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CircularMo+on Chapter5 Circularmo+on –Rota+ons– Describethemo-on: 1) Linearspeed:thedistance (meters,km,feet,..)travelled persecond(orminuteor…) 2) Rota-onal(orangular)speed: thenumberofrota+onsper second(orminuteor…) Todefinecircularmo-onwewillconcentrateonanglesinsteadofdistances. Physics140,Prof.M.Nikolic 2 Radians Measuringθindegrees(deginyourcalculator)turnsouttobeapoor choice. Radiansareamorenaturalchoiceofangularunit. 0 2π radians = 360 = 1 revolution = 1 rotation 3600 1 radian = = 57.30 2 ⋅ 3.14 s θ (rad) = r s–arclength r-radius Physics140,Prof.M.Nikolic 3 Conceptualques+on–Radians Q1 Awheelturnsthroughfivecompleterevolu+onsandthenonequarterofa revolu+on.Throughwhattotalangleinradianshasthewheelturned? A. B. C. D. 33radians. 50radians. 5radians. 66radians. 1revolu+on=2πradians →5+¼revolu+ons=(5+¼)x2πradians →10π+0.5π=10.5π=10.5x3.14=33radians Physics140,Prof.M.Nikolic 4 Angularposi+onanddisplacement y JustlikewedidinChapter3forlinearmo+on, wewilldefine: ! rf θf Angularposi+on–θ(radians) →measuredcounterclockwise(CCW)from ! ri thex-axis–posi+veangle θi x →θi–ini+alangularposi+on →θf–finalangularposi+on Angulardisplacement–Δθ(radians) Δθ = θ f − θ i Whenobjectmovesalongacircularpath,posi+onvectorsriandrfhave magnitudesequaltotheradiusofthecircle. Physics140,Prof.M.Nikolic 5 Angularvelocity JustlikewedidinChapter3forlinearmo+on,wewilldefine: Averageangularvelocity →angulardisplacementper+me Δθ θ f − θ i ω av = = Δt t f − ti Instantaneousangularvelocity Δθ ω = lim Δt →0 Δt Justlikelinearmo+onbutwithr→θ Unitsofangularvelocity: radianspersecond[rad/s] Counterclockwise(CCW)rota+onrepresentsposi+verota+on. Clockwiserota+on(CW)representsnega+verota+on. Physics140,Prof.M.Nikolic 6 Conceptualques+on–Ladybug Q2 Aladybugsitsattheouteredgeofamerry-go-round,andagentlemanbugsits halfwaybetweenherandtheaxisofrota+on.Themerry-go-roundmakesacomplete revolu+ononeachsecond.Thegentlemanbug’sangularvelocityis A. B. C. D. Halftheladybug’s. Thesameasladybug’s. Twicetheladybug’s. Itcan’tbedetermined. Bothbugscoverthesameangle(Δθ)overthe sameamountof+me→angularvelocityis thesameforbothbugs Physics140,Prof.M.Nikolic 7 Rela+onbetweenlinearandangularspeed y s r θf Anobjectmovingalongacircularpathduring angulardisplacement(Δθ)willtravelthedistance (s)equaltothearclength: s = r Δθ θi Averagelinearspeed(Chapter2): total distance s vav = = total time Δt vav = ⎛ Δθ ⎞ rΔθ = r⎜ ⎟ ⎝ Δt ⎠ Δt vav = rω av Physics140,Prof.M.Nikolic r–radiusofacircle 8 Conceptualques+on–Ladybug Q3 Aladybugsitsattheouteredgeofamerry-go-round,andagentlemanbugsits halfwaybetweenherandtheaxisofrota+on.Themerry-go-roundmakesacomplete revolu+ononeachsecond.Thegentlemanbug’slinearspeedis