Kinematics. Describing Motion. Reference Frames. Measurements of position, distance or speed must be with respect to a frame of reference.

Transcription

1 Kinematics Describing Mtin Reference Frames Measurements f psitin, distance r speed must be with respect t a frame f reference. What is the speed f a persn with respect t the grund if she walks tward the back f the train at 5km/h while the train mves frward at 40 km/h? Crdinate axes are used t represent the frame f reference 1

2 Displacement Defined as change in psitin What is the displacement f a persn wh walks 100 m East then 60 m West? Answer: = 40 m East Displacement has bth magnitude and directin it is a vectr Average Speed and Velcit Velcit is speed and directin Average speed = distance traveled time elapsed Average velcit = displacement time elapsed Nt alwas equal

3 Average Speed and Velcit Find average speed and velcit fr a trip 60m Nrth fllwed b 40m Suth in 10 secnds distance m 100m speed 10 m / s 10s displacement m Nrth 0m velcit m / s Nrth 10s Example During a fur secnd interval a runner s psitin changes frm x 1 = 50m t x = 10m. What was the average velcit? v av = Dx/Dt = (x x 1 )/4s = -40m/4s = -10 m/s What is the average speed? 10m/s Speed is alwas psitive! Cnvert this velcit t kilmeters per hur -10 m/s x 1 km/1000m x 3600 sec/ 1h = -36 km/h 3

4 Hw Far? Hw far culd a runner traveling at an average speed f 36 km/h g in 0. minutes? 0 min = 1/3 hur D x = v av Dt 36 km/h x 1/3h = 1 km r 1. x 10 4 m Instantaneus Velcit Velcit at a particular instant f time Defined as average velcit ver an infinitesimall shrt time interval v = lim(as Dt --> 0) Dx/Dt The slpe f the tangent line t a d vs t graph 4

5 Acceleratin Acceleratin is hw fast velcit changes Average acceleratin = change f velcit time elapsed a av = (v v 1 )/(t t 1 ) = Dv/Dt Example Find average acceleratin f a car that accelerates alng straight rad frm rest t 80 km/h in 5 secnds a av = (80 km/h 0 km/h)/5s = 16 km/h/s Challenge: Can an bject with zer velcit have nn zer acceleratin? 5

6 Instantaneus Acceleratin Acceleratin at a particular instant f time Defined as average acceleratin ver an infinitesimall shrt time interval Dv a lim D t 0 Dt Object Slwing Dwn Called deceleratin Find average acceleratin f a car mving t the right(+x directin) 15.0 m/s when driver brakes t 5.0 m/s in 5.0 s? a av =Dv/Dt = (5.0 m/s 15.0 m/s) 5.0 s = -.0 m/s (acceleratin negative) Wuld acceleratin still be negative if car was mving t the left? 6

7 Cnstant(Unifrm) Acceleratin Let t 1 = 0, t = t = elapsed time Re-name x 1 = x 0 ; v 1 = v 0 Average velcit v av = (x x 0 )/t a av = (v v 0 )/t Frm these it is pssible t derive (next slide) Fr Unifrm Acceleratin a v v at 1 x x vt at v v a( x x ) v v vavg Each equatin can be slved fr an f the variables Prblems can be slved mre than ne wa These need t be used when cnstant acceleratin is present in a given prblem 7

8 v v at Describes change in velcit under unifrm acceleratin. tells hw fast a particle will be ging at time t if at time zer its velcit was v 0 1 x x vt at Describes change in psitin under unifrm acceleratin Smetimes called "equatin f mtin." tells where a particle will be at time t, if at time zer it was at x 0 mving with velcit v 0 8

9 v v a( x x ) Velcit equatin Tells what velcit will be after a particle with initial velcit v 0 accelerates a distance x - x 0 with unifrm acceleratin a Take square rt t find v Simplifies t v = ad when v 0 = 0 and d = displacement v avg Average (mean) velcit f particle fr a trip under unifrm acceleratin Prvides shrtcut wa t slve certain prblems v v 9

10 Cncept Check (1) The velcit and acceleratin f an bject (a) must be in the same directin (b) must be in ppsite directins (c) can be in the same r ppsite directins (d) must be in the same directin r zer (c) Can be in the same r ppsite directins. Example: a rck thrwn upward Cncept Check() At the tp f its path the velcit and acceleratin f a brick thrwn upward are (a) bth nn zer (b) bth zer (c) velcit is zer acceleratin is nn zer (d) acceleratin is zer, velcit is nn zer (c) is crrect; v = 0, a = 9.80 m/s/s dwnward 10

