Circular Motionmotion AAA Powerpointpowerpointpowerpoint Presentationpresentationpresentation Bybyby Paulpaulpaul E.E.E

Home , Circular motion, Turn (angle)

ChapterChapter 10.10. UniformUniform CircularCircular MotionMotion AAA PowerPointPowerPointPowerPoint PresentationPresentationPresentation bybyby PaulPaulPaul E.E.E. Tippens,Tippens,Tippens, ProfessorProfessorProfessor ofofof PhysicsPhysicsPhysics SouthernSouthernSouthern PolytechnicPolytechnicPolytechnic StateStateState UniversityUniversityUniversity

•• ApplyApply youryour knowledgeknowledge ofof centripetalcentripetal accelerationacceleration andand centripetalcentripetal forceforce toto thethe solutionsolution ofof problemsproblems inin circularcircular motion.motion. •• DefineDefine andand applyapply conceptsconcepts ofof frequencyfrequency andand period,period, andand relaterelate themthem toto linearlinear speed.speed. •• SolveSolve problemsproblems involvinginvolving bankingbanking angles,angles, thethe conicalconical pendulum,pendulum, andand thethe verticalvertical circle.circle. UniformUniform CircularCircular MotionMotion

UniformUniform circularcircular motionmotion is motion along a circular path in which there is no change in speed, only a change in direction.

v Constant velocity Fc tangent to path. Constant force toward center.

Question: Is there an outward force on the ball? UniformUniform CircularCircular MotionMotion (Cont.)(Cont.)

TheThe questionquestion ofof anan outwardoutward forceforce cancan bebe resolvedresolved byby askingasking whatwhat happenshappens whenwhen thethe stringstring breaks!breaks! BallBall movesmoves tangenttangent toto v path,path, NOTNOT outwardoutward asas mightmight bebe expected.expected.

When central force is removed, ball continues in straight line.

CentripetalCentripetal forceforce isis neededneeded toto changechange direction.direction. ExamplesExamples ofof CentripetalCentripetal ForceForce YouYouYou areareare sittingsittingsitting ononon thethethe seatseatseat nextnextnext tototo thethethe outsideoutsideoutside door.door.door. WhatWhatWhat isisis thethethe directiondirectiondirection ofofof thethethe resultantresultantresultant forceforceforce ononon youyouyou asasas youyouyou turn?turn?turn? IsIsIs ititit awayawayaway fromfromfrom centercentercenter ororor towardtowardtoward centercentercenter ofofof thethethe turn?turn?turn?

•• CarCar goinggoing aroundaround aa FF c curve.curve.

ForceForce ONON youyou isis towardtoward thethe center.center. CarCar ExampleExample ContinuedContinued

Reaction F’ TheThe centripetalcentripetal Fc forceforce isis exertedexerted BYBY thethe doordoor ONON you.you. (Centrally)(Centrally)

ThereThereThere isisis ananan outwardoutwardoutward force,force,force, butbutbut ititit doesdoesdoes notnotnot actactact ONONON you.you.you. ItItIt isisis thethethe reactionreactionreaction forceforceforce exertedexertedexerted BYBYBY youyouyou ONONON thethethe door.door.door. ItItIt affectsaffectsaffects onlyonlyonly thethethe door.door.door. AnotherAnother ExampleExample

DisappearingDisappearing RR platformplatform atat fair.fair. FF c

WhatWhat exertsexerts thethe centripetalcentripetal forceforce inin thisthis exampleexample andand onon whatwhat doesdoes itit act?act?

TheTheThe centripetalcentripetalcentripetal forceforceforce isisis exertedexertedexerted BYBYBY thethethe wallwallwall ONONON thethethe man.man.man. AAA reactionreactionreaction forceforceforce isisis exertedexertedexerted bybyby thethethe manmanman ononon thethethe wall,wall,wall, butbutbut thatthatthat doesdoesdoes notnotnot determinedeterminedetermine thethethe motionmotionmotion ofofof thethethe man.man.man. SpinSpin CycleCycle onon aa WasherWasher

HowHow isis thethe waterwater removedremoved fromfrom clothesclothes duringduring thethe spinspin cyclecycle ofof aa washer?washer?

ThinkThink carefullycarefully beforebefore answeringanswering ...... DoesDoes thethe centripetalcentripetal forceforce throwthrow waterwater offoff thethe clothes?clothes?

