1

Lagrange EquationsUse kinetic and potential energy to solve for motion!

References

http://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)

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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 a dynamic 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

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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

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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 is the angular acceleration of the body (rad/s2).

jM

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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

z

y

x

r

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

2

2

trFdt

trdm

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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 mechanical systems.

• 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”

Energy Mechanical Electrical

Kinetic (Active)

Ke

Mass / Inertia

0.5 mv2 / 0.5 j2

Inductor

Potential

V

Gravity: mgh

Spring: 0.5 kx2

Capacitor

0.5 Cv2 = q2/2C

Dissipative

P

Damper / Friction

0.5 Bv2

Resistor

22

2

1

2

1qRRi

22

2

1

2

1qLLi

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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 theNote

equations. 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'

;2

1

energy) potential theis energy; kinetic theis (

2

i

iiii

e

ee

QP

Qq

P

q

L

q

L

dt

d

mghVmvK

VKVKL

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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, , ); 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 cos ; y = r sin

sin ; z = r cos ).

• 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 n degrees of freedom since it is characterized by the n

generalized 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

q

P

q

P

q

L

q

L

q

L

q

L

dt

d

.

.

.

.

.

. -

.

.1111

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Example of Linear Spring Mass System and Frictionless Table: The Steps

m

x

0 : togetherall Combine

; ; :sderivative theDo

0 :Equation sLagrang'

2

1

2

1 :Lagrangian 22

kxxmq

L

q

L

dt

d

kxq

Lxm

q

L

dt

dxm

q

L

q

L

q

L

dt

d

kxxmVKL

ii

iii

ii

e

k

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Mechanical Example: Mass-Spring Damper

2

22

2

2

2

12

1

2

12

12

1

xBP

xhmgKxxmVKL

xhmgKxV

xmKe

e

xBmgKxxm

xBmgKxxmdt

d

xBx

xhmgKxxmx

xhmgKxxmxdt

d

x

P

x

L

x

L

dt

df

fQxq

)()()(

)2

1())(

2

1

2

1(

)))(2

1

2

1((

:equation Lagrange the

write we, force applied with the thusand , coordinate dgeneralize thehave We

2

222

22

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Electrical Example: RLC Circuit

2

22

2

2

2

12

1

2

12

12

1

qRP

qC

qLVKL

qC

V

qLK

e

e

equation KVLjust is This capacitor. afor and

)(

)2

1()

2

1

2

1())

2

1

2

1((

have we, force applied with theand (charge), coordinate dgeneralize thehave We

22222

c

c

Cvqqi

Rivdt

diLqR

C

QqLqR

C

QqL

dt

d

qRq

qC

qLq

qC

qLqdt

d

q

P

q

L

q

L

dt

du

uQq

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Electromechanical System: Capacitor MicrophoneAbout them see: http://www.soundonsound.com/sos/feb98/articles/capacitor.html

2222

2222

o

2222

2

1

2

1

2

1

2

12

1

2

1 ;

2

1

2

1

separation plate theis - plate, theof area theis

(F/m),air theofconstant dielectric theis 2

1

2

1 ;

2

1

2

1

KxqxxA

xmqLL

xBqRPKxqxxA

V

xxAxx

AC

KxqC

VxmqLK

o

o

o

e

m)equilibriu from nt displaceme

and charge :mechanical and l(electrica

freedom of degrees twohas system This

x

q

20

vqxxA

qRqL

fA

qKxxBxm

qRq

P

A

qxx

q

LqL

q

L

xBx

KxA

q

x

Lxm

x

L

o

o

1 2

2

equations Lagrange twoobtain the Then we

; ;

P ;

2 ;

2

21

Robotic Example

rB

B

r

P

P

q

P

mgmr

mgr

r

L

L

q

L

rm

mr

rm

J

r

L

L

q

L

mgrrmJVKL

rBBP

mgrVrmJKmrJ

ff

Q

rθr

q

e

e

2

12

2

22

22

21

222

;)sin(

cos ;

sin2

1

2

12

1

2

1 :ndissipatiopower The

sin ;2

1

2

1 ;

force theis torque; theis component;each toforces Applicable

both vary) length; radial position;angular ( scoordinate dGeneralize

22

The Lagrange equation becomes

orinput vect theis ctor;gravity ve theis )(

vectorentripetalCoriolis/c theis ),( matrix; inertia theis )(

)(),()(

)sin(

)cos(2

0

0

)sin(

cos

2

2

12

2

12

2

QqG

qqVqM

QqGqqVqqM

fmg

mgr

r Bθmr

θmr B

r m

mr

rB

B

mgmr

mgr

rm

rmrmrQ

q

P

q

L

q

L

dt

dQ

23

Example: Two Mesh Electric Circuit

Ua(t)

