Transcript College Physics (II) Qingxu Li Tel: 62471347, Email:

College Physics (II)
Qingxu Li
Tel: 62471347, Email: [email protected]
Room 306, College of Mathematics and
Physics
“The most incomprehensible thing about the
universe is that it is comprehensible.”
—-Albert Einstein
About the Course
• College Physics (II)
• Textbook: General Physics, Bin Liang, et al.
• Contents: Mechanics, Oscillation and Wave, Optics;
Electromagnetism, Relativity, Quantum Physics, etc.
• Course grade: Final Exam (70%) + Performance (30%)
• Exercises and Exam are to be finished in English
(Chapter 2-7, 10)
Reference books
a. Principle of Physics, 3rd edition, Serway and Jewett
b. Feynman’ Lectures on Physics (Volume I), by R. P. Feynman
c. 物理学，马文蔚，高等教育出版社，第五版
……
A Brief Summary of Chapter 1
Units, Dimension, Significant Figures, Order of Magnitude, Vector
（单位，量纲，有效数字，数量级，矢量）
a. Units are indispensable for physical quantities.
b. Vectors are to be distinguished from scalars.
c. Properties of Vectors:
magnitude, direction, components,
equality, addition, dot product,
cross product, etc.
A
A   Ax , Ay , Az   AeA ; A  A  A  A ; eA 
A
A  B  AB cos AB  Ax Bx  Ay By  Az Bz ;
2
x
2
y
2
z
A  B  A  B e AB ; A  B  AB sin  AB
Position and Displacement Vectors
（位置矢量和位移矢量）
r  r ' r
path 路程,路线
locus 轨迹
distance 距离
Average Velocity and Instantaneous Velocity
（平均速度和瞬时速度）
r  r (t )  x(t )i  y (t ) j  z (t )k
r  rf  ri
r
the average velocity : v 
t
r dr
the instantaneous velocity : v  lim

t 0 t
dt
Fig 1.1 A particle moving in the xy plane
Alternative Expressions
（其他形式）
dr
v  v (t ) 
dt
v  (vx , v y , vz )  vev
dx
dy
dz
vx  , v y  , vz 
dt
dt
dt
v  v (t )，   x，y，z
Acceleration
（加速度）
The average acceleration of a particle over a time interval is
defined as:
v v f  vi
a

t t f  t i
And the instantaneous acceleration is defined as:
v dv
a  lim

t  0 t
dt
Alternative Expressions
dv (t )
a  a (t ) 
dt
a  (ax , a y , az )  aea
dv
a 
, a  a (t );   x, y, z
dt
v
Fig 1.2 The Velocity-Time diagram. The magnitude of acceleration
vector is the slope of the curve v—t.
Problems Related to Kinematics
position  displacement
velocity
r  r (t )  r  rf  ri
2
dv d r
a
 2
dt dt
acceleration
dr
v
dt
tf
r   vdt
ti
t
v (t )  v (t0 )   adt
t0
Mechanics
Kinematics
Dynamics
Part II Dynamics
The Laws of Motion
（运动定律）
Nature and nature’s laws lay hid in night.
God said: Let Newton be! and all was light.
--Alexander Pope
The Concept of Force
（力的概念）
The force is a vector quantity.
The unit of force is newton, which is defined as the force that,
when acting on a 1-kg mass, produces an acceleration of 1m/s2.
1 N  1 kg  m/s
The dimension of force is:
 F   ML / T
2
2
Newton’s First Law
（牛顿第一定律）
Newton’s first law of motion:
In the absence of external forces, an object at rest remains
at rest and an object in motion continues in motion with a
constant velocity (that is, with a constant speed in a straight
line)
（在没有外力的情况下，静止的物体会保持静止，而运动的物体则保
持运动速度不变，也就是说运动物体做匀速直线运动。）
In simpler terms, when no force acts on a body, its acceleration
is zero.
（简单地讲，如果没有外力作用，物体的加速度为零）
Comments on the First Law
1. The first law tell us that an object has a tendency to maintain its
original state of motion in the absence of the force. This tendency is
called inertia, and the first law sometimes called the law of inertia.
（牛顿第一定律告诉我们物体在不受外力的情况下有一个保持原来的运动状态的
趋势）-称为惯性，因而第一定律有时又被称为惯性定律）
2. Newton’s first law defines a special set of reference frames called
inertial frames. An inertial frame of reference is one in which the
first law is valid.
（利用牛顿第一定律可以定义一类特殊的参照系-惯性系：在惯性系中，第一定律
成立）
3. Inertial mass is the measure of an objects resistance to change in
motion in response to an external force. Inertial mass is different in
definition from gravitational mass, but they have the same value, so
we call them both simply mass.
（惯性质量是物体阻止运动状态发生改变能力即惯性大小的量度。惯性质量和引
力质量在定义上不同，但它们具有相同的数值，统称为质量）
Mass and Weight
（质量和重量）
Mass and weight are two different quantities, and should not be
confused with each other.
The magnitude of an object is equal to the magnitude of
the gravitational force exerted by the planet on which the objects
resides. While the mass of an object is the same everywhere. A
given object exhibits a fixed amount of resistance to changes in
motion regardless of its location.
E.g. A person of mass 60 kg on Earth also has a mass of 60 kg
on the moon. The same person weighs 588 Newton on Earth, but
weighs 98 Newton on the moon.
Newton’s Second Law
（牛顿第二定律）
Newton’s second law of motion:
The acceleration of an object is directly proportional to the
net force acting on it and inversely proportional to its mass.
（物体的加速度和它受到的合外力成正比和它的质量成反比）
F

