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).

4

• 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.

5

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?

6

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 =

9

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

zyx

r

space)in point material theofposition a indicating vector a is(r ),()(2

2

=

= 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.

11

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.

14

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.

15

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

==

++=++=++=

∂∂

+−∂∂

−−∂∂

=

∂∂

+∂∂

−

∂∂

=

=

&

&&&&&

&&

&&&

&&

19

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

Qq

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