1 1. The position vector and displacement Let the plane of the two –dimensional motion be the x-y plane of a Cartesian coordinate system. jtytty itxttx trttr rrr jtyitxtr if ? )]()([ ? )]()([ )()( ? )( ? )()( ?++ ?+= ?+= ?= += ? ? ? ? rr rrr r x y O s? §4.1 The position, velocity, and acceleration vectors in two dimensions 2 2. The path (trajectory) of a particle )( )( tyy txx = = Eliminating t )(xyy = Example 1: 1 sin)( cos)( ? sin ? cos)( 2 2 2 2 =+ == == += B y A x tBtyy tAtxx jtBitAtr ω ω ωω r §4.1 The position, velocity, and acceleration vectors in two dimensions §4.1 The position, velocity, and acceleration vectors in two dimensions 3. Speed and velocity vector a. Average speed and instantaneous speed t s t s v t s v t d d lim s0 ave == = → ? ? ? ? ? b. Average velocity and instantaneous velocity t trttr t r v ? ? ? ? )()( ave rrr r ?+ == x y O s? Note: aveave vv r ≠ 3 t tr t trttr tv t d )(d)()( lim)( s0 rrr r = ? ??+ = →? c. The components of the velocity in Cartesian coordinate system jtvitvj t ty i t tx t tr tv yx ? )( ? )( ? d )(d d )(d d )(d )( +=+== r r r §4.1 The position, velocity, and acceleration vectors in two dimensions Magnitude: t s tvtvtvv yx d d )]()([)( 2122 =+== r Direction: tangent to the path followed by the particle; or x y v v 1 tan ? =θ The angle between axisand ?xv r §4.1 The position, velocity, and acceleration vectors in two dimensions 4. Acceleration vector a. Average acceleration )(tv r )( ttv ?+ r )(tv r ? t tvttv t tv a ? ? ? ? )()()( ave rrr r ?+ == x y O )(tv r )( ttv ?+ r 4 §4.1 The position, velocity, and acceleration vectors in two dimensions jtaita j t tv i t tv t tv t tv ta yx y x t ? )( ? )( ? d )(d ? d )(d d )(d)( lim)( s0 += += == → rr r ? ? ? b. Instantaneous acceleration Magnitude: 2122 )]()([)( tatataa yx +== r Direction: x y a a 1 tan ? =θ The angle between axisand ?xa r )(tv r )( ttv ?+ r )(tv r ? c. The direction of acceleration： AB vvv rvr ?=? AB vv rr > A v r v r ? a rα g r v r A v r B v r AB vv rr < B v r v r ? a r g r α v r A v r AB vv rr = B v r v r ? a r α v r a r A v r B v r §4.1 The position, velocity, and acceleration vectors in two dimensions 5 Solution: (1)trajetory ? ? ? ?= = 2 2 2 ty tx (2) position vector： t = 0s，x = 0 y= 2 t = 2s，x = 4 y = -2 jir jr rr r r r 24 2 ?= ′ = Example 2: If we know the position vector of a particle jtitr rr r )2(2 2 ?+= (SI) Fined the trajectory of the particle; the position vector at t=s and t=2s; the velocity and the acceleration of the particle at instant t=2s. 4 2 2 x y ?= —parabola Eliminate t, we can get §4.1 The position, velocity, and acceleration vectors in two dimensions The magnitude： (m)47.4)2(4 (m)2 22 =?+=′=′ == rr rr r r o y x Q r r r′ r P2 -2 4 θ′ θ 4 2 2 x y ?= jir jr rr r r r 24 2 ?= ′ = r r The direction： 2326 4 2 arctg 90 0 2 arctg ′ ?= ? = ′ == o o θ θ The angle between and x-axis The angle between and x-axis r ′ r §4.1 The position, velocity, and acceleration vectors in two dimensions 6 (3)The velocity: 2 4 tantan 11 ? == ?? x y v v α The angle between velocity and x-axis: ( ) -1 2 222 2 sm47.4522 12 22 22 d d 22 ?=== +=+= ?== ?== ?+= vt tvvv tvv jti t r v jtitr yx yx rr r r rr r Magnitude: §4.1 The position, velocity, and acceleration vectors in two dimensions ( ) j t r t v a jti t r v jtitr ? 2 d d d d ? 2 ? 2 d d ? 2 ? 2 2 2 2 ?=== ?== ?+= rr r r r r (3)The acceleration: Direction of the acceleration: The magnitude: 2== aa r j ? ? §4.1 The position, velocity, and acceleration vectors in two dimensions 7 Exercise: The position of a small bumper car in an amusement park ride is described as a function of time by the coordinates )SI(0.20.100.1 )SI(5.00.52.0 2 2 ++?