a
m
net force
（合力）
Net Force
Ft   F
The net force is also known as:
① the resultant force
② the sum of the force
③ the total force
④ the unbalanced force
The Mathematical Form of the Second Law
（第二定律的数学形式）
F F

a

t
m
a
F


m
m
F

,   x, y , z
m
Comments on the Second Law
1. The Newton’s second law is the central rule of classical mechanics,
which bridges dynamics and kinematics and tells that force is the
cause of the change of motion (not motion!).
2. The second law has an alternative expressions:
dv dp
Ft  ma  m

dt dt
In special relativity, the mass of an object will vary with its velocity
and thus vary with time. The previous form is invalidated in this case
but the new form still holds. Of course, both form are equivalent for
non-relativistic cases.
2
3. The second law can also be expressed as:
d r
Ft  m 2
dt
Newton’s Third Law
（牛顿第三定律）
Newton’s third law of motion:
If two objects interact, the force exerted by object 1 on object
2 is equal in magnitude but opposite in direction to the force
exerted by object 2 on object 1.
（如果两个物体之间（存在）相互作用），则物体 1 作用到物
体 2 上的力和物体 2 作用到物体1上的力大小相等，方向相反）
Forces always occurs in pairs, i.e., that a single isolated force
cannot exist.
（力总是成对出现，也就是说，单个孤立的力是不能存在的）
Comments on the Third Law
The force that object 1 exerts on object 2 may be called action
force and the force of object 2 on object 1 the reaction force. The
action force is equal in magnitude to the reaction force and
opposite in direction. In all cases, the action and the reaction forces
act on different objects and must be of the same type.
（物体 1 作用在物体 2 上的力可以称作作用力，相应地我们称物体 2
作用在物体1上的力为反作用力。作用力和反作用力大小相等方向相反。
作用力和反作用力类型相同，并且作用在不同的物体上）
Applications of Newton’s Law
（牛顿定律的应用）
The Particle in Equilibrium
（处于平衡状态的质点）
Objects that are either at rest or moving with constant velocity are
said to be in equilibrium. From Newton’s second law, this condition
of equilibrium can be expressed as:
（我们称静止或作匀速直线运动的物体处于平衡状态。根据牛顿第二定律，物体
处于平衡状态的条件可以表达为：）
or:
Ft   F  0
Ft  0;   x, y, z
The Accelerating Particle
（加速质点）
When a nonzero net force is acting on a particle, the particle is
accelerating, and the second law tell us:
（如果质点受到一个非零的合外力，则质点加速运动，由第二定律可知）
F

