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Transcript
Physics110HJournal
GeneralPhysicsI‐Fall2013/Spring2014
Name Instructor Section Physics110ConstantsandEquationsSheet
AccelerationduetoGravityatEarth'sSurface
UniversalGravitationConstant
SpeedofLightinVacuum
MassofEarth
9.81 m s N m kg
6.67 10
3.00 10 m s 5.97 10 kg
6.37 10 m
1in 2.54cm
1mi 1609m
1ft 0.3048 m
1 mi⁄h 0.447 m/s
1lb 4.448N
0.454kg
1
2
2
1
2
2
∆
1
2
1
2
∙ 1
2
cos
/
∙
∆
∙
Physics 110H Journal
General Physics I Fall 2013/Spring 2014
DepartmentofPhysics
i
ii
Physics110HJournal‐2013‐2014
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 iii
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 iv
Physics110HJournal‐2013‐2014
CourseDescriptionandPolicies
CourseDescriptionandPolicies
Overview
majoroptions:Astronomy,LaserPhysics/Optics,SpacePhysics,andAppliedPhysicstopicssuchas
nuclearphysics.Eachphysicsmajoroptionrequires42credithoursofphysicsandmathematics
110/110H(GeneralPhysicsI)isthefirstinatwo‐partseriesofintroductorycalculus‐based
physicscourses,whichincludesNewtonianmechanicsandconservationofenergyandmomentum,
andisnormallytakenduringthefourth‐classyear.PHYSICS215/215H(GeneralPhysicsII)isthe
secondintheseriesofintroductorycalculus‐basedphysicscourseswhichemphasizes
electromagnetismandcircuits,andisnormallytakenduringthethird‐classyear.
aptitudeincalculusorhavingpreviouslytakenintroductoryphysicscoursesmaybeplacedin
honorsphysics.Honorsphysicsincludesenhancedcoverageoftheconceptscoveredintheregular
course,withmoreintegrateduseofcalculus,introductiontodifferentialequationsandrigorous
dataanalysistechniques.
CorePhysicsCourseDescriptions
Physics110,GeneralPhysicsI,isacalculus‐basedintroductiontoclassicalphysics,withemphasis
oncontemporaryapplications,inwhichyouwilllearntheconceptsandproblem‐solvingskills
requiredtounderstandandanalyzethemotionofobjects.Thefirsthalfofthecourseisasolid
foundationinkinematicsandNewton’slawsofmotion.Youwillthenbeintroducedtoseveral
conservationprinciples,whichareelegantwaysofvisualizingandunderstandingthemotionof
objects.Theseincludetheconservationofenergy,momentumandangularmomentum.Alongthe
way,youwillbeintroducedtoafewtopicsthatareimportanttoscientistsandengineers,including
orbitalmotion,rotationalmotionandoscillations.Labsandsimulationshighlightkeyphysics
concepts.
Physics215,GeneralPhysicsII,isanintroductorycalculus‐basedphysicscoursewithanemphasis
oncontemporaryapplications.Thecoursebeginswithafoundationinthebasicpropertiesof
electricchargeandworksuptodealingwiththesophisticatedconceptoftheelectricfield.Then,
simplecircuitsareanalyzed,relatingtheprinciplesofpotentialenergyandelectricpotentialtothe
electricfield.Next,magneticfieldsandelectromagneticinductionarestudied,culminatingina
completedescriptionofelectromagneticfields.Afterthat,lightwaves,thebendingoflightandthe
v
CourseDescriptionandPolicies
Physics110HJournal‐2013‐2014
interferencecausedbythewavenatureoflightarestudied.Finally,modernphysicsisintroduced
bystudyingquantizationandquantumuncertainty.Thiscourseutilizesvectorsandcalculusin
problemsolvingandincludesin‐classlaboratoriestohighlightkeyconcepts.
HonorsCorePhysicsCourses
Asa“techie”majorinthePhysicsHonorsCoursesequence,youcanexpectanumberofbenefits
comparedtotakingthestandardintroductorycourse.Perhapsthemostsignificantbenefitis
Youwillalsoseeenhancedcoverageoftopicsthatareimportantforscientistsandengineers,
including




amoreintegrateduseofcalculusthroughoutthecourse
anintroductiontotheuseofdifferentialequationsinsimpleharmonicmotion
enhanceddataanalysistechniques
somewhatmoreemphasisongraphicalandnumericaltechniques
Theemphasisonthesetopicswillmakeyourphysicsexperiencecomparabletowhatyourpeers
PhysicsHonorswillincludealargepercentageofcommonquestionstoallowagoodstatistical
WhatarethesimilaritiesanddifferencesbetweenPhysics“regular”andPhysicsHonors?
Bothcourseswillfollowthesamebasicsyllabusandusethesamecoursepolicies.Withfew
halfofthescheduledlabsarecommonbetweenthetwocourses.Theotherlabswillbemoreopen‐
endedforHonorsthanforthestandardcourseandwillrequireashort1‐2pagewrittenreport.To
Finally,classtimewillbeusedabitdifferentlyinHonors,withlesstimedevotedtocoveringthe
mostbasicmaterial.
CorePhysicsCoursePrerequisites
ForastudentenrolledinPhysics110,heorshemusthavecompletedorbeenrolledininMath142.
Importantmathconceptsrequired:
vi
Physics110HJournal‐2013‐2014




CourseDescriptionandPolicies
Algebraandtrigonometry
Vectoroperationsincludingdotproductandcrossproduct
Differentiationofpolynomialsandsimplefunctions
Integrationofpolynomialsandsimplefunctions
StudentsinPhysics215musthavecompletedPhysics110andMath142.
USAFAandCorePhysicsCourseOutcomes
Physicscorecourses(Physics110andPhysics215)areaprimarycontributortothedevelopment
andassessmentofthefollowingUSAFAoutcomes:quantitativeliteracy,criticalthinkingand
conceptsofmotion,Newton’sLaws,energy,momentum,electricity,magnetism,and
selectedtopicsinmodernphysics.
ofphysicaldataandinformation.
4. Cultivatehabitsofthemindconsistentwiththatofaneducated,scientificallyliterate
person.
CorePhysicsLearningGoals
Physics110isdesignedtoenhanceyourcriticalthinkingskillsandyourabilityto:
1. Describethemotionofobjectsusingkinematics
2. InterpretandsolvemotionproblemsusingNewton’sthreelaws
3. Analyzethemotionofobjectsusingconservationofenergy,momentumandangular
momentum
ThecourselearninggoalsforPhysics215are:
1. Identifyhowthefundamentalphysicalprinciplesofelectricity,magnetismapplyto
conceptualorquantitativeproblems.
2. Solveconceptualorquantitativephysicalproblemsinvolvingelectricity,magnetismand
modernphysics.
3. Applyexperimentalskillstoinvestigatethephysicalprinciplesgoverningelectricityand
magnetism.
4. Analyzeandexplainthephysicalprinciplesthatapplytotheoperationofelectro‐magnetic
systemsandcircuits.
vii
CourseDescriptionandPolicies
Physics110HJournal‐2013‐2014
Position
DirectorofCorePrograms
Physics110CourseDirector
Assistant110CourseDir
Physics110HCourseDirector
Assistant110HCourseDir
Physics215CourseDirector
Assistant215CourseDir
Physics215HCourseDirector
Assistant215HCourseDir
YourPhysicsInstructor
Name
ColKiziah
LtColNovotny
LtColKayser‐Cook
CaptSorensen
Dr.deLaHarpe
MajBuchanan
MajLane
Mrs.Lickiss
Dr.Kontur
LtColAnthonyDills
Office
2A33
2A27
2A43
2A101
2A219
2A109
2A153
2A149
2A107
2A25
Phone
333‐3510
333‐9248
333‐0357
333‐9733
333‐9719
333‐7707
333‐3615
333‐3412
333‐4224
333‐3272
RequiredCourseMaterials
ThefollowingmaterialsarerequiredforthiscourseandmustbeinyourpossessiononLesson1.
Failuretopossessyourpersonalcopyofeachofthefollowingisafailuretomeetcourse
requirements.QuestionsmaybedirectedtotheDirectorofCoreProgramsorthePhysics
eachclassperiod:yourtextbookandyourentireJournal(ina3‐ringbinder).
Physics110andPhysics110Honors:



TEXTBOOK.Wolfson,Richard,EssentialUniversityPhysics,2nded.,Vol1,SanFrancisco:
PearsonEducation,Inc.,2012.
JOURNAL.ThePhysics110Journalcontainscourseguidance,syllabus,learningobjectives,
questions,andproblems.
MASTERINGPHYSICS.MasteringPhysics®istheonlinehomeworksystemthataccompanies
thetextbook.Anaccountcanbepurchasedwiththetextbookorseparately,butisrequired
forthecourse.TopurchaseMasteringPhysics®separately,goto
www.masteringphysics.com,intheREGISTERblockclickontheSTUDENTSbuttonand
followtheinstructions.LeaveStudentIDblank.TheCourseIDislistedinthefollowing
table:
Course
MasteringPhysicsCourseID
Physics110 FALL2013PHYSICS110
Physics110H FALL2013PHYSICS110H
viii
Physics110HJournal‐2013‐2014

CourseDescriptionandPolicies
SUPPLEMENTALCOURSEMATERIAL.AllothercoursematerialisavailableonthePhysics
110SharePointsites:
Course
SharepointSites
Physics215andPhysics215Honors:



TEXTBOOK.Wolfson,Richard,EssentialUniversityPhysics,2nded.,Vol2,SanFrancisco:
PearsonEducation,Inc.,2012.
JOURNAL.ThePhysics215Journalcontainscourseguidance,syllabus,learningobjectives,
questions,andproblems.
MASTERINGPHYSICS.MasteringPhysics®istheonlinehomeworksystemthataccompanies
thetextbook.Anaccountcanbepurchasedwiththetextbookorseparately,butisrequired
forthecourse.TopurchaseMasteringPhysics®separately,goto
www.masteringphysics.com,intheREGISTERblockclickontheSTUDENTSbuttonand
followtheinstructions.LeaveStudentIDblank.TheCourseIDislistedinthefollowing
table:
Course
MasteringPhysicsCourseID
Physics215
FALL2013PHYSICS215
Physics215H FALL2013PHYSICS215H

SUPPLEMENTALCOURSEMATERIAL.AllothercoursematerialisavailableonthePhysics
215SharePointsites:
Course
SharepointSites
CoursePolicies
WORKEDEXAMPLES–CorePhysicsusestheWorkedExamplesapproachtolearning,which
requiresstudentstocometoclasspreparedtodiscusslessonmaterial.Forthisreason,class
preparationpointsareheavilyweighted(18‐20%)andincludejournalquestions,pre‐class
problems,andpreflightquestions.
JOURNALQUESTIONS–Journalquestionsareassignedforalllessons,exceptedasnotedoneach
yourlevelofpreparationforclass.Thegoalofthisassessmentistoevaluateyourhonest,
ix
CourseDescriptionandPolicies
Physics110HJournal‐2013‐2014
thoughtfuleffortatreasoningthroughtheproblems.Ifyougetstuckonaproblem,reviewthe
exampleproblemsinthechapterandnotehowtheconceptsandequationsinthesectionare
thesolutionasyouareable,listingspecificquestionsyouhaveandidentifyingpointsofconfusion.
3/3
solvethePre‐ClassProblem(s)inalogicalformat(IDEAformatisrecommended).
2/3
Problem(s)oralogicalapproachwasnotused.
1/3
orentirelyunfinished.
0/3
Lessthan50%oftheJournalQuestionsforthelessonwerecompleted.
lessons.Preflightsareintendedtobedoneafterthejournalquestions.Preflightquestions
mustbesubmittedonlinenolaterthan0700beforethestartofeachlesson.Theyaredesignedto
assessyourunderstandingofthelessonmaterialandprovidefeedbacktoyourinstructor
responsesintheJournal.
PRE‐CLASSPROBLEMSandHOMEWORKPROBLEMS–Pre‐classproblemsareselectedfromthe
understandthelesson.
HOMEWORKPROBLEMS–HomeworkshouldbecompletedinyourJournaltoprovideyou
referenceandstudymaterialforclass,quizzes,andexams.(Somequizzesmaybe“openJournal!”)
tobescored.
LABSandLABQUIZZES–Onlabdays,youwillcompletethedatacollectionandanalysesasa
group,handinthelabworksheetasagroup,andthentakeanindividual‐effortlabquiz.Ifyouare
x
Physics110HJournal‐2013‐2014
CourseDescriptionandPolicies
participatewithalabgroupandtakethelabquiz.Forexcusedabsences,youmustcompletemissed
labswithin3lessons(6classdays).ExemptionstothispolicymustbeapprovedbytheCourse
onindividuallessonpageswithintheJournal.TheHonorsCoursesmayberequiredtocompletea
labreportwhichisfurtherdefinedinAppendixA.
EXAM‐PREPQUIZZES–Exam‐prepquizzes(EPQ)consistofworkoutandmultiple‐choiceproblems
coursedirectorpolicyandwillbeannouncedpriortotheEPQ.
choicequestions,andseveralworkoutproblems.Youwillhave80minutestocompletetheexam.
numberwiththewrongphysics.Yourscoreisdeterminedbythesoundnessofthe
physicsconceptsandshoweachstepintheproblem‐solvingprocess(IDEAformatis
recommended).
ABSENCEandTARDINESS–
Action(SCA),youareresponsibletonotifyyourinstructoratleastTHREEDAYS(not
includingweekends)PRIORtothefirstofferingoftheexam.Ifyouaremorethan15
penalty.Ifyouarelessthan15minuteslate,youmaystilltaketheexamduringthe
scheduledtime.
(b)Ifyouwillmissalessonforanyreason,completeandturninacopyofthatlesson’s
examforanotherreason,workwithyourinstructortoscheduleatimetomakeuptheexamwithin
twolessons.
FINALEXAMandVALIDATION–TheFinalExamisacomprehensiveexaminationincluding
materialfromtheentirecourse.Thefinalexamisyouropportunitytodemonstrateproficiency;
therefore,validationoftheFinalExamisnotoffered.
RESOURCES:Anypublishedorunpublishedsource,websites,andanyindividuals.Forall
xi
CourseDescriptionandPolicies
Physics110HJournal‐2013‐2014
assignments,youmustproperlydocumentallassistanceandsourcesusedaccordingtothePhysics
Departmentpolicyletterondocumentationstandards(locatedontheSharePointsite).Thisdoes
notallowyoutosimplycopyresourcematerialortheworkofanotherstudent,pastor
accomplishedinthefooteroneachpageoftheJournal.
securityuntilreleasedbytheCourseDirector.DONOTdiscussthecontentsorthedifficultyof
CONSTANTSANDEQUATIONSSHEET–YouwillbegivenastandardizedConstantsandEquations
Understandingthephysicalconceptsgoverningtheuniversewillnotcomefromscanningan
equationsheetinsearchofvariablesthatfittheproblem.Youmustfullycomprehendthenatureof
theequations,themeaningsofthevariables,andtheconstraintsforusingeachequation.
EXTRAINSTRUCTION–ThesecondhourofclassformostlessonsisdedicatedtoExtraInstruction
(EI).Yourinstructorwillnotcovernewmaterialorholdreviewsessionsduringthistime,butheor
sheisavailabletohelpyou.Ifyouhaveotherperiodsfree,youmayseekEIinanyofthePhysics
classroomsduringsecondhourfromanyinstructorthatisteachingyourcourse.Donotexpect
one‐on‐oneEIifyoudonotseekEIduringthesecondhourofyourclass.
(MFR)explainingyourcase,attachittoyourexam,andsubmitittoyourinstructor.TheCourse
*AMemorandumforRecordistheAirForcestandardforofficialwrittencommunicationsandthe
formatisprovidedintheTongueandQuill,availableontheAirForceE‐Publishingwebsite.
xii
Physics110HJournal‐2013‐2014
CourseDescriptionandPolicies
Physics110HonorsCoursePointStructure
No.ofEvents/Points
JournalQuestions
PreflightQuestions
Pre‐LabQuestions
LabWorksheet
LabQuizzes
LabReports
Exam‐PrepQuizzes
CriticalThinking
Exercise
Homework
FinalExam
Total
Points Percentage
28 @
30 @
[email protected]
[email protected]
[email protected]
[email protected]
[email protected]
3 pointseach
5 pointseach
5 pointseach
10 pointseach
10 pointseach
25 pointseach
30 pointseach
84
150
30
30
30
75
120
5.6%
10.0%
2.0%
2.0%
2.0%
5.0%
8.0%
[email protected]
15 pointseach
45
3.0%
37 @
3 pointseach
3 @ 150 pointseach
1 @ 375 points
111
450
375
1500
7.4%
30.0%
25.0%
100.0% NOTE1:Approximately70%ofthecoursepointsareindividualeffort(labquizzes,exam‐prep
xiii
IDEAProblem‐SolvingStrategy
Physics110HJournal‐2013‐2014
IDEAProblem‐SolvingStrategy
SolvingProblemsUsingtheIDEAFormat
Physicsproblemscanbechallenging,butunderlyingallofphysicsisonlyahandfulofbasic
principles.Ifyoureallyunderstandthose,youcanapplytheminawidevarietyofsituations.Ifyou
approachphysicsasahodgepodgeofunrelatedlawsandequations,you’llmissthepointandmake
thingsdifficult.Butifyoulookforthebasicprinciplesandforconnectionsamongseemingly
unrelatedphenomena,thenyou’lldiscovertheunderlyingsimplicitythatreflectsthescopeand
powerofphysics.
Asystematicsolutionmethodhelpsdevelopcriticalthinkingandscientificmethodprinciples.One
suchapproachistheIDEAproblem‐solvingstrategy.Solvingaquantitativephysicsproblemalways
simplesteps—stepsthatmakeupacomprehensivestrategyfororganizingyourthoughts,clarifying
yourconceptualunderstanding,developingandexecutingplansforsolvingproblems,andassessing
Interpret
Identifythemainphysicsconceptusedtosolvetheproblem.
Develop
informationforwhichyouaresolving.
Evaluate
Solvetheproblemfrombasicprinciplesusingequationsrelatedtothemainphysics
concepts.Whenpossible,expressthesolutionsymbolicallybeforesubstituting
valuesintotheequations.Includeunitswithallnumericalvalues.
Assess
following:
a) Howdoesthesolutioncomparetoknownvalues?
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
xiv
Physics110HJournal‐2013‐2014
LearningObjectives
LearningObjectives
BlockI–Motion
Duringthisblockwewillstartwithstudyingthebasicconceptsofdisplacement,velocityand
acceleration.Wewillthenusetheequationsofmotiontoanalyzethemotionofobjects.
[Obj9]
ConvertphysicalmeasurementsfromvariousunitstothestandardSIunitsofmeters,
kilograms,andseconds.
multiplication,division,andexponentiationonthem.
Identifythenumberofsignificantfiguresgiveninaproblemstatement,andexpress
Explaintherelationsbetweenposition,displacement,speed,velocity,andacceleration
foranobjectmovinginoneandtwodimensions.
Constructandinterpretgraphsofposition,velocity,andaccelerationforanobject
movinginoneandtwodimensions.
Explainthedifferencebetweeninstantaneousandaveragevelocity,andbetween
instantaneousandaverageacceleration.
Usemathematicalandgraphicalmethodstocalculateinstantaneousandaverage
velocityandinstantaneousandaverageaccelerationinoneandtwodimensions.
Useequationsofmotiontosolveproblemsinvolvingmotionwithconstant
acceleration.
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantacceleration.
[Obj10]
Solveproblemsinvolvingfree‐fallmotionwithconstantgravitationalacceleration.
[Obj11]
Expressvectorsbothincomponentformandinmagnitude‐ directionform.
[Obj12]
subtraction,andscalarmultiplication.
Usevectorstorepresentposition,velocity,andacceleration.
[Obj1]
[Obj2]
[Obj3]
[Obj4]
[Obj5]
[Obj6]
[Obj7]
[Obj8]
[Obj13]
[Obj15]
Describehowtheeffectsofaccelerationdependuponthedirectionoftheacceleration
vectorrelativetothevelocityvector.
