Mastering Physics CH 12 HW College Physics I LCCC
Mastering Physics CH 12 HW College Physics I LCCC
Mastering Physics CH 12 HW College Physics I LCCC
Ch12HW
Ch12HW
Due:9:30amonMonday,May2,2016
Youwillreceivenocreditforitemsyoucompleteaftertheassignmentisdue.GradingPolicy
AngularMotionwithConstantAcceleration
LearningGoal:
Tounderstandthemeaningofthevariablesthatappearintheequationsforrotationalkinematicswithconstantangularacceleration.
Rotationalmotionwithaconstantnonzeroaccelerationisnotuncommonintheworldaroundus.Forinstance,manymachineshavespinningparts.When
themachineisturnedonoroff,thespinningpartstendtochangetherateoftheirrotationwithvirtuallyconstantangularacceleration.Manyintroductory
problemsinrotationalkinematicsinvolvemotionofaparticlewithconstant,nonzeroangularacceleration.Thekinematicequationsforsuchmotioncanbe
writtenas
(t) = 0 + 0 t +
and
(t) = 0 + t
1
2
Here,thesymbolsaredefinedasfollows:
istheangularpositionoftheparticleattimet .
istheinitialangularpositionoftheparticle.
(t) istheangularvelocityoftheparticleattime t .
0 istheinitialangularvelocityoftheparticle.
istheangularaccelerationoftheparticle.
t isthetimethathaselapsedsincetheparticlewaslocatedatitsinitialposition.
(t)
0
Inansweringthefollowingquestions,assumethattheangularaccelerationisconstantandnonzero:
PartA
Trueorfalse:Thequantityrepresentedby isafunctionoftime(i.e.,isnotconstant).
ANSWER:
true
false
Correct
PartB
Trueorfalse:Thequantityrepresentedby 0 isafunctionoftime(i.e.,isnotconstant).
ANSWER:
true
false
Correct
Keepinmindthat 0 representsaninitialvalue,notavariable.Itreferstotheangularpositionofanobjectatsomeinitialmoment.
PartC
Trueorfalse:Thequantityrepresentedby0 isafunctionoftime(i.e.,isnotconstant).
ANSWER:
true
false
Typesettingmath:56%
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Correct
PartD
Trueorfalse:Thequantityrepresentedbyisafunctionoftime(i.e.,isnotconstant).
ANSWER:
true
false
Correct
Theangularvelocityalwaysvarieswithtimewhentheangularaccelerationisnonzero.
PartE
Whichofthefollowingequationsisnotanexplicitfunctionoftimet ?Keepinmindthatanequationthatisanexplicitfunctionoftimeinvolvest asa
variable.
ANSWER:
= 0 + 0 t +
= 0 + t
2
0
1
2
+ 2( 0 )
Correct
Anequationthatisnotanexplicitfunctionoftimeisusefulwhenyoudonotknowordonotneedthetime.
PartF
Intheequation
= 0 + t
,whatdoesthetimevariablet represent?
Choosetheanswerthatisalwaystrue.Severalofthestatementsmaybetrueinaparticularproblem,butonlyoneisalwaystrue.
ANSWER:
themomentintimeatwhichtheangularvelocityequals0
themomentintimeatwhichtheangularvelocityequals
thetimeelapsedfromwhentheangularvelocityequals0 untiltheangularvelocityequals
Correct
ConsidertwoparticlesAandB.TheangularpositionofparticleA,withconstantangularacceleration,dependsontimeaccordingto
2
A (t) = 0 + 0 t +
t .Attime t = t 1 ,particleB,whichalsoundergoesconstantangularacceleration,hastwicetheangularacceleration,half
1
2
theangularvelocity,andthesameangularpositionthatparticleAhadattimet
= 0
PartG
WhichofthefollowingequationsdescribestheangularpositionofparticleB?
Hint1.Howtoapproachtheproblem
ThegeneralequationfortherotationofBcanbewrittenas
B (t) = B (t = t 1 ) + B (t = t 1 ) (t t 1 ) +
(t = t
(t = t
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1
2
B (t t 1 )
t
,
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where B (t
= t1)
t B = (t t 1 )
Battimet 1 .
Expressthequantitiesontherighthandsideofthisequationfor B (t) intermsofparticleA'svariablesandconstantsofmotion.
ANSWER:
B (t) = 0 + 2 0 t +
B (t) = 0 +
1
2
1
4
0 t + t
B (t) = 0 + 2 0 (t t 1 ) +
B (t) = 0 +
1
2
1
2
(t t 1 )
0 (t t 1 ) + (t t 1 )
B (t) = 0 + 2 0 (t + t 1 ) +
B (t) = 0 +
1
4
(t + t 1 )
0 (t + t 1 ) + (t + t 1 )
Correct
NotethatparticleBhasasmallerinitialangularvelocitybutgreaterangularacceleration.Also,ithasbeeninmotionforlesstimethanparticleA.
PartH
Howlongafterthetimet 1 doestheangularvelocityofparticleBequalthatofparticleA?
Hint1.Howtoapproachtheproblem
WriteexpressionsfortheangularvelocityofAandBasfunctionsoftime,eitherbycomparisonoftheaboveequationswiththegeneral
kinematicequationsorbydifferentiatingtheaboveequations.Thenequatethe2expressionsandsolvefort t 1 .
ANSWER:
0
4
0 +4t 1
2
0 +2t 1
2
Thetwoparticlesneverhavethesameangularvelocity.
Correct
AnExhaustedBicyclist
Anexhaustedbicyclistpedalssomewhaterraticallywhenexercisingonastaticbicycle.Theangularvelocityofthewheelsfollowstheequation
(t) = at bsin(ct)
for
t 0
PartA
Thereisaspotofpaintonthefrontwheelofthebicycle.Takethepositionofthespotattimet = 0tobeatangle = 0radianswithrespecttoan
axisparalleltotheground(andperpendiculartotheaxisofrotationofthetire)andmeasurepositiveanglesinthedirectionofthewheel'srotation.
Whatangulardisplacement hasthespotofpaintundergonebetweentime0and2seconds?
Expressyouranswerinradiansusingthreesignificantfigures.
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Hint1.Howtoapproachtheproblem
Thespotofpaintrotatestogetherwiththewheel.Therefore,theequationoftheangularvelocityofthewheelcanbeusedtodescribethe
rotationofthespotofpaint.Sincetheangularvelocityistherateofchangeoftheangulardisplacementwithtime,youneedtointegratethe
function(t) overthetimeintervalfrom0to2seconds.
Hint2.Findtheangulardisplacement
Giventheequationoftheangularvelocity(t) ,findanexpressionfortheangulardisplacement(t) asafunctionoftime.
Expressyouranswerasafunctionintermsoftime(t).
Hint1.Integration
Theangulardisplacement(t) ofthespotofpaintisgivenby
(t) =
Notetheinitialconditions(0)
and(0)
= 0
t
0
(t )dt .
= 0
ANSWER:
(t)
1
4
1
8
cos(2t)
1
8
ANSWER:
= 0.793 rad
Correct
PartB
Expresstheangulardisplacementundergonebythespotofpaintatt
shouldyouneedit.
secondsindegrees.RemembertousetheunroundedvaluefromPartA,
= 2
Expressyouranswerindegreesusingthreesignificantfigures.
Hint1.Radiansanddegrees
Recallthatthevalue d ofanangleindegreesis
d = r
180
where r isthevalueoftheanglemeasuredinradians.
ANSWER:
= 45.5
Correct
PartC
Whatdistanced hasthespotofpaintmovedin2secondsiftheradiusofthewheelis50centimeters?
Expressyouranswerincentimetersusingthreesignificantfigures.
