Vectors and Two Dimensional Kinematics

Transcription

1 C.1 - Vecor lgebra - I Chaper C Vecors and Two Dimensional Kinemaics linn College - Phsics Terr Honan Polar Coordinaes (, ) are he Caresian (or recangular) coordinaes of some poin on a plane. r and θ are he polar coordinaes; r = is he disance from he origin o he poin and θ is he angle measured counerclockwise from he posiive ais o he poin. The definiions of he rig funcions, for general angles, are given b sin θ = r, cos θ = and an θ = r. Using he above definiions i is a sraighforward maer o find he formulas for convering beween polar and recangular coordinaes. 5 (,) 4 3 r 2 θ = -2 and = 4 r = 4.47 and θ = Ineracive Figure r and θ and = r cos θ and = r sin θ and r and θ (C.1) an -1 r = and when > 0 θ = an -1 when < 0 (C.2) The suble in solving for he angle follows from he ideni: an θ = an(180 + θ). The range of he an -1 (also wrien arcan) funcion is -90 < θ < 90, which corresponds o he quadrans I and IV; his is equivalen o he condiion ha > 0. The case of quadrans II and III, when < 0, requires shifing he resul of he inverse angen b 180.

2 2 Chaper C - Vecors and Two Dimensional Kinemaics Eample C.1 - Polar Coordinaes Consider he poin (, ) = (-2, 4) in he Caresian plane. Find he polar coordinaes r and θ of his poin. This is a sraighforward applicaion of he formulas above, where = -2 and = 4. r = = 20 = 4.47 θ = an -1 = Vecor asics vecor is a quani wih boh a magniude and a direcion. scalar has onl a magniude; i is jus a real number. The magniude of a vecor is a non-negaive (posiive or zero) scalar. Veloci is a vecor quani and speed is is magniude. cceleraion, force and momenum are also vecors. Time, emperaure, mass and pressure are eamples of scalars. We will wrie a vecor variable b a smbol wih an " " over i. The magniude of a vecor is given b he smbol wihou he arrow or b appling he "/ 0" brackes o he vecor. If is some vecor hen = /0 is is magniude. We can represen vecors b arrows. Suppose someone walks from a saring poin P 1 o a sopping poin P 2. displacemen vecor s (or Δ r ) ma be viewed as an arrow wih is ail a P 1 and is ip a P 2. s P 2 P 1 posiion vecor r is a displacemen vecor wih is ail a he origin. For more general vecors we represen hem b arrows poining in he direcion of he vecors and he lengh of he arrow is proporional o he magniude of he vecor. For insance, if one veloci vecor has a magniude of 60 mi /hr and anoher has 30 mi /hr hen he arrow represening he firs should have wice he lengh of he second. vecor has no fied posiion. If a vecor arrow is moved keeping is lengh and direcion fied hen i sill is he same vecor. Componen Definiion - Posiion and Displacemen Vecors posiion vecor is a wa o label a posiion in he Caresian plane; i has is ail a he origin and is head a he posiion i labels. We will use an angled bracke noaion for vecors. The posiion vecor ha labels he poin (, ) will be wrien as r =,. (,) r =,

3 Chaper C - Vecors and Two Dimensional Kinemaics 3 We will define he magniude and direcion of a vecor so ha he polar coordinaes, r and θ, are he magniude and direcion of he vecor. In he Caresian plane we will denoe vecor from ( 1, 1 ) o ( 2, 2 ) b: s = 2-1, 2-1 This is called a displacemen vecor. posiion vecor is clearl a displacemen vecor wih is ail a he origin. P 2 = ( 2, 2 ) s = 2-1, 2-1 P 1 = ( 1, 1 ) Componen Definiion - General Vecors We will wrie a 2-dimensional vecor as a pair of real numbers and called componens. ( 3D vecor is a riple.) We will use he "angled-bracke" noaion for vecors. =, The componen has he inerpreaion as he amoun he vecor is in he -direcion and he -direcion. =, Magniude and Direcion ngle We define and as he componens of he vecor. is he par of in he direcion and similarl is he par. The Caresian coordinaes and are he componens of a posiion vecor r. For a wo dimensional vecor we can represen he direcion wih an angle, measured as in he polar coordinaes. To conver beween he magniude and direcion angle and he componens of a wo dimensional vecor we have analogous epressions o he ones for polar coordinaes. and θ and = cos θ and = sin θ and and θ (C.3) = and θ = an -1 when > an -1 when < 0 (C.4) Vecor ddiion Suppose a displacemen vecor s 1 corresponds o someone walking from P 1 o P 2. Suppose ha hen he person walks from P 2 o P 3 ; call his displacemen s 2. The ne displacemen is he vecor from P 1 o P 3 ; his is wha we will define as he sum of he wo displacemens s 1 + s 2. To generalize his o an vecors, we will define he sum of general vecors and. Draw he vecors as shown, wih he ail of a he ip of. The sum he vecors + is he vecor drawn from he ail of o he ip of.