Solveproblemsinvolvingprojectilemotionunderconstantgravitationalacceleration.
[Obj16]
Explainwhyuniformcircularmotioninvolvesacceleration.
[Obj17]
Solveproblemsinvolvinguniformandnonuniformcircularmotion.
[Obj14]
xv
LearningObjectives
Physics110HJournal‐2013‐2014
BlockII–Newton’sLaws
Inthisblock,westartbyintroducingNewton’sthreelawsofmotions.Wewillthenusetheselaws
tounderstandtheconceptofforce,todescribedifferenttypesofforces,andtoanalyzethemotion
ofobjectsinoneandtwodimensions.
[Obj18]
Explaintheconceptofforceandhowforcescausechangeinmotion.
[Obj19]
StateNewton’sthreelawsofmotionandgiveexamplesillustratingeachlaw.
[Obj20]
Explainthedifferencebetweenmassandweight.
[Obj21]
Constructfree‐bodydiagramsusingvectorstorepresentindividualforcesactingonan
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingona
singleobject.
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.
[Obj22]
[Obj23]
[Obj25]
Differentiatebetweentheforcesofstaticandkineticfrictionandsolveproblems
involvingbothtypesoffriction.
Describedragforcesqualitativelyand*quantitatively.
[Obj26]
Explainthephysicsconceptofwork.
[Obj27]
Evaluatetheworkdonebyconstantforcesandbyforcesthatvarywithposition.
[Obj24]
xvi
Physics110HJournal‐2013‐2014
LearningObjectives
BlockIII–EnergyandMomentum
Wewillstartthisblockbyintroducingtheconceptsofenergyandwork.Usingourunderstandingof
theseconcepts,wewilldeveloptheprincipleofconservationofenergywhichwillallowusto
analyzethecomplexmotionofobjectsincludingthoseinorbits.Wewillfinishthisblockdiscussing
collisionsandanotherconservationprinciple:conservationoflinearmomentum.
[Obj28]
Explaintheconceptofkineticenergyanditsrelationtowork.
[Obj29]
Explaintherelationbetweenenergyandpower.
[Obj30]
Explainthedifferencesbetweenconservativeandnonconservativeforces.
[Obj31]
Evaluatetheworkdonebybothconservativeandnonconservativeforces.
[Obj32]
Explaintheconceptofpotentialenergy.
[Obj33]
Evaluatethepotentialenergyassociatedwithaconservativeforce.
[Obj34]
Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanical
energy,orconservationofenergy.
[Obj35]
Describetherelationbetweenforceandpotentialenergyusingpotential‐energy
curves.
[Obj36]
Explaintheconceptofuniversalgravitation.
[Obj37]
Solveproblemsinvolvingthegravitationalforcebetweentwoobjects.
[Obj38]
Determinethespeed,acceleration,andperiodofanobjectincircularorbit.
[Obj39]
Solveproblemsinvolvingchangesingravitationalpotentialenergyoverlarge
distances.
[Obj40]
Usetheconceptofmechanicalenergytoexplainopenandclosedorbitsandescape
speed.
[Obj41]
Useconservationofmechanicalenergytosolveproblemsinvolvingorbitalmotion.
[Obj42]
Calculatethecenterofmassforsystemsofdiscreteparticlesandforcontinuousmass
distributions.
[Obj43]
Explaintheconceptoflinearmomentumofasystemofparticlesandexpress
Newton'ssecondlawofmotionintermsofthelinearmomentumofthesystem.
[Obj44]
Explainthelawofconservationoflinearmomentumandtheconditionunderwhichit
applies.
[Obj45]
Explaintheconceptofimpulseanditsrelationtoforce.
[Obj46]
Applyconservationoflinearmomentumtosolveproblemsinvolvingsystemsof
particles.
[Obj47]
Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.
[Obj48]
Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐
andtwo‐dimensions.
xvii
LearningObjectives
Physics110HJournal‐2013‐2014
BlockIV–RotationalMotionandSimpleHarmonicMotion
Duringthisblockwewillstudytherotationalandoscillatorymotionofobjects.Wewillstartby
exploringtherotationmotionofrigidobjects–discussingconceptsofangulardisplacement,
velocity,andacceleration.WewillthenrevisittheconceptsofNewton’sSecondLaw,conservation
ofenergy,andconservationofmomentumasappliedtoobjectsundergoingrotationalmotion.We
willendthecoursebyintroducingtheconceptofsimpleharmonicmotion.
[Obj49]
Explaintherelationbetweentherotationalmotionconceptsofangulardisplacement,
angularvelocity,andangularacceleration.
[Obj50]
Useequationsofmotionforconstantangularaccelerationtosolveproblemsinvolving
angulardisplacement,angularvelocity,andangularacceleration.
[Obj51]
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantangularacceleration.
[Obj52]
Explaintheconceptoftorqueandhowtorquescausechangeinrotationalmotion.
[Obj53]
Givenforcesactingonarigidobject,determinethenettorquevectorontheobject.
[Obj54]
Determinetherotationalinertiaforasystemofdiscreteparticles,rigidobjects,ora
combinationofboth.
[Obj55]
Compareandcontrasttheconceptsofmassandrotationalinertia.
[Obj56]
UseNewton’ssecondlawanditsrotationalanalogtosolveproblemsinvolving
translationalmotion,rotationalmotion,orboth.
[Obj57]
Solveproblemsinvolvingrotationalkineticenergyandexplainitsrelationtotorqueand
work.
[Obj58]
Explaintherelationbetweenlinearandangularspeedinrollingmotion.
[Obj59]
Useconservationofenergytosolveproblemsinvolvingrotatingorrollingmotion.
[Obj60]
Determinethedirectionsoftheangulardisplacement,angularvelocityandangular
accelerationvectorsforarotatingobject.
[Obj61]
Determinetheangularmomentumvectorfordiscreteparticlesandrotatingrigid
objects.
[Obj62]
Applyconservationofangularmomentumtosolveproblemsinvolvingrotatingsystems
changingrotationalinertiasandrotatingsystemsinvolvingtotallyinelasticcollisions.
[Obj63]
Definesimpleharmonicmotionandexplainwhyitissoprevalentinthephysicalworld.
[Obj64]
Determinetheperiodandfrequencyofasimpleharmonicoscillatorfromitsphysical
parameters,andcompletelyspecifyitsequationofmotion.
[Obj65]
Determinethevelocityandaccelerationofasimpleharmonicoscillatorfromits
equationofmotion.
[Obj66] Determinethepotentialandkineticenergiesofasimpleharmonicoscillatoratanypoint
initsmotion,anddescribethetimedependenceoftheseenergies.
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Physics110HJournal‐2013‐2014
Lesson1
Lesson1
Introduction
Chapter1,2.1,2.2
Examples
2.1,2.2
HomeworkProblems
1.16,1.24,2.51
LearningObjectives
[Obj1] [Obj2] [Obj3]
[Obj4] [Obj5] [Obj6] ConvertphysicalmeasurementsfromvariousunitstothestandardSIunitsofmeters,
kilograms,andseconds.
multiplication,division,andexponentiationonthem.
Identifythenumberofsignificantfiguresgiveninaproblemstatement,andexpress
Explaintherelationship betweenposition,displacement,speed,velocity,and
accelerationforanobjectmovinginoneandtwodimensions.
Constructandinterpretgraphsofposition,velocity,andaccelerationforanobject
movinginoneandtwodimensions.
Explainthedifferencebetweeninstantaneousandaveragevelocity,andbetween
instantaneousandaverageacceleration.
Notes
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Physics110HJournal‐2013‐2014
WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Atliftoff,theaccelerationofaspaceshuttleis29m/s2,anditspositionasafunctionoftimeis
definedas
, whereaistheaccelerationandtistime.a)Whatisthe(instantaneous)
velocity ofthespaceshuttleonesecondafterliftoff?b)Whatistheaveragevelocity ̅ overthe
firstminuteafterliftoff?
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 29m/s
2
1s
b) Average velocity over the first minute after liftoff (∆
̅
̅
58m/s
60s):
∆
∆
29m/s
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Lesson1
SELF‐EXPLANATIONPROMPTS
1. Whydoes∆ become
and∆ become whencalculatingaveragevelocity?
2. Describethemotionoftherocketasshownintheposition versustime graph.
3. Howdoyouexpecttheinstantaneousvelocityafteroneminutetocomparetotheaverage
velocitycalculatedinpart(b)?Calculatetheinstantaneousvelocityafteroneminute.
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Physics110HJournal‐2013‐2014
Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
knockingitoffitsoriginalpathata26°anglebutmaintainingitsoriginalspeed.Withthisnew
trajectory,howmuchlongerwillittakeforthemeteortoimpacttheMoon?
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Lesson1
PreflightQuestions
2. Solvethefollowingsystemofequationsfor and .
2
5
3
6
a)
b)
c)
d)
e)
3.40,
2.43,
3.40, 1.48,
1.00,
4.20
1.29
4.20
1.57
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.)
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HomeworkProblems
1.16
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Lesson1
1.24
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2.51
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Lesson2
Lesson2
Displacement,Velocity,andAcceleration
2.3,2.4
Examples
2.1‐ 2.3
HomeworkProblems
2.20,2.35,2.79
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj4]
[Obj5]
[Obj6]
[Obj7]
[Obj8]
[Obj9] Explaintherelationship between position,displacement,speed,velocity,and
accelerationforanobjectmovinginoneandtwodimensions.
Constructandinterpretgraphsofposition,velocity,andaccelerationforanobject
movinginoneandtwodimensions.
Explainthedifferencebetweeninstantaneousandaveragevelocity,andbetween
instantaneousandaverageacceleration.
Usemathematicalandgraphicalmethodstocalculateinstantaneousandaverage
velocityandinstantaneousandaverageaccelerationinoneandtwodimensions.
Useequationsofmotiontosolveproblemsinvolvingmotionwithconstant
acceleration.
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantacceleration.
Notes
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Lesson2
Physics110HJournal‐2013‐2014
WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
startsfromrestandhastocompletethe¼mile(402.3m)run.Thetimeisveryshortsothe
reactiontime(thetimeittakesthedrivertostartafterthegreenlightcomeson)isimportant.The
driverwiththeshortestoveralltime(runtime+reactiontime)isthewinner.
Thenationalrecordoveralltimeis4.42seconds.
a)Assumingthattheaccelerationwasconstant(togetasimpleestimateoftheacceleration),what
wastheaccelerationinthewinningrun?
b)Whatwastheaveragespeedofthedragster?
c)Againassumingconstantacceleration,whatwasthefinalspeedasthedragstercrossedthefinish
line?
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
∆
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Lesson2
CALCULATION
a) Acceleration of the winning run:
The initial velocity
is zero, so
̅ ∆
∙
Solving for acceleration gives
. .
b) Average speed of the dragster:
.
simplifies to
41m/s 402.3m
91m/s
4.42s
204mph
c) Final speed of the dragster (assuming constant acceleration):
Using the acceleration from part (a),
Since the dragster starts from rest,
41
∆
can be written as
∙ 4.42s
181m/s
∆
405mph
Note:Accelerationandfinalspeedarenotmeasuredindragraces;theaveragespeedduringthelast
20metersismeasured.Intherecordrun,theaveragespeedduringthelast20meterswas336mph.
SELF‐EXPLANATIONPROMPTS
1. Inyourtextbooklookupthederivationsoftheequationsusedinparts(a)and(b)aboveand
summarizeinyourownwordshowtheserelationsareobtained.
ifaccelerationisnotconstant?Gotothe
2. Isitvalidtousetherelation
equation.
3. Whatistherelationbetweeninitialspeedandfinalspeedduringatimeintervalwhenthe
accelerationisconstant?
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Lesson2
Physics110HJournal‐2013‐2014
Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
OnthewaytotheMoonthefirststageengineoftheSaturnVmoonrocketfiredfor156secondsto
liftthecraft38miles(61,155meters).WhatistheaverageaccelerationofSaturnVduringthis
stage?
TryIt!(1pt):WhatisthespeedoftheSaturnVattheendofthisstage?
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Lesson2
PreflightQuestions
2. Thefollowinggraphsdepictthevelocitiesoffourobjectsmovinginonedimension.Which
objecthasthegreatestdisplacementduringthetimeintervalshown?
Velocity, (m/s)
Velocity, (m/s)
Velocity, (m/s)
Velocity, (m/s)
10
10
10
10
5
5
5
5
10 20
10 20
10 20
10 20
Time, (s)
Time, (s)
Time, (s)
Time, (s)
a)
b)
c)
d)
3. Whichofthefollowingarrowscorrespondtoatimeatwhichtheinstantaneousvelocityis
greaterthantheaveragevelocityoverthetimeintervalshown?
B
Time,
Time,
Position, Time, Position, A
Position, Position, D
C
Time,
a)
b)
c)
d)
4. CRITICALTHINKING:Doesacarodometermeasuredisplacementordistance?Explain.
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HomeworkProblems
2.20
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Lesson2
2.35
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2.79
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Lesson3
Lesson3
Lab1–AccelerationDuetoGravity
2.5,Lab1Worksheet
Examples
2.6
HomeworkProblems
2.38,2.42,2.78
 ThereisaLABthislesson.
LearningObjectives
[Obj10]
Solveproblemsinvolvingfree‐fallmotionwithconstantgravitationalacceleration.
Notes
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Lesson3
Physics110HJournal‐2013‐2014
Pre‐LabQuestions
Score(5)
1. Brieflydescribethepurposeandgoalsofthislab.(Onetotwocompletesentences)
2. Whataretherelevantconceptsandequationsthatyouwillbeusinginthelab?
willyoudeterminetheangleoftheincline?
4. InPart1ofthelab,yourgroupwillmeasurethetimeittakesforanun‐weightedairtrackcart
totraveldifferentdistancesdownanincline.InPart2,yourgroupwillmeasurethetimeit
takesforaweightedairtrackcarttotravelthesamedistancesdownanincline.Howdoyou
expectthetimestocomparebetweentheweightedandun‐weightedcarts?Brieflyexplainyour
reasoning.
behindplottingthedatainsuchaway.
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Lesson3
LabNotes
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HomeworkProblems
2.38
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Lesson3
2.42
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2.78
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Lesson4
Lesson4
Two‐Dimensional&ProjectileMotion
Examples
HomeworkProblems
3.1– 3.5
3.3
3.34,3.53,3.54
LearningObjectives
[Obj11]
Expressvectorsbothincomponentformandinmagnitude‐directionform.
[Obj12]
subtraction,andscalarmultiplication.
Usevectorstorepresentposition,velocity,andacceleration.
[Obj13]
[Obj14]
Describehowtheeffectsofaccelerationdependuponthedirectionoftheacceleration
vectorrelativetothevelocityvector.
Notes
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Lesson4
Physics110HJournal‐2013‐2014
WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Anaircrafthasavelocityof
70 ̂ 50 ̂m/s.
Thewindpushestheaircraftwithavelocityof
45 ̂ 40 ̂m/s.Whatistheresultingfinal
?
velocityoftheplane,
STRATEGY
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
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Lesson4
SELF‐EXPLANATIONPROMPTS
componentsseparately?
2. Whencomponentsarecombined,aretheabsolutevaluesofthecomponentsusedordothe
componentsretaintheirnegativesignsiftheyhavethem?
accelerationoftheplaneinthisproblemandb)whatstepsyouwouldusetocalculatethe
accelerationoftheplane.
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Physics110HJournal‐2013‐2014
Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
70 ̂ 50 ̂m/s.Itexperiencesanaccelerationof
Anaircrafthasaninitialvelocityof
2.5 ̂ 2 ̂m/s astheresultofastrongwind.After20sinthiswind,whatisthenewvelocityof
theaircraft?
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120 ̂
10 ̂ m/s
Physics110HJournal‐2013‐2014
Lesson4
PreflightQuestions
2. Anobjectisinitiallymovinginthepositive ‐directionandthenexperiencesaccelerationinthe
positive ‐direction.Whichofthegraphsdepictsthe ‐and ‐positionsoftheobjectwhile
accelerating?
a)
b)
c)
d)
3. Theposition ofaparticleasafunctionoftime is
5 ̂
2 ̂.Which
statementistrueconcerningtheparticle?
a) Theparticleislocatedattheoriginat
b)
1
5m/s
c)
d)
e)
5 ̂
4
2
0.
2 ̂m/s
19/2m/s2
√
̂
2 ̂m/s f) Accelerationoftheparticleisconstant.
4. CRITICALTHINKING:Cananobjecthaveanorthwardvelocityandsouthwardacceleration?
Explain.
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HomeworkProblems
3.34
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Lesson4
3.53
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3.54
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Lesson5
Lesson5
ProjectileMotion
3.5
Examples
3.4
HomeworkProblems
3.55,3.70,MP
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
 ThereisanEXAM‐PREPQUIZthislesson.
LearningObjectives
[Obj14]
[Obj15]
Describehowtheeffectsofaccelerationdependuponthedirectionoftheacceleration
vectorrelativetothevelocityvector.
Solveproblemsinvolvingprojectilemotionunderconstantgravitationalacceleration.
Notes
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Lesson5
Physics110HJournal‐2013‐2014
WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Avintagebomberisparticipatinginanairshowandplanstoconductabombingrunusingan
“explosive”flourbag.Iftheaircraftisflyingat75m/sandreleasestheflourbomb1500meters
abovetheground,howfarbackfromthetargetmustthepilotreleasethisflourbomb(therelease
distancex)?Howfastisthebombmovinginthehorizontalandverticaldirectionswhenithitsthe
ground?
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.
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Lesson5
Now that we have the flight time, we will use
to solve for the
to determine the vertical speed of the bomb. There is no
release distance and
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

and
2 1500m / 9.8
17.5s..
Next, determine the how far back the bomb is released in the horizontal.
, we get ∆
Starting with
75
17.5s
0
1312m.
Finally, solve for the final bomb velocity in the vertical:
and
Starting with
0
9.8m/s
12.4s
171.5m/s
75m/s since there is no acceleration in the horizontal.
SELF‐EXPLANATIONPROMPT
1.Whyisthefinalposition,yf,anegativequantity?
2.Whatwouldbedifferentifweweretodesignatetheoriginatthegroundlevelbelowtherelease
point?
3.Whatwouldhappeniftheinitialvelocity,v0y,intheverticalwasnotzero?
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Lesson5
Physics110HJournal‐2013‐2014
Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Acannononaclifffiresatashipinapiratemovie.Theshipis200mfromthecliffandtheinitial
velocityofthelaunchedcannonballis
60 ̂ 20 ̂m/s.Ifthecannonballhitstheship,a)how
highisthecliff,andb)whatisthefiringangle?
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Lesson5
PreflightQuestions
aconstantspeed.Neglectingairresistance,thesnowballwillland
a) infrontofthesled.
b) onthesled.
c) behindthesled.
suchthatthehorizontalvelocitycomponentofthecannonballsareequal.Object1,ared
cannonball,isshotupwardatanangleof30°withrespecttothehorizontal.Object2,ablue
cannonball,isshotupwardsatanangleof60°withrespecttothehorizontal.Whichballwillhit
thegroundfurthestfromthecannon?
a) Theredcannonball.
b) Thebluecannonball.
c) Bothcannonballswillhitatthesamespot.
d) Thecannonballswillonlygoupanddown.
4. CRITICALTHINKING:Ahighjumperandalongjumperarebothhumanprojectiles,butwith
slightlydifferentgoals.Thehighjumperwantstotraveloverahighbarwithouttouchingit,and
Describehowthex‐andy‐componentsoftheinitialvelocityvectorshoulddifferbetweenthe
twotypesofjumpers.
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HomeworkProblems
3.55
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Lesson5
3.70
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MP
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Lesson6
Lesson6
Lab2–ProjectileMotion
Lab2Worksheet
Examples
3.4
HomeworkProblems
MP,3.58,MP
 ThereisaLABthislesson.
LearningObjectives
[Obj15]
Solveproblemsinvolvingprojectilemotionunderconstantgravitationalacceleration.
