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