4 4 Chaper C - Vecors and Two Dimensional Kinemaics + Wih our componen definiion vecor addiion akes he ver simple form: + =, +, = +, +. (C.5) Commuaive Proper n algebraic operaion is commuaive when changing he order of he iems doesn' affec he resul. For vecor addiion his akes he form. + = + The commuaive proper of he addiion of reals implies his for vecors. + ecause of he commuaive proper here are wo more was of adding vecors we can consider. In addiion o placing he ail of a he ip of, we can place he ail of a he ip of. lso here is he parallelogram rule: Draw he wo vecor ogeher ail o ail and complee he parallelogram; he sum i he vecor from he common ail of he vecors o he opposie corner. ssociaive Proper From he definiion of vecor addiion i is clear i saisfies C = + + C This proper is called associaivi. For an associaive operaion i is no necessar o use brackes, since he order of he operaion is unimporan. + C + C ( + )+ C = + ( + C ) Ideni The ideni vecor 0 is he vecor ha leaves an oher vecor unchanged under addiion. I is clear ha he zero vecor has zeros as componens. + 0 = 0 = 0, 0

5 Chaper C - Vecors and Two Dimensional Kinemaics 5 The magniude of he ideni vecor is 0, 0 = /00. Noe ha he direcion of 0 is undefined; in fac, i is he onl vecor wih an undefined direcion. ddiive Inverse. For an vecor here is an addiive inverse vecor - wih he proper: Clearl, his has he value and has he same magniude and is in he opposie direcion. + - = 0 - = -, - - Vecor Subracion We define vecor subracion b adding he inverse. - = + - In erms of componens we have - =, -, = -, -. The simples wa o view he vecor - is as he vecor ha when added o gives ; if he vecors are drawn ail o ail hen i is he vecor from he ip of o he ip of - Scalar Muliplicaion If is a vecor and c is a scalar hen we can define heir produc c as a vecor. c = c, c (C.6) I is clear ha is magniude is given b c = <c= /0, where <c= is he absolue value of he scalar. The direcion of c is he same as when c > 0 and opposie o when c < 0. When c = 0 we ge 0 = 0. Noe also ha 1 = and (-1) = -. The scalar muliplicaion operaion has he associaive and disribuive properies. ssociaive Proper (c d) = c d

6 6 Chaper C - Vecors and Two Dimensional Kinemaics Disribuive Properies (c + d) = c + d and c + = c + c Uni Vecors and Noaion uni vecor is a vecor of magniude one. We denoe uni vecors wih a " ^ " over is op. For an vecor we can simpl find he uni vecor in is direcion > b > =. /0 asis uni vecors are uni vecors along he coordinae aes. We will use > and > for he uni vecors in he and direcions. The more radiional noaion for hese uni vecors is o wrie hem as > i and > j. > = > i = 1, 0 and > = > j = 0, 1 n vecor can hen be wrien in erms of hese basis uni vecors =, = > + > =, > > > > 3D Vecors I is sraighforward o generalize our wo dimensional discussion o a hree dimensional one. We need o add z > = k > for he uni vecor in he z direcion. > = i > = 1, 0, 0, > = j > = 0, 1, 0 and z > = k > = 0, 0, 1 We can similarl wrie an vecor in erms or is componens and uni vecors The magniude of a vecor is =,, z = > + > + z z >. = /0 = z 2. The onl suble wih hree dimensions is specifing direcions as angles. I akes wo angles o specif a direcion is 3D; hese are he θ and ϕ of spherical coordinaes. To avoid hese issues he bes wa o represen a direcion for a hree dimensional vecor is o jus wrie a uni vecor in he direcion of ha vecor > =. /0 Eample C.2 - Vecor ddiion plane flies 500 km o he souh and hen 900 km in he direcion 50 norh of wes. Wha is he plane s ne displacemen? Wha are he magniude and direcion angle of he plane s ne displacemen?

