Chapter 3 Motion in a planeIn chapter 2 we studied the motion of objects along a line i.e. in one dimensionIn chapter 3 we will describe motion in two and three dimensions and introduce the following concepts:• position vector r, velocity v , acceleration a• projectile motion• uniform circular motion• relative motion (3-1)Px axisrttrajectoryO.Position Vector rThe position vector r(t) gives us the position of a moving object at tIn two dimensions (xy plane):In three dimensions:(3-2)Oy axistrajectoryPxy.t$$()rtxiyjzk=++r$$()rtxiyj=+r$Displacement vector ∆rThe displacement vector ∆r is defined as the difference between two position vectors at times t and t + ∆tt + ∆tt(3-3)()() , () , ()PQrrttrtrrtrrtt∆≡+∆−==+∆rrrrrrr()()avrttrtrvtt+∆−∆≡=∆∆rrrrlim as t0 P Q rdrvtdt∆≡=∆∆→→rrrtt + ∆tAverage velocity vav(3-4)Units: m/sInstantaneous velocity drvdt=rrOy axistrajectoryPxy.trij(3-5)$Determine the velocity Position vector ()The coordinates x and y are functions of t() and ()drvdtrtxiyjxxtyyt==+==rrr$$( )$$22 Speed xyxydrddxdyvxiyjijdtdtdtdtvvivjvvv==+=+=+→≡=+rr$$r$xvdxdt=yvdydt=(3-6)$( )222Example: The position vector of a moving object is given by:()()(sin)Find the velocity v. x(t) = , ()sin2sttxtyrrtAtBtiCeDtjAtBtytCeDtdxdvAtBtABtdtdtdydvCeDdtdtαααββ=++++=+==+=+==+rr$r( )( )( )incos sincosxtxddeexxdxxtdCeDtαααβββαβββα===+Relationship between the velocity vector v and the trajectory:The velocity vector v is tangent at each point P of the trajectorytrajectory(3-7)()()avvttvtvatt+∆−∆≡=∆∆rrrlim as t0 P Q vdvatdt∆≡=∆∆→→rrrtt + ∆tAverage acceleration aav(3-8)Units: m/s2Instantaneous acceleration v(t + ∆t)v(t) dvadt=rrOy axistrajectoryP.tv(t)j(3-9)$xyyDetermine the accelaration aVelocity ()The velocity components v and v are functions of time t() and v()xyxxydvdtvtvivjvvtvt==+==rrr$i$( )$$22 yxxyxyxydvdvdvdavivjijdtdtdtdtaaiajaaa==+=+=+→=+rr$$r$xaxdvdt=yaydvdt=(3-10)$( )( )yExample: The velocity of a moving object is given by:()(2)(cos)Find the accelaration a. () 2 , v()cos22costtxxxytyvvtABtiCeDtjvtABttCeDtdvdaABtBdtdtdvdaCeDtdtdtααααββαββαββ=+++=+=+==+===+=rr$r( )$( )( )2222 sin2sincossinttxxCeDtaBiCeDtjddeexxdxdxαααααββαβαββββ==−−=+−r$Geometric relationship between the velocity v, the acceleration a and the trajectory The velocity v is tangent to the trajectory at each point The acceleration a is not (3-11)(3-12)$Motion in the xy-plane with constant acceleration a , are given constantsWe apply the equations of kinematics (chapter 2) for uniformlyaccelerated motion along the x- and xyxyaaiajaa=+rr$22y-axis independentlyx-axis: 2y-axis: y2xxoxxooxyyoyyooxatvvatxxvtatvvatyyt=+=++=+=++Projectile motion: The motion of an object near the surface of the earth under the influence of gravity. Air resistance is omitted. Along the x-axis the acceleration is zero. The object moves with constant velocity Along the y-axis the object has constant acceleration -g g(3-13)g(3-14)( )( )2x-axis: 0 The velocity is constant along the x-axiscos , cosy-axis: Uniformly accelerated motionsin , sin2xxooooyyooooavvxvtaggtvvgtyvtθθθθ====−=−=−cossin00oxoooyoooovvvvxyθθ====Equation of the trajectory y = y(x)(3-15)( )( )( )( )( )2222122212cos cossin We substitute t in this equation2sincos2costan (equation of a parabola)2costan , 2ooooooooooooooooxxvttvgtyvtxgxyvvvgyxxCxCxvgCCθθθθθθθθθ=→==−=−=−=−==22cosoov θRange R of a projectileOP(3-16)212122221212The trajectory is described by the equation: Where tan , 2cosAt points O and P 0 0 ()0This equation has two solutions: 0 (point O) oooyCxCxgCCvyCxCxCCxxxθθ=−===→−=→−==12 and (point P)CxC=Range R of a projectile(3-17)112222222222 Where tan , and 2cossintancos2sincos 2cos2cossin2 R does not depend 2sincossinNote: 2 on the mas s m of theoooooooooooooooCgRCCC vvRgggvvvRgθθθθθθθθθθθθθ======== → projectile2sin2oovRgθ=π/23π/2ϕsinϕOMaximum rangeAt what angle θodo we have to shoot the projectile in order to achieve maximum range ?(vois fixed)R is maximum when sin2θo reaches its maximum value of 1. This happens when 2θo= 90° → θo= 45°Maximum range: (3-18)2sin2oovRgθ=2ovRg=Flight time TThe time it takes the projectile to go from point O to point B..vABOt2t(3-19)At point A (highest point of the trajectory) the velocity v is parallel to the x-axis. Thus 0 sin0 sin = the time it takes the projectile to go from O to AFlight tiyyoooovvvgtvtgθθ==−=→=r2sinme 2 oovTtTgθ=→=2sinoovTgθ=Maximum height hh..OAt(3-20)( )2222sin is the time it takes the progectile to go from O to Asinsinsin sin22sin2oooooooooooovtgvvgtgyvthvggvhgθθθθθθ==−→=−=22sin2oovhgθ=Uniform circular motionThe object moves on a circular orbit of radius r at a constant speed v This motion has the following characteristics:- The acceleration vector a ≠ 0 (even though the speed is constant. Remember: - The acceleration vector a points towards the center O of the circular orbit -- The magnitude of the accelerationO2var=(3-21) dvadt=rrCartesian coordinates: (x, y)Polar coordinates: (ρ, φ)How to convert Cartesian to polar coordinates and vice versaFrom triangle OAP we have:OAPB...s(2-22)22cos and ysin tanxyxyxρφρφρφ===+=Arc length in the equation above must be expressed iNon radiatse:nsρφφ=tt + ∆tConsider an object that is moving on a circle of radius R with speed v Angular speed ωAngular speed is telling us how fast the polar angle φ is changing with timeunits: rad/secddtφω =(3-23)()Relationship between and v: dsdRdvRRdtdtdtφφωω====vRω=Consider an object undergoing uniform circular motion on a circular orbit or radius r with velocity v Period T is defined as the time needed to complete one revolution. Unit: s(3-24)2Connection between and f. Period , 221 Frequency 2 2rTvrvrTffrTπωωππωωπωωπ==→====→=2fωπ=Frequency f is defined as the number of revolutions per second Unit: rev/sec = Hz1 fT=lim