 F  ma  a 
m
In practice, the above equation is broken into components, so
that two or three equations can be handled independently.
（上述方程在应用的时候通常分解为分量形式，这样就可以单独处理两个或
三个方程）
The Atwood Machine
（阿特伍德机）
E.g. 1.1 When two objects with unequal masses are hung vertically
over a light, frictionless pulley as in the figure, the arrangement is
called an Atwood machine. The device is sometimes used in the lab
to measure the free-fall acceleration. Calculate the magnitude of the
acceleration of the two objects and the tension in the string.
Fig 1.10 The Atwood machine.
Solution : Assuming the masses of the two objects are m1 , m 2 respectively,
and it is easily to find that both objects must have the same magnitude of
acceleration. We choose the upward direction is along the positive direction
of y axis.
So, when Newton's second law is applied to m1 , we find
Fty   Fy  T  m1 g  m1a
(1)
Similarly, for m 2 we find
Fty '   Fy '  T  m2 g   m2 a
(2)
From equation (1) and (2), we can obtain
 m2  m1  g   m2  m1  a
(3)
Solving the above equation, we get the acceleration
m2  m1 

a
g.
 m2  m1 
E.g. 1.2 Two blocks of masses m1 and m2 , wiht m1  m2 .
are placed in contact with each other on a frictionless,
horizontal surface, as in the figure. A constant horizontal
force F is applied to m1 as shown.
(a) Find the magnitudeof the acceleration of the system of
two blocks.
(b) Determine the magnitude of the contact force between
the two blocks.
Solution (a) Both blocks must experience the same acceleration
because they are in contact with each other and keep so. Thus we
model the system of both blocks as a particle under a net force.
For the force is horizontal, we have
Fx  F  ( m1  m2 )a
(1)
F
So the magnitude of the acceleration reads a 
.
(m1  m2 )
(b) It is obviously the contact force exerted on m2 is in the positive
x direction, and the net force acting on the m2 is
F2x  F2c
From Newton's second law, we get
m2 F
F2x  F2c  m2 a 
.
(m1  m2 )
(2)
Forces of Friction
（摩擦力）
When an object moves either on a surface or through a viscous
medium such as air or water, there is resistance to the motion. We
call such resistance a force of friction.
force of static friction
Force of friction
（静摩擦力）
force of kinetic friction
（动摩擦力）
Simplified model for force of friction
1. The magnitude of the force of static friction between any two
surfaces in contact can have the values
f s  s n
μs : the coefficient of static friction
n: the magnitude of normal force
2. The magnitude of the force of kinetic friction acting between
two surfaces is
f k  k n
μk : the coefficient of kinetic friction
3. The values ofμk andμs depend on the nature of the surfaces,
but the former is generally less than the latter.
4. The direction of the friction force on an object is opposite to the
actual motion or the impending motion of the object relative to
the surface with which it is in contact.
Fig 1.5 A graph of the magnitude of the friction force versus that
of the applied force.
The Gravitational Force:
Newton’s Law of Universal Gravitation
引力：牛顿万有引力定律
Newton’s Law of Universal Gravitation
Every particle in the Universe attracts every other particle with
a force that is directly proportional to the product of the masses of
the particles and inversely proportional to the square of the distance
between them.
m1m2
Fg  G 2
r
G  6.67 1011 N  m2 / kg 2 ,
is calle the universal gravitational constant.
The electrostatic force
Coulomb’s Law
The magnitude of the electrostatic force between two charged
particle separated by a distance r is:
q1q2
Fe  ke 2
r
ke  8.99 109 N  m 2  C2
The Coulomb constant
The Fundamental Forces of Nature
（自然界中的力）
① The Gravitational Force
（引力）
② The Electromagnetic Force
（电磁力）
③ The Strong Force (The Nuclear force)
（强力）
④ The Weak Force
（弱力）
electromagnetic force 
  electroweak force (1967,1984)
weak force 
Newton’s Second Law
Applied to a Particle in Uniform Circular Motion
A particle moving in a circular path with uniform speed experiences
a centripetal acceleration of magnitude:
v2
ac 
r
The acceleration vector is directed toward the centre of the circle and
is always perpendicular to its velocity.
Apply Newton’s second law to the particle along the radial direction:
centripetal force
（向心力）
mv
Ft  mac 
r
2
Non-uniform Circular Motion
For non-uniform circular motion, there is, in addition to the radial
component of acceleration, a tangential component, that is:
Ft
a  ar  at 
m
The total force exerted on the particle is
Ft  Ft _ r  Ft _ t
The first term in the RHS is directed toward the center of the circle
and is responsible for the centripetal acceleration; and the second term
is tangent to the circle and responsible for the tangential acceleration,
which causes the speed of the particle to change with time.
Energy of a System
（物理体系的能量）
kinetic energy
Energy
（动能）
potential energy
（势能）
Work（功）
The work done by a force on a system is defined as:
（作用在一个体系上的力对体系作功定义为）
dW  F  dr  Fdr cos
For a finite displacement,
（对于一个有限位移）
rf
rf
ri
ri
W   F  dr   Fdr cos 
Work Done by a Constant Force
（恒力作功）
For a constant force, the work reads:
rf
rf
ri
ri
W   F  dr   F cos  dr
 F  r  F r cos 
If the applied force is parallel to the direction of the displacement,
W  F r
And if the force is perpendicular to the displacement, then
W 0
From the definition of dot product, we get:
dW  F  dr   F dr

