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S.Widnall16.07Dynamics
Fall2009Version1.0
LectureL2- DegreesofFreedomandConstraints,RectilinearMotion
DegreesofFreedomDegreesoffreedomreferstothenumberofindependentspatialcoordinatesthatmustbespecifiedtodeterminethe
positionofabody. Ifthebody isapointmass, onlythree coordinates
arerequiredto determineitsposition.
Ontheotherhand,ifthebodyisextended,suchasanaircraft,threepositioncoordinatesandthreeangularcoordinatesarerequiredtocompletelyspecify
itspositionandorientationinspace.
KinematicConstraintsInmanysituationsthenumberofindependentcoordinateswillbereducedbelowthisnumber,eitherbecausethe
numberofspacial dimensions is reduced or because there are
relationshipsspecified among thespatialcoordinates.
Whensettingupproblemsforsolutionitisusefultothinkoftheserelationshipsasconstraints.For
example, if a point mass is constrained to move in a plane (two
dimensions) the number of
spatialcoordinatesnecessarytodescribeitsmotionistwo.
Ifinsteadofbeingapointmass,thisbodyhasextendeddimensions,suchasaflatplateconfinedtoaplane,itrequiresthreecoordinatestospecify
itspositionandorientation:
twopositioncoordinatesandoneangularcoordinate.
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If a particle is confined to move on acurve ineither two or
three dimensions, suchas a bead moving on
awire,thenumberofindependentcoordinatesnecessarytodescribeitsmotion
isone.
Anothersource
of
constraints
on
the
motion
of
particles
is
connections
between
them.
For
example,
the
two
particleconnectedbyacablepassingoverapulleyareconstrainedtomoveinequalandoppositedirections.More
complex arrangements are possible and can be analyzed using these
ideas. Two gears in contact
areconstraintomovetogetheraccordingtotheir individualgeometry.
Acylinderrollingonaplaneisconstrainedintwoways.
Contactwiththeplanereducesthetwo-dimensionalmotion to one spatial
coordinate along the plane, and the constraint of rolling provides
a relationship
betweentheangularcoordinatesandthespatialposition,resulting
inasingledegreeoffreedomsystem.
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InternalForce-BalanceConstraintsAnother type of constraint
occurs when we consider the of a system of particles and the
necessary forcebalancethatoccursbetweentheparts.
TheseconstraintsfollowdirectlyfromNewtonsthirdlaw:
theforceofactionandreactionbetweentwobodiesareequalinmagnitudeandoppositeindirection.
Wewillpursuethese ideas in greater depth later in the course. For
now, we will give a simple example to illustrate
theprinciple.Considerthesystemsshownina)andb).
Systema)consists oftwo massesm incontactrestingona
frictionlessplane inthepresence ofgravity. AforceF is applied to
mass 1, and it is obvious that the two masses will accelerate at a
=F/(2m). If welookatthetwomassesseparately,wecandeterminewhat
internal forcemustexistbetweenthemtocausethe motion. It is clear
that each mass feels a net force ofF/2, since its acceleration is
a=F/(2m). Thisnetforcearisesbecausebetweenthetwomassesthere
isanequalandopposite forceF/2actingacrosstheinterface.
Anotherwaytolookatthisisthattheinterfacebetweenthebodiesisabodyofzeromass,andthereforecanhavenonetforceactinguponitotherwise
itsaccelerationwouldbeinfinite.Systemb) is a bit more complex,
primarily because the forces between one mass and the mass above
itare shear forcesand must besuppliedby friction. Assumingthatthe
frictioncoefficient is largeenough
toacceleratethethreemassesanequalamountgivenbya=F/(3m),bythereasoningwehavediscussed,theforce
balance is as sketched inb): equal and opposite normal forcesF 2/3
on the vertical surfaces,
andequalandoppositeshearforcesF1/3onthehorizontalsurfaces.RectilinearMotionInmanycasewecangetanexactexpressionforthepositionofaparticleasafunctionoftime.
Westartbyconsideringthe simple motion of aparticle along astraight
line. Thepositionof particle Aatany
instantcanbespecifiedbythecoordinateswithoriginatsomefixedpointO.
