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S.Widnall,J.Peraire16.07Dynamics

Fall2008Version2.0

LectureL12- WorkandEnergySofarwehaveusedNewtonssecondlawF =matoestablishthe instantaneousrelationbetweenthesumoftheforcesactingonaparticleandtheaccelerationofthatparticle. Oncetheaccelerationis known, the velocity (or position) is obtained by integrating the expression of the acceleration(orvelocity). Newtons lawandtheconservationofmomentumgiveusavectordescriptionofthemotionofaparticle inthreedimensionsTherearetwosituations inwhichthecumulativeeffectsofunbalanced forcesactingonaparticleareof interesttous. These involve:

a) forces acting along the trajectory. In thiscase, integrationofthe forceswith respect tothedisplacement leadstotheprincipleofworkandenergy.

b) forcesactingoveratime interval. Inthis case, integration ofthe forces with respecttothetime leadstotheprincipleofimpulseandmomentumasdiscussed inLecture9.

It turns out that in many situations, these integrations can be carried beforehand to produceequations that relate the velocities at the initial and final integration points. In this way, thevelocitycanbeobtaineddirectly,thusmakingitunnecessarytosolvefortheacceleration. Weshallseethatthese integratedformsoftheequationsofmotionareveryuseful inthepracticalsolutionofdynamicsproblems.

MechanicalWorkConsider a force F acting on a particle that moves along a path. Let r be the position of theparticlemeasuredrelativetotheoriginO. TheworkdonebytheforceF whentheparticlemovesan infinitesimalamountdr isdefinedas

dW =F dr . (1)

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Thatis,theworkdonebytheforceF,overaninfinitesimaldisplacementdr,isthescalarproductofF anddr. Itfollowsthattheworkisascalarquantity. Usingthedefinitionofthescalarproduct,wehavethatdW =F dr=F dscos,wheredsisthemodulusofdr,andistheanglebetweenF anddr. Sincedr isparalleltothetangentvectortothepath,et,(i.e. dr=dset),wehavethatF et =Ft. Thus,

dW =Ftds , (2)which impliesthatonlythetangentialcomponentoftheforcedoeswork.During a finite increment in which the particle moves from position r1 to position r2, the totalworkdonebyF is

r2 s2W12 = F dr= Ftds . (3)

r1 s1Here,s1 ands2 arethepathcoordinatescorrespondingtor1 andr2.Note UnitsofWorkInthe internationalsystem,SI,theunitofwork istheJoule(J). We have that 1 J= 1 N m. IntheEnglishsystemtheunitofwork isthe ft-lb. Wenotethattheunitsofworkandmomentarethesame. Itiscustomarytouseft-lbforworkand lb-ftformomentstoavoidconfusion.

PrincipleofWorkandEnergyWenowaconsideraparticlemovingalongitspathfrompointr1 topointr2. Thepathcoordinatesatpoints1and2ares1 ands2,andthecorrespondingvelocitymagnitudesv1 andv2.

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Ifwestartfrom(3)anduseNewtonssecond law(F =ma)toexpressFt =mat,wehave

s2 s2 v2 dvds v2 1 1W12 = Ftds= matds= m

dsdtds= mvdv= 2 mv222mv12 . (4)s1 s1 v1 v1Here,wehaveusedtherelationshipatds=vdv,whichcanbeeasilyderivedfromat =v andv=s(see lectureD4).Definingthekineticenergy1,as

1 2T = mv ,2

wehavethat,W12 =T2T1 or T1+W12 =T2 . (5)

Theaboverelationshipisknownastheprincipleofworkandenergy,andstatesthatthemechanicalworkdoneonaparticle isequaltothechange inthekineticenergyoftheparticle.ExternalForcesSincethebodyisrigidandtheinternalforcesactinequalandoppositedirections,onlytheexternalforcesappliedtotherigidbodyarecapableofdoinganywork. Thus,thetotalworkdoneonthebodywillbe

n n (ri)2(Wi)12 = Fi dr ,

i=1 i=1 (ri)1whereFi isthesumofalltheexternalforcesactingonparticlei.WorkdonebycouplesIfthesumoftheexternalforcesactingontherigidbodyiszero,itisstillpossibletohavenon-zerowork. Consider, for instance, a moment M =F a acting on a rigid body. If the body undergoesa pure translation, it is clear that all the points in the body experience the same displacement,

1TheuseofT todenotethekineticenergy,insteadofK, iscustomary indynamicstextbooks

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and,hence,thetotalworkdonebyacouple iszero. Ontheotherhand, ifthebodyexperiencesarotationd,thentheworkdonebythecouple is

a adW =F d+F d=Fad=Md .

