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