Mechanics
Chapter 5 Angular momentum
5.3 Angular momentum theorem of a particle
质点的角动量定理
Section 14.7
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Chapter 5. Angular momentum
1
5.3 Angular momentum theorem of a particle
• Angular momentum theorem: the torque-angular
momentum relationship
• Angular impulse-momentum principle
• Conservation of angular momentum
2005年11月21日 8:00-9:50
Chapter 5. Angular momentum
2
Mechanics
Chapter 5 Angular momentum
5.3 Angular momentum theorem of a particle
5.3.1 Angular momentum theorem
5.3.2 Angular impulse-momentum principle
5.3.3 Conservation of angular momentum
2005年11月21日 8:00-9:50
Chapter 5. Angular momentum
3
5.3.1 Angular momentum theorem
Derive the relationship between torque and angular
momentum using Newton’s second law
Consider the motion of a particle of mass m in an inertial
frame
F: resultant force acting on the particle;
v: velocity of the particle
a: acceleration of the particle
dv d mv 
F  ma  m 
dt
dt
 Newton’s second law
O: a fixed point with respect to an inertial frame
r : the position vector of the particle measured from O
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Chapter 5. Angular momentum
4
5.3.1 Angular momentum theorem
L = r  mv = r  p
Differentiating this equation with respect to time



dL d 
 dr
  d mv 
 r  mv  
 mv  r 
dt dt
dt
dt
= v  mv = 0
 The angular
 dL
momentum
τ 
dt theorem about
a fixed point O
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F

The torque of the resultant
force acting on a particle
about a fixed point O in an
inertial frame equals to the
time derivative of the
particle’s angular momentum
about O
Chapter 5. Angular momentum
5
5.3.1 Angular momentum theorem
Rectangular components:
dLx
x 
dt
y 
dLy
dt
dLz
z 
dt
Angular momentum
theorem about z-axis
Note:
• The angular momentum theorem is valid only in inertial
reference frame. When applied in a non-inertial frame,
torque of inertial force must be added.
• The torque and angular momentum must be calculated
about the same reference point
• The reference point O is assumed to be fixed with
respect to the inertial frame. If O is moving, then dr/dt  v



 dL
 dL dr
τ
  mv 
dt dt
dt
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Chapter 5. Angular momentum
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5.3.1 Angular momentum theorem
The angular momentum theorem shows that the  is
related to the time derivative of L, but no to L. so that the
direction of  is generally not the same as L
For example: conical pendulum(圆锥摆)
• The forces acting on the bob are the weight
and the tension of the string. The torque of
the resultant force about O:

         
τ  r  T  W   r  T  r W  r W
  is parallel to v
• Angular momentum:
  
L  r  mv
 L is perpendicular to v and 
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Chapter 5. Angular momentum
L
o
T
A
W v,
7
5.3.1 Angular momentum theorem
• As the bob rotates, the angular momentum vector L
sweeps out a cone.


dL
L
L´
 lim
L
dt t 0 t
 Same direction as 
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Chapter 5. Angular momentum
L
8
Mechanics
Chapter 5 Angular momentum
5.3 Angular momentum theorem of a particle
5.3.1 Angular momentum theorem
5.3.2 Angular impulse-momentum principle
5.3.3 Conservation of angular momentum
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Chapter 5. Angular momentum
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5.3.2 Angular impulse-momentum principle
Angular impulse (moment of impulse):
The angular impulse of the resultant force F acting on a
particle about point O during the time interval t1 and t2 is
defined as
t2
t2
t1
t1
( AO )12   r  Fdt   τdt
Let O be the origin of a rectangular coordinate system,
then the rectangular components of (AO)1-2 are
( Ax )1 2 
( Ay )1 2 
( Az )1 2 
2005年11月21日 8:00-9:50

t2
t1

t2
t1

t2
t1
 x dt
 y dt
 z dt Angular impulse of F about the z-axis
Chapter 5. Angular momentum
10
5.3.2 Angular impulse-momentum principle
From the angular momentum theorem


 dL

τ
 τdt  dL
dt
Integrating over the time interval t1 to t2

t2
t1
 

t2 

τdt   dL  L2  L1  L
t1

 