Hint1.Distancetraveledbyaparticlethatrotatesaboutacertainaxis
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Consideraparticlethatrotatesaboutthezaxisinthexyplane.Theangulardiplacement traveledbytheparticleisgivenby
s
= 39.7 cm
Correct
PartD
Whichoneofthefollowingstatementsdescribesthemotionofthespotofpaintatt
seconds?
= 2.0
Hint1.Howtoapproachtheproblem
Asthebicyclistpedals,thewheelsrotateinthepositivedirectionaswehavedefinedit.Thismakestheangularvelocityofthewheelalways
nonnegative,andconsequentlytheangularvelocityofthespotofpaintisalwaysnonnegative.Inaddition,theangularvelocityofthespotof
paintincreasesifitsangularaccelerationispositive,anditdecreasesifitsangularaccelerationisnegative.
Hint2.Findtheangularaccelerationatt = 2 seconds
Whatistheangularaccelerationofthespotofpaintatt
seconds?
= 2
Expressyouranswerinradianspersecondpersecond.
Hint1.Findtheangularaccelerationofthespotofpaint
Writeanexpressionfortheangularacceleration(t)ofthespotofpaintasafunctionoftime.
Expressyouranswerasafunctionintermsoftime(t).
Hint1.Angularacceleration
Theangularaccelerationofaparticlethatrotateswithangularvelocityistherateofchangeofwithtimethatis,
d
dt
ANSWER:
(t)
1
2
1
2
cos(2t)
rad/sec 2
ANSWER:
= 0.827 rad/s 2
ANSWER:
Theangularaccelerationofthespotofpaintisconstantandthemagnitudeoftheangularspeedisdecreasing.
Theangularaccelerationofthespotofpaintisconstantandthemagnitudeoftheangularspeedisincreasing.
Theangularaccelerationofthespotofpaintispositiveandthemagnitudeoftheangularspeedisdecreasing.
Theangularaccelerationofthespotofpaintispositiveandthemagnitudeoftheangularspeedisincreasing.
Theangularaccelerationofthespotofpaintisnegativeandthemagnitudeoftheangularspeedisdecreasing.
Theangularaccelerationofthespotofpaintisnegativeandthemagnitudeoftheangularspeedisincreasing.
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Correct
LinearandRotationalKinematicsRankingTask
Thepulleyinrepresentsdifferentpulleysthatareattachedwithouterradiusandinnerradius
indicatedinthetable.Thehorizontalropeispulledtotherightataconstantlinearspeedthatis
thesameineachcase,andnoneofthetwoseparateropesslipsinitscontactwiththepulley.
PartA
Rankthesescenariosonthebasisofthelinearspeedoftheblock.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
Hint1.Relatingthetworopespeeds
Bypullingthehorizontalropeatconstantlinearspeed,thepulleyisgivenaconstantangularvelocity.Thisangularvelocityinturncausesthe
ropeattachedtotheblocktowindupataconstantlinearspeed.Sincebothropesareattachedtothesamepulley,eachoftheirspeedsmust
satisfytherelationship
v = r,
whereistheangularvelocityofthepulleyandr istheradiusfortherope.
ANSWER:
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Correct
Analysisofasinglepulley
Severalpointsonthepulleyareindicatedin.Eachletterdesignatesapointoneitherthepulley
oroneofthetworopesandpointDisatthecenteroftheaxle.Thehorizontalropeispulledto
therightataconstantlinearspeed,andneitherropeslipsinitscontactwiththepulley.
PartB
Rankthedesignatedpointsonthebasisoftheirlinearortangentialspeed.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
Hint1.Determiningspeed
Sinceeachpointislocatedonthesamepulley(orislocatedonaropeattachedtothesamepulley),eachpointslinearortangentialspeedis
determinedbyitsradialdistancefromtherotationaxisvia
v = r.
ANSWER:
Correct
PartC
Rankthedesignatedpointsinonthebasisofthemagnitudeoftheirlinearorradialacceleration.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
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Hint1.Constantspeed
Thehorizontalropeispulledatconstantlinearspeed,givingthepulleyaconstantangularspeed.Anypointonthepulley,however,is
undergoingcircularmotionandmusthaveanonzeroradialacceleration(exceptforthepointontheaxisofthepulley).Anypointnotonthe
pulleydoesnothavearadialacceleration,becausetheymoveinastraightlineatconstantspeed.
ANSWER:
Correct
Problem12.1
Askaterholdsherarmsoutstretchedasshespinsat100rpm.
PartA
Whatisthespeedofherhandsiftheyare140cmapart?
Expressyouranswerwiththeappropriateunits.
ANSWER:
7.33
m
s
Correct
Problem12.6
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Thethreemassesshowninthefigureareconnectedbymassless,rigidrods.Assumethatm1 =
220g andm2 =350g .
PartA
Whatisthex coordinateofthecenterofmass?
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
x
= 5.2cm
Correct
PartB
Whatistheycoordinateofthecenterofmass?
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
y
= 3.3cm
Correct
Problem12.15
Thethreemassesshowninthefigureareconnectedbymassless,rigidrods.
PartA
Findthecoordinatesofthecenterofmass.Findthex coordinate.
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Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
x cm
= 6.0cm
Correct
PartB
Findtheycoordinate.
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
y
cm
= 4.0cm
Correct
PartC
FindthemomentofinertiaaboutanaxisthatpassesthroughmassAandisperpendiculartothepage.
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
I
= 2.0103kgm2
Correct
PartD
FindthemomentofinertiaaboutanaxisthatpassesthroughmassesBandC.
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
I
= 1.3103kgm2
Correct
TorqueaboutthezAxis
LearningGoal:
Tounderstandtwodifferenttechniquesforcomputingthetorqueonanobjectduetoanappliedforce.
Imagineanobjectwithapivotpointpattheoriginofthecoordinatesystemshown.TheforcevectorF liesinthexyplane,andthisforceofmagnitude
F
ThetorqueontheobjectduetotheforceF isequaltothecrossproduct
= r F .When,asinthisproblem,theforcevectorandleverarmbothliein
thexyplaneofthepaperorcomputerscreen,onlythezcomponentoftorqueisnonzero.
Whenthetorquevectorisparalleltothezaxis( = ^
k),itiseasiesttofindthemagnitudeandsignofthetorque, ,intermsoftheangle betweenthe
positionandforcevectorsusingoneoftwosimplemethods:theTangentialComponentoftheForcemethodortheMomentArmoftheForcemethod.
Notethatinthisproblem,thepositivezdirectionisperpendiculartothecomputerscreenandpointstowardyou(givenbytherighthandrule^i
soapositivetorquewouldcausecounterclockwiserotationaboutthezaxis.
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^
^
j = k
),
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Tangentialcomponentoftheforce
PartA
DecomposetheforcevectorF intoradial(i.e.,parallelto
r )andtangential(perpendicularto
r )componentsasshown.Findthemagnitudeoftheradialandtangentialcomponents, F r
andFt .Youmayassumethat isbetweenzeroand90degrees.
Hint1.MagnitudeofF r
UsethegivenanglebetweentheforcevectorF anditsradialcomponent
F r tocomputethemagnitude F r .
ANSWER:
(F r , F t )
F cos, F sin
Correct
PartB
Isthefollowingstatementtrueorfalse?
Thetorqueaboutpointpisproportionaltothelengthr ofthepositionvectorr .
ANSWER:
true
false
Correct
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PartC
Isthefollowingstatementtrueorfalse?
BoththeradialandtangentialcomponentsofF generatetorqueaboutpointp.
ANSWER:
true
false
Correct
PartD
Isthefollowingstatementtrueorfalse?
