210-06 Kinetics of Particles

8/2/2019 210-06 Kinetics of Particles

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Introduction

DipakaidilingkungansendiriCurrentchapterintroducestwoadditionalmethodsofanalysis..amFr

Methodofimpulseandmomentum:directlyrelatesforce,Methodofworkandenergy:directlyrelatesforce,mass,=

r

mass,velocity,andtime.velocityanddisplacement.solvedthroughthefundamentalequationofmotion,Previously,problemsdealingwiththemotionofparticleswereDipakaidilingkungansendiriakademik.pesertakuliahDinamikaPartikelTMS-210yangdapatdiunduhdariportalBahanajarinidipakaidilingkungansendiridandisediakansecaragratisbagi2010.EngineeringMechanics:Dynamics,12Ed.,PrenticeHall,NewJersey,


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/3/.Hibbeler,R.C.EngineeringMechanics:Dynamics,6th.Ed.,JohnWiley,2008./2/.Meriam,J.L.;Kraige,L.G.MechanicsforEngineer:Dynamics,5thEd.,McGraw-Hill,NewYork,2008/1/.Beer,F.P.;Johnston,E.R.sepertitercantumberikutini:kepadapengalamanpenulissertamerujukkepadabeberapabukustandarmahasiswaJurusanTeknikMesinUniversitasAndalasyangberdasarkanBahanajarinidibuatuntukmemenuhikebutuhanbahanbacaanbagiparaUnandLDS6210TMSUnandLDS6210TMS

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ContentsMulyadiBur

DipakaidilingkungansendiriANDALASUNIVERSITYStructuralDynamicsLaboratory

DYNAMICSDipakaidilingkungansendiriSampleProblem13.3SampleProblem13.2SampleProblem13.1PowerandEfficiencyApplicationsofthePrincipleofWork&EnergyPrincipleofWork&EnergyWorkofaForceIntroductionENGINEERINGMECHANICSUnandLDS

6210TMSUnandLDS6210TMS()Workofaforceduringafinitedisplacement,WorkofaForce


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2

2s

a6/22

proportionaltodeflection,MagnitudeoftheforceexertedbyaspringisWorkofaForceplottedagainsts.Workisrepresentedbytheareaunderthe1==cosdsFdsF

AA21

=rdFr

r

2ssAA

tcurveofF

21


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U

DipakaidilingkungansendiriUnandLDS6210TMS

Workofaconstantforceinrectilinearmotion,5/22Dimensionsofworkareforce.length

WorkofaForce()()()J1.356lb1ftm1N1J1==joulemagnitudeandsignbutnotdirection.Workisascalarquantity,i.e.,ithascos()xFUD=dsF==rdFdU

ra8/22negativeofareaundercurveofFplottedagainstx,Workoftheforceexertedbythespringisequalto,i.e.,whenthespringisreturningtoWorkoftheforceexertedbyspringispositive22

2

1

12

21


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()xFFUD+-=itsundeformedposition.21Workoftheforceexertedbyspring,()

211x

-=-=2

21kxkxdxkxU

x

1

212


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Workoftheweightisequaltoproductof12()D-=--=yWyyWdyWdyWdzFdyFdxFdUWorkoftheforceofgravity,cosr

1y

2yzyx-=-=++=

21

21UUnitsareWorkoftheforceisWorkofaForceDipakaidilingkungansendiriDipakaidilingkungansendiriUnandLDS6210

TMSUnandLDS6210TMS=0):

WorkofaForceDipakaidilingkungansendiriForceswhichdonotdowork(ds=0orcosa12/22ForceswhichdonoworkareeliminatedfromAllquantitiesarescalarsandcanbeadded

expressionforaccelerationandintegrating.ForceactsnormaltopathanddoesnoWishtodeterminevelocityofpendulumbob10/22

Velocityfoundwithoutdetermining22

glv


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v

2gW21

=

2=+Wl

theproblem.02211=+directly.TUTPr

work.

.Considerwork&kineticenergy.2atA

DipakaidilingkungansendiriApplicationsofthePrincipleofWorkandEnergyhorizontally.weightofabodywhenitscenterofgravitymovesreactionatarollermovingalongitstrack,andmovesalongsurface,

reactionatfrictionlesssurfacewhenbodyincontactreactionatfrictionlesspinsupportingrotatingbody,UnandLDS6210TMSUnandLDS6210TMS

11/22TheworkoftheforceisequaltothechangeinenergykineticmvTTTU9/22

==JmNmUnitsofworkandkineticenergyarethesame:


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dsdvmv12rdtdvMmG==-=12

2mvmvdvvmdsFdvmvdsFdt2dssmkg

21dsdvr

kineticenergyoftheparticle.MmmttmmaFGdrdr

212,=====2

2

r

toAMm

tConsideraparticleofmassmacteduponby21IntegratingfromA


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2==mvT1

12pathshown),occupiesfixedpositionOwhileparticlemfollowsWorkofagravitationalforce(assumeparticleMsWorkofaForceDipakaidilingkungansendiriParticleKineticEnergy:PrincipleofWork&DipakaidilingkungansendiriUnandLDS6210TMS

UnandLDS6210TMSW746

slbft550hp1orsm

N1

dtrdFr

sJrPowerandEfficiencyDipakaidilingkungansendiri1(watt)W1=


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inputpower

outputpowerinputworkkoutputworefficiency===UnitsforpowerareDimensionsofpowerarework/timeorforce*velocity.vFr

=rdt

==dU=

===h16/22


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ft418=x14/22

()=-0lb1151lbft481000x

x

2211=+TUT()()()()xxlb11515sinlb4000lb1500-=+-=

21equalthekineticenergychange.

