Download pdf - −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

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Page 1: −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

PARTIIIProblemsProblem1Acomputerdiskstartsrotatingfromrestatconstantangularacceleration.Ifittakes0.750stocompleteitssecondrevolution:a)Howlongdoesittaketocompletethefirstcompleterevolution;b)Whatistheangularacceleration?ANSWER:a)1.81s;b) α =3.84rad i s−2 Problem 2 Amachinist turns the power on to a grindingwheel, at rest, at time t = 0 s. Thewheelacceleratesuniformlyfor10sandreachestheoperatingangularvelocityof42rad/s.Thewheelisrunatthatangularvelocityfor39sandthenpowerisshutoff.Thewheeldeceleratesuniformlyat2.6rad/s2untiltheitstops.Inthissituation,theangularaccelerationofthewheelbetweent=0sandt=10sisclosestto:a)5.9rad/s2b)7.6rad/s2c)4.2rad/s2d)5.0rad/s2e)6.7rad/s2

PartialSolution:Use

�

ωz = ω0z + α zt→α z = ωz −ω0z

t with

�

ω0z = 0and

�

ωz = 42 rads

Answer:cProblem3AlighttriangularplateOABisinahorizontalplane.Threeforces,F1=3N,F2=1N,andF3=9N,actontheplate,whichispivotedaboutaverticalaxesthroughpointO.InFigurebelow,themagnitudeofthetorqueduetoforceF1abouttheaxisthroughpointOisclosestto:

PartialSolution:Hereuse

!τ = !r ×

!F thatgivesτ1 = rF1 sin60

0 ,whereyoushouldbeabletodeducethattheanglebetween

!r and !F1 is 60! answere)

Problem4***Inthediagramonthenextpage,themassisreleasedfromrest.Asitdescendstheroperotatesthepulleyandthehollowspherewithoutslipping.T1isthetensioninthebottompartoftherope,andT2isthetensioninthetoppartoftherope,with

T1 ≠T2 .i) Drawafree-bodydiagramonm,anduseNewton’ssecondlaw(equation

5-1)toobtain,theequation,T1 −mg= −ma ,whereaistheacceleration.ii) Drawafree-bodydiagramonthepulleythatincludesthefourforces

actingonit,anduseNewton’ssecondlaw(equation10-45)forrotationtoobtaintheequationT1r −T2r = Iα p = I /r2( )α pr .

a)1.1N•mb)1.4N•mc)0.90N•md)1.8N•me)1.6N•m

Page 2: −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

iii) Drawafree-bodydiagramonthehollowthatincludesthethreeforcesactingonit,anduseNewton’ssecondlaw(equation10-45)forrotationtoobtaintheequationT2R = Isα s = 2/3( )MR2α s .

iv) Usetheno-slipconditions,a= rα p anda= Rα s ,tofindtheacceleration

ofm,andtheangularaccelerationofthesphereandpulley,aswellasthetensionT1andT2.Compareyouranswerswiththeonesyouobtainbytheenergymethod.ANSWER: a=1.2mi s−2 ;T1=36.9NandT2=8.8N.

Problem5Ablockofmass5.00kgisreleasedfromrest,slidesdownasurfaceinclinedat !9.36tothehorizontal.Thecoefficientofkineticfrictionis 1.0=kµ .AstringattachedtotheblockiswrappedaroundasolidcylindricalflywheelofM=25.0kgandR=0.15m.Thereisfrictionbetweentheropeandtheflywheelsothattheropedoesnotslipagainstthewheel.

A)Drawafreeenergydiagramontheblockshowingallforcesactingonit,andhencedeterminethekineticforceoffrictionontheblock.B)Usingtheconservationoftotalenergyorthework-energytheorem,determinethelinearspeedoftheblockafterithasslid1.1mdowntheincline(asshownindiagram).Don’tforgetthattheflywheelwillbespinning.C)Calculatetheangularvelocityoftheflywheelaftertheblockhasslid1.1m?ANSWER:A)fk =3.92N ;B)UseconservationofenergyWext = − fkd = ΔEmech = ΔU +ΔK = −mgh+ 1/2( )mv2 + 1/2( )Iω 2 ,

d=1.1m, h= dsin36.9! ,m=5kg,I = 1/2( )MR2 ,M=25kg,R=.15m,withno-slipconditionωR = v , v =1.1mi s−1 ;C)ω =7.441s−1 .

