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Admissible Variations/Virtual Displacements
2.003J/1.053JDynamicsandControlI,Spring 2007
PaulaEcheverri,ProfessorThomasPeacock
4/4/2007
Lecture
14
Lagrangian
Dynamics:
Virtual
Work
and
Generalized
Forces
Reading: Williams,Chapter5
L=T − V � � d ∂L ∂L dt ∂q˙i − ∂qi =Qi
Allqi arescalars.
qi: GeneralizedCoordinates
L: Lagrangian
Qi: GeneralizedForces
Admissible
Variations/Virtual
Displacements
VirtualDisplacement:
Admissiblevariations: hypothetical(notreal)smallchangefromone geometri
callyadmissiblestatetoa nearbygeometricallyadmissible state.
Bead
on
Wire
2
Admissible Variations/Virtual Displacements
Bothδxandδy areadmissiblevariations. Hypotheticalgeometricconfiguration
displacement. δ �=d δx �=dx dximpliestinvolved. y=f(x) df dy= · dx dx df(x) δy= · δx dx
Generalized
Coordinates
Minimal,complete,and independentset ofcoordinates
sis referredtoas complete: capableof describingallgeometric configurations
atall times.
s is referred to as independent: If all but one coordinate is fixed, there is a
continuousrangeofvaluesthatthefreeonecantake. Thatcorrespondsto the
admissiblesystemconfigurations.
Example: 2-Dimensional Rod
Figure2: 2D rodwith fixedtranslation inxand y but freeto rotate aboutθ.
� � � � 3 Virtual Work
Ifwefixxandy,wecanstillrotateina rangewithθ.
#degreesoffreedom=# ofgeneralizedcoordinates: Truefor2.003J.Truefor
HolonomicSystems.
Lagrange’sequationsworkforHolonomicsystems.
Virtual
Work
W = fi· dri← ActualWork
i
i=forcesactat thatlocation
δW = fi· δri ← VirtualWork
i
f =fapplied+fconstrained
i i i
Constrained: Friction in roll. Constraint to move on surface. Normal forces.
Tension,rigidbodyconstraints.
δw= fappi · δri=0 atequilibrium
i
Noworkdonebecausenomotionin directionofforce. Novirtualwork.
fi= 0
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Virtual Work
Example:
Hanging
Rigid
Bar
Figure3: Hanging rigid bar. Thebaris fixedtranslationallybut is subjectto
a force, F. It therefore can displace itself rotationally about its pivot point.
FigurebyMITOCW.
Displacement: δyA= −aδθjˆ δyB= −lδθjˆ Forces: = −F jˆ R=Rjˆ
Twoforcesapplied: i= 2
δw=F lδθ − Raδθ= 0 F l
R= at equilibrium
a
CouldalsohavetakenmomentsaboutO.
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Virtual Work
Example:
Tethered
Cart
Figure4: Tetheredcart. Thecartisattachedtoatetherthatisattachedtothe
wall. FigurebyMITOCW.
δw=F δyB− Rδxc= 0 yB =lsinθ df(x) Usingδy= dx δxc δyB=lcosθδθ δxc= −2lsinθδθ (−F lcosθ+ 2Rsinθ)δθ= 0 F
−F lcosθ+ 2Rsinθ= 0 ⇒ R= at equilibrium
2 tanθ
Figure 5: Application of Newton’s method to solve problem. The indicated
� �
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Generalized forces for Holonomic Systems
Generalized
forces
for
Holonomic
Systems
Inanholonomic system,thenumber of degreesoffreedom equals thenumber
ofgeneralizedcoordinates.
δw= fi· δri= Qiδqj i
i=number ofappliedforces: 1 ton
j =number ofgeneralizedcoordinates
ri =ri(q1, q2, . . . qj )
ri: Positionofpointwhereforce isapplied
m �∂ri δri= δqj ∂qj j Substitute: n m m� n � � �∂ri � � ∂ri fi · δqj = f ·∂qj ∂qj i∂qj i j j i n � ∂ri Qj = fi· GeneralizedForces ∂qj i fi=fNCi +fCONS.i
fCONS.: Gravity,Spring,andBuoyancyareexamples;PotentialFunctionExists.
i fCONS. = − ∂V ∂r Example: Vg =mgz,r=zjˆ fg= −mg∂zdzjˆ = −mgjˆ ∂ri ∂V ∂r ∂V ficons.· = − = − ∂q ∂r ∂qj ∂qj
Theconservativeforcesarealreadyaccountedforbythepotentialenergyterm
�
� �
Example: Cart with Pendulum, Springs, and Dashpots 7
n QN Cj = fN C ∂ri i · ∂q j i d ∂L ∂L − =QN C dt ∂q˙j ∂qj j Lagrange’sEquation
QN Cj =nonconservativegeneralizedforces
∂L contains ∂V .