Inthisproblem,thetangentialforcevectorwouldtendtoturnanobjectclockwisearoundpivotpointp.
ANSWER:
true
false
Correct
PartE
rF sin
Correct
Momentarmoftheforce
Inthefigure,thedashedlineextendingfromtheforcevectoriscalledthelineofactionofF .Theperpendiculardistance
r m fromthepivotpointptothe
lineofactioniscalledthemomentarmoftheforce.
PartF
Expressyouranswerintermsofr and .
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ANSWER:
rm
rsin
Correct
PartG
rF sin
Correct
Threeequivalentexpressionsforexpressingtorqueaboutthezaxishavebeendiscussedinthisproblem:
1.Torqueisdefinedasthecrossproductbetweenthepositionandforcevectors.WhenbothF and
r lieinthexyplane,onlythezcomponentof
torqueisnonzero,andthecrossproductsimplifiesto:
^
^
= r F = r F sin()k = k
Notethatapositivevaluefor indicatesacounterclockwisedirectionaboutthezaxis.
2.TorqueisgeneratedbythecomponentofF thatistangentialtothepositionvector
r (thetangentialcomponentofforce):
= r F t = r F sin()
3.Themagnitudeoftorqueistheproductoftheforceandtheperpendiculardistancebetweenthezaxisandthelineofactionofaforce,r m ,
calledthemomentarmoftheforce:
= r m F = r sin() F
Problem12.18
PartA
In,whatisthenettorqueabouttheaxle?
Expressyouranswerusingtwosignificantfiguresandincludetheappropriateunits.
ANSWER:
= 0.20Nm
Correct
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PSS12.1RotationalDynamicsProblems
LearningGoal:
TopracticeProblemSolvingStrategy12.1forrotationaldynamicsproblems.
Supposethatyouareholdingapencilbalancedonitspoint.Ifyoureleasethepencilanditbeginstofall,whatwillbetheangularaccelerationwhenithas
anangleof10.0degreesfromthevertical?Atypicalpencilhasanaveragelengthof15.0cmandanaveragemassof10.0g .Assumethetipofthe
pencildoesnotslipasitfalls.
PROBLEMSOLVINGSTRATEGY12.1Rotationaldynamicsproblems
MODEL: Modeltheobjectasasimpleshape.
VISUALIZE: Drawapictorialrepresentationtoclarifythesituation,definecoordinatesandsymbols,andlistknowninformation.
Identifytheaxisaboutwhichtheobjectrotates.
Identifyforces,anddeterminetheirdistancefromtheaxis.Formostproblems,itwillbeusefultodrawafreebodydiagram.
Identifyanytorquescausedbytheforcesandthesignsofthetorques.
SOLVE: ThemathematicalrepresentationisbasedonNewtonssecondlawforrotationalmotion:
or =
net = I
net
I
Findthemomentofinertiainthetableofcommonshapesofobjects,orifneeded,calculateitasanintegralorbyusingtheparallelaxis
theorem.
Userotationalkinematicstofindanglesandangularvelocities.
ASSESS: Checkthatyourresulthasthecorrectunits,isreasonable,andanswersthequestion.
Model
Therearetworeasonableapproximationstoconsiderforthepencilinthisproblem:acylinderandathinrod.However,inthisproblemwewilltreatthe
pencilasauniformthinrodoflength15.0cmandmass10.0g .
Visualize
PartA
Thepencilrotatesaboutanaxisperpendiculartotheplaneofthefigure.Whichofthelabeledpointsisthepointthattheaxisofrotationpasses
through?
ANSWER:
A
B
C
D
E
Correct
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PartB
Identifyanddrawtheforcesthatactonthepencil.Becertaintodraweachforceatthecorrectlocation(thepointatwhichtheforceactsonthe
pencil).Theblackdotrepresentsthecenterofmassofthepencil.
Drawthevectorsstartingatthepointsofapplicationofthecorrespondingforces.Thelocationandorientationofthevectorswillbegraded.
Thelengthofthevectorswillnotbegraded.
ANSWER:
Correct
Solve
PartC
Whatistheangularaccelerationofthepencilwhenitmakesanangleof10.0degreeswiththevertical?
Expressyouranswernumericallyinradianspersecondsquaredtothreesignificantfigures.
Hint1.Findthemomentofinertia
InPartA,youdeterminedthepointaboutwhichthepencilrotates.Todeterminetheangularaccelerationofthepencil,youwillneedtoknowits
momentofinertiaaboutthatpoint.WhatisthemomentofinertiaI ofthepencilaboutanend,givenourapproximationthatitisauniformthin
rodoflength15.0cmandmass10.0g ?
Expressyouranswernumericallyinkilogrammeterssquaredtofoursignificantfigures.
Hint1.Themomentofinertiaofathinrod
Theformulaformomentofinertiaofathinrodaboutanendis
1
3
mL
,wheremisthemassandListhelength.Youhavethemass
andlengthingramsandcentimeters,respectively.Besuretoconvertthesequantitiestotheappropriateunitsbeforeenteringtheminto
theformula.
ANSWER:
I
= 7.500105 kg m2
Hint2.Findthenettorque
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Hint1.Howtoapproachtheproblem
Recallthatthemagnitude ofatorqueisdefinedby = rF sin ,wherer isthedistancefromthepivotpointtothepointof
applicationfortheforce,F isthemagnitudeoftheforcethatexertsthetorque,and istheanglebetweentheforceandtheline
connectingthepivottothepointofapplication.
Hint2.Findthedistancefromtheaxisforeachforceactingonthepencil
Whatisthedistancer n betweenthepointofapplicationofnandtheaxisofrotation?Whatisthedistancer w betweenthepointof
applicationofwandtheaxis?Whatisthedistancer f betweenthepointofapplicationoff andtheaxis?Recallthattheaxisof
rotationinthisproblemishorizontal.
Expressyouranswersnumericallyinmeterstothreesignificantfiguresseparatedbyacomma.
ANSWER:
rn
,r w ,r f = 0,7.50102,0 m,m
Hint3.Findtheweightofthepencil
WhatisthemagnitudeFG oftheweightofthepencil?
Expressyouranswernumericallyinnewtonstofoursignificantfigures.
ANSWER:
FG
= 9.800102 N
ANSWER:
net
= 1.276103 N m
ANSWER:
= 17.0 rad/s 2
Correct
Assess
PartD
Calculatethetimet forthepenciltohittheground,assumingthatitfallsfromstandingperfectlyverticalandmaintainsthisangularacceleration.
Expressyouranswernumericallyinsecondstothreesignificantfigures.
Hint1.Howtoapproachtheproblem
Recallthatanglesmeasuredclockwisearenegative.Infallingfromverticaltostrikingtheground,thepenciltravelsclockwiseandgoesthrough
achangeinangleof90degrees,or/2 rad .UsetheangularaccelerationthatyoufoundinPartCalongwithoneoftheangular
kinematicsequationstofindthetimeforthepenciltotraverse/2 rad .
Hint2.Determinewhichequationtouse
Whichofthefollowingequationsformotionunderconstantangularacceleration\texttip{\alpha}{alpha}shouldyouusetofind\texttip{t}{t}?Let
\texttip{\omega_{\rm0}}{omega_0}betheinitialangularspeed,\texttip{\omega}{omega}theangularspeedattime\texttip{t}{t},and
\texttip{\Delta\theta}{Deltatheta}thechangeinangleduringtime\texttip{t}{t}.
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ANSWER:
\omega=\omega_0+\alphat
\large{\Delta\theta=\omega_0t+\frac{_1}{^2}\alphat^2}
\omega^2=\omega_0^2+2\alpha\Delta\theta
ANSWER:
\texttip{t}{t}= 0.430 \rms
Correct
Inthispart,youassumedthattheangularaccelerationfrompartwaythroughthefallistheconstantangularaccelerationthroughtheentirefall.
