设轮与斜面间的静、动滑动
摩擦系数为 f,f′，不计滚动摩阻，试分析轮的运动。
解,取轮为研究对象。
受力分析如图示。
运动分析：取直角坐标系 Oxy
aC y =0,aC x =aC,
一般情况下轮作平面运动。
根据平面运动微分方程，有
Fmgma C ?? as i n
Nmg ??? ac o s 0
FRI C ??
由 ?式得
aco smgN ?
?
?
?
?
?, ?两式中含有三个
未知数 aC, F,?,需
补充附加条件。
59
1,Assuming the surface of contact to be absolutely smooth we get
Because the wheel moves
from rest,?=0,the wheel slide downward.,c o n s t,0,s i n,0 ???? ??agaF C
?raC ?
3,Assume that the wheel both slides and rolls,Complementing the
equation F=f’ N and solving the equations above we get
Therefore,the wheel rolls without slipping if
,whence,
The formula above shows,
If f < (1/3)tga the answer is given by 3;
If f > (1/3)tga the answer is given by 2;
If f = 0 the answer is given by 1,
aa?aa co s,co s'2,)co s'( s i n mgfFR gfgfa C ?????
aa c o ss i n31 m a x f m gfNFmgF ???? atg31?f
2,Let the surface of contact be rough enough,then the wheel
rolls without slipping,Complementing the equation and
solving the equations above we obtain
aa?a s in31,s in3 2,s in32 mgFgRga C ???
60
1．设接触面绝对光滑。
因为轮由静止开始运动，故 ?＝ 0，轮沿斜面平动下滑。
常量。???? ??a,0,s i n,0 gaF C
aa?a s i n31 ; s i n3 2,s i n32 mgFgRga C ???
2．设接触面足够粗糙。轮作纯滚动,所以可解得,?raC ?
3．设轮与斜面间有滑动，轮又滚又滑。 F=f′N,可解得
aa?aa co s,co s'2,)co s'( s i n mgfFR gfgfa C ?????
轮作纯滚动的条件,
aa c o ss i n31 m a x f m gfNFmgF ????
atg31?f表明：当 时，解答 3适用；
当 时，解答 2适用； f =0 时解答 1适用。
atg31?f
atg31?f
61
1,Basic concepts,
1) Moment of momentum is a measure of the intensity of the
mechanical motion of a material body at any instant,
2) The moment of momentum for a particle is
3) The moment of momentum of a system of particles is
4) The moment of inertia is a measure of the inertia of a body
in rotational motion,
For a homogeneous thin and straight rod,a thin circular ring
and a thin circular disk,we should calculate the moments of
inertia with respect to an axis perpendicular to the symmetry
plane of the masses,
Exercises
vmrvmm O ??)(
? ?? iiiO vmrL
62
一．基本概念
1．动量矩,物体某瞬时机械运动强弱的一种度量。
2．质点的动量矩,
3．质点系的动量矩,
4．转动惯量,物体转动时惯性的度量。
vmrvmm O ??)(
? ?? iiiO vmrL
对于均匀直杆，细圆环，薄圆盘（圆柱）对过质心垂直于
质量对称平面的转轴的转动惯量要熟记。
第十三章 动量矩定理习题课
63
5) Calculation of the moment of momentum of a rigid body,
Translational motion,
Fixed-axis rotation,
Plane motion,
2.Theorem of moment momentum and the conservation
law of the moment of momentum of a particle,
1) Moment of momentum theorem of a particle,
2) Conservative law of the moment of momentum of a particle,
① If then Const.,
② If then Const,
)(,CzzCCO vmmLvmrL ???
??? zz IL
???? CCzz IvmmL )(
)()(o r )()]([ FmvmmdtdFmvmmdtd zzOO ??
0)( ?Fm O
0)( ?Fm z
?)( vmm O
?)( vmm z
64
5．刚体动量矩计算
平动,
定轴转动,
平面运动,
)(,CzzCCO vmmLvmrL ???
??? zz IL
???? CCzz IvmmL )(
二．质点的动量矩定理及守恒
1．质点的动量矩定理
)()( )()]([ FmvmmdtdFmvmmdtd zzOO ?? 或
2．质点的动量矩守恒
? 若,则 常矢量。
? 若,则 常量。
0)( ?Fm O
0)( ?Fm z
?)( vmm O
?)( vmm z
65
3,Moment of momentum theorem and its conservation law
of a system of particles,
?? ???? )()()()( )(o r )( ezezzeOeOO MFmdtdLMFmdtLd
2) The conversation law of moment of momentum of a system of
particles,
? If, then constant vector,
? If, then const,
0)( ?eOm
0)( ?ezm
?OL
?zL
)(
)( o r e
zC
zCe
C
C M
dt
dLM
dt
Ld ??
4,The moment of momentum theorem of a system with
respect to the center of mass of the system,
1)The moment of momentum theorem of a system of particles,
66
三．质点系的动量矩定理及守恒
1．质点系的动量矩定理
?? ???? )()()()( )( )( ezezzeOeOO MFmdtdLMFmdtLd 或
2．质点系的动量矩守恒
? 若,则 常矢量
? 若,则 常量
0)( ?eOm
0)( ?ezm
?OL
?zL
)(
)( e
zCzC
e
CC Mdt
dLM
dt
Ld ?? 或
四．质点系相对质心的动量矩定理
67
?? ?? )(o r )( z FmIFmI zzz ?? ??
5,Differential equations of fixed-axis motion and plane
motion of a rigid body,
1) Differential equations of fixed-axis motion of a rigid body,
2) Differential equations of plane motion of a rigid body,
or
?? Xma Cx
?? Yma Cy
)(?? FmI CC ?
?? Xxm C??
?? Yym C??
?? )( FmI CC ???
68
???? )( )( z FmIFmI zzz ?? ??或
五．刚体定轴转动微分方程和刚体平面运动微分方程
1．刚体定轴转动微分方程
2．刚体平面运动微分方程
或
?? Xma Cx
?? Yma Cy
)(?? FmI CC ?
?? Xxm C??
?? Yym C??
?? )( FmI CC ???
69
6,Application of the moment of momentum theorem,Applying
this moment of momentum theorem we may solve the following
problems (it is especially convenient for monaxial transmission
system),
1) Knowing the state of the rotational motion of a system,determine the
external forces or external torques acting on the system,
2) Knowing constant or time-dependent external torques acting on a system,
determine the angular velocity or the angular acceleration of the rigid body,
3) Knowing that the principal moment or the algebraic sum of the
projections on any axis of the moments of all the external forces acting on a
system is zero,applying the conversation law of the moment of momentum,
determine the angular displacement or angular velocity,
70
六．动量矩定理的应用
应用动量矩定理，一般可以处理下列一些问题：（对单轴
传动系统尤为方便）
1．已知质点系的转动运动，求系统所受的外力或外力矩。
2．已知质点系所受的外力矩是常力矩或时间的函数，求刚体
的角加速度或角速度的改变。
3．已知质点所受到的外力主矩或外力矩在某轴上的投影代数
和等于零，应用动量矩守恒定理求角速度或角位移。
71
7,Examples of application
[Example1] A circular homogeneous cylinder of radius r and weight Q is
placed in a corner as shown in the figure,The initial angular velocity is ?0
and the coefficient of kinetic friction of the contact at the wall and the floor
are f′,Neglecting rolling resistance determine the time until the rotation of
the cylinder stops,
Solution,Choose the cylinder to be investigated (noting that
it is a rigid body)
The force analysis is shown in the figure,Analysis of Motion,
The center of mass does not move,the rigid body rotates
around it,Applying the differential equations of plane
motion of a rigid body we get
)0,0( ?? CyCx aa
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
①
②
③
The complementary equations are
BBAA NfFNfF ' a n d ' ??