A. B. C. D. Halfastheladybug’s. Thesameastheladybug’s. Twicetheladybug’s. Itcan’tbedetermined. Bothbugshavethesameangularspeed (ω)butthelinearspeedsaredifferent →ladybug:vL=ωR →gentlemanbug:vG=ωR/2=½vL Physics140,Prof.M.Nikolic 9 Uniformcircularmo+on Uniformcircularmo+on è whenthespeedofapointmovinginacircleisconstant. PeriodandFrequency Thefrequency(f)isthenumberofcompleterevolu+onspersecond. Units:Hertz[Hz(rev/s)] Theperiod(T)isthe+meittakesapointtomakeonerevolu+on. T= Linearspeed: v= 1 f distance traveled 2π r = = 2π rf time T Physics140,Prof.M.Nikolic v 2π ω = = 2π f = r T 10 Exercise:Speedincentrifuge Acentrifugeisspinningat5400rpm. a)Findtheperiodandfrequencyofmo+on. 1rpm=1revolu+onsperminute→thisisfrequency ! rev $ ! 1min $ rev = 90 Hz First,convertrpmtorev/s=Hz: f = # 5400 &×# & = 90 " min % " 60s % s Period: T= 1 f T= 1 = 0.011 s 90 Hz b)Findangularspeed. ω= 2π = 2πf T Physics140,Prof.M.Nikolic ω = 2π rad ⋅ 90Hz = 180π rad rad = 565.2 s s 11 Radialaccelera+on Consideranobjectmovinginacircularpathof radiusratconstantspeed. r Linearvelocityisalwaysatangent tothemo+onpath. ! vi r ! vf ! Δv Direc+onofvectorvelocityischanging →velocityischanging→objecthasnon zeroaccelera+on ! ! ! Δv = v f − vi = aΔt →aisinthesamedirec+onasΔv WhenΔt→0,instantaneousaccelera+onpointstowardsthecenterofthecircle (radiallyinward)→andiscalledradialaccelera-onar. Physics140,Prof.M.Nikolic 12 Magnitudeofradialaccelera+on Similarly,to:s = rΔθ r r Δθ Wecanalsosay: s ! vf ! vi Δv = vΔθ Δθ ! Δv Δv = vωΔt ! ! vi = v f = v Magnitudeofradialaccelera+on: Δv ar = Δt Physics140,Prof.M.Nikolic v2 ar = vω = = ω 2 r r Units:[m/s2] 13 Q4 Conceptualques+on–Radialaccelera+on AnLPrecord,10cminradius,spinswithangularvelocityof20rad/s.Iftheangular speedoftherecordisdoubled,whathappenstoitsradialaccelera+onattherims? A. B. C. D. Radialaccelera+ondoubles. Radialaccelera+onquadruples. Radialaccelera+onquarters. Radialaccelera+onhalves. v2 ar = vω = = ω 2 r r ωdoubles→arquadruples Ormathema+cally: ar1 = ω12 r = (20rad / s)2 ⋅ 0.1m = 40 m/s2 ar 2 = ω 22 r = (40rad / s)2 ⋅ 0.1m = 160 m/s2 ar 2 = 4ar1 Physics140,Prof.M.Nikolic 14 ApplyingNewton’ssecondlaw Radialaccelera+onhasconstantmagnitudeandisdirectedtowardthecircle’scenter. ! ! Newton’ssecondlaw: F = ma net →Somethingmustprovidetheforce–Centripetalforce CentripetalforceisNOTaforceofnature.Anyoftheforceswediscussedin Chapter4canplayaroleofcentripetalforcewhenobject’smovingalongacircle. Tosolveproblemsinuniformcircularmo+on→applythesamesetofrules providedinChapter4 →setx-axisalongradialaccelera+on →therewillbenoaccelera+onydirec+onforuniformcircularmo+on 2 F = ma = m ω r ∑ x r Physics140,Prof.M.Nikolic and ∑F y =0 15 Exercise:Amusementparkride Therotorisanamusementparkridewherepeoplestandagainsttheinsideofa