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IntroductoryPhysics PHYS101 Dr RichardH.CyburtOfficeHours TRF9:30-11:00am AssistantProfessorofPhysics F12:30-2:00pm Myoffice:402cintheScienceBuilding Myphone:(304)384-6006 Meetingsmayalsobearrangedatothertimes, byappointment Myemail:[email protected] Inpersonoremailisthebestwaytogetahold Checkmyscheduleonmyofficedoor. ofme. PHYS101 PHYS101:IntroductoryPhysics 400 Lecture:8:00-9:15am,TRScienceBuilding Lab1:3:00-4:50pm,FScienceBuilding304 Lab2:1:30-3:20pm,MScienceBuilding304 Lab3:3:30-5:20pm,MScienceBuilding304 Lab20:6:00-7:50pm,MScienceBuilding304 PHYS101 MasteringPhysicsOnline GotoHYPERLINK"http://www.masteringphysics.com."www.masteringphysics.com. ◦ UnderRegisterNow,selectStudent. ◦ Confirmyouhavetheinformationneeded,thenselectOK!Registernow. RCYBURTPHYS101),andchooseContinue. ◦ Enteryourinstructor’sCourseID( ◦ EnteryourexistingPearsonaccountusername andpassword andselectSignin. ◦ YouhaveanaccountifyouhaveeverusedaPearsonMyLab &Masteringproduct,suchasMyMathLab,MyITLab,MySpanishLab,or MasteringChemistry. ◦ Ifyoudon’thaveanaccount,select Create andcompletetherequiredfields. ◦ Selectanaccessoption. ◦ Entertheaccesscodethatcamewithyourtextbookorwaspurchasedseparatelyfromthebookstore. PHYS101 Midterm1 Therewillbeasignoutsheetinmyoffice ◦ Youmustsigntogetyourexam Therewillbeabonusassignment,basedonyourexam ◦ ◦ ◦ ◦ Willearnyouextrapointsonyourexam Itwillbeonlineasahomework Youmustdobetteronthisassignment,thanyourtesttogetbonuspoints Bonus=30%x(Homework– Midterm) BonusHomeworkisOnline,dueSep26,12:59pm(justbeforelabsstartfortheday) PHYS101 Midterm2 Thursday,September29 CoveringChapters5-7(NolongerChapter8). ReviewSession,Wednesday,September28,7:00-9:00pmS300 Allowed1half-sheetofformulae/notes PHYS101 IntroductoryPhysics PHYS101 DouglasAdams Hitchhiker’sGuidetotheGalaxy PHYS101 You’realreadyknowphysics! Youjustdon’tnecessarilyknowtheterminologyand languageweuse!!! PhysicsofNASCAR PhysicsofAngerBirds PHYS101 CourtesyofmywifeandtheOED… Yourwordfortodayis: mathlete,n. mathlete, n. [‘ Apersonwhotakespartinamathematicscompetition,esp.oneorganizedforschoolchildren.Alsoinextendeduse.’] Pronunciation: Brit. /ˈmaθliːt/, U.S. /ˈmæθ(ə)ˌlit/ Origin:Formed withinEnglish,byblending.Etymons: mathematics n., math n.3, athlete n. Etymology:Blend of mathematics n. or math n.3 and athlete n. orig.andchiefly U.S. Apersonwhotakespartinamathematicscompetition,esp.oneorganizedforschoolchildren.Alsoinextended use.1933 TimesRecorder(Zanesville,Ohio) 5June 1/2 Cadel‘mathletes’tookfirst,fourthandfifthplaces[inarecentmathematicscontest withHarvard]. 1958 Waterloo(Iowa)DailyCourier 25Mar. 14/2 AtthelastmeetingoftheF.S.S.T.(FlyingSaucerSpottersofTerra)thefollowingGlossary ofSpaceAgetermswascompiledbythemembers... Physicist,mathlete. 1972 BucksCountyCourierTimes(Levittown,Pa.) 29Jan. 4/2 Moredittosheetsaredistributed,andthemathleteswaitforthesignalto turnthemandbeginontheproblem. 1995 Guardian 4May (OnlineSuppl.)1/1 The‘mathlete’finalistshadsecondstoanswerquestionsliketheonethatactuallyclinchedthe goldmedalforReifsnyder. 