Torquewillbeusedtoquantifyforcesappliedtobodieswhichcanrotate.Previously,intheparticlemodel,itdidn’tmatterwhereonabodyweappliedaforce.However,withrigidbodies,thatisnolongertrue.(Thingsarenolongerpointparticles.)
Torque
1.Introduction2.Torque
1.Leverarmchanges3.NetTorques4.MomentofRotationalInertia
1.MomentofInertiaforArbitraryShapes2.ParallelAxisTheorem
5.TaleofTwoEnergies6.Rollingandfriction7.ConservationofAngularMomentum
1.AngularMomentum-Particle8.Gyroscope
Introduction
Torque
1.Themagnitudeoftheforce2.Thedistancer,fromthepivotpointtowheretheforceisapplied3.Theangleatwhichtheforceisapplied
contributestorotatingthewrenchdoesnotcontributetotherotationatall.
Weknowfromexperiencethat
1.it’seasiertoopenadoorwhenyoupushatthepointthat’sfarthestfromthehinges2.youhavetopushinthedirectionyouwantittogo.
(let’sturnthatintomath)
F⊥
F∥
PHY 207 - torque - J. Hedberg - 2017
Page 1
Torque, willbegivenby:
Lever arm changes
Astheangleofthelineofactionchanges,thenthelengthofthe'leverarm'changes.
In general...
Eitherwayyousetitup,thetorqueisstillgivenbytheforceapplied,timesthedistanceawayfromthepivotpoint,andthesineoftheanglebetweentheforceandtheradiusofmotion.
or,
Quick Question 1Whichforcewillprovidethelargestmagnitudeoftorqueonthewrench?(Eiftorqueisthesameforeach.)
Net Torques
Justlikethenetforce,wecanaddupallthedifferenttorquesappliedtoanobjecttoseeif,andwhichway,itwillrotate.
τ
τ = rF⊥
τ = Fℓ
τ = r = rF sinϕF⊥
τ = Fℓ = Fr sinϕ
∑ τ = + + …τ1 τ2 τ3
PHY 207 - torque - J. Hedberg - 2017
Page 2
30 N
20 N
d = 4.0 cm
axis
Whatisthenettorquearoundthelabeledaxis?
Torque
Torque,likeforce,isavector
Quick Question 2Toopenthedoor,whichdirectionshouldthetorquevectorpoint?
Whatisthetorqueappliedtothenut?(useunitvectors)
Fulcrums and Levers
Let'sbalanceaboardonafulcrum.
Example Problem #1:
τ = r×F
a)ib)−ic)jd)−je)kf)−k
PHY 207 - torque - J. Hedberg - 2017
Page 3
m1
m2
Ameterstickofmass0.6kgsitsonafulcrumlocatedatthe30cmmarkatequilibrium.Attheendofthestick(0.0cm)hangsamassm.Whatism?
Moment of Rotational Inertia
Wesawbeforethataforcecausedalinearacceleration,withNewton’s2ndlaw:
Sinceatorqueisbasicallytherotationalequivalentofforce,thenitshouldcauseanangularacceleration.
However,weneedaconstantofproportionalitybetweenthetwoterms.
iscalledthe'momentofinertia'.Itwilldescribehoweasyorharditistorotatearigidbody.(justlikemasstold
ushowhardoreasyitwastomoveabody)InthecaseofF,andlinearmotion,ifweappliedthesameforcetotwodifferentmasses,thentheheaviermasswouldhaveasmalleracceleration.
But,inthecaseofrotatingobjects,it’snotjustthemassthataffectstheangularrotation,butalsohowthatmassisdistributed.
Inthiscase,the2ndlawforthisparticlewithmass,heldadistance fromthecenterwillbe:
butthetangentialaccelerationisrelatedtotheangularaccelerationby .Sowecanwrite:
Example Problem #2:
F = ma
τ → α
τ = Iα
I
m r
=aT
F
m
= αraT
α =F
mr
τ = rF⊥
PHY 207 - torque - J. Hedberg - 2017
Page 4
Fordiscretemasses: Forcontinuousmasses:
Usingalso ,wecanwritethisas:
Thuswehaveforaballrotatingaroundanaxis:
Forthisgeometry,themomentofinertia, ,isgivenby:
Thismakessense,ifthemassisheavier,thentheangularaccelerationwillbeless,andalsoifmassisfartherawayfromthecenter,theangularaccelerationwillbeless.
Ifwehadmultiplemasses:Wecansimplysumupallthemomentsofinertiaforeachlittleparticle:
Example
Calculatethemomentofinertiasforthisshapebasedonthetwoaxesshown.
