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1、University Physics,Chapter 1 Kinematics 运动学（质点运动学）,Key Words,kinematics 运动学,vector 矢量,scalar 标量,mass point/particle 质点,calculus 微积分,uniform 均匀的,definition 定义,atomic standard 原子标准,frame of reference 参考系,coordinate systems 坐标系,vacuum 真空,function 函数,meridian 子午线,axis/axes(pl.)(坐标)轴,origin 坐标原点,dimensio
2、n 维,mutually perpendicular 互相垂直,intersection 交点,magnitude 大小,unit 单位,direction 方向,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
3、 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: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。,1-1 Frame of
4、Reference Particle(质点）,1.Frame of Reference（参照系）,When we discuss the position and the velocity（速度）of an object（物体）,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（描述）the motion of
5、a given body,the indications（结果）will be different.,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（加速度）and orbit（轨道）of the object can be specifi
6、ed quantitatively（定量地）.Cartesian Coordinate system(直角坐标系）：,Quantitatively:定量地,2.Particles（质点）,Particle(质点)is an ideal model（模型）,in some circumstances（情况、形势）.We can treat a body as a particle,and concentrate on（集中）its translational motion(平动)and ignore（忽略）all the other motions.,质点：有质量无大小无体积,3.Time（时刻
7、）and time interval(时间）,Time t is a given instant（时刻）,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）is used in China,kg:千克 kilogram,m:米 meter,
8、s:秒 second,5.Scalar and Vector（标量和矢量）:,Two types of physical quantities（量）:,Scalars:mass,length,speed（速率）,temperature（温度）.,Vectors:velocity,acceleration,momentum（动量）.,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
9、）point P and denoted（标记）by,Addition（加）:The two vectorsA and B is added in followingway:C=A+B B A,C,A,B,In Cartesian coordinate system（直角坐标系）:,are unit vectors along ox,oy,oz.,In two dimension（维）:,Obviously(显然）:,In one dimensionIn two dimensionIn three dimension,In our teaching,we will mainly deal wi
10、th（涉及）two dimensional motions:motion in a plane.,Mechanical motions,（机械运动）,Frame of reference:Coordinate system（坐标系）,Body:a particle,Physical quantities:position,displacement,velocity,acceleration,energy,momentum,1-2 Position Vector and Displacement,1.Position Vector,Position vector is a vector that
11、 extents from the origin of the coordinate system to the particles position as shown in Figure,Magnitude:,Key Words,position vector 位置矢量displacement vector 位移矢量velocity 速度limit 极限average velocity 平均速度instantaneous velocity 瞬时速度segment 段，一段curved path 弯曲路径,magnitude and direction大小和方向,components 分量,a
12、cceleration 加速度,ratio 比值，比率,derivative 导数,tangent 相切、切线,coefficient 系数,differentiate v.微分、求导,In the two dimension:,Its two components（分量）,Path equation（轨迹方程）,2.Displacement(位移):,Displacement is introduced to describe the change in position during a given time interval:,That is,Its magnitude（大小),The
13、geometrical（几何）meaning of and the differences among them.,Note:,Example 1.1:A particle is located at at t1 and at at t2.Find the displacement in this time interval.,Solution:,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（
14、速率）:,（Instantaneous 瞬时）velocity at time t：,It is in the tangent(切线）of the path and points at the advance direction.,Direction:,Magnitude（大小）：,V-speed(瞬时）速率,时弧长等于弦长,In the coordinate system:,Magnitude of the velocity:,The angle formed between and+x direction is determined by,Example 1-1:A rabbit runs
15、 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,解,两边求导得,因,
16、选如图的坐标轴,即,2.Acceleration(加速度),Average acceleration:,Instantaneous acceleration,NOTE:,The directions of the position,velocity and acceleration vectors are not necessarily the same位置、速度和加速度矢量的方向未必相同,An object is not necessarily moving in the same direction as it is accelerating.物体的运动方向和加速度的方向未必相同。,Eit