④
72
七．应用举例
[例 1] 均质圆柱，半径为 r，重量为 Q，置圆柱于墙角。初始角
速度 ?0，墙面、地面与圆柱接触处的动滑动摩擦系数均为 f '，
滚阻不计，求使圆柱停止转动所需要的时间。
解,选取圆柱为研究对象。 (注意只是一个刚
体 )受力分析如图示。
运动分析：质心 C不动，刚体绕质心转动。
根据刚体平面运动微分方程 )0,0( ??
CyCx aa
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
?
?
?
补充方程,
BBAA NfFNfF ',' ?? ?
73
Substituting equation ④ into equations ① and ② we get
0)1'( 2 ??? QNf B
1'
',
1'
',
1'
',
1' 2
2
222 ???????? f
QfF
f
QfN
f
QfF
f
QN
AABB
Substituting the results above into equation③ we get
dtf frgfdr gff fdtd t?? ?????????? 0202 '1 '1'2,2''1 '1 0? ??
Solving it we obtain
)'1('2
)'1( 02
fgf
rft
?
?? ?
74
将 ?式代入 ?,?两式，有 0)1'( 2 ??? QNf B
1'
',
1'
',
1'
',
1' 2
2
222 ????????? f
QfF
f
QfN
f
QfF
f
QN
AABB
将上述结果代入 ?式，有
dtf frgfdr gff fdtd t?? ?????????? 0202 '1 '1'2,2''1 '1 0? ??
解得,
)'1('2
)'1( 02
fgf
rft
?
?? ?
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
?
?
?
补充方程,
BBAA NfFNfF ',' ?? ?
75
[Example 2] A rod formed by two thin homogeneous rods fixed together takes
the shape of the letter,T”,It rotates around a horizontal axis through the
center O,The angular velocity of the rod is ?=4rad/s when OA is in the
horizontal position,Determine the reaction of the bearing O at that moment,
Solution,Choose the rod to be investigated,
The force analysis is shown in the diagram,
Applying the differential equations of rotation
around a fixed axis we get
r a d /s 2 0,7 5
5.08.9825.08.98 5.081217
2
2
?
?????????
?
?
5.025.0 ???? mgmgI O ?
2222 121712131 mlmlmlmlI O ????
76
[例 2] 两根质量各为 8 kg的均质细杆固连成 T 字型，可绕通过 O
点的水平轴转动，当 OA处于水平位置时,T 形杆具有角速度 ?
=4rad/s 。求该瞬时轴承 O的反力。
解,选 T 字型杆为研究对象。
受力分析如图示。
r a d /s 2 0,7 5
5.08.9825.08.98 5.081217
2
2
?
?????????
?
?
5.025.0 ???? mgmgI O ?
2222 121712131 mlmlmlmlI O ????
由定轴转动微分方程
77
Applying the theorem of the motion of the center of mass we obtain
OxCxC Xmama ??? 21
mgmgYmama OyCyC ????? 21
N 96) 5.04 25.04( 8)( 2221 ??????????? xCxCO aamX
N 3.32 ) 5.075.20 25.075.20 ( 88.982 ????????OY
78
根据质心运动微分方程，得
OxCxC Xmama ??? 21
mgmgYmama OyCyC ????? 21
N 96) 5.04 25.04( 8)( 2221 ??????????? xCxCO aamX
N 3.32 ) 5.075.20 25.075.20 ( 88.982 ????????OY
79
[Example 3] Two circular homogeneous cylinders both are of
weight P and of radius r,An inextensible string is wounded
on the cylinder A,The other end of the of the string is wounded
on the cylinder B,Neglecting the mass of the string and the friction
of the axis O determine
1 the acceleration of the center of mass of the cylinder B when it falls,
2 if a counterclockwise torque M acts on the cylinder A,on which condition
the center of mass of the cylinder B will move upwards?
80
[例 3] 均质圆柱体 A和 B的重量均为 P，半径均为 r，一绳缠在
绕固定轴 O转动的圆柱 A上，绳的另一端绕在圆柱 B上，绳重
不计且不可伸长，不计轴 O处 摩擦。
求, ? 圆柱 B下落时质心的加速度。
? 若在圆柱体 A上作用一逆时针转向的转矩 M，试问在什么
条件下圆柱 B的质心将上升。
81
Take cylinder B as the object to be
investigated
rTrgP B '21 2 ??
'TPagP C ??
②
③
Applying the relation of kinematics,we have,
BAC rra ?? ??
④
TrrgP A ??221
①
Solution Choose cylinder A as the object to be investigated,
,52 rgBA ???? ga C 54 ?
From equation ① and ② we obtain
BA ?? ?
Substituting this equation into
equations ③ and ④ we obtain
82
选圆柱 B为研究对象
rTrgP B '21 2 ??
'TPagP C ??
?
?
运动学关系,
BAC rra ?? ??
?
TrrgP A ??221
?
解：选圆柱 A为研究对象
由 ?,?式得,
BA ?? ?
,52 rgBA ???? ga C 54 ?
代入 ?,?式得,
83
Applying the moment of momentum theorem with
respect to the axis O,we get
rPMMrgPrvgPrgPdtd eOBCA ???????? 2)222( )(22 ??
Choosing the whole system to be investigated we get
BCAO rg
Prv
g
Pr
g
PL ?? 22
222 ?????
rPMM eO 2)( ???
and substituting it into equation (1) we obtain
rPMrgPragPrgP BcA ?????? 2222 T h e r e f o r e 22 ??
(1)
BAC rra ?? ??
grP rPMarPMargPargP CCC ? ?????????? 5 )2(2 ; 222
Complementing the equation of motion
When M >2Pr,,hence,the center of mass
of cylinder B will move upwards,0?Ca
84
由动量矩定理,
rPMMrgPrvgPrgPdtd eOBCA ???????? 2)222( )(22 ??
rPMrgPragPrgP BcA ??????? 2222 22 ?? (1)
补充运动学关系式,BAC rra ?? ?? 代入 (1)式，得
grP rPMarPMargPargP CCC ? ?????????? 5 )2(2 ; 222
当 M >2Pr 时，,圆柱 B的质心将上升。 0?
Ca
再取系统为研究对象
BCAO rg
Prv
g
Pr
g
PL ?? 22
222 ?????
rPMM eO 2)( ???