W   F  dr   
rf
ri

rf 
ri
F dr   W

Work done by a Spring
Hooke’s Law
F  kx
k is force constant (spring constant or
stiffness), and x is the distance from
the equilibrium position.
The work done by the restoring force on a block connected with
a spring reads:
Ws  
xf
xi
1 2 1 2
Fs dx  kxi  kx f
2
2
restoring force 回复力
Kinetic Energy
（动能）
The work done on a system in motion:
rf
rf
ri
ri
W   Ft  dr   ma  dr
r f dr
dv
 m
 dr  m 
 dv
ri dt
ri dt
1 2 vf 1 2 1 2
 mv |vi  mv f  mvi
2
2
2
rf
Define the kinetic energy of a particle is:
1 2
K  mv
2
From the above definition, we get:
Wnet  K f  Ki  K
Work-kinetic energy theorem
（功能定理）
Work-kinetic energy theorem:
When work is done on a system and the only change in the
system is in its speed, the work done by the net force equals
the change in kinetic energy of the system.
（当外界对体系作的功给体系带来的只是体系运动速率的变化时，合
外力所做的功等于体系动能的增量）
E.g. 1.8 A 6.00 kg block initially at rest is pulled to the right along
a horizontal friction less surface by a constant, horizontal force of
12.0 N, as shown in the figure. Find the speed of the block after it
has moved 3.00m.
Solution : The work done by the horizontal force is
W  F x  12.0  3.00 J=36.0 J.
Using the work-kinetic theorem and noting that the initial
kinetic energy is zero, we find
1 2
W  K f  Ki  K f  mv f
2
2W
v f 
 3.46 m / s
m
Potential Energy
（势能）
Kinetic Energy: related to the motion of an object
Internal Energy: related to the temperature of a system
Potential Energy: related to the configuration of a system
Example: gravitational potential energy,
U g  mgy
configuration: 构型，结构
Conservative Forces
（保守力）
The work done by a conservative force does not depend on the
path followed by the members of the system, and depends on the
initial and final configurations of the system.
In other words, the work done by a conservative on an object
does not depend on the path of the object, but depends on its
initial and final position.
From above definition, it immediately follows that the work
done by a conservative force when an object is moved through
a closed path is equal to zero.
Conservative Forces and Potential Energy
rf
F dr  f (r , r )  U (r )  U (r )

Conservative force：
 F dr  0
f
ri
B
A
 F dr   F dr   F dr
  F dr   F dr  0
 F dr   F dr  U  U
p
A
B
B
p
A
B
A
p'
B
p'
A
B
p
A
p'
B
A
i
f
i
Define potential energy function as:
rf
U f    F dr  U i
ri
From above definition, we can get:
U  U f  U i    F dr  W
dU  F dr
and:
U U U
F  U  (
,
,
)
x y z
U
or F  
r
rf
ri
Gravitational Force
Consider a particle of mass m above the Earth’s surface:
The gravitational force on the particle due to the Earth reads:
GMm
Fg   2 er
r
The work done by the gravitational force
U f  
rf
ri
1 1
F dr  U i  GMm(  )  U i
rf ri
Assuming:
lim Ui  0
ri 
And we get:
GMm
Ur  
.
r
In summary, the gravitational potential energy for any pair of
particles varies as 1/r. Furthermore, the potential energy is negative
because the force is attractive and we have chosen the potential
energy to be zero when the particle separation is infinity.
GMm
U 
r
  
