1
Lagrange EquationsUse kinetic and potential energy to solve for motion!
Referenceshttp://widget.ecn.purdue.edu/~me563/Lectures/EOMs/Lagrange/In_Focus/page.html
System Modeling: The Lagrange Equations (Robert A. Paz: Klipsch School of Electrical and Computer Engineering)
Electromechanical Systems, Electric Machines, and Applied Mechatronics by Sergy E. Lyshevski, CRC, 1999.
Lagrange’s Equations, Massachusetts Institute of Technology @How, Deyst 2003 (Based on notes by Blair 2002)
2
We use Newton's laws to describe the motions of objects. It works well if the objects are undergoing constant acceleration but they can
become extremely difficult with varying accelerations. For such problems, we will find it easier to express the solutions with
the concepts of kinetic energy.
3
Modeling of Dynamic Systems
Modeling of dynamic systems may be done in several ways:
Use the standard equation of motion (Newton’s Law) for mechanical systems.
Use circuits theorems (Ohm’s law and Kirchhoff’s laws: KCL and KVL).
Today’s approach utilizes the notation of energy to model the dynamic system (Lagrange model).
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• Joseph-Louise Lagrange: 1736-1813.• Born in Italy and lived in Berlin and Paris.• Studied to be a lawyer.• Contemporary of Euler, Bernoulli, D’Alembert, Laplace, and Newton.• He was interested in math.• Contribution:
– Calculus of variations.– Calculus of probabilities.– Integration of differential equations– Number theory.
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Equations of Motion: Lagrange Equations
• There are different methods to derive the dynamic equations of adynamic system. As final result, all of them provide sets of equivalent equations, but their mathematical description differs with respect to their eligibility for computation and their ability to give insights into the underlying mechanical problem.
• Lagrangian method, depends on energy balances. The resulting equations can be calculated in closed form and allow an appropriate system analysis for most system applications.
• Why Lagrange:– Scalar not vector.– Eliminate solving for constraint forces (what holds the system together)– Avoid finding acceleration.– Uses extensively in robotics and many other fields.– Newton’s Law is good for simple systems but what about real systems?
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Mathematical Modeling and System DynamicsNewtonian Mechanics: Translational Motion
• The equations of motion of mechanical systems can be found using Newton’s second law of motion. F is the vector sum of all forces applied to the body; a is the vector of acceleration of the body with respect to an inertial reference frame; and m is the mass of the body.
• To apply Newton’s law, the free-body diagram (FBD) in the coordinate system used should be studied.
∑ = aF m
size". sticcharacteri" to problem thereduce tothese
eliminate then and forces constraint for the solve ma, F equate
,directions threeallin onsaccelerati find that werequiresapproach Newton
=
7
Translational Motion in Electromechanical Systems
• Consideration of friction is essential for understanding the operation of electromechanical systems.
• Friction is a very complex nonlinear phenomenon and is very difficult to model friction.
• The classical Coulomb friction is a retarding frictional force (for translational motion) or torque (for rotational motion) that changes its sign with the reversal of the direction of motion, and the amplitude of the frictional force or torque are constant.
• Viscous friction is a retarding force or torque that is a linear function of linear or angular velocity.
Fcoulomb :Force
8
Newtonian Mechanics: Translational Motion
• For one-dimensional rotational systems, Newton’s second law of motion is expressed as the following equation. M is the sum of all moments about the center of mass of a body (N-m); J is the moment of inertial about its center of mass (kg/m2); and α is the angular acceleration of the body (rad/s2).
αjM =
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Newton’s Second Law
The movement of a classical material point is described by the second law of Newton:
up. come willsderivative time thirdradiation, EM todueenergy losespoint material a If oo.equation t in theappear willsderivative first time
usually system in the dissipated isenergy if Indeed equation.in that appear shouldr of derivative timesecond aonly ion why justificat no is There effects. icrelativist and quantum
neglect to wasone ifeven course, of on,idealisatian isNewton of law second The fields. nalgravitatioor waves,
neticelectromag with nsinteractioor particles,other with nsinteractioaccount into by taking calculated bemay which field, force a represents t)F(r,Vector
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= trFdt
trdm
10
Energy in Mechanical and Electrical Systems• In the Lagrangian approach, energy is the key issue. Accordingly,
we look at various forms of energy for electrical and mechanicalsystems.