Notes
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Lesson6
Physics110HJournal‐2013‐2014
Pre‐LabQuestions
Score(5)
1. Brieflydescribethepurposeandgoalsofthislab.(Onetotwocompletesentences)
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
horizontalrange,Δ ,thattherocketwilltravel.
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Lesson6
LabNotes
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Lesson6
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HomeworkProblems
MP
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Lesson6
3.58
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MP
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Lesson7
Lesson7
AccelerationinCircularMotion
Examples
HomeworkProblems
3.6
3.7,3.8
MP,MP,3.80
LearningObjectives
[Obj16]
Explainwhyuniformcircularmotioninvolvesacceleration.
[Obj17]
Solveproblemsinvolvinguniformandnonuniformcircularmotion.
Notes
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Lesson7
Physics110HJournal‐2013‐2014
WorkedExample
Score(3)
thequestionsregardingtheproblem.
STATEMENTOFTHEPROBLEM
Anaircrafttravelingataconstant150m/smakesa360°turnata
constantaltitude(referredtoaslevelflight).Iftheaircraft’s
oftheturn?
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/s
1.5 9.8m/s
Notice that the units resolve as:
/
/
m
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Lesson7
SELF‐EXPLANATIONPROMPTS
1.Inthisproblem,theaircraftistravelingataconstantspeedof150m/s.Isthisaircraft(orany
objectexecutinguniformcircularmotion)undergoingacceleration?Explain.
2.Howdoyouknowthattheaccelerationintheverticaliszero?
3.Whatcausesthecenter‐directedacceleration?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
turnis50meters.Whataccelerationmustthefrictionofthetiresgenerateinordertoaccomplish
thisturn?Whatisthedirectionofthatacceleration?
TryIt!(1pt):Describeanddrawtheaccelerationvector
ifthecar’sspeedwasincreasingasitexecutedtheturn.
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centeroftheturn
Physics110HJournal‐2013‐2014
Lesson7
PreflightQuestions
directionofacceleration?
a) Thereisnoacceleration.
b) Thecarisacceleratingtowardthecenterofthecurve.
c) Thecarisacceleratingawayfromthecenterofthecurve.
d) Theaccelerationisinthesamedirectionthecaristraveling.
2
2 2
2
a)
b)
c)
d)
4. CRITICALTHINKING:Whenyourideinavehiclethatismakingaturnyourbodyfeelspushed
outward.Reconcilethisfactwiththephysicsstatementthattherealaccelerationofyourbody
isinwardtowardsthecenteroftheturn.
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HomeworkProblems
MP
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Lesson7
3.58
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MP
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Lesson8
Lesson8
LearningObjectives
[Obj9] ConvertphysicalmeasurementsfromvariousunitstothestandardSIunitsofmeters,
kilograms,andseconds.
multiplication,division,andexponentiationonthem.
Identifythenumberofsignificantfiguresgiveninaproblemstatement,andexpress
Explaintherelationship betweenposition,displacement,speed,velocity,and
accelerationforanobjectmovinginoneandtwodimensions.
Constructandinterpretgraphsofposition,velocity,andaccelerationforanobject
movinginoneandtwodimensions.
Explainthedifferencebetweeninstantaneousandaveragevelocity,andbetween
instantaneousandaverageacceleration.
Usemathematicalandgraphicalmethodstocalculateinstantaneousandaverage
velocityandinstantaneousandaverageaccelerationinoneandtwodimensions.
Useequationsofmotiontosolveproblemsinvolvingmotionwithconstant
acceleration.
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantacceleration.
[Obj10]
Solveproblemsinvolvingfree‐fallmotionwithconstantgravitationalacceleration.
[Obj11]
Expressvectorsbothincomponentformandinmagnitude‐directionform.
[Obj12]
subtraction,andscalarmultiplication.
[Obj13]
Usevectorstorepresentposition,velocity,andacceleration.
[Obj14]
[Obj15]
Describehowtheeffectsofaccelerationdependuponthedirectionoftheacceleration
vectorrelativetothevelocityvector.
Solveproblemsinvolvingprojectilemotionunderconstantgravitationalacceleration.
[Obj16]
Explainwhyuniformcircularmotioninvolvesacceleration.
[Obj17]
Solveproblemsinvolvinguniformandnonuniformcircularmotion.
[Obj1]
[Obj2]
[Obj3]
[Obj4]
[Obj5] [Obj6] [Obj7] [Obj8] Notes
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Susanisdrivingat50mphtopickupherfriendattheairport.Herfriend’sflightlandsin30
minutes,andsheis40miawayfromtheairport.WillSusanbeabletopickupherfriendontime?
Ifso,howlongwillittakeforhertoarriveathercurrentspeed?Ifnot,whatwillshehaveto
changeherspeedtoinordertoarriveattheairportontime?
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:
∆
∆
̅
̅
If
̅
̅
∆
If
̅
̅
∆
̅
̅
40mi
0.5hr
or
____________mi/hr
̅
(circle one)
̅
, how long will it take to arrive?
_____________hr
, what will Susan have to change her speed to?
SELF‐EXPLANATIONPROMPTS
1.WhatspeedwouldSusanneedtoarriveexactlyontime?
OptionalPracticeProblems:2.21,2.43,2.47
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
abowis76.2m/s.Ifsuchanarrowisshotstraightup,andairresistanceisneglected,howhigh
woulditgo?
STRATEGY(Fillintheblanks.)
In our case v = _______________, v0 = ________________,
a = __________ (watch the sign!), x – x0 is the
height.
Now, if we eliminate the variable t between
we get
½
and 2
CALCULATION(Fillintheblanks.)
height = ________________________= 296 m
That is almost 0.2 mile and probably unrealistic.
SELF‐EXPLANATIONPROMPTS
1.PerformthederivationintheSTRATEGYsection.
2.Whichquantitiesin
2
arepositive,whicharenegative?
OptionalPracticeProblems:2.37,2.51,2.61,2.69
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Twovectorsare
7̂
4 ̂ and
3̂
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“
OptionalPracticeProblems:3.11,3.14,3.31
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Avintagebomberparticipatinginanairshowwantstodropabombthatstaysintheairfor15
secondsbeforeimpact.Ifthehorizontalvelocityoftheplaneis75m/s,determinetherequired
launchaltitude.
STRATEGY(Fillintheblanks.)
Let’s set the ________ at the vertical point where the
bomb is __________.
written in the _______
We will use
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)
1
2
______m
0 _____s
1
2
9.8m/s
15s
1103m
SELF‐EXPLANATIONPROMPTS
1.Howaretheflighttimesinthehorizontalandtheverticaldirectionsconnected?
3.Howisthenegativedirectionofgravity’seffectaccountedforinordertoresultinapositive
valuefory0?
OptionalPracticeProblems:3.33,3.62
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Thishammerthrowerreleasesthehammerballwitha
tangentialspeedof21m/swhentheballis1.8mfromthe
centeroftheathlete’srotation.a)Whatisthecentripetal
accelerationoftheballattheinstantitisreleased?b)How
doesthisaccelerationcomparetotheaccelerationdueto
gravity?
STRATEGY(Fillintheblanks.)
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 _____________________.
SELF‐EXPLANATIONPROMPTS
1.Whatobjectprovidestheaccelerationofthehammerball?
2.Whatisthedirectionofthenetaccelerationofthehammerballjustbeforeitisreleased?
OptionalPracticeProblems:3.38,3.39,3.40
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Notes
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Notes
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Lesson9
Lesson9
ForcesandNewton’sLawsofMotion
Examples
HomeworkProblems
4.1– 4.4
4.1,4.2
4.15,4.26,4.60
LearningObjectives
[Obj18]
Explaintheconceptofforceandhowforcescausechangeinmotion.
[Obj19]
StateNewton’sthreelawsofmotionandgiveexamplesillustratingeachlaw.
[Obj20]
Explainthedifferencebetweenmassandweight.
Notes
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WorkedExamples
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Startingfromrest,thespeedincreasesto2m/sovera20metersdistance.Whatisthetensionin
therope?Assumefrictionisnegligible.
STRATEGY
Newton’s Second Law as applied to the car
states that the acceleration of the car
given by
is
. 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
1200kg
4
0 m /s
2 20m
120N
The force unit kg·m/s2 is called a newton, N, in honor of Isaac Newton.
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SELF‐EXPLANATIONPROMPTS
1.TheunitofforceisanewtonwhichisgiventhesymbolN.Expressthenewtonintermsofthe
fundamentalSIunits.
2.Findthedefinitionof“tensioninarope”inyourtextbookandrephraseitinyourownwords.
3.Whatchangewouldyoumakeinthecalculationifthetow‐ropewasdirectedatanangle?
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Pre‐ClassProblem
A45‐ggolfballatrestishitbyaclubwithaforceof5.0N.a)Whatistheball’sacceleration
immediatelyafteritishit?b)Howfardoestheballtravelinthefirsttenthofasecond?
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PreflightQuestions
2. A200‐kgrockisbeingpulledupwardwithanaccelerationof3m/s2.Thenetforceontherock
is
a) 200Nup
b) 200Ndown
c) Zero
d) 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?Usea
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HomeworkProblems
4.15
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4.26
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4.60
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Lesson10
Lesson10
UsingNewton’sLaws
4.5,4.6
Examples
4.3,4.4,4.5
HomeworkProblems
4.34,4.47,4.49
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj19]
StateNewton’sthreelawsofmotionandgiveexamplesillustratingeachlaw.
[Obj20]
Explainthedifferencebetweenmassandweight.
[Obj21]
Constructfree‐bodydiagrams usingvectorstorepresentindividualforcesacting onan
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingona
singleobject.
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.
[Obj22]
[Obj23]
Notes
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WorkedExamples
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Ateamofdogsispullingtwoconnectedsledswithaconstantaccelerationof2.3m/s2.The
passengersled,connectedtothedogsinfront,hasamassof96kg.Thecargosled,tiedtothefront
sled,hasamassof42kg.Fornow,weassumethattheretardingfrictionismuchsmallerthanthe
forceexertedbythedogs.
a)Howmuchistheforcethatthedogsexerton
thesledtrain?
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.
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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
.
CALCULATION
For each part we apply Newton’s Second Law
a) 2.3m/s
b) 2.3m/s .
320
in the forward direction
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.
SELF‐EXPLANATIONPROMPTS
1. RephraseNewton’sSecondLawinyourownwords.
2. Whatisthenetforceonthepassengersled?
3. Ifthecargosledwasremoved,howdoyouexpecttheforceappliedbythedogteamtochange
inordertoobtainthesameaccelerationof2.3m/s2forjustthepassengersled?Calculatethe
forceexertedbythedogteam forthisscenario.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A12‐kgchildisridingina4100‐kgelevatorwhichisacceleratingupwardataconstant1.3m/s2.
Whatistheforcethattheelevatorexertsonthechild?Whatistheforcethechildexertsonthe
elevator?
Free‐BodyDiagram(required)
Tryit!(1PFpt):Ifthechildwasstandingonascaleintheelevator,
and(b)acceleratingupwardat1.3m/s2?Showallyourwork.
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PreflightQuestions
2. Twoforcesofequalmagnitudeactonthesameobject.Whichofthefollowingmustbetrue?
a) Theobjectismoving.
b) Theobjectisaccelerating.
c) Iftheobjectisinitiallyatrest,itcannotremainatrest.
d) Thetwoforcesformathird‐lawpair.
e) Noneoftheabove.
3. Twoblocksarehangingmotionlessfromtheceilingasshowninthe
diagram.Whichofthefollowingistrue?
a)
b)
c)
onlyif
d)
4. CRITICALTHINKING:Theterm”weight”inphysicshasthefollowingveryspecificmeaning:
“Theweightofanobjectisthenamegiventoaparticularforce:thegravitationalforceexerted
bytheearthontheobject,givingitanaccelerationof9.8m/s2nearthesurfaceofEarth.”
Inordinaryspeechtheuseof“weight”isnowherenearlysoprecise.Explainwhetherthe
followingusagesarescientificallycorrect.
c) AnastronautorbitingEarthexperiencesweightlessness
d) Ifyoueattoomuchyoumaygainweight.
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HomeworkProblems
4.34
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4.47
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4.49
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Lesson11
Lesson11
Newton’sLawsinTwoDimensions
Examples
HomeworkProblems
5.1
5.1,5.2
5.16,MP,5.38
LearningObjectives
[Obj22]
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingona
singleobject.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
50kg
A50‐kgblockisonafrictionless30°ramp.Determinetheblock’s
accelerationdowntheramp.
STRATEGY
30°
Newton’s Second Law as applied to the block states that the acceleration of the block
given by
. In this problem, we have two forces acting on the block: weight
is
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
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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
SELF‐EXPLANATIONPROMPTS
wasdefinedasbeinguptheramp?
3.Whatwouldhappentothemagnitudeoftheblock’saccelerationiftheangleoftherampwas
increased?Whatisthemaximumaccelerationtheblockcanexperience?Whatistheminimum
accelerationtheblockcanexperience?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
CEILING
32°
68°
A3.0‐kgboxissuspendedfromaceilingasshown.What
arethemagnitudesofthetensionsexertedbytheropes
attachedtothebox?Assumetheropeshavenegligible
masscomparedtothebox.(Hint:LookatExample5.2in
m
coordinatesystemratherthantiltedforthisproblem?)
Free‐BodyDiagramofBox(required)
T2=11N
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Lesson11
PreflightQuestions
1.
2. A5‐kgblockispushedacrossahorizontalfloorwitha20‐Nforcedirected20°belowthe
horizontal.Whatisthemagnitudeofthenormalforceontheblock?
a)
b)
c)
d)
e)
3.
49N
6.8N
42N
56N
68N
20N
20°
5kg
IfRope1remainshorizontalandthepointatwhichRope2istiedis
a)
remainsthesameand increases.
b)
decreasesand increases.
Rope1
c) Both and remainthesame.
d) Both and increase.
4.
CRITICALTHINKING:Refertopreflightquestion3:IsitpossibletoattachRope2atpointCand
havebothropesparalleltotheground?Explain.
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HomeworkProblems
5.16
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MP
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5.38
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Lesson12
Lesson12
Newton’sLawswithMultipleObjects
Examples
HomeworkProblems
5.2
5.4
5.19,5.21,5.71
LearningObjectives
[Obj23]
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.
Notes
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WorkedExamples
Score(3)
problem.
STATEMENTOFTHEPROBLEM
A2,500‐kgtractorispullinga750‐kgcowoutofaravine,as
shown.Ifthetractorappliesaforceof20kN,determinethe
accelerationofthecowoutoftheravine.Assumetheropeand
pulleyaremasslessandtheropedoesnotstretch.
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.
CALCULATION
The net force on the tractor in the x-direction
is:
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Lesson12
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
750kg ∙ 9.8
2500kg
3.9m/s SELF‐EXPLANATIONPROMPTS
1. Explainwhytheaccelerationofthetractorinthex‐directionisthesameastheaccelerationof
thecowinthey‐direction.
2. Explainwhytheweightofthecowisanegativequantity.
3. Ifthetractorcouldonlyapplya2kNforce,calculatetheaccelerationofthecow.Describethe
motionofthecow+tractorsystemforthisscenario.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A10‐kgcartisconnectedbyastringtoa10‐kgweightoverapulley.
Assumingthatthemassesofthestringandthepulleycanbeneglected,
findtheaccelerationofthecartandthetensioninthestring.
Free‐BodyDiagramoftheCart(required)
Free‐bodyDiagramoftheWeight(required)
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Lesson12
PreflightQuestions
thetensionintheropecomparedtotheweightofthebucket?
a) Tensionislessthantheweightofthebucket.
b) Tensionisequaltotheweightofthebucket.
c) Tensionisgreaterthantheweightofthebucket.
d) Cannotbedeterminedfromthegiveninformation.
3. InCase1,BlockBacceleratesBlockAacrossafrictionlesstable.InCase2,aforceof98N
acceleratesBlockAacrossthesametable.TheaccelerationofBlockAis
A
A
a) zero.
10kg
10kg
b) greaterinCase1.
c) greaterinCase2.
d) thesameinbothcases.
Case1
Case2
B
10kg
4. CRITICALTHINKING:Whenyouareinanelevatoryouoftenfeelalittlelighterastheelevator
startstomovedownward.ExplainthisfeelingbasedonNewton’sLaws.
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98 N
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HomeworkProblems
5.19
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Lesson12
5.21
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5.71
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Lesson13
Lesson13
Lab3–Newton’sLaws
5.2,Lab3Handout
Examples
5.4
HomeworkProblems
MP,MP,MP
 ThereisaLABthislesson.
LearningObjectives
[Obj23]
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.
Notes
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Pre‐LabQuestions
Score(5)
1. Brieflydescribewithoneortwocompletesentencesthepurposeandgoalsofthis
lab.
2. Constructfree‐bodydiagramsform1andm2forthefollowingscenario.
Free‐BodyDiagram:Mass1
Free‐BodyDiagram:Mass2
3. UseNewton’ssecondlawtoderiveanexpressionfortheaccelerationofthemassesintermsof
m1,m2,θ,andg.
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Lesson13
LabNotes
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HomeworkProblems
MP
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MP
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MP
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Lesson14
Lesson14
Newton’sLawsinCircularMotion
5.3
Examples
5.5,5.6,5.7
HomeworkProblems
5.65,5.73,MP
 ThereisanEXAM‐PREPQUIZthislesson.
LearningObjectives
[Obj16]
Explainwhyuniformcircularmotioninvolvesacceleration.
[Obj17]
Solveproblemsinvolvinguniformandnonuniformcircularmotion.
[Obj22]
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingona
singleobject.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Anamusementparkrideconsistsofaverticalloopwhosediameteris
15mandasmall150‐kgcartthatrunsontheinsidetrackintheloop.
Thecartisgivenaninitialspeedatthebottomofthetrackandisnot
propelledfurther.Whenthecartclimbsverticallytothe90°point,its
speedis12.4m/s.Whatisthemagnitudeanddirectionofthenetforce
onthecartatthispoint?
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.
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CALCULATION
1.
3410
2. θ = 25.4 degrees below horizontal, to the left towards the center of the loop.
SELF‐EXPLANATIONPROMPTS
1. Drawfree‐bodydiagramsofthecartwhenitisatthebottomandtopofthetrack.
2. Doesthecarttravelaroundtheloopataconstantspeed?Explain.
3. Describehowtheweight,normalforceandnetforcechangeasthecartmovesaroundthetrack.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A50‐kgwrecker’sballishangingonan8‐mropethatcansupportamaximumforceof1000N.If
theballisswunginaverticalcircle,whatisfastestspeeditcanhaveatthelowestpointsuchthat
theropewon’tbreak?
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PreflightQuestions
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,describethemotionoftheballandyour
reasoning.
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HomeworkProblems
5.65
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5.73
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MP
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Lesson15
Lesson15
Newton’sLawswithFriction
Examples
HomeworkProblems
5.4,5.5
5.9,5.10,5.11
5.43,MP,5.57
LearningObjectives
[Obj24]
[Obj25]
Differentiatebetweentheforcesofstaticandkineticfrictionandsolveproblems
involvingbothtypesoffriction.
Describedragforcesqualitatively andquantitatively.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Ateamofdogsispullingtwoconnectedsledswithanaccelerationof2.3m/s2.Thepassengersled,
connectedtothedogsinfront,hasamassof96kg;thecargosled,tiedbehindthepassengersled,
hasamassof42kg.Thecoefficientofkinetic
frictionbetweenthesteelrunsonthesledsand
theiceisμ=0.007.Howmuchforcedoesthe
dogteamexertonthesledtrain?
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
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:
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Lesson15
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
42 kg
96
42 kg 2.3m/s
CALCULATION
0.007 96
F
F
320N in the forward direction.
SELF‐EXPLANATIONPROMPTS
1. InLesson10wesolvedthesameproblem,butwithoutfriction.Explainhowthemethod
changeswhenfrictionisincluded.
2. Explainwhykineticfrictionwasusedintheproblemratherthanstaticfriction.
3. Describethestepsusedtodeterminethekineticfriction.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A60‐kgblockisreleasedfromrestona45°rampwherethecoefficientoffrictionbetweenthe
blockandrampis0.4.Whatistheaccelerationoftheblock?
