大学物理(英文版)
多媒体课件
Introduction
Chapter 1 Kinematics
Chapter 2 Newton’s Laws of Motion
Chapter 3 Work and Energy
Chapter 4 Momentum
Chapter 5 Rotation of a rigid body
Chapter 6 The Kinetic Theory of Gases
Chapter 7 Fundamentals of Thermodynamics
Volume 1
Introduction
2001.9.11 Catastrophe(大灾难)
宇宙:约 1250亿个星系,
每个星系由数千亿个恒星
组成。
银河系
太阳系:地球,星星
看得见的:你我他它
分子
原子
原子核
基本粒子
相对论
天体物理
经典物理,力学,热等
量子力学
核物理
量子场论




银河系 相对论
天体物理
量子天体
物理学
史蒂芬,霍金
,时间简史,

The Galaxy
Sun,Earth,Planets
The body we can see
Molecules
atoms
nuclei
elementary particles


The general reletivity
astrophsics
Newton’s Mechanics
Heat,Thermodynamics …
Electromagnetic Theory
Quantum Mechanics
Nuclear Physics
Quantum Field Theory

TheoryO
ur
w
or
ld
an
d u
nive
rse
宇宙半径,~1026 m
地球 1024 kg
银河系,~1044 kg
我们的母亲:
地球
原子核半径,10-15 m
电子质量,10-31 kg
河外星系,1024 m
银河中心系,1020 m
太阳 1030 kg,1011 m
月亮,108 m
1969年 7月 16日
美国东部时间 9时 23分
阿波罗 11号发射升空。
三天后
阿姆斯特朗
奥尔德林
柯林斯
Mars(火星)
机遇号
The surface of Mars(火星表面)
Our world and Universe
Universe Elementary particles
The ancient physics
The classical physics
The modern physics
In the view of physics history:
主要讲授内容:
经典力学
相对论
热学
电磁学
波动光学振动与波动
量子论简介
日常生活
Physics
Chemistry
化学
Biology
生物学
Computer
计算机科学
Mechanics
机械学
Medicine
医学
Physics,fundamentals and methods.
References(参考书)
?张达宋, 物理学基本教程,
?李行一等,,物理学基本教程教学参考书,
?李行一等,,物理学基本教程, 习题分析与解答
?张三慧等,,大学物理学,
?Halliday et.al, Fundamentals of Physics》
?W,Sears et.al, University Physics》
?史蒂芬,霍金,,时间简史,
?盛正卯等,,物理学与人类文明,
?B.K.里德雷,《时间、空间和万物》
?…………..
Part One Mechanics
力学
Chapter 1 Kinematics
(运动学)质点运动学
第一章 质点运动学 (Kinematics)
§ 1-1 参考系 质点 Frame of reference particle
§ 1-2 位置矢量 位移 Position vector and displacement
§ 1- 3 速度 加速度 Velocity and acceleration
§ 1-4 两类运动学问题 Two types of Problems
§ 1-6 运动描述的相对性 Relative motion
§ 1-5 圆周运动及其描述 Circular motion
1,理解描述质点运动物理量的定义及其矢量性、相
对性和瞬时性;
2,掌握运动方程的物理意义,会用微积分方法求解
运动学两类问题;
3,掌握平面抛体运动和圆周运动的规律;
4,理解运动描述的相对性,会用速度合成定理和加
速度合成定理解题。
教 学基本 要 求
重要历史人物
伽利略 Galileo Galilei,
1564~1642意大利物理学
家、数学家、天文学家,
近代实验科学的创始人。
主要贡献:
?发明了望远镜,维护、坚持和发展了哥白尼学
说,发现木星的四个卫星;
?摆的等时性、惯性定律、落体运动定律;
?运动的合成原理和独立性原理,相对性原理;
?方法:实验科学。
§ 1-1 Frame of Reference Particle(质点)
1,Frame of Reference(参照系)
When we discuss the position
and the velocity( 速度 ) of an
object,we must answer the
questions:
“position with respect
to(相对于) what?” and
“Velocity with respect
to what?”
If we choose different
objects as the reference
frames to describe the
motion of a given body,the
indications(结果 ) will be
different.
It is convenient to
take the earth’s surface
as our frame of
reference in most cases
in this course.( What
cases?)
Coordinate system ( 坐标
系 ), fixed on the frame,
relative to which position,
velocity,acceleration and
orbit of the object can be
specified quantitatively.