= ++= tty ttx Find (a) the position vector at t=1.0 s and t=3.0 s. (b) the displacement vector between these time. (c) average velocity over the period from 1.0 s to 3.0 s, and the velocity at t=3.0 s. (d) the magnitude and direction of the acceleration at t=1.0 s and t=3.0 s. §4.1 The position, velocity, and acceleration vectors in two dimensions 5. Perpendicular motions are independent of each other §4.1 The position, velocity, and acceleration vectors in two dimensions We can analyze the motion along each coordinate axis separately. Experimental result 8 §4.2 Two dimensional motion with a constant acceleration (self-study) §4.3 motion in three dimension §4.3 motion in three dimension 2 2 aveaveave d d d d , d d , d d ,, ? )( ? )( ? )()( ? )( ? )( ? )()( t r t v a t r v t s v t v a t r v t s v ktzjtyitxtr ktzjtyitxtr rr r r r r r r r r r ==== ? ? = ? ? = ? ? = ?+?+?=? ++= §4.4 relative velocity addition and accelerations 1. Reference frame A coordinate system with a set of synchronized clocks, all ticking at the same rate, is called reference system. Different choices for the reference frame lead to different description of motion, but the underlying physics is nonetheless the same. Coordinate +synchronized clocks 9 §4.4 relative velocity addition and accelerations 2. relative velocity addition and accelerations rRr ′ += r r OOOPPO OOOPPO aaa vvv ′′ ′′ += += rrr rrr t r t R t r d d d d d d ′ += r r r OPPO OO aa v ′ ′ = = rr r then constant,if O r r r ′ r R r P O′ §4.4 relative velocity addition and accelerations 3. Some topics of discussion 1Is there really inertial reference frame? 2Are space and time absolute or not? 3The measurement of the time interval; 4The measurement of the space distance. 10 Exercise 1: A train travels due south at 28 m/s(relative to the ground) in a rain that is blown to the south by the wind. The path of each raindrop makes an angle of 64°with the vertical, as measured by an observer stationary on the earth. An observer on the train, however, see perfectly vertical tracks of rain on the windowpane. Determine the speed of the rain drops relative to the ground. §4.4 relative velocity addition and accelerations Solution: rt v r tg v r rg v r tg v r o 64 o 64 rg v r rt v r From the problem, we have m/s2864sin == tgrg vv o m/s3.33 84.0 28 64sin === o tg rg v v §4.4 relative velocity addition and accelerations tgrtrg vvv rrr += 11 Exercise 2: The compass in an airplane indicates that it is headed due east; its air speed indicator reads 215km/h. A steady wind of 65 km/h is blowing due north. (a)What is the velocity of the plane with respect to the ground? (b) If the pilot whishes to fly due east, what must be the heading? That is, what must the compass read? §4.4 relative velocity addition and accelerations Solution: (a) km/h6.224 65215 22 22 = += += AGPAPG vvv o 7.18 215 65 tantan 11 === ?? PA AG v v α β β cos sin PAPG PAAG vv vv = = (b) o 55.19 215 65 sinsin 11 === ?? PA AG V v β km/h9.20455.19cos215 == o PG v §4.4 relative velocity addition and accelerations AGPAPG vvv rrr += 12 Exercise 3: A boat must traverse a river 150 m wide. The river has a current of 3 km/h, and the boat can be rowed through the water with a uniform speed of 4 km/h. Set up a convenient coordinate system, express the position vector of the boat at the time t , assuming that the boat leaves the dock at the angle θ with respect to a point moving with the water, as shown in fig. (b). Calculate θ such that the boat lands at a point exactly opposite the starting point. How long will be the trip take? (a) Observer at O is fixed at shore (b) Observer at O’ is moving with water §4.4 relative velocity addition and accelerations Solution: ji vvv iv jiv wGbwbG wG bw ? sin)km/h4( ? ]3cos)km/h4[( ? km/h3 ? sin)km/h4( ? cos)km/h4( θθ θθ ++?= += = +?= rrr r r jtit rrr wGbwbG ? sin4 ? ]3cos4[( θθ ++?= += rrr The position of the boat as seen by the observer on the dock: The velocity of the boat and water §4.4 relative velocity addition and accelerations 13 From the problem we can know that 0= bG x That means 3min.h06.0 4sin41 0.15 km15.0sin4 41 75.0 km/h4 km/h3 cos 03cos4 ? sin4 ? ]3cos4[(in ≈== == = == =+?= ++?