∂qj ∂qj
Example:
Cart
with
Pendulum,
Springs,
and
Dashpots
Figure6: Thesystem containsa cartthat has a spring(k) and a dashpot(c)
attachedtoit. Onthecartisapendulumthathasatorsionalspring(kt)anda
torsionaldashpot (ct). There isaforce appliedto mthatis afunction oftime
F =F(t) Wewill model thesystemas 2 particlesin 2 dimensions. Figureby
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Example: Cart with Pendulum, Springs, and Dashpots
4degreesoffreedom: 2constraints. Cart movesin only1 direction. Rodfixes
distanceofthe2particles.
Thus,thereareanet2degreesoffreedom. For2.003J,allsystemsareholonomic
(thenumberofdegreesoffreedomequalsthenumberofgeneralizedcoordinates).
q1=x
q2=θ
Figure7: Forcesfeltbycartsystem. Figure byMITOCW.
F1: DamperandSpring in −xdirection
−(kx+cx˙ )ˆı F2: Twotorques: τ= −(ktθ+ctθ˙kˆ F3: F3=F0 sinωtˆı rA=xˆı=q1ˆı ← r1
rB=rA +rB/A= (x+lsinθ)ˆı − lcosθjˆ← r3
=θkˆ (Torquecreatesangulardisplacement) =q2kˆ
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Example: Cart with Pendulum, Springs, and Dashpots
Q1: ∂r1 ∂q1 =1ˆı, ∂r2 ∂q1 =0, ∂r3 ∂q1 =1ˆı ∂r1 ∂q2 =0, ∂r2 ∂q2 =1ˆk, ∂r3 ∂q2 Q1= −cq˙1 +F0 sinωt =lcosq2ˆı+lsinq2jˆ Q2= −ctq˙2 +F0 sinωt · lcosq2
Withthegeneralizedforces,wecanwritetheequationsofmotion.
Kinematics
M: m: rM ˙ rM ¨ rM =xˆı =xˆ˙ı =x¨ˆı rm=(x+lsinθ)ˆı − lcosθˆj ˙ rm=(x˙ +lcosθ · θ˙)ˆı+lsinθθ˙ˆj ¨rm=(¨x+l(cosθ)θ¨ − l(sinθ)θ˙2)ˆı+(l(sinθ)θ¨+l(cosθ)θ˙2)ˆj
GeneralizedCoordinates: q1=xandq2=θ.
Lagrangian
TM = L=T − V T =TM+Tm 1 2M(r˙M · r˙M )= 1 2Mx˙ 2 Tm= = 1 2m(r˙m· r˙m) 1 2m(x˙ 2+2lx˙cosθθ˙+l2θ˙2) (1) (2)10
Example: Cart with Pendulum, Springs, and Dashpots
T =1 2Mx˙ 2+1 2m(x˙ 2+2lx˙cosθθ˙+l2θ˙2) V =VM,g+MM,k+Vm,g+Vm,kt =M g(0)+1 2k(r˙M · r˙M)+mg(−lcosθ)+ 1 2ktθ 2 (3) (4) Symbol VM,g VM,k Vm,g Vm,kt PotentialEnergy GravityonM SpringonM Gravityonm TorsionalSpringonm V = 1 2kx 2+(−mglcosθ)+1 2ktθ 2 SubstituteinL=T− V L=1 2Mx˙ 2+1 2m(x˙ 2+2l˙ xθ˙cosθ+l2θ˙2) − 1 2kx 2+ mglcosθ − 1 2ktθ 2
Equations
of
Motion
Use d dt � ∂L ∂q˙i � − � ∂L ∂qi �=Ξi toderivetheequationsofmotion. Ξi=Qi.
Frombefore,Ξx=F0 sinω0t − cx˙ andΞθ=F0(sinωt)lcosθ − ctθ˙.
For Generalized Coordinate x
δx �=0 andδθ=0. UnitsofForce.
� � ∂L ∂x = −kx ∂L ∂x˙ =(M +m)x˙+ml(cosθ)θ˙ d dt � ∂L ∂x˙ � =(M+m)¨x+mlθ¨cosθ+mL(− sinθ)θ˙2 d ∂L ∂L
− = (M+m)¨x+mlθ¨(cosθ) +ml(− sinθ)θ˙2+kx=F0 sinωt − cx˙
� �
� �
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Example: Cart with Pendulum, Springs, and Dashpots
For Generalize Coordinate θ
δx=0 andδθ �=0. UnitsofTorque.
∂L =mlx˙θ˙(− sinθ) − mglsinθ − ktθ ∂θ ∂L =mlx˙ cosθ+ml2θ˙ ∂θ˙ d ∂L =mlx˙ (− sinθ)θ˙ +mlx¨ cosθ+ml2θ¨ dt ∂θ˙ d ∂L ∂L
− =mlx˙θ˙(− sinθ)+mlx¨ cosθ+ml2θ¨−mlx˙θ˙(− sinθ)+mglsinθ+ktθ
dt ∂θ˙ ∂θ
d ∂L ∂L
− = mlx¨ cosθ+ml2θ¨ +mglsinθ+ktθ=F0(sinωt)lcosθ − ctθ