F  U  ( , , )U
x y z
   GMm
GMm
GMm
Fg  ( , , )
  3  x, y, z    2 er
x y z
r
r
r
E.g. A particle of mass m is displaced through a small vertical distance
Δy near the Earth’s surface. Show that expression for the change in
gravitational potential energy reduces to the familiar relationship:
ΔUg=mg Δy.
Solution : We can express the potential energy change as:
rf  ri
1 1
U  GMm(  )  GMm
rf ri
rf ri
If both initial and final positions of the particle are close to the earth's
surface, then rf  ri  y and rf ri  RE2 . The change in potential energy
y
GM
therefore, becomes: U  GMm 2  mg y, where g  2 .
RE
RE
Electrostatic Force
（静电力）
The electrostatic force between two charged particles reads:
q1q2
Fe  ke 2 er
r
In a way similar to gravitational force case, we can get electric
potential energy function（电势能）:
q1q2
U e  ke
r
ke q1q2
   ke q1q2 ke q1q2
Fe  ( , , )

 x, y, z   2 er
3
x y z
r
r
r
The Force of a Spring
According to Hook’s law, a block connected to a spring
experiences a force:
Fs  kx
Therefore, the potential energy stored in a block-spring system is:
U f  
xf
xi
1
2
2
kxdx  U i  k ( x f  xi )  U i
2
If the initial position of the block is xi=0, Ui is always chosen as
zero, Then,
d
Fs   U   kx
dx
1 2
U f  kx
2
elastic potential energy
（弹性势能）
rf
U f    F dr  U i
ri
势能函数
保守力
U U U
F  U  (
,
,
)
x y z
Mechanical Energy
（机械能）
The sum of kinetic and potential energy is defined as mechanical
energy:
（体系的动能和势能之和定义为体系的机械能）
Emech  K  U
If in an isolated system there are only conservative forces which
do work, the mechanical energy will keep unchanged, as is called
conservation of mechanical energy.
（如果孤立体系中只有保守力做功，则系统的机械能保持不变，称体系
的机械能守恒）
The conservation of mechanical energy in a system can be
expressed as:
Emech  K i  U i  K f  U f  Const
The conservation of energy in an isolated system can be
expressed as:
Et  Ki  Ui  Eiint  K f  U f  E int
f  Const
Stability of Equilibrium
（平衡的稳定性）
Energy diagram: An energy diagram shows the potential energy of
the system as a function of the position of one of members of the
system.
Stable equilibrium: When the system locates such a position that
any movement away from this position results in a force directed back
toward the position. (this type of force is called restoring force.)
In general, positions of stable equilibrium correspond to those
positions for which the potential energy function has a relative
minimum value on an energy diagram. And positions of unstable
equilibrium correspond to those positions for which the potential
energy has a relative maximum value.
Energy Transfer
（能量转移）
system and its environment
（系统和环境）
isolated systems Vs non-isolated systems
（孤立系统 Vs 非孤立系统）
Work is one means of energy transfer between the system and
its environment. If positive work is done on the system, energy is
transferred from the system to the environment, whereas negative
work indicates that energy is transferred from the system to the
environment.
Heat and Thermal Conduction
（热和热传递）
Except for work, energy can also be transferred through thermal
conduction.
The work done on a system may also increase its internal energy,
in addition to change its kinetic energy. The internal energy of an
object is associated with its temperature. And it’s well known that
heat can be transferred from a warmer object to another object. The
energy transfer caused by a temperature difference between two
regions in space is called thermal conduction.
internal energy 内能
thermal conduction 热传递
heat 热，热量
Mechanical Wave
（机械波）
Mechanical wave are a means of transferring energy by
allowing a disturbance to propagate through into air or
another medium. This is the method by which energy
leaves a radio through the loudspeaker-sound-and by which
energy enters your ears to stimulate the hearing process.
disturbance 扰动
medium 介质，媒介
propagate 传播
Principle of Conservation of Energy
（能量守恒定律）
We can neither create nor destroy energy – energy is conserved.
（能量既不能创生，也不能消灭——能量是守恒的）
If the amount of energy in a system changes, it can only be due to
the fact that energy has crossed the boundary by a transfer
mechanism.
（如果体系的能量发生了变化，只可能是体系的一部分能量通过一种能量
转移机制穿出了体系的边界）
isolated and non-isolated systems
For isolated systems, energy is always conserved, so:
E  K  U  E
int
0
While for non-isolated systems, we have:
E   H
Work
Heat
matter transfer
Power（功率）
Power: the time rate of energy transfer.
（功率：能量转化的快慢）
Average power: If the work done by a force is W in the time
interval Δt , then the average power during this time interval
is defined as:
W
p
t
The instantaneous power:
W dW
p  lim