DeterminethedistancerequiredfortheworktoSampleProblem13.1Dipakaidilingkungansendiri()()lbft481000882.324000sft88U2

s3600h

mi

21ft5280


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1v

UnandLDS6210TMSUnandLDS6210TMS

15/22requiredfortheworktoequal13/22thekineticenergychange.,

DeterminethedistanceEvaluatethechangein2

3W

kineticenergy.lgl

2

SOLUTION:g=+=Wl22SampleProblem13.1v

WP

gW=-WPnn=amF

AsthebobpassesthroughAapplicationofNewtonssecondlaw.


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ofworkandenergywithanrequiressupplementingthemethodCalculatingthetensioninthecordaccelerationofthependulumbob.beappliedtodirectlydeterminetheinclineataspeedofPrincipleofworkandenergycannot

Dipakaidilingkungansendiriautomobileasitcomestoastop.Determinethedistancetraveledbythecausingaconstanttotalbreakingforce60mi/hwhenthebrakesareapplieddrivendowna5oAnautomobileweighing4000lbisDipakaidilingkungansendiri=

2of1500lb.glv2ApplicationsofthePrincipleofWorkandEnergyUnand

LDS6210TMSUnandLDS6210TMS2

2()()()()kg200m2N490m2vF2

12AAC1()=-()()m2m20


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vmFFN490N196225.02211()()2

:

TUTWNFN1962sm81.9kg200C=-+

=+AkAkAm

m()()======AWtoblocksAandB.ApplytheprincipleofworkandenergyseparatelySOLUTION:SampleProblem13.2

forthevelocity.cableforcescancel.Solvecombined,theworkoftheWhenthetworelationsareblocksAandB.andenergyseparatelytoApplytheprincipleofworkSOLUTION:SampleProblem13.2Dipakaidilingkungansendiriweightlessandfrictionless.

=0.25andthatthepulleyism20/22unknownintherelationistheandenergyforthereboundofcompressedandthevelocityisandenergybetweentheinitial18/222velocityatthefinalposition.


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thepackage.TheonlyApplytheprincipleofworktherelationisthefrictionzero.TheonlyunknowninwhichthespringisfullypositionandthepointatApplytheprincipleofwork2()()()()kg300m2N2940m2vF2

1

2BBc1

coefficient.

SOLUTION:()()m2m20vmWF2

SampleProblem13.3:

2211TUTN2940sm81.9kg300c

=+-=+-

=+==BW

Dipakaidilingkungansendiripassesagainthroughthepositionshown.and(b)thevelocityofthepackageasitfrictionbetweenthepackageandsurfaceDetermine(a)thecoefficientofkineticis40mm.andthemaximumdeflectionofthespringavelocityof2.5m/sinthepositionshowncompressed120mm.Thepackagehas


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andisheldbycablessothatitisinitiallyThespringhasaconstantk=20kN/mwhichisslidingonahorizontalsurface.Aspringisusedtostopa60kgpackageDipakaidilingkungansendiriUnandLDS6210TMSUnandLDS6210TMS

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sm43.4=v()()()()()v

Whenthetworelationsarecombined,thework17/22

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()v

2kg500J490012kg300kg200m2N490m2N294012()()()()

kg300m2N2940m2vF12()()()()kg200m2N490m2vF1

=+=-=-


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c=+-Cvelocity.ofthecableforcescancel.SolvefortheSampleProblem13.2Dipakaidilingkungansendirik

theplaneiscoefficientoffrictionbetweenblockAandAafterithasmoved2m.Assumethatthefromrest,determinethevelocityofblockcableasshown.IfthesystemisreleasedTwoblocksarejoinedbyaninextensibleUnandLDS6210TMSUnandLDS6

210TMS()()()J112J37732

ef()2

1

32

21

+=J36.5k323232+-=+=


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UUUm32kg600vmvTT===reboundofthepackage.ApplytheprincipleofworkandenergyfortheSampleProblem13.3

()()()()kkm2()()0J5.187sm5.2kg602

21

k12m-=-=J377m640.0sm81.9kg602


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-=msm103.13=v

32

()v

2kg60J5.360

:

1

=+

3322=+TUT

k()()()J112J377ef212121--=+=UUUmLDS

20.0=km


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=-()()J0.112m040.0N3200N2400maxmin()()()()N3200m160.0mkN20k

()m22/22

DipakaidilingkungansendiriUnand6210TMS

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0J112J377-J5.187:

2211()()N2400m120.0mkN202-=+-=12D+-=1

=+21

TUTe()f


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()xWU1

210max==D+=0min===21====TmvTcompressed.positionandthepointatwhichspringisfully

xPPUxxkPkxP

ApplyprincipleofworkandenergybetweeninitialSOLUTION:SampleProblem13.3DipakaidilingkungansendiriUnand

LDS6210TMS

210-06 Kinetics of Particles

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