大学物理英文版课件

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,UNIVERSITY PHYSICS 1,UNIVERSITY PHYSICS 1,1,大学物理（英文版）,多媒体课件,大学物理（英文版）,2,Introduction,Chapter 1 Kinematics,Chapter 2 Newtons 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,IntroductionChapter 1 Kinemat,3,Introduction,Introduction,4,2001.9.11 Catastrophe（大灾难）,2001.9.11 Catastrophe（大灾难）,5,宇宙：约1250亿个星系，每个星系由数千亿个恒星组成。,宇宙：约1250亿个星系，每个星系由数千亿个恒星组成。,6,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,物质世界,银河系相对论物质世界,7,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,银河系相对论量子天体史蒂芬.霍金,8,The Galaxy,Sun: Earth,Planets,The body we can see,Molecules,atoms,nuclei,elementary particles,The general reletivity,astrophsics,Newtons Mechanics,Heat,Thermodynamics ,Electromagnetic Theory,Quantum Mechanics,Nuclear Physics,Quantum Field Theory,Theory,Our world and universe, TheoryO,9,宇宙半径：10,26,m,地球10,24,kg,银河系：10,44,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：10,24,m,银河中心系：10,20,m,太阳10,30,kg：10,11,m,月亮：10,8,m,宇宙半径：1026 m地球1024 kg银河系：1044,10,1969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,1969年7月16日三天后,11,大学物理英文版课件,12,Mars(火星）,Mars(火星）,13,机遇号,机遇号,14,The surface of Mars(火星表面）,The surface of Mars(火星表面）,15,Our world and Universe,大学物理英文版课件,16,大学物理英文版课件,17,Universe,Elementary particles,UniverseElementary particles,18,The ancient physics,The classical physics,The modern physics,In the view of physics history:,The ancient physicsThe classic,19,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,主要讲授内容：经典力学热学电磁学波动光学振动与波动量子论简介,20,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.,PhysicsChemistryBiologyCompute,21,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,.,References(参考书）张达宋 物理学基本教程,22,Part One Mechanics,力学,Part One Mechanics,23,Chapter 1 Kinematics,（运动学）质点运动学,Chapter 1 Kinematics,24,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,13 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,第一章 质点运动学(Kinematics)1-1 参,25,1. 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,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,26,重要历史人物,伽利略Galileo Galilei: 15641642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。,重要历史人物伽利略Galileo Galilei: 1564,27,1-1 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?”,1-1 Frame of Reference Partic,28,If we choose different objects as the reference frames to describe the motion of a given body,the indications(,结果） will be different.,If we choose different obj,29,It is convenient to take the earths 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,It is convenient to take,30,2. 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,2. Particles（质点） Particle(质,31,3.,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）,3. Time（时刻）and time interval,32,4.Units(,单位）,International System of Units(SI: Syst,me International dUnits 法语）,is used in China,kg:,千克 kilogram,m,length,m:,米 meter,L,Time,t,s:,秒 second,mass,4.Units(单位）International Syste,33,5. 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,5. Scalar and vector（标量和矢量）:Tw,34,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）:,are unit vectors along OX,OY,OZ,O,X,Y,Z,Addition（加）: The two vectorsC,35,In two dimension（维）:,O,X,Y,If,and, we have:,Obviously(,显然）:,In two dimension（维）:OXYIf and,36,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）,In one dimensionIn our teachin,37,1-2,Position Vector and Displacement,P,(,x,y,z,),z,Y,X,1. Position Vector,Position vector is a vector that extents from the origin of the coordinate system to the particles position as shown in Figure,Magnitude:,1-2 Position Vector and Displ,38,In the two dimension:,Its two components,（分量）,Path equation(轨迹方程）,eliminating,消去,In the two dimension:Its two c,39,2. Displacement(,位移):,Displacement is introduced to describe the change in position during a given time interval:,That is,2. Displacement(位移):Displaceme,40,Its magnitude（大小),The geometrical（几何） meaning of and the differences among them.,Note:,Its magnitude（大小)The geometric,41,Example 1.1: A particle is located at at,and at at . Find the displacement in this time interval.,Solution:,Example 1.1: A particle is loc,42,1-3 Velocity（速度） and Acceleration（加速度）,Average（平均） velocity：,1.Velocity,which has a direction as same as that of,Average speed(,速率）:,1-3 Velocity（速度） and Accelera,43,（Instantaneous 瞬时） velocity at time t：,It is in the tangent(,切线） of the path and points at the advance direction.,Direction:,（Instantaneous 瞬时） velocity at,44,Magnitude(,大小）：,V-speed(瞬时）速率,时弧长等于弦长,In the coordinate system:,Magnitude(大小）：V-speed(瞬时）速率时,45,Magnitude of the velocity:,The angle,formed between and +x direction is determined by,Magnitude of the velocity:The,46,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 rabbits position as function of time t are given by:,with t in seconds and x and y in meters. Find its velocity at t=0.50s.,Solution:,The rabbits velocity at t=0.50s is equal to(,等于）,Example 1-2: A rabbit runs acr,47,2.Acceleration(加速度),Average acceleration:,Instantaneous acceleration,2.Acceleration(加速度)Average ac,48,In the coordinate system:,Its magnitude and direction:,指向曲线,凹的一方,In the coordinate system:Its m,49,Example 1.3: The of a Particle is,where, and are constants. Find the velocity and,acceleration.,Note: 微分，,细心，再细心！