Notes for torque, J. Hedberg ©...

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∥

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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

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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

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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⊥

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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

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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

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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∘

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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%

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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

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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

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+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

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+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∘

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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

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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

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Thus,theangularprecessionrate canbegivenby:

Ldϕ = dL = mgrdt

Ω= dϕ/dt

Ω=mgr

Iω

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