cylinder.Oncethecylinderisspinningfastenough,thefloordropsout. (a)Whatforcekeepsthepeoplefromfallingoutthebotomofthecylinder? First,drawafreebodydiagram(FBD)ofonepersonstandingagainstthewall. y ar fs N Makesurethat: →Weightpointsstraightdown →Normalforceisperpendiculartothesurface →Fric+onforceisparalleltothesurfaceandopposesthemo+on x w Xdirec+on:fsx=0 Nx=N Wx=0 Ydirec+on:fsy=fs Ny=0 Wy=-W=-mg Theforceofsta+cfric+onkeepspeople fromfallingoutthebotom. Physics140,Prof.M.Nikolic 16 Exercise:Amusementparkride Therotorisanamusementparkridewherepeoplestandagainsttheinsideofa cylinder.Oncethecylinderisspinningfastenough,thefloordropsout. (b)Ifμs=0.40andthecylinderhasr=2.5m,whatistheminimumangularspeedof thecylindersothatthepeopledon’tfallout? y ApplyNewton’s2ndLaw: fs N ∑ F = ma ∑F = 0 x y x r N = mω 2 r f s −W = 0 fs = W w Aslongasforceofsta+cfric+onis largerorequaltoforceofgravity (weight)èpeoplewillnotfallout µ s N = µ s mω 2 r = mg 9.8 m/s2 ω= = = 3.13 rad/s µs r (0.40)(2.5 m) g Physics140,Prof.M.Nikolic 17 Takingexams Fromnowon,Iwillprovideeveryonewith equa+onsnecessaryfortheexam. Iwillposttheequa+onsheetonCanvasaweek beforetheexamsoeveryonecanseeitango overit. YouareNOTallowedtobringanythingtothe exam(exceptforpen/pencilandcalculator).The equa+onsheetwillbeatachedtoyouexam. Physics140,Prof.M.Nikolic 18 Exercise:Slippingcoin A20-gcoinisplacedonarecordthatisrota+ngat33.3rpm.Ifμs=0.1,howfar fromthecenteroftherecordcanthecoinbeplacedwithouthavingitslipoff? DrawanFBDforthecoin. y ar N Xdirec+on:fsx=fs Nx=0 Wx=0 Ydirec+on:fsy=0 Ny=N Wy=-W=-mg Forceofsta+cfric+onkeepsthecoinfromslippingoff. fs ApplyNewton’s2ndLaw: x ∑F x w = mar fs = mω 2 r Weneedtofindfsfirstanduseittofindtheradius. ∑F y =0 N −W = 0 fs = µ s N ay=0(Thereisnoaccelera+oninver+caldirec+on). N = W = mg N = 20 ×10 −3 kg ⋅ 9.8 m/s2 = 0.196 N 19 Exercise:Slippingcoin A20-gcoinisplacedonarecordthatisrota+ngat33.3rpm.Ifμs=0.1,howfar fromthecenteroftherecordcanthecoinbeplacedwithouthavingitslipoff? ar y fs = µ s N = 0.1⋅ 0.196 N = 0.0196 N N fs = mω 2 r fs fs r= mω 2 Youaregivenfrequencyf=33.3rpm→convertittoHz→andthen calculateangularspeedω=2πf x ω = 2πf = 33.3 w r= rev ⎛ 2π rad ⎞⎛ 1 min ⎞ ⎜ ⎟⎜ ⎟ = 3.5 rad/s min ⎝ 1 rev ⎠⎝ 60 sec ⎠ 0.0196 kg m/s2 −3 20 ×10 kg ⋅ (3.50 rad/s) 2 = 0.08 m Notethatthesameprincipleapplieswhenyoudriveacarinacircularpathalonganunbankedroad. 