,Carefully!,Solution:,Example 1.3: The of a Par,50,Example 1. 4 已知质点运动方程为,x,=2,t,y,=19,2,t,2, 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）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,（1）,(2),Example 1. 4 已知质点运动方程为x=2t,51,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,加速度大小和方向：,与y轴正向相反,将时间t=1s代入速度和加速度分量式(1)、(2)中，求出时,52,Example 1-5 离水平面高为,h,的岸边，有人用绳以恒定速率,V,0,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：,解：,在如图所示的坐标系中,，船的位矢为：,因为：,Example 1-5 离水平面高为h 的岸边，有人用绳以,53, 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., 1-4 Two Types Problems in Ki,54,解：整理和分离变量可得下面方程,做积分：,Example1.6:,某物体的运动规律为,式中k为常数，t=0，初速度为,求 .,得：,请同学们完成积分,解：整理和分离变量可得下面方程做积分：Example1.6:,55,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.,Solution: From, we can obtain:,Using , we have,Example1.7: A particle moves i,56,Using,and the initial condition(0,0), we have,Using and the initial conditio,57,Example1.8: The acceleration of a particle is given by,where, and are constant. The initial conditions are,Find its velocity and position vector.,Solution: Its velocity is,The position vector of the particle is,Example1.8: The acceleration o,58,1-5,Circular Motion,1.The importance of Circular motion,(1) The movements of Sun, Earth, Planets , Electron, ., are related to circular motion;,1-5 Circular Motion1.The imp,59,(2) There are parts of instruments associated with the circular motion: clock, car, .,(2) There are parts of instrum,60,(3) The knowledge on circular motion is the base to study the general curvilinear motion（曲线运动）.,(3) The knowledge on circular,61,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.,You can accept(采用） the abo,62,2.tangential（切向） & normal（法向） components of acceleration,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.,A,o,2.tangential（切向） & normal（法向）,63,Hence(,所以）, the acceleration of particle is :,(,1-23,),Using and ,the velocity can be expressed as（表示成）:,Obviously（显然）, we have,(1-24).,A,o,Hence(所以）, the acceleration of,64,It is easy to prove the rate of the tangential unit vector to be equal to,t,o,t+,t,Prove: when, we have,It is easy to prove the rate o,65,To summarize（总结）, we have,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,To summarize（总结）, we have a,66,Angle :,Magnitude of,:,Changes the magnitude of the velocity;,Changes the direction of the velocity.,Angle :Magnitude of,67,(2)Uniform circular motion,(匀速),:,In this case， the magnitude of velocity is a constant， that is,which means that velocity changes only in direction. is usually called the,centripetal acceleration（向心）.,v,v,Therefore, we have,(2)Uniform circular motion (匀速,68,3.General curvilinear motion:,(1-30),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,A,C,3.General curvilinear motion:(,69,4. Angular variables(,角量） in circular motion, Angular position,Position function,Angular displacement,4. Angular variables(角量） in ci,70, angular velocity & angular acceleration,rad,rad.s,-1,rad.s,-2,Units:,角速度：,角加速度：, angular velocity & angular a,71,Relation between linear（线） & angular variables：,Counterclockwise(,反时针）: positive direction,Clockwise（顺时针） : negative direction,Two directions:,r,请同学自己推导！,Relation between linear（线） &,72,试根据： 的不同，讨论相应的运动。,For example:,匀加速直线运动,A,C,试根据： 的不,73,Example 1.9: The of a Particle is,where, and are constants. Find the tangential and,normal accelerations at time t.,Using, we have:,请同学完成,According to,can be obtained,请同学完成,Solution:,Example 1.9: The of a Par,74,1.10 一质点沿半径为,R,的圆周按,规律运动，,V,0,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,？,速度方向与圆周相切并指向前方，,（2,),由,得,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。,1.10 一质点沿半径为R 的圆周按规律运动，V0, b 是,75,1-6,Relative motion,1. Relative motion(,相对运动）,1-6 Relative motion1. Relati,76,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,The values of the positio,77,2. Theorems（定理） of velocity addition（相加） & acceleration addition,Let and we have:,(1),2. Theorems（定理） of velocity ad,78,Assuming(,假设） that O and O coincide at t=0 and moves along the x-axis at speed of u, we have,Galilean transformations,(伽利略变换）,Assuming(假设） that O and O coi,79,Taking time derivative of (1), we have,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 (1),80,Taking time derivative of velocity equation, we have,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 .,Taking time derivative of velo,81,1.12 某人骑摩托车向东前进， 其速率为10m.s,1,时觉得有南风， 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：,故有：,1.12 某人骑摩托车向东前进， 其速率为10m.s1,82,从上面的几何关系可得：,（3）由题意有：,(=10m/s),(=15m/s),东,北,45,从上面的几何关系可得：（3）由题意有：(=10m/s)(=,83,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,Example 1.13: 一飞机相对空气的速度为200Km,84,