11 Prblem Slving Tips Read and re-read the prblem Make a diagram with all given inf Ask urself what is prblem asking? Ask which phsics principles appl Lk fr mst applicable equatins Be sure prblem lies within their range f validit Special Tricks Break prblem up int parts, like up and dwn part f path f an bject thrwn up Use smmetr Chse reference frame that makes prblem easiest 11

12 Prblem slving D algebraic calculatins Be aware u ma have t slve equatins simultaneusl D arithmetic at end Check units and significant figures Ask, is answer reasnable? Yur Chice Yu ma chse up psitive r negative; same with left r right Yu ma put the zer f crdinates anwhere u chse Generall make chices that minimize number f negative quantities 1

13 Examples What time is required fr a car t travel 30.0 m while accelerating frm rest at a unifrm.00 m/s t? v 0 m / s x 30.0 m a.00 m / s 1 x vt at (.00) t 30 t t sec Examples A car traveling 8 m/s brakes at -6.0 m/s. What distance is required t stp. x? v 8 m / s v 0 m / s a 6.0 m / s v v ax v v ax v v x a 0 8 x 65m ( 6.0) 13

14 Free Fall Which falls faster, an elephant r a muse? Galile s Discver All bjects accelerate t earth equall (regardless f mass) Air resistance must be neglected fr this t be true Acceleratin due t gravit a = g=9.80m/s 14

15 Useful Frm f Kinematics Equatins fr Free Fall v = v 0 + gt = 0 +v 0 t + 1/gt v = v 0 + g(- 0 ) v av = (v 0 + v)/ (a) (b) (c) (d) Here assuming up is psitive g = m/s *Free Fall is mtin under the influence f gravit alne Distance Fallen Frm Rest Distance fallen in t secnds with n initial velcit time 1 5 distance d 1 gt

16 Example: Ball Thrwn Upward Persn thrws ball upward with v = 15.0 m/s. (a) Hw high des it g? (b) Hw lng is it in the air? a v m s a m s v m s ) 15.0 / 9.80 / 0 /? v v a v v a v v a m ( 9.8) b v m s a m s v m s ) 15.0 / 9.80 / 0 / t? v v at v v at v v t a s 9.8 Puzzlers What is the velcit f a ball thrwn upward at its maximum height? 0 m/s What is its acceleratin at that height? Are velcit and acceleratin alwas in the same directin n -9.8 m/s 16

17 Graphing Mtin Psitin Graph (x r vs. t) Slpe is velcit If curved slpe defined as slpe f tangent t the curve at that pint Velcit Graph Slpe is acceleratin Area under graph is displacement Acceleratin Graph Cnstant and nn zer fr unifrm acceleratin Cncept Check (3) Which f these culd be the velcit graph f a rck thrwn upward, then falling dwnward? (Assume up is psitive) (a) (b) (c) (d) (e) 17

18 Assignment pg. 5 #1,3,10,19 pg #9,13,16,1,6,38,44,56,63 18

19 Prjectiles Curtes darren/web/range/range.html Tw Dimensinal Mtin Curtes Phsics Lessns.cm 3.1 Cmpnents f Mtin N Vectrs are drawn as arrws n the crdinate plane. - the length f the arrw crrespnds t the magnitude - the vectr pints t its directin W S E Vectr A has a magnitude f 3 meters and a directin f 60 abve the psitive x-axis. B 3.0 m m A Vectr B has a magnitude f 3 meters and a directin f β abve the negative x-axis. 180 β 60 x 0 Vectr C has a magnitude f 1.5 meters and a directin f m 70 C 19

20 - cmpnent 1/13/ Cmpnents f Mtin T analze mtin, a vectr can be brken dwn int x- and - cmpnents. 5 units Given a vectr v with a magnitude f a directed an angle Ө abve the hrizntal v x = a cs Ө v = a sin Ө Ө = 53 x- cmpnent Nte: The magnitude f the vectr v is x example: Find the cmpnents f v if its magnitude is 5 units and Ө = 53. v x = a cs Ө = 5 cs 53 = 3 units v = a sin Ө = 5 sin 53 = 4 units v = v x + v Wh? 3.1 Cmpnents f Mtin Kinematics prblems in tw-dimensins can be slved b: reslving the displacement, velcit, and acceleratin vectrs int their respective cmpnents. using the three cnstant acceleratin equatins separatel fr the x and directins. using the Pthagrean therem t find the magnitude f ur resultant vectr. using an inverse trig functin t find the angle f ur resultant vectr 0