( Ao )12  L2  L1  L
The principle of angular impulse and angular momentum:
The change in angular momentum is equal to
the angular impulse of the resultant force
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Chapter 5. Angular momentum
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5.3.2 Angular impulse-momentum principle
Rectangular components:
( Ax )1 2  Lx 2  Lx1  Lx
( Ay )1 2  Ly 2  Ly1  Ly
( Az )1 2  Lz 2  Lz1  Lz
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Chapter 5. Angular momentum
12
Mechanics
Chapter 5 Angular momentum
5.3 Angular momentum theorem of a particle
5.3.1 Angular momentum theorem
5.3.2 Angular impulse-momentum principle
5.3.3 Conservation of angular momentum
2005年11月21日 8:00-9:50
Chapter 5. Angular momentum
13
5.3.3 Conservation of angular momentum
From the angular impulse-momentum principle:

 

t2 
AO   τdt  L2  L1  L
t1
If AO = 0, then
L1 = L2
or L = 0
 The principle of conservation of angular momentum:
If the angular impulse about O is zero during a
given time interval, the angular momentum of
the particle about O will be conserved during
that interval
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Chapter 5. Angular momentum
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5.3.3 Conservation of angular momentum
Note:
This principle is valid only if the angular impulse of the
resultant force acting on the particle is zero:
• The toque of the resultant force acting on the
particle is zero;
• The toque of the resultant force is not zero, but
the angular impulse of the force is zero.

This is a vector equation.
 We can apply it to any direction in which
there is no angular impulse applied.
e.g. if Ax = 0, then the angular momentum in xdirection is conserved
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Chapter 5. Angular momentum
15
5.3.3 Conservation of angular momentum
Because the angular momentum and torque are both
dependent of the reference point O, the angular
momentum conservation is also dependent on the
choice of O
For example: the motion of a planet around the sun
O: the center of the sun
 = r  F = 0  L conserved
O´: the center of the ellipse path
F r
r´
O´ O
F
´= r´  F  0  L not conserved
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Chapter 5. Angular momentum
16
5.3.3 Conservation of angular momentum
补充例题：计算行星的俘获截面，如果有一航天器在远方以初速度v0射向某一行
星（设航天器不带动力）。航天器计划在行星上登陆。以b 表示v0与行星的垂直
距离，b称为瞄准距离，求b最大值为多少时，航天器可以在行星上着陆，行星质
量为M，半径为R。

v0
解：无引力： b  R, s  R 2
A
m
有引力：航天器在行星处轨道向行星弯曲。
故b可以大于R，可选择b的大小，
使航天器轨道正好与行星表面相切。
b

r0

R
B

v
O
取O点为参考点，航天器在行星表面处的速度为v ，取z轴通过O点垂直纸面，
由于航天器所受的引力总是指向O点，故力矩为0, Lz守恒。
在A点：Lz  r0v0 sin   bv0

在B点: v  R Lz  Rv
 bv0  Rv (1)
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Chapter 5. Angular momentum
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5.3.3 Conservation of angular momentum
作用在航天器上的力为保守力（万有引力），机械能守恒。
1
2
mv02  12 mv 2  G
Mm
R
(2)
由（1）和（2）式得
 GMm



Ep
R
b  R 1 
  R 1
1
2
E

mv0 
2


GMm
1 2 Ep
Ep  
E  mv0
0
R
2
E
俘获截面为：
 Ep 
  R 2
s  b  R 1 
E 

2
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2
Chapter 5. Angular momentum
18
5.3.3 Conservation of angular momentum
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Chapter 5. Angular momentum
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