Tryit!(1pt):Determinethespeedoftheblockatthebottomoftheramp
ifitstartsfromrestatthetopofthe3‐mlongramp.Showyourwork.
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PreflightQuestions
2. Whichstatementconcerningfrictionistrue?
a) Staticfrictionisalwaysoppositethedirectionofmotion.
b) Kineticfrictionisalwaysoppositethedirectionofmotion.
c) Bothstaticandkineticfrictionarealwaysoppositethedirectionofmotion.
d) Neitherisalwaysoppositethedirectionofmotion.
3. Aboxisatrestontheflatbedofamovingtruck.Dawn
appliesthebrakesabruptlyandtheboxbeginstoslide.
Whichfree‐bodydiagramcorrectlydepictstheforces
actingontheboxanditsresultingmotion?
a)
c)
b)
d)
4. CRITICALTHINKING:Describe,inyourownwords,thedifferencebetweenstaticfrictionforces
andkineticfrictionforces.
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HomeworkProblems
5.43
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MP
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Lesson16
Lesson16
CriticalThinking:Newton’sLaws
withNon‐constantMass
Examples
HomeworkProblems
9.3Application,Handout
None
5.30,5.62,6.54
LearningObjectives
[Obj18]
Explaintheconceptofforceandhowforcescausechangeinmotion.
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Lesson16
PreflightQuestions
resultsfrom
a) theexhaustgasespushingagainsttheground.
b) theexhaustgasespushingagainsttheair.
c) thecombustiongasespushingagainsttherocket.
d) theequalandoppositereactiontogravitypullingdown.
e) thegravitationalenergyreleasedbyburningfuel.
3. Atsomepointbeyondatmosphericspaceshuttleflight,the3‐mainenginesstopproviding
thrustandthentheboostertankSEPARATESfromthecraft.Whentheconnectionbetweenthe
twoobjectsissevered,thevelocityoftheshuttle
a) increases.
b) decreases.
c) remainsunchanged.
theaccelerationofthespaceshuttleassembly.
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HomeworkProblems
5.30
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5.62
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Lesson17
Lesson17
LearningObjectives
[Obj18]
Explaintheconceptofforceandhowforcescausechangeinmotion.
[Obj19]
StateNewton’sthreelawsofmotionandgiveexamplesillustratingeachlaw.
[Obj20]
Explainthedifferencebetweenmassandweight.
[Obj21]
Constructfree‐bodydiagrams usingvectorstorepresentindividualforcesacting onan
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleforcesactingona
singleobject.
UseNewton’slawsofmotiontosolveproblemsinvolvingmultipleobjects.
[Obj22]
[Obj23]
[Obj24]
[Obj25]
Differentiatebetweentheforcesofstaticandkineticfrictionandsolveproblems
involvingbothtypesoffriction.
Describedragforcesqualitatively andquantitatively.
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
A1000‐kgcaristravelingat10m/swhenabrakingforceof500Nisapplied.Howmuchtimedoes
elapsebeforethecarcomestoacompletestop?
STRATEGY(Fillintheblanks.)
Newton’s Second Law as applied to the car states that the acceleration of the car
by
is given
. 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
SELF‐EXPLANATIONPROMPTS
1.Comparethisexampletothetow‐truckexample,stepbystep.
car’sacceleration?
OptionalPracticeProblems:4.13,4.15,4.23
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Lesson17
regardingtheproblem.
STATEMENTOFTHEPROBLEM
generated34.5x106Nofthrust.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 .
SELF‐EXPLANATIONPROMPTS
1.Whatwouldbetheaccelerationofarocketofthesamemassifitstartedfromrestinempty
space,awayfromobjectsthatexertgravitationalforceslikeEarth?
2.Furtherintothelift‐off,wouldyouexpecttheSaturn’saccelerationtoincrease,decrease,or
remainthesame?
3.WhatmagnitudeofthrustwouldmaketheSaturnjusthover,withnoacceleration?
OptionalPracticeProblems:4.27,4.37,4.45
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
58°
A5‐kgballissuspendedfromthreeropesasshowninthepicture.Whatistheforce
exertedonthewallbythehorizontalrope?
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.IsitpossibletosuspendtheballinthisexampleinsuchawaythatbothforcesT1andT2have
horizontalcomponentsonly?Explain.
2.Arethemagnitudesofanyofthetensionsinthethreeropeslargerthantheweightoftheball?
OptionalPracticeProblems:5.15,5.33,5.36
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Twoballswithmasses,M1andM2,areconnectedbyaropewhichpasses
overapulleyasshown.Findtheaccelerationsoftheballsastheyare
releasedfromrest.Assumethattheropedoesnotstretchandthemasses
oftheropeandthepulleyarenegligiblecomparedtothemassesofthe
balls.
STRATEGY(Fillintheblanks.)
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
CALCULATION(Fillintheblanks.)
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
SELF‐EXPLANATIONPROMPTS
1.Explaininyourownwordswhythemagnitudesoftheaccelerationsofthetwoballsarethesame.
2.Whyisthemagnitudeofthetensionintheropeontheleftsideofthepulleythesameasthe
magnitudeontherightsideofthepulley?
OptionalPracticeProblems:5.18,5.19,5.20
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
0.80,andthecoefficinetofkineticfrictionis
0.40.
Whatisthenetforceofthecar,magnitudeanddirection?
STRATEGY(Fillintheblanks.)
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.
. and
The net force is the vector sum of
.
CALCULATION(Fillintheblanks.)
1. Tangential force is
2.
Force of friction is
3. Net force is
__________ .
√_____
___________ _____
The direction is _____________________________.
SELF‐EXPLANATIONPROMPTS
1.Whydidwenotneedcoefficientoffrictionforthisproblem?
2.Whatisthemagnitudeanddirectionofthenetforceifthecarroundsthecurveatconstant
speed?
OptionalPracticeProblems:5.27,5.37,5.41
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Amanispushinga50‐kgcartthatacceleratesat1.3m/s2onlevelgroundwherethecoefficientof
frictionbetweenthewheelsandthegroundis0.03.Howmuchresistancefromthecartdoeshe
feel?
STRATEGY(Fillintheblanks.)
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.
SELF‐EXPLANATIONPROMPTS
1.Whydon’ttheforceonthecartandtheforceonthemancancelout?Thatis,whydoesthe
mathematicallycorrectstatement, 79.7 – 79.7 0,notimplythatthenetforceintheabove
scenarioiszero?
2.Whattypeoffrictionalforceactsonthecart:kineticorstatic?Explain.
OptionalPracticeProblems:5.29,5.43,5.49
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Lesson18
Lesson18
WorkwithConstantandVaryingForces
6.1,6.2
Examples
6.1– 6.5
HomeworkProblems
6.18,6.20,6.52
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj26]
Explainthephysicsconceptofwork.
[Obj27]
Evaluatetheworkdonebyconstantforcesandbyforcesthatvarywithposition.
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Amandragsa50‐kgcrate10macrossaroughhorizontalsurface,wherethecoefficientofkinetic
friction,μ ,betweenthecrateandthesurfaceis0.3.Hepullsatconstantspeedanddirectshis
pullingforce20°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
Free‐BodyDiagramofCrate
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
friction
and the force of kinetic
.
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,
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Solving for Fm we get
cosθ
The work done by Fm is then
∙∆
Δ
CALCULATION
SELF‐EXPLANATIONPROMPTS
1.Startwiththedefinition
zero.
cosθ
∙∆
Δ
1325J
∆ cos andexplainhowWcanbepositive,negative,or
2.Explainwhatitmeanstohave(a)positiveWand(b)negativeW.
3.Intheexample,youaretoldthatthenormalforceis:
neededtoobtainthenormalforceandthenshowthecalculation.
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.Describethesteps
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Acranelowersa120‐kgrockatconstantspeedthroughaverticaldistanceof5meters.Howmuch
workdoesthecraneperform?
Free‐BodyDiagramofRock(required)
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Lesson18
PreflightQuestions
2. Solveforwork andrankorderfromsmallest(negative)tolargest(positive)theworkdonein
thefollowingcases:
2m
2m
10N
10N
32°
112°
2m
10 N
10 N
4m
CaseA
CaseB
CaseC
CaseD
RankOrder:Smallest(1)_____(2)_____(3)_____(4)_____Largest
3. Twoidenticalobjectsareeachdisplacedthesamedistance,oneby
aforce pushinginthedirectionofmotionandtheotherbya
force2 pushingatanangle relativetothedirectionofmotion.
Theworkdonebythetwoforcesisthesame.Whatistheangle ?
(Hint:SeeGOTIT?6.1.)
2
a)
b)
c)
d)
0°
30°
45°
60°
4. CRITICALTHINKING:Aweightlifterpicksupabarbelland(1)liftsitchesthigh,(2)holdsitfor
30seconds,and(3)putsitdownslowly(butdoesnotdropit).Rankorderfromsmallestto
largestthework theweightlifterperformsduringthesethreeoperations.Labelthe
quantitiesas , ,and .Justifyyourrankingorder.
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HomeworkProblems
6.18
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6.20
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Lesson19
Lesson19
KineticEnergyandPower
Examples
HomeworkProblems
6.3,6.4
6.6,6.7,6.9
6.29,6.64,6.71
LearningObjectives
[Obj28]
Explaintheconceptofkineticenergyanditsrelationtowork.
[Obj29]
Explaintherelationbetweenenergyandpower.
[Obj34]
Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanical
energy,orconservationofenergy.
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
A3,000‐kgsailboatistravellingat25m/swhenaconstantnetforceof1200Nstartsactingonit,in
thedirectionofmotion.Whatisthespeedoftheboatafterithastravelled200mundertheaction
ofthisforce?
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
1
2
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
1
2
∆
1
2
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.
IMPLEMENTATION
1. Net work:
2.
3.
∙ Δ
∆ cos
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CALCULATION
1200N 200m cos 0°
240,000Nm
480,000J
240,000J joules 3000kg 25
3000kg
m
s
m
28 s
SELF‐EXPLANATIONPROMPTS
1. Statethework‐energytheoreminyourownwords.
equals
.Fromkinematics,weknowthatifanobjectstartsfromrestandaccelerateswith
;andfromNewton’sSecondLaw,weknowthat
.Combinethethreeequationsandshowthattheworkdonebytheforceequals
.
3. Usethesameprocedureasabovetoshowthattheworkdonetoincreasethespeedofmassm
fromv1tov2isequaltothechangeinitskineticenergy.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Galileoissaidtohavedroppedtwoobjectsofdifferentmassfromatalltowertoshowthatall
objectsfallwiththesamespeed.Ifyoudroptwomasses,m1andm2,fromthesameheighth,do
theyreachthegroundwiththesamekineticenergy?
CalculatethedifferenceintheirkineticenergiesΔK.
∆
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Lesson19
PreflightQuestions
2. Twocars,onefourtimesasheavyastheother,areatrestonafrictionlesshorizontaltrack.
lightercarwillbe_______thekineticenergyoftheheaviercar.
a) one‐quarter
b) one‐half
c) equalto
d) twice
e) fourtimes
4. Whichofthefollowingistrue?
a) NeitherΔKnorWnetcaneverbenegative.
b) Wnetcanneverbenegative,butΔKcanbenegativeorpositive.
c) ΔKcanneverbenegative,butWnetcanbenegativeorpositive.
d) ΔKandWnetcanbenegativeorpositive.
5. CRITICALTHINKING:OnMondayyourunupthestairstothetopfloorofatallbuilding.You
runataconstantspeed.OnTuesdayyouwalktothetop,alsoatconstantspeed.On
Wednesdayyoutakeaconstantspeedelevator.Howdotheamountsofworkyoudidgettingto
thetopofthebuildingeachdaycompare?Howdoesthepowercompare?
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HomeworkProblems
6.29
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6.64
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Lesson20
Lesson20
PotentialEnergy
Examples
HomeworkProblems
7.1,7.2
7.1,7.2
7.14,7.31,7.42
LearningObjectives
[Obj30]
Explainthedifferences betweenconservativeandnonconservativeforces.
[Obj31]
Evaluatetheworkdonebybothconservativeandnonconservativeforces.
[Obj32]
Explaintheconceptofpotentialenergy.
[Obj33]
Evaluatethepotentialenergyassociatedwithaconservativeforce.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Averticalspringwithaspringconstantk=150N/miscompresseddown1.5m.A2‐kgballis
placedonthecompressedspringandreleasedfromrest.Whatheightdoestheballreachafteritis
released?
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
1
2
Δ
1
2
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
1
2
1
2
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CALCULATION
First, we are given the following quantities:
y1 = 0 m
y2 = − 1.5 m
m = 2 kg
k = 150 N/m
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
1
2
1
2
8.6m
The ball rises 8.6 meters above the top of compressed spring.
SELF‐EXPLANATIONPROMPTS
1.Thepotentialenergystoredinacompressedspringcomesfromtheworkdonebycompressing
thespringagainstitsrestoringforceF=−ky.Calculatethatworkandverifytheaboveexpression
ofthespringpotentialenergyUs.
2.DothesameforthegravitationalpotentialenergyUg.
3.Insolvingtheproblemweignoredthemassofthespring.Includingthemassofthespringis
changetheoutcomeofthecalculation?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A2‐kgballisreleasedfromrest3metersaboveanunstretchedspringofwhosespringconstantis
150N/m.Howmuchdoesitcompressthespringbeforeitcomestorest?(Beforeyoustart
calculating,carefullydrawthecoordinatesystemandcarefullyidentifyalltherelevantvertical
Tryit!(1PFpt):Howhighwouldtheballneedtobereleasedifyou
wantedtodoubletheamountthatthespringiscompressed?
Showyourwork.
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Lesson20
PreflightQuestions
2. Theenergystoredinacompressedspringdependsontheamountofcompression.Agiven
springrequires10.0Jforacompressionof10.0cm.Howmuchtotalenergywouldbestoredifit
a) 22.5J
b) 12.5J
c) 5.00J
d) 1.25J
e) Cannotbedeterminedfromthegiveninformation.
3. Atrunkofmass isliftedalongacurvedpathoflength toaheight .
Anothertrunkwithtwicethemassisslidacrossalevelfloor(
0.5)
alongacurvedpathalsohavinglength .Whichisgreater,theworkdone
againstfrictionortheworkdoneagainstgravity?
a) Moreworkisdoneagainstfriction.
b) Moreworkisdoneagainstgravity.
c) Theworkdoneagainstfrictionisthesameastheworkdone
againstgravity.
d) Cannotbedeterminedfromthegiveninformation.
2
4. CRITICALTHINKING:Whycan’twedefinepotentialenergyforfriction?Explain.
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HomeworkProblems
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7.31
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Lesson21
Lesson21
ConservationofMechanicalEnergy
7.3,7.4
Examples
7.4,7.5,7.6
HomeworkProblems
7.24,7.25,7.55
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj34]
[Obj35]
Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanical
energy,orconservationofenergy.
Describetherelationbetweenforceandpotentialenergyusingpotential‐energy
curves.
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Note:Beforeworkingwiththisexample,revisitLesson14,whichhasthesameproblemneglecting
friction.
Anamusementparkrideconsistsofaverticalloopwhosediameteris15mandasmall150‐kgcart
loopifitistonegotiatethetopoftheloopsafely(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.
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Lesson21
IMPLEMENTATION
1
2
1
2
If we set the potential energy to be zero at the bottom of the track
2
The energy conservation equation then becomes
1
2
2
0
CALCULATION
1
2
SELF‐EXPLANATIONPROMPTS
2.5
86,000J
1. Stateinyourownwordswhatwemeanby“conservationprinciple.”
2. Whycanthezeropointofpotentialenergybechosenarbitrarily?
3. Sketchanenergybarchart(similartothoseinFigure7.8inyourtextbook)forthecartat(a)
thetopofthetrackand(b)atthebottomofthetrack.
Energy→
Energy→
0
0
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A50‐kgwrecker’sballishangingonan8‐mropethatcansupportamaximumforceof1000N.If
theballisswunginverticalcircle,whatisfastestspeeditcanhaveatthelowestpoint,suchthatthe
ropewon’tbreak?
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Lesson21
PreflightQuestions
2. Abottledroppedfromabalconystrikesthegroundwithaparticularspeed.To
doublethespeedatimpact,youwouldhavetodropthebottlefromabalconythat
is
a) twiceashigh.
b) threetimesashigh.
c) fourtimesashigh.
d) eighttimesashigh.
3. Atruckinitiallyatrestatthetopofahillisallowedtorolldown.Atthebottom,itsspeedis
14m/s.Next,thetruckisagainrolleddownthehill,butthistimeitdoesnotstartfromrest.It
hasaninitialspeedof14m/satthetopbeforeitstartsrollingdownthehill.Howfastisitgoing
whenitgetstothebottom?
a)
b)
c)
d)
e)
14m/s
17m/s
20m/s
24m/s
28m/s
speed.Describewhatishappeningtoherkineticenergy,herpotentialenergyandhertotal
mechanicalenergyasshefalls.Isanyworkbeingdone?Ifyes,wheredoesitgo?
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HomeworkProblems
7.24
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7.25
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Lesson22
Lesson22
Lab4‐ConservationofEnergy
7.3,Lab4Worksheet
Examples
7.5
HomeworkProblems
7.56,7.59,7.63
 ThereisaLABthislesson.
LearningObjectives
[Obj34]
Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanical
energy,orconservationofenergy.
Notes
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Score(5)
Pre‐LabQuestions
perfectlyhorizontalairtrack.Thespeedofthecart,v,ismeasuredsomedistancedowntheair
trackandusedtocalculatethespringconstant,k(refertothelabhandoutandExample7.4inthe
textbook).
1.
Usetheprincipleofconservationofmechanicalenergytofindanexpressionforthespeedof
thecartasafunctionofthecompressiondistance.
a)
b)
c)
d)
2
2
plotandexplainwhyitmakessensetoplotthedatainsuchaway.
3. Afterplottingvvs.x,yourgroupdeterminesthattheslopeofthebest‐fitlinethroughthe
datapointsis50s‐1.Iftheaircarthasamassof0.50kg,thespringconstantkis
a) 25N/m
b) 50N/m
c) 1250N/m
d) Cannotbedeterminedwiththegiveninformation.
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Lesson22
LabNotes
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HomeworkProblems
7.56
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Lesson22
7.59
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7.63
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Lesson23
Lesson23
OrbitalMotion
8.1– 8.3
Examples
8.1,8.2,8.3
HomeworkProblems
8.17,8.39,MP
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj36]
Explaintheconceptofuniversalgravitation.
[Obj37]
Solveproblemsinvolvingthegravitationalforcebetweentwoobjects.
[Obj38]
Determinethespeed,acceleration,andperiodofanobjectincircularorbit.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Aspacecraftisorbiting200kmabovethesurfaceoftheplanetMars.Oneoftheastronautson
boarddropsapen.HowfastdoesthepenfallrelativetothesurfaceofMars?Howfastdoesthe
penfallrelativetotherestofthespacecraft?
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,
law,
; with Newton’s Second
, 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.
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Lesson23
SELF‐EXPLANATIONPROMPTS
1.Whichdirectionisthespacecraftacceleratingandtowhichobjectsdoesthevalueofacceleration,
3.32 ,apply?
2.Howisthequantity“r”definedintheequationforuniversalgravitation?Useyourdefinitionto
intheexample.
justifywhy
3.Ifthepenisaccelerating,explainwhyitisconsideredtobein“freefall”.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Itiscommonlybelievedthatyouexperienceabrieffeelingofweightlessnesswhenyouareriding
anelevator.Consideranelevatorwhichhasafinalspeedof2.3m/s.Inorderforyoutoexperience
afeelingof“weightless”,howlongmustittaketheelevatortogofromresttoitsfinalspeed
(assumingconstantacceleration)?Doesthishappenwhentheelevatorisgoinguporgoingdown?