Thisissimilartothecalculationusingtheaverageangularacceleration.Therefore,youshouldgetavaluesimilarto(i.e.,withthesameorderof
magnitudeas)whatyoufindinreallife.Youcalculatedatimeofaroundhalfasecond.Thisiscertainlyreasonable,basedonyourexperience
withpencils.Ifyouhadfoundananswersuchas50secondsor1/100second,thenyouwouldneedtorethinkyouranswerandcheckoveryour
work.
AccelerationofaPulley
Astringiswrappedaroundauniformsolidcylinderofradius\texttip{r}{r},asshowninthefigure.
Thecylindercanrotatefreelyaboutitsaxis.Thelooseendofthestringisattachedtoablock.
Theblockandcylindereachhavemass\texttip{m}{m}.Notethatthepositiveydirectionis
downwardandcounterclockwisetorquesarepositive.
PartA
Findthemagnitude\texttip{\alpha}{alpha}oftheangularaccelerationofthecylinderastheblockdescends.
Expressyouranswerintermsofthecylinder'sradius\texttip{r}{r}andthemagnitudeoftheaccelerationduetogravity\texttip{g}{g}.
Hint1.Howtoapproachtheproblem
1.Theblockdoesnotrotate.Toanalyzeitsmotion,youshoulduseNewton'ssecondlawinitslinearform:F=ma.
2.Thepulleyrotates.Toanalyzeitsmotion,youshoulduseNewton'ssecondlawinitsangularform:\tau=I\alpha.
3.Usingthegeometryofthesituation,youneedtofindtherelationshipbetween\texttip{a}{a}and\texttip{\alpha}{alpha}.
4.Finally,solvethesystemofthreeequationstoobtainanexpressionfor\texttip{\alpha}{alpha}.
Hint2.Findthenetforceontheblock
Theblockhastwoforcesactingonit:thetensionofthestringanditsownweight.Whatisthenetforce\texttip{F}{F}actingontheblock?Use
thecoordinatesystemshowninthefigure.
Expressyouranswerintermsof\texttip{m}{m},\texttip{g}{g}(themagnitudeoftheaccelerationduetogravity),and\texttip{T}{T}(the
tensioninthestring).
ANSWER:
F=ma== mgT
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Hint3.Findthenettorqueonthepulley
Thetension\texttip{T}{T}inthestringproducesatorquethatactsonthepulley.Whatisthetorque?
Expressyouranswerintermsofthecylinder'sradius\texttip{r}{r}andthetension\texttip{T}{T}inthestring.
Hint1.Formulafortorque
Recallthat\tau=rF\sin(\phi),where\texttip{F}{F}istheforcecausingthetorque,\texttip{r}{r}isthedistancefromthepivottothepointat
whichtheforceacts,and\texttip{\phi}{phi}istheanglebetweenthepositionvectorofthepointmentionedaboveandtheforcevector.
ANSWER:
I\alpha== Tr
Hint4.Relatelinearandangularacceleration
Thestringdoesnotstretch.Therefore,thereisageometricconstraintbetweenthelinearacceleration\texttip{a}{a}andtheangularacceleration
\texttip{\alpha}{alpha}.Whatisthecylinder'sangularacceleration\texttip{\alpha}{alpha}intermsofthelinearacceleration\texttip{a}{a}ofthe
block?
Expressyouranswerintermsof\texttip{a}{a}and\texttip{r}{r}.Becarefulwithyoursigns.
ANSWER:
\texttip{\alpha}{alpha}= \large{\frac{a}{r}}
Hint5.Puttingittogether
Solvethesystemofequationstoeliminate\texttip{T}{T}andobtainanexpressionfor\texttip{\alpha}{alpha}.
ANSWER:
\texttip{\alpha}{alpha}= \large{\frac{g}{1.5r}}
Correct
Notethatthemagnitudeofthelinearaccelerationoftheblockis\large{\frac{2}{3}g},whichdoesnotdependonthevalueof\texttip{r}{r}.
PSS12.2StaticEquilibriumProblems
LearningGoal:
TopracticeProblemSolvingStrategy12.2forstaticequilibriumproblems.
TweedledumandTweedledeearecarryingauniformwoodenboardthatis\texttip{L}{L}=3.00{\rmm}longandhasamass\texttip{M}{M}=15.0{\rmkg}.
IfTweedledumappliesanupwardforceofmagnitude\texttip{F_{\rm1}}{F_1}=60.0{\rmN}attheleftendoftheboard,atwhatpointandwithwhat
magnitude\texttip{F_{\rm2}}{F_2}offorcedoesTweedledeehavetoliftfortheboardtobecarried?
PROBLEMSOLVINGSTRATEGY12.2Staticequilibriumproblems
MODEL: Modeltheobjectasasimpleshape.
VISUALIZE: Drawapictorialrepresentationshowingallforcesanddistances.Listknowninformation.
Pickanypointyouwishasapivotpoint.Thenettorqueaboutthispointiszero.
Determinethemomentarmsofallforcesaboutthispivotpoint.
Determinethesignofeachtorqueaboutthispivotpoint.
SOLVE: Themathematicalrepresentationisbasedonthefactthatanobjectintotalequilibriumhasnonetforceandnonettorque:
\vec{F}_{\rmnet}=\vec{0}and\tau_{\rmnet}=0.
Writeequationsfor\sumF_x=0,\sumF_y=0,and\sum\tau=0.
Solvethethreesimultaneousequations.
ASSESS: Checkthatyourresultisreasonableandanswersthequestion.
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Model
Modeltheboardasauniformrod,withitscenterofmassatthecenter.NotethatinorderforTweedledumandTweedledeetocarrytheboardsafely,the
boardmustbeinstaticequilibrium.Iftheboardwerenotinequilibrium,thenetforceactingontheboardwouldexertatorqueonit,causingtheboardto
rotate,likelyhittingoneofthecarriers.
Visualize
PartA
Whichofthefollowingdiagramsrepresentsthecorrectfreebodydiagramforthissituation?
ANSWER:
A
B
C
D
Correct
PartB
Taketheleftendoftheboardasapivotpoint.Sorttheforcesactingontheboardaccordingtowhethertheyexertapositive,negative,orzerotorque
aboutthispoint.
Dragtheappropriateitemstotheirrespectivebins.
Hint1.Thedefinitionofthepositivesenseofrotation
Byconvention,thepositivesenseofrotationiscounterclockwise.
ANSWER:
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Correct
Nowthatyouhaveidentifiedthecorrectfreebodydiagram,completeyourpictorialrepresentationoftheproblembydefiningsymbolsrepresenting
theimportantvariables,suchasdistancesandmagnitudesofforces.Makesureyouidentifywhattheproblemistryingtofind.Here,wewillcall
\texttip{d}{d}thedistancebetweentheleftendoftheboardandthepointatwhichTweedledeeexertsaforceofmagnitude\texttip{F_{\rm2}}{F_2}.
Therefore,yourtargetvariablesare\texttip{d}{d}and\texttip{F_{\rm2}}{F_2}.
Acompletepictorialrepresentationforthisproblemshouldlooklikethesketchshownbelow.Keepinmindthatbyconventionthepositivesense
ofrotationiscounterclockwise.
Solve
PartC
IfTweedledumappliesaforceof\texttip{F_{\rm1}}{F_1}attheleftendoftheboard,atwhatdistance\texttip{d}{d}fromtheleftendandwithwhat
magnitudeofforce\texttip{F_{\rm2}}{F_2}doesTweedledeehavetoliftfortheboardtobecarried?