85
When investigating the problems of dynamics for a rigid body in
plane motion,we have to find the dependence between the motion of
the center of mass and the rotational motion of the rigid body,
When writing equations according to the moment of momentum
theorem with respect to an axis,we have to care for the identity of the
definitions of the sign,
86
研究刚体平面运动的动力学问题，一定要建立补充方程，找
出质心运动与刚体转动之间的联系。
应用动量矩定理列方程时,要特别注意正负号的规定的一致
性。
87
88
Theoretical Mechanics
2
3
§ 13–1 Moment of momentum
§ 13–2 moment of momentum theorem
§ 13–3 Differential equations for the rotation of a rigid body
around a fixed-axis
§ 13–4 Moment of inertia of a rigid body with respect to an axis
§ 13–5 moment of momentum theorem for a system with respect
to its center of mass,differential equations of plane motion
of a rigid body
Exercises
Chapter 13,Moment of Momentum Theorem
4
§ 13–1 动量矩
§ 13–2 动量矩定理
§ 13–3 刚体定轴转动微分方程
§ 13–4 刚体对轴的转动惯量
§ 13–5 质点系相对于质心的动量矩定理,
刚体平面运动微分方程
习题课
第十三章 动量矩定理
5
Theorem of momentum,
The change of the momentum of a particle (or of a system of
particles) is the result of external forces (the principal vector of an
external force system),
Theorem of motion of the center of mass,
The motion of the center of mass is the result of external forces
(the principal vector of an external force system),
When the center of mass coincides with a certain point of a fixed
axis,then the momentum of a rotating body is always zero because
Vc=0,But in this case the system is still subjected to the action of
external forces,The moment of momentum theorem establishes
the dependence between the change of the moment of momentum
of a particle or a system with respect to a center (or a fixed axis)
and the torque given by all external forces acting on the particle or
the system with respect to the same center (or axis),
6
质点
质点系 动量定理,动量的改变 — ?外力（外力系主矢）
若当质心为固定轴上一点时,vC=0，则其动量恒等于零,
质心无运动，可是质点系确受外力的作用。 动量矩定理建立了
质点和质点系相对于某固定点（固定轴）的动量矩的改变与外
力对同一点（轴）之矩两者之间的关系。
质心运动定理,质心的运动 — ?外力（外力系主矢）
7
§ 13-1 Moment of momentum
Moment of momentum of a particle with respect to a center O is
,it is a vector,
Moment of momentum of a particle with respect to an axis Z is
,it is an algebraic quantity,
vmrvmm O ??)(
)()( xyOz vmmvmm ?
O A Bvmm O ?2)( ?
''2)( BOAvmm z ???The definition of the sign of the
moment of momentum with respect
to an axis is the same as that of the
moment of a force with respect to an
axis,Looking from the positive end
of the axis,it’s positive if it is
counterclockwise,and it is negative if
it is clockwise,
8
§ 13-1 动量矩
一．质点的动量矩
质点对点 O的动量矩,矢量
质点对轴 z 的动量矩,代数量
vmrvmm O ??)(
)()( xyOz vmmvmm ?
O A Bvmm O ?2)( ?
''2)( BOAvmm z ???
正负号规定与力对轴矩的规定相同
对着轴看：顺时针为负
逆时针为正
9
The relation between the moment of momentum of a particle with
respect to an axis and a center is
2,Moment of momentum of a system of particles,
The moment of momentum of a system with respect to a center O is
The moment of momentum of a system with respect to an axis Z is iiiiiOO vmrvmmL ??? ?? )(
? ? zOiizz LvmmL )( ?? ?
kg·ｍ 2/s,
Moment of momentum measures the intensity of rotation of a
body rotating around a fixed center or an axis at any instant,
? ? )( )( vmmvmm zzO ?
Calculation of moment of momentum of a rigid body,
1) Rigid body in translational motion
CCCOO vmrvmmL ??? )(
)( CCCiiiii vmrvrmvmr ????? ?? )( Czz vmmL ?
The moment of momentum of a rigid body in translational motion
with respect to a fixed center (axis) is equal to the moment of
momentum of the center of mass of the rigid body with respect to
that center (axis),
10
质点对点 O的动量矩与对轴 z 的动量矩之间的关系,
二．质点系的动量矩
质系对点 O动量矩,
质系对轴 z 动量矩,
iiiiiOO vmrvmmL ??? ?? )(
? ? zOiizz LvmmL )( ?? ?
kg·ｍ 2/s。 动量矩度量物体在任一瞬时绕固定点 (轴 )转动的强弱
? ? )( )( vmmvmm zzO ?
刚体动量矩计算,
1．平动刚体 CCCOO vmrvmmL ??? )(
)( CCCiiiii vmrvrmvmr ????? ??
)( Czz vmmL ?
平动刚体对固定点（轴）的动量矩等于刚体质心的动量
对该点（轴）的动量矩。
11
3) Rigid body in plane motion,
The moment of momentum of a rigid body in plane motion with
respect to an axis perpendicular to the symmetrical plane of mass is
equal to the sum of the moment of momentum of the center of mass
of the rigid body in translational motion together with the center of
mass about that axis and the moment of momentum of the rigid
body when rotating around the center of mass with respect to a
parallel axis through the center of mass of the body,
?? ?????? ziiiizz IrmvmmL 2)(
???? CCzz IvmmL )(
2) Rigid body in fixed-axis rotation,
The moment of momentum of a rigid body in fixed-axis rotation
with respect to the axis of rotation is equal to the product of the
moment of inertia of the rigid body with respect to that axis and its
angular velocity,
12
3．平面运动刚体
平面运动刚体对垂直于质量对称平面的固定轴的动量矩，等于
刚体随同质心作平动时质心的动量对该轴的动量矩与绕质心轴
作转动时的动量矩之和。
?? ?????? ziiiizz IrmvmmL 2)(
???? CCzz IvmmL )(
2．定轴转动刚体
定轴转动刚体对转轴的动量矩等于刚体对该轴转动惯量与角速
度的乘积。
13
112223 2
1 ?? RRvv ???
.)( 32322
2
2
2
2
1 vRmm
R
I
R
IL
O ????
OCOBOAO LLLL ???
2332222211 )( RvmRvmII ???? ??
Solution,
[Example 1] Pulley A,m1,R1,R1=2R2,I1
Pulley B,m2,R2,I2 ；
Material body C,m3
Determine the moment of momentum of the
system with respect to the axis O,
14
112223 2
1 ?? RRvv ???
32322
2
2
2
2
1 )( vRmm
R
I
R
IL
O ????
OCOBOAO LLLL ???
2332222211 )( RvmRvmII ???? ??
解,
[例 1] 滑轮 A,m1,R1,R1=2R2,I1
滑轮 B,m2,R2,I2 ；物体 C,m3
求 系统对 O轴的动量矩。
15
§ 13-2 Moment of momentum theorem
1,Moment of momentum theorem of a particle,
vmdt rdvmrdtddt vmdr ????? )()(
,)( a n d 0 B u t,FmFrvmvvmdt rd O??????
)()]([,)( FmvmmdtdFrvmrdtd OO ????
The time-derivative of the moment of momentum of a particle
with respect to any fixed center with respect to time is equal to the
moment of the force acting on the particle with respect to the same
center,This is the moment of momentum theorem of a particle,
Therefore,
Fdt vmd ?)(
Frdt vmdr ??? )(
r
, Now multiplying both sides of
this equation by the radius vector according
to vector multiplication,we get,
The left-hand side can be expressed in the form
16
Fdt vmd ?)(
§ 13-2 动量矩定理
一．质点的动量矩定理
两边叉乘矢径,有 Fr
dt
vmdr ??? )(r
左边可写成
vmdt rdvmrdtddt vmdr ????? )()(
,)(,0 FmFrvmvvmdt rd O??????而
)()]([,)( FmvmmdtdFrvmrdtd OO ????