Free‐BodyDiagram(required)
goingdown
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Lesson23
PreflightQuestions
2. What is the approximate force that the Moon exerts on you when it is directly overhead?
a)
b)
c)
d)
e)
2 10
2 10
2 10
2 10
2N
N
N
N
N
3. Themagnitudeoftheforceofgravitybetweentwoidenticalobjectsis .Ifthemassofeach
objectandthedistancearedoubled,whatisthenewforceofgravitybetweentheobjects?
a) b)4 c)8 d)
e)
4. CRITICALTHINKING:Ageosynchronousorbitisonewheretheorbitalobjectstaysbasically
overthesameplaceonearthallthetime.Theobjectstaysrelativelymotionlessinthesky
above.ThemassofEarthis5.97x1024kg,andtheperiodisthesameasthatofEarthat
T=23hr,56min,4sec.Describehowyouwoulddeterminethealtitudeforgeosynchronous
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HomeworkProblems
8.17
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8.39
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Lesson24
Lesson24
GravitationalEnergy
Examples
HomeworkProblems
8.4
8.4,8.5
8.27,8.52,MP
LearningObjectives
[Obj39]
[Obj40]
[Obj41]
Solveproblemsinvolvingchangesingravitationalpotentialenergyoverlarge
distances.
Usetheconceptofmechanicalenergytoexplainopenandclosedorbitsandescape
speed.
Useconservationofmechanicalenergytosolveproblemsinvolvingorbitalmotion.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
A120‐kgsatelliteisinacircularorbit100kmabovethesurfaceoftheEarth.Howwouldthetotal
energyofthesatellitechangeifitweremovedtoahigherorbit200kmabovethesurfaceofthe
Earth?
STRATEGY
The total energy (potential and kinetic) of a satellite
in a circular orbit about the Earth is
1
2
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
1
2
–
1
2
6.37
10
100
10
6.37
10
200
10
1
1
2
1
CALCULATION
Δ
1
6.67
2
10
Nm /kg
5.97
Δ
6.47
10 m
6.57
10 m
10 kg 120kg
3600
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1
1
6.47 10 m
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Lesson24
SELF‐EXPLANATIONPROMPTS
thisderivation.
2.Thegravitationalenergyequationincludesgravitationalpotentialenergy.Whereisthe
gravitationalpotentialenergyzero?
3.Explainwhyisthetotalenergynegative.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
A1200‐kgsatelliteisinanellipticalorbitaround
theEarth.Atperigee,thealtitudeofthesatellite
is1,000kmabovethesurface,andatapogeethe
altitudeis10,000kmabovethesurface.
perigee
apogee
Ifthesatelliteistravelingat8.6km/satperigee,
whatisitsspeedatapogee?
Tryit!(1PFpt):Determinethetotalenergy(kinetic+potential)ofthe
satelliteatbothperigeeandapogee.Showyourwork.
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Lesson24
PreflightQuestions
2. ConsideraspacecraftorbitingtheSuninacircularorbit.The
Sun’sgravity.
Comparethetotalenergy forthecircularorbit ,elliptical
orbit ,andparabolictrajectory .
a)
b)
c)
3. Supposeanobjectismovingalonganyoneofthegivenorbitalpaths.Whatistrueregardingthe
orbitsdepicted?
a) Thekineticenergyisconstantinalltheorbits,whilethepotentialenergychangeswith
distancefromtheSun.
b) Thepotentialenergyisconstantforallpointsinanyoneoftheorbits.
c) Totalenergydecreasesfromthecircularorbit untilitequalszerofortheparabolic
trajectory .
d) Totalenergyisconstantforanypointalonganyoneoftheorbits.
4. CRITICALTHINKING:Aphysicsbookclaimsthat,“Moon‐boundspacecrafthavespeedsjust
,sothatifanythinggoeswrong(aswithApollo13),theywillreturntoEarth.”
under
velocityequationandwhetheraspacecraftcangettotheMoonwithoutescapingtheEarth.
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HomeworkProblems
8.27
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8.52
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Lesson25
Lesson25
CriticalThinking:OrbitalEnergies
8.4
Examples
8.5
HomeworkProblems
MP,8.61,8.67
 ThereisanEXAM‐PREPQUIZthislesson.
LearningObjectives
[Obj39]
Solveproblemsinvolvingchangesingravitationalpotentialenergyoverlarge
distances.
[Obj40]
Usetheconceptofmechanicalenergytoexplainopenandclosedorbitsandescape
speed.
Notes
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PreflightQuestions
2. Thetotalenergyofasatelliteinaparticularboundorbit
a) variesdependingonthesatellite’spositioninthatorbit.
b) isalwayspositive.
c) isalwaysnegative.
d) isalwaysexactlyzero.
.Ifthedistanceisdoubled,whatisthenewgravitationalpotentialenergy?
a)
b)4
c)8
d)
e)
4. CRITICALTHINKING:TheInternationalSpaceStation(ISS)
(http://www.nasa.gov/mission_pages/station/main/index.html)
orbitsEarthatanaltitudeof350km.Sincethestationandthe
astronautsinsideareinfreefalltogether,theyfloataroundinsidethe
ISSmodules.Ifthestationwerestationaryatthataltitude,howwould
theastronauts’weightscomparetotheirweightsatthesurfaceof
Earth?
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Lesson25
HomeworkProblems
MP
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8.61
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Lesson25
8.67
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Notes
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Lesson26
Lesson26
CenterofMass
Examples
HomeworkProblems
9.1
9.1,9.2,9.3
9.16,9.37,9.89
LearningObjectives
[Obj42]
Calculatethecenterofmassforsystemsofdiscreteparticlesandforcontinuousmass
distributions.
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WorkedExample
Score(3)
problem.
1kg
STATEMENTOFTHEPROBLEM
1m
Foursmallmassesm1,m2,m3,andm4aretiedtogetherwithrigidrods
sothattheyformasquareofside1m,asshowninthefigure.Wewant
towriteNewton’sSecondLawfortheentiresystemasifallthemass
wereconcentratedatasinglepoint,thatis
1kg
,where
isthenetexternalforceonthesystem(i.e.the
vectorsumofalltheexternalforces)and istheaccelerationofthe
system.Whatisthelocationofsuchapoint?Considerthemassofthe
connectingrodstobeverysmallcomparedtothemassesonthe
corners.
STRATEGY 2kg
The point described above is called the center-ofmass 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 centerof-mass equation. Any coordinate system will work. We choose the origin of our system to
be the center of the square.
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Lesson26
CALCULATION
1kg
1kg
0.5m
0.5m
1kg
1kg
0.5m
0.5m
2kg
6kg
2kg
6kg
0.5m
0.5m
2kg
2kg
0.5m
0.5m
0
1
m
6
The center of mass is 1/6 meters under the origin on the y-axis.
SELF‐EXPLANATIONPROMPTS
1.Intuitively,thecenterofmasscanbethoughtofasthepointatwhichtheassemblycouldbe
stablysupported.Usingthisapproach,wherewouldyouexpectthecenterofmassofthetwo1‐kg
massestobe?
3.Justify,usingsymmetry,whythecenterofmassisonthey‐axisandbelowthex‐axis?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Findthelocationofthecenter‐of‐massofasystemcomprisedofthree1‐kgmasseslocatedatthree
cornersofasquarewhosesideis1m.(Hint:Drawapictureandmarkthecenter‐of‐mass.)
coordinatesystemwiththeoriginat
thecenterofthesquare
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Lesson26
PreflightQuestions
2. (True/False)A4.8tonelephantisstandingina15tonrailcarthatisatrestonafrictionless
track.Theelephantbeginstowalktowardstheotherendofthecar.Foreverymeterthe
elephantmoves,thecarmoves1meterintheoppositedirection.
a) True
b) False
3. (True/False)Accordingtotheequationsofmotionforaprojectile,afirecrackerfollowsa
parabolicpath,neglectingairresistance.Afteritexplodes,thecenterofmassofthepiecesstill
followsaparabolictrajectory.
a) True
b) False
Thecanoesareatrestonaplacidlake.Apassengerintheheaviercanoepushesthecanoes
apart,stretchingthebungeecord.Describewhathappenstothecenterofmassofthesystem
andexplainyourreasoning.
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HomeworkProblems
9.16
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9.37
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9.89
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Lesson27
Lesson27
ConservationofLinearMomentum&Collisions
Examples
HomeworkProblems
9.1– 9.5
CE9.1,9.4,9.5, 9.7
9.38,MP,MP
LearningObjectives
[Obj46]
ExplaintheconceptoflinearmomentumofasystemofparticlesandexpressNewton's
secondlawofmotionintermsofthelinearmomentumofthesystem.
Explainthelawofconservationoflinearmomentumandtheconditionunderwhichit
applies.
Applyconservationoflinearmomentumtosolveproblemsinvolvingsystems of
particles.
Explaintheconceptofimpulseanditsrelationtoforce.
[Obj47]
Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.
[Obj48]
Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐ and
two‐dimensions.
[Obj43]
[Obj44]
[Obj45]
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
A1‐kgball,m1,collideswitha3‐kgball,m2,asshown.The
3m/sand
ballshaveinitialvelocitiesof 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 ∑
specific to this problem,
,and,
∑
.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.
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Lesson27
CALCULATION
Starting with:
It becomes:
m
s Now substituting:
1kg
3
m
s
3kg
3
3kg
m
s
3kg
2
4m/s
SELF‐EXPLANATIONPROMPTS
1.Whyisitimportanttoensurethatnonetexternalforceactsontheobjects?
2.Whycanyounotusetheabsolutevaluesofthevelocitiesintheabovecalculations?
3.Usingthecalculatedfinalvelocityofball1,showthatlinearmomentumwasindeedconservedin
thecollision.
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Pre‐ClassProblem
Thediagramontheleftshowsacollisionbetweenaseriesof
departthecollisionsasonecoupledgroup,whatwillbethe
finalvelocityoftheassembly?
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Lesson27
PreflightQuestions
2. Linearmomentumofasystemisconservedif
a)
b)
c)
d)
thenetexternalforceiszero.
theenergyofthesystemisconserved.
thenetworkdoneispositive.
onlyconservativeforcesaredoingwork.
3. A500‐gfireworkrocketismovingat60m/sstraightupwardwhenitexplodes.Thesumofall
themomentumvectorsoftherocketfragmentsimmediatelyaftertheexplosionis
a) zero.
b) 30kgm/sstraightup.
c) 30kgm/sinmultipledirections.
4. CRITICALTHINKING:Considerarubberbulletandanaluminumbullet;bothhavethesamesize,
speedandmass.Eachbulletisfiredatablockofwood.Therubberbulletbouncesback,the
aluminumbulletpenetratestheblock.Whichismostlikelytoknocktheblockover?Explain.
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HomeworkProblems
9.38
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MP
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Lesson28
Lesson28
Lab5–1‐DCollisions
9.5,9.6,Lab5Worksheet
Examples
None
HomeworkProblems
9.28,9.44,9.61
 ThereisaLABthislesson.
LearningObjectives
[Obj46]
Explaintheconceptofimpulseanditsrelationtoforce.
[Obj47]
Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.
[Obj48]
Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐ and
two‐dimensions.
Notes
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JournalQuestions
Score(5)
1. Brieflydescribethepurposeandgoalsofthislab.(Onetotwocompletesentences)
2. InPartIofthelab,aheavycart(massm1)andastationarylightcart(massm2)willundergoa
one‐dimensionalcollisiononafrictionlessairtrack.Assumingthecollisioniselastic,writethe
expressionforthefinalvelocityofcart2,v2f,intermsoftheinitialvelocityofcart1,v1i.
3. InPartIIofthelab,aheavycart(massm1)andastationarylightcart(massm2)willundergoa
totallyinelasticone‐dimensionalcollisiononafrictionlessairtrack.Deriveasimilar
expressionforthefinalvelocityofthejoinedcarts(massesm1+m2),vf,intermsoftheinitial
velocityofcart1,v1i,startingfromtheequationforconservationofmomentum.
4. Supposeyoumakeaplotofthefinalvelocityofcart2,v2f,versustheinitialvelocityofcart1,v1i,
fortheelasticcollision.Whatwouldtheplotlooklike?Writeanexpression,intermsofm1and
m2,fortheslopeassociatedwiththisplot.
5. Ifyouplottedthefinalvelocityofthejoinedcarts,vf,versustheinitialvelocityofcart1,v1i,for
comparetotheslopefortheelasticcollision?Explain.
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LabNotes
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HomeworkProblems
9.28
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9.44
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9.61
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Lesson29
Lesson29
CollisionsandConservationofEnergy:
WheredoestheEnergyGo?
Examples
HomeworkProblems
9.3,9.4
9.10
MP,9.68,9.78
LearningObjectives
[Obj46]
Explaintheconceptofimpulseanditsrelationtoforce.
[Obj47]
Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.
[Obj48]
Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐ and
two‐dimensions.
Notes
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WorkedExample
Score(3)
problem.
STATEMENTOFTHEPROBLEM
A1.0‐kgpuck(puck1)isslidingat45°abovethex‐axis
withaspeedof1.0m/s.Another1.0‐kgpuck(puck2)is
slidingwithaspeedof0.50m/sat45°belowthex‐axis.
Thepuckscollide,andpuck2fliesoffat45°belowthex‐
axis,at0.80m/s.
1
m1=1.0kg
v1i=1.0m/s
x
a)Whatisthevelocityofpuck1afterthecollision?
m2=1.0kg
v2i=0.5m/s
b)Wasthiscollisionelastic?
2
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 ycomponents 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
1
2
1
2
? 1
2
1
2
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Lesson29
CALCULATION
a) Before the collision, the x- and y- components of the total momentum are:
1.0kg 1.0
1.0kg 1.0
m
∙ cos45°
s
m
∙ sin45°
s
1.0kg 0.50
m
∙ cos45°
s
1.0kg ∙ 0.50
∙ sin45°
1.06
kg ∙ m
s
0.354
kg ∙ m
s
and after collision they are:
1.0kg
1.0kg
1.0kg 0.80
1.0kg 0.80
m
∙ cos45°
s
m
∙ sin45°
s
1.06
kg ∙ m
s
0.354
Solving for the speed of the puck 1 and the direction angle θ we get:
kg ∙ m
s
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.
SELF‐EXPLANATIONPROMPTS
1. Solvethemomentumequationsforthespeedanddirectionofmotionofpuck1(i.e.,fillinthe
stepsomittedabove).
2. Calculatethetotalkineticenergyofthetwopucksbeforeandafterthecollision(i.e.,fillinthe
stepsomittedabove),andconfirmthatthecollisionisinelastic.
3. Wheredidtheenergygoduringtheinelasticcollision?
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Pre‐ClassProblem
Twomasses,m1andm2,moveatrightangles,meetattheoriginandflyof,stickingtogether.Their
initialspeedsarethesame.Ifm1=3m2,whataretheirspeedanddirectionaftercollision?
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Lesson29
PreflightQuestions
a) Themagnitudeanddirectionof maychange.
b) Themagnitudeof remainsconstantbutthedirectionmaychange.
c) Themagnitudeof maychangebutthedirectionremainsconstant.
d) Themagnitudeanddirectionof remainconstant.
3. Matchthediagramtothetypeofcollisionbetweenobjectsofequalmass.
CaseA
_____
_____
_____
CaseC
CaseB
Elasticcollision
Inelasticcollision
Totallyinelasticcollision
4. CRITICALTHINKING:Birdstrikesareasignificantflightsafety
hazard.ConsideranF‐16birdstrikewhereagooseimpactsthe
canopy.TheF‐16canopydeformsduringthecollisionandthe
birdpartsdeflectawayfromtheaircraft.Whattypeofcollisionis
this?Explain.
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CaseD
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HomeworkProblems
MP
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9.68
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9.78
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Lesson30
Lesson30
LearningObjectives
[Obj26]
Explainthephysicsconceptofwork.
[Obj27]
Evaluatetheworkdonebyconstantforcesandbyforcesthatvarywithposition.
[Obj28]
Explaintheconceptofkineticenergyanditsrelationtowork.
[Obj29]
Explaintherelationbetweenenergyandpower.
[Obj30]
Explainthedifferencesbetweenconservativeandnonconservativeforces.
[Obj31]
Evaluatetheworkdonebybothconservativeandnonconservativeforces.
[Obj32]
Explaintheconceptofpotentialenergy.
[Obj33]
Evaluatethepotentialenergyassociated withaconservativeforce.
[Obj34]
[Obj36]
Solveproblemsbyapplyingthework‐energytheorem,conservationofmechanical
energy,orconservationofenergy.
Describetherelationbetweenforceandpotentialenergyusingpotential‐energy
curves.
Explaintheconceptofuniversalgravitation.
[Obj37]
Solveproblemsinvolvingthegravitationalforcebetweentwoobjects.
[Obj38]
Determinethespeed,acceleration,andperiodofanobjectincircularorbit.
[Obj39]
Solveproblemsinvolvingchangesingravitationalpotentialenergyoverlarge
distances.
Usetheconceptofmechanicalenergytoexplainopenandclosedorbitsandescape
speed.
Useconservationofmechanicalenergytosolveproblemsinvolvingorbitalmotion.
[Obj35]
[Obj40]
[Obj41]
[Obj46]
Calculatethecenterofmassforsystemsofdiscreteparticlesandforcontinuousmass
distributions.
ExplaintheconceptoflinearmomentumofasystemofparticlesandexpressNewton's
secondlawofmotionintermsofthelinearmomentumofthesystem.
Explainthelawofconservationoflinearmomentumandtheconditionunderwhichit
applies.
Applyconservationoflinearmomentumtosolveproblemsinvolvingsystemsof
particles.
Explaintheconceptofimpulseanditsrelationtoforce.
[Obj47]
Explainthedifferencesbetweenelastic,inelastic,andtotallyinelasticcollisions.
[Obj48]
Applyappropriateconservationlawstosolveproblemsinvolvingcollisionsinone‐ and
two‐dimensions.
[Obj42]
[Obj43]
[Obj44]
[Obj45]
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Amandragsa50‐kgcrate10macrossaroughhorizontalsurface,wherethecoefficientoffriction
betweenthecrateandsurfaceis0.3.Hepullsataconstantspeedandhedirectshispullingforce
20°downwardfromthehorizontal.Howmuchworkdoesheperform?
STRATEGY(Fillintheblanks.)
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
____________
CALCULATION(Fillintheblanks.)
cosθ
Δ
1328J
SELF‐EXPLANATIONPROMPTS
1.Theworkdoneinthiscaseis3JmorethanintheworkedexampleforLesson18.Explainwhy
theworkincreased.
2.Whatworkdoesfrictiondo?
OptionalPracticeProblems:6.13,6.19,6.21
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Lesson30
regardingtheproblem.
STATEMENTOFTHEPROBLEM
A1500‐kgcaristravellingat26.8m/s.Thedriverappliesasmallbrakingforceof800N.Howfar
doesthecartravelbeforeitslowsdownto13.4m/s?
STRATEGY(Fillintheblanks.)
Apply the __________________
theorem and solve the
resulting equation for the unknown displacement.
CALCULATION(Fillintheblanks.)
∙ Δ
_________________
_____________
Δ
505m
SELF‐EXPLANATIONPROMPTS
1. Whathappenstothestoppingdistanceifthespeedofthecarisdoubled,assumingthesame
brakingforce?
2. Wheredoesthe“brakingforce”comefrom?
OptionalPracticeProblems:6.27,6.39
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Anobjectofmassmisreleasedatrestfromaheighthaboveground.Whatisthespeedoftheobject
justbeforeitreachestheground?
STRATEGY(Fillintheblanks.)
The object has positive potential energy
relative to the ground. As it falls, that
energy gets converted into kinetic.
Symbolically,
∆
⇒ _____________
CALCULATION(Fillintheblanks.)
_____________
½mv2
Setting the kinetic energy equal to ΔUg
and solving for the speed v we get
2
SELF‐EXPLANATIONPROMPTS
dropsout.
energydoesnotdependonthemass?
3.Thisproblemcanalsobesolvedusing1‐Dkinematics.Usethismethodandcomparetheresults.
OptionalPracticeProblems:7.13,7.17,7.30
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
A60‐kgskierstartsfromrestandskisdownazig‐zagtrail.Whenshereachesthebottomofthetrail
shehasdescendedaverticalelevationof600m.Ifsheloses12%ofherenergytofriction,whatis
herspeedatthebottomofthetrail?