20 Bankedcurves Topreventcarsfromgoingintoskid èroadsarebanked(+ltedataslightangle) y N ApplyNewton’s2ndLaw: ∑F ar y x ∑F x =0 N cosθ = mg = max v2 N sin θ = mar = m r v2 mg tan θ = m r v2 tan θ = gr v 2 = gr tan θ Fric+onforcenegligiblecomparedhorizontalcomponentofthenormalforce. Physics140,Prof.M.Nikolic 21 Circularorbits ConsiderasatelliteinacircularorbitabouttheEarth. Butthereisnostringjoiningthesatellitetothe Earthnoristhereanythingtohavefric+onagainst Gravity Whatforceisholdingthesatelliteinacircularorbit? Gms M e It’sgravity: Fg = r2 ApplyNewton’s2ndlaw Fg = ms ar Physics140,Prof.M.Nikolic Gms M e v2 = ms 2 r r GM e v= r Thespeedofasatelliteina circularorbitdoesnotdepend onmassofthesatellite 22 Kepler’s1stLawofPlanetaryMo+on Thelawoforbits:Allplanetsmoveinellip+calorbits,withtheSunatonefocus Luckilyforus,wewillonlyworkwith perfectcirclesinthisclass. Physics140,Prof.M.Nikolic 23 Kepler’s2ndLawofPlanetaryMo+on Thelawofareas:AlinethatconnectsaplanettotheSunsweepsout equalareasintheplaneoftheplanet’sorbitinequal+mes. EartharoundSum -closestdistanceè v=30.3km/s -furthestdistanceè v=29.3km/s PlanetsaremovingfasterwhentheyareclosertotheSun GSCI101,Prof.M.Nikolic 24 Kepler’s3rdLawofPlanetaryMo+on Thelawofperiods:Thesquareoftheperiodofanyplanetis propor+onaltothecubeofthesemimajoraxisofitsorbit. OrinEnglish:Outerplanetshavefurthertogoand movemoreslowlyintheirorbitsaroundSun 2π 3 / 2 T= r GM GSCI101,Prof.M.Nikolic or 2 4 π T2 = r3 GM 25 Exercise:Kepler’slaws TheHubbleSpaceTelescopeorbitsEarth613kmaboveEarth’ssurface.Whatisthe periodofthetelescope’sorbit? Whatisgiven: MassoftheEarth:ME=5.98x1024kg RadiusoftheEarth:RE=6371km h=613km G=6.67x10-11Nm2/kg 2 4 π T2 = r3 GM E DistancerismeasuredfromthecenteroftheEarthtothetelescope:r=RE+handhastobe convertedtometers: →r=6371km+613km=6984km=6984x103m 4π 2 T= ( Re + h)3 GM E 4 ⋅ 3.14 2 ⋅ (6984 ×10 3 m)3 7 T= = 4 . 2 × 10 s −11 2 2 24 6.67 ×10 Nm /kg ⋅ 5.98 ×10 kg Physics140,Prof.M.Nikolic 26 Nonuniformcircularmo+on Whathappenswhenthespeedincircularmo+onisnotconstant? Then,whenΔt→0,Δvdoesnotpointtowardsthecenterofthecircle →Thereisnowanothercomponenttoaccelera+ontangenttothe pathofthecircle→tangen-alaccelera-onat a Thenetaccelera+onis: at 2 a = ar + at 2 atchangesthemagnitudeofv archangesthedirec-onofv ar v AndNewton’s2ndlaws+llapplies: ∑ F = ma ∑ F = ma r Physics140,Prof.M.Nikolic r t t 27 Exercise:Childonaswing A35-kgchildswingsonaropewithalengthof6.5mthatishangingfromatree.At thebotomoftheswing,thechildismovingataspeedof4.2m/s.Whatisthe tensionintherope? DrawanFBDforthechildatthebotomoftheswing. y T ar at →radialaccelera+onpointstowardsthecenterofthecircle (alongyaxis) →tangen+alaccelera+onistangenttothemo+on ApplyNewton’s2ndLaw: ∑F x r w Xdirec+on: Ydirec+on: Tx=0 Ty=T Wx=0 Wy=-W=-mg Physics140,Prof.M.Nikolic = mar v2 T − W = mω r = m r v2 T = mg + m r 2 (4.2m / s) 2 T = 35kg ⋅ 9.8m / s + 35kg = 438 N 6.5m 2 28 Tangen+alandangularaccelera+on Duringnonuniformcircularmo+onangularvelocity(ω)ischanging →thereshouldbeanangularaccelera-on(α) Averageangularaccelera-on Δω ω f − ω i α av = = Δt t f − ti Instantaneousangularaccelera-on Δω α = lim Δt →0 Δt