Cartesian Coordinate
system(直角 坐标系):
o X
Z Figure 1-2
Quantitatively:
定量地
2,Particles( 质点)
Particle(质点 ) is an ideal model,in some
circumstances( 情况, 形势 ), We can treat a
body as a particle,and concentrate on its
translational motion(平动 ) and ignore( 忽略 )
all the other motions,点:
有质量
无大小
无体积
3,Time(时刻) and time interval(时间)
Time t is a given instant,and time
interval(间隔 ) △ t is the difference of two
given instants,We use the former to describe
( 描述 ) the state of the object,the latter
to describe the process.( 过程 )
4.Units(单位)
International System of Units(SI,Syst?me
Internationald’Unit?s 法语 ) is used in China
kg:千克 kilogramm
length m:米 meterL
Time t s:秒 second
mass
5,Scalar and vector(标量和矢量),
Two types of physical quantities(量),
Scalars,mass,length,speed,temperature…..
Vectors,velocity,acceleration,momentum...
Vector A( black), its magnitude(大小 ) and
direction( 方向 ) may be represented by a
line OP directed from the initial point O to the
terminal( 终 ) point P and denoted( 标记 )
by
OP
o
PA
Addition(加), The two vectors
A and B is added in following
way:
C=A+B B A
C
A
B
In Cartesian coordinate system(直角坐标系),
kAjAiAA zyx ??? ???
are unit vectors along OX,OY,OZkandji ???,
O X
Y
Z
In two dimension(维),
jAiAA yx ?? ??
O X
Y
xA
yA
?
x
y
A
A
tg ??
jBAiBAABjCiCC yyxxyx ???? )()( ????????
If
jAiAA yx ?? ??
and
jBiBB yx ?? ??,we have:
xxx BAC ?? yyy BAC ??
Obviously(显然),
In one dimension
In two dimension
In three dimension
In our teaching,we will mainly deal with(涉及)
two dimensional motions,motion in a plane.
Rectilinear motion(直线)
Curvilinear motion(曲线)
Circular motion(圆周)
Mechanical motions
(机械运动)
§ 1-2 Position Vector and Displacement
P(x,y,z)
z
r?
YX
1,Position Vector
Position vector is a vector
that extents from the origin
of the coordinate system to
the particle’s position as
shown in Figure
r?
kzjyixr ???? ???
222 zyxrr ???? ?Magnitude:
r
x??c o s
r
z??c o s
r
y??co s
In the two dimension:
jtyitxtrr ???? )()()( ???
Its two components
(分量)
)(
)(
tyy
txx
?
?
)( xyy ?
Path equation(轨迹方程)
eliminating消去
x
y
o
P
r?
)y,x(
2,Displacement(位移 ):
Displacement is
introduced to describe
the change in position
during a given time
interval:
r??
x
y
o
P
1r
?
1t
2t
2r
?
r??
12 rrr
??? ???
That is
jyyixx
jyixjyixr
??
?????
)()( 1212
1122
????
?????
x
y
o
P
1r
?
1t
2t
2r
?
r??
Its magnitude(大小 )
212212 )()( yyxxr ????? ?
The geometrical(几何) meaning of
and the differences among them,rrs
???,,
r?
s
Note:
Example 1.1,A particle is located at at jir ??? 751 ??
and at at, Find the displacement
in this time interval.
jir ??? 532 ???
1t 2t
Solution:
ji
ji
rrr
??
??
??
28
7553
12
???
??????
???
)()(
§ 1-3 Velocity(速度) and Acceleration(加速度)
Average(平均) velocity:
1.Velocity
12
12
tt
rr
t
rV
?
??
?
?? ????
which has a direction
as same as that of r??
Average speed(速率),
t
sV
?
???
所用的时间
走过的路程
x
y
o
P
1r
?
1t
2t
2r
?
r??
s?
x
y
o
P
r?
t
2t
2r
?
r??
V?
( Instantaneous 瞬时)
velocity at time t:
t
r
t
rtV
t d
d
0
???
????
??
lim
It is in the tangent(切线 ) of the path and
points at the advance direction.
Direction:
x
y
o
P
r?
t
V?
?
Magnitude(大小):
Vtstrt
t
?????
?? d
dV
0
??
l i m
V---speed(瞬时)速率
时弧长等于弦长0??tIn the coordinate system:
jViV
j
t
y
i
t
x
t
r
V
yx
??
????
??
???
d
d
d
d
d
d
t
y
V
t
x
V
y
x
d
d
d
d
?