质点对任一固定点的动量矩对时间的导数，等于作用在质
点上的力对同一点之矩。这就是 质点对固定点的动量矩定理。
故,
17
Projecting the equation given above on the axes of a Cartesian
coordinate system through a fixed center O we obtain
)()( ),()( ),()( FmvmmdtdFmvmmdtdFmvmmdtd zzyyxx ???
These equations are the moment of momentum theorem of a
particle with respect to a fixed axis,They are also called the
projection forms of the principle of moments with respect to a fixed
axis,The derivative of the moment of momentum of a particle with
respect to any fixed axis is equal to the moment of the force acting
on the particle with respect to the same axis,
If
These results express the conservation law of the moment of
momentum of a particle,
)0)(( 0)( ?? FmFm zO then ?)( vmm O const,c o n s t ))(( ?vmm z
18
将上式在通过固定点 O的三个直角坐标轴上投影，得
)()( ),()( ),()( FmvmmdtdFmvmmdtdFmvmmdtd zzyyxx ???
上式称 质点对固定轴的动量矩定理,也称为质点动量矩定
理的投影形式。即质点对任一固定轴的动量矩对时间的导数，
等于作用在质点上的力对同一轴之矩。
称为 质点的动量矩守恒 。
若 )0)(( 0)( ?? FmFm
zO 则 ?)( vmm O
常矢量 ))(( 常量?vmm z
19
The force analysis is shown in figure,
By the analysis of motion we obtain
?? ?? 2)( mllmlvmm O ??
OMlv ??,??
Applying the moment of momentum theorem we have
,hence,
When the pendulum swing with small amplitude we can use,
Let,then,Solving the differential equation and
taking at into account,we obtain the equation of
motion of the simple pendulum for small amplitude as
The period of the pendulum is
)()( Fmvmmdtd OO ? 0s i n,s i n)( 2 ???? ???? lgm g lmldtd ???
sin ?? ?
l
g
n ?
2? 02 ?? ??? n??
tlgc o s0?? ?
l
gT ?2?
?s i n)()()( m g lgmmTmFm OOO ????
Solution Choose the small ball m as a particle,
[Example 2] A pendulum of mass m and length l is
released from ?= ?0 at t =0,Deduce its equation of
motion,
0,,0 00 ??? ??? ?t
20
运动分析,。 ?? ?? 2)( mllmlvmm
O ??OMlv ??,??
由动量矩定理
即
)()( Fmvmmdtd OO ?
0s i n,s i n)( 2 ???? ???? lgm g lmldtd ???
微幅摆动时,并令,则,sin ?? ?
l
g
n ?
2? 02 ?? ??? n??
解微分方程,并代入初始条件 则运动方程 )0,,0(
00 ??? ??? ?t
tlgc o s0?? ?,摆动周期
l
gT ?2?
?s i n)()()( m g lgmmTmFm OOO ????
解,将小球视为质点。
受力分析；受力图如图示。
[例 2] 单摆 已知 m,l,t =0时 ?= ?0，从静止
开始释放。 求 单摆的运动规律。
21
Note,When we calculate a moment of momentum and a
moment of force,the definition of their signs should be
identical (here we call a moment positive if it tends to rotate
the pendulum counterclockwise),
Apply the moment of momentum theorem of
a particle when
1) the particle is subjected to a central force,
2) the particle rotates around a center (an axis),
22
注：计算动量矩与力矩时，符号规定应一致（本题规定逆时
针转向为正）
质点动量矩定理的应用,
?在质点受有心力的作用时。
?质点绕某心（轴）转动的问题。
23
2,Moment of momentum theorem of a system of
particles,
On the left-hand side we exchange the order of the operations
(summation and derivative) and get
hence,,
,0)( ),( )( ?? ?? iiOiiOO FmvmmL
? ?? )()( )( eOeiOO MFmdtLd
This is the moment of momentum
theorem of a system with respect to
any fixed center,
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ?????? ???
For the system of particles,
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ??????For the particle Mi,,
24
二．质点系的动量矩定理
左边交换求和与导数运算的顺序，而
则,0)( ),( )( ?? ?? iiOiiOO FmvmmL
? ?? )()( )( eOeiOO MFmdtLd 一 质点系对固定点的动量矩定理
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ?????? ???对质点系，有
),,3,2,1( )()()( )()( niFmFmvmmdtd eiOiiOiiO ??????对质点 Mi,
25
The time-derivative of the total moment of momentum of a system
with respect to any fixed center is equal to the geometric sum (the
principal moment )of the moments of all the external forces acting
on the same system with respect to the same center,
??? ?????? )()()()()()( )(,)(,)( ezeizzeyeiyyexeixx MFmdtdLMFmdtdLMFmdtdL
Projecting the equation above on the axes of a Cartesian
coordinate system through a fixed center O we obtain
26
质点系对任一固定点的动量矩对时间的
导数，等于作用在质点系上所有外力对同一
点之矩的矢量和（外力系的主矩）。
??? ?????? )()()()()()( )(,)(,)( ezeizzeyeiyyexeixx MFmdtdLMFmdtdLMFmdtdL
将上式在通过固定点 O的三个直角坐标轴上投影，得,
27
The formulae above are called the moment of momentum theorem of
a system of particles with respect to any fixed axis, The time-
derivative of the total moment of momentum of a system with respect
to any fixed axis is equal to the algebraic sum of the moments of all
the external forces acting on the system with respect to the same axis
(the principal moment of external force system with respect to the
same axis),
The theorem states that only external forces can change the moment
of momentum of the system,internal forces ca not do this,
The conversation law of the total moment of momentum of a system,
① If then constant vector,
② If then const,
0)( ?eOM
0)( ?ezM
?OL
?zL
28
上式称为 质点系对固定轴的动量矩定理 。即质点系对任一固
定轴的动量矩对时间的导数，等于作用在质点系上所有外力对同
一固定轴之矩的代数和（外力系对同一轴的主矩）。
质点系的动量矩守恒
?当 时,常矢量。
?当 时,常量。
0)( ?eOM
0)( ?ezM
?OL
?zL
定理说明内力不会改变质点系的动量矩，只有外力才能改
变质点系的动量矩。
29
[Example3], D et er m i n e,k n o w n ar e an d,?rPPP
BA ?
Solution Take the whole system as the object to
be investigated,The force analysis is shown in
the diagram,The analysis of motion gives
v =ｒ ?,
rPPrPrPM BABAeO )()( ????
?OBAO IrvgPrvgPL ?????
)2(
2 P
PPgrL BAO ??? ?
Applying the moment of momentum theorem we obtain
rPPPPPgrdtd BABA )()]2([ 2 ?????
2/PPP
PP
r
g
dt
d
BA
BA ?? ????? ??
2
2
1 r
g
PI
O ?
Substituting into the equation we get
30
解, 取整个系统为研究对象,
受力分析如图示。
运动分析,v =ｒ ?
rPPrPrPM BABAeO )()( ????
?OBAO IrvgPrvgPL ?????
)2(,21 22 PPPgrLrgPI BAOO ???? ?得代入将
由动量矩定理,
rPPPPPgrdtd BABA )()]2([ 2 ?????