(m1=1.0kg,m2=1.5kgandm3=1.0kg)
10 cmm1 m2
m3
10 cmm1 m2
m3
a
b
Forshapesofarbitrarydimensions,thingswouldgetalittlemorecomplicated.Evenforaseeminglysimpleshapelikeathinrodbeingrotatedaroundoneofitsends,we'dhavetodosomeintegralcalculustofigureoutthemomentofinertia:
τ = rF⊥
α =τ
mr2
α =τ
mr2
I
I = mr2
I = + + …= ∑ mm1r21 m2r22 m3r23 r2
Example Problem #3:
I = ∑ mi r2i I = ∫ dmr2
Play/Pause Play/Pause
PHY 207 - torque - J. Hedberg - 2017
Page 5
Moment of Inertia for Arbitrary Shapes
We'llbeginbyintegratingthemassoftheobject:
Quick Question 3Hereisauniformrodoftotalmass .Expressthemasselement intermsof :
Rotate a rod:
Tocalculate forthisgeometry:
We'llneedtoexpress intermsof (position.)
Wecanintegrateoverdx
Afterintegratingandevaluating:
I = ∫ dmr2
M dmdx
a)
dm = dxL
M
b)
dm = dxM
L
c)
dm = ML dx
d)
dm =LM
dx
I
I = ∫ dmr2
dm x
dm = dxM
L
dx∫ x=+L/2
x=−L/2x2
M
L
I = M112
L2
PHY 207 - torque - J. Hedberg - 2017
Page 6
Parallel Axis Theorem
Thiswillbeeasyforsimpleshapes,butinthecaseofmorecomplexshapes,orlessobviousaxesofrotation,wemightneedtheparallelaxistheorem.
Here, isthemomentofinertiathroughthecenterofmass,and isthedistancebetweenthataxisandouraxisofinterest.
Rotate a rod (around and end)
Inthecaseofarotationaxislocatedatoneoftheendsoftherod,wecanusetheparallelaxistheorem.(Thenewaxisattheendisparalleltothecomaxis.
Weknowthe thoughthecenterofmass:
Andthedistancetotheendis :
Moment of inertia for various geometries
Theengineinaplanecandeliver500Nmtothepropeller.Thepropellerhasamassof40kgandis2.0meterslong(diameter).Howlongdoesittaketoreach2000rpm?
Atelephonepolefallsoverinastorm.Itis7.0meterstallandhasamassof260kg.Estimatetheangularaccelerationofthepolewhenithasfallenby fromthevertical.
I = +MIcom h2
Icom h
I = +MIcom h2
I
I = M +M112
L2 h2
L/2
I = M +M = M112
L2 ( )L
2
2 13
L2
Example Problem #4:
⋅
Example Problem #5:
25∘
PHY 207 - torque - J. Hedberg - 2017
Page 7
Energy of Rotation
Thetotalmechanicalenergyofasystemwasthekineticpluspotential
However, needstoincludeboththetranslationalenergy: andtherotationalenergy:
thus
Quick Question 4Whichobjectwinstherace?
Quick Question 5Whichcanwinstherace?
Tale of Two Energies
Thiscombinationofmotiondemandsthatweaccountforbothlinearandrotationalenergywhendescribingarollingobject:
Justtobemorespecific
We'lloftenneedthegeometricalconstraint: todealwiththesetwoterms.
Quick Question 6Abowlingballisrollingwithoutslippingatconstantspeedtowardthepinsonalane.Whatpercentageoftheball’stotalkineticenergyistranslationalkineticenergy?
= KE +UEmech
KE m12
v2
K = IErot12
ω2
= m + I +mghEmech12
v212
ω2
a)TheSolidCylinderb)TheEmptyHoopc)It'llbeatie
a)ChickenBrothb)Cream'OChickenc)It'llbeatie
= I + MKrollingw.o.slipping12
ω2 12
v2
= + MKrollingw.o.slipping12
Icomω2 12
v2com
v = ωr
a)50%b)71%c)46%d)29%e)33%
PHY 207 - torque - J. Hedberg - 2017
Page 8
R
θ
R
θ
Quick Question 7Whichdrawingbelowshowstheforcesthatwillpreventtheladderfromslipping?
Rolling and friction
Wecanmakeafreebodydiagramofarollingball.Theforcesactingontheball:
1.Weight(gravity)2.Normalforce(perptoramp)3.staticfriction(pointsuptheramp)
Quick Question 8Whichforcecreatesanon-zerotorque?
What's the torque?
StartingwiththerotationalversionofNewton'slaw:
wecanconsidertheforcesthatwillcontributetoarotationofthedisk.
Theonlyforceactingwithaperpendicularcomponentisfriction.
Since
a)StaticFrictionb)Gravityc)NormalForced)Theyallcreatetorques
τ = Iα
R = τ = Iαfs
α =− /Racom
=−s comacom
PHY 207 - torque - J. Hedberg - 2017
Page 9
Magnitude
Direction
Conservation:
But,wecanalsoconsiderthesumofforcesalongthesurfaceoftheramp:
Plugsomestuffin:
Rearrangefor :
Asphericalshellrollsdownarampstartingatheight .Whatisthespeedofthesphereatthebottom?