Expressyouranswersnumericallyinnewtonsandmeterstothreesignificantfiguresseparatedbycommas.
Hint1.Setuptheequilibriumconditionforthenetforce
Fortheboardtobeinequilibrium,thereshouldbenotranslationalmotion,i.e.,thenetforceactingontheboardmustbezero.Thisequilibrium
conditionbecomes\SigmaF_y=0inthisproblembecausealltheforcesactingontheboardarevertical.
Writeanexpressionfor\SigmaF_y,thesumoftheverticalcomponentsofallforcesactingontheboard.Assumethattheverticalaxisis
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positiveintheupwardsdirection.
Expressyouranswerintermsofsomeorallofthevariables\texttip{F_{\rm1}}{F_1},\texttip{F_{\rm2}}{F_2},\texttip{L}{L},and
\texttip{M}{M}.Use\texttip{g}{g}fortheaccelerationduetogravity.
ANSWER:
\SigmaF_y=0== F_{1}+F_{2}Mg
Hint2.Setuptheequilibriumconditionforthenettorque
Becausetheboardisnotrotating,thenettorque\tau_{\rmnet}=\Sigma\tauaboutanypointmustbezero.Belowisalistofequationsfor
\Sigma\taufordifferentpivotpoints.Ineachcase,thexcoordinateofthepivotpointisgivenassumingthatthexaxisisparalleltotheboard
withtheoriginlocatedattheleftendoftheboard.
Whichofthefollowingisthecorrectequationfor\Sigma\tauforthegivenpivotpoint?Torquesthatwouldcauseacounterclockwiserotationare
positive.Theremaybemorethanonecorrectanswer.
Pivotpoint
\Sigma\tau
x=0
\large{\frac{1}{2}MgL+F_2d=0}
x=L
\large{\frac{1}{2}MgLF_2(Ld)F_1L=0}
x=0
\large{\frac{1}{2}MgL+F_2d=0}
x=d
\large{F_1dF_2\left(d\frac{1}{2}L\right)=0}
\large{x=\frac{1}{2}L}
\large{\frac{1}{2}F_1L+F_2\left(d\frac{1}{2}L\right)=0}
\large{x=\frac{1}{2}L}
\large{\frac{1}{2}F_1L+F_2\left(d\frac{1}{2}L\right)=0}
Entertheletter(s)ofthecorrectequation(s)inalphabeticalorder.Forinstance,ifyouthoughtthatchoicesA,B,andDwerecorrect,
youwouldenterABD.
Hint1.Findingthetorque
Thetorque\texttip{\tau}{tau}duetoaforceistheproductofthemagnitude\texttip{F}{F}oftheforceandthemomentarm\texttip{l}{l}of
theforce:
\tau=F\cdotl.
Asshowninthefigure,themomentarmoftheforceistheperpendiculardistancebetweentheaxisofrotationandthelineofactionof
theforce.
Ifthelineofactionoftheforcegoesthroughtheaxisofrotationthatyouhave
chosen,themomentarmoftheforceisequaltozero.Hence,thisforcewill
notcauseatorqueaboutthechosenaxis.
ANSWER:
BCE
ANSWER:
\texttip{F_{\rm2}}{F_2},\texttip{d}{d}= 87.0,2.53 {\rmN},{\rmm}
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Correct
Assess
PartD
Now,assumethatTweedledumappliesanupwardforceatthefarleftendoftheboard(\texttip{F_{\rmleft}}{F_left})andthatTweedledeeappliesan
upwardforceatthefarrightendoftheboard(\texttip{F_{\rmright}}{F_right}).Withwhatforcedoeseachpersonlift?Recallthatthemassoftheboard
is\texttip{M}{M}and\texttip{g}{g}istheaccelerationduetogravity.
ANSWER:
F_{\rmleft}=2MgF_{\rmright}=2Mg
F_{\rmleft}=0F_{\rmright}=Mg
\large{F_{\rmleft}=\frac{1}{2}Mg}\large{F_{\rmright}=\frac{1}{2}Mg}
\large{F_{\rmleft}=\frac{1}{2}Mg}\large{F_{\rmright}=\frac{3}{2}Mg}
F_{\rmleft}=MgF_{\rmright}=0
Correct
Whentwopeopleholdtheboardatequaldistancesfromtheboard'scenterofmass,theymusteachexertthesameforceontheboard,thatis,
F_{\rmleft}=F_{\rmright}.IfF_{\rmleft}\neqF_{\rmright},theboardwouldrotate.Sincethenetforceontheboardmustbezero,F_{\rmleft}+
F_{\rmright}=Mg.Hence,eachpersonmustexertaforceontheboardequaltoMg/2.
Considertheoriginalsituation,inwhichthepersonontheright(Tweedledee)heldtheboardclosertothecenterofmassthanthepersonholding
theleftendoftheboard(Tweedledum).Ifyouconsiderarotationaxisthroughthecenterofmass,thenettorqueaboutthataxismustbezero.
Sincetorqueisdirectlyproportionaltobothforceandmomentarm,whenonepersonholdstheboardclosertothecenterofmass(smaller
momentarm),thatpersonmustexertmoreforcetocreateatorquewiththesamemagnitudeasthepersonholdingtheboardfartherfromthe
centerofmass.Yourcalculationsconfirmedthisfact.
ABarSuspendedbyTwoVerticalStrings
Arigid,uniform,horizontalbarofmass\texttip{m_{\rm1}}{m_1}andlength\texttip{L}{L}issupportedbytwoidenticalmasslessstrings.Bothstringsare
vertical.StringAisattachedatadistanced<L/2fromtheleftendofthebarandisconnectedto
theceilingstringBisattachedtotheleftendofthebarandisconnectedtothefloor.Asmall
blockofmass\texttip{m_{\rm2}}{m_2}issupportedagainstgravitybythebaratadistance
\texttip{x}{x}fromtheleftendofthebar,asshowninthefigure.
Throughoutthisproblempositivetorqueisthatwhichspinsanobjectcounterclockwise.Use
\texttip{g}{g}forthemagnitudeoftheaccelerationduetogravity.
PartA
Find\texttip{T_{\mitA}}{T_A},thetensioninstringA.
ExpressthetensioninstringAintermsof\texttip{g}{g},\texttip{m_{\rm1}}{m_1},\texttip{L}{L},\texttip{d}{d},\texttip{m_{\rm2}}{m_2},and
\texttip{x}{x}.
Hint1.Choosinganaxis
Choosearotationaxisp,aboutwhichtoapplytherequirement\sum\tau_p=0.Sincethesystemisinstaticequilibrium,thechoiceofrotation
axisisarbitraryhowever,thereisaconvenientchoiceofptofind\texttip{T_{\mitA}}{T_A}byeliminatingthetorquefromanunknownforce.
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Hint2.Findthetorquearoundthebestaxis
ItisconvenienttochoosetherotationaxistobethroughthepointwherestringBisattachedtothebar.Thiseliminatesanytorquefromthe
tensioninstringB.Findthetotaltorqueaboutthispoint.
Answerintermsof\texttip{T_{\mitA}}{T_A},\texttip{m_{\rm1}}{m_1},\texttip{m_{\rm2}}{m_2},\texttip{L}{L},\texttip{x}{x},\texttip{d}
{d},and\texttip{g}{g}.
ANSWER:
\sum\tau_B= \large{T_{A}d\frac{m_{1}gL}{2}m_{2}gx}
Hint3.Summingthetorques
\sum\tau_p=0forastaticsystem.Solvefor\texttip{T_{\mitA}}{T_A}.