STRATEGY(Fillintheblanks.)
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.
CALCULATION(Fillintheblanks.)
___________
___________
310,464J
___________
101m/s
SELF‐EXPLANATIONPROMPTS
1. Inyourownwordsdescribethemodifiedconservationequationweused.
2. Whydoesthecontouroftheterrainnotmatterinthiscalculation?
3. Howwouldtheproblemneedtobechangedsothatwecouldusetheforceoffrictionformula,
?
OptionalPracticeProblems:7.19,7.53,7.57
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Manymoviescontainsceneswheretheactorsexperienceweightlessness.Thesescenesareusually
filmedinanairplanethatisundergoingcarefullychoreographedmaneuverswhichsimulate
weightlessness.Considerthefollowingflightpath:Fromthestartofthemaneuvertot=120s,the
airplane’sheightisdescribedby
225 .Duringthenextpartofthemaneuver,whichlasts60
seconds,theairplanesheightisgivenby
4.9
1176
43560.Forthefinal15secondsof
23000.Duringwhattimeperiodisthe
themaneuver,theairplane’saltitudeisgivenby
airplane“weightless”?
STRATEGY(Fillintheblanks.)
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.Whatpartofthepositionequationdeterminestheplaneisfreefallinthe2ndpartofthe
maneuver?Whatpartsoftheequationdoesnotmatter?
2.Howcouldyoumodifythepositionequationforthethirdpartofthemaneuvertomaketheplane
beinfreefall?
OptionalPracticeProblems:8.35,8.19,
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Ifa300‐kgsatellite,inacircularorbit150kmabovethesurfaceoftheEarthcrashestotheground
(orburnsup),howmuchenergyislost?
STRATEGY(Fillintheblanks.)
The total energy (potential and kinetic) of a satellite in a
circular orbit about the Earth is
1
2
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
CALCULATION(Fillintheblanks.)
Δ
_____________________
6.37
10
1
1
2
150
10
1
6.52 10 m
__________________________
216
10 J
SELF‐EXPLANATIONPROMPTS
1.Thegravitationalenergyequationincludesthegravitationalpotentialenergy.Whereisthe
gravitationalpotentialenergyzero?
2.Completethecalculation.
OptionalPracticeProblems:8.30,8.58
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Whereisthecenterofmassofa2‐mbarbellwitha1‐kgmassontheleftanda3‐kgmassonthe
right?Considerthemassofthebartobeverysmall(negligible)comparedtothemassesonthe
corners.
3kg
1kg
2m
STRATEGY(Fillintheblanks.)
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.
CALCULATION(Fillintheblanks.)
1kg 3kg 0.5m
SELF‐EXPLANATIONPROMPTS
1.Doestheresultagreewithyourintuition?
2.Convinceyourselfthatthechoiceofthecoordinatesystemdoesnotmatter.Choosetheoriginof
thecoordinatesystemattheleftend,atthelocationofthe1kgmass,andshowthatyougetthe
andagainascalculatedusingthenewcoordinatesystem.
OptionalPracticeProblems:9.12,9.38,9.49
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
A100‐kgclownislaunchedfroma500‐kgcircuscannon.AfterthefiringofBozo,hehasaspeedof
15m/s.Assumingthathisshoesandcostumearesohighlypolishedthatthereisnofrictionashe
movesoutofthecannonbarrelandthatthecannonandclowninitiallyareatrest,whatisthefinal
velocityofthecannon?
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 ____________________.
CALCULATION(Fillintheblanks.)
First:∑
Now: ∑_____________
_______
With numbers: _____ _____
100kg _____m/s ___
500kg _____100kg
___ 100kg ____m/s /_____kg
_______m/s 3m/s
SELF‐EXPLANATIONPROMPTS
2. Istheinitialcondition(clowninsidecannon)thesameasifthetwoweresittingatrestnextto
oneanother?Explain.
OptionalPracticeProblems:9.18,9.19,9.20,9.42
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
Twoidenticalmasses,M,approachtheoriginwiththesamespeed
v,at45degreesfromthehorizontal.
Theycollideandsticktogether.Whatarethespeedanddirectionof
motionaftercollision?
STRATEGY(Fillintheblanks.)
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 _____________
CALCULATION(Fillintheblanks.)
2
45
________________
45 inthepositivex‐direction.
SELF‐EXPLANATIONPROMPTS
1.Isthiscollisionelastic?
energy.
OptionalPracticeProblems:9.43,9.68,9.77
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Notes
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Notes
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Lesson31
Lesson31
RotationalMotion
10.1,10.2
Examples
10.1,10.2,10.3
HomeworkProblems
10.19,10.23,10.45
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj49]
[Obj50]
[Obj51]
Explaintherelationbetweentherotationalmotionconceptsofangulardisplacement,
angularvelocity,andangularacceleration.
Useequationsofmotionforconstantangularaccelerationtosolveproblemsinvolving
angulardisplacement,angularvelocity,andangularacceleration.
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantangular
acceleration.
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WorkedExamples
Score(3)
STATEMENTOFTHEPROBLEM
TheSmokyHillsWindFarmnearSalina,Kansasemploy140,000poundVestasV801.8‐megawatt
(revolutionsperminute.)
revolutionsdoesitturn?
STRATEGY


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
To obtain the tangential speed, we use
2
centripetal acceleration is then
. The
.
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
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Lesson31
CALCULATION
a) Maximum angular velocity:
b) Maximum linear speed of the tip:
40m
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
Note: Compare
for rotational motion to
2
4revolutions
2
for
linear motion.
SELF‐EXPLANATIONPROMPTS
3.Ifanobjectisrotatingwithanon‐zeroangularvelocityωandzeroangularaccelerationα,isthere
acentripetalacceleration?Ifanobjectisrotatingwithanon‐zeroangularvelocityωandnon‐zero
angularaccelerationα,whatisthetotallinearacceleration?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
taketheflywheeltoreach12rpm(revolutionsperminute)ifitstartsfromrest?
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PreflightQuestions
a) 30°
b) 60°
c) 90°
d) 180°
3. Twoantscrawlontothesurfaceofacompactdisc.AntAisfartherfromthecenterofthedisc
thanAntB.Thecompactdiscbeginstospin.Whichofthefollowingstatementsistrue?
a)
b)
c)
d)
AntAexperiencesagreatertangentialaccelerationthanAntB.
AntAexperiencesagreaterangularaccelerationthanAntB.
Neitherstatementistrue.
Bothstatementsaretrue.
4. CRITICALTHINKING:Whatistheapproximateangularspeedoftheearthrevolvingaroundthe
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HomeworkProblems
10.19
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Lesson31
10.32
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10.45
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Lesson32
Lesson32
RotationalInertia&Torque
10.2,10.3
Examples
10.4,10.5
HomeworkProblems
10.30,10.28,10.52
LearningObjectives
[Obj52]
Explaintheconceptoftorqueandhowtorquescausechangeinrotationalmotion.
[Obj53]
Givenforcesactingonarigidobject,determinethenettorquevectorontheobject.
[Obj54]
Determinetherotationalinertiaforasystemofdiscreteparticles,rigidobjects,ora
combinationofboth.
Compareandcontrasttheconceptsofmassandrotationalinertia.
[Obj55]
Notes
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WorkedExamples
Score(3)
problem.
STATEMENTOFTHEPROBLEM
Youhaveaflattireonyourcarand,inordertochangethetire,youneedtoremovethelugnutsthat
securethewheeltothecar.Ifthe30‐cmlongwrenchyouareusingtoremovethenutsisata55°
angletothehorizontalandyouapplyaforceof120Ndirectlydownontheendofthewrench,what
isthemagnitudeofthetorqueyouexertonthelugnut?
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).
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SELF‐EXPLANATIONPROMPTS
1.Inyourownwords,explainhowtorquediffersfromforce.
2.Whydidweuse125°fortheangleoftheappliedforceandnot55°?
3.Explainhowthemagnitudeofthetorqueexertedonthelugnutswouldchangeiftheforcewas
appliedinthemiddleofthehandleratherthantheend.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
ofthetorque,andb)themagnitudeoftheangularaccelerationifthedoor’srotationalinertia,I,is
30kgm2.
Tryit!(1PFpt):Calculatethemagnitudeofthe
torqueiftheforcewasappliedatanangleof25°
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Lesson32
PreflightQuestions
2. Rankorderthemagnitudeofthetorquesfromsmallesttolargest.Eachrodis50‐cmlongfrom
thepivot().
2N
2N
(A)
(B)
(C)
4N
45°
2N
(D)
(E)
4N
RankOrder:Smallest(1)_____(2)_____(3)_____(4)_____(5)_____Largest
inertia.Shecouldachievethatby
a) stretchingherarmsfartherawayfromtheverticalrotationaxis.
b) bringingherarmsclosertoherbody.
c) loweringherbodybybendingherkneesandsquattingdown.
d) bendingforwardatherwaistsoherbodyisL‐shaped.
e) Rotationalinertiacanonlydecreaseifhermassdecreases.
ofthebookwilldependupontheaxischosen.RankthechoicesAtoCaboveinorderof
increasingmomentsofinertiaandexplainyourranking.
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HomeworkProblems
10.30
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10.28
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10.52
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Lesson33
Lesson33
RotationalAnalogtoNewton’sSecondLaw
10.3
Examples
10.8,10.9
HomeworkProblems
10.56,10.57,MP
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj55]
Compareandcontrasttheconceptsof massandrotationalinertia.
[Obj56]
UseNewton’ssecondlawanditsrotationalanalogtosolveproblemsinvolving
translationalmotion,rotationalmotion,orboth.
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WorkedExample
Score(3)
STATEMENTOFTHEPROBLEM
A50‐kgblockanda100‐kgweightareconnectedwitharope,
passingoverapulleyasshown.The50‐kgblockisona30°ramp
is0.2mandwhosemomentofinertiais2kgm2.Theropedoes
notstretch.
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:
The normal force
– 30°
.
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
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Lesson33
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
– 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‐EXPLANATIONPROMPTS
1.Justifywhyeachcoordinatesystemonthefree‐bodydiagramswasused.
2.Justifythenegativesignusedfortheaccelerationintheequationofmotionforthe100‐kgweight.
3.Explaininyourownwordswhythetensionsonthetwosidesofthepulleyaredifferent.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Whatmagnitudeoftorquehastobeappliedtoa2.3‐kg,18‐cmdiameter,soliddiskrotatingat
800rpmtostopitin10seconds?
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Lesson33
PreflightQuestions
2. Whichstatementiscorrect?
a) Iftorqueincreases,rotationalinertiamustincrease.
b) Therotationalinertiaofanobjectdoesnotdependonthelocationofitsaxisofrotation.
c) Anobjectwithmoremasshasahigherrotationalinertiathananobjectwithlessmass.
d) Rotationalinertiameasuresanobject'sresistancetochangesinitsrotationalmotion.
1N
2kg 2m
CaseA
3. Rankordertheangularacceleration ofeachcase.
2kg
1N
Theobjectsareconnectedwithmasslessrodsof
lengthsshown.Theforcesshownaretheonlyforces
2N
2kg 2m
2kg
point().
2N
a)
4kg
1N
4m
CaseC
b)
1N
4kg
c)
2N
2kg
d)
4m
30°
CaseD 30°
2kg
2N
4. CRITICALTHINKING:Thetwoblocksfromtheworkedexampleproblem,mand
M,arenowhungdirectlydownfromthepulleyasshown.Describehowthe
equationfortheaccelerationoftheblockswouldchangeforthisscenario.
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HomeworkProblems
10.56
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10.57
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MP
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Lesson34
Lesson34
RotationalEnergyandRollingMotion
Examples
HomeworkProblems
10.4,10.5
10.10‐10.12,CE10.1
10.60,10.62,10.68
LearningObjectives
[Obj58]
Solveproblemsinvolvingrotationalkineticenergyandexplainitsrelationtotorque
andwork.
Explaintherelationbetweenlinearandangularspeedinrollingmotion.
[Obj59]
Useconservationofenergytosolveproblemsinvolvingrotatingorrollingmotion.
[Obj57]
Notes
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WorkedExample
Score(3)
Aboulderontopofahillbreaksfreeandbeginstorolldownthehillwithoutslipping.
bottomofthehillafterithasundergoneaverticaldisplacementof100m?
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
final state becomes
1
2
in
the
1
2
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 1
2
1
2
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
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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
relationship between angular speed and translational speed
and the
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:
10
7
10 ∙ 9.8 m
s
7
∙ 100m
37 m⁄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“no
workisdonebyfrictionontheboulder”.
2.Explainwhytheboulderhasonlyrotationandtranslationkineticenergyatthebottomofthehill.
rolling.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Inapinballmachine,asolidmetal0.050‐kgballisreleasedfromaspringandrollsaroundthe
machinehittingvarioustargets.Ifthespringhasaspringconstantkof410N/mandiscompressed
b)Whatisthetranslationalkineticenergyoftheballimmediatelyafterrelease?
Tryit!(1PFpt):Calculatethetranslationkineticenergyofthe
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PreflightQuestions
)andasolidsteel
)startfromthesamepositionandroll
downarampwithoutsliding.Atthebottomoftheramp,
a) thealuminumcylinderhasgreatertotalkineticenergy.
b) thesteelcylinderhasgreatertotalkineticenergy.
c) thecylindershavethesametotalkineticenergy.
3. A4.5‐kgbicycletire(
0.6kgm ,
37cm)isspinningonamechanic’sstandatthesame
rateasifitwererollingatalinearspeed
10m/s.Themechanicappliesthebrake
5m/srollingspeed.Whatisthework
supplyingaforcetoslowtherotationequivalentto
donebythebrake?
a)
16.4mJ
b) 164J
c) 164J
d) 16.4mJ
4. CRITICALTHINKING:An8‐kgwheelhasamomentofinertiaIof0.1kgm2.Thewheelisrolling
alongwithoutslipping.Whatistheratioofitstranslationalkineticenergytoitsrotational
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10.60
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10.62
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Lesson35
RotationalVectorsandAngularMomentum
Examples
HomeworkProblems
11.1– 11.3
11.1
11.16,11.17,11.21
LearningObjectives
[Obj60]
[Obj61]
Determinethedirectionsoftheangulardisplacement,angularvelocityandangular
accelerationvectorsforarotatingobject.
Determinetheangularmomentumvectorfordiscreteparticlesandrotatingrigid
objects.
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WorkedExample
Score(3)
Acarisdrivingclockwisearoundacircularracetrack.Thetiresonthecarrotate50timesevery
second.a)Whatisthecar’sangularvelocityasitasittravelsduenorthanddueeast?b)Whatis
theaverageangularaccelerationofthecarduringthe10secondsittakestogofromtravelingdue
northtotravelingdueeast.
STRATEGY
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:
∆
velocity, we apply the right hand rule.
. To find the direction of the angular
CALCULATION
a) Angular velocity:
50
rotations
second
2
rotation
314
s
b) Average angular acceleration:
s
1
10s
31.4
s
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 righthand 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.
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SELF‐EXPLANATIONPROMPTS
tothetailofanothervector.Explainhowthisapproachisconsistentwiththeabovestatement
thattheaverageaccelerationvectorgoesfromthetipofinitialvectortothetipofthefinal
vector.
2. Whichdirectionisthevelocityvectorwhenthecaristravelingwest?South?
3. Howwouldyoudescribethedisplacementvectorofthecar?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Achildisdoingtrickswitharemote‐controlledairplane.Initiallythepropellersontheairplaneare
spinningat1200rpmastheplanedivesstraighttowardtheground.Threesecondslater,the
airplaneisinlevel‐flight,flyingnorthandthepropellersarespinningat1800rpm.Whatwasthe
averageangularaccelerationofthepropellers?
20.9
,
34°abovelevelflight
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PreflightQuestions
2. (True/False)Thenettorque
andangularacceleration alwayspointinthesamedirection.
a) True
b) False
3. (True/False)Theangularacceleration andangularvelocity alwayspointinthesame
direction.
a) True
b) False
4. CRITICALTHINKING:Howcanaparticlewithlinearvelocityhaveangularmomentum?
Explain.
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11.16
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11.17
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Lesson36
ConservationofAngularMomentum
Examples
HomeworkProblems
11.4
CE11.1
11.26,11.27,11.43
LearningObjectives
[Obj62]
Applyconservationofangularmomentumtosolveproblemsinvolvingrotating
systemschangingrotationalinertiasandrotatingsystemsinvolvingtotallyinelastic
collisions.
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WorkedExample
STATEMENTOFTHEPROBLEM
A2.0‐kgprojectilewithaspeedof5.0m/sstrikesafinonawheel
asshownthefigure.Theprojectilestrikesatapoint1.48mtothe
rightoftheaxisofrotation.Aftertheprojectilecollideswiththe
wheelitstickstothefinatthepointofimpact.Ifthewheelhasa
rotationalinertiaofI=100kgm2,whatwillbetheangularvelocity
ofthewheel+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/s
100kgm
2kg 1.48m
Notice that the result depends on the positioning of the launcher relative to the axle of the
wheel.
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SELF‐EXPLANATIONPROMPTS
1.Howdoestheresultchangeifyoumovelaunchersothatthepointofimpactisatagreater
distancefromtheaxleofthewheel?
2.Canyoutellifthecollisioniselasticorinelastic?Explainhowyouknow,orwhyyoucannottell.
3.Ifyouthinkit'sinelastic,howmuchenergyislostinthecollision?Ifyouthinkit'selasticcheckto
seeifyou'recorrect.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
blockofclayontothewheelwithavelocityof0.75m/sinthesamedirectionasthewheel.How
fastisthewheelspinningimmediatelyaftertheclaylandsonthepotter’swheel,inrpm?
or7.33rpm
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PreflightQuestions
2. Ifanettorqueisappliedtoarigidobject,whichofthefollowingisnottrue?
a)
b)
c)
d)
Theangularmomentumoftheobjectwillchange.
Thekineticenergyoftheobjectwillchange.
Theobjectwillexperienceanangularacceleration.
Therotationalinertiaoftheobjectwillchange.
closetoherbody,her
a) angularmomentumremainsthesame.
b) angularvelocityincreases.
c) kineticenergyincreases.
d) Alloftheabovearetrue.
4. CRITICALTHINKING:Explainwhyhelicoptersmusthavetworotorstofunctionproperly.Your
explanationshouldinvolveangularmomentumconcepts.
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HomeworkProblems
11.26
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11.27
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11.43
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Lesson37
CriticalThinking:Energy&AngularMomentum
Chapter10&11
Examples
None
HomeworkProblems
11.46,11.49,MP
 ThereisanEXAM‐PREPQUIZthislesson.
LearningObjectives
[Obj62]
Applyconservationofangularmomentumtosolveproblemsinvolvingrotating
systemschangingrotationalinertiasandrotatingsystemsinvolvingtotallyinelastic
collisions.
Notes
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PreflightQuestions
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.
rev/45days,whatistherotationrateoftheneutronstar?(Treatthestarasasolidspherewith
.)
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HomeworkProblems
11.46
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11.49
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MP
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Lesson38
Lesson38
Lab6–ConservationofAngularMomentum
11.4,Lab6Worksheet
Examples
11.2
HomeworkProblems
11.45,12.69,12.87
 ThereisaLABthislesson.
LearningObjectives
[Obj62]
Applyconservationofangularmomentumtosolveproblemsinvolvingrotating
systemschangingrotationalinertiasandrotatingsystemsinvolvingtotallyinelastic
collisions.
Notes
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JournalQuestions
Score(5)
1. Brieflydescribethepurposeandgoalsofthislab.(Onetotwocompletesentences)
RefertoConceptualExample11.1inyourtextbookforthefollowingquestions.
2. Whentheboyjumpsontothemerry‐go‐round,
a)
b)
c)
d)
thetotalrotationalinertiaoftheplatformchanges.
thetotalangularmomentumoftheplatformchanges.
Both(a)and(b)arecorrect.
Neither(a)nor(b)iscorrect.
3. Whenthegirljumpsontothemerry‐go‐round,
a)
b)
c)
d)
thetotalrotationalinertiaoftheplatformchanges.
thetotalangularmomentumoftheplatformchanges.