21 3.1 Cmpnents f Mtin Example 1: A bat travels with a speed f 5.0 m/s in a straight path n a still lake. Suddenl, a stead wind pushes the bat perpendicularl t its straight line path with a speed f 3.0 m/s fr 5.0 s. Relative t its psitin just when the wind started t blw, where is the bat at the end f this time? Slutin: List givens fr x and directins separatel. x-directin -directin Unknwns: x and v = 5.0 m/s v = 3.0 m/s a = 0 a = 0 Cmmn t bth: t = 5.0 s Analsis: Bth mtins are mtin with cnstant velcit. Chse the straight path f the bat as the x axis and the directin f the wind as the axis. Sketch: wind d bat Ө x x-directin -directin Unknwns: x and v = 5.0 m/s v = 3.0 m/s a = 0 a = 0 wind Cmmn t bth: t = 5.0 s d bat Ө x x-cmpnent: (use equatin 1) x = v t + ½ at = (5.0 m/s)(5.0 s) + 0 = 5 m -cmpnent: (use equatin 1) = v t + ½ at = (3.0 m/s)(5.0 s) + 0 = 15 m Nw, d = x + = (5 m) + (15 m) = 9 m And Ө = tan -1 = 15m = 31 5 m 1

22 3. Vectr Additin and Subtractin vectr additin cmbining vectr quantities b using ne f several techniques gemetric methds (triangle, parallelgram, plgn) vectr cmpnents and the analtical cmpnent methd vectr subtractin a special case f vectr additin A - B = A + (-B) subtracting a vectr is the same as adding a negative vectr resultant the verall effect f cmbining vectrs; the vectr sum Example: Add vectrs A + B + C = R Slutin: Use tip-t-tail methd 3. Vectr Additin and Subtractin Gemetric (Graphical) Methds use a cnvenient scale (ex: 1 cm = 10 meters) T add vectrs A and B, draw them tip-t-tail, starting at the rigin. The vectr that extends frm the tail f A t the tip f B, cmpleting the triangle, is the resultant R = A + B. When drawn t scale, the magnitude f R can be fund b measuring R and using the scale cnversin. Use a prtractr t measure the directin f the resultant, Ө R. Yu ma add n as man vectrs as u like; the rder desn t matter! Example: Add three vectrs graphicall. Slutin:

23 3. Vectr Additin and Subtractin Analtical (Trignmetr) Methd When vectrs make a right triangle with each ther, use the Pthagrean therem t find the magnitude f ur resultant vectr. use the inverse trig functin t find the angle f ur resultant vectr. A = side adjacent t angle Ө O = side ppsite t angle Ө H = hptenuse f triangle SOH CAH - TOA Ө H A O sin Ө = O cs Ө = A tan Ө = O A H A Ө = sin -1 O Ө = cs -1 A Ө = tan -1 O A H A 3. Vectr Additin and Subtractin Example: Eric leaves the base camp and hikes 11 km, nrth and then hikes 11 km east. Determine Eric's resulting displacement and express in the magnitude-angle frm. Slutin: Magnitude f displacement = 16 km 3

24 3. Vectr Additin and Subtractin 3. Vectr Additin and Subtractin 4

25 What s a Prjectile? An bject mving in tw dimensins under the influence f gravit alne. Yu Predict Tw identical balls leave the surface f a table at the same time, ne essentiall drpped the ther mving hrizntall with a gd speed. Which hits the grund first? The hit at the same time. The have the same acceleratin in the directin and the same initial velcit in the directin. Curtes f mms/staff/hand/drp.jpg 5

26 Galile s Analsis Hrizntal and vertical mtins can be analzed separatel Ball accelerates dwnward with unifrm acceleratin g Ball mves hrizntall with n acceleratin S ball with hrizntal velcit reaches grund at same time as ne merel drpped. Zer Launch angle v A zer launch angle implies a perfectl hrizntal launch. 6

27 Equatins Hrizntal Vertical v x v x v v gt x x0 vxt 1 0 vt gt v v g Assuming psitive up; a x = 0, a = g = m/s Sample Prblem The Zambezi River flws ver Victria Falls in Africa. The falls are apprximatel 108 m high. If the river is flwing hrizntall at 13.6 m/s just befre ging ver the falls, what is the speed f the water when it hits the bttm? Assume the water is in freefall as it drps. Onl accelerating verticall v v 0 (9.8)(108) 46 m / s add in the hrizntal cmpnent v v v m / s x 7

28 Sample Prblem An astrnaut n the planet Zircn tsses a rck hrizntall with a speed f 6.75 m/s. The rck falls a distance f 1.0 m and lands a hrizntal distance f 8.95 m frm the astrnaut. What is the acceleratin due t gravit n Zircn? x v t x x 8.95 t 1.33 s v 6.75 x 1 gt gt n initial vertical velcit since it is thrwn hrizntal (1.0) g 1.36 m / s t (1.33) General launch angle v Prjectile mtin is mre cmplicated when the launch angle is nt straight up r dwn (90 r 90 ), r perfectl hrizntal (0 ). 8