• For objects in motion, we have kinetic energy Ke which is always a scalar quantity and not a vector.
• The potential energy of a mass m at a height h in a gravitational field with constant g is given in the next table. Only differences in potential energy are meaningful. For mechanical systems with springs, compressed a distance x, and a spring constant k, the potential energy is also given in the next table.
• We also have dissipated energy P in the system. For mechanical system, energy is usually dissipated in sliding friction. In electrical systems, energy is dissipated in resistors.
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Electrical and Mechanical Counterparts“Energy”
ResistorDamper / Friction0.5 Bv2
DissipativeP
Capacitor0.5 Cv2 = q2/2C
Gravity: mghSpring: 0.5 kx2
PotentialV
InductorMass / Inertia0.5 mv2 / 0.5 jω2
Kinetic (Active) Ke
ElectricalMechanicalEnergy
22
21
21 qRRi &=
22
21
21 qLLi &=
12
LagrangianThe principle of Lagrange’s equation is based on a quantity called
“Lagrangian” which states the following: For a dynamic system in which a work of all forces is accounted for in the Lagrangian, an admissible motion between specific configurations of the system at time t1 and t2
in a natural motion if , and only if, the energy of the system remains constant.
The Lagrangian is a quantity that describes the balance between no dissipative energies.
equation aldifferentiorder -second a isequation above that theNoteequations. threebe will theres,coordinate dgeneralize threeare thereIf system on the acting (forces)
inputs external dgeneralize are );dissipated isenergy at which rate (halffunction power is
:Equation sLagrange'
;21
energy) potential theis energy; kinetic theis (
2
i
iiii
e
ee
QP
QqP
qL
qL
dtd
mghVmvK
VKVKL
=∂∂
+∂∂
−
∂∂
==
−=
&&
13
Generalized Coordinates• In order to introduce the Lagrange equation, it is important to first
consider the degrees of freedom (DOF = number of coordinates-number of constraints) of a system. Assume a particle in a space: number of coordinates = 3 (x, y, z or r, θ, φ); number of constrants= 0; DOF = 3 - 0 = 3.
• These are the number of independent quantities that must be specified if the state of the system is to be uniquely defined. These are generally state variables of the system, but not all of them.
• For mechanical systems: masses or inertias will serve as generalized coordinates.
• For electrical systems: electrical charges may also serve as appropriate coordinates.
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Cont..
• Use a coordinate transformation to convert between sets of generalized coordinates (x = r sin θ cos φ ; y = r sin θsin φ ; z = r cos θ ).
• Let a set of q1, q2,.., qn of independent variables be identified, from which the position of all elements of the system can be determined. These variables are called generalized coordinates, and their time derivatives are generalized velocities. The system is said to have ndegrees of freedom since it is characterized by the ngeneralized coordinates.
• Use the word generalized, frees us from abiding to any coordinate system so we can chose whatever parameter that is convenient to describe the dynamics of the system.
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For a large class of problems, Lagrange equations can be written in standard matrix form
=
∂∂
∂∂
+
∂∂
∂∂
∂∂
∂∂
n
nnn
f
f
qP
qP
qL
qL
qL
qL
dtd
.
.
.
.
.
. -
.