2000 Sacramento(Calif.)Bee(Nexis) 5Aug. b1 Herstudentsrepeatedlywonfirstplaceinmonthly‘mathletes’competitioninSacramento. PHYS101 Wife’sRequest:MindShudderMagazine Weareanonlinemagazinethatpublishesweirdanddarkfiction. Wearesearchingfornewauthorsofshortandflashfiction.Weencourageyou tosubmityourworkforpossiblepublication. Wewantthebestofyourflashfictionandshortstories.Youcanwriteabout anythingthatyouconsiderweirdordarkfiction. Weareinourinfancyandwillacceptsubmissionsallyearround.Onceourfirst issueispublished,itwillbemadeavailabletothepublic. Themagazineiscurrentlyfreetosubmit. www.mindshuddermagazine.com PHYS101 Inclass!! PHYS101 Thislecturewillhelpyouunderstand: TorqueReview Centerofgravity RotationalDynamics&MomentofInertia UsingNewton’s2nd LawforRotation RollingMotion PHYS101 Torque Torque(τ)istherotationalequivalentofforce. Torqueunitsarenewton-meters,abbreviatedN× m. ©2015PearsonEducation,Inc. Torque Theradialline istheline startingatthepivotand extendingthroughthepoint whereforceisapplied. Theangle ϕ is measuredfromthe radiallinetothe directionoftheforce. ©2015PearsonEducation,Inc. Torque Theradialline istheline startingatthepivotand extendingthroughthe pointwhereforceis applied. Theangle ϕ ismeasured fromtheradiallinetothe directionoftheforce. Torqueisdependenton theperpendicularcomponent oftheforcebeingapplied. ©2015PearsonEducation,Inc. Torque Analternatewaytocalculate torqueisintermsofthe momentarm. Themomentarm (orlever arm)istheperpendicular distancefromthelineof action tothepivot. Thelineofaction istheline thatisinthedirectionofthe forceandpassesthroughthe pointatwhichtheforceacts. ©2015PearsonEducation,Inc. Torque Theequivalentexpressionfor torqueis Forbothmethodsfor calculatingtorque,the resultingexpressionis thesame: ©2015PearsonEducation,Inc. Section7.4Gravitational Torque andtheCenterofGravity ©2015PearsonEducation,Inc. GravitationalTorqueandtheCenterof Gravity Gravitypullsdownwardon everyparticle thatmakesup anobject(likethegymnast). Eachparticleexperiencesa torqueduetotheforceof gravity. ©2015PearsonEducation,Inc. GravitationalTorqueandtheCenterof Gravity Thegravitationaltorquecan becalculatedbyassuming thatthenetforceofgravity (theobject’sweight)actsas asinglepoint. Thatsinglepointiscalledthe centerofgravity. ©2015PearsonEducation,Inc. Example7.12Thetorqueonaflagpole A3.2kgflagpoleextends fromawallatanangleof 25° fromthehorizontal. Itscenterofgravityis 1.6mfromthepoint wherethepoleisattached tothewall.Whatisthegravitationaltorqueontheflagpoleaboutthepointof attachment? PREPARE FIGURE7.26showsthesituation.Forthepurposeofcalculatingtorque, wecanconsidertheentireweightofthepoleasactingatthecenterofgravity. Becausethemomentarmr⊥ issimpletovisualizehere,we’lluseEquation7.11 forthetorque. ©2015PearsonEducation,Inc. Example7.12Thetorqueonaflagpole (cont.) SOLVE FromFigure7.26, weseethatthemomentarm isr⊥ =(1.6m)cos25° = 1.45m.Thusthegravitational torqueontheflagpole,about thepointwhereitattachesto thewall,is Weinsertedtheminussignbecausethetorquetriestorotatethepoleinaclockwise direction. ASSESS Ifthepolewereattachedtothewallbyahinge,thegravitationaltorquewould causethepoletofall.However,theactualrigidconnectionprovidesacounteracting (positive)torquetothepolethatpreventsthis.Thenettorqueiszero. ©2015PearsonEducation,Inc. GravitationalTorqueandtheCenterof Gravity Anobjectthatisfreeto rotateaboutapivotwill cometorestwiththe centerofgravity belowthepivotpoint. Ifyouholdarulerby oneendandallowitto rotate,itwillstoprotating whenthecenterofgravity isdirectlyaboveorbelowthepivotpoint.Thereisnotorqueactingatthese positions. ©2015PearsonEducation,Inc. QuickCheck7.12 Whichpointcouldbethecenterofgravityofthis L-shapedpiece? D. A. B. C. ©2015PearsonEducation,Inc. QuickCheck7.12 Whichpointcouldbethecenterofgravityofthis L-shapedpiece? D. A. B. C. ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity Thetorqueduetogravitywhenthepivotisat thecenterofgravity iszero. Wecanusethistofindanexpressionforthepositionofthecenter ofgravity. ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity [InsertFigure7.29] Forthedumbbellto balance,thepivotmust beatthecenterof gravity. Wecalculatethetorque oneithersideofthe pivot,whichislocated atthepositionxcg. ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity [InsertFigure7.29(repeated)] Thetorqueduetothe weightontheleftside ofthepivotis Thetorqueduetothe weightontherightside ofthepivotis ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity Thetotaltorqueis Thelocationofthecenterofgravityis ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity Becausethecenterof gravitydependson distanceandmassfrom thepivotpoint,objects withlargemassescount moreheavily. Thecenterofgravity tendstolieclosertothe heavierobjectsorparticles thatmakeuptheobject. ©2015PearsonEducation,Inc. CalculatingthePositionoftheCenterof Gravity Text:p.204 ©2015PearsonEducation,Inc. Section7.5Rotational Dynamics andMomentofInertia ©2015PearsonEducation,Inc. RotationalDynamicsandMomentof Inertia Atorquecausesanangular acceleration. Thetangentialandangular accelerationsare ©2015PearsonEducation,Inc. RotationalDynamicsandMomentof Inertia Wecomparewithtorque: Wefindtherelationship withangularacceleration: ©2015PearsonEducation,Inc. Newton’sSecondLawforRotational Motion Forarigidbody rotatingabouta fixedaxis,wecan thinkoftheobject asconsistingof multipleparticles. Wecancalculate thetorqueoneach particle. Becausetheobjectrotates together,eachparticlehasthe same angularacceleration. ©2015PearsonEducation,Inc. Newton’sSecondLawforRotational Motion Thetorqueforeach “particle”is Thenet torqueis ©2015PearsonEducation,Inc. Newton’sSecondLawforRotational Motion ThequantityΣmr2 in Equation7.20,which istheproportionality constantbetweenangular accelerationandnettorque, iscalledtheobject’s momentofinertiaI: Theunitsofmomentofinertiaarekg× m2. Themomentofinertiadependsontheaxisofrotation. ©2015PearsonEducation,Inc. Newton’sSecondLawforRotational Motion Anettorqueisthecauseofangularacceleration. ©2015PearsonEducation,Inc. InterpretingtheMomentofInertia Themomentofinertia istherotational equivalentofmass. Anobject’smomentof inertiadependsnotonly ontheobject’smassbut alsoonhowthemassis distributed aroundthe rotationaxis. ©2015PearsonEducation,Inc. InterpretingtheMomentofInertia Themomentofinertiais therotationalequivalent ofmass. Itismoredifficulttospin themerry-go-roundwhen peoplesitfarfromthe centerbecauseithasa higherinertiathanwhen peoplesitclosetothe center. ©2015PearsonEducation,Inc. InterpretingtheMomentofInertia Text:p.208 ©2015PearsonEducation,Inc. Example7.15Calculatingthemomentof inertia Yourfriendiscreatingan abstractsculpturethat consistsofthreesmall, heavyspheresattached byverylightweight 10-cm-longrodsasshown inFIGURE7.36.The sphereshavemasses m1 = 1.0kg,m2 = 1.5kg,andm3 = 1.0kg.Whatistheobject’smomentofinertiaifitis rotatedaboutaxisA?AboutaxisB? PREPARE We’lluseEquation7.21forthemomentofinertia: I = m1r12 + m2r22 + m3r32 Inthisexpression,r1,r2,andr3 arethedistancesofeachparticlefromtheaxisof rotation,sotheydependontheaxischosen. ©2015PearsonEducation,Inc. Example7.15Calculatingthemomentof inertia(cont.) Particle1liesonboth axes,sor1 = 0cmin bothcases. Particle2 lies10cm(0.10m)from bothaxes. Particle3is 10cmfromaxisAbut fartherfromaxisB. We canfindr3 foraxisBbyusingthe Pythagoreantheorem,whichgivesr3 = 14.1cm. Thesedistancesareindicatedinthefigure. ©2015PearsonEducation,Inc. Example7.15Calculatingthemomentof inertia(cont.) [InsertFigure7.36(repeated)] SOLVE Foreachaxis, wecanprepareatable ofthevaluesofr,m, andmr 2 foreach particle,thenaddthe valuesofmr 2.For axisAwehave ©2015PearsonEducation,Inc. Example7.15Calculatingthemomentof inertia(cont.) ForaxisBwehave ©2015PearsonEducation,Inc. Example7.15Calculatingthemomentof inertia(cont.) ASSESS We’vealready notedthatthemomentof inertiaofanobjectis higherwhenitsmassis distributedfartherfrom theaxisofrotation. Here,m3 isfartherfrom axisBthanfromaxisA,leadingtoahighermomentofinertiaaboutthataxis. ©2015PearsonEducation,Inc. TheMomentsofInertiaofCommon Shapes ©2015PearsonEducation,Inc. Section7.6Using Newton’sSecondLaw forRotation ©2015PearsonEducation,Inc. UsingNewton’sSecondLawforRotation Text:p.211 ©2015PearsonEducation,Inc. Example7.16Angularaccelerationofa fallingpole Inthecabertoss,acontestof strengthandskillthatispart ofScottishgames,contestants tossaheavyuniformpole, landingitonitsend.A 5.9-m-tallpolewithamass of79kghasjustlandedon itsend.Itistippedby25° fromtheverticalandis startingtorotateaboutthe endthattouchestheground. Estimatetheangular acceleration. ©2015PearsonEducation,Inc. Example7.16Angularaccelerationofa fallingpole(cont.) PREPARE Thesituationisshown inFIGURE7.37,wherewe defineoursymbolsandlistthe knowninformation.Twoforces areactingonthepole:thepole’s weight whichactsatthe centerofgravity,andtheforce ofthegroundonthepole(notshown).Thissecondforceexertsnotorque becauseitactsattheaxisofrotation.Thetorqueonthepoleisthusdue onlytogravity.Fromthefigureweseethatthistorquetendstorotatethe poleinacounterclockwisedirection,sothetorqueispositive. ©2015PearsonEducation,Inc. Example7.16Angularaccelerationofa fallingpole(cont.) SOLVE We’llmodelthepoleasa uniformthinrodrotatingabout oneend.Itscenterofgravity isatitscenter,adistanceL/2 fromtheaxis.Youcansee fromthefigurethatthe perpendicularcomponentof isw⊥ = w sinθ.Thusthetorqueduetogravityis ©2015PearsonEducation,Inc. Example7.16Angularaccelerationofa fallingpole(cont.) FromTable7.1,themoment ofinertiaofathinrod rotatedaboutitsendis Thus,from Newton’ssecondlawfor rotationalmotion,the angularaccelerationis ©2015PearsonEducation,Inc. Example7.16Angularaccelerationofa fallingpole(cont.) ASSESS Thefinalresultforthe angularaccelerationdidnot dependonthemass,aswe mightexpectgiventheanalogy withfree-fallproblems.And thefinalvaluefortheangular accelerationisquitemodest. Thisisreasonable:Youcanseethattheangularaccelerationis inverselyproportionaltothelengthofthepole,andit’salong pole.Themodestvalueofangularaccelerationisfortunate—the caberisprettyheavy,andfolksneedsometimetogetoutofthe waywhenittopples! ©2015PearsonEducation,Inc. Example7.18Startinganairplaneengine Theengineinasmallair-planeis specifiedtohaveatorqueof 500N× m.Thisenginedrivesa 2.0-m-long,40kgsingle-blade propeller.Onstart-up,howlong doesittakethepropellertoreach 2000rpm? ©2015PearsonEducation,Inc. Example7.18Startinganairplaneengine (cont.) PREPARE Thepropellercanbe modeledasarodthatrotates aboutitscenter.Theengine exertsatorqueonthepropeller. FIGURE7.38showsthepropeller andtherotationaxis. ©2015PearsonEducation,Inc. Example7.18Startinganairplaneengine (cont.) SOLVE Themomentofinertia ofarodrotatingaboutits centerisfoundinTable7.1: The500N× mtorqueofthe enginecausesanangular accelerationof ©2015PearsonEducation,Inc. Example7.18Startinganairplaneengine (cont.) Thetimeneededtoreach ωf = 2000rpm= 33.3rev/s= 209rad/sis ©2015PearsonEducation,Inc. Example7.18Startinganairplaneengine (cont.) ASSESS We’veassumedaconstant angularacceleration,whichis reasonableforthefirstfewseconds whilethepropellerisstillturning slowly.Eventually,airresistance andfrictionwillcauseopposing torquesandtheangularacceleration willdecrease.Atfullspeed,the negativetorqueduetoairresistance andfrictioncancelsthetorqueoftheengine.Then andthepropellerturnsatconstantangularvelocitywithnoangularacceleration. ©2015PearsonEducation,Inc. ConstraintsDuetoRopesandPulleys Ifthepulleyturnswithouttherope slippingonit thentherope’sspeedmust exactlymatchthespeedoftherimofthepulley. Theattachedobjectmusthavethesamespeed andaccelerationastherope. ©2015PearsonEducation,Inc. Section7.7Rolling Motion ©2015PearsonEducation,Inc. RollingMotion Rollingisacombinationmotion inwhichanobjectrotatesaboutanaxisthatis movingalongastraight-linetrajectory. ©2015PearsonEducation,Inc. RollingMotion Thefigureaboveshowsexactlyonerevolutionforawheelorsphere thatrollsforwardwithoutslipping. Theoverallpositionismeasuredattheobject’scenter. ©2015PearsonEducation,Inc. RollingMotion Inonerevolution,thecentermovesforwardbyexactlyone circumference(Δx=2πR). ©2015PearsonEducation,Inc. RollingMotion Since2π/T istheangularvelocity,wefind Thisistherollingconstraint,thebasiclinkbetweentranslationandrotationfor objectsthatrollwithoutslipping. ©2015PearsonEducation,Inc. RollingMotion Thepointatthebottomofthewheelhasatranslationalvelocityanda rotationalvelocityinoppositedirections,whichcanceleachother. Thepointonthebottomofarollingobjectisinstantaneouslyatrest. Thisistheideabehind“rollingwithoutslipping.” ©2015PearsonEducation,Inc. Example7.20Rotatingyourtires Thediameterofyourtiresis0.60m.Youtakea60miletripataspeedof45mph. a. Duringthistrip,whatwasyourtires’angularspeed? b. Howmanytimesdidtheyrevolve? ©2015PearsonEducation,Inc. Example7.20Rotatingyourtires(cont.) PREPARE Theangularspeedisrelatedtothespeedofawheel’scenterbyEquation 7.25: ν = ω R.Becausethecenterofthewheelturnsonanaxlefixedtothecar, thespeedv ofthewheel’scenteristhesameasthatofthecar.Weprepareby convertingthecar’sspeedtoSIunits: Onceweknowtheangularspeed,wecanfindthenumber oftimesthetiresturnedfromtherotational-kinematicequationΔθ = ω Δt.We’ll needtofindthetimetraveled Δt fromν = Δx/Δt. ©2015PearsonEducation,Inc. Example7.20Rotatingyourtires(cont.) SOLVE a.FromEquation7.25wehave b.Thetimeofthetripis ©2015PearsonEducation,Inc. Example7.20Rotatingyourtires(cont.) Thusthetotalanglethroughwhichthetiresturnis Becauseeachturnofthewheelis2π rad,thenumberofturnsis ©2015PearsonEducation,Inc. Example7.20Rotatingyourtires(cont.) ASSESS Youprobablyknowfromseeingtiresonpassingcarsthatatirerotates severaltimesasecondat45mph.Becausethereare3600sinanhour,andyour 60miletripat45mphisgoingtotakeoveranhour—say,≈ 5000s—youwould expectthetiretomakemanythousandsofrevolutions.So51,000turnsseems tobeareasonableanswer.Youcanseethatyourtiresrotateroughlyathousand timespermile.Duringthelifetimeofatire,about50,000miles,itwillrotate about50milliontimes! ©2015PearsonEducation,Inc.