Both(a)and(b)arecorrect.
Neither(a)nor(b)iscorrect.
4. Intheexample,thegirljumpsinthesamedirectionastheplatformisrotating.Suppose,
platformiscountertoitsrotation.Describehowyouwouldmathematicallyaccountforthis
changewhensolvingforthefinalangularspeedofthemerry‐go‐round.
5. Supposethegirldoesnotjumpdirectlyinthetangentialdirection,butatanangleθtothe
tangentialdirection.Describehowyouwouldmathematicallyaccountforthischangewhen
solvingforthefinalangularspeedofthemerry‐go‐round.
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LabNotes
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HomeworkProblems
11.45
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12.69
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12.87
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Lesson39
Lesson39
SimpleHarmonicMotion
13.1,13.2,13.3
Example
13.3
HomeworkProblems
13.22,13.67,13.43
 ThereisanoptionalEquationDictionaryentryinAppendixDforthislesson(1PFpt).
LearningObjectives
[Obj63]
[Obj64]
[Obj65]
Definesimpleharmonicmotionandexplainwhyitissoprevalentinthephysical
world.
Determinetheperiodand frequencyofasimpleharmonicoscillatorfromitsphysical
parameters,andcompletelyspecifyitsequationofmotion.
Determinethevelocityandaccelerationofasimpleharmonicoscillatorfromits
equationofmotion.
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WorkedExample
Score(3)
STATEMENTOFTHEPROBLEM
Anidealgrandfatherclockconsistsofasimplependulumwhichswingsbackandforthonceevery
second.Whatisa)theoscillationfrequency,b)theangularfrequencyandc)howfarfromtheend
oftherodshouldthemasssit?
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. The angular frequency of the pendulum is given by
is length is
0.248m.
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/ . Therefore the pendulum
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Lesson39
SELF‐EXPLANATIONPROMPTS
1.Inyourownwords,describethedifferencebetweentheoscillationfrequencyandtheangular
frequency.
2.Inyourownwords,explainwhytheperiodofthependulumisnotdependentonthemassofthe
pendulum.
3.Inyourownwords,describewhytheperiodofthependulumisinverselyproportionaltothe
lengthofthependulum.
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
successfullymetalltheothercriteria,thereisoneremainingtest:isthegravitationalpullofthe
planetwithin30%ofEarth’snormalgravity?Theprobecontainsasimplependulumwhichituses
todeterminethegravitationalconstantofthatplanet.Thecompactpendulumisonly5.0cmlong
andtakes0.33secondstomovefromtheleftmostpartofitsswingtothecenterofitsswing.Isthe
planetsuitableforcolonization?
gnew=1.14m/s2
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PreflightQuestions
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
comparisontothechild,willnecessarilyswingwith
a) amuchgreaterperiod
b) amuchgreaterfrequency
c) thesameperiod
d) thesameamplitude
4. CRITICALTHINKING:Anoscillationisaphysicalphenomenoncharacterizedbythefactthatthe
configurationofthephysicalsystemrepeatsitselfoverandoveragain.Simpleharmonic
oscillationsareaspecialcase.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.
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HomeworkProblems
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13.67
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Lesson40
Lesson40
EnergyinSimpleHarmonicMotion
Examples
HomeworkProblems
13.5
13.5
13.29,13.63,13.73
LearningObjectives
[Obj66]
Determinethepotentialandkineticenergiesofasimpleharmonicoscillatoratany
pointinitsmotion,anddescribethetimedependenceoftheseenergies.
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WorkedExample
Score(3)
regardingtheproblem.
STATEMENTOFTHEPROBLEM
Amotionlessmassisconnectedtoaspring(withaspringconstantof85N/m)
whichiscompressed30cmfromitsequilibriumposition.Themass,whichis
restingonafrictionlesssurface,isthenreleased.Atwhatpositionwillthe
kineticenergyofthesystembeequaltoexactlyhalfthepotentialenergyofthe
system?
STRATEGY
30cm
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
position gives ∆
∆
/
/
. Solving for
24.5cm.
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SELF‐EXPLANATIONPROMPTS
1.Whydon’tyouneedtoexplicitlycalculatethekineticenergyofthesystem?
3.Whydoesthepointwherethepotentialenergyisequalto2/3itsinitialvaluecorrespondtothe
pointwherethekineticenergyishalfthepotentialenergy?
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Pre‐ClassProblem
STATEMENTOFTHEPROBLEM
Amotionlessmassisconnectedtoaspringwhichisstretched45cmfromitsequilibriumposition.
Themass,whichisrestingonafrictionlesssurface,isthenreleased.Themaximumkineticenergy
ofthesystemis10.6J.Whatisthespringconstantofthespring?
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PreflightQuestions
2. For thesimpleharmonicmotionofamassonaspringwithoutfriction,itistruethat
a) theenergyisindependentoftheamplitude
b) theenergyisindependentoftheperiod
c) 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
B
C
D
a)
b)
c)
d)
GraphAisthepotentialenergy,graphCisthekineticenergy.
GraphCisthepotentialenergy,graphAisthekineticenergy.
GraphBisthepotentialenergy,graphDisthekineticenergy.
GraphDisthepotentialenergy,graphBisthekineticenergy.
4. CRITICALTHINKING:Foragivenharmonicoscillator,ifthespringconstantandthemassare
bothdoubledbuttheamplituderemainsthesame,explainwhathappenstothemechanical
energyoftheoscillator.
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HomeworkProblems
13.29
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Block4Review
Block4Review
LearningObjectives
[Obj49]
Explaintherelationbetweentherotationalmotionconceptsofangulardisplacement,
angularvelocity,andangularacceleration.
[Obj50]
Useequationsofmotionforconstantangularaccelerationtosolveproblemsinvolving
angulardisplacement,angularvelocity,andangularacceleration.
[Obj51]
Usecalculustosolveproblemsinvolvingmotionwithnon‐constantangular
acceleration.
[Obj52]
Explaintheconceptoftorqueandhowtorquescausechangeinrotationalmotion.
[Obj53]
Givenforcesactingonarigidobject,determinethenettorquevectorontheobject.
[Obj54]
Determinetherotationalinertiaforasystemofdiscreteparticles,rigidobjects,ora
combinationofboth.
[Obj55]
Compareandcontrasttheconceptsofmassandrotationalinertia.
[Obj56]
UseNewton’ssecondlawanditsrotationalanalogtosolveproblemsinvolving
translationalmotion,rotationalmotion,orboth.
[Obj57]
Solveproblemsinvolvingrotationalkineticenergyandexplainitsrelationtotorque
andwork.
[Obj58]
Explaintherelationbetweenlinearandangularspeedinrollingmotion.
[Obj59]
Useconservationofenergytosolveproblemsinvolvingrotatingorrollingmotion.
[Obj60]
Determinethedirectionsoftheangulardisplacement,angularvelocityandangular
accelerationvectorsforarotatingobject.
[Obj61]
Determinetheangularmomentumvectorfordiscreteparticlesandrotatingrigid
objects.
[Obj62]
Applyconservationofangularmomentumtosolveproblemsinvolvingrotating
systemschangingrotationalinertiasandrotatingsystemsinvolvingtotallyinelastic
collisions.
[Obj63]
Definesimpleharmonicmotionandexplainwhyitissoprevalentinthephysical
world.
[Obj64]
Determinetheperiodandfrequencyofasimpleharmonicoscillatorfromitsphysical
parameters,andcompletelyspecifyitsequationofmotion.
[Obj65]
Determinethevelocityandaccelerationofasimpleharmonicoscillatorfromits
equationofmotion.
[Obj66]
Determinethepotentialandkineticenergiesofasimpleharmonicoscillatoratany
pointinitsmotion,anddescribethetimedependenceoftheseenergies.
Notes
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regardingtheproblem.
STATEMENTOFTHEPROBLEM
linearaccelerationattherimofthewheelattheinstantwhenitsangularvelocityis12rpm?
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
For the equation
/ . In terms of angular velocity
to be valid, the angular velocity has to be in radians.
To convert 120 rpm into radians we write
12
The centripetal acceleration is then _______x ________ = 2.5
m/s2.
1.3
.
/ .
The magnitude of the total acceleration is 5.1 m/s2.
The direction of the total acceleration is 60.9 degrees.
SELF‐EXPLANATIONPROMPTS
1.Explaininyourownwordswhytheangularspeedofarigidrotatingobjectisthesameforall
2.Justifytherelationac=ω2r.
3.Isthecentripetalaccelerationinanywayrelatedtotheangularacceleration?
OptionalPracticeProblems:10.13,10.18,10.41
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Block4Review
regardingtheproblem.
STATEMENTOFTHEPROBLEM
Twoweightsofmasses2mandmareattachedtoeitherendofathinrodoflengthL.Calculatethe
oftherod.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
SELF‐EXPLANATIONPROMPTS
1.Inyourownwords,explainrotationalinertiaofanobject.
2.Inthisproblem,whywasthedistancefromtherotationaxisrequaltoL/2forbothweights?
3.Wouldtherotationalinertiaoftheobjectincrease,decrease,orstaythesameiftherotationaxis
OptionalPracticeProblems:10.22,10.24,10.29
DocumentationStatement:
309
Block4Review
Physics110HJournal‐2013‐2014
regardingtheproblem.
STATEMENTOFTHEPROBLEM
A50‐kgbucketishangingfromarope,whichiswoundarounda20‐kgsoliddisc.Thediameterof
thediskis50cm.Themassoftheropeisnegligiblecomparedtotheothermassesintheproblem.
Whatistheaccelerationofthefallingbucket?
STRATEGY(Fillintheblanks.)
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.
CALCULATION(Fillintheblanks.)
Newton’sLawforthefallingbucket:
_____________
Newton’sLawfortherotatingdisc:
_______________
Eliminating
½
from the two equations and solving for the acceleration we get
8.2m/s
1
2
SELF‐EXPLANATIONPROMPTS
1.Supplythemissingalgebra.Writedownthetwoequationsofmotion,eliminatethetensionand
solvefortheacceleration.
OptionalPracticeProblems:10.32,10.59
DocumentationStatement:
310
Physics110HJournal‐2013‐2014
Block4Review
regardingtheproblem.
STATEMENTOFTHEPROBLEM
decideitwillbeeasiertorollthebarrelsdowntherampatthebackoftheaircraftandofftheplane.
Ifabarrelhasaspeedof0.5m/swhenitreachestheramp,whatisitsspeedafterithasrolled
downtherampwithoutslippingandofftheplane?Theverticalheightoftherampis1.5meters.
STRATEGY(Fillintheblanks.)
We will use the principle of conservation of _________________ to solve for the speed of the
barrel at the end of the ramp.
CALCULATION(Fillintheblanks.)
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
because the barrel is not
Now, we replace ω with
____________ and put rotational inertia in terms of mass
(Approximating the barrel as a solid cylinder, its rotational inertia is
1
2
1
2
1 2 )
Solving for the final speed of the barrel at the end of the ramp gives:
√_______________
4.5 / SELF‐EXPLANATIONPROMPTS
hollow,woulditreachthebottomoftherampearlierorlaterthanifitwassolid?Assumethesame
initialspeed.
OptionalPracticeProblems:10.38,10.36,10.61
DocumentationStatement:
311
Block4Review
Physics110HJournal‐2013‐2014
regardingtheproblem.
STATEMENTOFTHEPROBLEM
Whatisthemagnitudeanddirectionoftheangularmomentumofa10‐kgsoliddisc,60cmin
diameter,rotatingcounter‐clockwiseat120rpmarounditscentralaxis?
STRATEGY(Fillintheblanks.)
We will use the relation,
___
, and the ____________
CALCULATION(Fillintheblanks.)
First, we need to convert rotational speed from rpm to
______
For a solid disc rotating around it central axis
_____ _______ 0.45kgm Now substitute to get:
5.66kg
m
s
The direction, from the right-hand rule, is __________________________.
SELF‐EXPLANATIONPROMPTS
OptionalPracticeProblems:11.15,11.19,11.22
DocumentationStatement:
312
Physics110HJournal‐2013‐2014
Block4Review
regardingtheproblem.
STATEMENTOFTHEPROBLEM
Twodisksarerotatingatdifferentspeedsalongthesameaxisasshown.The
finalangularspeedofthecombineddisks?
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).
CALCULATION(Fillintheblanks.)
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:
_________________;
_________________ SELF‐EXPLANATIONPROMPTS
1.Showthatkineticenergyisnotconversedintheexample.
2.Examplewhyangularmomentumisconversedbutkineticenergyisnot.
OptionalPracticeProblems:11.25,11.28
DocumentationStatement:
313
Block4Review
Physics110HJournal‐2013‐2014
regardingtheproblem.
STATEMENTOFTHEPROBLEM
Ayounggirldecidestobuildasimplependulumtoknockoveratoycar.Todoso,shetiesoneend
ofa40‐cmstringtotherailingandarubberballtotheotherendofthestring.Thesetupis
designedsothattheballwillhitthetoycarwhenthependulumisatthelowestpointofitsarc.The
girlpositionsthependulumsothatthestringistightandtheballis10cmofftheground.Howlong
doesittakefortheballstrikethetoycar?
STRATEGY(Fillintheblanks.)
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.
CALCULATION(Fillintheblanks.)
______
The time it takes the pendulum to strike the car is
0.32s.
SELF‐EXPLANATIONPROMPTS
2.Whydoesn’ttheinitialheightofthependulumaffecttheperiodoftheswing?
OptionalPracticeProblems:13.31,13.22,13.25
DocumentationStatement:
314
Physics110HJournal‐2013‐2014
Block4Review
regardingtheproblem.
STATEMENTOFTHEPROBLEM
A2.0‐kgmassisattachedtoaverticallyorientedspringwhichhasaspringconstantof25N/m.The
springiscompressed31cmrelativetoitsequilibriumpoint.Whatisthespeedofthemassatthe
equilibriumpointofthespring?
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.
CALCULATION(Fillintheblanks.)
For convenience, we choose the equilibrium point of
the spring to be the reference point.
______=6.99J
The gravitational potential energy is
The spring potential energy is
At the equilibrium point,
______=
Solving for velocity we find that
_______=2.40 J
.
3.1 m/s.
SELF‐EXPLANATIONPROMPTS
1.Whyistheequilibriumpointaconvenientchoiceoforigin?
2.Explain,inyourownwords,howyoufoundthekineticenergyattheequilibriumpoint.
3.Whycanthespringenergybecombinedwiththegravitationalenergy?
OptionalPracticeProblems:13.41,13.43,13.77
DocumentationStatement:
315
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Physics110HJournal‐2013‐2014
Notes
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Notes
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317
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Notes
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Notes
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Notes
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320
Physics110HJournal‐2013‐2014
AppendixA:LabReportTemplate
AppendixA:LabReportTemplate
Purpose
Aformallabreportisessentialtothescientificprocess.Itisthemostcommonwaythattheresults
ofascientificstudyarecommunicatedtothescientificcommunity.Youmayassumeyouraudience
isscientificallyliteratebuthasnotperformedtheexperimentinquestion.
Format
Usethefollowingguidancetoformatyourreport:












Setthepagebordersto1”
Spacelineswithinthesameparagraphat1.0or1.15
Refertoyourexperimentand/orcalculationsinthepasttense
Usethethird‐person(i.e.,avoidusing“I”or“we”)
Usescientificnotationwhereappropriateandincludeallunits(e.g.,1.1 10 m)
Citeanyoutsidesources(otherthanyourtextbook)usingMLAformat
underneathit.Youmayneatlyhand‐drawfigures.
Graphsorplotsmaybeusedtosummarizedataandshowanalysis.Theymustincludea
titleandlabeledaxesandmustnotbedonebyhand.Ensurethegraphorplotislarge
cross‐referencethegraphorplotandincludeitasanattachmentattheendofyourreport.
Sections
Usethefollowingformattocreateyourlabreport.Rememberthereisabalancebetweentoolittle
informationandtoomuchinformation.Youwantyourreporttoincludewhatisrelevantwithout
writtendocument,whichmeansheorshewillneverlearnofyourresultsorfindings.
TitlePage
Useaseparatetitlepage.Ensurethetitleofyourreportiscenterednearthetopofthefirstpage.
Listallcontributorstothelabreportunderneaththetitle.Alsoincludethecoursenameand
numberandthedate.Atthebottomofthetitlepage,neatlyincludeyourdocumentationstatement
foranyoutsidehelpyouobtained(butnotoutsidereferences).
xix
AppendixA:LabReportTemplate
Physics110HJournal‐2013‐2014
Introduction
importanceofyourwork.Youshouldalsobrieflysummarizeanypertinentmaterial,including
relevantequationsorconcepts.
ExperimentalMethods
Inthissection,youmustsuccinctlydescribethemethodsyouusedtoobtainyourdata.Ingeneral,
keepingthissectioncompletebutconcise.Graphicsandfiguresshouldbeusedsparinglyinthis
section.
ResultsandDiscussion
Thisisanextremelyimportantsectionofyourreport.Hereyoushouldcommunicatewhatyou
foundanddrawpertinentconclusionsbasedoninterpretationofyourdata.Graphics,figuresand
Excelplotsmaybeusedtoeffectivelycommunicateyourresults.FollowtheguidanceintheFormat
section.Makesureyourresultsarecorroboratedorjustifiedbythedatayouobtained.Keepin
mindthatnotalldataisnumerical.Forthelabsinthiscourse,youmaybeabletomakesome
powerfulconclusionsbasedonqualitativeobservations.
Conclusion
Hereyouwillsummarizeyourresults.Ensureyoudonotintroduceanynewinformationinthis
section.
References
Ifyoucitedanyreferences,includethemhereusingMLAformat.
Appendixes(ifneeded)
Usethissectiontoplaceanylargeorcomplicatedgraphsordataplots.
xx
Physics110HJournal‐2013‐2014
AppendixB:SignificantFigures,UncertaintyandErrorPropagation
AppendixB:SignificantFigures,Uncertaintyand
ErrorPropagation
References:
[1] P.R.Bevington,andD.K.Robinson,DataReductionandErrorAnalysis,3rded.,McGraw‐Hill,NewYork,2003.
Numericcalculationsandexperimentalmeasurementsareonlyasaccurate(orreliable)astheleast
precisemeasurement.Everyphysicalmeasurementhasuncertaintyandlaboratorymeasurements
donotyieldexactresults.Errorsanduncertaintiesinphysicalexperimentsmustbereduced
throughexperimentaltechniquesandrepeatedmeasurements–remaininguncertaintymustbe
estimatedandreportedtoestablishthevalidityoftheresult.
Thetermerrorisdefinedasthedifferencebetweenobserved(orcalculated)valueandthe“true”
value.Inlaboratorymeasurementswerarelyknowthetruevalue,thereforewemustestablish
systematicmeansofdeterminingthevalidityofourexperimentalresults.Errorsthatoriginatefrom
mistakesinmeasurementareknownasillegitimate(gross)errors,andarecorrectedthrough
attentionandcarefulrepeatedmeasurements.Inourexperimentsweareconcernedwith
uncertaintiesintroducedbyrandomfluctuationsinmeasurementsandsystematicerrorsthatlimit
theprecisionandaccuracyofourresults.Randomerrorsarefluctuationsthatoccurinobservations
eachtimeameasurementisrepeated.Randomerrormaybereduced
throughlaboratorytechniqueorrepeatedobservations.Systemic
errorsaredifficulttodetectandmaymakeallourresultsvarywith
reproduciblediscrepancy.Systemicerrormayresultfrompoorly
calibratedequipmentorbiasbytheobserver.
(a)(b)
Accuracyisameasureofhowclosetheresultistothetruevalue.
Precisionisameasureofhowwelltheresulthasbeendetermined
(withoutregardtoagreementwithtruevalue).Precisionisalsoa
measureofanexperiment’sreproducibility.FigureB1illustratesthe
differencebetweenaccuracyandprecision.
xxi
FIG.B1.Accuracyand
precisiondemonstrated
throughtargetpractice.