7 Chaper C - Vecors and Two Dimensional Kinemaics 7-600km R Norh 200km Eas 900km 500km km Ineracive Figure Here we are adding wo vecors, saring wih magniude and direcion informaion. To find he componens using equaion (C.3) we mus idenif he direcion angles which are measured counerclockwise from he posiive direcion, which we ake o be Eas. Vecor is o he Souh so i has direcion angle -90 and i has magniude 500 km. lhough we could use (C.3) find and i is easier o see ha since i is purel in he negaive direcion: =, = 0, -500 km. The direcion angle for vecor is θ = = 130. (See he ineracive diagram above.) The magniude of is = 900 km so using (C.3) we ge =, = cos θ, sin θ = , km. The ne displacemen is sum of hese wo vecors, he resulan vecor R = +. R = + = R, R = 0, -500 km , km = , km Using (C.4) we can find he Magniude and direcion angle of he ne displacemen. R = R 2 + R 2 = 609 km (Magniude) θ = an -1 R R = (Direcion ngle) The ne displacemen is jus R. R = -579, 189 km (Ne Displacemen) C.2 - Kinemaics in 2D and 3D The General Problem Posiion as a Funcion of Time There are wo equivalen was of describing posiion in wo dimensions. One is b labeling he coordinaes (, ). The oher is b giving he posiion vecor r of he posiion. The wo are relaed; he coordinaes are he componens of he posiion vecor. To label posiion as a funcion of ime we can consider and as separae funcions of ime or as r as a funcion of ime.

8 8 Chaper C - Vecors and Two Dimensional Kinemaics () and () r () = (), () The acual pah followed b he bod is called he rajecor. I is represened b a plo of a pah in he plane. r Ineracive Figure verage Veloci The average veloci, as we saw in he one dimensional case, refers o wo imes. i he posiion vecor is r i = r ( i ) and a f i is r f = r f. The displacemen is he difference of hese wo posiions Δ r = r f - r i = f - i, f - i. The average veloci vecor is hen defined as v = Δ r Δ = Δ Δ, Δ Δ = v, v. (C.7) f r f Δr v i r i i f i f f i i f v = Δ Δ = slope v = Δ Δ = slope Ineracive Figure

9 Chaper C - Vecors and Two Dimensional Kinemaics 9 Insananeous Veloci The insananeous veloci is defined as he limi of he average veloci as Δ approaches zero; i is he ime derivaive of he posiion vecor. v = d r d = lim Δ r Δ 0 Δ = d d, d d = v, v. (C.8) The magniude of he veloci is called he speed. v = / v 0 = v 2 + v 2 = speed (C.9) Δr v v 0 0 Ineracive Figure verage cceleraion s wih he one dimensional case acceleraion is o veloci as he veloci is o posiion. a = Δ v Δ = Δ v Δ, Δ v Δ = a, a. (C.10) Insananeous cceleraion The insananeous acceleraion is similarl defined as a limi of he average acceleraion or simpl as he ime derivaive of he veloci. Consan Veloci and cceleraion a = d v d = lim Δ v Δ 0 Δ = d v d, d v d = a, a. The cases of consan veloci and acceleraion follows he one dimensional eample. Since we are now dealing wih vecors he arbirar consans inroduced in anidiffereniaion become vecor quaniies. cons v a = 0 and r () = r 0 + v cons a v() = v 0 + a and r () = r v a 2 (C.11) Projecile Moion n imporan case of moion wih consan acceleraion is ha of projecile moion. Projecile moion is wo dimensional moion of bod

10 10 Chaper C - Vecors and Two Dimensional Kinemaics under he influence of onl gravi. Insising on onl gravi means ha we ignore air resisance. Spin effecs like a curving baseball or a slicing golf ball are associaed wih air resisance and will also be ignored. For 2D moion wih consan acceleraion we can modif he four consan acceleraion equaions ino wo ses of four equaions. Equaions v = v 0 + a Δ = 1 2 (v + v 0 ) Equaions v = v 0 + a Δ = 1 2 v + v 0 Δ = v a 2 Δ = v a 2 v 2 = v a Δ v 2 = v a Δ. For projecile moion we will ake as he horizonal direcion and as he verical direcion. The acceleraion due o gravi is downward and of magniude g. Wriing his as a vecor gives Insering hese componens ino he wo ses of equaions above gives: a = -g > = 0, -g or a = 0 and a = -g Horizonal Equaions v = v 0 Δ = v 0 Verical Equaions v = v 0 - g Δ = 1 2 v + v 0 Δ = v g 2 v 2 = v g Δ. (C.12) The horizonal moion is simple. Since here is no horizonal (he direcion) acceleraion, he componen of he veloci is consan. The verical par (he direcion) of he moion is equivalen o free fall. The ke o solving projecile moion problems is keeping he wo pars separae. If a projecile is launched a an iniial angle of θ wih an iniial speed v 0 hen he componens of he iniial veloci are given b v 0 = v 0 cos θ and v 0 = v 0 sin θ. Eample C.3 - Puned Fooball puner kicks a fooball from ground level wih an iniial speed of 27 m /s a an inial angle of 63. (Neglec all air resisance.) ma R/2 R (a) Wha is he maimum heigh reached b he fooball? R/2 R ma T/2 T T/2 T egin wih idenifing wha is given. We know he iniial speed and iniial angle. We can calculae he componens of he iniial veloci. v 0 = 27 m /s and θ = 63 v 0 = v 0 cos θ = m /s and v 0 = v 0 sin θ = m /s The maimum heigh depends onl on v 0. he highes poin he veloci is purel horizonal, so v = 0. We are looking for ma in he graphs above. ma = Δ when v = 0. We need he fourh of he verical projecile equaions (C.12).