.1111
&
&
&
&
16
Example of Linear Spring Mass System and Frictionless Table: The Steps
m
x
0 : togetherall Combine
; ; :sderivative theDo
0 :Equation sLagrang'
21
21 :Lagrangian 22
=+=∂∂
−
∂∂
−=∂∂
=
∂∂
=∂∂
=∂∂
−
∂∂
−=−=
kxxmqL
qL
dtd
kxqLxm
qL
dtdxm
qL
qL
qL
dtd
kxxmVKL
ii
iii
ii
e
&&&
&&&
&&
&
&
k
17
Mechanical Example: Mass-Spring Damper
( )
( )
2
22
2
2
21
21
21
2121
xBP
xhmgKxxmVKL
xhmgKxV
xmKe
e
&
&
&
=
+−−=−=
++=
=
xBmgKxxm
xBmgKxxmdtd
xBx
xhmgKxxmx
xhmgKxxmxdt
dxP
xL
xL
dtdf
fQxq
&&&
&&
&&
&
&&
&&
+++=
+−−−=
∂∂
++−−∂∂
−
+−−∂∂
=
∂∂
+∂∂
−
∂∂
=
==
)()()(
)21())(
21
21(
)))(21
21((
:equation Lagrange the write we, force applied with the thusand , coordinate dgeneralize thehave We
2
222
22
18
Electrical Example: RLC Circuit
2
22
2
2
21
21
21
2121
qRP
qC
qLVKL
qC
V
qLK
e
e
&
&
&
=
−=−=
=
=
equation KVLjust is This capacitor. afor and
)(
)21()
21
21())
21
21((
have we, force applied with theand (charge), coordinate dgeneralize thehave We
22222
c
c
Cvqqi
RivdtdiLqR
CQqLqR
CQqL
dtd
qRq
qC
qLq
qC
qLqdt
dqP
qL
qL
dtdu
uQq
==
++=++=++=
∂∂
+−∂∂
−−∂∂
=
∂∂
+∂∂
−
∂∂
=
=
&
&&&&&
&&
&&&
&&
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Electromechanical System: Capacitor MicrophoneAbout them see: http://www.soundonsound.com/sos/feb98/articles/capacitor.html
( )
( ) 2222
2222
o
2222
21
21
21
21
21
21 ;
21
21
separation plate theis - plate, theof area theis (F/m),air theofconstant dielectric theis
21
21 ;
21
21
KxqxxA
xmqLL
xBqRPKxqxxA
V
xxAxxAC
KxqC
VxmqLK
o
o
o
e
−−−+=
+=+−=
−
=
+=+=
ε
ε
εε
&&
&&
&&
m)equilibriu from nt displaceme and charge :mechanical and l(electrica
freedom of degrees twohas system This
xq
20
( )
( ) vqxxA
qRqL
fA
qKxxBxm
qRqP
Aqxx
qLqL
qL
xBx
KxA
qxLxm
xL
o
o
=−++
=−++
=∂∂−
=∂∂
=∂∂
=∂∂
−=∂∂
=∂∂
1
22
equations Lagrange twoobtain the Then we
; ;
P ;2
;2
ε
ε
ε
ε
&&&
&&&
&&
&&
&&
&&
21
Robotic Example
( )
( )
( )
=
∂∂∂∂
=∂∂
−
−=
∂∂∂∂
=∂∂
=
=
∂∂∂∂
=∂∂
−+=−=
+=
=+==
=
=
rBB
rP
P
qP
mgmr
mgr
rL
L
qL
rmmr
rmJ
rL
L
qL
mgrrmJVKL
rBBP
mgrVrmJKmrJ
ff
Q
rθr
q
e
e
&
&
&
&
&&&
&
&
&
&
&
&&
&&
&&
2
12
2
22
22
21
222
;)sin(
cos ;
sin21
21
21
21 :ndissipatiopower The
sin ;21
21 ;
force theis torque; theis component;each toforces Applicable
both vary) length; radial position;angular ( scoordinate dGeneralize
θθθθ
θθθθθ
θθ
θ
θθ
ττ
θ
22
