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CH 4 

Motion in two and three Dimensions 

I. PositionandDisplacement:

A. Position:

1. Thepositionofaparticlecanbedescribedbyapositionvector,withrespecttoareferenceorigin.

 

                            

B. DisplacementVector: r

1. DisplacementVector:Thedisplacementofaparticleisthechangeofthepositionvectorduringacertaintime.

r

 

           

 

 

  

 

 

 

 

 

 

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C. Exampleof2Dmotion

 

 

1. Solution

 

 

t = 15 sec 

SO: 

 

 

 

 

 

 

 

 

 

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2. Graphs

II. AverageVelocityandInstantaneousVelocity

A. Ifaparticlemovesthroughadisplacementofrinttime,thentheaveragevelocityis:

 

 

 

                     

B. Inthelimitthatthettimeshrinkstoasinglepointintime,theaveragevelocityisapproachesinstantaneousvelocity.Thisvelocityisthederivativeofdisplacementwithrespecttotime.

 

   

 

 

 

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C. Fortheprevioussampleproblem,let’sfindthevelocityoftherabbitat15sec:

 

1. Solution

 

 

t = 15 sec 

 

 

 

 

 

 

 

 

 

 

 

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2. Graph

III. AverageandInstantaneousAccelerations 

A. Followingthesamedefinitionasinaveragevelocity, 

 

 

 

                     

B. Ifweshrinkttozero,thentheaverageaccelerationvalueapproachestotheinstantaccelerationvalue,whichisthederivativeofvelocitywithrespecttotime:

 

   

 

 

 

 

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C. Fortheprevioussampleproblem,let’sfindtheaccelerationoftherabbitat15sec:

 

1. Solution

 

 

t = 15 sec 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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2. Graph

 

IV. ExampleProblem 

A. AParticlemovesintheXYplane.Itscoordinatesvarywithtimeas3

3( ) 1.00 ( ) 32.0 ( )

m mx t t t

s s and 2

2( ) 5.00 ( ) 12.0 ( )

m my t t t

s s .

Findtheposition,velocity,andaccelerationatt=3sec.

1. Solution 

 

 

 

 

 

 

 

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V. Nextclass(Friday),PROJECTILEMOTION. 

Read pages 70‐75 

 

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VI. PROJECTILEMOTION.

A. Aprojectileis

1. Aparticlemovingintheverticalplane

2. Withsomeinitialvelocity

3. Whoseaccelerationisalwaysfree‐fallacceleration(g)

B. Themovementofaprojectileisprojectilemotion,withtheonlyaccelerationequaltothefreefallacceleration,g.

 

C. Inprojectilemotion,thehorizontalmotionandtheverticalmotionare__________________________________________;thatisneithermotionisaffectedbytheother.

 

D. Theinitialvelocityoftheprojectileis: 0v

 

 

1. Here, 0xv and 0 yv  

 

 

Examples in sports:

Tennis

Baseball

Football

Lacrosse

Racquetball

Soccer………….

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E. Projectilemotionanalyzed,assumingnoexternalforcesotherthantheweight:

1. Followmyformatanditwillmaketheseproblemssimpler.

2. Startbyalwaysdrawingthefollowinggrid   

Vfy = V0y + DV V0y + ayt   

   

   

   

Vertical  Horizontal 

ay= ‐g =‐ 9.8 m/s2  ax=0 

Dy (height) =  Dx (range) = Voy = VoSinq=  Vox = VoCosq= Vfy =    

t =  t = 

 

3. Firstrule:TimeistheONLYInterloper!NothingelsecancrossbetweenVerticalandHorizontal

4. Thenlet’sfillitinwithourbig5equationsontheleftabovethewordvertical.

5. Afterwehavethebig5equationsontheleft,wewritethefirstequationontherightwithxsubscripts.Noticethatax=0sodrawalinethroughthe+½axt2becausethistermgoestozero.

6. Nowwillfillinourgivendatafromthewordproblem.

7. Thenwecirclewhichvariabletheywantustofind.

8. Thenwelookatourequationsandseewhichoneswewillusetonavigatethroughthisproblem.

 We will do a sample problem shortly to make sure we understand this process.  However, what if we wanted to plot the projectiles position or know its range. Let’s derive two equations which could help us determine these values more rapidly.  

 

 

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F. Projectile’sPath

1. From

 

a) ThensolveEquation1fortime. 