?
x
y
o
P
r?
t
V??
xV
?
yV
?
Magnitude of the velocity:
22
yx VVV ??
The angle ? formed
between and +x
direction is determined
by
V?
x
y
V
V1?
? t an?
Example 1-2,A rabbit runs across a parking lot(近路 )
on which a set of coordinate axes has,strangely enough,
been draw,The coordinates of the rabbit’s position as
function of time t are given by:
3019220
2827310
2
2
???
????
tty
ttx
..
..
with t in seconds and x and y in meters,Find its
velocity at t=0.50s.
Solution,
jtitjtyitxV ????? )..()..( 1944027620dddd ???????
The rabbit’s velocity at t=0.50s is equal to(等于)
jiV ??? 9896,,??
2.Acceleration(加速度 )
x
y
o
P
1t
1V
?
2t
2V
?
1V
?
2V
?
V??
Average acceleration:
12
12
tt
VV
t
Va
?
??
?
?? ????
Instantaneous acceleration
2
2
0 d
d
d
d
t
r
t
V
t
Vta
t
????
?????
??
l i m
In the coordinate system:
x
y
o
Pt
a?
jaia
j
t
y
i
t
x
j
t
V
i
t
V
a
yx
yx
??
??
???
??
??
??
2
2
2
2
d
d
d
d
d
d
d
d
Its magnitude and direction:
22
yx aaa ??
xa?
ya?
?
x
y
a
a1?? t an?
指向曲线
凹的一方
Example 1.3,The of a Particle is r? jtitr ??? 322 ?? ??
where ? and ? are constants,Find the velocity and
acceleration.
jtitV ??? 234 ?? ?? 4222 916 ttV ?? ??
222 3616 ta ?? ??jtia ??? ?? 64 ??
Note,微分,细心,再细心!!
Carefully!!
Solution:
Example 1,4 已知质点运动方程为 x=2t,y=19?2t2,式中 x,y以米计,
t 以秒计,试求:( 1)轨道方程;( 2) t=1s 时的速度和加速度。
2
2
119 xy ??
( 2)对运动方程求导,得到任意时刻的速度
对速度求导,得到任意时刻的加速度:
解:( 1)运动方程联立,消去时间 t得到轨道方程
??
?
??
?
2219
2
ty
tx
( 1)
?
?
?
??
?
tV
V
y
x
4
2
(2)
?
?
?
??
?
4
0
y
x
a
a
将时间 t=1s代入速度和加速度分量式 (1),(2)中,求出时间 t=1s
对应的速度和加速度:
速度大小和与 x 轴夹角
jitV ??? 421 ??? )(
jta ?? 41 ??? )(
122 474 ???? msVVV yx,
????? ? 4632 41,t a n?
加速度大小和方向:
24 ??? msa 与 y轴正向相反
Example 1-5 离水平面高为 h 的岸边,有人用绳以恒定速率 V0拉船
靠岸。试求:船靠岸的速度,加速度随船至岸边距离变化的关系式?
对时间求导得到速度和加速度:
由题意知:
o
y
r
?
h
x
x0
v解,在如图所示的坐标系中,船的位矢为:
jhixr ??? ??
i
t
x
a
i
t
x
V
??
??
2
2
d
d
d
d
?
?
t
rV
d
d??
?
因为:
222 hxr ??
i
x
hV
a
i
x
h
VV
??
??
3
22
0
2
0 1
??
?
?
?
?
?
?
???
§ 1-4 Two Types Problems in Kinematics
( 2) Given acceleration(or velocity) and initial
condition,find the velocity and position vector by
means of vector integration method 积分法,
In general,there are two kinds of problems to
be solved:
(1) Given position vector,find the velocity and
acce-leration by using derivation method 微分法,
See the examples above.
解:整理和分离变量可得下面方程
k t d tVdV ??? 2
做积分:
Example1.6,某物体的运动规律为 tkVdtdV 2??
式中 k为常数,t=0,初速度为
0V
求,)(tV
?? ???
tV
V
k t d tVdV
0
2
0
得:
请同学们完成积分
Example1.7,A particle moves in a plane with an acce-
leration,where g is constant,When t=0,its
velocity is at a initial point
(0,0),Find its velocity at time t and path equation.
jga ?? ??
jViVV ??? ?? s i nc o s ??? ??
Solution,From jga ?? ??,we can obtain:
dtjgVd ?? ??
jgtViVVjgtV ????? )s i n(c o s ?????? ?? 000
???? VjgtV
???