Target(a)isaccurate,butnot
precise,while(b)isprecise
butnotparticularlyaccurate.
AppendixB:SignificantFigures,UncertaintyandErrorPropagation
Physics110HJournal‐2013‐2014
SignificantFigures
Thenumberofdigitsinreportinganexperimentalresultimpliestheprecisionofameasurement
anduncertaintyshouldbereportedspecificallywitheachnumericresult.
Rulesfornumberofsignificantfigures:
1.
2.
3.
4.
Leftmostnonzerodigitisthemostsignificantdigit.
Ifthereisnodecimalpoint,therightmostnonzerodigitistheleastsignificantdigit.
Thenumberofdigitsbetweenthemostandleastsignificantdigitcountareknownastheas
significantfigures.
Rulesforsignificantfiguresincalculatingnumbers:
1. Multiplication/division.Thenumericresultcannothavemoresignificantfiguresthananyof
theoriginalnumbers.
decimalpointthananyoftheoriginalnumbers.
3. Roundingresults.Insignificantdigitsaredroppedfromtheresultandthelastdigitis
roundedforbestaccuracy.
Uncertainty
Everyphysicalmeasurementhasuncertaintyduetotheaccuracyorprecisionoflaboratory
equipmentandtherandomdistributionofourdata.Sincewedonotnormallyknowtheactualerror
(discrepancyfromthetruevalue)inexperimentalresults,weseektodevelopamethodof
thebestvalue
anduncertaintyestimate ,andreporttheresultas
.
Uncertaintyinexperimentalmeasurementscanbeestimatedinanumberofways,including
repeatedmeasurements.
xxii
(B1)
Physics110HJournal‐2013‐2014
AppendixB:SignificantFigures,UncertaintyandErrorPropagation
UncertaintyinAnalogMeasurements
cylinders,etc.)isestimatedasonehalfthesmallestdivision
markedonthedevice.
Example:Alengthmeasurementistakenusingarulermarkedin
incrementsof1mmasshowninFigureB2.Uncertaintyisonehalfthe
smallestincrement,or0.5mm.Thelengthisreportedas26 0.5mm
accordingtoEqn.B1.
FIG.B2.Analogmeasurement
usingaruler.
UncertaintyinDigitalMeasurements
Manymodernlaboratorydevicesaredigital(scales,timers,
multimeters,etc.).Systematicerrorisreducedifthedeviceis
properlycalibrated.Estimateduncertaintyistheleastsignificant
mustbetakenandothermethodsofestimatinguncertaintymustbe
used.
FIG.B3.Digitalmeasurement
usingphotogatetimer.
one‐thousandthofasecondasshowninFigureB3.Uncertaintyistheleast
significantdigit,or0.001s.Thetimemeasurementisrecordedas1.673 0.001saccordingtoEqn.B1.
UncertaintyinRepeatedMeasurements
Repeatedmeasurementshelpusextractthebestvalueofourexperimentalresultsanddetermine
theestimatederrorwithconfidence.Aswetakemoremeasurements,weexpectapatternto
emergewithdatapointsdistributedaroundthecorrectvalue(assumingwecorrectforsystematic
errors).
Supposeduringanexperiment,wetakeasampleof measurementsofaquantity .
Thearithmeticmean ̅ oftheexperimentaldistributionisgivenas
̅
∑
.
(B2)
Theexpressionforthestandarddeviation ofthesamplepopulationisgivenby
∑
̅ (B3)
whichrepresentsthebestestimateforthedeviationsquaredoftheparentdistribution(asifwe
tookandinfinitenumberofmeasurements)basedonthesmallersampledistribution.Thestandard
deviation representsaquantitativemeasureoftheuncertaintyinanysinglemeasurement.Ifwe
xxiii
13.6%
13.6%
weretotakeanothersamplemeasurementthereisa68.2%
chanceitwillbewithin ̅
,a95.4%chanceitwillbewithin
̅ 2 ,anda99.7%chanceitwillbewithin ̅ 3 ,asshown
inFigureB4.
34.1%
Physics110HJournal‐2013‐2014
34.1%
AppendixB:SignificantFigures,UncertaintyandErrorPropagation
Givenrepeatedtrialsandcalculationsofthemean,itis
2
̅
2 possibletodeterminevariationinthevalueofthemean.
FIG.B4.Normal(Gaussian)distribution
Whendeterminingexperimentalresultswithalargesample
withstandarddeviation .
size,weseekaquantitativemeasureofthestandarddeviation
ofthemean,orthestandarddeviationinthesamplemeanrelativetothetrue(mean)value.
∑
√
̅ (B4)
Experimentalresultsbasedonthissampledistributionarereportedas
̅
.
(B5)
Forsmallsamplesets(threeorfewertrials),wemayestimateuncertaintyusingtheexpression
.
(B6)
ErrorPropagation
Experimentalquantitiesderivedfrommeasuredvalueswithuncertaintywillinturnhave
uncertainty.Estimateduncertaintyiscalculatedbasedonthemathematicaloperationsusedinthe
derivation.Supposewemeasurevalues( , , , , , )withuncertainty( , , , , , ).We
seektheuncertaintyinacalculatedvalue .
If
then
.
MultiplicationorDivisionwithUncertainty
If
xxiv
(B7)
Physics110HJournal‐2013‐2014
AppendixB:SignificantFigures,UncertaintyandErrorPropagation
then
| |
.
(B8)
MultiplicationbyaKnownValuewithUncertainty
IfQiscalculatedbymultiplyingbyaknownvalue (e.g.
2 or
)byaquantity with
uncertainty,
then
| |
(B9)
orequivalently
| |
.
(B10)
| |
UncertaintywithExponents
If isanexactnumberand
then
| |
| |
.
| |
(B11)
ReportingExperimentalValues
Itshouldbeemphasizedthatuncertaintyestimatesareonlyestimatesandvaluesshouldbe
presentedwithappropriateprecision.
Rulesforreportingexperimentalvalues:
1. Theleastsignificantfigureinanyreportedvalueshouldbethesameorderofmagnitude
(samedecimalposition)astheuncertainty.
2. Estimateduncertaintyisnormallyroundedtoonesignificantfigure.
xxv
AppendixC:MathematicsReference
Physics110HJournal‐2013‐2014
AppendixC:MathematicsReference
√
2
4
CoordinateSystems
Conventiondictatesright‐handedcoordinatesystems.Alternatecoordinatesystemsmaybeused–
besuretoclearlyindicatechosencoordinateaxes.Twocommoncoordinatesystemsusedin
physicsareshownbelow.
CartesianCoordinateSystem
, ,
̂
̂
̂
̂
SphericalCoordinateSystem
, ,
0
0
0
sin cos
̂
sin
sin
2 xxvi
̂
sin sin
sin
̂
̂
cos
Physics110HJournal‐2013‐2014
AppendixC:MathematicsReference
Trigonometry
Opposite
sin
Opp
Hyp
csc
1
sin
Hyp
Opp
cos
Hyp
sec
1
cos
Hyp
sin
cos
tan
Opp
1
tan
cot
cos
sin
Opp
sin
LawofSines
sin
LawofCosines
sin
2
cos TrigonometricIdentities
sin
sin
sin 2
sin
sin cos
cos
1
1
2 sin cos sin cos
cos tan
cos 2
sec
cos
cos sin sin
cos
1
1
cot
2 sin
cos cos
sin
sin
2 sin
cos
cos
cos
2 cos
cos
cos
cos
2 sin
sin
xxvii
∓
csc
2 cos
1
∓ sin sin AppendixC:MathematicsReference
Physics110HJournal‐2013‐2014
Vectors
Givenvectors
̂
̂
and
̂
̂
,
cos ∙
∙
̂
sin ̂
ExponentialsandLogarithms
ln
≅ 2.71828 …
ln
ln
ln ln
ln
ln ln ln
ln
ln 10
log
ln 1 ≡ 0
1
DerivativesandIntegrals
1
,
sin
where isaconstant.
cos
cos
sin
1
1
cos
cos
ln
/
/
xxviii
1.
sin
1
1
sin
ln
,
1
√
√
Physics110HJournal‐2013‐2014
AppendixC:MathematicsReference
TaylorSeriesExpansionsandApproximations
ATaylorseriesexpansionofarealfunction
isgivenby
⋯
2!
.
!
SeriesExpansionsofCommonFunctions(for| |
1)
For| | ≪ 1
1
sin
cos
1
2!
3!
5!
7!
2!
4!
6!
ln 1
1
2
3
2!
⋯
cos
⋯
⋯
sin
⋯
4
1
1
1
⋯
3!
xxix
1
2
ln 1
1
1
AppendixD:EquationDictionary
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
Oncertainlessons,youwillhavetheoptiontocompleteaworksheetonaparticularequationfor
pre‐flightpoints.Theequationdictionaryisdesigned:(1)toallowyoutobecomemorefamiliar
withanequation,and(2)toenableyoutocreateahighlyorganizedandeasilyaccessiblestudy
guideforexampreparation.Themoretimeyouspendcreatingmeaningfulentriestoyourequation
dictionary,themorepreparedyouwillbeforexamsinthecourse.
Inthewhiteboxontheupperleftcorner,youwillfindtheequationreferencenumber.Intheblack
boxintheupperrightcorner,youwillfindthelessonnumberwheretheequationisfirst
introduced.Anexampleofahigh‐qualityequationdictionaryentryisshownbelow.
xxx
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
#1
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxi
AppendixD:EquationDictionary
Physics110HJournal‐2013‐2014
1
2
#2
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxii
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
#3
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxiii
AppendixD:EquationDictionary
Physics110HJournal‐2013‐2014
#4
∙
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxiv
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
#5
∆
/
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxv
AppendixD:EquationDictionary
Physics110HJournal‐2013‐2014
#6
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxvi
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
1
2
#7
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxvii
AppendixD:EquationDictionary
Physics110HJournal‐2013‐2014
#8
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxviii
Physics110HJournal‐2013‐2014
AppendixD:EquationDictionary
cos
#9
PhysicsConcept
Variables(includeallunits)
DescriptionandNotes
Diagram
xxxix
AppendixE:RotationalInertiasandAstrophysicalData
Physics110HJournal‐2013‐2014
AppendixE:RotationalInertiasandAstrophysicalData
Table10.2RotationalInertias
AstrophysicalData
Earth Mass
OrbitalPeriod
SurfaceGravity
5.97 10 kg
6.37 10 m
3.16 10 9.81 m s Moon Mass
OrbitalPeriod
SurfaceGravity
7.35 10 kg
1.74 10 m
2.36 10 1.62 m s Sun Mass
OrbitalPeriod
SurfaceGravity
1.99 10 kg
6.96 10 m
6 10 274 m s xl
Physics110HJournal‐2013‐2014
AppendixF:UnitsandConversions
AppendixF:UnitsandConversions
Ref:http://wwwppd.nrl.navy.mil/nrlformulary/NRL_FORMULARY_07.pdf
PhysicalQuantity
Dimension
Length
Mass
Time
ElectricCurrent
Temperature
AmountOfSubstance
SIUnits
BASEUNIT
BASEUNIT
ElectricPotential
∙
3
10 3
∙
/
1
3
1
3
∙
/
/
/
∙
∙
,
Inductance
1
,
MagneticField
MagneticFlux Φ Momentum,Impulse
Permeability
Pressure,Stress Tension SpecificHeatCapacity
∙
∙
∙
∙
∙
/
‐‐‐‐‐
36
10 ‐‐‐‐‐
10 10
Work Energy ∙
10 ∙
/
10 10 ∙
∙
∙
xli
∙
∙ ∙
10 ∙
10 ∙
∙
1
4
∙
10
10 ∙
10 1
∙
∙
∙
VectorPotential
∙
10
10 ∙
/
1
∙
/
/
∙
∙
∙
ThermalConductivity
∙
∙
/
10 1
1
9
1
9
∙
∙
∙
∙
10 10 10 1
Permittivity
10 10 ∙
1
Impedance,Resistance
Viscosity
10 Frequency
Velocity
3
12
,ϕ 10 10 /
9
/
BASEUNIT
/
ElectricCharge
Torque
/
/
GaussianUnits
1
∙
∙
/
Displacement
Power
Conversion
Factor
CurrentDensity
Force
SISymbol
1
ElectricField
SIUnits
10 1
Entropy
InTermsof
OtherSIUnits
AngularVelocity
10 GaussianUnits
1
3
PhysicalQuantity
Density
10 BASEUNIT
Dimensions
SI
Gaussian
Current
10 Capacitance
BASEUNIT
Acceleration
Conversion
Factor
SISymbol
10
10 ∙
Physics110HJournal‐2013‐2014
Ref:http://physics.nist.gov/cuu/Constants/index.html
PhysicalConstant
Symbol
Value Uncertainty
AtomicMassUnit
1.660 538 782(83) 10-27 kg
AtomicUnitOfCharge
1.602 176 487(40)
BohrMagneton
C
927.400 915(23) 10-26 J T-1
5.788 381 7555(79) 10-5 eV T-1
0.529 177 208 59(36) 10-10 m
0.529 177 208 59(36) Å
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
0.510 998 910(13) MeV
5.485 799 0943(23) 10-4 amu
8.187 104 38(41) 10-14 J
1.073 544 188
8.065 544 65(20)
ComptonWavelength
2.426 310 2175(33) 10-12 m
ElectronMass
9.109 382 15(45) 10-31 kg
1.602 176 487(40)
ElectronVolt
Fine‐StructureConstant
Impedance Vacuum J
7.297 352 537 6(50) 10-3
1.112 650 056 10-17 kg
1.380 6504(24)
InverseCentimeter
MolarGasConstant
10-23
J
6.241 509 65(16) 1018 eV
5.034 117 47(25) 1022 cm-1
8.617 343(15)
0.695 035 6(12) cm-1
10-5
eV
2.417 989 454 1014 Hz
6.022 141 79(30) 1026 amu
5.609 589 12(14) 1035 eV
8.987 551 787 1016 J
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
1.008 664 915 97(43) amu
1.505 349 505(75) 10-10 J
8.314 472(15) J mol-1 K-1
(@ 273.15 K, 101.325 kPa)
1.674 927 211(84) 10-27 kg
939.565 346(23) MeV
⁄2
5.050 783 24(13)
3.152 451 2326(45)
12.566 370 614 10-7 N A-2
Permeability MagneticConstant 10-27
J
T-1
Permittivity ElectricConstant 8.854 187 817 10-12 F m-1
PlanckConstant
6.626 068 96(33) 10-34 J s
ProtonMass
RydbergConstant

Stefan‐BoltzmannConstant
NUMBERSandAPPROXIMATIONS
10-8
eV
1.054 571 628(53)
Js
6.582 118 99(16) 10-16 eV s
1.672 621 637(83)
10-27
kg
938.272 013(23) MeV
10 973 731.568 527(73)
101 325 Pa
(@ 273.15 K)
2.542 623 616(64)
10-4
cm-1
T-1
4.135 667 33(10) 10-15 eV s
10-34
m-1
T-1
4 π 10-7 H m-1
109 737.315 8
299 792 458 m s-1
StdAtmosphere
StdAccelerationOfGravity
cm-1
 1 / 137.035 999 679(94)
22.413 996(39) 10-3 m3 mol-1
NeutronMass
SpeedOfLight Vacuum amu
103
376.730 313 461 Ω
Kilogram
PlanckConstant/2π
10-9
∙ 13.996 246 04(35) 109 Hz T-1
6.674 28(67) 10-11 m3 s-2 kg-1
Kelvin
NuclearMagneton
Joule
MolarVolumeOfIdealGas
10-19
∙ 0.466 864 515(12) cm-1 T-1
96 485.339 9(24) C mol-1
⁄
GravitationalConstant
1.492 417 830(74) 10-10 J
⁄2
⁄
7.513 006 671(11) 1012 cm-1
6.022 141 79(30) 1023 mol-1
BoltzmannConstant
931.494 028(23) MeV
10-19
cm-1
1.007 276 466 77(10) amu
1.503 277 359(75) 10-10 J
(
(
) 13.605 691 93(34) eV
) 2.179 871 97(11) 10-18 J
/
29.92 inHg
14.696 psi
760 Torr
9.806 65 m s-2
5.670 400(40) 10-8 W m-2 K-4
π 3.141592653589793…
e
2.718281828459045…
xlii
≅1973.27 eVÅ
≅0.511MeV
Physics110HJournal‐2013‐2014
Physics215ConstantsandEquationsSheet
ElectricForce(Coulomb)Const
ElectricConst(Permittivity)
MagneticConst(permeability)
ElementaryCharge
8.99
10 N m C
8.85
10
4
10
1.60
10
4
C N
m
Conversions:
NA 1G
C 1eV
ElectronMass
9.11
10
kg ProtonMass
1.67
10
kg PlanckConst
6.63
10
J s 10
T 1.60
10
MaxwellEquations
Gaussfor ∙
∙
Φ
Gaussfor ∙
Ampère
∙
0 Φ
⋅
∆
∆
sin
sin
1.22
Φ
Ɛ
⋅
̂
∆
∙
Φ
̂
Φ
∙
xliii
∆ ∆
2
J Physics110HJournal‐2013‐2014
Physics110HCourseSyllabus
KEY: –Doubleperiod;
CE–ConceptualExample;
EPQUIZ–Exam‐PrepQuiz; CTE‐CriticalThinkingExercise
Lab– LabExercise;
Learning
Objectives
LessonTitle
1 Introduction
PHYSICSTESTING
2 Displacement,Velocity,andAcceleration
3 Lab1–AccelerationDuetoGravity
4 Two–Dimensional&ProjectileMotion
5 ProjectileMotion
EPQUIZ
6 Lab2–ProjectileMotion
7 AccelerationinCircularMotion
Examples
Homework
Problems
2.1,2.2
1.16,1.24,2.51
4–9
Chapter1.
2.1,2.2
2.3,2.4
2.1–2.3
2.20,2.35,2.79
10
2.5,Lab1
2.6
2.38,2.42,2.78
11–14
3.1–3.5
3.3
3.34,3.53,3.54
14,15
3.5
3.4
3.55,3.70,MP
15
Lab2
3.4
MP,3.58,MP
16,17
3.6
3.7,3.8
MP,MP,3.80
4.15,4.26,4.60
1–6
18–20
4.1–4.4
4.1,4.2
10 UsingNewton’sLaws
9 ForcesandNewton’sLawsofMotion
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
13 Lab3–Newton’sLaws
14 Newton’sLawsinCircularMotion
EPQUIZ
23
5.2
5.4
5.19,5.21,5.71
23
5.2,Lab3
5.4
MP,MP,MP
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
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
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
22 Lab4–ConservationofEnergy
25 CTE:OrbitalEnergies
39,40
8.4
8.5
MP,8.61,8.67
26 CenterofMass
EPQUIZ
42
9.1
9.16,9.37,9.89
27 ConservationofLinearMomentum&Collisions
43–48
9.1–9.5
46–48
9.5,9.6,Lab5
9.1,9.2,9.3
CE9.1,9.4,9.5,
9.7
None
9.28,9.44,9.61
47,48
9.3,9.4
9.10
MP,9.68,9.78
49–51
10.1,10.2
10.1,10.2,10.3
10.19,10.32,10.45
52–55
10.2,10.3
10.4,10.5
10.30,10.28,10.52
55–56
10.3
10.56,10.57,MP
34 RotationalEnergyandRollingMotion
57–59
10.4,10.5
35 RotationalVectorsandAngularMomentum
60,61
11.1–11.3
10.8,10.9
10.10,10.11,
10.12,CE10.1
11.1
28 Lab5–Collisions
29 CollisionsandConservationofEnergy
9.38,MP,MP
31 RotationalMotion
32 Torque&RotationalInertia PHYSICSTESTING
33 RotationalAnalogofNewton’sSecondLaw
36 ConservationofAngularMomentum
10.60,10.62,10.68
11.16,11.17,11.21
62
11.4
CE11.1
11.26,11.27,11.43
62
Chpt10&11
None
11.46,11.49,MP
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
37 CTE:EnergyandAngularMomentum
EPQUIZ
38 Lab6–ConservationofAngularMomentum
FINALEXAM
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