36.9°

L

Solidcylinderwithmomentofinertia

L=1.1m

HangingBoxm=4.3kgPulley, ,r=0.099m

HollowSphere, ,

M=11kg,R=0.820

Page 3: −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

Problem6***In the diagram below, a uniform spherical boulder starts from rest and rolls down a 50.0 m high hill. The top half of the hill is rough so that the boulder rolls without slipping, but the bottom half is covered with ice, and so is frictionless. What is the translational speed of the boulder when it reaches the bottom of the hill? ANSWER: 29 m/s

Problem7Indiagrambelowa2kgrockisatpointPtravelinghorizontallywithaspeedof12m/s.Atthisinstantwhatisthemagnitudeanddirectionoftheangularmomentum?Iftheonlyforceactingontherockisitsweight,whatistherateofchange(magnitudeanddirection)oftheangularmomentum?AngularMomentum

!L = !r × !p = m!r × !v validforpointparticlew.r.t.pointO

!r ,r=8m

36.9°143.1°

!v Forthemagnitude

L =

!L = mvr sin143.1" = 2kg ×12m / s × 8m × .6

L=115.2kg-m2/s.

25.0m

25.0m

Solidsphere(table9.2)

+ Clockwiseispositive

Point1

Point2

Point3

RoughRollwithoutslipping

SmoothSphereslips

M–massofsphereR–radiusofsphere

Directionperpendiculartox-yplaneindicates+zoutofthepageindicates–zintopage

Also+xright+yup

Usingtherighthandruleonitiseasytoseethatthe

directionof is–zorintothepage

Page 4: −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

Torqueduetogravityonparticlew.r.t.pointO.

!τ = !r ×

!Fg ,Fg = mg = 2kg × 9.8m / s

2 = 19.6N !r ,r=8m

53.1°36.9°

!Fg

Sincetherateofchangeofangularmomentum !τ = d

!L / dt hasoppositedirection

(+z)comparedtothedirectionofthecurrentangularmomentum !L = !r × !p = m!r × !v

(-z),theangularmomentumisdecreasing.Canyouseethesimilaritywithourmuchearlierdiscussiononlinearkinetics?FINALCOMMENTANDADVICE:Angularmomentum

!L andTorque

!τ dependon

theorigin(O).Forexample, !L = 0 and

!τ = 0maybezerofororiginO,butnonzero

!L ≠ 0 and

!τ ≠ 0 inanotheroriginO/.Studyangularmomentumproblemsand

staticequilibriumproblems.Problem8A carousel has a radius of 3.0m and amoment of inertia of IC = 8000kg •m2 , forrotation about axis perpendicular to the its center. The carousel is rotatingunpoweredandwithoutfrictionwithanangularvelocityof1.2rad/s.An80-kgmanruns with a velocity of 5.0 m/s, on a line tangent to the rim of the carousel,overtakingit.Themanrunsontothecarouselandgrabsholdofapoleontherim.

a)Beforethecollision,whatisthemagnitudeoftheangularmomentumoftherotatingcarousel,

!LC ,withrespecttothecenterofthecarousel?Whatisthe

directionof !LC ?ANSWER:

LC = 9600 kg im2

s,direction ⊙ outofpage,+z,or

!LC = 9600kg im2

s⎛

⎝⎜⎞

⎠⎟k̂ b)Beforethecollision,whatisthemagnitudeoftheangular

momentoftherunning80-kgman, !LM ,withrespecttothecenterofthecarousel?

Usingtherighthandruleon itiseasytoseethatthedirectionof is+zoroutthepage.

Thiscanbeexpressedinadifferentunit

Usingsecondlawforrotationintermsofangularmomentum

.Hencethenettorqueistherateofchangeofangular

momentum

+y

+x

Directionperpendiculartox-yplaneindicates+zoutofthepageindicates–zintopage

USETHISCONVENTIONTOINDICATETHEDIRECTIONOFANGULARMOMENTUM

Page 5: −ω →α t - alinhana.lakeheadu.caalinhana.lakeheadu.ca/PracticeExamF2016PartIII.pdf... , the magnitude of the torque due to force F ... τ=! r ×! Fthat gives τ 1= ... 2 is the

Whatisthedirectionof !LM ?ANSWER:

LM = 1200 kg im2

s,direction ⊙ outofpage

(+z),or

!LC = 1200kg im2

s⎛

⎝⎜⎞

⎠⎟k̂ .c)Afterthecollisionwhenthemanisonthe

carousel,whatisthemagnitudeofthefinalangularvelocityofthecarousel(withthemanonit),ω fC ?Whatisthedirectionofthefinalangularvelocity

!ω fC ?Note:

Itotal = IC +mR2 .ANSWER:ω fc = 1.24

radsoutofpage ⊙ or+z,

!ω fc = 1.24

radsk̂


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