ANSWER:
\texttip{T_{\mitA}}{T_A}= \large{{\frac{g}{d}}\left({\frac{m_{1}L}{2}}+m_{2}x\right)}
Correct
PartB
Find\texttip{T_{\mitB}}{T_B},themagnitudeofthetensioninstringB.
ExpressthemagnitudeofthetensioninstringBintermsof\texttip{T_{\mitA}}{T_A},\texttip{m_{\rm1}}{m_1},\texttip{m_{\rm2}}{m_2},and
\texttip{g}{g}.
Hint1.Twodifferentmethodstofind\texttip{T_{\mitB}}{T_B}
Therearetwoequivalentwaystofind\texttip{T_{\mitB}}{T_B}.Onewayistobalancethetorquesaswasdoneinthecalculationof
\texttip{T_{\mitA}}{T_A},exceptusingadifferentrotationaxis.Inthiscase,aconvenientaxisisthroughthepointwherestringAisattachedto
thebar.Thesecond,andeasier,methodistousethesecondequationforstaticequilibrium,\sum\vec{F}=0.
Hint2.Directionofforces
Sincebothstringsarevertical,allforcesonthebarthetensionforcesandtheweightsofthebarandblockactvertically.Thus,onlyvertical
componentsofforcesneedbeconsidered.
ANSWER:
\texttip{T_{\mitB}}{T_B}= T_{A}\left(m_{1}+m_{2}\right)g
Correct
PartC
Ifthebarandblockaretooheavythestringsmaybreak.Whichofthetwoidenticalstringswillbreakfirst?
ANSWER:
stringA
stringB
Correct
PartD
Ifthemassoftheblockistoolargeandtheblockistooclosetotheleftendofthebar(nearstringB)thenthehorizontalbarmaybecomeunstable
(i.e.,thebarmaynolongerremainhorizontal).
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Whatisthesmallestpossiblevalueof\texttip{x}{x}suchthatthebarremainsstable(callit\texttip{x_{\rmcritical}}{x_critical})?
Expressyouranswerfor\texttip{x_{\rmcritical}}{x_critical}intermsof\texttip{m_{\rm1}}{m_1},\texttip{m_{\rm2}}{m_2},\texttip{d}{d},and
\texttip{L}{L}.
Hint1.Natureoftheunstablemotion
Whenthebarbecomesunstablethereareonlytwopointsaboutwhichthebarcanrotate:thepointswherethestringsattachtothebar.About
whichpointwillthebarrotatewhenx<x_{\rmcritical}?
ANSWER:
ThepointwherestringAisattachedtothebar
ThepointwherestringBisattachedtothebar
Hint2.TensioninstringBatthecriticalpoint
ThetensioninstringBcounteractstheclockwiserotationofthebaraboutthepointwherestringAisattachedtothebar.As\texttip{x}{x}is
decreased,\texttip{T_{\mitB}}{T_B}islikewisedecreasedbecausetheclockwisetorqueaboutthispointdecreases.Thecriticalvalue
\texttip{x_{\rmcritical}}{x_critical}correspondstowhenT_B=0.If\texttip{x}{x}isdecreasedfurther,\texttip{T_{\mitB}}{T_B}willcontinuetobe
zeroandthecounterclockwisetorqueduetotheweightoftheblockwillbegreaterthantheclockwisetorqueduetotheweightofthebar,
causingthesystemtorotate.
Hint3.Calculatethetorques
AddupthetotaltorqueaboutthepointinwhichstringAattachestothebarwhenthemass\texttip{m_{\rm2}}{m_2}isat\texttip{x_{\rmcritical}}
{x_critical}.Rememberthat\texttip{T_{\mitB}}{T_B}hasaspecialvalueatthispointandthat,owingtothechoiceoforigin,thetorquedueto
stringAis0.Remembertopayattentiontothedirectionofthetorques.
Answerintermsof\texttip{m_{\rm2}}{m_2},\texttip{m_{\rm1}}{m_1},\texttip{d}{d},\texttip{L}{L},\texttip{g}{g},and\texttip{x_{\rm
critical}}{x_critical}.
Hint1.FindthedistanceofthecenterofmassofthebarfromstringA
Whatisthedistance\texttip{d_{\rm1}}{d_1}ofthecenterofmassofthebarfromstringA?
Answerintermsofthegivenvariables.
ANSWER:
\texttip{d_{\rm1}}{d_1}= \large{\frac{L}{2}d}
Hint2.Findthedistanceof\texttip{m_{\rm2}}{m_2}fromthestringA
Whatisthedistance\texttip{d_{\rm2}}{d_2}of\texttip{m_{\rm2}}{m_2}fromthestringA?
Answerintermsofthegivenvariables.
ANSWER:
\texttip{d_{\rm2}}{d_2}= dx_{\rm{critical}}
ANSWER:
\sum\tau_A=0= \large{m_{2}g\left(dx_{\rm{critical}}\right)m_{1}g\left(\frac{L}{2}d\right)}
ANSWER:
\texttip{x_{\rmcritical}}{x_critical}= \large{d\frac{m_{1}}{m_{2}}\left({\frac{L}{2}}d\right)}
Correct
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PartE
Notethat\texttip{x_{\rmcritical}}{x_critical},ascomputedinthepreviouspart,isnotnecessarilypositive.Ifx_{\rmcritical}<0,thebarwillbestable
nomatterwheretheblockofmass\texttip{m_{\rm2}}{m_2}isplacedonit.
Assumingthat\texttip{m_{\rm1}}{m_1},\texttip{d}{d},and\texttip{L}{L}areheldfixed,whatisthemaximumblockmass\texttip{m_{\rmmax}}{m_max}
forwhichthebarwillalwaysbestable?Inotherwords,whatisthemaximumblockmasssuchthatx_{\rmcritical}\leq0?
Answerintermsof\texttip{m_{\rm1}}{m_1},\texttip{d}{d},and\texttip{L}{L}.
Hint1.Requirementofstability
If\texttip{x}{x}iscalculatedtobelessthanzero,thesolutionisunphysical.(Thebardoesnotextendtheretosupportit!)Theminimumvalue
that\texttip{x}{x}canhaveisobviouslyzero.If\texttip{m}{m}islessthanthemassthatwouldgivex_{\rmcritical}=0thenthebarwillbestable
foranyphysicalvalueof\texttip{x}{x}.
ANSWER:
\texttip{m_{\rmmax}}{m_max}= \large{\frac{m_{1}\frac{L}{2}m_{1}d}{d}}
Correct
AnUnfairRace
Thisappletshowstheresultsofreleasingafrictionlessblockandarollingdiskwithequalmassesfromthetopofidenticalinclinedplanes.
PartA
Whichofthefollowingisthebestexplanationoftheresultsshownintheapplet?
ANSWER:
Thedisklosesenergytofrictionasitrolls,buttheboxisfrictionlessandsoitspeedsupmorequicklyandgetstothebottomfirst.
Thepotentialenergyofthediskisconvertedintotranslationalandrotationalkineticenergy,sothetranslationalspeedgrowsmoreslowly
thanthatofthebox,whichhasnorotationalenergy.
Thenetforcesonthetwoobjectsareequal,buttheforceonthediskgetspartiallyusedupincreatingthetorquenecessarytomakeitroll.
Thenetforcesonthetwoobjectsareequal,buttheforceonthediskisnotdirectedparalleltotheramp,andsodoesnotcreateasgreatan
accelerationdowntheramp.
Correct
Thisappletshowsthesamesituation,butitalsoshows,throughbargraphsthatchangewithtime,thewaythattheenergyistransformedastheboxand
thediskgodowntheinclinedplane.