The Lagrange equation becomes
( )
orinput vect theis ctor;gravity ve theis )( vectorentripetalCoriolis/c theis ),( matrix; inertia theis )(
)(),()(
)sin()cos(2
00
)sin(
cos
2
2
12
2
12
2
QqGqqVqM
QqGqqVqqM
fmgmgr
r Bθmr
θmr Br m
mr
rBB
mgmr
mgr
rmrmrmrQ
qP
qL
qL
dtdQ
&
&&&
&
&
&
&
&&
&&
&
&
&&&
&&&&
&&
=++
=
+
−+
+
−
−−
+=
∂∂
+∂∂
−
∂∂
=
τθθθθ
θθθ
θθθ
23
Example: Two Mesh Electric Circuit
Ua(t)
R1 L1 L2
L12
R2
C1
C2q1q2
222
22112
211
12
21
12211
1
2
121
21)(
21
21
:is energy) (kineticenergy magnetic totalThe
).( ; ; ; ; : thatknow should We
as denoted is system the toapplied force dgeneralize The loop. second in the charge electric theis and loopfirst in the charge
electric theis wheres,coordinate dgeneralizet independen theas and Assume
qLqqLqLK
tUQsiq
siqqiqi
qqq
e
a
&&&&
&&
+−+=
=====
24
( )
( )
222
111
222
211
2
2
21
1
12
22
1
21
112212222
212112111
and ;21
21 :dissipatedenergy heat totalThe
and ;21
21
energy) (potentialenergy electric totalfor theequation theUse
;0
;0
qRqPqR
qPqRqRP
Cq
qV
Cq
qV
Cq
CqV
qLqLLqK
qK
qLqLLqK
qK
ee
ee
&&
&&
&&
&&&
&&&
=∂∂
=∂∂
+=
=∂∂
=∂∂
+=
−+=∂∂
=∂∂
−+=∂∂
=∂∂
−−
+=
++−−
+=
=++++=++−+
=∂∂
+∂∂
+∂∂
−∂∂
=∂∂
+∂∂
+∂∂
−∂∂
222
2112
122221211
1
1
1211
2
2222122112
1
1112121121
22221
1111
)(1 ;
)(1
0)(- ;)(
0)( ;)(
qRCqqL
LLqUqLqR
Cq
LLq
CqqRqLLqLU
CqqRqLqLL
qV
qP
qK
qK
dtdQ
qV
qP
qK
qK
dtd
a
a
eeee
&&&&&&&&&&
&&&&&&&&&&
&&&&
25
Another Example
Ua(t)
R L
RL
Cq1 q2
ia(t) iL(t)
ucuL
22
222
111
22
121
21
; ;0
0 ;0 ;0 ;21
)( ; ; :scoordinate dgeneralizet independen theas and Use
qLqK
dtdqL
qK
qK
qK
dtd
qK
qK
qLK
Qtuqiqiqq
eee
eeee
aLa
&&&
&&
&&&
&&
=
∂∂
=∂
∂=
∂∂
=
∂∂
=∂∂
=∂∂
=
===
26
( )
( )LLcLa
Lcc
La
La
eeee
L
L
iRuLdt
diR
tui
Ru
Cdtdu
CqqqR
Lqu
Cqq
Rq
CqqqRq Lu
CqqqR
qV
qP
qK
qK
dtdQ
qV
qP
qK
qK
dtd
qRqPqR
qP
qRqRP
Cqq
qV
Cqq
qV
CqqV
−=
+−−=
−+−=
++−
=
=+−
++=−
+
=∂∂
+∂∂
+∂∂
−
∂∂
=∂∂
+∂∂
+∂∂
−
∂∂
=∂∂
=∂∂
+=
+−=
∂∂−
=∂∂
−=
1 ;)(1
get welaw, sKirchhoff' usingBy
1 ;1
0 ;
0 ;
and
21
21 :isenergy dissipated totalThe
and
21 :isenergy potential totalThe
2122
211
2122
211
22221
1111
22
11
22
21
21
2
21
1
221
&&&&
&&&&
&&&&
&&
&&
&&
27
Directly-Driven Servo-System
Lrs
rrs
rrs
TQuQuQqiqiq
qsiq
si
q
−======
===
321
321
321
; ;; ; ;
; ; ;
ω
θ
&&&
Rotor
Stator
Load
Ls
Rs
LrRr
ur
us
ir
is
TL
ωr, Te
θr
torqueLoad : torqueneticelectromag :
