南开大学：《力学》课程教学资源（PPT课件，英文讲稿）Chapter 1 Kinematics

Chapter.1Kinematics

Chapter 1 Kinematics

$ 1.IntroductionKinematicsMath (Definition & Description)MechanicsDynamicsPhysics(Reason of motion )Particle (mass point)ModelSystem of particlesRigid bodySpaceFrame of referencecoordinateTimeCartesian coordinatesSpacePlane polar coordinatesNatural coordinates

§1.Introduction Kinematics Math (Definition & Description ) Dynamics Physics(Reason of motion ) Particle (mass point) System of particles Rigid body coordinate Space Time Cartesian coordinates Plane polar coordinates Natural coordinates Mechanics Model Frame of reference Space

S 2.kinematical equations2-1.Position vector and displacementFosition of a particle in 3-D*Position vector= xi +j + zk=-Abstract(Independent ofcoordinates)*Dlisplacement:In Rectangular coordinates =(x2 -x)i +(y2 -y)j+(z2 -z)kp2(x2, y2,22)pi(xi,Ji,z)ry

§2.kinematical equations 2-1.Position vector and displacement Position of a particle in 3-D *Position vector: *Displacement: r xi yj zk     = + + 2 1 s r r    = − Abstract(Independent of coordinates) In Rectangular coordinates s x x i y y j z z k     ( ) ( ) ( ) z = 2 − 1 + 2 − 1 + 2 − 1 y x 2 r  1 r  s  ( , , ) 2 2 2 2 p x y z ( , , ) 1 1 1 1 p x y z

2-2.velocity and Acceleration*Instantaneous velocitydrn-n=lim.= limAtN0tdtA1->0In Rectangular coordinatesdxi+dj+dk=vi+vj+v.kdtdtdtExampleFinding from , the position is givenr= A(ei+e-"j)drAα(eai-e-αj)VdtSpeedmagnitude isv=[|= /v +v,2 =αA(e2a +e-2a)2

2-2.velocity and Acceleration *Instantaneous velocity In Rectangular coordinates lim lim . 0 2 1 0 dt dr t r t r r v t t      =   =  − =  →  → k v i v j v k dt dz j dt dy i dt dx v x y z        = + + = + + Example Finding from , the position is given v  r  r A(e i e j) t t    − = + A (e i e j) dt dr v t t       − = = − Speed magnitude is 2 1 2 2 2 2 ( ) t t x y v v v v A e e    − = = + = + 

Accelerationddd-xd'yi+adt?dtdtdt4We could continue to form mew vectors by taking higher derivativesof r.but we shall see in our study of dynamics that r, y and a areof chief interestExample :Uniform circular motionCentripetalaccelerationr = r(cos wti + sin wti)drV= rw(- sin wti + cos wti)VEdtv = rw(sin - wt +cos* wt) = rwd?rot= rw(- cos wti - sin wti)=-rw?dr?OJX

*Acceleration k dt d z j dt d y i dt d x dt d r dt dv a      2 2 2 2 2 2 2 2 = = = + + We could continue to form mew vectors by taking higher derivatives of .but we shall see in our study of dynamics that and are of chief interest. r  r v   , a  Example :Uniform circular motion r r(coswti sin wtj)    = + rw( sin wti coswtj) dt dr v    = = − + v = rw(sin wt + cos wt) = rw 2 2  = = − − = − = − r r v rw wti wtj rw dt d r a 2 2 2 2 2 ( cos sin )    Centripetal acceleration y x i j r O t

2-3 KinematicalequationsDefinition of v, a.Differential equationsd(t)a(t)d(t) = adtdtdr(t)' dv(t) = f'adt(t)?dtIntegral equationsv(t) - v(to)= ['adt(t)=v(to)+ I a(t)dtr(0)=r(0)+ I' (i)dt(0)=l=-0Initial condition福(t) = f(t)+CFinding C.V, =v,(to)+fa,idt x=x(to)+ f'vedt

2-3 Kinematical equations Definition of v, a.   ( ) ( ) ( ) ( ) v t dt dr t a t dt dv t     = = Integral equations   = + = + t t t t r t r t v t dt v t v t a t dt 0 0 ' ' 0 0 ( ) ( ) ( ) ( ) ( ) ( )       v t f t C ( ) ( ) = + Initial condition Finding C. 0 ( ) 0 t t v t v = =   ' ' 0 0 v v (t ) a t dt t t x = x +  x dv t adt   ( ) =   = t t t t dv t adt 0 0 ( )   − = t t v t v t adt 0 ( ) ( ) 0   = + t t x x t vx dt 0 ( ) 0 Differential equations

ExampleFinding velocity from Acceleration。 =(0,5,-3)m / s, a =(0,0,-2)m / s2Find its velocity after5s(t)= Vo + [ adt = V + [ adtv, =vox +fa,(t)di =0+ ['odtV, = 5m/ sv, = Vo +[a,(t)dt =-3+[(-2)dt =-3-2x5= -13(m / s)

Finding velocity from Acceleration (0,5, 3) / , 0 v = − m s  2 a = (0,0,−2)m/s    = + = + 5 0 0 0 0 v(t) v adt v adt t t        = + = + t t x x x v v a t dt dt 0 ' 0 ' ' 0 ( ) 0 0 v m s y = 5 / Example: Find its velocity after 5s. ( ) 3 ( 2) 3 2 5 5 0 ' 0 ' ' = 0 + = − + − = − −    v v a t dt dt t z z z = −13(m/s)

Example:Motion in a uniform gravitational fieldZa=-gkVx= Vox,V, = VojV, = Vo. + f(-g)dt= Voz - gt0Xx =Xo + Voxt18z = Zo + Vozt - J, gtdt = zo + Vot2

Motion in a uniform gravitational field. . z O x v0  a = −g k  x x y y v v v v 0 0 = , =  = + − t z z v v g dt 0 ' 0 ( ) v gt = 0z − x x v t = 0 + 0x  = + − = + − t z z v z t gtdt z v z t gt 0 2 0 0 0 0 2 1 Example:

ExampleFinding u and a of that boat(v is uniform)x? = 12 - h?1x=v2-h?一工2(-h)3.212iudixDu:-Jp-h?1dl=-v!)dt1y?h?dua:dtx

Finding u and a of that boat(v is uniform) v h L x u X 2 2 2 x = l − h 2 2 x = l − h vi l h l u   2 2 − = − i x v h dt du a    3 2 2 = = − i dt dl i l h l dt dx u    ( ) 2 2 1 2 1 2 2 = = −  − Example: ( !) dl v dt = −

S3.Motion in plane polarcoordinates3-1.Polar coordinatesCartesian coordinates-----motion in a straight linePolar coordinates---circular motionA(cylindrical system, z is nonzero)Aθ r are orthogonal.r=icoso+jsin0(perpendicular)0=-isin0+jcos00yj=rsin0+0cos01i=rcos-sin0Position vector0r=xi+yj=rrX

§3.Motion in plane polar coordinates 3-1.Polar coordinates Cartesian coordinates-motion in a straight line Polar coordinates-circular motion (cylindrical system, z is nonzero)      sin cos cos sin       = − + = + i j r i j       cos sin sin cos       = − = + i r j r  r   Position vector    r = x i+ y j = rr     r are orthogonal. (perpendicular) y X 0 i j r 