Justlikelinearmo+onbutwithv→ω Unitsofangularaccelera+on: radianspersecond[rad/s2] Physics140,Prof.M.Nikolic 29 Angularmo+onvs.linearmo+on Tangen+alaccelera+on: Δv Δω at = =r = rα Δt Δt Linearmo-on Δvx = v fx − vix = a x Δt 1 Δx = x f − xi = vix Δt + ax Δt 2 2 v 2fx − vix2 = 2ax Δx Δx = 1 (vix + v fx )Δt 2 Physics140,Prof.M.Nikolic v = rω Angularmo-on Δω = ω f − ω i = αΔt 1 Δθ = θ f − θ i = ω i Δt + αΔt 2 2 ω 2f − ωi2 = 2αΔθ Δθ = 1 (ωi + ω f )Δt 2 30 Exercise:StoppingtheEarth SupposetheEarthstartedtoundergoanangularaccelera+onof-1.2×10-10rad/s2in theoppositedirec+ontoitscurrentrota+on. a)Azerhowlongwoulditcometoastop(inrota+on)? Whatisgiven: α=-1.2×10-10rad/s2–TheEarthisslowingdown→accelera+onisnega+ve ωf=0–theEarthstopped Δω = ω f − ω i = αΔt Wedonotknowini+al(current)angularvelocitybut,hopefully, everyoneknowstheperiodofEarth’srota+on: →T=1day=24h=86400s ωi = ω f − ωi Δt = α Physics140,Prof.M.Nikolic 2π T ωi = 2 ⋅ 3.14 rad = 7.28 ×10−5 86400 s s 0 − ωi 0 − 7.28 ×10 −5 rad / s Δt = = = 6.1×10 5 s ≈ 7 days −10 2 α −1.2 ×10 rad / s 31 Exercise:StoppingtheEarth SupposetheEarthstartedtoundergoanangularaccelera+onof1.2×10-10rad/s2in theoppositedirec+ontoitscurrentrota+on. b)Howmanyrevolu+onswouldtheEarthmakebeforeitstopped? Whatisgivenoralreadyfound: α=-1.2×10-10rad/s2–TheEarthisslowingdown→accelera+onisnega+ve ωf=0–theEarthstopped ωi=7.28×10-5rad/s Δt=6.1x105s Revolu+onsareexpressedinangleθ→1revolu+on=2πradians Δθ = Δθ = 1 (ωi + ω f )Δt 2 1 7.28 ×10 −5 rad/s + 0 ⋅ 6.1 ×10 5 s = 22.2 rad 2 ( ) ⎛1 revolution ⎞ ⎟ = 3.54 revolutions Δθ = 22.2 rad × ⎜ ⎝ 2π rad ⎠ Physics140,Prof.M.Nikolic 32 Apparentweightandar+ficialgravity Trytorecallproblemswiththeapparentandtrueweight(personinanelevator) →personfeelsweightlesswhena=g Similarsitua+onforastronautsinspace →Wecansimulategravitybyrota+ngthespacesta+on →astronauthitsthewalls →normalforcemakeshimrotatewiththespacesta+onwith radialaccelera+onar=mω2r Foraspacesta+onwewant: ar = g N Physics140,Prof.M.Nikolic 2 v Fr = mar = mω 2 r = m = mg r 33 Exercise:Ar+ficialgravity Ifawashingmachine’sdrumwithradiusof25cmproducesar+ficialgravityonthe clothesof16g,whatisthefrequencyinrevolu+onsperminuteofthedrum? Whatisgiven: r=25cm=0.25m ar=16g=16x9.8m/s2 ar = ω 2 r 16 g ω= r 16 ⋅ 9.8m / s 2 ω= = 25 rad/s 0.25m Tofindthefrequency: f = ω 2π Physics140,Prof.M.Nikolic f = ⎛ 60s ⎞ 25 rad/s ⎟ = 240 rpm = 4 Hz = 4 rev/s × ⎜ ⎝ 6.28 1min ⎠ 34 Hollywoodmoviesbusted StarTrek:USSEnterprise–notpossible 2001:AspaceOdyssey–possible StarWars:Millenniumfalcon–notpossible Physics140,Prof.M.Nikolic 35 WeightlessintheISS TheInterna+onalSpaceSta+onorbitstheEarthevery91minutesata distanceof353kmabovethesurfaceoftheEarth Interna+onalspacesta+ondoesnotrotateènoapparentgravity Physics140,Prof.M.Nikolic 36