= o o r t ty x jtitr bG bG bG θ θ θ θ θθ §4.4 relative velocity addition and accelerations §4.5 The circular motion 1. Uniform circular motion v r v r θ? r r A circular motion at constant speed The motion of a particle about a point We can demonstrate that the magnitude of centripetal acceleration is r v a c 2 = Experiment: the direction of the acceleration 14 §4.5 The circular motion 2. How to describe the circular motion? jrir jtyitxtr ? sin ? cos ? )( ? )()( θθ += += r t t r t r t s t ts v t t d )(d lim lim d )(d 0s 0s θ ? θ? ? ? ? ? == == → → 1Angular position (coordinate) )(tθ 3speed and position of the particle 2 Angular displacement )(tθ? θ?? rs = Arc length v r v r θ? r r θ x y §4.5 The circular motion 4angular speed and angular velocity vector Right-hand rule for k ? Define: the time rate of change of the angular coordinate of the particle is the angular speed t t t z d )(d )( θ ω = r r θ x y z ktt z ? )()( ωω = r v r k t t ktt z ? d )(d ? )()( θ ωω == r angular velocity vector 15 §4.5 The circular motion r r θ x y z ktt z ? )()( ωω = r v r 5The relation of )(and)(),( trttv rrr ω )()()( trttv rrr ×=ω )()( trtv ω= 6Centripetal acceleration and the angular speed of uniform circular motion jtritrtr ? )(sin ? )(cos)( θθ += r ? ? ? ? ? ? +?= == jt t rit t r t t tr tt tr ta c ? )](cos d d [ ? )](sin d d [ d d ] d )(d [ d d d )(d )( 2 2 θ θ θ θ rr r §4.5 The circular motion {} )( ? )](sin[ ? )](cos[) d d ( ? )](cos d d [ ? )](sin d d [ d d )( 2 2 tr jtritr t jt t rit t r t ta z c r r ω θθ θ θ θ θ θ ?= +?= ? ? ? ? ? ? +?= Magnitude: 2 zc ra ω= Direction: opposite to r r )()( d )(d )( )]()([ d d d )(d )( tvt t tr t trt tt tv ta c rr r r rr r r ×=×= ×== ωω ω r r θ x y z ktt z ? )()( ωω = r v r a r 16 §4.5 The circular motion 3. Nonuniform circular motion and the angular acceleration )()()()( d )(d )()( d )(d )]()([ d d d )(d )( tvttrt t tr ttr t t trt tt tv ta rrrr r rr r rr r r ×+×= ×+×= ×== ωα ω ω ω Angular acceleration centripetal acceleration 1 angular acceleration kt t t t z ? )( d )(d )( α ω α = = r r ω r α r z n r v tr dt d nvtr tvttrtta z ? ? ? ? )()()()()( 2 +=+= ×+×= ω ωα ωα rrrrr 3tangential acceleration §4.5 The circular motion 2 The direction of angular acceleration ω r ω r α r α r Speeding up: is parallel to . Slowing down: is antiparallel to . )(tα r )(tω r )(tα r )(tω r 17 §4.5 The circular motion ω r α r t ar =× rr α r r )()()( trtta t rrr ×=α Define: tangential acceleration Speeding up: is parallel to . Slowing down: is antiparallel to . )(ta t r )(tv r )(ta t r )(tv r 4. Nonuniform circular motion with a constant angular acceleration From costant d d == t z z ω α Integrate in both sides of the equation ttt zzz t zz z z αωωαω ω ω +== ∫∫ 0 0 )(,dd 0 From t t z d (t)d )( θ ω = Integrate in both sides of the equation §4.5 The circular motion 18 2 00 0 0 0 2 1 )( d)(dd 0 ttt ttt zz t zz t z αωθθ αωωθ θ θ ++= +== ∫∫∫ costant d d == t z z ω α 2 00 2 1 )( ttt zz αωθθ ++= tt zzz αωω += 0 )( costant d d == t v a x x 2 00 2 1 )( tatvxtx xx ++= tavtv xxx += 0 )( Comparison : nonounoform circular motion with a constant angular acceleration and the rectilinear motion with a constant acceleration §4.5 The circular motion Exercise 1: Your antique stereo turntable of radius 13.7 cm, initially spinning at 33.0 revolutions per minute, is shut off. The turntable coasts to a stop after 120 s. Assume a constant angular acceleration. Calculate the angular acceleration of the turntable and the number of revolutions through which it spins as it stops. §4.5 The circular motion 19 Exercise 2: A particle begins at rest on a circular track and is subjected to a constant angular acceleration of magnitude α beginning when t = 0 s. (a) Show that the magnitudes of the tangential and centripetal accelerations of the particle are equal when independent of the radius of the circular track. (b) What is the angle that the total acceleration vector makes with the radial direction at this time? 21 ) 1 ( α =t §4.5 The circular motion