NO.NO.NO. Actually,Actually,Actually, ititit isisis thethethe LACKLACKLACK ofofof aaa forceforceforce thatthatthat allowsallowsallows thethethe waterwaterwater tototo leaveleaveleave thethethe clothesclothesclothes throughthroughthrough holesholesholes ininin thethethe circularcircularcircular wallwallwall ofofof thethethe rotatingrotatingrotating washer.washer.washer. CentripetalCentripetal AccelerationAcceleration ConsiderConsider ballball movingmoving atat constantconstant speedspeed vv inin aa horizontalhorizontal circlecircle ofof radiusradius RR atat endend ofof stringstring tiedtied toto pegpeg onon centercenter ofof table.table. (Assume(Assume zerozero friction.)friction.) n Fc v W R ForceForce FF c andand accelerationacceleration aac towardtoward center.center. WW == nn DerivingDeriving CentralCentral AccelerationAcceleration

Consider initial velocity at A and final velocity at B:

vf vf B -vo v s v vo o R R A DerivingDeriving AccelerationAcceleration (Cont.)(Cont.) v v f Definition: ac = -v t o v s vo Similar v s R = Triangles v R mass m v vs vv a = = = c t Rt R

Centripetal vm22v Centripetal aFma; acceleration:acceleration: cccR R ExampleExample 1:1: AA 33--kgkg rockrock swingsswings inin aa circlecircle ofof radiusradius 55 mm.. IfIf itsits constantconstant speedspeed isis 88 m/sm/s,, whatwhat isis thethe centripetalcentripetal acceleration?acceleration? v v2 m a  m = 3 kg c R R R = 5 m; v = 8 m/s (8 m/s)2 a 12.8 m/s2 c 5 m F = (3 kg)(12.8 m/s2) mv2 Fmacc F = 38.4 N R Fcc = 38.4 N ExampleExample 2:2: AA skaterskater movesmoves withwith 1515 m/sm/s inin aa circlecircle ofof radiusradius 3030 mm.. TheThe iceice exertsexerts aa centralcentral forceforce ofof 450450 NN.. WhatWhat isis thethe massmass ofof thethe skater?skater? DrawDraw andand labellabel sketchsketch mv2 FR Fm; c v = 15 m/s c R v2 (450 N)(30 m) Fc R m  2 450 N 30 m (15 m/s) m=? mm == 60.060.0 kgkg Speed skater ExampleExample 3.3. TheThe wallwall exertsexerts aa 600600 NN forceforce onon anan 8080--kgkg personperson movingmoving atat 44 m/sm/s onon aa circularcircular platform.platform. WhatWhat isis thethe radiusradius ofof thethe circularcircular path?path? DrawDraw andand labellabel sketchsketch NewtonNewton’’ss 2nd2nd lawlaw m = 80 kg; forfor circularcircular motion:motion: v = 4 m/s2 F = 600 N mv22 mv c Fr; rF r = ? (80 kg)(4 m/s)2 r  r = 2.13 m 600 N r = 2.13 m CarCar NegotiatingNegotiating aa FlatFlat TurnTurn v FF c

R WhatWhat isis thethe directiondirection ofof thethe forceforce ON thethe car?car? Ans.Ans. TowardToward CenterCenter ThisThis centralcentral forceforce isis exertedexerted BYBY thethe roadroad ONON thethe car.car. CarCar NegotiatingNegotiating aa FlatFlat TurnTurn v FF c

R IsIs therethere alsoalso anan outwardoutward forceforce actingacting ONON thethe car?car?

Ans.Ans. No,No, butbut thethe carcar doesdoes exertexert aa outwardoutward reactionreaction forceforce ONON thethe road.road. CarCar NegotiatingNegotiating aa FlatFlat TurnTurn

TheThe centripetalcentripetal forceforce FF c isis thatthat ofof staticstatic frictionfriction ff s :: n F = f Fc R c s m fs v R mg TheThe centralcentral forceforce FF andand thethe frictionfriction forceforce ff The central force FCC and the friction force fss areareare notnotnot twotwotwo differentdifferentdifferent forcesforcesforces thatthatthat areareare equal.equal.equal. ThereThereThere isisis justjustjust oneoneone forceforceforce ononon thethethe car.car.car. TheTheThe naturenaturenature ofofof thisthisthis centralcentralcentral forceforceforce isisis staticstaticstatic friction.friction.friction. FindingFinding thethe maximummaximum speedspeed forfor negotiatingnegotiating aa turnturn withoutwithout slipping.slipping.

n Fc = fs fs Fc R R m v mg

TheThe carcar isis onon thethe vergeverge ofof slippingslipping whenwhen FF C isis equalequal toto thethe maximummaximum forceforce ofof staticstatic frictionfriction ff s. . mv2 f =  mg Fc = fs Fc = s s R MaximumMaximum speedspeed withoutwithout slippingslipping (Cont.)(Cont.)