R1 L1 L2

L12

R2

C1

C2q1

q2

222

22112

211

12

21

12211

1

2

121

2

1)(

2

1

2

1

: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

tUQs

iq

s

iqqiqi

Q

q

qqq

e

a

24

222

111

222

211

2

2

21

1

12

22

1

21

112212222

212112111

and ;2

1

2

1 :dissipatedenergy heat totalThe

and ;2

1

2

1

energy) (potentialenergy electric totalfor theequation theUse

;0

;0

qRq

PqR

q

PqRqRP

C

q

q

V

C

q

q

V

C

q

C

qV

qLqLLq

K

q

K

qLqLLq

K

q

K

ee

ee

222

2112

122221211

1

1

1211

2

2222122112

1

1112121121

22221

1111

)(

1 ;

)(

1

0)(- ;)(

0)( ;)(

qRC

qqL

LLqUqLqR

C

q

LLq

C

qqRqLLqLU

C

qqRqLqLL

q

V

q

P

q

K

q

K

dt

dQ

q

V

q

P

q

K

q

K

dt

d

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 ;2

1

)( ; ;

:scoordinate dgeneralizet independen theas and Use

qLq

K

dt

dqL

q

K

q

K

q

K

dt

d

q

K

q

KqLK

Qtuqiqi

qq

eee

eeee

aLa

26

LLcLa

Lcc

La

La

eeee

L

L

iRuLdt

di

R

tui

R

u

Cdt

du

C

qqqR

Lqu

C

qq

Rq

C

qqqRq Lu

C

qqqR

q

V

q

P

q

K

q

K

dt

dQ

q

V

q

P

q

K

q

K

dt

d

qRq

PqR

q

P

qRqRP

C

qq

q

V

C

qq

q

VC

qqV

1 ;

)(1

get welaw, sKirchhoff' usingBy

1 ;

1

0 ;

0 ;

and

2

1

2

1 :isenergy dissipated totalThe

and

2

1 :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

TQuQuQ

qiqiq

qs

iq

s

iq

321

321

321

; ;

; ; ;

; ; ;

Rotor

Stator

Load

Ls

Rs

LrRr

ur

us

ir

is

TL

r, Te

r

torqueLoad :

torqueneticelectromag :

L

e

T

T

28

The Lagrange equations are expressed in terms of each independent coordinate

33333

22222

11111

)(

)(

)(

Qq

V

q

P

q

K

q

K

dt

d

Qq

V

q

P

q

K

q

K

dt

d

Qq

V

q

P

q

K

q

K

dt

d

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

2

1

2

1cos

2

1

)reluctance gmagnetizin is( coscos)(

)90( ;

)()( :inductance Mutual

2

1

2

1

2

1

l)(Mechanica 2

1 l);(Electrica

2

1

2

1

qJq

KqqqL

q

KqLqqL

q

K

q

K

qqLqLq

K

q

K

qJqLqqqLqLK

LqLLL

NNLL

NNL

qJqLqqLqLKKK

qJKqLqqLqLK

eM

erM

ee

Msee

rMse

MMrMrsr

m

rssrM

rm

rsrsr

rsrsemeee

emrsrsee

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

dBiiL

dt

dJ

uiRdt

diL

dt

diL

dt

diL

uiRdt

diL

dt

diL

dt

diL

qBq

PqR

q

PqR

q

P

qBqRqRP

qBPqRqRP

PPP

qkq

V

q

V

q

V

qkVk

sin

sincos

sincos

system-servofor equations aldifferenti threehave we values,original thengSubstituti

and ; ;

2

1

2

1

2

12

1 ;

2

1

2

1

:as expressed is dissipatedenergy heat totalThe

;0 ;0

2

1: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

iiLT

T

TkBiiLJdt

ddt

d

TkBiiLJdt

d

uLuLiLiLRiLLiLRLLLdt

di

uLuLLLiLRiLiLRLLLdt

di

ωdt

dθ

sin

:developed torqueneticelectromag for the expression obtain thecan We

)sin(1

:equation third thegConsiderin

)sin(1

cos2sin2

1sin

2

1

cos

1

cossincos2sin2

1

cos

1

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