Abucketisattachedtoastringthatiswrappedaroundacylinderasshown.Ifthebucketisreleasedfromrest1meterabovetheground,howlongwillittaketohitthefloor?(Massofthebucket=2.0kg,massofcylinder=1kg,radiusofcylinder=2.0cm,massofstring=0kg)
Conservation of Angular Momentum
=−fs Icomacom
R2
Σ = +friction = mF∥ FG∥ acom
Mg sin θ− = mIcomacom
R2acom
acom
=acomg sin θ
1 + /MIcom R2
Example Problem #6:
h
Example Problem #7:
L = Iω
L = Iω
=Li Lf
PHY 207 - torque - J. Hedberg - 2017
Page 10
+x
+z
Akid(mass=36kg)standsatthecenterofarotatingdisk(a.k.amerry-go-round).Itsmassis200kgandrotatesonceevery2.5seconds.Ifthekidwalks2.0metersawayfromthecentertoreachtheedge,whatwilltheperiodofrotationbewhenhereachestheedge?
Quick Question 9Whichdirectionisangularmomentumofthewheel/person/stoolsystem?Thewheelisspinningclockwisewhenlookingdown.
Angularmomentumcanbeconsideredtherotationalanalogueoflinearmomentum.Itsmagnitudeisgivenbythemomentofrotationalinertiatimestheangularvelocity.
Justlikelinearmomentum, ,angularmomentumisalsoavector: .Itsdirectionwillpointinthesamedirectionastheangularvelocityvector.(i.e.perpendiculartotheplaneofrotation,andfollowingtheRHR.)
Anotherconservationlaw:Inanisolatedsystem,intheabsenceofexternaltorques,thetotalangularmomentumdoesnotchange.
L = Iω
p L
=Li Lf
Example Problem #8:
a)+xb)−xc)+yd)−ye)+zf)−z
PHY 207 - torque - J. Hedberg - 2017
Page 11
+x
+z
Quick Question 10Now,weflipthewheelupsidedown.Whichwayistheangularmomentumofthewheel/person/stoolsystem?
Angular Momentum if a Particle
Ifaparticleshasalinearmomentum,thewemightexpectittoalsohaveanangularmomentum?
Thisquantityisconserved.Here, isthedistanceawayfromthe'origin'and isthemomentumvector.
Quick Question 11Theparticleismovinginthe+ydirection.Basedonit'scurrentposition,whichisgivenby:
whichwouldbetherightapplicationoftheangularmomentumequation?(w.r.ttheorigin)
a)+xb)−xc)+yd)−ye)+zf)−z
L = r×p
r p
r = +A −A +0i j k
a)L = |r||p| sin( )90∘
b)L = |r||p| sin( )55∘
c)L = |r||p| sin( )135∘
d)L = |r||p| cos( )0∘
PHY 207 - torque - J. Hedberg - 2017
Page 12
Quick Question 12Theparticleismovinginthe+ydirection.It'scurrentpositionisgivenby:
Whatdirectionistheangularmomentumvectorpointing?(w.r.ttheorigin)
Angular Momentum - Particle
Angular momentum of a satellite
Ballistic Pendulum
A2.0kgblockhangsfromtheendofa1.5kg,1.0meterlongrod,togetherformingapendulumthatswingsfromafrictionlesspivotatthetopendoftherod.A10gbulletisfiredhorizontallyintotheblock,whereitsticks,causingthependulumtoswingouttoa30°angle.Whatwasthespeedofthebullet?(Solveusingang.momentum)
The 2nd Law
Earlier,werephrased as
Wecanconsidertherotationalequivalenttothislaw:
r = +A −A +0i j k
a)+xb)−xc)+yd)−ye)+zf)−z
Aparticleismovinginthe+ydirection.
Example Problem #9:
F= ma F= dpdt
=τ netdLdt
PHY 207 - torque - J. Hedberg - 2017
Page 13
Gyroscope
Wesawthatthesimplegyroscopeingeneralstaysupright.
Conservationofangularmomentumsaysthatintheabsenceofexternaltorques,theangularmomentumvectorwillnotchange.
A fallen Gyro
Inthecaseofanon-spinninggyroscope,thetorqueduegravitywillcausethegryotofallover,withoutdelay.
A spinning gyro
Ifhowever,thegyroscopeisspinning.
Now,thereisanexternaltorque(duetogravity)soitwillchangetheangularmomentum.But,wecan'tusethetorquetochangethe ofthegyroscope.(it's to )Thus,themagnitudeofthe won'tchange,butthedirectionwill.
Precession
Thischangeinthedirectionof (and )iscalledthegryoscopicprecession.
Therateofprecessioncanbecalculatedbyconsideringthetorque:
But,thetorqueisjust
ω⊥ ω L
L ω
= ⇒ d = dtτ dL
dtL τ
mgr
PHY 207 - torque - J. Hedberg - 2017
Page 14
Thus,theangularprecessionrate canbegivenby:
Ldϕ = dL = mgrdt
Ω= dϕ/dt
Ω=mgr
Iω
PHY 207 - torque - J. Hedberg - 2017
Page 15
Top Related