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Theinstantaneousvelocityisds
v= =v . (1)dt
We willbeusingthe dot notation, to indicate time derivative,
e.g. ()d/dt. Here, apositive v meansthattheparticle
ismovinginthedirectionof increasings,whereasanegativev,
indicatesthattheparticleismovingintheoppositedirection.
Theacceleration is
dv d2sa= =v = = s . (2)
dt dt2The above expression allows us to calculate the speed and
the acceleration if s and/or v are given as afunctionoft,i.e.
s(t)andv(t).
Inmostcaseshowever,wewillknowtheaccelerationandthen,thevelocityandthepositionwillhavetobedeterminedfromtheaboveexpressionsbyintegration.
Determiningthevelocity fromtheaccelerationFrom
a(t)Iftheaccelerationisgivenasafunctionoft,a(t),thenthevelocitycanbedeterminedbysimpleintegrationofequation(2),
t
v(t) =v0 + a(t)dt. (3)t0
Here,v0 isthevelocityattimet0,which isdeterminedbythe
initialconditions.From
a(v)Iftheaccelerationisgivenasafunctionofvelocitya(v),then,wecanstilluseequation(2),butinthiscasewewillsolveforthetimeasafunctionofvelocity,
v dvt(v) =t0 + . (4)
a(v)v0Once the relationship t(v) has been obtained, we can, in
principle, solve for the velocity to obtain v(t).A typical example
in which the acceleration is known as a function of velocity is
when aerodynamic dragforcesarepresent.
Dragforcescauseanaccelerationwhichopposesthemotionand
istypicallyoftheforma(v)
v2 (the sign meansproportional to, that is, a(v) =v2 for some ,
which is not a function ofvelocity).From a(s)Whenthe acceleration
isgivenas a function of s then, we needtouse a combinationof
equations (1) and(2),tosolvetheproblem. From
dv dv ds dva=
dt = ds dt =v ds (5)4
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wecanwriteads=vdv . (6)
Thisequationcannowbeusedtodeterminev asafunctionofs,
v2
(s)=
v20 +2
ss0 a(s)
ds
.
(7)
where,v0,isthevelocityoftheparticleatpoints0.
Here,wehaveusedthefactthat,v v 2 2 2
vdv= d(v2
) = v2
v
20 .v0
v0Aclassicalexampleofanaccelerationdependentonthespatialcoordinates,isthatinducedbyadeformedlinearspring.
Inthiscase,theacceleration isoftheforma(s)s.Of course, when the
acceleration is constant, any of the above expressions (3, 4, 7),
can be employed. Inthiscaseweobtain,
v=v0 +a(tt0), or v2 =v02 +
2a(ss0).Ifa=g,thisreducestothefamiliar
v=v0 +g(tt0), or v2 =v02 + 2g(ss0) .
Determiningtheposition fromthevelocityOnce we know the velocity,
the position can be found by integrating ds = vdt from equation
(1). Thus,whenthevelocityisknownasafunctionoftimewehave, t
s=s0 + v(t)dt. (8)t0
Ifthevelocityisknownasafunctionofposition,thens ds
t=t0 + . (9)v(s)s0
Here,s0 isthepositionattimet0.It
isworthpointingoutthatequation(6),canalsobeusedtoderiveanexpressionforv(s),givena(v),
s vv
ss0 = ds= a(v)dv . (10)s0 v0This equation canbeused
wheneverequation(4) isapplicable and gives v(s) insteadof t(v). For
thecaseofconstantacceleration,eitherofequations(8,9),canbeusedtoobtain,
1s=s0 +v0(tt0) + a(tt0)2 .
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Inmanypracticalsituations,itmaynotbepossibletocarryouttheaboveintegrationsanalyticallyinwhichcase,numerical
integration isrequired. Usually,numerical
integrationwillalsoberequiredwheneitherthevelocityortheaccelerationdependonmorethanonevariable,i.e.
v(s, t),or,a(s, v).
Example
Reentry,Ballistic
Coefficient,
Terminal
Velocity
TerminalVelocityTerminal velocity occurs when the acceleration
becomes zero and the velocity Consider an
air-droppedpayloadstartingfromrest.