2. Andsince  

3. PlugthatvalueoftbackintoEquationabove: 

 

 

 

 

 

 

 

 

 

4. FinallywegettheEquationofPath.

 

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G. HorizontalRange,assumingnoexternalforces:

1. Thehorizontalrangeofaprojectileisthehorizontaldistancewhenitreturnstoitslaunchingheight

 

2. Thedistanceequationsinthex‐andy‐directionsrespectively:

 

3. Eliminatingt: 

 

 

 

 

 

 

 

 

4. ThusR=

5. Caution: 

6.

 

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H. Example,projectilemotion:                                                                         

                                                                        

 

 

 

 

 

 During the civil war, a cannon placed on top of a 40 m high hill is fired with a velocity of 200 m/s at an 

angle of 30∞ from horizontal.  How close did the cannon operators have to the enemy get in order to 

ensure a hit? 

1. Solution:

               

Vertical  Horizontal 

ay= ‐g =‐ 9.8 m/s2  ax=0 

Dy (height) =  Dx (range) = Voy = VoSinq=  Vox = VoCosq= Vfy =    

t =  t = 

 

 

 

 

 

 

 

Pleasant Hill – 40m  

 

 

April 9, 1864 

30 ±

200 m/s

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I. ExampleII,projectilemotion:

1. Uponspottinganinsectonatwigoverhangingwater,anarcherfishsquirtswaterdropsattheinsecttoknockitintothewater(Fig.below).Althoughthefishseestheinsectalongastraight‐linepathatangleanddistanced,adropmustbelaunchedatadifferentangle0ifitsparabolicpathistointersecttheinsect.If=36andd=0.900m,whatlaunchangle0isrequiredforthedroptobeatthetopoftheparabolicpathwhenitreachestheinsect?

 

 

a) Solution:∆∆

2

 

1/2  

 

0  

 

    thus 

 and 

2   thus 

22

2

 

2 2  

212

 

2 55.46° 

Tuesday’s lesson (Monday’s a Holiday) will be Uniform Circular Motion read pages 76‐79 

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VII. UniformCircularMotion

A. Aparticleisin“UniformCircularMotion”ifittravelsaroundacircleoracirculararcataconstant(uniform)speed.Note:althoughthespeedisnotchanging,theparticleisacceleratingbecausethevelocityischangingdirection.

 

B. Drawing 

 

 

 

 

 

 

 

 

 

 

C. Centripetalacceleration(center‐seeking) 

 

 

 

 

 

 

1. Proofisinthebookonpage77(proofdoesnotneedtobememorized). 

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D. Period(orPeriodofrevolution)SymbolT.

1. Tisdefinedasthetimetomakeonerevolution.

2. T=

3. ac= 

E. Side‐notes(willbeusefulwhenwegettochapter10,butonlyinterestingrightnow)

1. AngularVelocitynotmeterspersecond,butnumberofradianspersecond.Symbolw.w=2p/T

2. Velocity=w*r

3. ac=w2*r

4. Frequency=revolutionspersecond=1/T.UnitsHertzorrev/secorcyclespersec.

F. SampleProblem:UniformCircularMotion

1. Arotatingfancompletes1200revolutionseveryminute.Considerthetipofablade,ataradiusof0.15m.(a)Throughwhatdistancedoesthetipmoveinonerevolution?Whatare(b)thetip'sspeedand(c)themagnitudeofitsacceleration?(d)Whatistheperiodofthemotion?

 

a) Solution 

 

 

 

 

 

 

 

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VIII. Relativemotioninone‐dimension

A. Thevelocityofaparticledependsonthereferenceframeofwhoeverisobservingthevelocity.

1. SupposeAlex(A)isattheoriginofframeA(asinFig.4‐18),watchingcarP(the“particle”)speedpast.

2. SupposeBarbara(B)isattheoriginofframeB,andisdrivingalongthehighwayatconstantspeed,alsowatchingcarP.Supposethattheybothmeasurethepositionofthecaratagivenmoment.Then:

 

wherexPAisthepositionofPasmeasuredbyA.

3. Graph

 

4. Consequently,

 

5. Also,

a) SincevBAisconstant,thelasttermiszeroandwehave

  aPA = aPB. 

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6. Example,relativemotion,1‐D:

a)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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Watch the following video about frames of reference it is very interesting and informative (it is a classic): 

http://www.youtube.com/watch?v=Y75kEf8xLxI  

IX. Relativemotionintwo‐dimensions

A. AandB,thetwoobservers,arewatchingP,themovingparticle,fromtheiroriginsofreference.BmovesataconstantvelocitywithrespecttoA,whilethecorrespondingaxesofthetwoframesremainparallel.rPAreferstothepositionofPasobservedbyA,andsoon.Fromthesituation,itisconcluded: