Physics 110H 2013-2014 Journal FINALoap.nmsu.edu/JiTT_NMSU_workshop/CHAT_06_rsrc/110H_2013_14_Jo… · physics courses, which includes Newtonian mechanics and conservation of energy

Physics110HJournalGeneralPhysicsI‐Fall2013/Spring2014

USAFADepartmentofPhysics,CorePhysicsPublication

Name Instructor

Section

Physics110ConstantsandEquationsSheet

12

12

2 2

∆ /

∙

12

∙ 12

∆ ∙ 12

cos

AccelerationduetoGravityatEarth'sSurface 9.81 m s 1in 2.54cm

UniversalGravitationConstant 6.67 10 N m kg 1mi 1609m

SpeedofLightinVacuum 3.00 10 m s 1ft 0.3048 mMassofEarth 5.97 10 kg 1mi h⁄ 0.447 m/s

RadiusofEarth 6.37 10 m 1lb 4.448N 0.454kg

i

Physics 110H Journal

Genera l Phys ics I - Fa l l 2013/Spr ing 2014

DepartmentofPhysics

UnitedStatesAirForceAcademy

ii

PagereservedforPublisher’sCopyrightinformation

Physics110HJournal‐2013‐2014

iii

Contents

Physics 110 Constants and Equations Sheet ................................................. ii

Course Description and Policies .................................................................... v

IDEA Problem‐Solving Strategy ...................................................................xiv

Learning Objectives ..................................................................................... xv

Lesson 1 ........................................................................................................ 1

Lesson 2 ........................................................................................................ 9

Lesson 3 ...................................................................................................... 17

Lesson 4 ...................................................................................................... 23

Lesson 5 ...................................................................................................... 31

Lesson 6 ...................................................................................................... 39

Lesson 7 ...................................................................................................... 45

Lesson 8 ...................................................................................................... 53

Lesson 9 ...................................................................................................... 61

Lesson 10 .................................................................................................... 69

Lesson 11 .................................................................................................... 77

Lesson 12 .................................................................................................... 85

Lesson 13 .................................................................................................... 93

Lesson 14 .................................................................................................... 99

Lesson 15 .................................................................................................. 107

Lesson 16 .................................................................................................. 115

Lesson 17 .................................................................................................. 121

Lesson 18 .................................................................................................. 129

Lesson 19 .................................................................................................. 137

Lesson 20 .................................................................................................. 145

Lesson 21 .................................................................................................. 153

Lesson 22 .................................................................................................. 161

Lesson 23 .................................................................................................. 167

Lesson 24 .................................................................................................. 175

Lesson 25 .................................................................................................. 183

Lesson 26 .................................................................................................. 189

Lesson 27 .................................................................................................. 197

Lesson 28 .................................................................................................. 205

Lesson 29 .................................................................................................. 211

Lesson 30 .................................................................................................. 219

Lesson 31 .................................................................................................. 231

Lesson 32 .................................................................................................. 239

Lesson 33 .................................................................................................. 247

Lesson 34 .................................................................................................. 255

Lesson 35 .................................................................................................. 263

Lesson 36 .................................................................................................. 271

Lesson 37 .................................................................................................. 279

Lesson 38 .................................................................................................. 285

iv

Lesson 39 ................................................................................................... 291

Lesson 40 ................................................................................................... 299

Block 4 Review........................................................................................... 307

Appendix A: Lab Report Template ............................................................. xix

Appendix B: Significant Figures, Uncertainty and Error Propagation ........ xxi

Appendix C: Mathematics Reference .......................................................xxvi

Appendix D: Equation Dictionary ............................................................. xxx

Appendix E: Rotational Inertias and Astrophysical Data ............................. xl

Appendix F: Units and Conversions ............................................................ xli

Appendix F: Physical Constants ................................................................. xlii

Physics 215 Constants and Equations Sheet ............................................. xliii

Physics 110H Course Syllabus.................................................................... xliv

Physics110HJournal‐2013‐2014 CourseDescriptionandPolicies

v

CourseDescriptionandPolicies

Overview

TheUSAirForceAcademyoffersabroadgeneralphysicscurriculumwithfourspecializedphysicsmajoroptions:Astronomy,LaserPhysics/Optics,SpacePhysics,andAppliedPhysicstopicssuchasnuclearphysics.Eachphysicsmajoroptionrequires42credithoursofphysicsandmathematicscoursesinadditiontocoreacademicrequirements,includingafaculty‐directedcapstonephysicsresearchproject.AftergraduationphysicsmajorssucceedinawidevarietyofoperationalAirForceassignmentsorcompleteanadvancedacademicdegreeatgraduateschool.

TheUSAFADepartmentofPhysicsofferstwocorecourses,eachwithanhonorsoption.PHYSICS110/110H(GeneralPhysicsI)isthefirstinatwo‐partseriesofintroductorycalculus‐basedphysicscourses,whichincludesNewtonianmechanicsandconservationofenergyandmomentum,andisnormallytakenduringthefourth‐classyear.PHYSICS215/215H(GeneralPhysicsII)isthesecondintheseriesofintroductorycalculus‐basedphysicscourseswhichemphasizeselectromagnetismandcircuits,andisnormallytakenduringthethird‐classyear.

HonorsphysicscoursesaredesignedtobetteraddresstheneedsoftechnicalmajorsattheUSAirForceAcademyandmeettheneedsofanincreasinglytechnicalAirForce.Cadetsdemonstratingaptitudeincalculusorhavingpreviouslytakenintroductoryphysicscoursesmaybeplacedinhonorsphysics.Honorsphysicsincludesenhancedcoverageoftheconceptscoveredintheregularcourse,withmoreintegrateduseofcalculus,introductiontodifferentialequationsandrigorousdataanalysistechniques.

CorePhysicsCourseDescriptions

Physics110,GeneralPhysicsI,isacalculus‐basedintroductiontoclassicalphysics,withemphasisoncontemporaryapplications,inwhichyouwilllearntheconceptsandproblem‐solvingskillsrequiredtounderstandandanalyzethemotionofobjects.ThefirsthalfofthecourseisasolidfoundationinkinematicsandNewton’slawsofmotion.Youwillthenbeintroducedtoseveralconservationprinciples,whichareelegantwaysofvisualizingandunderstandingthemotionofobjects.Theseincludetheconservationofenergy,momentumandangularmomentum.Alongtheway,youwillbeintroducedtoafewtopicsthatareimportanttoscientistsandengineers,includingorbitalmotion,rotationalmotionandoscillations.Labsandsimulationshighlightkeyphysicsconcepts.

Physics215,GeneralPhysicsII,isanintroductorycalculus‐basedphysicscoursewithanemphasisoncontemporaryapplications.Thecoursebeginswithafoundationinthebasicpropertiesofelectricchargeandworksuptodealingwiththesophisticatedconceptoftheelectricfield.Then,simplecircuitsareanalyzed,relatingtheprinciplesofpotentialenergyandelectricpotentialtotheelectricfield.Next,magneticfieldsandelectromagneticinductionarestudied,culminatinginacompletedescriptionofelectromagneticfields.Afterthat,lightwaves,thebendingoflightandthe

CourseDescriptionandPolicies Physics110HJournal‐2013‐2014

vi

interferencecausedbythewavenatureoflightarestudied.Finally,modernphysicsisintroducedbystudyingquantizationandquantumuncertainty.Thiscourseutilizesvectorsandcalculusinproblemsolvingandincludesin‐classlaboratoriestohighlightkeyconcepts.

HonorsCorePhysicsCourses

Asa“techie”majorinthePhysicsHonorsCoursesequence,youcanexpectanumberofbenefitscomparedtotakingthestandardintroductorycourse.Perhapsthemostsignificantbenefitislearningphysicsmoreefficientlyandmoreenjoyablyamidstudentsofsimilaracademicabilities.Youwillalsoseeenhancedcoverageoftopicsthatareimportantforscientistsandengineers,including

amoreintegrateduseofcalculusthroughoutthecourse anintroductiontotheuseofdifferentialequationsinsimpleharmonicmotion enhanceddataanalysistechniques somewhatmoreemphasisongraphicalandnumericaltechniques

Theemphasisonthesetopicswillmakeyourphysicsexperiencecomparabletowhatyourpeerswouldreceiveatacivilianuniversitywhentakingaphysicscourseforscientistsandengineers.

Toallaypossibleconcernsaboutyourgrade,DFPwillensurethatyouarenotpenalizedfortakingPhysicsHonorsinplaceofstandardPhysicsCourse.ThegradedreviewsforPhysics“regular”andPhysicsHonorswillincludealargepercentageofcommonquestionstoallowagoodstatisticalcomparisonofthetwocourses,sothatyourfinalgradewillnotdependonwhichversionofthecourseyoutake.Wehavealsobalancedtheoverallworkloadsothatstudentsineithercourse,onaverage,havethesamenumberofhomeworkproblems,journalquestions,etc.,tocomplete.

WhatarethesimilaritiesanddifferencesbetweenPhysics“regular”andPhysicsHonors?

Bothcourseswillfollowthesamebasicsyllabusandusethesamecoursepolicies.Withfewexceptions,studentsinbothcourseswillstudythesametextbookexamplesandanswerthesamejournalandpreflightquestions.Some(about35%)ofthehomeworkproblemsaredifferenttomakebetteruseofcalculusandothermathskillsortohighlightdifferentphysicsconcepts.Abouthalfofthescheduledlabsarecommonbetweenthetwocourses.Theotherlabswillbemoreopen‐endedforHonorsthanforthestandardcourseandwillrequireashort1‐2pagewrittenreport.Tobalancetheworkload,HonorsstudentswillbeexcusedfromthelabquizzesforthesethreelabsaswellasallofthecomputersimulationexercisesinPhysics110.Thegradedreviewswillbeverysimilar;abouttwo‐thirdsofeachGRwillbequestionsandproblemscommontobothcourses.Finally,classtimewillbeusedabitdifferentlyinHonors,withlesstimedevotedtocoveringthemostbasicmaterial.

CorePhysicsCoursePrerequisites

ForastudentenrolledinPhysics110,heorshemusthavecompletedorbeenrolledininMath142.Importantmathconceptsrequired:


vii

Algebraandtrigonometry Vectoroperationsincludingdotproductandcrossproduct Differentiationofpolynomialsandsimplefunctions Integrationofpolynomialsandsimplefunctions

StudentsinPhysics215musthavecompletedPhysics110andMath142.USAFAandCorePhysicsCourseOutcomes

Physicscorecourses(Physics110andPhysics215)areaprimarycontributortothedevelopmentandassessmentofthefollowingUSAFAoutcomes:quantitativeliteracy,criticalthinkingandprinciplesofscience,andthescientificmethod.Additionally,thesecoursesaredesignedforyouto:

1. Developadeeper,moreintegratedunderstandingofphysicalconcepts,withafocusontheconceptsofmotion,Newton’sLaws,energy,momentum,electricity,magnetism,andselectedtopicsinmodernphysics.

2. Applythinkingandproblem‐solvingskillstomakeinformedconclusionsaboutthemeaningofphysicaldataandinformation.

3. Applyexperimentalskillsandreadingcomprehensiontoinvestigateprinciplesofnature.4. Cultivatehabitsofthemindconsistentwiththatofaneducated,scientificallyliterate

person.

CorePhysicsLearningGoals

Physics110isdesignedtoenhanceyourcriticalthinkingskillsandyourabilityto:

1. Describethemotionofobjectsusingkinematics2. InterpretandsolvemotionproblemsusingNewton’sthreelaws3. Analyzethemotionofobjectsusingconservationofenergy,momentumandangular

momentum4. Developvalidphysics‐basedconclusionsaboutreal‐worldproblemsandapplications

ThecourselearninggoalsforPhysics215are:

1. Identifyhowthefundamentalphysicalprinciplesofelectricity,magnetismapplytoconceptualorquantitativeproblems.

2. Solveconceptualorquantitativephysicalproblemsinvolvingelectricity,magnetismandmodernphysics.

3. Applyexperimentalskillstoinvestigatethephysicalprinciplesgoverningelectricityandmagnetism.

4. Analyzeandexplainthephysicalprinciplesthatapplytotheoperationofelectro‐magneticsystemsandcircuits.


viii

PhysicsCoreCourseAdministration

Position Name Office PhonePhysicsDepartmentHead ColKiziah 2A33 333‐3510DirectorofCorePrograms LtColNovotny 2A27 333‐9248Physics110CourseDirector LtColKayser‐Cook 2A43 333‐0357Assistant110CourseDir CaptSorensen 2A101 333‐9733Physics110HCourseDirector Dr.deLaHarpe 2A219 333‐9719Assistant110HCourseDir MajBuchanan 2A109 333‐7707Physics215CourseDirector MajLane 2A153 333‐3615Assistant215CourseDir Mrs.Lickiss 2A149 333‐3412Physics215HCourseDirector Dr.Kontur 2A107 333‐4224Assistant215HCourseDir LtColAnthonyDills 2A25 333‐3272

YourPhysicsInstructor

RequiredCourseMaterials

ThefollowingmaterialsarerequiredforthiscourseandmustbeinyourpossessiononLesson1.Failuretopossessyourpersonalcopyofeachofthefollowingisafailuretomeetcourserequirements.QuestionsmaybedirectedtotheDirectorofCoreProgramsorthePhysicsDepartmentHead.Inadditiontothefirstdayofclass,youarerequiredtobringthefollowingtoeachclassperiod:yourtextbookandyourentireJournal(ina3‐ringbinder).

Physics110andPhysics110Honors:

TEXTBOOK.Wolfson,Richard,EssentialUniversityPhysics,2nded.,Vol1,SanFrancisco:PearsonEducation,Inc.,2012.

JOURNAL.ThePhysics110Journalcontainscourseguidance,syllabus,learningobjectives,questions,andproblems.

MASTERINGPHYSICS.MasteringPhysics®istheonlinehomeworksystemthataccompaniesthetextbook.Anaccountcanbepurchasedwiththetextbookorseparately,butisrequiredforthecourse.TopurchaseMasteringPhysics®separately,gotowww.masteringphysics.com,intheREGISTERblockclickontheSTUDENTSbuttonandfollowtheinstructions.LeaveStudentIDblank.TheCourseIDislistedinthefollowingtable:

Course MasteringPhysicsCourseIDPhysics110 FALL2013PHYSICS110Physics110H FALL2013PHYSICS110H


ix

SUPPLEMENTALCOURSEMATERIAL.AllothercoursematerialisavailableonthePhysics110SharePointsites:

Course SharepointSitesPhysics110 https://eis.usafa.edu/academics/physics/110/default.aspxPhysics110H https://eis.usafa.edu/academics/physics/110H/default.aspx

Physics215andPhysics215Honors:

TEXTBOOK.Wolfson,Richard,EssentialUniversityPhysics,2nded.,Vol2,SanFrancisco:PearsonEducation,Inc.,2012.

JOURNAL.ThePhysics215Journalcontainscourseguidance,syllabus,learningobjectives,questions,andproblems.

MASTERINGPHYSICS.MasteringPhysics®istheonlinehomeworksystemthataccompaniesthetextbook.Anaccountcanbepurchasedwiththetextbookorseparately,butisrequiredforthecourse.TopurchaseMasteringPhysics®separately,gotowww.masteringphysics.com,intheREGISTERblockclickontheSTUDENTSbuttonandfollowtheinstructions.LeaveStudentIDblank.TheCourseIDislistedinthefollowingtable:

Course MasteringPhysicsCourseIDPhysics215 FALL2013PHYSICS215Physics215H FALL2013PHYSICS215H

SUPPLEMENTALCOURSEMATERIAL.AllothercoursematerialisavailableonthePhysics215SharePointsites:

Course SharepointSitesPhysics215 https://eis.usafa.edu/academics/physics/215/default.aspxPhysics215H https://eis.usafa.edu/academics/physics/215H/default.aspx

CoursePolicies

WORKEDEXAMPLES–CorePhysicsusestheWorkedExamplesapproachtolearning,whichrequiresstudentstocometoclasspreparedtodiscusslessonmaterial.Forthisreason,classpreparationpointsareheavilyweighted(18‐20%)andincludejournalquestions,pre‐classproblems,andpreflightquestions.

JOURNALQUESTIONS–Journalquestionsareassignedforalllessons,exceptedasnotedoneachlessonpageinyourjournal.Readtheselectionfromthetextbook,studytheassignedexamples,andanswerthequestionsbasedonthoseexamples.Givecompleteanswersandjustifyasyouwouldonanexam‐prepquizorgradedreview.YourinstructorwillgradeyourJournaleachlessontoassessyourlevelofpreparationforclass.Thegoalofthisassessmentistoevaluateyourhonest,


x

thoughtfuleffortatreasoningthroughtheproblems.Ifyougetstuckonaproblem,reviewtheexampleproblemsinthechapterandnotehowtheconceptsandequationsinthesectionareapplied.Ifyouarestillstuck,youcanreceivecreditforyourJournalbywritingdownasmuchofthesolutionasyouareable,listingspecificquestionsyouhaveandidentifyingpointsofconfusion.Journalswillbegradedbasedonthefollowingguidelines:

3/3 GoodeffortwasmadetoanswerallJournalQuestions.GoodeffortwasmadetosolvethePre‐ClassProblem(s)inalogicalformat(IDEAformatisrecommended).

2/3 OneortwoJournalQuestionswerenotansweredorpooreffortwasmadetoanswerseveraloftheJournalQuestions.PooreffortwasmadetosolvethePre‐ClassProblem(s)oralogicalapproachwasnotused.

1/3 MultipleJournalQuestionswerenotansweredorthePre‐ClassProblemwasmostlyorentirelyunfinished.

0/3 Lessthan50%oftheJournalQuestionsforthelessonwerecompleted.

Ifyouaremorethan15minuteslateunexcused,youwillreceiveazerofortheJournalgrade.

PREFLIGHTQUESTIONS–Preflightquestionsareassignedeachlessonexceptgradedreviewlessons.Preflightsareintendedtobedoneafterthejournalquestions.Preflightquestionsmustbesubmittedonlinenolaterthan0700beforethestartofeachlesson.Theyaredesignedtoassessyourunderstandingofthelessonmaterialandprovidefeedbacktoyourinstructorbeforeclass.AnswerthepreflightsinyourJournalthenenteryourresponsesintheJust‐In‐TimeTeaching(JiTT)applicationathttp://dfp‐usafas‐computer.usafa.edu/usafa/login.php.Yourusernameisbasedonyoure‐mailaccount,e.g.C14Joe.Smith,andthedefaultpasswordisfall2013.ResponsesaregradedthroughtheJiTTapplication.YourinstructorwillnotgradewrittenpreflightresponsesintheJournal.

PRE‐CLASSPROBLEMSandHOMEWORKPROBLEMS–Pre‐classproblemsareselectedfromthetextbookoruniquelydesignedforthelesson.Pre‐classproblemsaregradedaspartofeachlesson’sJournalgrade.Pre‐classproblemsarechosentogiveyoupracticedevelopingessentialskillstounderstandthelesson.

HOMEWORKPROBLEMS–HomeworkshouldbecompletedinyourJournaltoprovideyoureferenceandstudymaterialforclass,quizzes,andexams.(Somequizzesmaybe“openJournal!”)Oncethehomeworkproblemsarecompleted,youshouldenteryouranswerintoMasteringPhysicstobescored.

LABSandLABQUIZZES–Onlabdays,youwillcompletethedatacollectionandanalysesasagroup,handinthelabworksheetasagroup,andthentakeanindividual‐effortlabquiz.Ifyouaremorethan15minuteslateunexcused,youwillreceiveazeroforthelabworksheet.Youmay


xi

participatewithalabgroupandtakethelabquiz.Forexcusedabsences,youmustcompletemissedlabswithin3lessons(6classdays).ExemptionstothispolicymustbeapprovedbytheCourseDirector.LabworksheetsareavailableontheCourse’sSharePointsite.AdditionalinstructionsareonindividuallessonpageswithintheJournal.TheHonorsCoursesmayberequiredtocompletealabreportwhichisfurtherdefinedinAppendixA.

EXAM‐PREPQUIZZES–Exam‐prepquizzes(EPQ)consistofworkoutandmultiple‐choiceproblemssimilartothoseongradedreviewsandthefinalexam.Youshouldusethesequizzestogaugeyourunderstandingofthematerialbeforetheexams.AdditionalresourcesmaybeuseddependinguponcoursedirectorpolicyandwillbeannouncedpriortotheEPQ.

GRADEDREVIEWADMINISTRATION–GradedReviews(GRs)normallyconsistoftenmultiple‐choicequestions,andseveralworkoutproblems.Youwillhave80minutestocompletetheexam.

GRADING–Physicsisnota“plug‐and‐chug”subject.Submittinganumericallycorrectanswerforaworkoutproblemdoesnotguaranteecredit.Itispossibletogettherightnumberwiththewrongphysics.Yourscoreisdeterminedbythesoundnessofthereasoningthatledtoyouranswer.Inordertoreceivefullcredityoumustidentifythemainphysicsconceptsandshoweachstepintheproblem‐solvingprocess(IDEAformatisrecommended).

ABSENCEandTARDINESS–

(a)IfyouwillbeabsentduringaGradedReviewduetoaUSAFASchedulingCommitteeAction(SCA),youareresponsibletonotifyyourinstructoratleastTHREEDAYS(notincludingweekends)PRIORtothefirstofferingoftheexam.Ifyouaremorethan15minuteslateunexcusedforaGradedReview,youmusttakeamakeupexamwitha25%penalty.Ifyouarelessthan15minuteslate,youmaystilltaketheexamduringthescheduledtime.

(b)Ifyouwillmissalessonforanyreason,completeandturninacopyofthatlesson’sgradedworkbeforeyouleaveorsenditwithanotherstudenttoturninontime.

MAKEUPEXAMS(GRsandQuizzes)–Ifyouaretravelingwithanathleticteamorcadetclub,thepreferredoptionistotaketheexamontheroad.Ifthisisnotanoptionorifyouhavemissedtheexamforanotherreason,workwithyourinstructortoscheduleatimetomakeuptheexamwithintwolessons.

FINALEXAMandVALIDATION–TheFinalExamisacomprehensiveexaminationincludingmaterialfromtheentirecourse.Thefinalexamisyouropportunitytodemonstrateproficiency;therefore,validationoftheFinalExamisnotoffered.

DOCUMENTATION–ClearlydocumentallhelpreceivedongradedworkfromsourcesotherthanyourWolfsontextbook.Pleasefeelfreetoseekhelpfromotherinstructors,students,orothertextsatanytime.Forallgradedworkoutsideofclass,youmayusethefollowingAUTHORIZEDRESOURCES:Anypublishedorunpublishedsource,websites,andanyindividuals.Forall


xii

assignments,youmustproperlydocumentallassistanceandsourcesusedaccordingtothePhysicsDepartmentpolicyletterondocumentationstandards(locatedontheSharePointsite).Thisdoesnotallowyoutosimplycopyresourcematerialortheworkofanotherstudent,pastorpresent,anddocumentthesource.Thereisnoacademiccreditforcopiedwork.Youmustalsoindicatewhethernohelpwasreceived.Documentationforalloutside‐of‐classwork—isaccomplishedinthefooteroneachpageoftheJournal.

ACADEMICSECURITY–Allexam‐prepquizzesandgradedreviewsremainunderacademicsecurityuntilreleasedbytheCourseDirector.DONOTdiscussthecontentsorthedifficultyofthematerialwithanyoneexceptyourinstructoruntilafteritisreleasedfromacademicsecurity.

CONSTANTSANDEQUATIONSSHEET–YouwillbegivenastandardizedConstantsandEquationsSheetforuseonalllabquizzes,exam‐prepquizzes,gradedreviews,andthefinalexam.Understandingthephysicalconceptsgoverningtheuniversewillnotcomefromscanninganequationsheetinsearchofvariablesthatfittheproblem.Youmustfullycomprehendthenatureoftheequations,themeaningsofthevariables,andtheconstraintsforusingeachequation.

EXTRAINSTRUCTION–ThesecondhourofclassformostlessonsisdedicatedtoExtraInstruction(EI).Yourinstructorwillnotcovernewmaterialorholdreviewsessionsduringthistime,butheorsheisavailabletohelpyou.Ifyouhaveotherperiodsfree,youmayseekEIinanyofthePhysicsclassroomsduringsecondhourfromanyinstructorthatisteachingyourcourse.Donotexpectone‐on‐oneEIifyoudonotseekEIduringthesecondhourofyourclass.

RE‐GRADES–Re‐gradingofquizzesandlabsisconsideredonanindividualbasisbyyourinstructor.Ifyoudesireare‐gradeonagradedreview,firstshowyourinstructoryourworkandheorshewillletyouknowifare‐gradeiswarranted.Ifitiswarranted,typeaMemoforRecord*(MFR)explainingyourcase,attachittoyourexam,andsubmitittoyourinstructor.TheCourseDirectorwillre‐gradeyourwork.Youcouldalsolosepoints,sincetheentireproblemwillbere‐graded.Youhavesevencalendardaysfromthedateagradedeventisreturnedtorequestare‐grade.

*AMemorandumforRecordistheAirForcestandardforofficialwrittencommunicationsandtheformatisprovidedintheTongueandQuill,availableontheAirForceE‐Publishingwebsite.


xiii

Physics110HonorsCoursePointStructure

GradedEvent Points Percentage

JournalQuestions 28 @ 3 pointseach 84 5.6%PreflightQuestions 30 @ 5 pointseach 150 10.0%Pre‐LabQuestions 6 @ 5 pointseach 30 2.0%LabWorksheet 3 @ 10 pointseach 30 2.0%LabQuizzes 3 @ 10 pointseach 30 2.0%LabReports 3 @ 25 pointseach 75 5.0%Exam‐PrepQuizzes 4 @ 30 pointseach 120 8.0%CriticalThinkingExercise

3 @ 15 pointseach 45 3.0%

Homework 37 @ 3 pointseach 111 7.4%GradedReviews 3 @ 150 pointseach 450 30.0%FinalExam 1 @ 375 points 375 25.0%Total 1500 100.0%

No.ofEvents/Points

NOTE1:Approximately70%ofthecoursepointsareindividualeffort(labquizzes,exam‐prepquizzes,gradedreviewsandthefinalexam).

NOTE2:Asufficientlylowgradeonthefinalexamcouldresultinfailureofthecourseregardlessoftheoverallscore,atthediscretionofthePhysicsDepartmentHead.

IDEAProblem‐SolvingStrategy Physics110HJournal‐2013‐2014

xiv

IDEAProblem‐SolvingStrategy

SolvingProblemsUsingtheIDEAFormat

Physicsproblemscanbechallenging,butunderlyingallofphysicsisonlyahandfulofbasicprinciples.Ifyoureallyunderstandthose,youcanapplytheminawidevarietyofsituations.Ifyouapproachphysicsasahodgepodgeofunrelatedlawsandequations,you’llmissthepointandmakethingsdifficult.Butifyoulookforthebasicprinciplesandforconnectionsamongseeminglyunrelatedphenomena,thenyou’lldiscovertheunderlyingsimplicitythatreflectsthescopeandpowerofphysics.

Asystematicsolutionmethodhelpsdevelopcriticalthinkingandscientificmethodprinciples.OnesuchapproachistheIDEAproblem‐solvingstrategy.Solvingaquantitativephysicsproblemalwaysstartswithbasicprinciplesorconceptsandendswithapreciseanswerexpressedaseitheranumericalquantityoranalgebraicexpression.Thepathfromprincipletoanswerfollowsfoursimplesteps—stepsthatmakeupacomprehensivestrategyfororganizingyourthoughts,clarifyingyourconceptualunderstanding,developingandexecutingplansforsolvingproblems,andassessingyouranswers.

Interpret Identifythemainphysicsconceptusedtosolvetheproblem.

Develop Drawadiagramdepictingthesituation.Labelthegiveninformationandidentifytheinformationforwhichyouaresolving.

Evaluate Solvetheproblemfrombasicprinciplesusingequationsrelatedtothemainphysicsconcepts.Whenpossible,expressthesolutionsymbolicallybeforesubstitutingvaluesintotheequations.Includeunitswithallnumericalvalues.

Assess Criticallyassessthevalidityofthesolutionbyansweringquestionssimilartothefollowing:

a) Howdoesthesolutioncomparetoknownvalues?b) Howwouldtheanswerchangeifthevalueofoneofthevariableschanged?c) Isthesolutionphysicallypossible?Explain

“Even for the physicist the description in plain language will be a criterion of the degree of understanding that has been reached.”

Werner Heisenberg, Physics and Philosophy

Physics110HJournal‐2013‐2014 LearningObjectives

xv

LearningObjectives

BlockI–Motion

Duringthisblockwewillstartwithstudyingthebasicconceptsofdisplacement,velocityandacceleration.Wewillthenusetheequationsofmotiontoanalyzethemotionofobjects.

[Obj1] ConvertphysicalmeasurementsfromvariousunitstothestandardSIunitsofmeters,kilograms,andseconds.

[Obj2] Expressquantitiesusingscientificnotationandperformaddition,subtraction,multiplication,division,andexponentiationonthem.

[Obj3] Identifythenumberofsignificantfiguresgiveninaproblemstatement,andexpresstheanswerusingthecorrectnumberofsignificantfigures.

[Obj4] Explaintherelationsbetweenposition,displacement,speed,velocity,andaccelerationforanobjectmovinginoneandtwodimensions.

[Obj5] Constructandinterpretgraphsofposition,velocity,andaccelerationforanobjectmovinginoneandtwodimensions.

[Obj6] Explainthedifferencebetweeninstantaneousandaveragevelocity,andbetweeninstantaneousandaverageacceleration.

[Obj7] Usemathematicalandgraphicalmethodstocalculateinstantaneousandaveragevelocityandinstantaneousandaverageaccelerationinoneandtwodimensions.

[Obj8] Useequationsofmotiontosolveproblemsinvolvingmotionwithconstantacceleration.

[Obj9] Usecalculustosolveproblemsinvolvingmotionwithnon‐constantacceleration.

[Obj10] Solveproblemsinvolvingfree‐fallmotionwithconstantgravitationalacceleration.

[Obj11] Expressvectorsbothincomponentformandinmagnitude‐ directionform.

[Obj12] Usemathematicalandgraphicalmethodstoperformvectoraddition,vectorsubtraction,andscalarmultiplication.

[Obj13] Usevectorstorepresentposition,velocity,andacceleration.

[Obj14] Describehowtheeffectsofaccelerationdependuponthedirectionoftheaccelerationvectorrelativetothevelocityvector.

[Obj15] Solveproblemsinvolvingprojectilemotionunderconstantgravitationalacceleration.

[Obj16] Explainwhyuniformcircularmotioninvolvesacceleration.

[Obj17] Solveproblemsinvolvinguniformandnonuniformcircularmotion.

LearningObjectives Physics110HJournal‐2013‐2014

xvi

BlockII–Newton’sLaws

Inthisblock,westartbyintroducingNewton’sthreelawsofmotions.Wewillthenusetheselawstounderstandtheconceptofforce,todescribedifferenttypesofforces,andtoanalyzethemotionofobjectsinoneandtwodimensions.

[Obj18] Explaintheconceptofforceandhowforcescausechangeinmotion.

[Obj19] StateNewton’sthreelawsofmotionandgiveexamplesillustratingeachlaw.

[Obj20] Explainthedifferencebetweenmassandweight.

[Obj21] Constructfree‐bodydiagramsusingvectorstorepresentindividualforcesactingonanobject,andevaluatethenetforceusingvectoraddition.

[Obj22] UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingonasingleobject.

[Obj23] UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.

[Obj24] Differentiatebetweentheforcesofstaticandkineticfrictionandsolveproblemsinvolvingbothtypesoffriction.

[Obj25] Describedragforcesqualitativelyand*quantitatively.

[Obj26] Explainthephysicsconceptofwork.

[Obj27] Evaluatetheworkdonebyconstantforcesandbyforcesthatvarywithposition.

Physics110HJournal‐2013‐2014 LearningObjectives

xvii

BlockIII–EnergyandMomentum

Wewillstartthisblockbyintroducingtheconceptsofenergyandwork.Usingourunderstandingoftheseconcepts,wewilldeveloptheprincipleofconservationofenergywhichwillallowustoanalyzethecomplexmotionofobjectsincludingthoseinorbits.Wewillfinishthisblockdiscussingcollisionsandanotherconservationprinciple:conservationoflinearmomentum.

[Obj28] Explaintheconceptofkineticenergyanditsrelationtowork.

[Obj29] Explaintherelationbetweenenergyandpower.

[Obj30] Explainthedifferencesbetweenconservativeandnonconservativeforces.

[Obj31] Evaluatetheworkdonebybothconservativeandnonconservativeforces.

[Obj32] Explaintheconceptofpotentialenergy.

[Obj33] Evaluatethepotentialenergyassociatedwithaconservativeforce.

[Obj34] Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanicalenergy,orconservationofenergy.

[Obj35] Describetherelationbetweenforceandpotentialenergyusingpotential‐energycurves.

[Obj36] Explaintheconceptofuniversalgravitation.

[Obj37] Solveproblemsinvolvingthegravitationalforcebetweentwoobjects.

[Obj38] Determinethespeed,acceleration,andperiodofanobjectincircularorbit.

[Obj39] Solveproblemsinvolvingchangesingravitationalpotentialenergyoverlargedistances.

[Obj40] Usetheconceptofmechanicalenergytoexplainopenandclosedorbitsandescapespeed.

[Obj41] Useconservationofmechanicalenergytosolveproblemsinvolvingorbitalmotion.

[Obj42] Calculatethecenterofmassforsystemsofdiscreteparticlesandforcontinuousmassdistributions.

[Obj43] ExplaintheconceptoflinearmomentumofasystemofparticlesandexpressNewton'ssecondlawofmotionintermsofthelinearmomentumofthesystem.

[Obj44] Explainthelawofconservationoflinearmomentumandtheconditionunderwhichitapplies.

[Obj45] Explaintheconceptofimpulseanditsrelationtoforce.

[Obj46] Applyconservationoflinearmomentumtosolveproblemsinvolvingsystemsofparticles.

[Obj47] Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.

[Obj48] Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐andtwo‐dimensions.

LearningObjectives Physics110HJournal‐2013‐2014

xviii

BlockIV–RotationalMotionandSimpleHarmonicMotion

Duringthisblockwewillstudytherotationalandoscillatorymotionofobjects.Wewillstartbyexploringtherotationmotionofrigidobjects–discussingconceptsofangulardisplacement,velocity,andacceleration.WewillthenrevisittheconceptsofNewton’sSecondLaw,conservationofenergy,andconservationofmomentumasappliedtoobjectsundergoingrotationalmotion.Wewillendthecoursebyintroducingtheconceptofsimpleharmonicmotion.

[Obj49] Explaintherelationbetweentherotationalmotionconceptsofangulardisplacement,angularvelocity,andangularacceleration.

[Obj50] Useequationsofmotionforconstantangularaccelerationtosolveproblemsinvolvingangulardisplacement,angularvelocity,andangularacceleration.

[Obj51] Usecalculustosolveproblemsinvolvingmotionwithnon‐constantangularacceleration.

[Obj52] Explaintheconceptoftorqueandhowtorquescausechangeinrotationalmotion.

[Obj53] Givenforcesactingonarigidobject,determinethenettorquevectorontheobject.

[Obj54] Determinetherotationalinertiaforasystemofdiscreteparticles,rigidobjects,oracombinationofboth.

[Obj55] Compareandcontrasttheconceptsofmassandrotationalinertia.

[Obj56] UseNewton’ssecondlawanditsrotationalanalogtosolveproblemsinvolvingtranslationalmotion,rotationalmotion,orboth.

[Obj57] Solveproblemsinvolvingrotationalkineticenergyandexplainitsrelationtotorqueandwork.

[Obj58] Explaintherelationbetweenlinearandangularspeedinrollingmotion.

[Obj59] Useconservationofenergytosolveproblemsinvolvingrotatingorrollingmotion.

[Obj60] Determinethedirectionsoftheangulardisplacement,angularvelocityandangularaccelerationvectorsforarotatingobject.

[Obj61] Determinetheangularmomentumvectorfordiscreteparticlesandrotatingrigidobjects.

[Obj62] Applyconservationofangularmomentumtosolveproblemsinvolvingrotatingsystemschangingrotationalinertiasandrotatingsystemsinvolvingtotallyinelasticcollisions.

[Obj63] Definesimpleharmonicmotionandexplainwhyitissoprevalentinthephysicalworld.

[Obj64] Determinetheperiodandfrequencyofasimpleharmonicoscillatorfromitsphysicalparameters,andcompletelyspecifyitsequationofmotion.

[Obj65] Determinethevelocityandaccelerationofasimpleharmonicoscillatorfromitsequationofmotion.

[Obj66] Determinethepotentialandkineticenergiesofasimpleharmonicoscillatoratanypointinitsmotion,anddescribethetimedependenceoftheseenergies.

Physics110HJournal‐2013‐2014 Lesson1

DocumentationStatement:

1

Lesson1

Introduction

Reading Chapter1,2.1,2.2Examples 2.1,2.2HomeworkProblems 1.16,1.24,2.51

Thereisanon‐gradedPHYSICSKNOWLEDGEASSESSMENTTESTthislesson.

LearningObjectives




[Obj4] Explaintherelationship betweenposition,displacement,speed,velocity,andaccelerationforanobjectmovinginoneandtwodimensions.



Notes

Lesson1 Physics110HJournal‐2013‐2014


2

WorkedExample

Studythegivenproblemandsolution,thenanswerthequestionsregardingtheproblem.

STATEMENTOFTHEPROBLEM

Atliftoff,theaccelerationofaspaceshuttleis29m/s2,anditspositionasafunctionoftimeisdefinedas , whereaistheaccelerationandtistime.a)Whatisthe(instantaneous)velocity ofthespaceshuttleonesecondafterliftoff?b)Whatistheaveragevelocity overthefirstminuteafterliftoff?

STRATEGY

We interpret this as a problem involving the relationship between position, velocity, and acceleration. Additionally, we are interested in the difference between instantaneous velocity and average velocity. IMPLEMENTATION

We are given the position equation, so in order to arrive at a value for velocity at one given instant, we will need to take the derivative of the equation with respect to time ( / ). To find the average velocity, we calculate the change in position over a given length of time ( ∆ /∆ ).

CALCULATION

a) Velocity 1 second after liftoff ( 1s):

2

2 29m/s 1s 58m/s

b) Average velocity over the first minute after liftoff (∆ 60s):

∆∆

29m/s 60s 1740m/s

Score(3)



3

SELF‐EXPLANATIONPROMPTS

1. Whydoes∆ become and∆ become whencalculatingaveragevelocity?

2. Describethemotionoftherocketasshownintheposition versustime graph.

3. Howdoyouexpecttheinstantaneousvelocityafteroneminutetocomparetotheaveragevelocitycalculatedinpart(b)?Calculatetheinstantaneousvelocityafteroneminute.



4

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMAlargemeteoris9700kmawayandheadingstraighttowardstheMoon.ItistravellingataspeedsuchthatitwouldimpacttheMoonin15minutes,butinsteaditcollideswithasmallermeteor,knockingitoffitsoriginalpathata26°anglebutmaintainingitsoriginalspeed.Withthisnewtrajectory,howmuchlongerwillittakeforthemeteortoimpacttheMoon?

Answer:~100seconds



5

PreflightQuestions1. Whattopicsdidyoufindmostchallengingfromthereading?

2. Solvethefollowingsystemofequationsfor and .

2 5 3 6

a) 3.40, 4.20b) 2.43, 1.29c) 3.40, 4.20d) 1.48, 1.57e) 1.00, 3.00

3. Itispossibleforanobjecttohave,atthesametime…

a) …bothzerovelocityandnon‐zeroacceleration.b) …bothnon‐zerovelocityandzeroacceleration.c) Both(a)and(b)arepossible.d) Neither(a)nor(b)arepossible.

4. CRITICALTHINKING:Explainthedifferencebetweenaverageandinstantaneous

speed/velocity/acceleration.(Hint:Youshouldconsiderthequantityoftime.)



6

HomeworkProblems

1.16



7

1.24



8

2.51



9

Lesson2

Displacement,Velocity,andAcceleration

Reading 2.3,2.4Examples 2.1‐ 2.3HomeworkProblems 2.20,2.35,2.79

ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).

LearningObjectives

[Obj4] Explaintherelationship between position,displacement,speed,velocity,andaccelerationforanobjectmovinginoneandtwodimensions.






Notes



10

WorkedExample



Dragracingisanaccelerationcompetitionthattakesplaceoveralevel¼miletrack.Adragsterstartsfromrestandhastocompletethe¼mile(402.3m)run.Thetimeisveryshortsothereactiontime(thetimeittakesthedrivertostartafterthegreenlightcomeson)isimportant.Thedriverwiththeshortestoveralltime(runtime+reactiontime)isthewinner.

Thenationalrecordoveralltimeis4.42seconds.

a)Assumingthattheaccelerationwasconstant(togetasimpleestimateoftheacceleration),whatwastheaccelerationinthewinningrun?

b)Whatwastheaveragespeedofthedragster?

c)Againassumingconstantacceleration,whatwasthefinalspeedasthedragstercrossedthefinishline?

STRATEGY

This problem assumes constant acceleration and asks us to relate the given time to the speed and acceleration of the dragster. We will use the definition of average speed and equations of motion to solve this problem. IMPLEMENTATION

For part (a) we apply the relation .

For part (b) we apply the relation for average speed

For part (c) we use the equation for average acceleration

0

∆

Score(3)



11

CALCULATION

a) Acceleration of the winning run:

The initial velocity is zero, so simplifies to . Solving for acceleration gives ∙ .

. 41m/s

b) Average speed of the dragster:

∆

402.3m4.42s

91m/s 204mph c) Final speed of the dragster (assuming constant acceleration): Using the acceleration from part (a),

∆ can be written as ∆

Since the dragster starts from rest, 41 ∙ 4.42s 181m/s 405mph Note:Accelerationandfinalspeedarenotmeasuredindragraces;theaveragespeedduringthelast20metersismeasured.Intherecordrun,theaveragespeedduringthelast20meterswas336mph.


1. Inyourtextbooklookupthederivationsoftheequationsusedinparts(a)and(b)aboveandsummarizeinyourownwordshowtheserelationsareobtained.

2. Isitvalidtousetherelation ifaccelerationisnotconstant?Gotothe

assignedreadinginthetextbookandfindwhatassumptionwasmadeinthederivationoftheequation.

3. Whatistherelationbetweeninitialspeedandfinalspeedduringatimeintervalwhentheaccelerationisconstant?



12

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMOnthewaytotheMoonthefirststageengineoftheSaturnVmoonrocketfiredfor156secondstoliftthecraft38miles(61,155meters).WhatistheaverageaccelerationofSaturnVduringthisstage?

Answer:5m/s2TryIt!(1pt):WhatisthespeedoftheSaturnVattheendofthisstage?



13

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Thefollowinggraphsdepictthevelocitiesoffourobjectsmovinginonedimension.Which

objecthasthegreatestdisplacementduringthetimeintervalshown?

3. Whichofthefollowingarrowscorrespondtoatimeatwhichtheinstantaneousvelocityis

greaterthantheaveragevelocityoverthetimeintervalshown?

4. CRITICALTHINKING:Doesacarodometermeasuredisplacementordistance?Explain.

DC

B

A

Position,

Time, Position,

Time, Position,

Time, Position,

Time,

a) b) c) d)

a)

Time, (s)

Velocity,(m/s)

5

10

10 20

d)

Time, (s)

Velocity,(m/s)

5

10

10 20

c)

Time, (s)Velocity,(m/s)

5

10

10 20

b)

Time, (s)

Velocity,(m/s)

5

10

10 20



14

HomeworkProblems

2.20



15

2.35



16

2.79



17

Lesson3

Lab1–AccelerationDuetoGravity

Reading 2.5,Lab1WorksheetExamples 2.6HomeworkProblems 2.38,2.42,2.78

ThereisaLABthislesson.

LearningObjectives


Notes



18

Pre‐LabQuestions

1. Brieflydescribethepurposeandgoalsofthislab.(Onetotwocompletesentences)

2. Whataretherelevantconceptsandequationsthatyouwillbeusinginthelab?

3. InthesetupofPart1ofthelab,youareaskedtomeasuretheangleoftheinclinedtrack.Howwillyoudeterminetheangleoftheincline?

4. InPart1ofthelab,yourgroupwillmeasurethetimeittakesforanun‐weightedairtrackcarttotraveldifferentdistancesdownanincline.InPart2,yourgroupwillmeasurethetimeittakesforaweightedairtrackcarttotravelthesamedistancesdownanincline.Howdoyouexpectthetimestocomparebetweentheweightedandun‐weightedcarts?Brieflyexplainyourreasoning.

5. WhengraphingthedatainpartIandII,youareaskedtoplot vs.x.Explainthereasonsbehindplottingthedatainsuchaway.

Score(5)



19

LabNotes



20

HomeworkProblems

2.38



21

2.42



22

2.78



23

Lesson4

Two‐Dimensional&ProjectileMotion

Reading 3.1– 3.5Examples 3.3HomeworkProblems 3.34,3.53,3.54

LearningObjectives

[Obj11] Expressvectorsbothincomponentformandinmagnitude‐directionform.

[Obj12] Usemathematicalandgraphicalmethodstoperformvectoraddition,vectorsubtraction,andscalarmultiplication.



Notes



24

WorkedExample


STATEMENTOFTHEPROBLEMAnaircrafthasavelocityof 70 50 m/s.Thewindpushestheaircraftwithavelocityof

45 40 m/s.Whatistheresultingfinalvelocityoftheplane, ?

STRATEGY

Adding vectors is done by adding the x-components and the y-components to construct the net velocity vector. IMPLEMENTATION

Add the components of the two vectors to build the final vector as: (x-component total) +(y-component total) CALCULATION

70 45 m/s 50 40 m/s 115 10 m/s

Score(3)



25

SELF‐EXPLANATIONPROMPTS1. Superpositionofvectorsistheprocessofaddingvectors.Whydoyouaddthex‐andy‐

componentsseparately?2. Whencomponentsarecombined,aretheabsolutevaluesofthecomponentsusedordothe

componentsretaintheirnegativesignsiftheyhavethem?3. Describea)whatadditionalinformationyouwouldneedtobegiventodeterminethe

accelerationoftheplaneinthisproblemandb)whatstepsyouwouldusetocalculatetheaccelerationoftheplane.



26

Pre‐ClassProblem


Anaircrafthasaninitialvelocityof 70 50 m/s.Itexperiencesanaccelerationof2.5 2 m/s astheresultofastrongwind.After20sinthiswind,whatisthenewvelocityof

theaircraft?

Answer: 120 10 m/s



27

PreflightQuestions1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Anobjectisinitiallymovinginthepositive ‐directionandthenexperiencesaccelerationinthe

positive ‐direction.Whichofthegraphsdepictsthe ‐and ‐positionsoftheobjectwhileaccelerating?

3. Theposition ofaparticleasafunctionoftime is 5 2 .Which

statementistrueconcerningtheparticle?

a) Theparticleislocatedattheoriginat 0.b) 1 5m/s

c) 5 2 2 m/s

d) 4 19/2m/s2

e) √ 2 m/s

f) Accelerationoftheparticleisconstant.

4. CRITICALTHINKING:Cananobjecthaveanorthwardvelocityandsouthwardacceleration?Explain.

a) d)b) c)



28

HomeworkProblems

3.34



29

3.53



30

3.54



31

Lesson5

ProjectileMotion

Reading 3.5Examples 3.4HomeworkProblems 3.55,3.70,MP

ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt). ThereisanEXAM‐PREPQUIZthislesson.

LearningObjectives



Notes



32

WorkedExample



Avintagebomberisparticipatinginanairshowandplanstoconductabombingrunusingan“explosive”flourbag.Iftheaircraftisflyingat75m/sandreleasestheflourbomb1500metersabovetheground,howfarbackfromthetargetmustthepilotreleasethisflourbomb(thereleasedistancex)?Howfastisthebombmovinginthehorizontalandverticaldirectionswhenithitstheground?STRATEGY

This is a projectile motion problem where the only acceleration affecting the motion is assumed to be due to gravity, acting vertically downward. We need to apply the basic kinematics equations separately for the motion in the horizontal, x-direction, and the vertical, y-direction. In this problem there is no horizontal acceleration, so the release distance will be the horizontal velocity times the flight time. The flight time will come from analyzing the vertical motion - knowing the total distance and the bomb’s initial vertical velocity. The kinematics equations we will need are and . These equations can be written for both motion in the x- and y-directions with the flight time t being a common variable. IMPLEMENTATION

First, we need to establish an origin and coordinate system. Let’s set the origin at the point of release of the bomb with the x-axis pointed to the right and the y-axis pointed up (in a standard configuration). Now, we will determine the flight time (i.e. the time that the flour bomb travels from release to impact). The aircraft is flying in level flight, so the initial velocity in the y-direction v0y is zero. Also, the only acceleration is due to gravity, acting in a downward direction (g=-9.8 m/s2). We will manipulate and solve for flight time. Note that using our origin set at the point of release, the final position (yf) will be a negative 1500 m.

Score(3)



33

Now that we have the flight time, we will use to solve for the release distance and to determine the vertical speed of the bomb. There is no horizontal acceleration, so the horizontal velocity of the bomb is the same as when it was part of the aircraft. CALCULATION

First, determine the flight time in the vertical direction. Starting with ∆ , we get 1500m 0m 0

9.8m/s 2 1500m / 9.8 and 17.5s..

Next, determine the how far back the bomb is released in the horizontal. Starting with , we get ∆ 75 17.5s 0 1312m.

Finally, solve for the final bomb velocity in the vertical: Starting with and 0 9.8m/s 12.4s 171.5m/s

In addition, 75m/s since there is no acceleration in the horizontal. SELF‐EXPLANATIONPROMPT

1.Whyisthefinalposition,yf,anegativequantity?2.Whatwouldbedifferentifweweretodesignatetheoriginatthegroundlevelbelowthereleasepoint?3.Whatwouldhappeniftheinitialvelocity,v0y,intheverticalwasnotzero?



34

Pre‐ClassProblem


Acannononaclifffiresatashipinapiratemovie.Theshipis200mfromthecliffandtheinitialvelocityofthelaunchedcannonballis 60 20 m/s.Ifthecannonballhitstheship,a)howhighisthecliff,andb)whatisthefiringangle?

Answers:12m,18.4°



35

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Asnowballisthrownverticallyupwardfromamovingsledtravelingonastraight,levelroadat

aconstantspeed.Neglectingairresistance,thesnowballwillland

a) infrontofthesled.b) onthesled.c) behindthesled.d) Theanswerdependsonthespeedofthesled.

3. Twoidenticalmassesareshotoutofacannonsittingonaflatsurface.Thecannonisadjusted

suchthatthehorizontalvelocitycomponentofthecannonballsareequal.Object1,aredcannonball,isshotupwardatanangleof30°withrespecttothehorizontal.Object2,abluecannonball,isshotupwardsatanangleof60°withrespecttothehorizontal.Whichballwillhitthegroundfurthestfromthecannon?

a) Theredcannonball.b) Thebluecannonball.c) Bothcannonballswillhitatthesamespot.d) Thecannonballswillonlygoupanddown.e) Theanswercannotbedeterminedfromthegivendata.

4. CRITICALTHINKING:Ahighjumperandalongjumperarebothhumanprojectiles,butwith

slightlydifferentgoals.Thehighjumperwantstotraveloverahighbarwithouttouchingit,andthelongjumperwantstotravelasgreatadistanceaspossiblewithouttouchingtheground.Describehowthex‐andy‐componentsoftheinitialvelocityvectorshoulddifferbetweenthetwotypesofjumpers.



36

HomeworkProblems

3.55



37

3.70



38

MP



39

Lesson6

Lab2–ProjectileMotion

Reading Lab2WorksheetExamples 3.4HomeworkProblems MP,3.58,MP


LearningObjectives


Notes



40

Pre‐LabQuestions


2. Anobjectislaunchedhorizontallyfromaheight withvelocity .Howmuchtime doesittake

fortheobjecttoreachthelevelgroundbelow?

a

b

c

d)

3. Youwillbelaunchingasmallaircompressionrocketforthislab.First,youwilllaunchthe

rocketverticallyandmeasurethetimeofflight.Deriveanequationthatrelatestimeofflight,t,toinitialvelocity,v0,fortherocket.

4. Inthesecondpartofthelab,youwilllaunchtherocketatanangle,θ,andaheight,h,above

levelground.Usingθ,h,andinitialvelocityv0asknownquantities,deriveanexpressionforthehorizontalrange,Δ ,thattherocketwilltravel.

Score(5)



41

LabNotes



42

HomeworkProblems

MP



43

3.58



44

MP



45

Lesson7

AccelerationinCircularMotion

Reading 3.6Examples 3.7,3.8HomeworkProblems MP,MP,3.80

LearningObjectives



Notes



46

WorkedExample



Anaircrafttravelingataconstant150m/smakesa360°turnataconstantaltitude(referredtoaslevelflight).Iftheaircraft’saccelerationtowardthecenteroftheturnis1.5g,whatistheradiusoftheturn?

STRATEGY

This is a problem involving uniform circular motion (UCM), where several things in the horizontal are uniform: radius (r), tangential speed (vtan), and the center-directed acceleration (acentripetal). The aircraft experiences NO acceleration in the vertical direction. The center-directed acceleration (acentripetal) is related to the tangential velocity by the UCM basic relationship: IMPLEMENTATION

First, we need to determine the magnitude of the center-directed acceleration. We are given that it is 1.5 g. This means 1.5 times the acceleration due to gravity (9.8 m/s2). Next, we will manipulate the UCM basic relationship so that r is alone on the left side of the equation. We then solve for the radius. CALCULATION

First, manipulate the UCM basic relationship to solve for r :

becomes

Now, substitute and solve:

150m/s1.5 9.8m/s

1530m

Notice that the units resolve as: /

/m

Score(3)



47

SELF‐EXPLANATIONPROMPTS1.Inthisproblem,theaircraftistravelingataconstantspeedof150m/s.Isthisaircraft(oranyobjectexecutinguniformcircularmotion)undergoingacceleration?Explain.2.Howdoyouknowthattheaccelerationintheverticaliszero?3.Whatcausesthecenter‐directedacceleration?



48

Pre‐ClassProblem


A650‐kgFormulaOneracecarexecutesaportionofacircularturnat20m/s.Theradiusoftheturnis50meters.Whataccelerationmustthefrictionofthetiresgenerateinordertoaccomplishthisturn?Whatisthedirectionofthatacceleration?

Answer:8m/s2,towardsthecenteroftheturn

TryIt!(1pt):Describeanddrawtheaccelerationvectorifthecar’sspeedwasincreasingasitexecutedtheturn.



49

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Whenacartravelingataconstantspeedgoesaroundacurveonalevelroad,whatisthe

directionofacceleration?

a) Thereisnoacceleration.b) Thecarisacceleratingtowardthecenterofthecurve.c) Thecarisacceleratingawayfromthecenterofthecurve.d) Theaccelerationisinthesamedirectionthecaristraveling.

3. Rankinordertheradialaccelerationsofthefollowingobjectsfromlargesttosmallest.

a) b) c) d)

4. CRITICALTHINKING:Whenyourideinavehiclethatismakingaturnyourbodyfeelspushed

outward.Reconcilethisfactwiththephysicsstatementthattherealaccelerationofyourbodyisinwardtowardsthecenteroftheturn.

2 2

2

2



50

HomeworkProblems

MP



51

3.58



52

MP



53

Lesson8

GRADEDREVIEW1

LearningObjectives




[Obj4] Explaintherelationship betweenposition,displacement,speed,velocity,andaccelerationforanobjectmovinginoneandtwodimensions.







[Obj11] Expressvectorsbothincomponentformandinmagnitude‐directionform.

[Obj12]

Usemathematicalandgraphicalmethodstoperformvectoraddition,vectorsubtraction,andscalarmultiplication.






Notes



54

Lesson1:“Areyouready?”

Readtheproblembelowandworkthroughtheguidedsolution,thenanswerthequestionsregardingtheproblem.


Susanisdrivingat50mphtopickupherfriendattheairport.Herfriend’sflightlandsin30minutes,andsheis40miawayfromtheairport.WillSusanbeabletopickupherfriendontime?Ifso,howlongwillittakeforhertoarriveathercurrentspeed?Ifnot,whatwillshehavetochangeherspeedtoinordertoarriveattheairportontime?

STRATEGY(Fillintheblanks.)

We will need to first determine if Susan’s current speed is sufficient to allow her to arrive within 30 minutes. We can calculate the speed necessary to cover the given remaining distance and compare it to her current speed. If her current speed is greater than the needed speed, then she will be able to arrive on time. If her current speed is less than the needed speed, then she will need to modify her current speed. CALCULATION(Fillintheblanks.)

Needed speed based on remaining distance:

∆∆

40mi0.5hr

____________mi/hr

or (circle one)

If , how long will it take to arrive?

∆∆

_____________hr

If , what will Susan have to change her speed to?


1.WhatspeedwouldSusanneedtoarriveexactlyontime?

OptionalPracticeProblems:2.21,2.43,2.47



55




Theaccelerationduetogravityinfreefallisabout9.8m/s2.Atypicalspeedforanarrowshotfromabowis76.2m/s.Ifsuchanarrowisshotstraightup,andairresistanceisneglected,howhighwoulditgo?


In our case v = _______________, v0 = ________________,

a = __________ (watch the sign!), x – x0 is the height.

Now, if we eliminate the variable t between

½ and

we get 2

CALCULATION(Fillintheblanks.)

height = ________________________= 296 m

That is almost 0.2 mile and probably unrealistic.


1.PerformthederivationintheSTRATEGYsection.

2.Whichquantitiesin 2 arepositive,whicharenegative?

OptionalPracticeProblems:2.37,2.51,2.61,2.69



56




Twovectorsare 7 4 and 3 2 .Whatisthevector ? STRATEGY(Fillintheblanks.)

Perform this subtraction by dealing with the x-components and the y-components separately. IMPLEMENTATION(Fillintheblanks.)

The Ax and Bx components are: Ax =___________ Bx = ____________ The Ay and By components: Ay =___________ By = ____________ CALCULATION(Supplytheneededsigns,ornumbers)

______3 7and 2_______4

___________ ___________ SELF‐EXPLANATIONPROMPTS

1.Whatisthemagnitudeofthevector ?2.Howdowehandlethesubtractionofa“negative”component,likethe“ 4/3 ”?OptionalPracticeProblems:3.11,3.14,3.31



57




Avintagebomberparticipatinginanairshowwantstodropabombthatstaysintheairfor15secondsbeforeimpact.Ifthehorizontalvelocityoftheplaneis75m/s,determinetherequiredlaunchaltitude.


Let’s set the ________ at the vertical point where the bomb is __________. We will use written in the _______ dimension, using _______ for the time of flight. The v0 will still be ______ in the vertical, and the __________ will still be ____ 9.8 m/s2. CALCULATION(Fillintheblanks)

12

______m 0 _____s12

9.8m/s 15s 1103m


1.Howaretheflighttimesinthehorizontalandtheverticaldirectionsconnected?2.Howwouldaninitialvelocityintheverticalaffecttheanswerinthisproblem?3.Howisthenegativedirectionofgravity’seffectaccountedforinordertoresultinapositivevaluefory0?OptionalPracticeProblems:3.33,3.62



58




Thishammerthrowerreleasesthehammerballwithatangentialspeedof21m/swhentheballis1.8mfromthecenteroftheathlete’srotation.a)Whatisthecentripetalaccelerationoftheballattheinstantitisreleased?b)Howdoesthisaccelerationcomparetotheaccelerationduetogravity?


To solve this problem we use the UCM basic relationship acentripetal =___________________. CALCULATION(Fillintheblanks.)

acentripetal = _____________________________ = 245 m/s2

acentripetal is directed ________________ and is __________ times larger than the acceleration due to gravity, which is directed _____________________.


1.Whatobjectprovidestheaccelerationofthehammerball?

2.Whatisthedirectionofthenetaccelerationofthehammerballjustbeforeitisreleased?




59

Notes



60

Notes



61

Lesson9

ForcesandNewton’sLawsofMotion

Reading 4.1– 4.4Examples 4.1,4.2HomeworkProblems 4.15,4.26,4.60

LearningObjectives




Notes



62

WorkedExamples



Atowtruckispullingadisabled1200‐kgcaralongalevelroad.Thetow‐ropeisparalleltotheroad.Startingfromrest,thespeedincreasesto2m/sovera20metersdistance.Whatisthetensionintherope?Assumefrictionisnegligible.

STRATEGY

Newton’s Second Law as applied to the car states that the acceleration of the car is given by . We know the mass of the car, and we can use kinematics to find the acceleration of the car. Newton’s Second Law can then be used to obtain the net force. IMPLEMENTATION

To get the net force, we multiply the acceleration of the car, obtained from the kinematics equation 2 , by the mass of the car. CALCULATION

2

1200kg4 0 m /s2 20m

120N

The force unit kg·m/s2 is called a newton, N, in honor of Isaac Newton.

Score(3)



63

SELF‐EXPLANATIONPROMPTS1.TheunitofforceisanewtonwhichisgiventhesymbolN.ExpressthenewtonintermsofthefundamentalSIunits.

2.Findthedefinitionof“tensioninarope”inyourtextbookandrephraseitinyourownwords.

3.Whatchangewouldyoumakeinthecalculationifthetow‐ropewasdirectedatanangle?



64

Pre‐ClassProblem

A45‐ggolfballatrestishitbyaclubwithaforceof5.0N.a)Whatistheball’saccelerationimmediatelyafteritishit?b)Howfardoestheballtravelinthefirsttenthofasecond?

Answer:110m/s2,0.56m



65

PreflightQuestions

1. Whattopicsdidyoufindmostchallengingfromthereading?2. A200‐kgrockisbeingpulledupwardwithanaccelerationof3m/s2.Thenetforceontherock

isa) 200Nupb) 200Ndownc) Zerod) Noneoftheabove.

3. Thenetforcevectorforanobjectinmotionis

a) alwaysinthesamedirectionastheobject'saccelerationvector.b) sometimesinthesamedirectionastheobject'saccelerationvector.c) alwaysinthesamedirectionastheobject'svelocityvector.d) alwaysinthesamedirectionastheobject'sdisplacementvector.

4. CRITICALTHINKING:Thetake‐offmassofanF‐16is16,875kg.Itsenginecanexertaforceof

105,840N.IfyoumountedtheF‐16engineonacar,whataccelerationwouldyouget?Useareasonableestimateforthemassofacarandexplainhowyouobtainedyouranswer.



66

HomeworkProblems

4.15



67

4.26



68

4.60



69

Lesson10

UsingNewton’sLaws

Reading 4.5,4.6Examples 4.3,4.4,4.5HomeworkProblems 4.34,4.47,4.49


LearningObjectives



[Obj21] Constructfree‐bodydiagrams usingvectorstorepresentindividualforcesacting onanobject,andevaluatethenetforceusingvectoraddition.



Notes



70

WorkedExamples



Ateamofdogsispullingtwoconnectedsledswithaconstantaccelerationof2.3m/s2.Thepassengersled,connectedtothedogsinfront,hasamassof96kg.Thecargosled,tiedtothefrontsled,hasamassof42kg.Fornow,weassumethattheretardingfrictionismuchsmallerthantheforceexertedbythedogs.a)Howmuchistheforcethatthedogsexertonthesledtrain?b)Withwhatforceisthecargosledpullingbackonthepassengersled?

STRATEGY

The accelerating sleds are subject to Newton’s Second Law, which states that the acceleration of an object is proportional to the applied force and inversely proportional to the mass of the accelerating object. To answer part (a), we apply the law to the sled train with the combined mass of 138 kg and solve the resulting equation for the unknown applied force. Newton’s Third Law states that when two objects are connected and the first one exerts a force on the second one, the second one responds with a reaction force of the same magnitude, acting back on the first one. Since we know the mass and the acceleration of the cargo sled, we can determine the applied force exerted on the cargo sled by the passenger sled. It is the passenger sled that pulls the cargo sled, not the dogs directly. The reaction force exerted by the cargo sled on the passenger sled has the same magnitude as the force exerted by the passenger sled on the cargo sled and is pulling back on it.

Score(3)



71

IMPLEMENTATION

Let’s label the force exerted by the dog team on the sled team. Let’s label the force exerted by the passenger sled on the cargo sled . Let’s label the force exerted by the cargo sled on the passenger sled . CALCULATIONFor each part we apply Newton’s Second Law .a) 2.3m/s

320 in the forward direction

b) 2.3m/s 97 in the forward direction

The dogs pull the sled train forward with a force of 320 N. The cargo sled pulls back on the passenger sled with a force of 97N.


1. RephraseNewton’sSecondLawinyourownwords.2. Whatisthenetforceonthepassengersled?3. Ifthecargosledwasremoved,howdoyouexpecttheforceappliedbythedogteamtochange

inordertoobtainthesameaccelerationof2.3m/s2forjustthepassengersled?Calculatetheforceexertedbythedogteam forthisscenario.



72

Pre‐ClassProblem


A12‐kgchildisridingina4100‐kgelevatorwhichisacceleratingupwardataconstant1.3m/s2.Whatistheforcethattheelevatorexertsonthechild?Whatistheforcethechildexertsontheelevator?

Free‐BodyDiagram(required)

Answer:133N,‐133N

Tryit!(1PFpt):Ifthechildwasstandingonascaleintheelevator,whatwouldthescalereadwhentheelevatorwas(a)stationaryand(b)acceleratingupwardat1.3m/s2?Showallyourwork.



73

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Twoforcesofequalmagnitudeactonthesameobject.Whichofthefollowingmustbetrue?

a) Theobjectismoving.b) Theobjectisaccelerating.c) Iftheobjectisinitiallyatrest,itcannotremainatrest.d) Thetwoforcesformathird‐lawpair.e) Noneoftheabove.

3. Twoblocksarehangingmotionlessfromtheceilingasshowninthe

diagram.Whichofthefollowingistrue?

a) b) c) d) onlyif

4. CRITICALTHINKING:Theterm”weight”inphysicshasthefollowingveryspecificmeaning:

“Theweightofanobjectisthenamegiventoaparticularforce:thegravitationalforceexertedbytheearthontheobject,givingitanaccelerationof9.8m/s2nearthesurfaceofEarth.”Inordinaryspeechtheuseof“weight”isnowherenearlysoprecise.Explainwhetherthefollowingusagesarescientificallycorrect.

a) A3‐kgobjecthasaweightofabout30NatthesurfaceofEarth.b) A120‐lbpersonweighsabout55kg.c) AnastronautorbitingEarthexperiencesweightlessnessd) Ifyoueattoomuchyoumaygainweight.



74

HomeworkProblems

4.34



75

4.47



76

4.49



77

Lesson11

Newton’sLawsinTwoDimensions

Reading 5.1Examples 5.1,5.2HomeworkProblems 5.16,MP,5.38

LearningObjectives


Notes



78

30°

50kg

WorkedExample



A50‐kgblockisonafrictionless30°ramp.Determinetheblock’saccelerationdowntheramp.STRATEGY

Newton’s Second Law as applied to the block states that the acceleration of the block is given by . In this problem, we have two forces acting on the block: weight and the normal force . These forces act in the x– and y-directions, so we need to separate each force into its components. Once in component form, we can sum the forces in each direction and apply Newton’s Second Law to find acceleration. Because the motion of the block is along the incline, we “tilt” the coordinate system of our free-body diagram to align with the incline of the ramp and the normal force that is acting on the block. IMPLEMENTATION

Let’s draw a free-body diagram for our object of interest: the block. There are two forces acting on the block, weight and normal force, that are included in the diagram. Since the block is moving down the ramp, we use a tilted coordinate system. The net force on the block in the x-direction is:

sin

The net force on the block in the y-direction is:

cos

Score(3)



79

CALCULATION

The acceleration in the y-direction (perpendicular to the incline as defined by our coordinate system) is zero. To find the acceleration of the block, we need to solve for the acceleration in the x-direction . Cancelling mass in the net force equation above gives:

sin 9.8m/s sin 30° 4.9m/s


1.Explainwhytiltingthecoordinatesystemsimplifiedtheproblem.Thinkabouthowtheprocedurewouldhavechangedhadtraditionalx‐ycoordinatesbeenused.

2.Wouldtheanswerhavechangedhadthecoordinatesystembeenswitched,sothepositivex‐axiswasdefinedasbeinguptheramp?

3.Whatwouldhappentothemagnitudeoftheblock’saccelerationiftheangleoftherampwasincreased?Whatisthemaximumaccelerationtheblockcanexperience?Whatistheminimumaccelerationtheblockcanexperience?



80

Pre‐ClassProblem


A3.0‐kgboxissuspendedfromaceilingasshown.Whatarethemagnitudesofthetensionsexertedbytheropesattachedtothebox?Assumetheropeshavenegligiblemasscomparedtothebox.(Hint:LookatExample5.2inthetextbook.Whyisiteasiertouseatraditionalx—ycoordinatesystemratherthantiltedforthisproblem?)

Answer:T1=25N

T2=11N

Free‐BodyDiagramofBox(required)

m

68° 32°CEILING



81

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?

2. A5‐kgblockispushedacrossahorizontalfloorwitha20‐Nforcedirected20°belowthehorizontal.Whatisthemagnitudeofthenormalforceontheblock?

a) 49Nb) 6.8Nc) 42Nd) 56Ne) 68N

3. IfRope1remainshorizontalandthepointatwhichRope2istiedismovedfrom to ,whatistrueaboutthetensionintheropes?

a) remainsthesameand increases.

b) decreasesand increases.

c) Both and remainthesame.

d) Both and increase.4. CRITICALTHINKING:Refertopreflightquestion3:IsitpossibletoattachRope2atpointCand

havebothropesparalleltotheground?Explain.

Rope1

5kg

20N

20°



82

HomeworkProblems

5.16



83

MP



84

5.38



85

Lesson12

Newton’sLawswithMultipleObjects

Reading 5.2Examples 5.4HomeworkProblems 5.19,5.21,5.71

LearningObjectives


Notes



86

WorkedExamples



A2,500‐kgtractorispullinga750‐kgcowoutofaravine,asshown.Ifthetractorappliesaforceof20kN,determinetheaccelerationofthecowoutoftheravine.Assumetheropeandpulleyaremasslessandtheropedoesnotstretch.

STRATEGY

There are multiple components in this problem (tractor, rope, pulley, and cow), so we need first to determine which objects are of interest. Once we have identified the objects of interest, we will draw free-body diagrams for each and apply Newton’s Second Law (N2L).

IMPLEMENTATION

For this problem, we are only interested in the tractor and the cow since the rope and pulley are massless. Let’s draw free-body diagrams for each object and apply Newton’s Second Law, summing the forces acting on each object. This operation will give us separate equations that include forces and accelerations. Since the objects are connected by a massless rope that does not stretch, the magnitudes of the tensions and accelerations are the same. We can then solve for the unknown acceleration. CALCULATIONThe net force on the tractor in the x-direction is:

Score(3)



87

The net force on the cow in the y-direction is:

Combining these two equations gives:

Solving for acceleration:

Substituting in values gives an acceleration of:

20000N 750kg ∙ 9.8

750kg 2500kg3.9m/s


1. Explainwhytheaccelerationofthetractorinthex‐directionisthesameastheaccelerationofthecowinthey‐direction.

2. Explainwhytheweightofthecowisanegativequantity.

3. Ifthetractorcouldonlyapplya2kNforce,calculatetheaccelerationofthecow.Describethemotionofthecow+tractorsystemforthisscenario.



88

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA10‐kgcartisconnectedbyastringtoa10‐kgweightoverapulley.Assumingthatthemassesofthestringandthepulleycanbeneglected,findtheaccelerationofthecartandthetensioninthestring.

Free‐BodyDiagramoftheCart(required)

Free‐bodyDiagramoftheWeight(required)

Answers:4.9m/s2,49N



89

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Abucketattachedtoaropeisraisedoutofawellataconstantspeed.Whatcanbesaidabout

thetensionintheropecomparedtotheweightofthebucket?

a) Tensionislessthantheweightofthebucket.b) Tensionisequaltotheweightofthebucket.c) Tensionisgreaterthantheweightofthebucket.d) Cannotbedeterminedfromthegiveninformation.

3. InCase1,BlockBacceleratesBlockAacrossafrictionlesstable.InCase2,aforceof98N

acceleratesBlockAacrossthesametable.TheaccelerationofBlockAis

a) zero.b) greaterinCase1.c) greaterinCase2.d) thesameinbothcases.

4. CRITICALTHINKING:Whenyouareinanelevatoryouoftenfeelalittlelighterastheelevator

startstomovedownward.ExplainthisfeelingbasedonNewton’sLaws.

A10kg

B10kg

A10kg

Case1 Case2

98 N



90

HomeworkProblems

5.19



91

5.21



92

5.71



93

Lesson13

Lab3–Newton’sLaws

Reading 5.2,Lab3HandoutExamples 5.4HomeworkProblems MP,MP,MP


LearningObjectives


Notes



94

Pre‐LabQuestions

1. Brieflydescribewithoneortwocompletesentencesthepurposeandgoalsofthislab.

2. Constructfree‐bodydiagramsform1andm2forthefollowingscenario.

Free‐BodyDiagram:Mass1 Free‐BodyDiagram:Mass2

3. UseNewton’ssecondlawtoderiveanexpressionfortheaccelerationofthemassesintermsofm1,m2,θ,andg.

Score(5)



95

LabNotes



96

HomeworkProblems

MP



97

MP



98

MP



99

Lesson14

Newton’sLawsinCircularMotion

Reading 5.3Examples 5.5,5.6,5.7HomeworkProblems 5.65,5.73,MP

ThereisanEXAM‐PREPQUIZthislesson.

LearningObjectives




Notes



100

WorkedExample



Anamusementparkrideconsistsofaverticalloopwhosediameteris15mandasmall150‐kgcartthatrunsontheinsidetrackintheloop.Therideisdesignedtocarryamaximumloadof320kg.

Thecartisgivenaninitialspeedatthebottomofthetrackandisnotpropelledfurther.Whenthecartclimbsverticallytothe90°point,itsspeedis12.4m/s.Whatisthemagnitudeanddirectionofthenetforceonthecartatthispoint?

STRATEGY

The cart is subject to the force of gravity, which is equal to its mass times the acceleration due to gravity, 9.8 m/s2 vertically down. The cart also is subject to a normal force from the track that is directed towards the center of the loop and acts like a centripetal force. We add the two force vectors to obtain the net force. IMPLEMENTATION

Normal force: directed horizontally to the left. Force due to gravity (weight) directed vertically downward. 1. The magnitude of the net force is

2. The direction of the net force is at an angle

tan tan

below the horizontal. The net force is causing the cart to slow down as it climbs to track.

Score(3)



101

CALCULATION

1. 3410

2. θ = 25.4 degrees below horizontal, to the left towards the center of the loop.


1. Drawfree‐bodydiagramsofthecartwhenitisatthebottomandtopofthetrack.

2. Doesthecarttravelaroundtheloopataconstantspeed?Explain.

3. Describehowtheweight,normalforceandnetforcechangeasthecartmovesaroundthetrack.



102

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA50‐kgwrecker’sballishangingonan8‐mropethatcansupportamaximumforceof1000N.Iftheballisswunginaverticalcircle,whatisfastestspeeditcanhaveatthelowestpointsuchthattheropewon’tbreak?

Answer:9.0m/s



103

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Anobjectmovesataconstantspeedinacircularpath.Theinstantaneousvelocityandthe

instantaneousaccelerationvectorsare

a) bothtangenttothecircularpath.b) bothperpendiculartothecircularpath.c) perpendiculartoeachother.d) oppositetoeachother.e) noneoftheabove.

3. Aballonastringmovesaroundaverticalcircle.Atthebottomofthecircle,thetensioninthestring

a) isgreaterthantheweightoftheball.b) islessthantheweightoftheball.c) isequaltotheweightoftheball.d) maybegreaterorlessthantheweightoftheball.

4. CRITICALTHINKING:Thefigureshownisaviewlookingdownonahorizontaltabletop.Aball

rollsalongthegraybarrierwhichexertsaforceontheball,guidingitsmotioninacircularpath.Aftertheballceasescontactwiththebarrier,describethemotionoftheballandyourreasoning.



104

HomeworkProblems

5.65



105

5.73



106

MP



107

Lesson15

Newton’sLawswithFriction

Reading 5.4,5.5Examples 5.9,5.10,5.11HomeworkProblems 5.43,MP,5.57

LearningObjectives


[Obj25] Describedragforcesqualitatively andquantitatively.

Notes



108

WorkedExample



Ateamofdogsispullingtwoconnectedsledswithanaccelerationof2.3m/s2.Thepassengersled,connectedtothedogsinfront,hasamassof96kg;thecargosled,tiedbehindthepassengersled,hasamassof42kg.Thecoefficientofkineticfrictionbetweenthesteelrunsonthesledsandtheiceisμ=0.007.Howmuchforcedoesthedogteamexertonthesledtrain?

STRATEGY

The accelerating sleds are subject to Newton’s Second Law (N2L), which states that the acceleration of an object is proportional to the applied force and inversely proportional to the mass of the accelerating object. We apply N2L to the sled train with the combined mass of 138 kg and solve the resulting equation for the unknown applied force, including the frictional force which acts opposite the direction of motion. IMPLEMENTATION

Let’s label the force exerted by the dog team d. Let’s label the force exerted on the cargo sled by passenger sled cp. Let’s label the force exerted on the passenger sled by cargo sled pc. Let’s label the force exerted by the kinetic friction k on both sleds. CALCULATION

The net force on the sled team in the x-direction is:

The net force on the sled team in the y-direction is:

Score(3)



109

The acceleration in the y-direction is zero, so the normal force is:

The kinetic force is given by: f μn

Substituting into the equation of the net force in the x-directions gives:

F μw m a CALCULATION

F 0.007 96 42 kg 96 42 kg 2.3m/s F 320N in the forward direction.


1. InLesson10wesolvedthesameproblem,butwithoutfriction.Explainhowthemethodchangeswhenfrictionisincluded.

2. Explainwhykineticfrictionwasusedintheproblemratherthanstaticfriction.

3. Describethestepsusedtodeterminethekineticfriction.



110

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA60‐kgblockisreleasedfromrestona45°rampwherethecoefficientoffrictionbetweentheblockandrampis0.4.Whatistheaccelerationoftheblock?

Answer:4.2m/s2

Tryit!(1pt):Determinethespeedoftheblockatthebottomoftherampifitstartsfromrestatthetopofthe3‐mlongramp.Showyourwork.



111

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Whichstatementconcerningfrictionistrue?

a) Staticfrictionisalwaysoppositethedirectionofmotion.b) Kineticfrictionisalwaysoppositethedirectionofmotion.c) Bothstaticandkineticfrictionarealwaysoppositethedirectionofmotion.d) Neitherisalwaysoppositethedirectionofmotion.

3. Aboxisatrestontheflatbedofamovingtruck.Dawnappliesthebrakesabruptlyandtheboxbeginstoslide.Whichfree‐bodydiagramcorrectlydepictstheforcesactingontheboxanditsresultingmotion?

4. CRITICALTHINKING:Describe,inyourownwords,thedifferencebetweenstaticfrictionforcesandkineticfrictionforces.

a)

b)

c) d)



112

HomeworkProblems

5.43



113

MP



114

5.57



115

Lesson16

CriticalThinking:Newton’sLawswithNon‐constantMass

Reading 9.3Application,HandoutExamples NoneHomeworkProblems 5.30,5.62,6.54

LearningObjectives


Notes



116

Notes



117

PreflightQuestions

1. Whattopicsdidyoufindmostchallengingfromthereading?2. Arocketliftsofffromthelaunchpadandrisesmajesticallyonitsflight.Thethrustoftherocket

resultsfrom

a) theexhaustgasespushingagainsttheground.b) theexhaustgasespushingagainsttheair.c) thecombustiongasespushingagainsttherocket.d) theequalandoppositereactiontogravitypullingdown.e) thegravitationalenergyreleasedbyburningfuel.

3. Atsomepointbeyondatmosphericspaceshuttleflight,the3‐mainenginesstopproviding

thrustandthentheboostertankSEPARATESfromthecraft.Whentheconnectionbetweenthetwoobjectsissevered,thevelocityoftheshuttle

a) increases.b) decreases.c) remainsunchanged.d) Theanswerdependsonthemassofthebooster.

4. CRITICALTHINKING:Thespaceshuttleassemblyonthelaunchpadhasamassofabout2

millionkg.Theexhaustvelocityofthepropellantgasesisabout4000m/s.Thegasesarestreamingoutofthenozzlesattherateofabout18,000kg/s.Giventhisinformation,estimatetheaccelerationofthespaceshuttleassembly.



118

HomeworkProblems

5.30



119

5.62



120

MP



121

Lesson17

GRADEDREVIEW2

LearningObjectives




[Obj21] Constructfree‐bodydiagrams usingvectorstorepresentindividualforcesacting onanobject,andevaluatethenetforceusingvectoraddition.




[Obj25] Describedragforcesqualitatively andquantitatively.

Notes



122




A1000‐kgcaristravelingat10m/swhenabrakingforceof500Nisapplied.Howmuchtimedoeselapsebeforethecarcomestoacompletestop?


Newton’s Second Law as applied to the car states that the acceleration of the car is given by . We know the mass of the car and the net force, so we can get the deceleration of the car applying Newton’s Second Law. We can then use kinematics to find the stopping time. CALCULATION(Fillintheblanks.)

0.5 .

___________

20


1.Comparethisexampletothetow‐truckexample,stepbystep.2.Wecalculatedtheaccelerationtobe–0.5m/s2.Whatdoestheminussignindicateaboutthecar’sacceleration?OptionalPracticeProblems:4.13,4.15,4.23



123




OnJuly16,1969,aSaturnVrocketliftedoffthepadinFloridaonmankind’sfirsttriptothesurfaceoftheMoon.Thefully‐loadedrockethadamassof2.8x106kg.Topropelitselfupwarditgenerated34.5x106Nofthrust.Whatwastheinitialaccelerationoftherocket?STRATEGY(Fillintheblanks.)

We calculate the acceleration by dividing the net force on the rocket by its mass, / . There is an upward force on the rocket from the thrust of its engines, and a downward force, the weight of the rocket, from gravity acting on the rocket. The net upward force is therefore thrust minus weight. CALCULATION(Fillintheblanks.)

The weight of the rocket is ________ 27,440,000N The net upward force on the rocket is

__________ __________ 7,060,000N The initial acceleration of the rocket is

2.52m/s .


1.Whatwouldbetheaccelerationofarocketofthesamemassifitstartedfromrestinemptyspace,awayfromobjectsthatexertgravitationalforceslikeEarth?2.Furtherintothelift‐off,wouldyouexpecttheSaturn’saccelerationtoincrease,decrease,orremainthesame?3.WhatmagnitudeofthrustwouldmaketheSaturnjusthover,withnoacceleration?OptionalPracticeProblems:4.27,4.37,4.45



124




A5‐kgballissuspendedfromthreeropesasshowninthepicture.Whatistheforceexertedonthewallbythehorizontalrope?STRATEGY(Fillintheblanks.)

Apply Newton’s Second Law to the junction of the three ropes. The system is not accelerating, so the vector sum of the forces is zero. Decompose the forces into components and solve for the unknown force.

IMPLEMENTATION(Fillintheblanks.)

The forces acting on the vertical rope are: ________, _______, and _______. The net force in the x-direction is:

_____________________

The net force in the y-direction is:

____________________

Since _______ = 0, the equation relating the forces , , and the weight of the ball is: _______ ________ 0

CALCULATION

Solving the two equations gives us T1 = 8.00 N SELF‐EXPLANATIONPROMPTS

1.IsitpossibletosuspendtheballinthisexampleinsuchawaythatbothforcesT1andT2havehorizontalcomponentsonly?Explain.2.Arethemagnitudesofanyofthetensionsinthethreeropeslargerthantheweightoftheball?


58°



125




Twoballswithmasses,M1andM2,areconnectedbyaropewhichpassesoverapulleyasshown.Findtheaccelerationsoftheballsastheyarereleasedfromrest.Assumethattheropedoesnotstretchandthemassesoftheropeandthepulleyarenegligiblecomparedtothemassesoftheballs.


We draw free-body diagrams for the two balls and apply Newton’s Second Law to each. Since the rope does not stretch, the magnitudes of the balls’ accelerations are the same


The net force in the x-direction is for M1 is :

______________________ The net force in the y-direction for M2 is:

∑

_____________________

3. The eqns in 1 and 2 above have two unknowns: a and T. Combining these equations and eliminating tensions, gives


1.Explaininyourownwordswhythemagnitudesoftheaccelerationsofthetwoballsarethesame.

2.Whyisthemagnitudeofthetensionintheropeontheleftsideofthepulleythesameasthemagnitudeontherightsideofthepulley?




126




A1300‐kgcarisroundingacurveonaflathorizontalroadway.Thecaristravelingat13.4m/sandslowingdownat2m/s2.Theradiusofthecurveis30meters.Thecoefficientofstaticfrictionis

0.80,andthecoefficinetofkineticfrictionis 0.40.

Whatisthenetforceofthecar,magnitudeanddirection?


The car is slowing down which means it has a force directed opposite to its motion. Since this direction is tangent to the road, it is called the tangential force .

The car is also changing direction which means it has a radial force caused by friction between the tires and the road, . This force is directed towards the center of the curve.

The net force is the vector sum of . and . CALCULATION(Fillintheblanks.)

1. Tangential force is __________ .

2. Force of friction is ___________

3. Net force is √_____ _____

The direction is _____________________________.


1.Whydidwenotneedcoefficientoffrictionforthisproblem?

2.Whatisthemagnitudeanddirectionofthenetforceifthecarroundsthecurveatconstantspeed?




127




Amanispushinga50‐kgcartthatacceleratesat1.3m/s2onlevelgroundwherethecoefficientoffrictionbetweenthewheelsandthegroundis0.03.Howmuchresistancefromthecartdoeshefeel?


First, we apply Newton’s Second Law to determine the force needed to accelerate the cart. Then, we use Newton’s Third Law to determine the frictional force. CALCULATION

The net force in the x-direction for the cart is:

_____________________

The net force in the y-direction for the cart is:

______________________

The force exerted on the cart by the man = _____________ x _____________ = 79.7 N in the forward direction. The force exerted on the man by the cart = _____________ in the ____________ direction.


1.Whydon’ttheforceonthecartandtheforceonthemancancelout?Thatis,whydoesthe

mathematicallycorrectstatement, 79.7 – 79.7 0,notimplythatthenetforceintheabovescenarioiszero?

2.Whattypeoffrictionalforceactsonthecart:kineticorstatic?Explain.




128

Notes



129

Lesson18

WorkwithConstantandVaryingForces

Reading 6.1,6.2Examples 6.1– 6.5HomeworkProblems 6.18,6.20,6.52


LearningObjectives



Notes



130

WorkedExample



Amandragsa50‐kgcrate10macrossaroughhorizontalsurface,wherethecoefficientofkineticfriction,μ ,betweenthecrateandthesurfaceis0.3.Hepullsatconstantspeedanddirectshispullingforce20°upwardfromthehorizontal.Howmuchworkdoesheperform?STRATEGY

Work done by a force is defined as the dot product of the applied force and the displacement: ⋅ ∆ ∆ ,where θ is the angle between the direction of the force vector and the direction of the displacement vector.

To find the work done by the man, we find the force he applies to the crate, the displacement, and θ, then compute the dot product between work and displacement. IMPLEMENTATION

Since the crate is moving at constant speed, (acceleration is zero), the net force on the crate must be zero. The net force on the crate is the vector sum of the force applied by the man and the force of kinetic friction .

Note that the force of friction depends on the direction of the man’s force because the man’s force affects the normal force (unless he pulls horizontally.)

The force of kinetic friction = (coefficient of friction) (normal force)

Since there is no acceleration, the x-component of Fm must equal fk, Thus,

Score(3)

Free‐BodyDiagramofCrate



131

Solving for Fm we get

cosθ

The work done by Fm is then ∙ ∆ Δ

CALCULATION

cosθ

Δ 1325J


1.Startwiththedefinition ∙ ∆ ∆ cos andexplainhowWcanbepositive,negative,orzero.

2.Explainwhatitmeanstohave(a)positiveWand(b)negativeW.

3.Intheexample,youaretoldthatthenormalforceis: .Describethestepsneededtoobtainthenormalforceandthenshowthecalculation.



132

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMAcranelowersa120‐kgrockatconstantspeedthroughaverticaldistanceof5meters.Howmuchworkdoesthecraneperform?

Free‐BodyDiagramofRock(required)

Answer:‐5880J



133

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Solveforwork andrankorderfromsmallest(negative)tolargest(positive)theworkdonein

thefollowingcases:

RankOrder:Smallest(1)_____(2)_____(3)_____(4)_____Largest

3. Twoidenticalobjectsareeachdisplacedthesamedistance,oneby

aforce pushinginthedirectionofmotionandtheotherbya

force2 pushingatanangle relativetothedirectionofmotion.Theworkdonebythetwoforcesisthesame.Whatistheangle ?(Hint:SeeGOTIT?6.1.)

a) 0°b) 30°c) 45°d) 60°

4. CRITICALTHINKING:Aweightlifterpicksupabarbelland(1)liftsitchesthigh,(2)holdsitfor

30seconds,and(3)putsitdownslowly(butdoesnotdropit).Rankorderfromsmallesttolargestthework theweightlifterperformsduringthesethreeoperations.Labelthequantitiesas , ,and .Justifyyourrankingorder.

2

112°

10N

4 m

2 m

10 N

2 m

10 N

32°

2m

10N

CaseA CaseB CaseC CaseD



134

HomeworkProblems

6.18



135

6.20



136

6.52



137

Lesson19

KineticEnergyandPower


LearningObjectives




Notes



138

WorkedExample



A3,000‐kgsailboatistravellingat25m/swhenaconstantnetforceof1200Nstartsactingonit,inthedirectionofmotion.Whatisthespeedoftheboatafterithastravelled200mundertheactionofthisforce?

STRATEGY

When a force acts on a moving object, work is done on the object. The work done on the object results in the change of the object’s kinetic energy K, defined as

12

where m is the mass of the object and v is its speed. The net work done on the object and the change in the kinetic energy are related by the work-energy theorem

∆12

12

To find the answer to the question posed in the problem: 1. we find the net work done on the boat, 2. set it equal to the change in kinetic energy of the boat, and 3. solve the resulting equation for the unknown final speed. IMPLEMENTATION1. Net work: ∙ Δ ∆ cos

2.

3.

Score(3)



139

CALCULATION

1200N 200m cos 0° 240,000Nm 240,000J joules

480,000J 3000kg 25

ms

3000kg28

ms


1. Statethework‐energytheoreminyourownwords.

2. TheworkdonebyaconstantforceF,actingalongthedirectionofmotionoveradistanceΔxequals .Fromkinematics,weknowthatifanobjectstartsfromrestandaccelerateswithaccelerationaoveradistanceΔx,2 ∆ ;andfromNewton’sSecondLaw,weknowthat

.Combinethethreeequationsandshowthattheworkdonebytheforceequals .

3. Usethesameprocedureasabovetoshowthattheworkdonetoincreasethespeedofmassmfromv1tov2isequaltothechangeinitskineticenergy.



140

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMGalileoissaidtohavedroppedtwoobjectsofdifferentmassfromatalltowertoshowthatallobjectsfallwiththesamespeed.Ifyoudroptwomasses,m1andm2,fromthesameheighth,dotheyreachthegroundwiththesamekineticenergy?CalculatethedifferenceintheirkineticenergiesΔK.

Answer:∆



141

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Twocars,onefourtimesasheavyastheother,areatrestonafrictionlesshorizontaltrack.

Equalforcesactoneachofthesecarsforadistanceofexactly5m.Thekineticenergyofthelightercarwillbe_______thekineticenergyoftheheaviercar.

a) one‐quarterb) one‐halfc) equaltod) twicee) fourtimes

4. Whichofthefollowingistrue?

a) NeitherΔKnorWnetcaneverbenegative.b) Wnetcanneverbenegative,butΔKcanbenegativeorpositive.c) ΔKcanneverbenegative,butWnetcanbenegativeorpositive.d) ΔKandWnetcanbenegativeorpositive.

5. CRITICALTHINKING:OnMondayyourunupthestairstothetopfloorofatallbuilding.You

runataconstantspeed.OnTuesdayyouwalktothetop,alsoatconstantspeed.OnWednesdayyoutakeaconstantspeedelevator.Howdotheamountsofworkyoudidgettingtothetopofthebuildingeachdaycompare?Howdoesthepowercompare?



142

HomeworkProblems

6.29



143

6.64



144

6.71



145

Lesson20

PotentialEnergy

Reading 7.1,7.2Examples 7.1,7.2HomeworkProblems 7.14,7.31,7.42

LearningObjectives

[Obj30] Explainthedifferences betweenconservativeandnonconservativeforces.



[Obj33] Evaluatethepotentialenergyassociatedwithaconservativeforce.

Notes



146

WorkedExample


STATEMENTOFTHEPROBLEMAverticalspringwithaspringconstantk=150N/miscompresseddown1.5m.A2‐kgballisplacedonthecompressedspringandreleasedfromrest.Whatheightdoestheballreachafteritisreleased?STRATEGY

This problem involves two examples of potential energy: the elastic energy of a compressed (or stretched) spring and the gravitational potential energy as an object moves from one elevation to another. An object is said to possess potential energy if, because of its condition, it can generate kinetic energy. A ball on a spring, for example, can be propelled by the force of the spring and gain kinetic energy. A ball can be dropped from a height and be propelled by the force of gravity and gain kinetic energy. Potential energy is traditionally denoted by the symbol U.

The change in the elastic potential energy as a spring’s vertical extension changes from y1 to y2 is given by

Δ12

12

The change in the gravitational potential energy as an object of mass m moves from height y2 to height y3 is given by

Δ

We solve the problem by comparing the energy imparted to the ball by the compressed spring to the energy lost by the ball as it climbs against the force of gravity. Symbolically

Δ ⇒ ⇒ Δ IMPLEMENTATION

12

12

Score(3)



147

CALCULATION

First, we are given the following quantities: y1 = 0 m y2 = − 1.5 m k = 150 N/m m = 2 kg g = 9.8 m/s2 Let’s set up a vertical coordinate axis with y1 = 0 at the position of the unstretched spring. Now, we solve for y3 – y2 in

12

12

8.6m

The ball rises 8.6 meters above the top of compressed spring. SELF‐EXPLANATIONPROMPTS1.ThepotentialenergystoredinacompressedspringcomesfromtheworkdonebycompressingthespringagainstitsrestoringforceF=−ky.CalculatethatworkandverifytheaboveexpressionofthespringpotentialenergyUs.2.DothesameforthegravitationalpotentialenergyUg.3.Insolvingtheproblemweignoredthemassofthespring.Includingthemassofthespringismessy,butansweringthefollowingquestionisnot.Howwouldincludingthemassofthespringchangetheoutcomeofthecalculation?



148

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA2‐kgballisreleasedfromrest3metersaboveanunstretchedspringofwhosespringconstantis150N/m.Howmuchdoesitcompressthespringbeforeitcomestorest?(Beforeyoustartcalculating,carefullydrawthecoordinatesystemandcarefullyidentifyalltherelevantverticalcoordinates.Whenyouequatethetwopotentialenergychangesyouwillgetaquadraticequation!)

Answer:1m

Tryit!(1PFpt):Howhighwouldtheballneedtobereleasedifyouwantedtodoubletheamountthatthespringiscompressed?Showyourwork.



149

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Theenergystoredinacompressedspringdependsontheamountofcompression.Agiven

springrequires10.0Jforacompressionof10.0cm.Howmuchtotalenergywouldbestoredifitwerecompressedanadditional5.00cm?

a) 22.5Jb) 12.5Jc) 5.00Jd) 1.25Je) Cannotbedeterminedfromthegiveninformation.

3. Atrunkofmass isliftedalongacurvedpathoflength toaheight .Anothertrunkwithtwicethemassisslidacrossalevelfloor( 0.5)alongacurvedpathalsohavinglength .Whichisgreater,theworkdoneagainstfrictionortheworkdoneagainstgravity?

a) Moreworkisdoneagainstfriction.b) Moreworkisdoneagainstgravity.c) Theworkdoneagainstfrictionisthesameastheworkdone

againstgravity.d) Cannotbedeterminedfromthegiveninformation.

4. CRITICALTHINKING:Whycan’twedefinepotentialenergyforfriction?Explain.

2



150

HomeworkProblems

7.14



151

7.31



152

7.42



153

Lesson21

ConservationofMechanicalEnergy



LearningObjectives



Notes



154

WorkedExample



Note:Beforeworkingwiththisexample,revisitLesson14,whichhasthesameproblemneglectingfriction.

Anamusementparkrideconsistsofaverticalloopwhosediameteris15mandasmall150‐kgcartthatrunsontheinsidetrackintheloop.Therideisdesignedtocarryamaximumloadof320kg.

Ifthecartiscarryingitsmaximumload,howmuchkineticenergymustithaveatthebottomoftheloopifitistonegotiatethetopoftheloopsafely(upsidedown)withoutleavingthetrack?

STRATEGY

In order not to leave the track at the top of the loop, the cart needs to go fast enough so that its weight provides the centripetal force necessary to just keep it on the track.

becomes

As the cart climbs up the loop it loses kinetic energy and gains potential energy. The kinetic energies at the bottom and the top are related to the potential energies by

Since we know the minimal required kinetic energy at top, we can use the energy conservation equation to find KE at the bottom.

Score(3)



155

IMPLEMENTATION

12

12

If we set the potential energy to be zero at the bottom of the track

2

The energy conservation equation then becomes

12

2 0

CALCULATION

12

2.5 86,000J


1. Stateinyourownwordswhatwemeanby“conservationprinciple.”

2. Whycanthezeropointofpotentialenergybechosenarbitrarily?

3. Sketchanenergybarchart(similartothoseinFigure7.8inyourtextbook)forthecartat(a)thetopofthetrackand(b)atthebottomofthetrack.

Energy→

0

Energy→

0



156

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA50‐kgwrecker’sballishangingonan8‐mropethatcansupportamaximumforceof1000N.Iftheballisswunginverticalcircle,whatisfastestspeeditcanhaveatthelowestpoint,suchthattheropewon’tbreak?

Answer:9m/s



157

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Abottledroppedfromabalconystrikesthegroundwithaparticularspeed.To

doublethespeedatimpact,youwouldhavetodropthebottlefromabalconythatis

a) twiceashigh.b) threetimesashigh.c) fourtimesashigh.d) eighttimesashigh.

3. Atruckinitiallyatrestatthetopofahillisallowedtorolldown.Atthebottom,itsspeedis14m/s.Next,thetruckisagainrolleddownthehill,butthistimeitdoesnotstartfromrest.Ithasaninitialspeedof14m/satthetopbeforeitstartsrollingdownthehill.Howfastisitgoingwhenitgetstothebottom?

a) 14m/sb) 17m/sc) 20m/sd) 24m/se) 28m/s

4. CRITICALTHINKING:Askydiverwhoseparachuteisfullydeployedisdescendingatconstant

speed.Describewhatishappeningtoherkineticenergy,herpotentialenergyandhertotalmechanicalenergyasshefalls.Isanyworkbeingdone?Ifyes,wheredoesitgo?



158

HomeworkProblems

7.24



159

7.25



160

7.55



161

Lesson22

Lab4‐ConservationofEnergy



LearningObjectives


Notes



162

Pre‐LabQuestions

Inthislab,aspringiscompressedadistancexandusedtolaunchacartofmassMalongaperfectlyhorizontalairtrack.Thespeedofthecart,v,ismeasuredsomedistancedowntheairtrackandusedtocalculatethespringconstant,k(refertothelabhandoutandExample7.4inthetextbook).

1. Usetheprincipleofconservationofmechanicalenergytofindanexpressionforthespeedofthecartasafunctionofthecompressiondistance.

a)

b)

c) 2

d) 2

2. WhengraphingthedatainPartII,youareaskedtoplotvvs.x.Describetheshapeoftheplotandexplainwhyitmakessensetoplotthedatainsuchaway.

3. Afterplottingvvs.x,yourgroupdeterminesthattheslopeofthebest‐fitlinethroughthedatapointsis50s‐1.Iftheaircarthasamassof0.50kg,thespringconstantkisa) 25N/mb) 50N/mc) 1250N/md) Cannotbedeterminedwiththegiveninformation.

Score(5)



163

LabNotes



164

HomeworkProblems

7.56



165

7.59



166

7.63



167

Lesson23

OrbitalMotion

Reading 8.1– 8.3Examples 8.1,8.2,8.3HomeworkProblems 8.17,8.39,MP


LearningObjectives




Notes



168

WorkedExample



Aspacecraftisorbiting200kmabovethesurfaceoftheplanetMars.Oneoftheastronautsonboarddropsapen.HowfastdoesthepenfallrelativetothesurfaceofMars?Howfastdoesthepenfallrelativetotherestofthespacecraft?

STRATEGY

The spacecraft is in orbit about mars, meaning that it is traveling in a circular path 200 km above the surface of Mars. Since the orbit is circular, the motion of the spacecraft (and everything on board it including the pen) is undergoing centripetal motion; the acceleration is therefore centripetal acceleration. IMPLEMENTATION

1. What is the orbital radius of the spacecraft?

2. What is the gravitational force?

By combining the gravitational force with Newton’s Second Law, we can find the acceleration of the spacecraft and everything on board. CALCULATION

1. 3,389km 200km 3589km

2. Combining the gravitational force, ; with Newton’s Second

law, , we find the magnitude acceleration, 3.32 .

The entire space craft and everything inside is accelerating at the same rate, so the pen will not appear to fall.

Score(3)



169


1.Whichdirectionisthespacecraftacceleratingandtowhichobjectsdoesthevalueofacceleration,

3.32 ,apply?

2.Howisthequantity“r”definedintheequationforuniversalgravitation?Useyourdefinitiontojustifywhy intheexample.

3.Ifthepenisaccelerating,explainwhyitisconsideredtobein“freefall”.



170

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMItiscommonlybelievedthatyouexperienceabrieffeelingofweightlessnesswhenyouareridinganelevator.Consideranelevatorwhichhasafinalspeedof2.3m/s.Inorderforyoutoexperienceafeelingof“weightless”,howlongmustittaketheelevatortogofromresttoitsfinalspeed(assumingconstantacceleration)?Doesthishappenwhentheelevatorisgoinguporgoingdown?

Free‐BodyDiagram(required)

Answer:0.23s,goingdown



171

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. What is the approximate force that the Moon exerts on you when it is directly overhead?

(Hint:YouwillneeddatafromAppendixEtoanswerthisquestion)

a) 2 10 Nb) 2 10 Nc) 2 10 Nd) 2 10 Ne) 2N

3. Themagnitudeoftheforceofgravitybetweentwoidenticalobjectsis .Ifthemassofeach

objectandthedistancearedoubled,whatisthenewforceofgravitybetweentheobjects?

a) b)4 c)8 d) e) 4. CRITICALTHINKING:Ageosynchronousorbitisonewheretheorbitalobjectstaysbasically

overthesameplaceonearthallthetime.Theobjectstaysrelativelymotionlessintheskyabove.ThemassofEarthis5.97x1024kg,andtheperiodisthesameasthatofEarthatT=23hr,56min,4sec.Describehowyouwoulddeterminethealtitudeforgeosynchronousorbit.(Hint:HowareorbitalperiodTandorbitalradiusrelated?)



172

HomeworkProblems

8.17



173

8.39



174

MP



175

Lesson24

GravitationalEnergy

Reading 8.4Examples 8.4,8.5HomeworkProblems 8.27,8.52,MP

LearningObjectives




Notes



176

WorkedExample



A120‐kgsatelliteisinacircularorbit100kmabovethesurfaceoftheEarth.Howwouldthetotalenergyofthesatellitechangeifitweremovedtoahigherorbit200kmabovethesurfaceoftheEarth?

STRATEGY

The total energy (potential and kinetic) of a satellite in a circular orbit about the Earth is

12

where G = 6.67 x 10-11 Nm2/kg2 is the universal gravitational constant m is the mass of the satellite M is the mass of the Earth = 5.97 x 1024 kg r is the radius of the orbit (radius of the Earth + altitude) IMPLEMENTATION

We will calculate the energy in each of the two orbits and subtract to get the change in energy between the orbits.

Δ 12

–12

12

1

1

CALCULATION

6.37 10 100 10 6.47 10 m

6.37 10 200 10 6.57 10 m

Δ126.67 10 Nm /kg 5.97 10 kg 120kg

16.47 10 m

1

6.57 10 m

Δ 3600 10 J

Score(3)



177


1.Howisthegravitationalenergyformuladerived?Lookinthetextandsummarizethestepsforthisderivation.

2.Thegravitationalenergyequationincludesgravitationalpotentialenergy.Whereisthegravitationalpotentialenergyzero?

3.Explainwhyisthetotalenergynegative.



178

apogeeperigee

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA1200‐kgsatelliteisinanellipticalorbitaroundtheEarth.Atperigee,thealtitudeofthesatelliteis1,000kmabovethesurface,andatapogeethealtitudeis10,000kmabovethesurface.

Ifthesatelliteistravelingat8.6km/satperigee,whatisitsspeedatapogee?

Answer:3.89m/sTryit!(1PFpt):Determinethetotalenergy(kinetic+potential)ofthesatelliteatbothperigeeandapogee.Showyourwork.



179

PreflightQuestions


2. ConsideraspacecraftorbitingtheSuninacircularorbit.ThespacecraftfiresitsenginesaddingenergyuntilitescapestheSun’sgravity.Comparethetotalenergy forthecircularorbit ,ellipticalorbit ,andparabolictrajectory .

a) b) c)

3. Supposeanobjectismovingalonganyoneofthegivenorbitalpaths.Whatistrueregardingthe

orbitsdepicted?

a) Thekineticenergyisconstantinalltheorbits,whilethepotentialenergychangeswithdistancefromtheSun.

b) Thepotentialenergyisconstantforallpointsinanyoneoftheorbits.c) Totalenergydecreasesfromthecircularorbit untilitequalszerofortheparabolic

trajectory .d) Totalenergyisconstantforanypointalonganyoneoftheorbits.

4. CRITICALTHINKING:Aphysicsbookclaimsthat,“Moon‐boundspacecrafthavespeedsjustunder ,sothatifanythinggoeswrong(aswithApollo13),theywillreturntoEarth.”Explainwhythisstatementiscorrectorincorrect.ThinkaboutthederivationoftheescapevelocityequationandwhetheraspacecraftcangettotheMoonwithoutescapingtheEarth.



180

HomeworkProblems

8.27



181

8.52



182

MP



183

Lesson25

CriticalThinking:OrbitalEnergies

Reading 8.4Examples 8.5HomeworkProblems MP,8.61,8.67


LearningObjectives



Notes



184

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass? 2. Thetotalenergyofasatelliteinaparticularboundorbit

a) variesdependingonthesatellite’spositioninthatorbit.b) isalwayspositive.c) isalwaysnegative.d) isalwaysexactlyzero.

3. Fortwoobjectsseparatedbyadistance ,themagnitudeofthegravitationalpotentialenergyis

.Ifthedistanceisdoubled,whatisthenewgravitationalpotentialenergy?

a) b)4 c)8 d) e)

4. CRITICALTHINKING:TheInternationalSpaceStation(ISS)(http://www.nasa.gov/mission_pages/station/main/index.html)orbitsEarthatanaltitudeof350km.Sincethestationandtheastronautsinsideareinfreefalltogether,theyfloataroundinsidetheISSmodules.Ifthestationwerestationaryatthataltitude,howwouldtheastronauts’weightscomparetotheirweightsatthesurfaceofEarth?



185

HomeworkProblems

MP



186

8.61



187

8.67



188

Notes



189

Lesson26

CenterofMass

Reading 9.1Examples 9.1,9.2,9.3HomeworkProblems 9.16,9.37,9.89

LearningObjectives


Notes



190

WorkedExample



Foursmallmassesm1,m2,m3,andm4aretiedtogetherwithrigidrodssothattheyformasquareofside1m,asshowninthefigure.WewanttowriteNewton’sSecondLawfortheentiresystemasifallthemass

wereconcentratedatasinglepoint,thatis

,where isthenetexternalforceonthesystem(i.e.thevectorsumofalltheexternalforces)and istheaccelerationofthesystem.Whatisthelocationofsuchapoint?Considerthemassoftheconnectingrodstobeverysmallcomparedtothemassesonthecorners.STRATEGY

The point described above is called the center-of-mass of the system. Under the action of external forces the assembly of the masses moves as if it were a single mass. For example, in projectile motion the center of mass of the four masses will follow a parabola.

The location of the center of mass point is given by

⋯ ⋯

IMPLEMENTATION

We choose a coordinate system for the assembly of the four masses and apply the center-of-mass equation. Any coordinate system will work. We choose the origin of our system to be the center of the square.

Score(3)

1kg 1kg

2kg 2kg

1m



191

CALCULATION

1kg 0.5m 1kg 0.5m 2kg 0.5m 2kg 0.5m

6kg0

1kg 0.5m 1kg 0.5m 2kg 0.5m 2kg 0.5m

6kg16m

The center of mass is 1/6 meters under the origin on the y-axis. SELF‐EXPLANATIONPROMPTS

1.Intuitively,thecenterofmasscanbethoughtofasthepointatwhichtheassemblycouldbestablysupported.Usingthisapproach,wherewouldyouexpectthecenterofmassofthetwo1‐kgmassestobe?

2.Whataboutthecenterofmassofthetwo2‐kgmasses?

3.Justify,usingsymmetry,whythecenterofmassisonthey‐axisandbelowthex‐axis?



192

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMFindthelocationofthecenter‐of‐massofasystemcomprisedofthree1‐kgmasseslocatedatthreecornersofasquarewhosesideis1m.(Hint:Drawapictureandmarkthecenter‐of‐mass.)

Answer:x=−0.17m,y=−0.17minacoordinatesystemwiththeoriginat

thecenterofthesquare



193

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass? 2. (True/False)A4.8tonelephantisstandingina15tonrailcarthatisatrestonafrictionless

track.Theelephantbeginstowalktowardstheotherendofthecar.Foreverymetertheelephantmoves,thecarmoves1meterintheoppositedirection.

a) Trueb) False

3. (True/False)Accordingtotheequationsofmotionforaprojectile,afirecrackerfollowsa

parabolicpath,neglectingairresistance.Afteritexplodes,thecenterofmassofthepiecesstillfollowsaparabolictrajectory.

a) Trueb) False

4. CRITICALTHINKING:Afully‐loadedcanoeisattachedtoanemptycanoewithabungeecord.

Thecanoesareatrestonaplacidlake.Apassengerintheheaviercanoepushesthecanoesapart,stretchingthebungeecord.Describewhathappenstothecenterofmassofthesystemandexplainyourreasoning.



194

HomeworkProblems

9.16



195

9.37



196

9.89



197

Lesson27

ConservationofLinearMomentum&Collisions

Reading 9.1– 9.5Examples CE9.1,9.4,9.5, 9.7HomeworkProblems 9.38,MP,MP

LearningObjectives



[Obj45] Applyconservationoflinearmomentumtosolveproblemsinvolvingsystems ofparticles.



[Obj48] Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐ andtwo‐dimensions.

Notes



198

WorkedExample



A1‐kgball,m1,collideswitha3‐kgball,m2,asshown.Theballshaveinitialvelocitiesof 3m/sand

3m/s.Immediatelyafterthecollision

2m/s,whatisv1final?

STRATEGY

This is a one-dimensional Conservation of Linear Momentum problem. To use the concept of conservation of momentum, we must ensure that there is no net external force acting on the system. Since 1) we are only interested in what happens immediately before and after the collision, and 2) the collision is brief, we can assume that any external forces acting on the balls are negligible. Because of these conditions, we say that linear momentum is conserved in collisions. (Note that linear momentum can be conserved during other interactions as long as the condition of no net external forces is met.) The Conservation of Linear Momentum equation is ∑ ∑ ,and, specific to this problem, .As the figure shows it is possible for one of the balls to have a negative velocity (oppositely directed) after or prior to the collision.

IMPLEMENTATION

We will designate a standard x-y coordinate system as shown. We will use the conservation of linear momentum equation and solve it for v1final.

Score(3)



199

CALCULATION

Starting with:

It becomes:

Now substituting:

1kg 3ms 3kg 3

ms 3kg 2

ms

3kg

4m/s


1.Whyisitimportanttoensurethatnonetexternalforceactsontheobjects?

2.Whycanyounotusetheabsolutevaluesofthevelocitiesintheabovecalculations?

3.Usingthecalculatedfinalvelocityofball1,showthatlinearmomentumwasindeedconservedinthecollision.



200

Pre‐ClassProblem

Thediagramontheleftshowsacollisionbetweenaseriesofrailroadcars.Ifthecars,eachhavingamassof3000kg,departthecollisionsasonecoupledgroup,whatwillbethefinalvelocityoftheassembly?

Answer:2.5m/s



201

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Linearmomentumofasystemisconservedif

a) thenetexternalforceiszero.b) theenergyofthesystemisconserved.c) thenetworkdoneispositive.d) onlyconservativeforcesaredoingwork.

3. A500‐gfireworkrocketismovingat60m/sstraightupwardwhenitexplodes.Thesumofall

themomentumvectorsoftherocketfragmentsimmediatelyaftertheexplosionis

a) zero.b) 30kgm/sstraightup.c) 30kgm/sinmultipledirections.d) morethan30kgm/sbecauseoftheenergyaddedbytheexplosion.

4. CRITICALTHINKING:Considerarubberbulletandanaluminumbullet;bothhavethesamesize,speedandmass.Eachbulletisfiredatablockofwood.Therubberbulletbouncesback,thealuminumbulletpenetratestheblock.Whichismostlikelytoknocktheblockover?Explain.



202

HomeworkProblems

9.38



203

MP



204

MP



205

Lesson28

Lab5–1‐DCollisions

Reading 9.5,9.6,Lab5WorksheetExamples NoneHomeworkProblems 9.28,9.44,9.61


LearningObjectives




Notes



206

JournalQuestions


2. InPartIofthelab,aheavycart(massm1)andastationarylightcart(massm2)willundergoaone‐dimensionalcollisiononafrictionlessairtrack.Assumingthecollisioniselastic,writetheexpressionforthefinalvelocityofcart2,v2f,intermsoftheinitialvelocityofcart1,v1i.

3. InPartIIofthelab,aheavycart(massm1)andastationarylightcart(massm2)willundergoatotallyinelasticone‐dimensionalcollisiononafrictionlessairtrack.Deriveasimilarexpressionforthefinalvelocityofthejoinedcarts(massesm1+m2),vf,intermsoftheinitialvelocityofcart1,v1i,startingfromtheequationforconservationofmomentum.

4. Supposeyoumakeaplotofthefinalvelocityofcart2,v2f,versustheinitialvelocityofcart1,v1i,fortheelasticcollision.Whatwouldtheplotlooklike?Writeanexpression,intermsofm1andm2,fortheslopeassociatedwiththisplot.

5. Ifyouplottedthefinalvelocityofthejoinedcarts,vf,versustheinitialvelocityofcart1,v1i,forthetotallyinelasticcollisioninstead,howwouldtheslopeforthetotallyinelasticcollisioncomparetotheslopefortheelasticcollision?Explain.

Score(5)



207

LabNotes



208

HomeworkProblems

9.28



209

9.44



210

9.61



211

Lesson29

CollisionsandConservationofEnergy:

WheredoestheEnergyGo?

Reading 9.3,9.4Examples 9.10HomeworkProblems MP,9.68,9.78

LearningObjectives




Notes



212

WorkedExample



A1.0‐kgpuck(puck1)isslidingat45°abovethex‐axiswithaspeedof1.0m/s.Another1.0‐kgpuck(puck2)isslidingwithaspeedof0.50m/sat45°belowthex‐axis.Thepuckscollide,andpuck2fliesoffat45°belowthex‐axis,at0.80m/s.

a)Whatisthevelocityofpuck1afterthecollision?

b)Wasthiscollisionelastic?

STRATEGY

Since total linear momentum ( ∑ ) is conserved in any collision, we can use the conservation of linear momentum to obtain the set of equations we need to solve for the velocity of puck 1. We can then compare the kinetic energies before and after the collision. If the kinetic energies before and after the collision are the same (conserved), the collision was elastic.

IMPLEMENTATION

Since total momentum is conserved we have:

=

Writing this expression in terms of x- and y- components gives us two equations with two unknowns – the magnitude and direction of the velocity of the first puck –which we will solve. To see if the collision was elastic we compare the kinetic energies before and after the collision

12

12

12

12

Score(3)

m1=1.0kg

v1i=1.0m/s

m2=1.0kg

v2i=0.5m/s

1

2

x

?



213

CALCULATION

a) Before the collision, the x- and y- components of the total momentum are:

1.0kg 1.0ms∙ cos45° 1.0kg 0.50

ms∙ cos45° 1.06

kg ∙ ms

1.0kg 1.0ms∙ sin45° 1.0kg ∙ 0.50 ∙ sin45° 0.354

kg ∙ ms

and after collision they are:

1.0kg 1.0kg 0.80ms∙ cos45° 1.06

kg ∙ ms

1.0kg 1.0kg 0.80ms∙ sin45° 0.354

kg ∙ ms

Solving for the speed of the puck 1 and the direction angle θ we get: 0.54 and 23°, above the positive x-axis.

b) Calculating the total kinetic energy of the two pucks before the collision, we get 0.62 J; after the collision the kinetic energy is 0.46 J. The collision was not elastic, but it was not totally inelastic.


1. Solvethemomentumequationsforthespeedanddirectionofmotionofpuck1(i.e.,fillinthestepsomittedabove).

2. Calculatethetotalkineticenergyofthetwopucksbeforeandafterthecollision(i.e.,fillinthestepsomittedabove),andconfirmthatthecollisionisinelastic.

3. Wheredidtheenergygoduringtheinelasticcollision?



214

Pre‐ClassProblem

Twomasses,m1andm2,moveatrightangles,meetattheoriginandflyof,stickingtogether.Theirinitialspeedsarethesame.Ifm1=3m2,whataretheirspeedanddirectionaftercollision?

Answer:0.79v,18.4°



215

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Ifthenetexternalforceactingonanobjectisconstant,whatistrueaboutitsmomentum ?

a) Themagnitudeanddirectionof maychange.b) Themagnitudeof remainsconstantbutthedirectionmaychange.c) Themagnitudeof maychangebutthedirectionremainsconstant.d) Themagnitudeanddirectionof remainconstant.

3. Matchthediagramtothetypeofcollisionbetweenobjectsofequalmass.

_____ Elasticcollision_____ Inelasticcollision_____ Totallyinelasticcollision

4. CRITICALTHINKING:Birdstrikesareasignificantflightsafety

hazard.ConsideranF‐16birdstrikewhereagooseimpactsthecanopy.TheF‐16canopydeformsduringthecollisionandthebirdpartsdeflectawayfromtheaircraft.Whattypeofcollisionisthis?Explain.

CaseA CaseB CaseC CaseD



216

HomeworkProblems

MP



217

9.68



218

9.78



219

Lesson30

GRADEDREVIEW3

LearningObjectives





[Obj30] Explainthedifferencesbetweenconservativeandnonconservativeforces.



[Obj33] Evaluatethepotentialenergyassociated withaconservativeforce.












[Obj45] Applyconservationoflinearmomentumtosolveproblemsinvolvingsystemsofparticles.




Notes



220




Amandragsa50‐kgcrate10macrossaroughhorizontalsurface,wherethecoefficientoffrictionbetweenthecrateandsurfaceis0.3.Hepullsataconstantspeedandhedirectshispullingforce20°downwardfromthehorizontal.Howmuchworkdoesheperform?


The strategy is the same as used for the worked example in Lesson 18, the only difference is that the force of friction is now

The force exerted by the man is now

____________


cosθ

Δ 1328J


1.Theworkdoneinthiscaseis3JmorethanintheworkedexampleforLesson18.Explainwhytheworkincreased.

2.Whatworkdoesfrictiondo?

3.Howwouldtheanswerchangeifthesurfacewasfrictionlessinstead?




221




A1500‐kgcaristravellingat26.8m/s.Thedriverappliesasmallbrakingforceof800N.Howfardoesthecartravelbeforeitslowsdownto13.4m/s?


Apply the __________________ theorem and solve the resulting equation for the unknown displacement. CALCULATION(Fillintheblanks.)

∙ Δ _____________ _________________

Δ

505m


1. Whathappenstothestoppingdistanceifthespeedofthecarisdoubled,assumingthesamebrakingforce?

2. Wheredoesthe“brakingforce”comefrom?

OptionalPracticeProblems:6.27,6.39



222




Anobjectofmassmisreleasedatrestfromaheighthaboveground.Whatisthespeedoftheobjectjustbeforeitreachestheground?STRATEGY(Fillintheblanks.)

The object has positive potential energy relative to the ground. As it falls, that energy gets converted into kinetic. Symbolically,

∆ ⇒ _____________


_____________ ½mv2

Setting the kinetic energy equal to ΔUg and solving for the speed v we get

2

SELF‐EXPLANATIONPROMPTS 1.Whyistherenomassminthefinalanswer?Reviewthecalculationandshowwherethemassdropsout.2.Doesthefactthatthereisnomassminthefinalanswertellusthatthegravitationalpotentialenergydoesnotdependonthemass?3.Thisproblemcanalsobesolvedusing1‐Dkinematics.Usethismethodandcomparetheresults.OptionalPracticeProblems:7.13,7.17,7.30



223




A60‐kgskierstartsfromrestandskisdownazig‐zagtrail.Whenshereachesthebottomofthetrailshehasdescendedaverticalelevationof600m.Ifsheloses12%ofherenergytofriction,whatisherspeedatthebottomofthetrail?


We apply a modified energy conservation equation, accounting for the energy lost to friction

0.88

And taking PE to be zero at the bottom of the trail.


___________ ___________

310,464J

___________

101m/s


1. Inyourownwordsdescribethemodifiedconservationequationweused.

2. Whydoesthecontouroftheterrainnotmatterinthiscalculation?

3. Howwouldtheproblemneedtobechangedsothatwecouldusetheforceoffrictionformula,

?




224




Manymoviescontainsceneswheretheactorsexperienceweightlessness.Thesescenesareusuallyfilmedinanairplanethatisundergoingcarefullychoreographedmaneuverswhichsimulateweightlessness.Considerthefollowingflightpath:Fromthestartofthemaneuvertot=120s,theairplane’sheightisdescribedby 225 .Duringthenextpartofthemaneuver,whichlasts60seconds,theairplanesheightisgivenby 4.9 1176 43560.Forthefinal15secondsofthemaneuver,theairplane’saltitudeisgivenby 23000.Duringwhattimeperiodistheairplane“weightless”?


The back of the airplane will appear to be weightless (aka “free fall”) when the airplane accelerates at same rate as all the objects in the plane. We will use the derivative to find the times when the acceleration is 9.8 m/s2. CALCULATION

1. During the first part of the maneuver, we take the derivative of position to find the velocity, which is ____________. By taking the derivative of velocity, we find the acceleration to be 0 m/s2. The plane is not in free fall.

2. During the second part of the maneuver, find that the velocity is given by __________. We take the derivative of velocity to find that the acceleration is given by ____________. The plane is therefore in free fall.

3. During the third part of the maneuver, we find that the velocity is _________ and the acceleration is ___________. The plane is not in free fall. SELF‐EXPLANATIONPROMPTS

1.Whatpartofthepositionequationdeterminestheplaneisfreefallinthe2ndpartofthemaneuver?Whatpartsoftheequationdoesnotmatter?

2.Howcouldyoumodifythepositionequationforthethirdpartofthemaneuvertomaketheplanebeinfreefall?

OptionalPracticeProblems:8.35,8.19,



225




Ifa300‐kgsatellite,inacircularorbit150kmabovethesurfaceoftheEarthcrashestotheground(orburnsup),howmuchenergyislost?STRATEGY(Fillintheblanks.)

The total energy (potential and kinetic) of a satellite in a circular orbit about the Earth is

12

where G = 6.67 x 10-11 Nm2/kg2 is the universal gravitational constant m is the mass of the satellite M is the mass of the Earth = 5.97 x 1024 kg r is the radius of the orbit, i.e. radius of the Earth + altitude


Δ _____________________12

1

1

6.37 10 150 10 6.52 10 m

__________________________

216 10 J


1.Thegravitationalenergyequationincludesthegravitationalpotentialenergy.Whereisthegravitationalpotentialenergyzero?

2.Completethecalculation.




226




Whereisthecenterofmassofa2‐mbarbellwitha1‐kgmassontheleftanda3‐kgmassontheright?Considerthemassofthebartobeverysmall(negligible)comparedtothemassesonthecorners.


Choose a coordinate system with the origin at the center of the bar.

Apply the center-of-mass equation and solve for the x-coordinate.


1kg 3kg

0.5m


1.Doestheresultagreewithyourintuition?

2.Convinceyourselfthatthechoiceofthecoordinatesystemdoesnotmatter.Choosetheoriginofthecoordinatesystemattheleftend,atthelocationofthe1kgmass,andshowthatyougetthesameresult.(Drawadiagramandmarkthelocationofthecenter‐of‐massascalculatedoriginallyandagainascalculatedusingthenewcoordinatesystem.


1kg 3kg2m



227




A100‐kgclownislaunchedfroma500‐kgcircuscannon.AfterthefiringofBozo,hehasaspeedof15m/s.Assumingthathisshoesandcostumearesohighlypolishedthatthereisnofrictionashemovesoutofthecannonbarrelandthatthecannonandclowninitiallyareatrest,whatisthefinalvelocityofthecannon?STRATEGY(Fillintheblanks.)

We will use a standard x-y orientation for this _______ dimensional problem involving conservation of linear ____________. The statement “that there is no friction” allows us to meet the condition of no ______ _______ acting on the objects during the event. The event in which momentum is conserved in this case is an explosion where the cannon and the clown are considered initially as _________ stationary object that then becomes two objects with individual __________. We will start with the Conservation of Momentum relation and solve for the ____________________.


First:∑ ∑_____________

Now: _______ _____ ___

With numbers: _____ 100kg _____m/s 500kg _____100kg _______m/s

___ 100kg ____m/s /_____kg 3m/s


1. Whatdoes“atrest”implyabouttheinitialvelocitiesofthecannonandtheclown?

2. Istheinitialcondition(clowninsidecannon)thesameasifthetwoweresittingatrestnexttooneanother?Explain.

OptionalPracticeProblems:9.18,9.19,9.20,9.42



228




Twoidenticalmasses,M,approachtheoriginwiththesamespeedv,at45degreesfromthehorizontal.

Theycollideandsticktogether.Whatarethespeedanddirectionofmotionaftercollision?


We apply the conservation of _______________________ to determine the motion after collision.

The y-component of the momentum after collision must be zero because __________.

The x-component of the momentum before collision is _____________


2 45 ________________

45 inthepositivex‐direction.


1.Isthiscollisionelastic?

Ifyouranswerisyes,explainyourreasoning?Ifyouanswerisno,calculatethechangeinkineticenergy.




229

Notes



230

Notes



231

Lesson31

RotationalMotion



LearningObjectives




Notes



232

WorkedExamples



TheSmokyHillsWindFarmnearSalina,Kansasemploy140,000poundVestasV801.8‐megawattwindturbines.Thethree‐bladeturbineshaveadiameterof80mandoperateat15.5to16.8rpm(revolutionsperminute.)

a)Whatisthelinearspeedofthebladetipatmaximumrotationalspeed?

b)Whatisthecentripetalaccelerationatthetipofthebladeatthemaximumspeed?

c)Ifthebladeslowsdownfrommaximumspeedtorestin30seconds,throughhowmanyrevolutionsdoesitturn? STRATEGY

We use the relation:linearquantity=(radius)times(correspondingangularquantity). arclength=(radius)times(anglesubtended)

linearspeedalongthearc=(radius)times(angularspeed)

tangentialacceleration=(radius)times(angularacceleration)

The above relations are valid if the angles are measured in radians. The tangential acceleration at is non-zero if the angular speed is changing. Whenever the angular speed is non-zero, there is always a centripetal acceleration, , responsible for changing the direction of the tangential velocity. IMPLEMENTATION

The angular speed is given in revolutions per minute. Since there are 2π radians in a revolution and 60 seconds in a minute, we multiply rpms by 2π/60 to get the angular speed in radians per second.

To obtain the tangential speed, we use . The centripetal acceleration is then 2.

The kinematics equations for angular quantities mimic kinematics equations for linear motion. For constant angular acceleration α, the relation between θ, ω, α, and time is:

2 .

Score(3)



233

CALCULATION

a) Maximum angular velocity:

. 1.76rad/sec

b) Maximum linear speed of the tip: 40m 1.76rad/ sec 70.4m/s c) Centripetal acceleration at the tip:

123.9m/s d) During the 30 seconds slow-down the blade undergoes an angular deceleration

0.06rad/s and it turns through 25.8radians 4revolutions

Note: Compare 2 for rotational motion to 2 for linear motion.


1.Showtheconversionof15.5rpmtorad/s.2.Theangularmeasureradianisdefinedastheratioofthearclengthtotheradius.Convert1degreetoradians.Convert1radiantodegrees.3.Ifanobjectisrotatingwithanon‐zeroangularvelocityωandzeroangularaccelerationα,isthereacentripetalacceleration?Ifanobjectisrotatingwithanon‐zeroangularvelocityωandnon‐zeroangularaccelerationα,whatisthetotallinearacceleration?



234

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA3‐mdiameterflywheelisspinningupwithanangularaccelerationof3rad/s2.Howlongdoesittaketheflywheeltoreach12rpm(revolutionsperminute)ifitstartsfromrest?

Answer:0.4seconds



235

PreflightQuestions

2. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Whichofthefollowingistheclosesttooneradian?

a) 30°b) 60°c) 90°d) 180°

3. Twoantscrawlontothesurfaceofacompactdisc.AntAisfartherfromthecenterofthediscthanAntB.Thecompactdiscbeginstospin.Whichofthefollowingstatementsistrue?

a) AntAexperiencesagreatertangentialaccelerationthanAntB.b) AntAexperiencesagreaterangularaccelerationthanAntB.c) Neitherstatementistrue.d) Bothstatementsaretrue.

4. CRITICALTHINKING:Whatistheapproximateangularspeedoftheearthrevolvingaroundthe

Sun,inrad/day? Explainthereasoningyouusedindeterminingyouranswer.



236

HomeworkProblems

10.19



237

10.32



238

10.45



239

Lesson32

RotationalInertia&Torque

Reading 10.2,10.3Examples 10.4,10.5HomeworkProblems 10.30,10.28,10.52

Thereisanon‐gradedPHYSICSKNOWLEDGEASSESSMENTTESTthislesson.

LearningObjectives





Notes



240

WorkedExamples



Youhaveaflattireonyourcarand,inordertochangethetire,youneedtoremovethelugnutsthatsecurethewheeltothecar.Ifthe30‐cmlongwrenchyouareusingtoremovethenutsisata55°angletothehorizontalandyouapplyaforceof120Ndirectlydownontheendofthewrench,whatisthemagnitudeofthetorqueyouexertonthelugnut?

STRATEGY

For this problem, we are interested in the applied torque τ. Torque is the rotational analog to force; it is the effectiveness of a force to cause an object to rotate about a pivot point. To solve for torque, we need to consider 1) the magnitude of the applied force F, 2) how far from the pivot point the force is applied r, and 3) the angle that the force is applied θ. IMPLEMENTATION

Let’s draw a diagram and label the applied force vector , the vector that goes from the pivot point to the where the force is applied, and the angle θ between and . Note that the angle θ is not 55°, but (180° - 55°) = 125°.

The relation between τ, r, F, and θ is: sin

CALCULATION

The magnitude of the torque exerted on the lug nuts is

sin 0.30m 120N 125 29Nm.

The units of torque are newton-meters (N m).

Score(3)



241


1.Inyourownwords,explainhowtorquediffersfromforce.

2.Whydidweuse125°fortheangleoftheappliedforceandnot55°?

3.Explainhowthemagnitudeofthetorqueexertedonthelugnutswouldchangeiftheforcewasappliedinthemiddleofthehandleratherthantheend.



242

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMIfyouapplya45‐Nforceperpendicularlytoadooratdistancesof1m,a)determinethemagnitudeofthetorque,andb)themagnitudeoftheangularaccelerationifthedoor’srotationalinertia,I,is30kgm2.

Answer:(a)45Nm;(b)1.5rad/s2

Tryit!(1PFpt):Calculatethemagnitudeofthetorqueiftheforcewasappliedatanangleof25°instead.



243

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Rankorderthemagnitudeofthetorquesfromsmallesttolargest.Eachrodis50‐cmlongfrom

thepivot(). RankOrder:Smallest(1)_____(2)_____(3)_____(4)_____(5)_____Largest

3. Inordertospinfasteraboutaverticalaxis,aniceskaterneedstodecreaseherrotationalinertia.Shecouldachievethatby

a) stretchingherarmsfartherawayfromtheverticalrotationaxis.b) bringingherarmsclosertoherbody.c) loweringherbodybybendingherkneesandsquattingdown.d) bendingforwardatherwaistsoherbodyisL‐shaped.e) Rotationalinertiacanonlydecreaseifhermassdecreases.

4. CRITICALTHINKING:Abookcanberotatedaboutmanydifferentaxes.Themomentofinertiaofthebookwilldependupontheaxischosen.RankthechoicesAtoCaboveinorderofincreasingmomentsofinertiaandexplainyourranking.

(A) (B) (C)

(D) (E)45°

2N 2N

4N

2N

4 N



244

HomeworkProblems

10.30



245

10.28



246

10.52



247

Lesson33

RotationalAnalogtoNewton’sSecondLaw

Reading 10.3Examples 10.8,10.9HomeworkProblems 10.56,10.57,MP


LearningObjectives

[Obj55] Compareandcontrasttheconceptsof massandrotationalinertia.


Notes



248

WorkedExample


STATEMENTOFTHEPROBLEMA50‐kgblockanda100‐kgweightareconnectedwitharope,passingoverapulleyasshown.The50‐kgblockisona30°rampwherefrictionisnegligible.Thepulleyisasoliddiscwhoseradiusis0.2mandwhosemomentofinertiais2kgm2.Theropedoesnotstretch.

Whenreleasedfromrest,whatistheaccelerationofthesystem,includingdirection?

STRATEGY

First, we draw free-body diagrams and apply Newton’s Second Law for each of the two masses and Newton’s Law for rotational motion for the pulley. We then solve the system of three equations for the common acceleration. The system of equations has three unknowns, the acceleration and the two tensions. Since the inertia of the pulley is not negligible, the tension on the left side of the pulley is not the same as the tension on the right side of the pulley. IMPLEMENTATION

The 50-kg block, m: The net force on the block is: – 30° . The normal force equals the component of the weight perpendicular to the ramp,

The 100-kg block, M: The net force on the block is: – .

The lengths of the arrows do not indicate the magnitudes of the forces since we don’t know those until we make the calculations. Note the negative sign for the acceleration, to be consistent with the direction chosen for the 50-kg mass.

The pulley: Newton’s Law for rotation states that . I is the moment of inertia, α is the angular acceleration and τ is the

Score(3)



249

torque, defined as the applied force multiplied by the perpendicular distance to the axis of rotation from the application point of the force.

The net torque on the pulley is: – The Newton’s Law equation for the pulley reads: –

CALCULATION

The three equations now read: – 30° – –

The angular acceleration and the linear acceleration are related by a = αr. Solving for the acceleration we get:

– 30°

2.2 /

The sign of the calculated acceleration is positive. That means that the 50-kg mass is accelerating up the ramp and the 100-kg weight is accelerating down.

SELF‐EXPLANATIONPROMPTS1.Justifywhyeachcoordinatesystemonthefree‐bodydiagramswasused.

2.Justifythenegativesignusedfortheaccelerationintheequationofmotionforthe100‐kgweight.

3.Explaininyourownwordswhythetensionsonthetwosidesofthepulleyaredifferent.



250

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMWhatmagnitudeoftorquehastobeappliedtoa2.3‐kg,18‐cmdiameter,soliddiskrotatingat800rpmtostopitin10seconds?

Answer:0.078Nm



251

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Whichstatementiscorrect?

a) Iftorqueincreases,rotationalinertiamustincrease.b) Therotationalinertiaofanobjectdoesnotdependonthelocationofitsaxisofrotation.c) Anobjectwithmoremasshasahigherrotationalinertiathananobjectwithlessmass.d) Rotationalinertiameasuresanobject'sresistancetochangesinitsrotationalmotion.

3. Rankordertheangularacceleration ofeachcase.

Theobjectsareconnectedwithmasslessrodsoflengthsshown.Theforcesshownaretheonlyforcesactingontheobjectscausingrotationaboutthepivotpoint().

a) b) c) d)

4. CRITICALTHINKING:Thetwoblocksfromtheworkedexampleproblem,mandM,arenowhungdirectlydownfromthepulleyasshown.Describehowtheequationfortheaccelerationoftheblockswouldchangeforthisscenario.

2kg

2kg1N

1N

2kg

2kg2N

2N

4kg

4kg1N

1N

2kg

2kg2N

2N

30°30°

2m

2m

4m

4m

CaseA

CaseB

CaseC

CaseD



252

HomeworkProblems

10.56



253

10.57



254

MP



255

Lesson34

RotationalEnergyandRollingMotion

Reading 10.4,10.5Examples 10.10‐10.12,CE10.1HomeworkProblems 10.60,10.62,10.68

LearningObjectives




Notes



256

WorkedExample


Aboulderontopofahillbreaksfreeandbeginstorolldownthehillwithoutslipping.Approximatingtheboulderasasolidspherewithradius3m,whatisthespeedoftheboulderatthebottomofthehillafterithasundergoneaverticaldisplacementof100m?

STRATEGY

We will use the principle of conservation of mechanical energy to solve for the speed of the boulder at the bottom of the hill.

Conservation of mechanical energy applies to this problem, because, although frictional force is acting on the boulder causing it to roll, no work is done by friction on the boulder.

IMPLEMENTATION

First, we need to determine the types of mechanical energy in both the initial (top of the hill) and the final (bottom of the hill) states. Since the boulder is initially at rest, it has only gravitational potential energy. The total mechanical energy at the top of the hill is given by

After the boulder has undergone a vertical displacement of 100 m, the gravitational potential energy has been converted to translational and rotational kinetic energy. In the final state, the boulder will have a combination of gravitational potential energy, translational kinetic energy and rotational kinetic energy. If we take the bottom of the hill to be where the gravitational potential energy is zero, the total mechanic energy in the final state becomes

12

12

The total mechanical energy at the top and at the bottom of the hill is the same (conserved), so our conservation of mechanical energy equation becomes

12

12

Now we can solve for the translational speed of the boulder at the bottom of the hill. The rotational kinetic energy is dependent on the rotational inertia and angular velocity of the

Score(3)



257

boulder. We are told that the boulder is (a) a solid sphere and (b) that is not slipping as it rolls – this means that we can use the rotational inertia of a solid sphere and the relationship between angular speed and translational speed to put rotational kinetic energy in terms of the mass, radius, and translational speed of the boulder.

This equation can be simplified further as mass of the boulder appears on both sides of the equation and can be cancelled.

CALCULATION

Solving for the translational speed of the boulder at the bottom of the hill becomes:

107

10 ∙ 9.8m s ∙ 100m

737m s⁄

Note: The speed of the boulder is independent of both the mass and the radius of the boulder. SELF‐EXPLANATIONPROMPTS

1.Frictionalforceisneededforthebouldertoroll,andnotslide,downthehill.Explainwhy“noworkisdonebyfrictionontheboulder”.

2.Explainwhytheboulderhasonlyrotationandtranslationkineticenergyatthebottomofthehill.

3.Howwouldthefinalspeedchangeiftheboulderwasslidingdownthehillinsteadofrolling?Deriveanexpressionforthespeedoftheboulderatthebottomofthehillifitwasslidinginsteadofrolling.



258

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMInapinballmachine,asolidmetal0.050‐kgballisreleasedfromaspringandrollsaroundthemachinehittingvarioustargets.Ifthespringhasaspringconstantkof410N/mandiscompressedadistancexof22cm,a)whatistherotationalkineticenergyoftheballimmediatelyafterrelease?b)Whatisthetranslationalkineticenergyoftheballimmediatelyafterrelease?

Answer:0.047J;0.12J

Tryit!(1PFpt):Calculatethetranslationkineticenergyoftheballifitwasslidinginsteadofrolling.Showyouwork.



259

PreflightQuestions


2. Asolidaluminumcylinder(mass ,radius ,rotationalinertia )andasolidsteel

cylinder(mass2 ,radius ,rotationalinertia )startfromthesamepositionandroll

downarampwithoutsliding.Atthebottomoftheramp,

a) thealuminumcylinderhasgreatertotalkineticenergy.b) thesteelcylinderhasgreatertotalkineticenergy.c) thecylindershavethesametotalkineticenergy.

3. A4.5‐kgbicycletire( 0.6kgm , 37cm)isspinningonamechanic’sstandatthesamerateasifitwererollingatalinearspeed 10m/s.Themechanicappliesthebrakesupplyingaforcetoslowtherotationequivalentto 5m/srollingspeed.Whatisthework

donebythebrake?

a) 16.4mJb) 164Jc) 164Jd) 16.4mJ

4. CRITICALTHINKING:An8‐kgwheelhasamomentofinertiaIof0.1kgm2.Thewheelisrolling

alongwithoutslipping.Whatistheratioofitstranslationalkineticenergytoitsrotationalkineticenergy?Explainhowyouobtainedyouranswer.



260

HomeworkProblems

10.60



261

10.62



262

MP



263

Lesson35

RotationalVectorsandAngularMomentum

Reading 11.1– 11.3Examples 11.1HomeworkProblems 11.16,11.17,11.21

LearningObjectives



Notes



264

WorkedExample


Acarisdrivingclockwisearoundacircularracetrack.Thetiresonthecarrotate50timeseverysecond.a)Whatisthecar’sangularvelocityasitasittravelsduenorthanddueeast?b)Whatistheaverageangularaccelerationofthecarduringthe10secondsittakestogofromtravelingduenorthtotravelingdueeast.

STRATEGY

The problem asks about vector quantities, thus the answers have both a magnitude and a direction component which can be considered separately. First, we find the magnitude of the angular velocity and use this to find the magnitude of the average acceleration. To find the direction of the angular velocity and acceleration, we will use the right hand rule to find the velocity direction and from that, deduce the direction of the angular acceleration. IMPLEMENTATION

To find the magnitude of the velocity, we apply unit analysis. We find the magnitude of the angular acceleration using the relation: ∆

∆. To find the direction of the angular

velocity, we apply the right hand rule. CALCULATION

a) Angular velocity:

50rotationssecond

2radiansrotation

314rads

b) Average angular acceleration:

314radianss

110s

31.4rads

c) The wheel is rotating forward, so if the fingers of our right hand point wrap forward and down – mimicking the motion of the wheel - then our thumb points to the left (west) which is the direction of the angular velocity. When the car is traveling east, the right-hand rule gives us a thumb pointing towards the top of the page (north). To find the direction of the acceleration vector, we draw a vector going from the tip of the west arrow to the tip of the north velocity vector. Thus the acceleration vector is to the north east.

Score(3)



265


1. Vectorscanbeaddedpictoriallybydrawingthevectorssuchthatthetailofonevectorconnectstothetailofanothervector.Explainhowthisapproachisconsistentwiththeabovestatementthattheaverageaccelerationvectorgoesfromthetipofinitialvectortothetipofthefinalvector.

2. Whichdirectionisthevelocityvectorwhenthecaristravelingwest?South?

3. Howwouldyoudescribethedisplacementvectorofthecar?



266

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMAchildisdoingtrickswitharemote‐controlledairplane.Initiallythepropellersontheairplanearespinningat1200rpmastheplanedivesstraighttowardtheground.Threesecondslater,theairplaneisinlevel‐flight,flyingnorthandthepropellersarespinningat1800rpm.Whatwastheaverageangularaccelerationofthepropellers?

Answer: 20.9 ,

34°abovelevelflight



267

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. (True/False)Thenettorque andangularacceleration alwayspointinthesamedirection.

a) Trueb) False

3. (True/False)Theangularacceleration andangularvelocity alwayspointinthesamedirection.

a) Trueb) False

4. CRITICALTHINKING:Howcanaparticlewithlinearvelocityhaveangularmomentum?

Explain.



268

HomeworkProblems

11.16



269

11.17



270

11.21



271

Lesson36

ConservationofAngularMomentum

Reading 11.4Examples CE11.1HomeworkProblems 11.26,11.27,11.43

LearningObjectives


Notes



272

WorkedExample



A2.0‐kgprojectilewithaspeedof5.0m/sstrikesafinonawheelasshownthefigure.Theprojectilestrikesatapoint1.48mtotherightoftheaxisofrotation.Aftertheprojectilecollideswiththewheelitstickstothefinatthepointofimpact.IfthewheelhasarotationalinertiaofI=100kgm2,whatwillbetheangularvelocityofthewheel+projectilecombinationafterwards?STRATEGY

This problem is an example of a rotational collision. If the wheel spins freely, there is no net torque acting on the system as a whole, so long as the system includes both the wheel and the projectile. In this case, the total angular momentum cannot change (see N2LRot).

IMPLEMENTATION

The initial angular momentum is that of the projectile:

The final angular momentum is that of the wheel plus the projectile attached to the fin:

Solving for the final angular velocity we get:

CALCULATION

1.48m 2kg 5m/s100kgm 2kg 1.48m

0.14rad/s

Notice that the result depends on the positioning of the launcher relative to the axle of the wheel.

Score(3)



273


1.Howdoestheresultchangeifyoumovelaunchersothatthepointofimpactisatagreaterdistancefromtheaxleofthewheel?

2.Canyoutellifthecollisioniselasticorinelastic?Explainhowyouknow,orwhyyoucannottell.

3.Ifyouthinkit'sinelastic,howmuchenergyislostinthecollision?Ifyouthinkit'selasticchecktoseeifyou'recorrect.



274

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMA12‐kgpotter’swheelisspinningat5.0rpmandhasaradiusof0.5m.Thepotterthrowsa2.0‐kgblockofclayontothewheelwithavelocityof0.75m/sinthesamedirectionasthewheel.Howfastisthewheelspinningimmediatelyaftertheclaylandsonthepotter’swheel,inrpm?

Answer:0.77rad/sor7.33rpm



275

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Ifanettorqueisappliedtoarigidobject,whichofthefollowingisnottrue?

a) Theangularmomentumoftheobjectwillchange.b) Thekineticenergyoftheobjectwillchange.c) Theobjectwillexperienceanangularacceleration.d) Therotationalinertiaoftheobjectwillchange.

3. Aniceskaterisspinningat2rad/secwithherarmsoutstretched.Ifshenowpullsherarmsin

closetoherbody,her

a) angularmomentumremainsthesame.b) angularvelocityincreases.c) kineticenergyincreases.d) Alloftheabovearetrue.

4. CRITICALTHINKING:Explainwhyhelicoptersmusthavetworotorstofunctionproperly.Your

explanationshouldinvolveangularmomentumconcepts.



276

HomeworkProblems

11.26



277

11.27



278

11.43



279

Lesson37

CriticalThinking:Energy&AngularMomentum

Reading Chapter10&11Examples NoneHomeworkProblems 11.46,11.49,MP


LearningObjectives


Notes



280

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Aplatformdiverjumpsoffthedivingtowerandperformsatwistmaneuver.Whileintheair,he

cannotchangehis

a) rotationalenergy.b) rotationalspeed.c) rotationalinertia.d) angularmomentum.

3. Whatconditionmustbetrueinorderfortheangularmomentumofanobjecttobeconserved?

a) Nonetexternalforceactsontheobject.b) Nonetexternaltorqueactsontheobject.c) Both(a)and(b)aretrue.

4. CRITICALTHINKING:Attheendofitslife,astargoessupernova.Itscore(radius=20Mm)

collapsestoformaneutronstar(radius=6.0km).Iftheinitialrotationrateofthestarwas1rev/45days,whatistherotationrateoftheneutronstar?(Treatthestarasasolidspherewith

.)



281

HomeworkProblems

11.46



282

11.49



283

MP



284

Notes



285

Lesson38

Lab6–ConservationofAngularMomentum



LearningObjectives


Notes



286

JournalQuestions


RefertoConceptualExample11.1inyourtextbookforthefollowingquestions.

2. Whentheboyjumpsontothemerry‐go‐round,

a) thetotalrotationalinertiaoftheplatformchanges.b) thetotalangularmomentumoftheplatformchanges.c) Both(a)and(b)arecorrect.d) Neither(a)nor(b)iscorrect.

3. Whenthegirljumpsontothemerry‐go‐round,

a) thetotalrotationalinertiaoftheplatformchanges.b) thetotalangularmomentumoftheplatformchanges.c) Both(a)and(b)arecorrect.d) Neither(a)nor(b)iscorrect.

4. Intheexample,thegirljumpsinthesamedirectionastheplatformisrotating.Suppose,instead,thatshejumpsintheoppositedirection,sothathervelocityjustbeforelandingontheplatformiscountertoitsrotation.Describehowyouwouldmathematicallyaccountforthischangewhensolvingforthefinalangularspeedofthemerry‐go‐round.

5. Supposethegirldoesnotjumpdirectlyinthetangentialdirection,butatanangleθtothetangentialdirection.Describehowyouwouldmathematicallyaccountforthischangewhensolvingforthefinalangularspeedofthemerry‐go‐round.

Score(5)



287

LabNotes



288

HomeworkProblems

11.45



289

12.69



290

12.87



291

Lesson39

SimpleHarmonicMotion

Reading 13.1,13.2,13.3Example 13.3HomeworkProblems 13.22,13.67,13.43


LearningObjectives


[Obj64] Determinetheperiodand frequencyofasimpleharmonicoscillatorfromitsphysicalparameters,andcompletelyspecifyitsequationofmotion.


Notes



292

WorkedExample



Anidealgrandfatherclockconsistsofasimplependulumwhichswingsbackandforthonceeverysecond.Whatisa)theoscillationfrequency,b)theangularfrequencyandc)howfarfromtheendoftherodshouldthemasssit?

STRATEGY

First, connect the period to the frequency and angular frequency; then we can find the length of the pendulum associated with that angular frequency.

IMPLEMENTATION

1. In order to answer part (a) and (b), we need to consider how the frequency of simple harmonic motion relates to the period and also how the oscillation frequency of the pendulum depends on the angular frequency.

2. How does the length of the pendulum relate to the frequency of the pendulum? CALCULATION

1. The period of the spring is inversely related to the oscillation frequency of the spring by 1Hz. The angular frequency is related to the oscillation frequency by 2

2 radians/s.

2. The angular frequency of the pendulum is given by / . Therefore the pendulum is length is 0.248m.

Score(3)



293


1.Inyourownwords,describethedifferencebetweentheoscillationfrequencyandtheangularfrequency.

2.Inyourownwords,explainwhytheperiodofthependulumisnotdependentonthemassofthependulum.

3.Inyourownwords,describewhytheperiodofthependulumisinverselyproportionaltothelengthofthependulum.



294

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMAspaceprobeissenttoadistantplanettodetermineifitissuitableforcolonization.Afterhavingsuccessfullymetalltheothercriteria,thereisoneremainingtest:isthegravitationalpulloftheplanetwithin30%ofEarth’snormalgravity?Theprobecontainsasimplependulumwhichitusestodeterminethegravitationalconstantofthatplanet.Thecompactpendulumisonly5.0cmlongandtakes0.33secondstomovefromtheleftmostpartofitsswingtothecenterofitsswing.Istheplanetsuitableforcolonization?

Answer:No,sincegnew=1.14m/s2



295

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. Astronauts in space took a coiled spring of known spring constant k, attached a bob (small mass) to it,

and set it oscillating. Measuring the period, they could determine

a) the time of day the acceleration due to gravity b) the mass of the bob c) the weight of the bob

3. Anadultandachildaresittingonadjacentidenticalswings.Oncetheygetmoving,theadult,by

comparisontothechild,willnecessarilyswingwith

a) amuchgreaterperiodb) amuchgreaterfrequencyc) thesameperiodd) thesameamplitude

4. CRITICALTHINKING:Anoscillationisaphysicalphenomenoncharacterizedbythefactthatthe

configurationofthephysicalsystemrepeatsitselfoverandoveragain.Simpleharmonicoscillationsareaspecialcase.An oscillation is simple harmonic if the period does not depend on the amplitude. In the following set, identify the oscillations that are simple harmonic, the ones that are approximately simple harmonic, and the ones that are not simple harmonic. Briefly explain your reasoning for each.

a) Thependuluminagrandfatherclock.b) Aboatinwaterpusheddownandreleased.c) Achildonaswing.d) Amasshangingfromanidealspring.e) Apingpongbouncingonthefloor.



296

HomeworkProblems

13.22



297

13.67



298

13.43



299

Lesson40

EnergyinSimpleHarmonicMotion

Reading 13.5Examples 13.5HomeworkProblems 13.29,13.63,13.73

LearningObjectives


Notes



300

WorkedExample



Amotionlessmassisconnectedtoaspring(withaspringconstantof85N/m)whichiscompressed30cmfromitsequilibriumposition.Themass,whichisrestingonafrictionlesssurface,isthenreleased.Atwhatpositionwillthekineticenergyofthesystembeequaltoexactlyhalfthepotentialenergyofthesystem?STRATEGY

Since the spring starts at rest, the system has potential energy, but no kinetic energy. When the spring is released, the total mechanical energy of the system is conserved. This means that when the initial potential energy is equal to 2/3 of it’s initial value, the kinetic energy will be half the potential energy of the system.

IMPLEMENTATION

1. What is the spring potential energy of the system as a function of position?

2. What is the total mechanical energy of the system?

3. We then solve for the position where the potential energy is equal to 2/3 of its initial value.CALCULATION

1. ∆ 3.83J

2.Since the system is initially at rest, the total mechanical energy is equal to the initial potential energy.

3. 2.55J ∆ / . Solving for position gives ∆ / 24.5cm.

Score(3)

30cm



301


1.Whydon’tyouneedtoexplicitlycalculatethekineticenergyofthesystem?

2.Whatpointisthefinalanswerfordisplacementrelativeto?

3.Whydoesthepointwherethepotentialenergyisequalto2/3itsinitialvaluecorrespondtothepointwherethekineticenergyishalfthepotentialenergy?



302

Pre‐ClassProblem

STATEMENTOFTHEPROBLEMAmotionlessmassisconnectedtoaspringwhichisstretched45cmfromitsequilibriumposition.Themass,whichisrestingonafrictionlesssurface,isthenreleased.Themaximumkineticenergyofthesystemis10.6J.Whatisthespringconstantofthespring?

Answer:132N/m



303

PreflightQuestions

1. Whattopicfromthereadingwouldyouliketodiscussduringclass?2. For thesimpleharmonicmotionofamassonaspringwithoutfriction,itistruethat

a) theenergyisindependentoftheamplitudeb) theenergyisindependentoftheperiodc) both(a)and(b)d) neither(a)nor(b)

3. The position x(t) of a simple harmonic oscillator is shown to the right

as a function of time. Which of the graph sets below correctly represent the kinetic and potential energies of the oscillator ?

a) GraphAisthepotentialenergy,graphCisthekineticenergy.b) GraphCisthepotentialenergy,graphAisthekineticenergy.c) GraphBisthepotentialenergy,graphDisthekineticenergy.d) GraphDisthepotentialenergy,graphBisthekineticenergy.

4. CRITICALTHINKING:Foragivenharmonicoscillator,ifthespringconstantandthemassare

bothdoubledbuttheamplituderemainsthesame,explainwhathappenstothemechanicalenergyoftheoscillator.

A B

C D



304

HomeworkProblems

13.29



305

13.63



306

13.73

Physics110HJournal‐2013‐2014 Block4Review


307

Block4Review

LearningObjectives
















[Obj64] Determinetheperiodandfrequencyofasimpleharmonicoscillatorfromitsphysicalparameters,andcompletelyspecifyitsequationofmotion.



Notes

Block4Review Physics110HJournal‐2013‐2014


308




A3‐mdiameterflywheelisspinningupwithanangularaccelerationof3rad/s2.Whatisthetotallinearaccelerationattherimofthewheelattheinstantwhenitsangularvelocityis12rpm?STRATEGY(Fillintheblank)

Since the wheel has an angular acceleration, there is tangential linear acceleration . We note that the linear acceleration depends on the radius, i.e. points farther from the center have larger linear accelerations (as well as larger linear velocities.) The rotating points also have a centripetal acceleration, directed at the center of rotation. We calculate both accelerations and add the two vectors, which are perpendicular to each other. CALCULATION

The tangential acceleration at = _____ x _____ = 4.5 m/s2 The centripetal acceleration is equal to / . In terms of angular velocity . For the equation to be valid, the angular velocity has to be in radians. To convert 120 rpm into radians we write 12

1.3 / .

The centripetal acceleration is then _______x ________ = 2.5 m/s2. The magnitude of the total acceleration is 5.1 m/s2. The direction of the total acceleration is 60.9 degrees. SELF‐EXPLANATIONPROMPTS

1.Explaininyourownwordswhytheangularspeedofarigidrotatingobjectisthesameforallpartsoftheobjectwhilethelinearspeedsofdifferentpartsoftheobjectvarywiththeradius.

2.Justifytherelationac=ω2r.

3.Isthecentripetalaccelerationinanywayrelatedtotheangularacceleration?




309




Twoweightsofmasses2mandmareattachedtoeitherendofathinrodoflengthL.Calculatetherotationalinertiaofthemass‐rodsystemaboutaperpendicularrotationalaxisthroughthecenteroftherod.Assumethethinrodhasnegligiblemass.STRATEGY(Fillintheblanks.)

The rotational inertia of an object that consists of multiple discrete masses depends on how those discrete masses are spatially distributed relative to the axis of rotation.

For this problem, the object consists of three components: _______________, _______________, and ___________ connecting the two weights. The thin rod has negligible mass, so it does not contribute to the rotational inertia of the system. To determine the rotational inertia of the system, we will sum the rotational inertia of each component. CALCULATION(Fillintheblanks.)

The rotational inertia of the object is determined by summing the individual rotational inertias for each discrete mass.

When the rotation axis is through the center of the rod, the rotational inertia is:

∑ ____________+ ___________ 3 4


1.Inyourownwords,explainrotationalinertiaofanobject.

2.Inthisproblem,whywasthedistancefromtherotationaxisrequaltoL/2forbothweights?

3.Wouldtherotationalinertiaoftheobjectincrease,decrease,orstaythesameiftherotationaxiswasatoneendinsteadofthroughthecenteroftherod?




310




A50‐kgbucketishangingfromarope,whichiswoundarounda20‐kgsoliddisc.Thediameterofthediskis50cm.Themassoftheropeisnegligiblecomparedtotheothermassesintheproblem.Whatistheaccelerationofthefallingbucket?


We will draw free-body diagrams for the two objects of interest: the bucket and disk. We will then apply Newton’s Second Law to each object and the system of two equations for the unknown acceleration.


Newton’sLawforthefallingbucket:

_____________

Newton’sLawfortherotatingdisc:

_______________ ½

Eliminating from the two equations and solving for the acceleration we get

12

8.2m/s


1.Supplythemissingalgebra.Writedownthetwoequationsofmotion,eliminatethetensionandsolvefortheacceleration.

2.Inafewsentencestrytoexplainwhytheradiusofthediscdoesnotaffectthefinalresult.




311




YouarehelpingtounloadcargooffaC‐130Herculesaircraft.Thecargoispackedinbarrels,soyoudecideitwillbeeasiertorollthebarrelsdowntherampatthebackoftheaircraftandofftheplane.Ifabarrelhasaspeedof0.5m/swhenitreachestheramp,whatisitsspeedafterithasrolleddowntherampwithoutslippingandofftheplane?Theverticalheightoftherampis1.5meters.


We will use the principle of conservation of _________________ to solve for the speed of the barrel at the end of the ramp.


Starting with conservation of _________________,

we substitute in the types of energy in the initial (at the top of the ramp) and final (at the bottom of the ramp) states.

______ + (______ + _______)0 =( )f Now, we replace ω with because the barrel is not ____________ and put rotational inertia in terms of mass and radius of the barrel. (Approximating the barrel as a solid cylinder, its rotational inertia is )

12

12

12

Solving for the final speed of the barrel at the end of the ramp gives: √_______________ 4.5 /


1.Comparetherotationalinertiaofahollowcylindertoasolidcylinder.Ifthebarrelwereinsteadhollow,woulditreachthebottomoftherampearlierorlaterthanifitwassolid?Assumethesameinitialspeed.

2.Doesthefinalanswerdependonthemassortheradiusofthebarrel?Explain.




312




Whatisthemagnitudeanddirectionoftheangularmomentumofa10‐kgsoliddisc,60cmindiameter,rotatingcounter‐clockwiseat120rpmarounditscentralaxis?


We will use the relation, ___ , and the ____________ rule to find the answer. CALCULATION(Fillintheblanks.)

First, we need to convert rotational speed from rpm to rad/s

______

For a solid disc rotating around it central axis

_____ _______ 0.45kgm

Now substitute to get:

5.66kgms

The direction, from the right-hand rule, is __________________________.


1.Howwouldtheanswerchangeisthediscwasahoop?

2.Whyaretherenounitsofradiansinthefinalanswer?




313




Twodisksarerotatingatdifferentspeedsalongthesameaxisasshown.Thetopdiskis5kgandrotatingat1.0rad/s;thebottomdiskis10kgandrotatingat2.0rad/s.Ifthetopdiskisreleasedandlandsonthebottomdisk,whatisthefinalangularspeedofthecombineddisks?STRATEGY(Fillintheblanks.)

This problem is an example of a _______________________. There is no net torque acting on the system, so the total angular momentum does not change (conserved).


The initial angular momentum of the top disk is:

I ___________________

The initial angular momentum of the bottom disk is:

I ___________________

The final angular momentum of the combined disks can be determined by ___________________________________.

Solving for the final angular velocity, we get:

_________________; _________________


1.Showthatkineticenergyisnotconversedintheexample.

2.Examplewhyangularmomentumisconversedbutkineticenergyisnot.




314




Ayounggirldecidestobuildasimplependulumtoknockoveratoycar.Todoso,shetiesoneendofa40‐cmstringtotherailingandarubberballtotheotherendofthestring.Thesetupisdesignedsothattheballwillhitthetoycarwhenthependulumisatthelowestpointofitsarc.Thegirlpositionsthependulumsothatthestringistightandtheballis10cmofftheground.Howlongdoesittakefortheballstrikethetoycar?


First, we find the period of the pendulum. The time it takes the ball to swing from its initial position to the collision point is ¼ the period.


______

The time it takes the pendulum to strike the car is 0.32s.


1.Whyisthefinalansweronly¼oftheperiod?

2.Whydoesn’ttheinitialheightofthependulumaffecttheperiodoftheswing?




315




A2.0‐kgmassisattachedtoaverticallyorientedspringwhichhasaspringconstantof25N/m.Thespringiscompressed31cmrelativetoitsequilibriumpoint.Whatisthespeedofthemassattheequilibriumpointofthespring?STRATEGY(Fillintheblanks.)

This is a conservation of energy problem. The mass initially has both gravitation potential energy and spring potential energy. At the equilibrium point, all of this energy is converted to kinetic energy.


For convenience, we choose the equilibrium point of the spring to be the reference point.

The gravitational potential energy is ______=6.99J

The spring potential energy is _______=2.40 J

At the equilibrium point, ______= .

Solving for velocity we find that 3.1 m/s.


1.Whyistheequilibriumpointaconvenientchoiceoforigin?

2.Explain,inyourownwords,howyoufoundthekineticenergyattheequilibriumpoint.

3.Whycanthespringenergybecombinedwiththegravitationalenergy?




316

Notes



317

Notes



318

Notes



319

Notes



320

Notes

Physics110HJournal‐2013‐2014 AppendixA:LabReportTemplate

xix

AppendixA:LabReportTemplate

Purpose

Aformallabreportisessentialtothescientificprocess.Itisthemostcommonwaythattheresultsofascientificstudyarecommunicatedtothescientificcommunity.Youmayassumeyouraudienceisscientificallyliteratebuthasnotperformedtheexperimentinquestion.

Format

Usethefollowingguidancetoformatyourreport:

Useaclearlyreadable12‐pointfont Setthepagebordersto1” Spacelineswithinthesameparagraphat1.0or1.15 Separateparagraphswithadoublespace Usesectionheadingstoidentifytransitionsbetweensections Refertoyourexperimentand/orcalculationsinthepasttense Usethethird‐person(i.e.,avoidusing“I”or“we”) Usescientificnotationwhereappropriateandincludeallunits(e.g.,1.1 10 m) Citeanyoutsidesources(otherthanyourtextbook)usingMLAformat Ifyouincludeafigure,centeritinthepageandensureithasadescriptivecaption

underneathit.Youmayneatlyhand‐drawfigures. Graphsorplotsmaybeusedtosummarizedataandshowanalysis.Theymustincludea

titleandlabeledaxesandmustnotbedonebyhand.Ensurethegraphorplotislargeenoughtobeclearlyreadandinterpretedbyyourreader.Forclarity,youmaychoosetocross‐referencethegraphorplotandincludeitasanattachmentattheendofyourreport.

Yourinstructormayprovideadditionalguidance.

Sections

Usethefollowingformattocreateyourlabreport.Rememberthereisabalancebetweentoolittleinformationandtoomuchinformation.Youwantyourreporttoincludewhatisrelevantwithoutbecomingtoolong,complicated,orconfusing.Yourreaderwilllikelynotstrugglethroughapoorlywrittendocument,whichmeansheorshewillneverlearnofyourresultsorfindings.

TitlePage

Useaseparatetitlepage.Ensurethetitleofyourreportiscenterednearthetopofthefirstpage.Listallcontributorstothelabreportunderneaththetitle.Alsoincludethecoursenameandnumberandthedate.Atthebottomofthetitlepage,neatlyincludeyourdocumentationstatementforanyoutsidehelpyouobtained(butnotoutsidereferences).

AppendixA:LabReportTemplate Physics110HJournal‐2013‐2014

xx

Introduction

Inthissection,youwillprovideabriefintroductiontoyourreaderaboutyourpurposeandtheimportanceofyourwork.Youshouldalsobrieflysummarizeanypertinentmaterial,includingrelevantequationsorconcepts.

ExperimentalMethods

Inthissection,youmustsuccinctlydescribethemethodsyouusedtoobtainyourdata.Ingeneral,readerswillbemostinterestedinreadingaboutyourdata,results,andconclusions,butifyourresultsareinteresting,areaderwillalsobeinterestedinhowyouobtainedyourdata.Focusonkeepingthissectioncompletebutconcise.Graphicsandfiguresshouldbeusedsparinglyinthissection.

ResultsandDiscussion

Thisisanextremelyimportantsectionofyourreport.Hereyoushouldcommunicatewhatyoufoundanddrawpertinentconclusionsbasedoninterpretationofyourdata.Graphics,figuresandExcelplotsmaybeusedtoeffectivelycommunicateyourresults.FollowtheguidanceintheFormatsection.Makesureyourresultsarecorroboratedorjustifiedbythedatayouobtained.Keepinmindthatnotalldataisnumerical.Forthelabsinthiscourse,youmaybeabletomakesomepowerfulconclusionsbasedonqualitativeobservations.

Conclusion

Hereyouwillsummarizeyourresults.Ensureyoudonotintroduceanynewinformationinthissection.

References

Ifyoucitedanyreferences,includethemhereusingMLAformat.

Appendixes(ifneeded)

Usethissectiontoplaceanylargeorcomplicatedgraphsordataplots.

Grading

Seetheindividuallablessonsfortherubricyourinstructorwillusetogradeyourreport.Youwillbegradedontheappearanceandqualityofyourlabreport.

Physics110HJournal‐2013‐2014 AppendixB:SignificantFigures,UncertaintyandErrorPropagation

xxi

AppendixB:SignificantFigures,UncertaintyandErrorPropagation

References:

[1] P.R.Bevington,andD.K.Robinson,DataReductionandErrorAnalysis,3rded.,McGraw‐Hill,NewYork,2003.[2] USAirForceAcademy,CoreChemistry/PhysicsLaboratoryDataAnalysisGuide.

Numericcalculationsandexperimentalmeasurementsareonlyasaccurate(orreliable)astheleastprecisemeasurement.Everyphysicalmeasurementhasuncertaintyandlaboratorymeasurementsdonotyieldexactresults.Errorsanduncertaintiesinphysicalexperimentsmustbereducedthroughexperimentaltechniquesandrepeatedmeasurements–remaininguncertaintymustbeestimatedandreportedtoestablishthevalidityoftheresult.

Thetermerrorisdefinedasthedifferencebetweenobserved(orcalculated)valueandthe“true”value.Inlaboratorymeasurementswerarelyknowthetruevalue,thereforewemustestablishsystematicmeansofdeterminingthevalidityofourexperimentalresults.Errorsthatoriginatefrommistakesinmeasurementareknownasillegitimate(gross)errors,andarecorrectedthroughattentionandcarefulrepeatedmeasurements.Inourexperimentsweareconcernedwithuncertaintiesintroducedbyrandomfluctuationsinmeasurementsandsystematicerrorsthatlimittheprecisionandaccuracyofourresults.Randomerrorsarefluctuationsthatoccurinobservationseachtimeameasurementisrepeated.Randomerrormaybereducedthroughlaboratorytechniqueorrepeatedobservations.Systemicerrorsaredifficulttodetectandmaymakeallourresultsvarywithreproduciblediscrepancy.Systemicerrormayresultfrompoorlycalibratedequipmentorbiasbytheobserver.

Accuracyisameasureofhowclosetheresultistothetruevalue.Precisionisameasureofhowwelltheresulthasbeendetermined(withoutregardtoagreementwithtruevalue).Precisionisalsoameasureofanexperiment’sreproducibility.FigureB1illustratesthedifferencebetweenaccuracyandprecision.

FIG.B1.Accuracyandprecisiondemonstratedthroughtargetpractice.Target(a)isaccurate,butnotprecise,while(b)isprecisebutnotparticularlyaccurate.

(a)(b)

AppendixB:SignificantFigures,UncertaintyandErrorPropagation Physics110HJournal‐2013‐2014

xxii

SignificantFigures

Thenumberofdigitsinreportinganexperimentalresultimpliestheprecisionofameasurementanduncertaintyshouldbereportedspecificallywitheachnumericresult.

Rulesfornumberofsignificantfigures:

1. Leftmostnonzerodigitisthemostsignificantdigit.2. Ifthereisnodecimalpoint,therightmostnonzerodigitistheleastsignificantdigit.3. Ifthereisadecimalpoint,therightmostdigitistheleastsignificantdigit,evenifitisazero.4. Thenumberofdigitsbetweenthemostandleastsignificantdigitcountareknownastheas

significantfigures.

Rulesforsignificantfiguresincalculatingnumbers:

1. Multiplication/division.Thenumericresultcannothavemoresignificantfiguresthananyoftheoriginalnumbers.

2. Addition/subtraction.Theresultcannothavemoresignificantdigitstotherightofthedecimalpointthananyoftheoriginalnumbers.

3. Roundingresults.Insignificantdigitsaredroppedfromtheresultandthelastdigitisroundedforbestaccuracy.

Uncertainty

Everyphysicalmeasurementhasuncertaintyduetotheaccuracyorprecisionoflaboratoryequipmentandtherandomdistributionofourdata.Sincewedonotnormallyknowtheactualerror(discrepancyfromthetruevalue)inexperimentalresults,weseektodevelopamethodofdeterminingtheestimatederror.Analysisofthedistributionofrepeatedmeasurementscanleadtoanunderstandingoftheexperimentalerror,reportedasthespreadofthedistribution.Determinethebestvalue anduncertaintyestimate ,andreporttheresultas

. (B1)

Uncertaintyinexperimentalmeasurementscanbeestimatedinanumberofways,includingstandardreadingofanalogordigitalinstruments,orstatisticalanalysisofthedistributionofrepeatedmeasurements.


xxiii

UncertaintyinAnalogMeasurements

Uncertaintyinreadinganalogdevices(rulers,balances,graduatedcylinders,etc.)isestimatedasonehalfthesmallestdivisionmarkedonthedevice.

Example:Alengthmeasurementistakenusingarulermarkedinincrementsof1mmasshowninFigureB2.Uncertaintyisonehalfthesmallestincrement,or0.5mm.Thelengthisreportedas26 0.5mmaccordingtoEqn.B1.

UncertaintyinDigitalMeasurements

Manymodernlaboratorydevicesaredigital(scales,timers,multimeters,etc.).Systematicerrorisreducedifthedeviceisproperlycalibrated.Estimateduncertaintyistheleastsignificantdigitthatcanbedisplayedifthereadingisconstant(i.e.notfluctuating).Ifthereadingisfluctuating,repeatedmeasurementsmustbetakenandothermethodsofestimatinguncertaintymustbeused.

Example:Atimemeasurementistakenusingaphotogatetimerreadingtoone‐thousandthofasecondasshowninFigureB3.Uncertaintyistheleastsignificantdigit,or0.001s.Thetimemeasurementisrecordedas1.673 0.001saccordingtoEqn.B1.

UncertaintyinRepeatedMeasurements

Repeatedmeasurementshelpusextractthebestvalueofourexperimentalresultsanddeterminetheestimatederrorwithconfidence.Aswetakemoremeasurements,weexpectapatterntoemergewithdatapointsdistributedaroundthecorrectvalue(assumingwecorrectforsystematicerrors).

Supposeduringanexperiment,wetakeasampleof measurementsofaquantity .

Thearithmeticmean oftheexperimentaldistributionisgivenas

∑ . (B2)

Theexpressionforthestandarddeviation ofthesamplepopulationisgivenby

∑ (B3)

whichrepresentsthebestestimateforthedeviationsquaredoftheparentdistribution(asifwetookandinfinitenumberofmeasurements)basedonthesmallersampledistribution.Thestandarddeviation representsaquantitativemeasureoftheuncertaintyinanysinglemeasurement.Ifwe

FIG.B2.Analogmeasurementusingaruler.

FIG.B3.Digitalmeasurementusingphotogatetimer.

AppendixB:SignificantFigures,UncertaintyandErrorPropagation Physics110HJournal‐2013‐2014

xxiv

weretotakeanothersamplemeasurementthereisa68.2%chanceitwillbewithin ,a95.4%chanceitwillbewithin 2 ,anda99.7%chanceitwillbewithin 3 ,asshowninFigureB4.

Givenrepeatedtrialsandcalculationsofthemean,itispossibletodeterminevariationinthevalueofthemean.Whendeterminingexperimentalresultswithalargesamplesize,weseekaquantitativemeasureofthestandarddeviationofthemean,orthestandarddeviationinthesamplemeanrelativetothetrue(mean)value.

√

∑ (B4)

Experimentalresultsbasedonthissampledistributionarereportedas

. (B5)

Forsmallsamplesets(threeorfewertrials),wemayestimateuncertaintyusingtheexpression

. (B6)

ErrorPropagation

Experimentalquantitiesderivedfrommeasuredvalueswithuncertaintywillinturnhaveuncertainty.Estimateduncertaintyiscalculatedbasedonthemathematicaloperationsusedinthederivation.Supposewemeasurevalues( , , , , , )withuncertainty( , , , , , ).Weseektheuncertaintyinacalculatedvalue .

AdditionorSubtractionwithUncertainty

If

then

. (B7)

MultiplicationorDivisionwithUncertainty

If

2 2

34.1%

34.1%

13.6%

13.6%

FIG.B4.Normal(Gaussian)distributionwithstandarddeviation .


xxv

then

| |

. (B8)

MultiplicationbyaKnownValuewithUncertainty

IfQiscalculatedbymultiplyingbyaknownvalue (e.g. 2 or )byaquantity with

uncertainty,

then

| | (B9)

orequivalently

| | | |

. (B10)

UncertaintywithExponents

If isanexactnumberand

then

| |

| || |. (B11)

ReportingExperimentalValues

Itshouldbeemphasizedthatuncertaintyestimatesareonlyestimatesandvaluesshouldbepresentedwithappropriateprecision.

Rulesforreportingexperimentalvalues:

1. Theleastsignificantfigureinanyreportedvalueshouldbethesameorderofmagnitude(samedecimalposition)astheuncertainty.

2. Estimateduncertaintyisnormallyroundedtoonesignificantfigure.

AppendixC:MathematicsReference Physics110HJournal‐2013‐2014

xxvi

AppendixC:MathematicsReference

QuadraticFormula

Solutionsofthequadraticequation 0aregivenbythequadraticformula.

QuadraticFormula √ 42

CoordinateSystems

Conventiondictatesright‐handedcoordinatesystems.Alternatecoordinatesystemsmaybeused–besuretoclearlyindicatechosencoordinateaxes.Twocommoncoordinatesystemsusedinphysicsareshownbelow.

CartesianCoordinateSystem

SphericalCoordinateSystem

00 0 2

sin cos sin sin cos

sin sin sin

, ,

, ,

Physics110HJournal‐2013‐2014 AppendixC:MathematicsReference

xxvii

Trigonometry

TrigonometricIdentities

sin sin cos cos

sin cos 1 1 tan sec 1 cot csc

sin 2 2 sin cos cos 2 cos sin 1 2 sin 2 cos 1

sin sin cos cos sin cos cos cos ∓ sin sin

sin sin 2 sin cos ∓

cos cos 2 cos cos

cos cos 2 sin sin

sinOppHyp

csc1

sinHypOpp

cosAdjHyp

sec1

cosHypAdj

tansincos

OppAdj

cot1

tancossin

AdjOpp

LawofSinessin sin sin

LawofCosines 2 cos

Opposite

Adjacent

AppendixC:MathematicsReference Physics110HJournal‐2013‐2014

xxviii

Vectors

Givenvectors and ,

∙ cos ∙

sin

ExponentialsandLogarithms

ln ln ln

ln ln ln ln

ln ln loglnln 10

≅ 2.71828… ln 1 ≡ 0 1

DerivativesandIntegrals

1

, where isaconstant. 1, 1.

sin cos sin1cos

cos sin cos1sin

1

ln1 ln

/ √

/ √

Physics110HJournal‐2013‐2014 AppendixC:MathematicsReference

xxix

TaylorSeriesExpansionsandApproximations

ATaylorseriesexpansionofarealfunction aboutapoint isgivenby

2!⋯

!.

SeriesExpansionsofCommonFunctions(for| | 1)

For| | ≪ 1

12! 3!

⋯ 1

sin3! 5! 7!

⋯ sin

cos 12! 4! 6!

⋯ cos 12

ln 12 3 4

⋯ ln 1

1 11

2!⋯ 1 1

AppendixD:EquationDictionary Physics110HJournal‐2013‐2014

xxx

AppendixD:EquationDictionary

Oncertainlessons,youwillhavetheoptiontocompleteaworksheetonaparticularequationforpre‐flightpoints.Theequationdictionaryisdesigned:(1)toallowyoutobecomemorefamiliarwithanequation,and(2)toenableyoutocreateahighlyorganizedandeasilyaccessiblestudyguideforexampreparation.Themoretimeyouspendcreatingmeaningfulentriestoyourequationdictionary,themorepreparedyouwillbeforexamsinthecourse.

Inthewhiteboxontheupperleftcorner,youwillfindtheequationreferencenumber.Intheblackboxintheupperrightcorner,youwillfindthelessonnumberwheretheequationisfirstintroduced.Anexampleofahigh‐qualityequationdictionaryentryisshownbelow.

Physics110HJournal‐2013‐2014 AppendixD:EquationDictionary

xxxi

Variables(includeallunits)

DescriptionandNotes

Diagram

PhysicsConcept

#1


xxxii


DescriptionandNotes

Diagram

12

PhysicsConcept

#2


xxxiii


DescriptionandNotes

Diagram

PhysicsConcept

#3


xxxiv


DescriptionandNotes

Diagram

∙

PhysicsConcept

#4


xxxv


DescriptionandNotes

Diagram

∆ /

PhysicsConcept

#5


xxxvi


DescriptionandNotes

Diagram

PhysicsConcept

#6


xxxvii


DescriptionandNotes

Diagram

12

PhysicsConcept

#7


xxxviii


DescriptionandNotes

Diagram

PhysicsConcept

#8


xxxix


DescriptionandNotes

Diagram

cos

PhysicsConcept

#9

AppendixE:RotationalInertiasandAstrophysicalData Physics110HJournal‐2013‐2014

xl

AppendixE:RotationalInertiasandAstrophysicalDataTable10.2RotationalInertias

AstrophysicalData

Earth Mass 5.97 10 kg

Meanradius 6.37 10 mOrbitalPeriod 3.16 10 SurfaceGravity 9.81 m s

Moon Mass 7.35 10 kg

Meanradius 1.74 10 mOrbitalPeriod 2.36 10 SurfaceGravity 1.62 m s

Sun Mass 1.99 10 kg

Meanradius 6.96 10 mOrbitalPeriod 6 10 SurfaceGravity 274 m s

Physics110HJournal‐2013‐2014 AppendixF:UnitsandConversions

xli

AppendixF:UnitsandConversionsRef:http://wwwppd.nrl.navy.mil/nrlformulary/NRL_FORMULARY_07.pdf

PhysicalQuantity Dimension SIUnits SISymbolConversionFactor GaussianUnits

Length BASEUNIT 10

Mass BASEUNIT 10

Time BASEUNIT 1

ElectricCurrent BASEUNIT 3 10

Temperature

AmountOfSubstance

PhysicalQuantityDimensions

SIUnits SISymbolInTermsofOtherSIUnits

ConversionFactor GaussianUnits

SI Gaussian

Acceleration 10

AngularVelocity 1

11

Capacitance

∙∙

9 10

Current / /

BASEUNIT 3 10

CurrentDensity

/

/3 10

Density 10

Displacement /

/12 10

ElectricCharge ∙ 3 10

ElectricField

/

/

∙∙

13

10

ElectricPotential , ϕ

/ / ∙∙

13

10

Entropy 10

Force

∙10

Frequency , 1

1 11

Impedance,Resistance ,

∙

∙∙

19

10

Inductance

∙ ∙∙

∙∙

19

10

MagneticField

/

/

∙∙ ∙

10

MagneticFlux Φ

/ /

∙∙ ∙ ∙

∙ 10

Momentum,Impulse ∙

∙10

∙

Permeability 1

14

10 ‐‐‐‐‐

Permittivity 1 36 10 ‐‐‐‐‐

Power

∙ ∙ 10

Pressure,Stress Tension

∙10

SpecificHeatCapacity ∙ ∙

ThermalConductivity

∙10

∙ ∙

Torque ∙

∙10

VectorPotential

/ /10 ∙

Velocity 10

Viscosity

∙∙ 10

Work Energy ∙

∙10


xlii

Ref:http://physics.nist.gov/cuu/Constants/index.htmlPhysicalConstant Symbol Value Uncertainty

AtomicMassUnit 1.660 538 782(83) 10-27 kg 931.494 028(23) MeV 7.513 006 671(11) 1012 cm-1 1.492 417 830(74) 10-10 J

AtomicUnitOfCharge 1.602 176 487(40) 10-19 C

AvogadroConstant 6.022 141 79(30) 1023 mol-1

BohrMagneton 2⁄ 927.400 915(23) 10-26 J T-1 5.788 381 7555(79) 10-5 eV T-1 ∙ 0.466 864 515(12) cm-1 T-1 ∙ 13.996 246 04(35) 109 Hz T-1

BohrRadius 0.529 177 208 59(36) 10-10 m 0.529 177 208 59(36) Å

BoltzmannConstant ⁄ 1.380 6504(24) 10-23 J K-1 8.617 343(15) 10-5 eV K-1 0.695 035 6(12) cm-1 K-1 2.083 664 4(36) 1010 Hz K-1

ComptonWavelength 2.426 310 2175(33) 10-12 m

ElectronMass 9.109 382 15(45) 10-31 kg 0.510 998 910(13) MeV 5.485 799 0943(23) 10-4 amu 8.187 104 38(41) 10-14 J

ElectronVolt 1.602 176 487(40) 10-19 J 1.073 544 188 10-9 amu 8.065 544 65(20) 103 cm-1 2.417 989 454 1014 Hz

FaradayConstant 96 485.339 9(24) C mol-1

Fine‐StructureConstant ⁄ 7.297 352 537 6(50) 10-3 1 / 137.035 999 679(94)

GravitationalConstant 6.674 28(67) 10-11 m3 s-2 kg-1

Impedance Vacuum 376.730 313 461 Ω

Joule 1.112 650 056 10-17 kg 6.241 509 65(16) 1018 eV 5.034 117 47(25) 1022 cm-1

Kelvin 1.380 6504(24) 10-23 J 8.617 343(15) 10-5 eV 0.695 035 6(12) cm-1

Kilogram 6.022 141 79(30) 1026 amu 5.609 589 12(14) 1035 eV 8.987 551 787 1016 J

InverseCentimeter 1.986 445 501(99) 10-23 J 1.239 841 875(31) 10-4 eV 1.331 025 0394(19) 10-13 amu 2.997 924 58 1010 Hz

MolarGasConstant 8.314 472(15) J mol-1 K-1

MolarVolumeOfIdealGas 22.413 996(39) 10-3 m3 mol-1 (@ 273.15 K, 101.325 kPa)

NeutronMass 1.674 927 211(84) 10-27 kg 939.565 346(23) MeV 1.008 664 915 97(43) amu 1.505 349 505(75) 10-10 J

NuclearMagneton 2⁄ 5.050 783 24(13) 10-27 J T-1 3.152 451 2326(45) 10-8 eV T-1 2.542 623 616(64) 10-4 cm-1 T-1

Permeability MagneticConstant 12.566 370 614 10-7 N A-2 4 π 10-7 H m-1

Permittivity ElectricConstant 8.854 187 817 10-12 F m-1

PlanckConstant 6.626 068 96(33) 10-34 J s 4.135 667 33(10) 10-15 eV s

PlanckConstant/2π 1.054 571 628(53) 10-34 J s 6.582 118 99(16) 10-16 eV s

ProtonMass 1.672 621 637(83) 10-27 kg 938.272 013(23) MeV 1.007 276 466 77(10) amu 1.503 277 359(75) 10-10 J

RydbergConstant 10 973 731.568 527(73) m-1 109 737.315 8 cm-1 ( ) 13.605 691 93(34) eV ( ) 2.179 871 97(11) 10-18 J

SpeedOfLight Vacuum 299 792 458 m s-1 /

StdAtmosphere 101 325 Pa (@ 273.15 K) 29.92 inHg 14.696 psi 760 Torr

StdAccelerationOfGravity 9.806 65 m s-2

Stefan‐BoltzmannConstant 5.670 400(40) 10-8 W m-2 K-4 NUMBERSandAPPROXIMATIONS π 3.141592653589793… e 2.718281828459045… ≅1973.27 eVÅ ≅0.511MeV


xliii

Physics215ConstantsandEquationsSheet

MaxwellEquations

Gaussfor ∙ Gaussfor ∙ 0

Faraday ∙Φ

Ampère ∙Φ

Φ ∙

⋅ sin

∆ ⋅ sin

∆∆

1.22

ƐΦ

Φ ∙ ∆ ∆2

ElectricForce(Coulomb)Const 8.99 10 N m C 4

ElectricConst(Permittivity) 8.85 10 C N m Conversions:

MagneticConst(permeability) 4 10 N A 1G 10 T

ElementaryCharge 1.60 10 C 1eV 1.60 10 J

ElectronMass 9.11 10 kg

ProtonMass 1.67 10 kg

PlanckConst 6.63 10 J s


Physics110HCourseSyllabus

KEY: –Doubleperiod;EPQUIZ–Exam‐PrepQuiz;

CE–ConceptualExample;CTE‐CriticalThinkingExercise

Lab– LabExercise;

LessonTitleLearningObjectives

Reading ExamplesHomeworkProblems

1 Introduction PHYSICSTESTING 1–6Chapter1.2.1,2.2

2.1,2.2 1.16,1.24,2.51

2 Displacement,Velocity,andAcceleration 4–9 2.3,2.4 2.1–2.3 2.20,2.35,2.79

3 Lab1–AccelerationDuetoGravity 10 2.5,Lab1 2.6 2.38,2.42,2.78

4 Two–Dimensional&ProjectileMotion 11–14 3.1–3.5 3.3 3.34,3.53,3.54

5 ProjectileMotion EPQUIZ 14,15 3.5 3.4 3.55,3.70,MP

6 Lab2–ProjectileMotion 15 Lab2 3.4 MP,3.58,MP

7 AccelerationinCircularMotion 16,17 3.6 3.7,3.8 MP,MP,3.80

8 GRADEDREVIEW1

9 ForcesandNewton’sLawsofMotion 18–20 4.1–4.4 4.1,4.2 4.15,4.26,4.60

10 UsingNewton’sLaws 19–23 4.5,4.6 4.3,4.4,4.5 4.34,4.47,4.49

11 Newton’sLawsinTwoDimensions 22 5.1 5.1,5.2 5.16,MP,5.38

12 Newton’sLawswithMultipleObjects 23 5.2 5.4 5.19,5.21,5.71

13 Lab3–Newton’sLaws 23 5.2,Lab3 5.4 MP,MP,MP

14 Newton’sLawsinCircularMotion EPQUIZ 16,17,22 5.3 5.5,5.6,5.7 5.65,5.73,MP

15 Newton’sLawswithFriction 24,25 5.4,5.5 5.9,5.10,5.11 5.43,MP.5.57

16 CTE:Newton’sLawswithNon‐constantMass 18 9.3,Handout None 5.30,5.62,6.54

17 GRADEDREVIEW2

18 WorkwithConstantandVaryingForces 26,27 6.1,6.2 6.1–6.5 6.18,6.20,6.52

19 KineticEnergyandPower 28–29,34 6.3,6.4 6.6,6.7,6.9 6.29,6.64,6.71

20 PotentialEnergy 30–33 7.1,7.2 7.1,7.2 7.14,7.31,7.42

21 ConservationofMechanicalEnergy 34–35 7.3,7.4 7.4,7.5,7.6 7.24,7.25,7.55

22 Lab4–ConservationofEnergy 34 7.3,Lab4 7.5 7.56,7.59,7.63

23 OrbitalMotion 36–38 8.1–8.3 8.1,8.2,8.3 8.17,8.39,MP

24 GravitationalEnergy 39–41 8.4 8.4 8.27,8.52,13.41

25 CTE:OrbitalEnergies EPQUIZ 39,40 8.4 8.5 MP,8.61,8.67

26 CenterofMass 42 9.1 9.1,9.2,9.3 9.16,9.37,9.89

27 ConservationofLinearMomentum&Collisions 43–48 9.1–9.5CE9.1,9.4,9.5,9.7

9.38,MP,MP

28 Lab5–Collisions 46–48 9.5,9.6,Lab5 None 9.28,9.44,9.61

29 CollisionsandConservationofEnergy 47,48 9.3,9.4 9.10 MP,9.68,9.78

30 GRADEDREVIEW3

31 RotationalMotion 49–51 10.1,10.2 10.1,10.2,10.3 10.19,10.32,10.45

32 Torque&RotationalInertia PHYSICSTESTING 52–55 10.2,10.3 10.4,10.5 10.30,10.28,10.52

33 RotationalAnalogofNewton’sSecondLaw 55–56 10.3 10.8,10.9 10.56,10.57,MP

34 RotationalEnergyandRollingMotion 57–59 10.4,10.510.10,10.11,10.12,CE10.1

10.60,10.62,10.68

35 RotationalVectorsandAngularMomentum 60,61 11.1–11.3 11.1 11.16,11.17,11.21

36 ConservationofAngularMomentum 62 11.4 CE11.1 11.26,11.27,11.43

37 CTE:EnergyandAngularMomentum EPQUIZ 62 Chpt10&11 None 11.46,11.49,MP

38 Lab6–ConservationofAngularMomentum 62 11.4,Lab6 11.2 11.45,12.69,12.87

39 SimpleHarmonicMotion 63–65 13.1–13.3 13.1,13.3 13.22,13.67,13.43

40 ApplicationsofSimpleHarmonicMotion 66 13.5 13.5 13.29,13.63,13.73

FINALEXAM

Physics 110H 2013-2014 Journal FINALoap.nmsu.edu/JiTT_NMSU_workshop/CHAT_06_rsrc/110H_2013_14_Jo… · physics courses, which includes Newtonian mechanics and conservation of energy

Documents