Theforceonthebodyisacombinationofgravityandairdragandhastheform
1F =mg v2 (11)CDA
2ApplyingNewtons lawandsolvingfortheaccelerationaweobtain
a=g2
1v2CD
mA
(12)Thequantity CDmA characterizes the combined effect of
bodyshape and mass on the acceleration; it
isanimportantparameterinthestudyofreentry;itiscalledtheBallisticCoefficient.
Itisdefinedas=
CDmA.Unlikemanycoefficientsthatappearinaerospaceproblems,theBallisticCoefficientisnotnon-dimensional,buthasunitsofmass/length2
orkg/meters2 inmksunits.
Also,insomeapplications,theballisticcoefficientisdefinedastheinverse,B=
CDmA,so itpaystobecareful initsapplication.
Equation13thenbecomes
1a=g v2/ (13)
2Terminalvelocityoccurswhentheforceofgravityequalsthedragontheobjectresultinginzeroacceleration.Thisbalancegivestheterminalvelocityas
vterminal = 2gm
CDA =2g
(14)For the Earth, atmospheric density at sea level is =
1.225kg/m3; we shall deal with the variation
ofatmosphericdensitywithaltitudewhenweconsideratmosphericreentryofspacevehicles.
Typicalvalueof rangefrom=1(AssumingCD =.5,atennisballhas=35.)
to=1000forareentryvehicle.
Asanexample,consideratypicalcaseof=225,wherethevariousparametersthengivethefollowingexpressionfortheacceleration
a=g0.002725v2m/s.Here g = 9.81m/s2, is the acceleration due to
gravity andv isthe downwardvelocity. It is clear
fromthisexpressionthatinitiallytheaccelerationwillbeg.
Therefore,thevelocitywillstarttoincreaseandkeeponincreasinguntila=0,atwhichpointthevelocitywillstayconstant.
Theterminalvelocityisthengivenby,
0 =g0.002725v2 or vf =60m/s .f
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Todeterminethevelocityasafunctionoftime,theaccelerationcanbere-writtenintroducingtheterminalvelocityas,a=g(1(v/vf)2)
. Wethenuseexpression(4),andwrite
1 v dv 1 v 1/2 1/2 vf vf +vt= = ( + )d v= ln .
g 0 1(v/vf)2 g 0 1 + (v/vf) 1(v/vf) 2g vfvSolvingforv weobtain,
2gt/vf
v=vf e2gt/vf
1m/s.
e + 1Wecaneasilyverifythat for larget,v=vf. Wecanalsofindouthow
longdoes
ittakeforthepayloadtoreach,say,95%oftheterminalvelocity,
vf 1.95t= ln = 11.21s .
2g 0.05To obtain an expression for the velocity as a function of
the traveled distance we can use expression (10)andwrite
v 21 v dv vs= g 1(v/vf)2 =2
fgln(1(v/vf)2).0
Solvingforv weobtain v=vf 1e2gs/v2 m/s .f
We see that for, say, v = 0.95vf, s = 427.57m. This is the
distance traveled by the payload in
11.21s,whichcanbecomparedwiththedistancethatwouldbetraveledinthesametime
ifweweretoneglectairresistance,sno drag =gt2/2=615.75m.
Example Spring-masssystemHere, we consider a mass allowed to
move without friction on a horizontal slider and subject to the
forceexertedbyalinearspring.
Initiallythesystemisinequilibrium(noforceonthespring)ats=0.
Suddenly,themass isgivenavelocityv0 andthen thesystem is left
freeto oscillate. Weknow that the effectofthespring
istocauseanaccelerationtothebody,opposingthemotion,oftheforma=s,where
>0 isaconstant.
Usingequation(7),wehavev2 =v
0
2s2 .Thedisplacementcannowbeobtainedusingexpression(9),
s ds 1 st= = arcsin ,
20 v0s2 v0
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whichgives,s=v0
sin
t .Finally,thevelocityasafunctionoftime issimply,v=v0cost.
Werecognizethismotionasthatofanundampedharmonicoscillator.
References[1] R.Dugas, AHistoryofMechanics,Dover,1988.
ADDITIONALREADINGJ.L.MeriamandL.G.Kraige,EngineeringMechanics,DYNAMICS,5thEdition
1/1,1/2,1/3,1/4,1/5(EffectorAltitudeonly),1/6,1/7,3/2
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16.07 Dynamics
Fall 2009
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