2 2

If M is constant, the work is simply W12 = M(21). In other words, the couples do workwhichresults inthekineticenergyofrotation.ExternalandInternalForcesInatypicaldynamicalsystem,theforceF iscomposedoftwoterms: anexternalforceFE andaninternalforceFI. Theexternalforceresultsfromanexternalactorapplyinganarbitraryforceofhischoicetothesystem. Theexternal forcedoeswork andchangesthe energyofthesystem atthe

whim

of

the

actor.

When

the

external

force

is

removed,

the

system

may

oscillate

or

otherwise

movesubjectto internal forces. Internal forcesareoftwotypes: conservative internal forcessuchasgravitywhichconserveenergybut forexamplecantransformkinetic intopotentialenergyandviceversa;andfrictionwhichactsinternallytodissipateenergyfromthesystem. Initialconditionsareoftensetbyappliedexternal forcestothesystem,suchasdoingworkbymovingapendulumthrough an initial angle 0. When they are removed, the system oscillates perhaps conservingenergy iffriction isabsent. Wewillconsiderconservativeforcesshortly.

Example Blockonan inclineConsider a block resting on an incline at position 1 in the presence of gravity. The gravitationalforce acting in the vertical direction is mg. The block is supported on the plane by a normalforce FN = mgsin. We desire to push the block up to position 2. To do this, we must applyanexternal forcetoovercomethecomponentofgravity(internal force)alongtheplaneplusthe

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friction force (internal force)acting to oppose the motion. In this process, the normal force doesnowork.

Thetotaltangentialforceactingontheblock isthenFT =FE +FI (6)

withFE =mgsinmgcos (7)FI = +mgsin+mgcos (8)

for a total force, internal plus external, of zero. Under these assumptions, the block arrives atposition2withzerovelocity. Therefore,by equation(5), nowork isdone. Atfirst,thisdoesnotagree with our intuition. We certainly felt we did work in pushing the block up the plane. Buttheworkdonebythefrictionforceandgravitationalforceexactlycanceledthiswork. Wealsofeelthat having raised the block to a higher position, there is some inherent gain in energy whichcould

be

collected

in

the

future.

This

is

true!

This

result

that

the

total

force

is

zero

resulting

in

noworkandnokineticenergybeinggainedisduetotheexternal applicationof force. Theseexternal forces exactly canceled the internal forces of gravity and friction, driving the total forcetozeroandresulting innototalworkInotherwords,webecameanactor insteadofanobserver.These issueswillbeconsideredfurtherwhenweconsiderconservativeforcesandpotentials

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Example Blockslidingdownan inclineHavingappliedanexternalforceFE tomovetheblocktoahigherpositionontherampswhereitrestswithnokineticenergy,wenowbecomeanobserverandreleaseitwithonlytheinternalforcesof gravity and friction acting. The coefficient of kinetic friction between the surface of the rampand the block is . We want to determine the velocity of the block as a function of the distancetraveledontheramp,s.

Theforcesontheblockare: theweight,mg,thenormalforce,N,and,thefrictionforce,N. WehavethatFn =man andFt =mat. SinceFn =Nmgcosandan =0,wehaveN =mgcos.Thus,Ft =mgsinN =mgsinmgcos,which isconstant. Ifweapplytheprincipleofwork and energy between the position (1), when the block is at rest at the top of the ramp, andtheposition(2),whentheblockhastraveledadistances,wehaveT1 =0,T2 = (mv2)/2,andtheworkdonebyFt issimplyW12 =Fts. Thus,

T1+W12 =T2 , or, mg(sincos)s=2

1mv2 .

Fromwhichweobtain,forthevelocity,v= 2g(sincos)s .

We make two observations: first, the normal force, N, does no work since it is, at all times,perpendiculartothepath,andsecond,wehaveobtainedthevelocityoftheblockdirectlywithouthaving tocarryout any integrations. Notethatanalternative, longerapproachwouldhavebeentodirectlyuseF =ma,andintegratethecorrespondingexpressionfortheacceleration.

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RollingCylinder,FrictionForces,WorkThecylinderrollingonaflatplaneisaverybasicconfigurationindynamics. AsnotedinLecture2, it is a single degree of freedom system with a definite relationship between the position of thecenter ofthecylinder,x0(t)andtherotationangle(t): (t) =x0(t)/R. Thekinematicsoftherolling cylinder are shown in a). Consider a mass point at the edge of the cylinder. The dashedcurveshowsthepathtakenbythismasspoint. Ofsignificanceisthebehaviorneartheplane. ThepointO isaninstantaneouscenterofrotation;thetangentialvelocityofthecylinderiszeroaboutthispoint. Theaccelerationofthemasspoint isnotzero,nor is itsverticalvelocity.