t 0 t
dt
For work done by a varying force:
dW
dr
p
F
 F v
dt
dt
In general, power is defined for any type of energy transfer and
the most general expression for power is, therefore
dE
p
dt
Unit of Power
（功率的单位）
Unit of Power: Watt
1 Watt  1J / s  1kg  m / s
2
Other mostly used units of power:
1hp  746W
1kWh  3.60 106 J
3
Solution : As shown in the figure, the car is the system and the only
force we must consider in the calculation is the horizontal force of
kinetic friction (denoted by Fk ). Thus we can get in the two cases:
Case 1. The initial speed is v (assuming the mass of the car is m) :
1 2
 Fk d  0  mv
2
Case2. Given the distance the car slides is x, then:
1
2
 Fk x  0  m  2v 
2
From above two equations, it can be easily obtained:
x
 4  x  4d .
d
Solution : As shown in the figure, the car is the system and the only
force we must consider in the calculation is the horizontal force of
kinetic friction (denoted by Fk ). Thus we can get in the two cases:
Case 1. The initial speed is v (assuming the mass of the car is m) :
1 2
 Fk d  0  mv
2
Case2. Given the distance the car slides is x, then:
1
2
 Fk x  0  m  2v 
2
From above two equations, it can be easily obtained:
x
 4  x  4d .
d
Momentum and Impulse
（动量和冲量）
kinematics
mechanics (force)
motion
energy
momentum
Momentum
（动量）
The (linear) momentum of an object of mass m moving with a
velocity v is defined to be the product of the mass and velocity:
p  mv
Momentum is a vector quantity and its direction is the same as
that for velocity; And it has dimension ML/T. In SI system, the
momentum has the units kg·m/s.
p  mv ,   x, y, z
Momentum and Force
As pointed out before, the Newton’s second law can be rewritten
as:
dp
F
dt
From above equation, we see that if the net force on an object is
zero, the time derivative of the momentum is zero, and therefore
the momentum of the object must be constant. Of course, if the
particle is isolated, then no forces act on it and the momentum
remains unchanged——this is Newton’s first law.
Momentum and Isolated Systems
The total momentum of an isolated system remains constant.
（孤立体系的动量是一个常数）
The total momentum
pt   pi
i
for an isolated system
Ft  0
dpt
Thus, we have
 0  pt   pi =Const
dt
i
pt  Const,   x, y, z.
The law of conservation of linear momentum!
（动量守恒定律）
Impulse and Momentum
（冲量和动量）
Assuming a net varying force acts on a particle, then we get:
dp
 Ft  dp  Ft dt
dt
thus the change in the momentum of the particle during the time
interval Δt = tf − ti reads:
tf
p  p f  pi   Ft dt
ti
The Impulse of a force is defined as:
tf
I   Ft dt  p
ti
impulse-momentum theorem
（冲量-动量定理）
Also valid for a system of particles（对于质点系也成立）
Impulse is an interaction between the system and its environment.
As a result of this interaction, the momentum of the system changes.
（冲量是体系和环境之间的一种相互作用，它带来体系动量的变化）
The impulse approximation: We assume that one of the forces
exerted on a particle acts for a short time but is much greater than
other force present. this simplification model allows us to ignore
the effects of other forces, because these effects are be small during
the short time during which the large force acts.
（冲量近似：如果有一个力短时间作用于一个质点，并且作用
过程中这个力比该质点所受到的其它力要大很多，这时可以忽
略其它力带来的效应。）
Collisions
（碰撞）
When two objects collide, it is a good approximate in many cases to assume
that the forces due to the collision are much larger than any external forces
present, so we can use the simplification model: the impulse approximation.
Collisions
Elastic collision （弹性碰撞）
Inelastic collision（非弹性碰撞）
Perfectly inelastic collision
Momentum is conserved in all cases, but kinetic energy is
conserved only in elastic collisions.
（动量在所有的碰撞过程中守恒，而动能仅在弹性碰撞中守恒）
x component: m1v1i  0  m1v1 f cos   m2v2 f cos 
y component: 0  0  m1v1 f sin   m2v2 f sin 
For the collision is elastic, we get the third equation for
conservation of kinetic energy:
1
1
1
2
2
2
m1v1i  m1v1 f  m2 v2 f
2
2
2
Solve the set of equations composed of above three
equations, we obtain:
v1 f sin 
tan  
v1 f cos   v1i
2
m
1
1
m1v12i  m1v12f  1  v12i  v12f  2v1i v1 f cos  
2
2
2m2
The Centre of Mass
（质心）
The center of mass of a system of particles is defined as:
rCM 