L
e
TT
28
The Lagrange equations are expressed in terms of each independent coordinate
33333
22222
11111
)(
)(
)(
QqV
qP
qK
qK
dtd
QqV
qP
qK
qK
dtd
QqV
qP
qK
qK
dtd
ee
ee
ee
=∂∂
+∂∂
+∂∂
−∂∂
=∂∂
+∂∂
+∂∂
−∂∂
=∂∂
+∂∂
+∂∂
−∂∂
&&
&&
&&
29
The total kinetic energy is the sum of the total electrical (magnetic) and mechanical (moment of inertia) energies
33
3213
23122
32111
23
22321
21
3
0max
23
2221
21
23
2221
21
;sin;cos;0
cos;0 :result sderivative partial following The
21
21cos
21
)reluctance gmagnetizin is( coscos)()90(
;)(
)( :inductance Mutual
21
21
21
l)(Mechanica 21 l);(Electrica
21
21
qJqK
qqqLqK
qLqqLqK
qK
qqLqLqK
qK
qJqLqqqLqLK
LqLLL
NNLL
NNL
qJqLqqLqLKKK
qJKqLqqLqLK
eM
erM
ee
Msee
rMse
MMrMrsr
m
rssrM
rm
rsrsr
rsrsemeee
emrsrsee
&&
&&&&&
&&&
&&&&&
&&&&&
&&&&&
=∂∂
−=∂∂
+=∂∂
=∂∂
+=∂∂
=∂∂
+++=
==ℜ
==ℜ
=
+++=+=
=++=
θθ
θθ
30
We have only a mechanical potential energy: Spring with a constant ks
Lrsr
mrrsMr
rrrr
rsMs
rMr
r
sssr
rrMr
rMs
s
mrs
mrs
mMrsE
ME
s
ss
Tkdt
dBiiLdt
dJ
uiRdt
diLdtdi
LdtdiL
uiRdt
diLdtdiL
dtdi
L
qBqPqR
qPqR
qP
qBqRqRP
qBPqRqRP
PPP
qkqV
qV
qV
qkVk
−=+++
=+−+
=+−+
=∂∂
=∂∂
=∂∂
++=
=+=
+=
=∂∂
=∂∂
=∂∂
=
θθ
θθ
θθθ
θθθ
sin
sincos
sincos
system-servofor equations aldifferenti threehave we values,original thengSubstituti
and ; ;
21
21
21
21 ;
21
21
:as expressed is dissipatedenergy heat totalThe
;0 ;0
21:constant with spring theofenergy potential The
2
2
33
22
11
23
22
21
23
22
21
3321
23
&&
&&
&&
&&&
&&&
31
rrsMe
e
LrsrmrrsMr
rr
LrsrmrrsMr
rssrMrrMrsrrrsMssMsrMrs
r
rrMsrrrMrrrMrrrsMsrsrMrs
s
r
iiLTT
TkBiiLJdt
ddt
d
TkBiiLJdt
d
uLuLiLiLRiLLiLRLLLdt
di
uLuLLLiLRiLiLRLLLdt
di
ωdt
dθ
θ
θωθω
ωθ
θωθω
θθωθωθ
θθωθθωθ
sin:developed torqueneticelectromag for the expression obtain thecan We
)sin(1 :equation third thegConsiderin
)sin(1
cos2sin21sin
21
cos1
cossincos2sin21
cos1
variablesstate asposition and locity,angular ve current,rotor andcurrent stator using Also,
).(locity angular verotor of in terms written be shouldequation last The
222
222
−=
−−−−=
=
−−−−=
+−−−−−
−=
−+++−−
−=
=
32
More ApplicationApplication of Lagrange equations of motion in the modeling of two-
phase induction motor and generator.
Application of Lagrange equations of motion in the modeling of permanent-magnet synchronous machines.
Transducers