Assumethattheboxanddiskeachhavemass\texttip{m}{m},thetopoftheinclineisatheight\texttip{h}{h},andtheanglebetweentheinclineandthe
groundis\texttip{\theta}{theta}(i.e.,theinclineisatanangle\texttip{\theta}{theta}abovethehorizontal).Also,lettheradiusofthediskbe\texttip{R}{R}.
PartB
Howmuchsoonerdoestheboxreachthebottomoftheinclinethanthedisk?
Expressyouranswerintermsofsomeorallofthevariables\texttip{m}{m},\texttip{h}{h},\texttip{\theta}{theta},and\texttip{R}{R},aswellas
theaccelerationduetogravity\texttip{g}{g}.
Hint1.Howtoapproachtheproblem
Youcanuseconservationofenergytodeterminethefinalspeedatthebottomoftherampforeachobject.Fromthat,youcandeterminethe
averagespeedandthenusekinematicstodeterminethetimetogettothebottom.Thedifferenceintimesforthetwoobjectstogettothe
bottomiswhatyouaretryingtofind.Ifyoutakethedirectiondowntheinclinedplaneasthepositivexaxis,thenthevelocitiesarealways
positiveandonedimensionalinthexdirection,sothevelocityandthespeedhavethesamesign.
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Hint2.Findthefinalspeedofthebox
Determinethespeed\texttip{v_{\rmbox}}{v_box}oftheboxatthebottomoftheramp.
Expressyouranswerintermsofsomeorallofthevariables\texttip{m}{m},\texttip{h}{h},\texttip{\theta}{theta},and\texttip{R}{R},as
wellastheaccelerationduetogravity\texttip{g}{g}.
Hint1.Usingconservationofenergy
Initially,theboxhasonlygravitationalpotentialenergymgh.Atthebottomoftheincline,allitspotentialenergyhasbeenconvertedto
kineticenergy\large{\frac12mv_{\rmbox}^2}.Thus,
\large{mgh=\frac12mv_{\rmbox}^2}.
Youcansolvethisequationtofind\texttip{v_{\rmbox}}{v_box}intermsofthegivenvariables.
ANSWER:
\texttip{v_{\rmbox}}{v_box}= \sqrt{2gh}
Hint3.Findthefinalspeedofthedisk
Determinethespeed\texttip{v_{\rmdisk}}{v_disk}ofthecenterofmassofthediskatthebottomoftheramp.Becertaintoaccountfor
translationalandrotationalkineticenergy.
Expressyouranswerintermsofsomeorallofthevariables\texttip{m}{m},\texttip{h}{h},\texttip{\theta}{theta},and\texttip{R}{R},as
wellastheaccelerationduetogravity\texttip{g}{g}.
Hint1.Usingconservationofenergy
Initially,thediskhasonlygravitationalpotentialenergymgh.Atthebottomoftheincline,thepotentialenergyhasbeenentirely
convertedtokineticenergy.However,thekineticenergyisnotsimply\large{\frac12mv_{\rmdisk}^2},becausethediskhasrotational
kineticenergyaswell.Thetotalkineticenergyis\large{\frac12mv_{\rmdisk}^2+\frac12I\omega^2},where\texttip{I}{I}isthemomentof
inertiaand\texttip{\omega}{omega}istheangularspeed.Thus,
\large{mgh=\frac12mv_{\rmdisk}^2+\frac12I\omega^2}.
Youneedtofindthevaluesof\texttip{I}{I}and\texttip{\omega}{omega}intermsofthegivenvariablesbeforeyousolvefor
\texttip{v_{\rmdisk}}{v_disk}.
Hint2.Momentofinertiaforadisk
Themomentofinertia\texttip{I}{I}forauniformdiskofradius\texttip{R}{R}andmass\texttip{m}{m}is
\large{I=\frac12mR^2}.
Hint3.Relatingangularspeedandthespeedofthecenterofmass
Recallthatyouareassumingthatthediskisrollingwithoutslipping.Whenadisk(orotherobjectwithcircularcrosssection)rolls
withoutslipping,itfollowstherelationv=\omegaR,where\texttip{v}{v}isthespeedofthecenterofmass,\texttip{R}{R}istheradius,
and\texttip{\omega}{omega}istheangularspeed.
ANSWER:
\texttip{v_{\rmdisk}}{v_disk}= \large{\sqrt{\frac{4gh}{3}}}
Hint4.Findingtheaveragespeed
Sincetheforcesremainconstantthroughoutthemotion,thismustbemotionwithconstantacceleration.Therefore,thevelocitymustchange
linearly.Recallthatinmotionwithconstantacceleration,theaveragevelocity\texttip{v_{\rmavg}}{v_avg}isgivenbytheformulav_{\rmavg}=
(v_{\rminitial}+v_{\rmfinal})/2,where\texttip{v_{\rminitial}}{v_initial}and\texttip{v_{\rmfinal}}{v_final}aretheinitialandfinalvelocities.
Hint5.Findingthetimefromtheaveragespeed
Recallthataveragevelocity\texttip{v_{\rmavg}}{v_avg}isdefinedby
\large{v_{\rmavg}=\frac{x_{\rmfinal}x_{\rminitial}}{t_{\rmfinal}t_{\rminitial}}},
where\texttip{x_{\rmfinal}}{x_final}and\texttip{x_{\rminitial}}{x_initial}arethefinalandinitialpositions,and\texttip{t_{\rmfinal}}{t_final}and
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\texttip{t_{\rminitial}}{t_initial}arethefinalandinitialtimes.Thechangeintimeiswhatyouwanttofind,andthechangeinpositionisjustthe
lengthoftheinclinedplane.
Hint6.Findthelengthoftheincline
Whatisthelength\texttip{L}{L}oftheinclinedplane?Thefiguremaybehelpful.
Expressyouranswerintermsofoneorbothofthevariables\texttip{h}{h}and
\texttip{\theta}{theta}.
ANSWER:
\texttip{L}{L}= \large{\frac{h}{{\sin}\left({\theta}\right)}}
ANSWER:
\large{\frac{\sqrt{\frac{3h}{g}}\sqrt{\frac{2h}{g}}}{{\sin}{\theta}}}
Correct
Youshouldlookatyouranswerandconsiderlimitingcases.Asimpleoneisthatthetimedifferenceshouldtendtozeroasthelengthoftheboard
shrinkstozero.Simplyexpresstheheightoftheboardintermsofthelengthoftheinclineandyou'llseethatyouranswerindeedbehavesthis
way.
Youransweralsopredictsthatthedifferenceintimegrowslongeras\texttip{\theta}{theta}shrinkstowardzerowhiletheheightremainsfixed(i.e.,
thedifferenceintimegrowslongerasthelengthoftheboardgrowslonger).Itmightnotbeimmediatelyobvioustoyouthatthisshouldhappen,
butitisnotinconceivable,andyoucandosomesimpleexperimentstoseethatitisactuallytrue.
As\texttip{\theta}{theta}growstoward\pi/2\{\rmrad}=90^\circ,youmightexpectthedifferenceintimetogotozero,becauseifyoudropadisk
andaboxtheyfallatthesamerate.However,recallthatyourderivationincludedtheassumptionthatthediskrollswithoutslipping,whichis
definitelynotthecaseifthediskissimplydroppedvertically.Therefore,thisformulashouldn'tapplytothecaseofsimplydroppingthediskand
box.Canyouthinkofasituationwithaverticaldropinwhichthediskwouldobeyv=\omegaR?
Problem12.46
PartA
Whatisthemagnitudeoftheangularmomentumofthe2.70{\rmkg},5.70cmdiameterrotatingdiskinthefigure?