11 Chaper C - Vecors and Two Dimensional Kinemaics 11 2 v 0 v 2 = v g Δ 0 = v g ma ma = 2 g = 29.5 m (b) Wha is he hang-ime of he pun? (Noe ha hang-ime is he oal ime he fooball is in he air, from he ground o he ground.) Since he fooball ends a he same elevaion where i began, so Δ = 0. We need o solve for ime so he relevan formula is he hird verical projecile equaion (C.12). 1 Δ = v 0-2 g = v 0-2 g 1 2 = v 0-2 g This gives wo soluions bu = 0 is riviall rue. We wan he oher. We will call his T as shown in he graphs above. 1 2 v 0 0 = v 0-2 g T T = g = s = 4.91 s (c) Wha is he horizonal range of he fooball? The range is he oal horizonal disance he fooball ravels in he air. So far we have no used an horizonal informaion or equaions. The range R is Δ when Δ = 0. Using he ime T we jus found in he second horizonal projecile equaions (C.12). R = Δ = v 0 = v 0 T = 60.2 m To solve for he rajecor we can choose, for simplici, ha he moion begins a he origin 0 = 0 = 0 giving Δ = and Δ =. Solving he horizonal equaion for ime gives = /v 0. Insering his ino he verical epression = v g 2 gives I is clear ha his is a parabola. v 0 = - v 0 The range R of a projecile is he oal horizonal disance raveled in he air when i reurns o is original level, = 0 in he epression above. We can hen facor ou an from he epression and solve for. g 2 2 v v 0 v 0 = 0 and =. g The = 0 soluion is rivial. Seing he oher o R, wriing he componens in erms of v 0 and θ, and using he ideni sin 2 θ = 2 sin θ cos θ gives an epression for he range. R = v 0 2 g sin 2 θ (C.13) Eample C.4 - Puned Fooball (coninued) (d) ppl he range formula (C.13) o he fooball eample o ge he same answer as in par (c). R = 2 v 0 sin 2 θ = 60.2 m g C.3 - Relaive Moion frame of reference is some coordinae ssem used o sud moion. Suppose here is a fied frame K and a moving frame K, ha moves wih a veloci v 0 wih respec o K. If we ake a moving bod o be a poin P. If r is he vecor from he origin of K o P, r is he vecor from he origin of K o P, and r 0 is he vecor from he origin of K o he origin of K.

12 12 Chaper C - Vecors and Two Dimensional Kinemaics P r z z r r 0 Moving Frame K I follows ha he hree posiion vecors are relaed b Fied Frame K r = r + r 0. We wan o relae he velociies of he moving bod wih respec o hese wo frames. The velociies wih respec o K and K are v and v, respecivel. We can relae he velociies in hese wo frames b aking he ime derivaive of he epression above: v = v + v 0, (C.14) where v 0 is he veloci of he moving frame. Eample C.5 - ubba ubba drives his pickup a 85 mi /h while hrowing his beer bole ou he window a 15 mi /h, relaive o he pickup. v 15mi/h v 15mi/h v v=? v 0 85mi/h v v=?? θ v 0 85mi/h (a) (c) and (d) (a) Wha is he speed of he bole, relaive o he road, if he bole is hrown forward? Take he fied frame K o be he frame of he road and K o be he ruck s frame. We are suding he moion of he bole wih respec o hese wo frames. For a one dimensional problem remember ha real numbers are one-dimensional vecors where he sign gives he direcion. We hen have (C.14) wihou he vecor arrows. v = v 0 + v = 85 mi /h + 15 mi /h = 100 mi /h (b) Wha is he speed of he bole, relaive o he road, if he bole is hrown backward? Now v is in he opposie direcion, so we will make i negaive. v = v 0 + v = 85 mi /h + (-15 mi /h) = 70 mi /h (c) Wha is he speed of he bole, relaive o he road, if he bole is hrown direcl o ubba s lef? This is now wo-dimensional. v 0 and v are perpendicular and form wo sides of a righ riangle. v is he hpoenuse; o find is magniude use he Phagorean heorem. v = (85 mi /h) 2 + (15 mi /h) 2 = 86.3 mi /h

13 Chaper C - Vecors and Two Dimensional Kinemaics 13 (d) Relaive o he road, wha is he angle he bole makes from he ruck s direcion? From he righ riangle we m find θ using rig. 15 mi /h θ = arcan 85 mi /h = 10.0