F = f n c s fs R mv2 = smg R mg

v = sgR

F m c R VelocityVelocity vv isis maximummaximum v speedspeed forfor nono slipping.slipping. ExampleExample 4:4: AA carcar negotiatesnegotiates aa turnturn ofof radiusradius 7070 mm whenwhen thethe coefficientcoefficient ofof staticstatic frictionfriction isis 0.70.7.. WhatWhat isis thethe maximummaximum speedspeed toto avoidavoid slipping?slipping? mv2 Fc = fs = s mg m R Fc R From which: v = sgR v s = 0.7 g = 9.8 m/s2; R = 70 m

v = 21.9 m/s vgRs (0.7)(9.8)(70m) v = 21.9 m/s OptimumOptimum BankingBanking AngleAngle ByBy bankingbanking aa curvecurve atat thethe optimumoptimum angle,angle, thethe normalnormal F m c R forceforce nn cancan provideprovide thethe necessarynecessary centripetalcentripetal forceforce v withoutwithout thethe needneed forfor aa frictionfriction force.force. f fs = 0 s n n n f w s w  w   slowslow speedspeed fastfast speedspeed optimumoptimum FreeFree--bodybody DiagramDiagram

n AccelerationAcceleration aa isis towardtoward thethe xx center.center. SetSet xx axisaxis alongalong

thethe directiondirection ofof aac ,, i.i. e.,e., mg  horizontalhorizontal (left(left toto right).right). nn coscos  nn  nn  ++ aac nn sinsin  mgmg  mgmg OptimumOptimum BankingBanking AngleAngle (Cont.)(Cont.) nn coscos  n  n

nn sinsin  mg  mgmg

ApplyApply mvmv2 FF == mmaa nn sinsin   NewtonNewton’’ss 2nd2nd x c RR LawLaw toto xx andand yy axes.axes. FF y == 00 nn coscos == mgmg OptimumOptimum BankingBanking AngleAngle (Cont.)(Cont.) n cos  n n  nsin tan  n sin  ncos mg  mg

mv2 mv2 n sin   v2 R tan R mg gR n cos  = mg 1 OptimumOptimum BankingBanking AngleAngle (Cont.)(Cont.)

n cos  n  n

n sin  mg  mg

Optimum Banking v2 tan  Angle  gR ExampleExample 5:5: AA carcar negotiatesnegotiates aa turnturn ofof radiusradius 8080 m.m. WhatWhat isis thethe optimumoptimum bankingbanking angleangle forfor thisthis curvecurve ifif thethe speedspeed isis toto bebe equalequal toto 1212 m/s?m/s? n v2 (12 m/s)2 tan  = = gR (9.8 m/s2)(80 m) mg  n cos  tan = 0.184  = 10.40  n How might you find the mv2 n sin  centripetalF  force on the car, knowingC its mass? mg R TheThe ConicalConical PendulumPendulum A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.

T cos  T L   h T T sin  R mg Note:Note: TheThe inwardinward componentcomponent ofof tensiontension TT sinsin  givesgives thethe neededneeded centralcentral force.force. AngleAngle  andand velocityvelocity vv::

T cos  T L   h T T sin  R mg Solve two mv2 T sin   v2 equations R tan  = to find gR angle  T cos = mg ExampleExample 6:6: AA 22--kgkg massmass swingsswings inin aa horizontalhorizontal circlecircle atat thethe endend ofof aa cordcord ofof lengthlength 1010 mm.. WhatWhat isis thethe constantconstant speedspeed ofof thethe massmass ifif thethe roperope makesmakes anan angleangle ofof 30300 withwith thethe vertical?vertical?