2/PPP
PP
r
g
dt
d
BA
BA ?? ????? ??
[例 3] 已知, 。求。 ; ; ?rPPP
BA ?
31
Solution so the moment of
momentum of the system does not change,
? ?,0)( )( eO Fm
rvvmrvm ABAA )(0 ??? 2vv A ?
Therefore,the absolute velocities of the
monkeys A and B are identical,Both
are,
2v
[Example 4] The weight of two monkeys A an B are knowing,The
monkey B climbs upwards with a velocity v relative to the rope,while
the monkey A does not move relative to the rope,Determine
(1)the motion of the monkey A move when the monkey B climbs
upward,
(2)the velocity of the monkey A (neglecting the weight of the rope),
32
解, 系统的动量矩守恒。 ? ??,0)( )( e
O Fm
rvvmrvm ABAA )(0 ???
2
vv
A ?
猴 A与猴 B向上的绝对速度是一样的,
均为 。
2v
[例 4] 已知：猴子 A重 =猴子 B重，猴 B以相对绳速度
上爬，猴 A不动，问当猴 B向上爬时，猴 A将如何动？
动的速度多大？（轮重不计）
v
33
§ 13-3 Differential equations for
the rotation of a rigid body around a fixed-axis
For a rigid body in fixed-axis rotation we have ?
zz IL ?
)()( e
zz MIdt
d ??
)(
2
2)(
o r ezzezz MdtdIMI ?? ??
Substituting this equation into the formula of the moment of
momentum theorem,we get
This is the differential equation of
fixed-axis rotation for a rigid body,Two types of problems can be solved in this way,
(1) Knowing the moment of the external forces acting on a rigid
body we can determine the law of the rotational motion of the
rigid body,
(2) Knowing the law of the rotational motion of a rigid body we
can determine the external forces (or moment of force ) acting
on the body,but we can not determine the reactions at bearings,
For this problem,the theorem of motion of the center of mass
has to be applied,
34
§ 13-3 刚体定轴转动微分方程
对于 一个定轴转动刚体
代入质点系动量矩定理，有
?zz IL ?
)()( e
zz MIdt
d ??
)(
2
2)( e
zzezz Mdt
dIMI ??? ?? 或— 刚体定轴转动微分方程
解决两类问题,
?已知作用在刚体的外力矩，求刚体的转动规律。
?已知刚体的转动规律，求作用于刚体的外力（矩）。
但不能求出轴承处的约束反力，需用质心运动定理求解。
35
Special cases are,
? If,then const,the rigid body will
rotate uniformly or remain at rest,
? If const,then ? =const,the rigid body will be in uniform
rotation,Comparing with,we see that the moment
of inertia is a measure of a body’s inertia concerning its rotational
motion,
? ?? 0)( )()( ezez FmM ???,0
?)(ezM
)( ezz MI ?? Fam ?
zI
36
特殊情况,
? 若,则 恒量，刚体作匀速转动或
保持静止。
? 若 常量，则 ? =常量，刚体作匀变速转动。
将 与 比较，刚体的转动惯量 是刚体
转动惯性的度量。
? ?? 0)( )()( ezez FmM ?? ??,0
?)(ezM
)( ezz MI ?? Fam ?
zI
37
§ 13-4 Moment of inertia of
a rigid body with respect to an axis
1,Definition,
If the mass in a rigid body is
distributed continuously,then
?? 2iiz rmI
?? dmrI mz 2
The moment of inertia is a measure of a rigid body’s inertia
concerning its rotational motion around a fixed axis,Its
magnitude shows whether it is difficult or easy for the rotational
state of a rigid body to be changed,The moment of inertia is
always positive,The unit for it in the SI system is kg·m2,
38
§ 13-4 刚体对轴的转动惯量
一．定义,
若刚体的质量是连续分布，则
?? 2iiz rmI
?? dmrI mz 2
刚体的转动惯量是刚体对某轴转动惯性大小的度量，它的
大小表现了刚体转动状态改变的难易程度。
转动惯量恒为正值，国际单位制中单位 kg·m2 。
39
1)Method of integration
This method may be applied to any homogenous rigid body with
regular geometric shape,
[Example 1] There is a thin homogeneous rod
of length and mass m,Determine
(1) its moment of inertia IZ with respect to the axis Z and
(2) its moment of inertia IZ with respect to the axis Z,
2,Calculation of the moment of Inertia,
222
2 12
1 mldx
l
mxI l
lz ? ??
2
0
2' 31 mldxlmxI lz ? ???
Solution,
40
１．积分法 （具有规则几何形状的均匀刚体可采用）
[例 1] 匀质细直杆长为 l,质量为 m 。
求, ?对 z轴的转动惯量 ；
?对 z' 轴的转动惯量 。
zI
'zI
二．转动惯量的计算
222
2 12
1 mldx
l
mxI l
lz ? ??
2
0
2' 31 mldxlmxI lz ? ???
解,
41
2) Radius of gyration
The radius of gyration of a rigid body with respect to an axis Z is
defined by the equation,
and,
m
I z??
z?
2zz mI ??
For a homogeneous rigid body,the radius of gyration is given
only by its geometric shape,but not by its density,Two
homogenous rigid bodies with identical shapes,but made of
different materials (different densities),have the same radius of
gyration,
In the designing handbooks of mechanical engineering the moments
of inertia and the radius of gyration for parts with simple geometric
shapes can be found,The and of some common homogenous
rigid body are given in these books,too,
z?
zI z?
42
2,回转半径
由 所定义的长度 称为刚体对 z 轴的回转半径。
m
I z?? z?
2zz mI ??
对于均质刚体,仅与几何形状有关，与密度无关。对
于几何形状相同而材料不同（密度不同）的均质刚体，其回
转半径是相同的。
z?
在机械工程设计手册中，可以查阅到简单几何形状或已
标准化的零件的转动惯量和回转半径。书中列出几种常见均质
刚体的,以供参考。
zzI ?和
43
3) The parallel axis theorem,
The moments of inertia of a body with respect to different axes
are generally different,
2' mdII zCz ??
The moment of inertia of a rigid body with respect to any axis is
equal to the moment of inertia of the body with respect to a
parallel axis through the center of mass of the body plus the
product of the mass of the body with the square of the distance d
between the two axes,
44
3,平行移轴定理
同一个刚体对不同轴的转动惯量一般是不相同的。
2' mdII zCz ??
刚体对某轴的转动惯量等于刚体对通过质心且与该轴平行的
轴的转动惯量，加上刚体的质量与两轴间距离的平方之乘积。
45
Proof,Suppose that the center of mass of a rigid body of mass M is
C and, CzzO //''
In example 1 e.g.,the moment of inertia of the thin homogeneous rod
with respect to the axis Z′ is
2222 3141121)2(' mlmlmllmII zz ?????
Consequently,the moment of inertia is the smallest with respect to the
axis through the center of mass of a rigid body,
? ? ??? )( 222 iiiiizC yxmrmI
? ? ??? )''(' 222' iiiiiz yxmrmI
? ????
???
])([
','
22
' dyxmI
dyyxx
iiiz
iiii?
?
? ?
?
???
ii
iiii
ymd
dmyxm
2
)()( 222
? ? ?????? 2' 0,mdIImyymmm zCzCiii?