Using,we have jViVV ??? ?? s i nc o s
??? ??
Using
jdtdyidtdxV ??? ??
and the initial condition(0,0),we have
tVx ?co s0?
2
0 2
1s i n gttVy ?? ?
?? 220
2
c os2
1
v
gxtgxy ??
Example1.8,The acceleration of a particle is given by
jita ??? 0203 2,,??
where ? and ? are constant,The initial conditions are
iVr ??? 010 00,??Find its velocity and position vector.
Solution,Its velocity is
jtit
ijtitVtaV
t
Va
??
????????
0201
0102d
d
d
3
3
0
.).(
..
???
??????? ?
The position vector of the particle is
jtittrtVrtrV ?????
??
2
4
0 4dd
d ??????? ? )(
§ 1-5 Circular Motion
1.The importance of Circular motion
(1) The movements of Sun,Earth,Planets,
Electron,…,.,are related to circular motion;
(2) There are parts of instruments associated
with the circular motion,clock,car,……,
(3) The knowledge on circular motion is the base to
study the general curvilinear motion( 曲线运动 ),
You can accept(采
用 ) the above
method to study
Circular Motion。
A particle is in
circular motion if it
travels around a circle
or a circular arc( 弧 ),
Uniform circular
motion(匀速 ), around
a circle and at constant
speed.
o
),( yxP
r?
tc o n sr t a n?
2.tangential(切向) & normal(法向)
components of acceleration
⑴ The nature coordinate system(自然坐标系)
Two unit vector are
introduced to describe the
circular motion:
is an unit vector tangent
(相 切 )to the circle at A
directing to the advance
direction and an unit
vector normal to the circle
at a ( 法向 ) directing
toward the center o.
??
n?
??
n?
A
o
??? VV ?
Hence( 所以 ),the
acceleration of particle
is,
t
Va
d
d
?
?
?
(1-23)
Using and,the velocity can be expressed
as( 表示成 ),
?? n?
Obviously(显然),we have
tVt
VV
tt
Va
d
d
d
d
d
d
d
d ??? ???
??
???? )(
(1-24).
??
n?
A
o
It is easy to prove the rate of the tangential unit vector
to be equal to
??
nRVdtd ?
?
??
??
t
o
??
t+?t
??
???
n
R
tV
n
?
??
?
?
???? ?? 1
Prove,when,we have0??t
R
tV
R
s ??????
To summarize(总结),we have
nt
nt
aa
naan
R
V
dt
dV
a
??
?????
??
???? ??
2
ta?
na
?and are called the tangential acceleration( 切向加
速度 ) and normal acceleration ( 法 向 加 速 度 )
respectively,and their magnitudes are given by
R
V
a
t
V
a
2
n ?? d
d
t
22
nt aaa ??
Angle,
n
t
a
atg 1??α
Magnitude of,a?
?
a?
na
?
ta
?
ta?
Changes the magnitude of the velocity;
na
? Changes the direction of the velocity.
(2)Uniform circular motion (匀速 ),
In this case, the magnitude of velocity is a
constant,that is
0?tdd V
which means that velocity changes only in direction,is
usually called the centripetal acceleration( 向
心 ),
na
?
na
?
v
v
R
Vaa 2
n ??,t 0
Therefore,we have
3.General curvilinear motion:
ρ
V
a
t
V
a
2
?
?
n
d
d
t (1-30)
AC?ρ
is the radius of
curvature( 曲率 ) at A
and C is the center of
curvature circle ( 曲率
圆 ),
A small part of curvilinear path can be considered as
a part of a circle as shown in the below figure,We have
ta?
na
?
?
A
C a?
V?
o
),( yxP
tc o n sr t a n?
?
?? ??
4,Angular variables(角量) in circular motion
⑴ Angular position
Position function?
)(t?? ?
Angular displacement
??
⑵ angular velocity & angular acceleration
ttl i mt d
d ?
?
???
?
??
??
??? ??? tttlimt d
d
d
d 2??
?
???
?
? - rad
?- rad.s-1
?- rad.s-2
Units:
角速度:
角加速度:
o
),( yxP
tc o n sr t a n?
?
?? ??
t?
Relation between linear(线) & angular
variables:
?
?
?
?
?
??
?
??
?
??
?
????
ra,ra
rV
rS,rS
nt
Counterclockwise(反时针), positive direction
Clockwise(顺时针), negative direction
Two directions:
),(),,( ??oraaV nt
r
请同学自己推导!