17、her the acceleration is positive or negative doesnt necessarily mean the velocity is increasing or decreasing respectively.加速度的正负并不意味速度的增减。,In the coordinate system:,Its magnitude and direction:,指向曲线凹的一方,Example 1.3:The of a Particle is,where and are constants.Find the velocity and acceleration.,Not
18、e:微分，细心，再细心！Carefully!,Solution:,Example 1.4 已知质点运动方程为x=2t,y=192t2,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),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,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
19、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：,1-4 Two Types Problems in Kinematics,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.,（2）Given accelerat
20、ion(or velocity)and initial condition,find the velocity and position vector by means of vector integration method 积分法.,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,Example1.7:A particle moves in a plane with an acce-leration,where g is constant.When t=0,its velocity is at
21、 a initial point(0,0).Find its velocity at time t and path equation.,Using,we have,Using,and the initial condition(0,0),we have,Solution:Its velocity is,The position vector of the particle is,1-5 Circular Motion,1.The importance of Circular（圆周）motion,(1)The movements of Sun,Earth,Planets（行星）,Electro
22、n,.,are related to circular motion;,Key Words,circular motion 圆周运动uniform circular motion 匀速圆周运动circular motion with varying speed变速圆周运动arc length 弧长 center of a circle 圆心angular position 角位置 radius 半径radian(s)弧度 tangential acceleration 切向加速度rotate v.旋转 centripetal acceleration 向心加速度rotation n.旋转 no
23、rmal acceleration 法向加速度diameter 直径,(2)There are parts of instruments（仪器）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 circula
24、r motion if it travels around a circle or a circular arc（弧）.,Uniform circular motion(匀速）:around a circle and at constant speed.,2.Tangential（切向）and 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 vect
25、or 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.,Hence,the acceleration of particle is:,(1-23),Obviously（显然）,we have,Using and,the velocity can be expressed as（表示成）:,(1-24).,It is easy to prove the rate
26、 of the tangential unit vector to be equal to,t,o,t+t,Prove:when,we have,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,Angle:,Magnitude of:,Changes the magnitude of the velocity;,Changes the dire
27、ction of the velocity.,(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）.,Therefore,we have,3.General curvilinear motion:,(1-30),=AC is the radius of curv
28、ature（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,4.Angular variables(角量）in circular motion,Angular position,Position function,Angular displacement,angular velocity&angular acceler
29、ation,角速度：,角加速度：,Relation between linear（线）&angular variables：,Counterclockwise(逆时针）:positive directionClockwise（顺时针）:negative direction,Two directions:,与匀变速率直线运动类比,匀变速率圆周运动,Solution:,请同学完成,According to,We can be obtained,请同学完成,1.10 一质点沿半径为R 的圆周按,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？,速度方
30、向与圆周相切并指向前方.,解：（1）已知运动轨道的问题，选用自然坐标系。,(2)根据,讨论：运动的性质，过程，总加速度的方向如何？,得,时总加速度大小等于b。,1-6 Relative motion,1.Relative motion(相对运动）,物体运动的轨迹依赖于观察者所处的参考系,Key Words,relative velocity 相对速度relative acceleration 相对加速度observer 观察者differentiate 求微分Galilean transformation equation 伽利略变换(reference)frame 参考系relative t
31、o 相当于connected velocity 牵连速度,The values of the position,velocity and 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,3.Theorems（定理）of velocity addition（相加）and acceleration addition,Let
32、and we have:,(1),Assuming(假设）that O and O coincide at t=0 and moves along the x-axis at speed of u,we have,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:,(2)The difference between this theorem and superposition(叠加）of motion.,(1)It is a vector equation;,1.12 某人骑摩托车向东前进，其速率为10m.s1时觉得有南风，当速率增大到15m.s1时，又觉得有东南风。试求风的速度？,（2）风对地：，人对地：，风对人：,故有：,从上面的几何关系可得：,（3）由题意有：,