 1.1. DrawDraw && labellabel sketch.sketch. 2.2. RecallRecall formulaformula forfor pendulum.pendulum. L  h 2 T v tan  Find:Find: gR vv == ?? R 3.3. ToTo useuse thisthis formula,formula, wewe needneed toto findfind RR == ?? R = L sin 300 = (10 m)(0.5) R = 5 m ExampleExample 6(Cont.):6(Cont.): FindFind vv forfor  == 30300

4.4. UseUse givengiven infoinfo toto findfind thethe  0 velocityvelocity atat 3030 .. R = 5 m L  R = 5 m g = 9.8 m/s2 h R = 5 m g = 9.8 m/s T SolveSolve forfor v2 tan  R vv == ?? gR

vgR2  tan vgR tan

v  (9.8 m/s20 )(5 m) tan 30 vv == 5.325.32 m/sm/s ExampleExample 7:7: NowNow findfind thethe tensiontension TT inin thethe cordcord ifif mm == 22 kgkg,,  == 30300,, andand LL == 1010 mm..

T cos  T L   h T T sin  2 kg R mg

Fy = 0: T cos  - mg = 0; T cos  = mg

mg (2 kg)(9.8 m/s2) T = = TT == 22.622.6 NN cos  cos 300 ExampleExample 8:8: FindFind thethe centripetalcentripetal forceforce FF c forfor thethe previousprevious example.example.

0 T cos   == 3030 T

L   h T Fc T sin  2 kg R mg

m = 2 kg; v = 5.32 m/s; R = 5 m; T = 22.6 N mv2 F = or F = T sin 300 c c FFc == 11.311.3 NN R c SwingingSwinging SeatsSeats atat thethe FairFair b ThisThis problemproblem isis identicalidentical toto thethe otherother examplesexamples L  exceptexcept forfor findingfinding R.R. h T d R = d + b R R = L sin  + b

v2 tan  = and v = gR tan  gR ExampleExample 9.9. IfIf bb == 55 mm andand LL == 1010 mm,, whatwhat willwill bebe thethe speedspeed ifif thethe angleangle  == 26260?? v2 tan  = R = d + b gR b L  0  dd == (10(10 m)m) sinsin 2626 == 4.384.38 mm T d R = 4.38 m + 5 m = 9.38 m R vgR2  tan vgR tan v  (9.8 m/s20 )(9.38 m) tan 26 vv == 6.706.70 m/sm/s MotionMotion inin aa VerticalVertical CircleCircle v v ConsiderConsider thethe forcesforces onon aa + v ballball attachedattached toto aa stringstring asas v BottomBottomTmg + itit movesmoves inin aa verticalvertical loop.loop. T T

mg+ NoteNote alsoalso thatthat thethe positivepositive mg+ +mg T directiondirection isis alwaysalways alongalong v acceleration,acceleration, i.e.,i.e., towardtoward

TopTop ofof PathPath LeftLeft SideSide + TopTopTopTopmg RightRightRightRight thethe centercenter ofof thethe circle.circle. TensionTension isis WeightWeightMaximumMaximum hashas nono WeightWeightminimumminimum causeshascauseshas asas nono NoteNote changeschanges asas youyou clickclick tensiontensioneffecteffectBottomBottom on onT,T, TTWW smallsmallweightweighteffecteffect decreasedecrease helpsonhelpson TT thethe mousemouse toto showshow newnew opposesopposes FF inin F Ftensiontension c forceforce TTc positions.positions. . . _ _

. ? . ? 30 N 30 N _ _ 50 N 50 N __?___ __?___ 40 N 40 N T = T = T = T = T = T = T = T = W = 10 N W = 10 N

is required to is required to 40 N 40 N resultant is The tension T must resultant is = The tension T must = adjust so that central adjust so that central c c = 40 N As an exercise, assume As an exercise, assume that a central force of that a central force of F F maintain circular motion maintain circular motion of a ball and of a ball and = 40 N T T

v 10 N = 40 10 N = 40 – + + R – T T 10 N + 10 N +

T T

+ + 10 N 10 N

v At top: At top: Bottom: Bottom: MotionMotion inin aa VerticalVertical CircleCircle v Resultant force mv2 Fc = mg toward center R T R

Consider TOP of circle: v mv2 AT TOP: mg + T = R + mg mv2 T = - mg T R VerticalVertical Circle;Circle; MassMass atat bottombottom v Resultant force mv2 F = toward center c R R T Consider bottom of circle: v mv2 mg T - mg = AT Bottom: R T mv2 + T = + mg mg R VisualVisual Aid:Aid: AssumeAssume thatthat thethe centripetalcentripetal forceforce requiredrequired toto maintainmaintain circularcircular motionmotion isis 2020 NN.. FurtherFurther assumeassume thatthat thethe weightweight isis 55 NN..