46
? ? ??? )( 222 iiiiizC yxmrmI
? ? ??? )''(' 222' iiiiiz yxmrmI
? ????
???
])([
','
22
' dyxmI
dyyxx
iiiz
iiii?
?
? ?
?
???
ii
iiii
ymd
dmyxm
2
)()( 222
证明,设质量为 m的刚体，质心为 C,CzzO //''
? ? ?????? 2' 0,mdIImyymmm zCzCiii?
例如,对于例 1中均质细杆 z' 轴的转动惯量为
2222 3141121)2(' mlmlmllmII zz ?????
刚体对通过质心轴的转动惯量具有最小值 。
47
When a material body consists of few bodies of regular geometric
shape,computing the moments of inertia of every part (every
material body)and then adding them,we can obtain the moment of
inertia of the whole body,If a body has a hollow part,we should
treat the moment of inertia of that part as a negative quantity,
4) Combination method of the calculation of a moment of inertia,
d i s cr o d OOO III ?? 222221 )(2131 RlmRmlm ????
)423(2131 22221 lRlRmlm ????
Solution,
[Example 2] A clock pendulum is formed by
a homogeneous rod of mass m1and length l
and a circular homogeneous disc of mass
m2 and radius R,Determine IO,
48
当物体由几个规则几何形状的物体组成时，可先计算每一
部分 (物体 )的转动惯量,然后再加起来就是整个物体的转动惯量。
若物体有空心部分,要把此部分的转动惯量视为负值来处理。
4．计算转动惯量的组合法
盘杆 OOO III ?? 222221 )(2131 RlmRmlm ????
)423(2131 22221 lRlRmlm ????
解,
[例 2] 钟摆,均质直杆 m1,l ；
均质圆盘,m2,R 。 求 IO 。
49
[Example 3] In the crane the weights of the wheels A and B,
considered as circular homogeneous discs,are P1,P2 and their radii
are r1,r2,A constant moment of force M1 acts on the wheel,
Determine the acceleration when the load moves upward,
Considering the wheel B together with the load C
we obtain,
( 2 ) ')21( 232232222 rPrTvrgPrgPdtd ?????
Complementary conditions of kinematics give
112222,??? rarvr ???
Simplifying equation (1) we have,
Simplifying equation (2) we get
332 '2
2 PTa
g
PP ???
TrMagP ??
1
11
2
gPPP PrMa 22/
321
311 ?
??
??
( 1 ) 21 111211 TrMrgP ????
Solution,Taking the wheel A as the object
under consideration,we have
Therefore
50
[例 3] 提升装置中，轮 A,B的重量分别为 P1, P2,半径分别为
r1, r2,可视为均质圆盘 ; 物体 C 的重
量为 P3 ; 轮 A上作用常力矩 M1 。
求 物体 C上升的加速度。
取轮 B连同物体 C为研究对象
( 2 ) ')21( 232232222 rPrTvrgPrgPdtd ?????
补充运动学条件 112222,??? rarvr ???
化简 (1) 得,
化简 (2) 得,
332 '2
2 PTa
g
PP ???
TrMagP ??
1
11
2
gPPP PrMa 22/
321
311 ??? ???
( 1 ) 21 111211 TrMrgP ????解, 取轮 A为研究对象
51
§ 13-5 Moment of momentum theorem
for a system with respect to its center of mass,
differential equations of plane motion of a rigid body
1,The moment of momentum of a system of particles is
)( rCCrCCCO LLLvmrL ????
? ?? )()( )( eCeiCrC MFmdtLd
2,The moment of momentum theorem of a system of particles
with respect to its
center of mass is
The change of the moment of momentum of a system of particles with
respect to the center of mass only involves the external forces acting
on the system,not the internal forces,
The moment of momentum theorem of a system with respect to the
center of mass has the same mathematical form as that with respect to
a fixed center,for the moment of momentum to other moving points
this simple relation is not valid,in general,
52
§ 13-5 质点系相对于质心的
动量矩定理刚体平面运动微分方程
一．质点系动量矩 )(
rCCrCCCO LLLvmrL ????
质点系相对于质心和固定点的动量矩定理，具有完全相似
的数学形式，而对于质心以外的其它动点，一般并不存在这种
简单的关系 。
? ?? )()( )( eCeiCrC MFmdtLd二．质点系相对质心的动量矩定理
质点系相对于质心的动量矩的改变，只与作用在质点系上
的外力有关，而与内力无关。
53
3,Differential equations of plane motion of a rigid body,
Assuming that a rigid body in plane motion has a plane of symmetry
of its mass,then the force system may be simplified into a
force system in that plane,Denoting the plane of symmetry of the
mass by S,the center of mass is in the plane S,
nFFF,,,21 ???
Choosing the center of mass to coincide with the origin of the moving
coordinate system,the plane motion of the rigid body can be
decomposed into,
① a translation (all the points of the plane move
in the same way as the center of mass (xC,yC)),
② a rotation about the center of mass (?),
Employing the theorem of motion of the center
of mass and the moment of momentum theorem
with respect to the center of mass we have
??? ??CCrCCrC IIdtdLIL,???
?? ?? )(,t h e r e f o r e )( eCCC FmIFam ?
54
三．刚体平面运动微分方程
设有一平面运动刚体具有质量对称平面，力系
可以简化为该平面内的一个力系。取质量对称平面为平面图形 S，
质心一定位于 S内。
nFFF,,,21 ???
取质心 C为动系原点，则此平面运动可分解为
? 随质心 C的平动 (xC,yC)
? 绕质心 C的平动 （ ?）
可通过质心运动定理和相对质心的动量
矩定理来确定。
??? ??? CCrCCrC IIdtdLIL,???
?? ??? )(,)( eCCC FmIFam ?
55
The projected equations are
?? ? ??? )(,,)( eCCyCxC FmIYmaXma ?
or
?? ? ??? )(,,)( eCCCC FmIYymXxm ? ??????
The equations above are called the differential equations of
plane motion,
56
写成 投影形式
?? ? ??? )(,,)( eCCyCxC FmIYmaXma ?
或
?? ? ??? )(,,)( eCCCC FmIYymXxm ? ??????
上式称为 平面运动微分方程 。
57
[Example 4] A circular homogeneous wheel of mass m and radius R is
placed on a plane of inclination angle a,Under the action of gravity it start
to move,Assuming that the coefficients of the static and kinetic friction are
f and f’ respectively,Neglecting the rolling resistance analyze the motion of
the wheel,
Solution,Investigate the wheel,The force
analysis is shown in the figure,Analysis of
motion,Choosing the cartesian coordinate
system Oxy,shown in figure,we get aC y =0,
aC x =aC,Because the wheel is in plane motion
from the differential equations of plane
motion we obtain
From equation ② we get
,
The two equations ①
and ③ contain three
unknown quantities aC,
F,?,Additional
conditions have to be
complemented,
Fmgma C ?? as i n
Nmg ??? ac o s 0
FRI C ??
aco smgN ?
①
②
③
④
58
[例 4] 质量为 m半径为 R的均质圆轮置放于倾角为 a 的斜面上，
在重力作用下由静止开始运动。设轮与斜面间的静、动滑动
摩擦系数为 f,f′，不计滚动摩阻，试分析轮的运动。
解,取轮为研究对象。
受力分析如图示。
运动分析：取直角坐标系 Oxy
aC y =0,aC x =aC,
一般情况下轮作平面运动。
根据平面运动微分方程，有
Fmgma C ?? as i n
Nmg ??? ac o s 0
FRI C ??
由 ?式得
aco smgN ?
?
?
?
?
?, ?两式中含有三个
未知数 aC, F,?,需
补充附加条件。
59
1,Assuming the surface of contact to be absolutely smooth we get
Because the wheel moves
from rest,?=0,the wheel slide downward.,c o n s t,0,s i n,0 ???? ??agaF C
?raC ?
3,Assume that the wheel both slides and rolls,Complementing the
equation F=f’ N and solving the equations above we get
Therefore,the wheel rolls without slipping if
,whence,
The formula above shows,
If f < (1/3)tga the answer is given by 3;
If f > (1/3)tga the answer is given by 2;
If f = 0 the answer is given by 1,
aa?aa co s,co s'2,)co s'( s i n mgfFR gfgfa C ?????
aa c o ss i n31 m a x f m gfNFmgF ???? atg31?f
2,Let the surface of contact be rough enough,then the wheel
rolls without slipping,Complementing the equation and
solving the equations above we obtain
aa?a s in31,s in3 2,s in32 mgFgRga C ???
60
1．设接触面绝对光滑。
因为轮由静止开始运动，故 ?＝ 0，轮沿斜面平动下滑。
常量。???? ??a,0,s i n,0 gaF C
aa?a s i n31 ; s i n3 2,s i n32 mgFgRga C ???
2．设接触面足够粗糙。轮作纯滚动,所以可解得,?raC ?
3．设轮与斜面间有滑动，轮又滚又滑。 F=f′N,可解得
aa?aa co s,co s'2,)co s'( s i n mgfFR gfgfa C ?????
轮作纯滚动的条件,
aa c o ss i n31 m a x f m gfNFmgF ????
atg31?f表明：当 时，解答 3适用；
当 时，解答 2适用； f =0 时解答 1适用。
atg31?f
atg31?f
61
1,Basic concepts,
1) Moment of momentum is a measure of the intensity of the
mechanical motion of a material body at any instant,
2) The moment of momentum for a particle is
3) The moment of momentum of a system of particles is
4) The moment of inertia is a measure of the inertia of a body
in rotational motion,
For a homogeneous thin and straight rod,a thin circular ring
and a thin circular disk,we should calculate the moments of
inertia with respect to an axis perpendicular to the symmetry
plane of the masses,
Exercises
vmrvmm O ??)(
? ?? iiiO vmrL
62
一．基本概念
1．动量矩,物体某瞬时机械运动强弱的一种度量。
2．质点的动量矩,
3．质点系的动量矩,
4．转动惯量,物体转动时惯性的度量。
vmrvmm O ??)(
? ?? iiiO vmrL
对于均匀直杆，细圆环，薄圆盘（圆柱）对过质心垂直于
质量对称平面的转轴的转动惯量要熟记。
第十三章 动量矩定理习题课
63
5) Calculation of the moment of momentum of a rigid body,
Translational motion,
Fixed-axis rotation,
Plane motion,
2.Theorem of moment momentum and the conservation
law of the moment of momentum of a particle,
1) Moment of momentum theorem of a particle,
2) Conservative law of the moment of momentum of a particle,
① If then Const.,
② If then Const,
)(,CzzCCO vmmLvmrL ???
??? zz IL
???? CCzz IvmmL )(
)()(o r )()]([ FmvmmdtdFmvmmdtd zzOO ??
0)( ?Fm O
0)( ?Fm z
?)( vmm O
?)( vmm z
64
5．刚体动量矩计算
平动,
定轴转动,
平面运动,
)(,CzzCCO vmmLvmrL ???
??? zz IL
???? CCzz IvmmL )(
二．质点的动量矩定理及守恒
1．质点的动量矩定理
)()( )()]([ FmvmmdtdFmvmmdtd zzOO ?? 或
2．质点的动量矩守恒
? 若,则 常矢量。
? 若,则 常量。
0)( ?Fm O
0)( ?Fm z
?)( vmm O
?)( vmm z
65
3,Moment of momentum theorem and its conservation law
of a system of particles,
?? ???? )()()()( )(o r )( ezezzeOeOO MFmdtdLMFmdtLd
2) The conversation law of moment of momentum of a system of
particles,
? If, then constant vector,
? If, then const,
0)( ?eOm
0)( ?ezm
?OL
?zL
)(
)( o r e
zC
zCe
C
C M
dt
dLM
dt
Ld ??
4,The moment of momentum theorem of a system with
respect to the center of mass of the system,
1)The moment of momentum theorem of a system of particles,
66
三．质点系的动量矩定理及守恒
1．质点系的动量矩定理
?? ???? )()()()( )( )( ezezzeOeOO MFmdtdLMFmdtLd 或
2．质点系的动量矩守恒
? 若,则 常矢量
? 若,则 常量
0)( ?eOm
0)( ?ezm
?OL
?zL
)(
)( e
zCzC
e
CC Mdt
dLM
dt
Ld ?? 或
四．质点系相对质心的动量矩定理
67
?? ?? )(o r )( z FmIFmI zzz ?? ??
5,Differential equations of fixed-axis motion and plane
motion of a rigid body,
1) Differential equations of fixed-axis motion of a rigid body,
2) Differential equations of plane motion of a rigid body,
or
?? Xma Cx
?? Yma Cy
)(?? FmI CC ?
?? Xxm C??
?? Yym C??
?? )( FmI CC ???
68
???? )( )( z FmIFmI zzz ?? ??或
五．刚体定轴转动微分方程和刚体平面运动微分方程
1．刚体定轴转动微分方程
2．刚体平面运动微分方程
或
?? Xma Cx
?? Yma Cy
)(?? FmI CC ?
?? Xxm C??
?? Yym C??
?? )( FmI CC ???
69
6,Application of the moment of momentum theorem,Applying
this moment of momentum theorem we may solve the following
problems (it is especially convenient for monaxial transmission
system),
1) Knowing the state of the rotational motion of a system,determine the
external forces or external torques acting on the system,
2) Knowing constant or time-dependent external torques acting on a system,
determine the angular velocity or the angular acceleration of the rigid body,
3) Knowing that the principal moment or the algebraic sum of the
projections on any axis of the moments of all the external forces acting on a
system is zero,applying the conversation law of the moment of momentum,
determine the angular displacement or angular velocity,
70
六．动量矩定理的应用
应用动量矩定理，一般可以处理下列一些问题：（对单轴
传动系统尤为方便）
1．已知质点系的转动运动，求系统所受的外力或外力矩。
2．已知质点系所受的外力矩是常力矩或时间的函数，求刚体
的角加速度或角速度的改变。
3．已知质点所受到的外力主矩或外力矩在某轴上的投影代数
和等于零，应用动量矩守恒定理求角速度或角位移。
71
7,Examples of application
[Example1] A circular homogeneous cylinder of radius r and weight Q is
placed in a corner as shown in the figure,The initial angular velocity is ?0
and the coefficient of kinetic friction of the contact at the wall and the floor
are f′,Neglecting rolling resistance determine the time until the rotation of
the cylinder stops,
Solution,Choose the cylinder to be investigated (noting that
it is a rigid body)
The force analysis is shown in the figure,Analysis of Motion,
The center of mass does not move,the rigid body rotates
around it,Applying the differential equations of plane
motion of a rigid body we get
)0,0( ?? CyCx aa
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
①
②
③
The complementary equations are
BBAA NfFNfF ' a n d ' ??