m r
i i
i
 xCM i  yCM j  zCM k
M
i  mi xi  j  mi yi  k  mi zi
i
For an extended object reads:
i
i
M
rCM
1

rdm

M
The centre of mass of a homogeneous, symmetric body must
lie on an axis of symmetry.
The centre of mass of a system is different from its centre of
gravity. Each portion of a system is acted on by the gravitational
force. The net effect of all of these forces is equivalent to the
effect of a single force Mg acting at a special point called the
center of gravity. The centre of gravity is the average position of
the gravitational force on all parts of the object. If g is uniform
over the system, the centre of gravity coincides with the centre
of mass. In most cases, for objects or systems of reasonable size,
the two points can be considered to be coincident.
E.g. 0.1 A system consists of three particles located at
the corners of a right triangle as in the figure. Find the
centre of mass of the system.
E.g. 0.2 A rod of length 30.0 cm has a linear density:
  50.0 g/m  20.0 x g/m
2
where x is the distance from one end, measured in meters.
(a)What is the mass of the rod?
(b)How far from the x = 0 end is its center of mass?
Motion of a System of Particles
rCM 
MaCM
 mi ri
dri
mi

drCM
dt
i

 vCM
dt
M
i
M
dpt
  mi ai  Ft 
dt
i
质心运动定律
MvCM
dri
  mi
 pt
dt
i
E.g. 0.2 A rod of length 30.0 cm has a linear density:
  50.0 g/m  20.0 x g/m
2
where x is the distance from one end, measured in meters.
(a)What is the mass of the rod?
(b)How far from the x = 0 end is its center of mass?
Motion of a System of Particles
rCM 
MaCM
 mi ri
dri
mi

drCM
dt
i

 vCM
dt
M
i
M
dpt
  mi ai  Ft 
dt
i
质心运动定律
MvCM
dri
  mi
 pt
dt
i
Outline of Rotational Motion
Rigid body model: A rigid body is any system of particles in which
the particles remain fixed in position with respect to one another.
Rotation about a fixed axis: Every particles on a rigid body has
the same angular speed and the same angular acceleration.
 d 
  lim

t  t
dt
 d 
  lim

t  t
dt
rigid body 刚体
rotation about a fixed axis 定轴转动
Rotational kinematics
（转动学）
The rigid body under constant angular acceleration
（常角加速度转动的刚体）
 d 
  lim

t  t
dt
 d 
  lim

t  t
dt
Relations Between Rotational and Translational Quantities
（转动量和平移量之间的关系）
translational motion 平动
Rotational Kinetic Energy
（转动动能）
the moment of inertial
I   mi ri
2
i
rotational kinetic energy:
1 2
K R  I
2
For an extended system:
I  lim
mi 0
2
2
2
r

m

r
dm


r
i i 
 dV
i
The Rigid Body under a Net Torque
（力矩作用下的刚体）
The net torque acting on the rigid body is proportional to
its angular acceleration.
 t  I
(转动定律)
The Rigid Body in Equilibrium
（处于平衡状态的刚体）
The torque vector
（力矩）
  r F
Two conditions for complete equilibrium of an object:
Ft  0;  t  0
translational equilibrium
rotational equilibrium
Work and Energy in Rotational Motion
（转动中的作功与能量）
Work done by a torque
The power of a torque
f
f
i
i
W    d  
dW   d  Id
dW
 
dt
1 2 1 2
I  d  I  f  I i
2
2
Angular Momentum
（角动量）
The angular momentum of the particle relative to the origin is defined as:
Lrp
dL
dp
r

dt
dt
Conservation of Angular Momentum
（角动量守恒）
The total angular momentum of a system remains constant
if the net external torque acting on the system is zero.
一个刚体的运动，可以视为二种运动所组成，即质量
中心，受到所有外力作用所引起的运动，加上物体在外力作用
下，绕质心的转动。一个刚体的动能，系质心的平移运动的动
能，加上绕质心运动的转动动能。
引自《古典动力学》，吴大猷
刚体的运动总可以分解为质心的平动刚体的转动的合成。
A Brief Summary of Part I
① Kinematics of a Particle
② Newton’s Laws of Motion
③ Work and Energy
④ Momentum and Impulse
⑤ Motion of a System of Particles
⑥ Rotations of a Rigid Body about a Fixed Axis
The End of Part I