试根据,的不同,讨论相应的运动。
tn aa n da
For example:
caa tn ??,0
匀加速直线运动
ta?
na
?
?
A
C a?
V?
Example 1.9,The of a Particle is r?
jtitr ??? 322 ?? ??
where ? and ? are constants,Find the tangential and
normal accelerations at time t.
jtitV ??? 234 ?? ?? 4222 916 ttV ?? ??
222 3616 ta ?? ??jtia
??? ?? 64 ??
Using 4222 916 ttV ?? ??,we have:
?? dtdVa t
请同学完成
According to 22222
tnyx aaaaa ????
,
na can be obtained
?na 请同学完成
Solution:
1.10 一质点沿半径为 R 的圆周按 20 21 bttVs ??
规律运动, V0,b 是正值常数 。 求,( 1) t 时刻总加速度? ( 2)
t 为何值时总加速度大小等于 b?
btVtsV ??? 0dd
速度方向与圆周相切并指向前方,
n
R
btV
ba
R
btV
R
V
a
b
t
V
a
n
??? 20
2
0
2
d
d
)(
)(
?
????
?
?
?
??
?
?
?
??
???
?
?
b
R
btVbaaa
n ?
?????
2
4
0222 )(
?
( 2)由 得
b
Vt 0?
讨论:运动的性质,过程,总加速度的方向如何?
解:( 1)已知运动轨道的问题,选用自然坐标系。
§ 1-6 Relative motion
1,Relative motion(相对运动)
The values of the position,
velocity & acceleration of a
object depend on( 依赖 ) the
frame of reference in which the
quantities( 量 ) are measured.
2.Relativity of the description about a motion
观测者 1
观测者 2 观测对象
o
x
y
z
o?
x?
y?
z?
),,( aVr ??? ),,( aVr ??? ???
P
2,Theorems(定理) of velocity addition(相加) &
acceleration addition
ox
y
z
o?
x?
y?
z?
PBr?
PAr?
BAr?
Let and we have,zoozyooyxoox ?????? ||,||,||
PBBAPA rrr
??? ?? (1)
P
o
x
y
z
o? x?
y?
z?
),,(
),,(
zyx
zyx
???
u
Assuming(假设) that O and O’ coincide at t=0 and
moves along the x-axis at speed of u,we have zyxo ????
?
?
?
?
?
?
?
??
??
??
???
tt
zz
yy
utxx
Galilean transformations
(伽利略变换)
Taking time derivative of (1),we have
PBBAPA
PBBAPA VVV
t
r
t
r
t
r ?????? ?????
d
d
d
d
d
d
绝对速度 牵连速度 相对速度
which is called as Theorem of velocity addition,That
means( 意义为 ) that the velocity of P with respect to
( 相对于 ) A is equal to that with respect to B plus(加 )
the velocity of B with respect to A,
Note,(1) It is a vector equation;
(2) The difference between this theorem and
superposition(叠加 ) of motion.
Taking time derivative of velocity equation,we have
PBBAPA
PBBAPA aaa
t
V
t
V
t
V ???
???
?????
d
d
d
d
d
d
绝对加速度 牵连加速度 相对加速度
which is called as Theorem of acceleration addition,That
means that the acceleration of P with respect to A is equal
to that with respect to B plus(加 ) the acceleration of B
with respect to A,
1.12 某人骑摩托车向东前进,⑴ 其速率为 10m.s?1时觉得
有南风,⑵ 当速率增大到 15m.s?1时,又觉得有东南风。
试求风的速度?
解,(1)研究对象:风 速度, 观测者:地球坐标和
骑摩托车的人 。
( 2)风对地:,人对地:,风对人:
风地V
?
人地V
? 风人V?
故有:
风人人地风地 VVV
??? ??
从上面的几何关系可得:
smV /125510 22 ???风地
'34261051 ??? ?tg?
( 3)由题意有:
人地V
? (=10m/s)
风地V
?
?
人地V
? (=15m/s)


风地V
?
45
Example 1.13,一飞机相对空气的速度为 200Km/h,风
速为 56Km/h,方向从西向东,地面雷达测得飞机速度
大小为 192Km/h,方向是:
( A)南偏西 16.3 ° ( B)北偏东 16.3 °
( C)向正南或正北 ( D)西偏东 16.3 °
( E)东偏南 16.3°
风V
?
相V?