④
72
七．应用举例
[例 1] 均质圆柱，半径为 r，重量为 Q，置圆柱于墙角。初始角
速度 ?0，墙面、地面与圆柱接触处的动滑动摩擦系数均为 f '，
滚阻不计，求使圆柱停止转动所需要的时间。
解,选取圆柱为研究对象。 (注意只是一个刚
体 )受力分析如图示。
运动分析：质心 C不动，刚体绕质心转动。
根据刚体平面运动微分方程 )0,0( ??
CyCx aa
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
?
?
?
补充方程,
BBAA NfFNfF ',' ?? ?
73
Substituting equation ④ into equations ① and ② we get
0)1'( 2 ??? QNf B
1'
',
1'
',
1'
',
1' 2
2
222 ???????? f
QfF
f
QfN
f
QfF
f
QN
AABB
Substituting the results above into equation③ we get
dtf frgfdr gff fdtd t?? ?????????? 0202 '1 '1'2,2''1 '1 0? ??
Solving it we obtain
)'1('2
)'1( 02
fgf
rft
?
?? ?
74
将 ?式代入 ?,?两式，有 0)1'( 2 ??? QNf B
1'
',
1'
',
1'
',
1' 2
2
222 ????????? f
QfF
f
QfN
f
QfF
f
QN
AABB
将上述结果代入 ?式，有
dtf frgfdr gff fdtd t?? ?????????? 0202 '1 '1'2,2''1 '1 0? ??
解得,
)'1('2
)'1( 02
fgf
rft
?
?? ?
BA FN ??0
QNF BA ???0
rFrFdtdrgQ BA 21 2 ????
?
?
?
补充方程,
BBAA NfFNfF ',' ?? ?
75
[Example 2] A rod formed by two thin homogeneous rods fixed together takes
the shape of the letter,T”,It rotates around a horizontal axis through the
center O,The angular velocity of the rod is ?=4rad/s when OA is in the
horizontal position,Determine the reaction of the bearing O at that moment,
Solution,Choose the rod to be investigated,
The force analysis is shown in the diagram,
Applying the differential equations of rotation
around a fixed axis we get
r a d /s 2 0,7 5
5.08.9825.08.98 5.081217
2
2
?
?????????
?
?
5.025.0 ???? mgmgI O ?
2222 121712131 mlmlmlmlI O ????
76
[例 2] 两根质量各为 8 kg的均质细杆固连成 T 字型，可绕通过 O
点的水平轴转动，当 OA处于水平位置时,T 形杆具有角速度 ?
=4rad/s 。求该瞬时轴承 O的反力。
解,选 T 字型杆为研究对象。
受力分析如图示。
r a d /s 2 0,7 5
5.08.9825.08.98 5.081217
2
2
?
?????????
?
?
5.025.0 ???? mgmgI O ?
2222 121712131 mlmlmlmlI O ????
由定轴转动微分方程
77
Applying the theorem of the motion of the center of mass we obtain
OxCxC Xmama ??? 21
mgmgYmama OyCyC ????? 21
N 96) 5.04 25.04( 8)( 2221 ??????????? xCxCO aamX
N 3.32 ) 5.075.20 25.075.20 ( 88.982 ????????OY
78
根据质心运动微分方程，得
OxCxC Xmama ??? 21
mgmgYmama OyCyC ????? 21
N 96) 5.04 25.04( 8)( 2221 ??????????? xCxCO aamX
N 3.32 ) 5.075.20 25.075.20 ( 88.982 ????????OY
79
[Example 3] Two circular homogeneous cylinders both are of
weight P and of radius r,An inextensible string is wounded
on the cylinder A,The other end of the of the string is wounded
on the cylinder B,Neglecting the mass of the string and the friction
of the axis O determine
1 the acceleration of the center of mass of the cylinder B when it falls,
2 if a counterclockwise torque M acts on the cylinder A,on which condition
the center of mass of the cylinder B will move upwards?
80
[例 3] 均质圆柱体 A和 B的重量均为 P，半径均为 r，一绳缠在
绕固定轴 O转动的圆柱 A上，绳的另一端绕在圆柱 B上，绳重
不计且不可伸长，不计轴 O处 摩擦。
求, ? 圆柱 B下落时质心的加速度。
? 若在圆柱体 A上作用一逆时针转向的转矩 M，试问在什么
条件下圆柱 B的质心将上升。
81
Take cylinder B as the object to be
investigated
rTrgP B '21 2 ??
'TPagP C ??
②
③
Applying the relation of kinematics,we have,
BAC rra ?? ??
④
TrrgP A ??221
①
Solution Choose cylinder A as the object to be investigated,
,52 rgBA ???? ga C 54 ?
From equation ① and ② we obtain
BA ?? ?
Substituting this equation into
equations ③ and ④ we obtain
82
选圆柱 B为研究对象
rTrgP B '21 2 ??
'TPagP C ??
?
?
运动学关系,
BAC rra ?? ??
?
TrrgP A ??221
?
解：选圆柱 A为研究对象
由 ?,?式得,
BA ?? ?
,52 rgBA ???? ga C 54 ?
代入 ?,?式得,
83
Applying the moment of momentum theorem with
respect to the axis O,we get
rPMMrgPrvgPrgPdtd eOBCA ???????? 2)222( )(22 ??
Choosing the whole system to be investigated we get
BCAO rg
Prv
g
Pr
g
PL ?? 22
222 ?????
rPMM eO 2)( ???
and substituting it into equation (1) we obtain
rPMrgPragPrgP BcA ?????? 2222 T h e r e f o r e 22 ??
(1)
BAC rra ?? ??
grP rPMarPMargPargP CCC ? ?????????? 5 )2(2 ; 222
Complementing the equation of motion
When M >2Pr,,hence,the center of mass
of cylinder B will move upwards,0?Ca
84
由动量矩定理,
rPMMrgPrvgPrgPdtd eOBCA ???????? 2)222( )(22 ??
rPMrgPragPrgP BcA ??????? 2222 22 ?? (1)
补充运动学关系式,BAC rra ?? ?? 代入 (1)式，得
grP rPMarPMargPargP CCC ? ?????????? 5 )2(2 ; 222
当 M >2Pr 时，,圆柱 B的质心将上升。 0?
Ca
再取系统为研究对象
BCAO rg
Prv
g
Pr
g
PL ?? 22
222 ?????
rPMM eO 2)( ???
85
When investigating the problems of dynamics for a rigid body in
plane motion,we have to find the dependence between the motion of
the center of mass and the rotational motion of the rigid body,
When writing equations according to the moment of momentum
theorem with respect to an axis,we have to care for the identity of the
definitions of the sign,
86
研究刚体平面运动的动力学问题，一定要建立补充方程，找
出质心运动与刚体转动之间的联系。
应用动量矩定理列方程时,要特别注意正负号的规定的一致
性。
87
88