The dynamics of the rolling cylinder is shown in b). If the cylinder moves with constant velocity, nothing more need be said. However, if x0 is not zero, then =, the angular accelerationwill be nonzero. This will require a moment about the center of mass, which in the simplestconfiguration sketched, can only come from friction with the plane. FfR = IG. Since thepoint in contact with the plane is an instantaneous center of rotationdoes not movethis friction force does no work. Also shown in b) are a variety of configurations of rolling cylinders.The solid cylinder has a moment of inertia of IG = 1/2mR2; the cylinder whose mass is concentrated in the rim, has a moment of inertia IG = mr2; the cylinder whose mass point is concentrated in the center has a moment of inertia of zero. Therefore, no moment is required tochange itsangularvelocityand itbehaves likeamasspointmovingona frictionlesssurface. Thecollision of two mass points can easily be realized by the collision of two such rolling cylinders.Note Cylinderrollingdownaramp

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Inparallelwithourdiscussionofablockslidingdownaramp inthepresenceof friction,wenowconsideracylinderrollingdownaramp inthepresenceoffriction. Weassumethatfrictionforcesare large enough to keep the kinematic relationship between the velocity of the cylinder and theangular velocity of the cylinder intact. The cylinder is located at an initial position1 as shown,andisatrest. Itisreleasedandrollsdowntoposition2whereweobserveit. Theforcesactingonthecylinderaregravity,thenormalforceN,andthefrictionforceFf. Althoughthefrictionforceis necessary to keep the kinematic relationship intact, it does no work; as before, the normalforcedoesnowork. Therefore,theonlywork isdonebygravity. Thuswecanwrite

T1+W12 =T2 (9)Theinitialkineticenergyiszero. Theworkdonebygravityismgsins;thefinalkineticenergy,whichincludesboththekineticenergyduetotranslationandthekineticenergyduetorotationisT2 = 1/2mv2+ 1/2IG(v/R)2. Therefore,thefinalvelocity is

2gsinsv= (10)

1 +IG/R2

Note Alternativeexpressions for dWWehaveseeninexpression(2)thataconvenientsetofcoordinatestoexpressdW arethetangential-normal-binormalcoordinates. Alternativeexpressionscanbederivedforothercoordinatesystems.Forinstance,wecanexpressdW =F dr in:

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cartesiancoordinates,dW =Fxdx+Fydy+Fzdz ,

cylindrical(polar)coordinates,dW =Frdr+Frd+Fzdz ,

orsphericalcoordinates,dW =Frdr+Frcos d+Frd .

As an illustration, lets calculate the work done by a constant internal force, such as that due togravity. Theforceonaparticleofmassm isgivenbyF =mgk. Whentheparticlemoves frompositionr1 =x1i+y1j+z1ktopositionr2 =x2i+y2j+z2k,workisdone,andtheworkmaybewrittenas

r2 z2W12 = F dr= mgdz=mg(z2z1).

r1 z1

PowerIn

many

situations

it

is

useful

to

consider

the

rate

at

which

adevice

can

deliver

work.

The

work

perunittime iscalledthepower,P. Thus,

dW drP = =F =F v .

dt dt Theunitofpower intheSIsystem istheWatt(W).Wehavethat1W=1J/s. IntheEnglishsystemtheunitofpoweristheft-lb/s. Acommonunitofpowerisalsothehorsepower(hp),whichisequivalentto550 ft-lb/s,or746W.

Note EfficiencyTheratioofthepowerdeliveredoutofasystem,Pout,tothepowerdeliveredintothesystem,Pin,iscalledtheefficiency,e,ofthesystem.

Poute= .

Pin

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This definition assumes that the energy into andout of the system flows continuously and is notretainedwithinthesystem. Theefficiencyofanyrealmachineisalwayslessthanunitysincethereisalwayssomemechanicalenergydissipatedasheatduetofrictionforces.

ADDITIONALREADING J.B.Marion,S.T.ThorntonClassicalDynamicsofParticlesandSystems,HarcourtBrace,

NewYork,Section2.5 J.L.MeriamandL.G.Kraige,EngineeringMechanics,DYNAMICS,5thEdition

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16.07 Dynamics

Fall 2009

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