v mv2 F 20 N C R R Resultant central force FFC at every point in path!

v FF C == 2020 NN Weight vector WW is FFC == 2020 NN at top AND at bottom. downward at every point. W = 5 N, down VisualVisual Aid:Aid: TheThe resultantresultant forceforce ((2020 NN)) isis thethe vectorvector sumsum ofof TT andand WW atat ANYANY pointpoint inin path.path.

v Top: T + W = FC W + T + 5 N = 20 N T R T = 20 N - 5 N = 15 N

T + Bottom: W v T - W = F C

FFC == 2020 NN at top T - 5 N = 20 N AND at bottom. T = 20 N + 5 N = 25 N ForFor MotionMotion inin CircleCircle v AT TOP: mv2 R + T = - mg mg R T v

AT BOTTOM: 2 T mv + T = + mg R mg ExampleExample 10:10: AA 22--kgkg rockrock swingsswings inin aa verticalvertical circlecircle ofof radiusradius 88 mm.. TheThe speedspeed ofof thethe rockrock asas itit passespasses itsits highesthighest pointpoint isis 1010 m/sm/s.. WhatWhat isis tensiontension TT inin rope?rope? mv2 At Top: mg + T = v R mg mv2 T R T = - mg R (2 kg)(10 m/s)2 v T 2 kg(9.8 m/s2 ) 8 m

T = 25 N - 19.6 N TT == 5.405.40 NN ExampleExample 11:11: AA 22--kgkg rockrock swingsswings inin aa verticalvertical circlecircle ofof radiusradius 88 mm.. TheThe speedspeed ofof thethe rockrock asas itit passespasses itsits lowestlowest pointpoint isis 1010 m/sm/s.. WhatWhat isis tensiontension TT inin rope?rope? mv2 v At Bottom: T - mg = R mv2 R T = + mg R T v (2 kg)(10 m/s)2 T 2 kg(9.8 m/s2 ) mg 8 m

T = 25 N + 19.6 N TT == 44.644.6 NN ExampleExample 12:12: WhatWhat isis thethe criticalcritical speedspeed vv c atat thethe top,top, ifif thethe 22--kgkg massmass isis toto continuecontinue inin aa circlecircle ofof radiusradius 88 mm?? 0 mv2 v At Top: mg + T = mg R

T R vc occurs when T = 0 mv2 v mg = vc = gR R v = gR = (9.8 m/s2)(8 m) v = 8.85 m/s vcc = 8.85 m/s TheThe LoopLoop--thethe--LoopLoop SameSame asas cord,cord, nn replacesreplaces TT v AT TOP: mv2 + n = - mg R mg R

v n

AT BOTTOM: mv2 n n= + mg + R mg TheThe FerrisFerris WheelWheel v mv2 AT TOP: mg -n= n R R + mv2 n = mg - v mg R

AT BOTTOM: mv2 n n = + mg + R mg ExampleExample 13:13: WhatWhat isis thethe apparentapparent weightweight ofof aa 6060--kgkg n personperson asas sheshe movesmoves v + throughthrough thethe highesthighest pointpoint mg whenwhen RR == 4545 mm andand thethe R speedspeed atat thatthat pointpoint isis 66 m/s?m/s? v ApparentApparent weightweight willwill bebe thethe normalnormal forceforce atat thethe top:top: mv2 mv2 mg -n= n = mg - R R

2 2 (60kg)(6m/s) n 60 kg(9.8 m/s ) = 540 N 45 m nn = 540 N SummarySummary Centripetal vm22v Centripetal aFma; acceleration:acceleration: cccR R

v2 tan  = v = sgR gR

Conical v = gR tan  pendulum: Summary:Summary: MotionMotion inin CircleCircle v AT TOP: 2 + T = mv - mg R mg R T v AT BOTTOM: 2 T T = mv + mg + R mg Summary:Summary: FerrisFerris WheelWheel v mv2 AT TOP: mg -n= n R R + mv2 n = mg - v mg R

AT BOTTOM: mv2 n n = + mg + R mg CONCLUSION:CONCLUSION: ChapterChapter 1010 UniformUniform CircularCircular MotionMotion