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ANSWER:
6.89102 \rm{kg\,m^2/s}
Correct
PartB
Whatisitsdirection?
ANSWER:
xdirection
xdirection
ydirection
ydirection
zdirection
zdirection
Correct
Problem12.70
Blocksofmassm_1andm_2areconnectedbyamasslessstringthatpassesoverthepulleyinthefigure.Thepulleyturnsonfrictionlessbearings,and
massm_1slidesonahorizontal,frictionlesssurface.Massm_2isreleasedwhiletheblocksare
atrest.
PartA
Assumethepulleyismassless.Findtheaccelerationofm_1.
Expressyouranswerintermsofthegivenquantities.
ANSWER:
a_1= \large{\frac{m_{2}g}{m_{1}+m_{2}}}
Correct
PartB
Findthetensioninthestring.
Expressyouranswerintermsofthegivenquantities.
ANSWER:
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T= \large{\frac{m_{1}m_{2}g}{m_{1}+m_{2}}}
Correct
PartC
Supposethepulleyhasmassm_pandradiusR.Findtheaccelerationofm_1.VerifythatyouranswersagreewithpartAifyousetm_p=0.
Expressyouranswerintermsofthegivenquantities.
ANSWER:
a_1= \large{\frac{2m_{2}g}{2m_{1}+2m_{2}+m_{p}}}
Correct
PartD
Findthetensionintheupperportionofthestring.VerifythatyouranswersagreewithpartBifyousetm_p=0.
Expressyouranswerintermsofthegivenquantities.
ANSWER:
T_{upper}= \large{\frac{2m_{1}m_{2}g}{2m_{1}+2m_{2}+m_{p}}}
Correct
PartE
Findthetensioninthelowerportionsofthestring.VerifythatyouranswersagreewithpartBifyousetm_p=0.
Expressyouranswerintermsofthegivenquantities.
ANSWER:
T_{lower}= \large{\frac{2m_{1}m_{2}g+m_{p}m_{2}g}{2m_{1}+2m_{2}+m_{p}}}
Correct
Problem12.79
A10\rmgbullettravelingat430{\rmm/s}strikesa11{\rmkg},0.90\rmmwidedoorattheedgeoppositethehinge.Thebulletembedsitselfinthedoor,
causingthedoortoswingopen.
PartA
Whatistheangularvelocityofthedoorjustafterimpact?
Expressyouranswertotwosignificantfiguresandincludetheappropriateunits.
ANSWER:
w= 1.3\large{{\rm\frac{rad}{s}}}
Correct
RotationalKinematicsRankingTask
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Thefigureshowsangularpositionversustimegraphsforsixdifferentobjects.
PartA
Rankthesegraphsonthebasisoftheangularvelocityofeachobject.Rankpositiveangularvelocitiesaslargerthannegativeangularvelocities.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
Hint1.Determiningangularvelocityfromanangularpositionversustimegraph
Theslopeofanangularpositionversustimegraphisthe"rise"(changeinangularposition)overthe"run"(changeintime).Inphysics,theratio
ofchangeinangularpositiontochangeintimeisdefinedastheangularvelocity.Thus,theslopeofanangularpositionversustimegraph
givestheangularvelocityoftheobjectbeinggraphed.
ANSWER:
Correct
PartB
Rankthesegraphsonthebasisoftheangularaccelerationoftheobject.Rankpositiveangularaccelerationsaslargerthannegativeangular
accelerations.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
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Hint1.Determiningangularacceleration
Angularaccelerationisproportionaltothechangeinanobject'sangularvelocity.Sinceangularvelocitycanbedeterminedfromtheslopeofan
angularpositionversustimegraph,angularaccelerationcanbedeterminedfromchangesinslope.
ANSWER:
Correct
BalancingTorquesRankingTask
Asignistobehungfromtheendofathinpole,andthepolesupportedbyasinglecable.Yourdesignfirmbrainstormsthesixscenariosshownbelow.In
scenariosA,B,andD,thecableisattachedhalfwaybetweenthemidpointandendofthepole.InC,thecableisattachedtothemidpointofthepole.InE
andF,thecableisattachedtotheendofthepole.
PartA
Rankthedesignscenarios(AthroughF)onthebasisofthetensioninthesupportingcable.
Rankfromlargesttosmallest.Torankitemsasequivalent,overlapthem.
Hint1.Howtoapproachtheproblem
Inallcases,thepoleandsignareheldinequilibriumbythetorquesuppliedbythecable.Therefore,bysettingthetorqueduetothecableequal
tothetorqueduetothesignandthepole,youcandeterminetherelativetensioninthecable.Notethatthepoleandsignarethesameineach
case,sothetorqueduetothemisthesameforeachcase.
Hint2.Themathematicalrelationship
Thetorqueduetothecableisgivenby
\tau=rF\sin\theta,
where\texttip{r}{r}isthedistancefromthehingetotheforceand\texttip{\theta}{theta}istheangleoftheforcerelativetothepole.Sincethe
poleisinequilibrium,thetorqueduetothecableisequalinmagnitudetothenettorqueduetothesignandtheweightofthepole.Thus,
\tau_{\rmcable}=\tau_{\rmsign}+\tau_{\rmpole},
or
rF\sin\theta=\tau_{\rmsign}+\tau_{\rmpole},
andtherefore
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\large{F=\frac{\tau_{\rmsign}+\tau_{\rmpole}}{r\sin\theta}}.
Substitutingdifferentvaluesof\texttip{r}{r}and\texttip{\theta}{theta}intothisrelationshipcangiveyoutherelativesizesoftherequiredcable
forces.
ANSWER:
Correct
VideoTutor:BalancingaMeterStick
First,launchthevideobelow.Youwillbeaskedtouseyourknowledgeofphysicstopredicttheoutcomeofanexperiment.Then,closethevideowindow
andanswerthequestionatright.Youcanwatchthevideoagainatanypoint.
PartA
Supposewereplacethemassinthevideowithonethatisfourtimesheavier.Howfarfromthefreeendmustweplacethepivottokeepthemeter
stickinbalance?
Hint1.Howtoapproachtheproblem.
Forthemetersticktobeinequilibrium,thenettorqueonitmustbezero.Torquesaboutthefulcrummaybeexertedbythemasshangingfrom
theendofthestickandbythesticksownweight.
Usetheconditionthatthenettorquemustbeequaltozerotoobtainarelationshipinvolving1)thedistancebetweentheleftendofthestick
andthefulcrumand2)thedistancebetweenthecenterofmassofthestickandthefulcrum.Thesetwodistancesmustadduptoaconstant.
Youshouldgettwoequationsthatyoucansolveforthelocationofthefulcrum.
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ANSWER:
50cm(inthemiddle)
25cm
10cm
75cm(25cmfromtheweight)
90cm(10cmfromtheweight)
Correct
VideoTutor:Walkingtheplank
First,launchthevideobelow.Youwillbeaskedtouseyourknowledgeofphysicstopredicttheoutcomeofanexperiment.Then,closethevideowindow
andanswerthequestionatright.Youcanwatchthevideoagainatanypoint.
PartA
Inthevideo,thetorqueduetothemassoftheplankisusedinthecalculations.Forthisquestion,ignorethemassoftheboard.Rank,fromlargestto
smallest,themass\rmmneededtokeeptheboardfromtippingover.
Torankitemsasequivalent,overlapthem.
Hint1.Howtoapproachtheproblem
Inequilibrium,thesumoftheclockwisetorquesaboutanaxismustequalthesumofthecounterclockwisetorques.
ANSWER:
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Correct
ScoreSummary:
Yourscoreonthisassignmentis118%.
Youreceived103.37outofapossibletotalof105points,plus20.56pointsofextracredit.
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