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