29 General launch angle v Yu must begin prblems like this b reslving the velcit vectr int its cmpnents. Trajectr f a -D Prjectile x Definitin: The trajectr is the path traveled b an prjectile. It is pltted n an x- graph. 9

30 Trajectr f a -D Prjectile x Mathematicall, the path is defined b a parabla. Trajectr f a -D Prjectile x Fr a prjectile launched ver level grund, the smmetr is apparent. 30

31 Range f a -D Prjectile Range x Definitin: The RANGE f the prjectile is hw far it travels hrizntall. Maximum height f a prjectile Maximum Height Range The MAXIMUM HEIGHT f the prjectile ccurs when it stps mving upward. x 31

32 Maximum height f a prjectile Maximum Height Range The vertical velcit cmpnent is zer at maximum height. x Maximum height f a prjectile Maximum Height Range Fr a prjectile launched ver level grund, the maximum height ccurs halfwa thrugh the flight f the prjectile. x 3

33 Acceleratin f a prjectile g g g g g x Acceleratin pints dwn at 9.8 m/s fr the entire trajectr f all prjectiles. Velcit f a prjectile v v v v v f x Velcit is tangent t the path fr the entire trajectr. 33

34 Velcit f a prjectile v x v v x v v x v v x v v x x The velcit can be reslved int cmpnents all alng its path. Velcit f a prjectile v x v v x v v x v v x v v x x Ntice hw the vertical velcit changes while the hrizntal velcit remains cnstant. 34

35 Velcit f a prjectile v x v v x v v x v v x v v x x Maximum speed is attained at the beginning, and again at the end, f the trajectr if the prjectile is launched ver level grund. Velcit f a prjectile v Launch angle is smmetric with landing angle fr a prjectile launched ver level grund. v - 35

36 Time f flight fr a prjectile t t = 0 The prjectile spends half its time traveling upward Time f flight fr a prjectile t t = 0 t and the ther half traveling dwn. 36

37 Prblem Slving Read carefull Draw Diagram Chse x crdinates and rigin Analze hrizntal and vertical separatel Reslve initial velcit int cmpnents List knwns and unknwns Remember v x des nt change v = 0 at tp Upwardl Launched Prjectile With Velcit v and Angle v v x v v cs sin v sin Time in air g Range ( v cs ) t ( v sin ) Max height g ( v )(sin ) Trignmetric identit: sin cs sin g v 0 R 37

38 Sample prblem A sccer ball is kicked with a speed f 9.50 m/s at an angle f 5 abve the hrizntal. If the ball lands at the same level frm which is was kicked, hw lng was it in the air? Time in air depends n the cmpnent v v 9.50sin m / s 4.01 m / s because the path is smmetrical a g 9.8 m / s t? v v at ( 9.8) t t t 0.818s r Time in air Time in air v sin g Time in air 0.819s (9.50 m/ s)sin 5 (9.8 m/ s ) Questins What is the acceleratin vectr at maximum height? 9.80 m/s dwnward at all times Hw wuld u find the velcit at a given time? Use kinematics equatins fr v x and v, then find v b sqrt sum f squares f them. Hw wuld u find the height at an time? Use kinematics equatin fr Hw des the speed at launch cmpare Same with that just befre impact? 38

39 Lngest Range What angle f launch gives the lngest range and wh? Assume the prjectile returns t the height frm which launched. 45 degrees; must maximize sin ; maximum value f sine is ne, happens fr 90 0 ; then = 45 0 If a launch angle gives a certain range, what ther angle will give the same range and wh? Hint: R ges as sin cs Mving Launch Vehicle If ball is launched frm mving cart, where will it land? 39

40 A Punt Ftball kicked frm 1.00 m abve grund at 0.0 m/s at angle abve hrizntal 0 = Find range. Can we use range frmula? N. It desn t appl since 0 Let x 0 = 0 = 0 = m Punt, cntinued Find time f flight using = 0 + v 0 t 1/gt -1.00m = 0 + (1.0m/s)t (4.90 m/s )t (4.90 m/s )t - (1.0m/s)t (1.00m) = 0 use quadratic frmula t slve fr t t =.53s r s (impssible) x = v x0 t = v 0 cs37 0 t= (16.0 m/s)(.53s) = 40.5 m 40

41 Using Frmulas Be sure the frmula applies t the situatin that the prblem lies within its range f validit Make sure u understand what is ging n. Assignment pg #3,4,15 pg #,6,15,19,1,8,35,38,4,46 41