an introduction to three-dimensional, rigid body dynamics ......unit 6 lagrange’s equations,...

Copyright © James W. Kamman, 2017

An Introduction to

Three-Dimensional, Rigid Body Dynamics

James W. Kamman, PhD

Volume II: Kinetics

Unit 6

Lagrange’s Equations, d’Alembert’s Principle, and Kane’s Equations

for Systems with Constraints

Summary

This unit discusses the application of Lagrange’s equations, d’Alembert’s principle and Kane’s equations

to rigid body dynamic systems with constraints. In Units 4 and 5 of this volume, the applications of Lagrange’s

equations and d’Alembert’s principle were based on a set of independent generalized coordinates, and the

applications of Kane’s equations were based on a set of independent generalized speeds. There are systems

for which it is inconvenient or impossible to eliminate surplus generalized coordinates or simply inconvenient

to eliminate generalized speeds from the analysis. For these systems, the application of each of these three

methods can be supplemented with the use of Lagrange multipliers.

The resulting equations of motion are a set of differential and algebraic equations. The Lagrange

multipliers are algebraic unknowns related to the forces and torques required to maintain the constraints. An

Addendum to this unit explores this connection.

In the analysis that follows, constraint equations are assumed to be equality constraints, that is, some

function of the generalized coordinates and/or generalized speeds is equal to zero. The case of inequality

constraints is not considered.

Page Count Examples Suggested Exercises

30 4 7

Copyright © James W. Kamman, 2017 Volume II, Unit 6 – page: 1/30

Configuration Constraints for Mechanical Systems

Suppose the configuration of a mechanical system is defined by “n”

generalized coordinates, say ( 1, , )kq k n= . These coordinates may all be

independent, or they may form a dependent set. For example, consider the

single link pendulum. The coordinate set , ,G Gx y is a dependent set of

coordinates, because the following independent constraint equations can

be used to relate the coordinates

12

sin( )Gx L = 12

cos( )Gy L =

Hence, for this system only one generalized coordinate is required. Any set of two or more coordinates for this

system forms a dependent set. Note that care must be taken to include only independent constraint equations

in this process.

The types of constraints described above are referred to as configuration constraints. They directly relate

some or all the generalized coordinates. For a mechanical system described by “n” generalized coordinates

( 1, , )kq k n= with “m” independent configuration constraints, many constraints can be written in the form

1 2( , , , , ) 0j nf q q q t = ( 1, , )j m= (1)

Here, the functions ( 1, , )jf j m= and their derivatives through second order with respect to all variables

are assumed to be continuous.

Configuration constraints are most useful in a dynamic analysis when they are differentiated into a form

linear in the time derivatives of the generalized coordinates. Using concepts from multivariate calculus, the

time derivatives of the constraint functions jf can be written as

1

0n

j j j

k

k k

d f f fq

dt q t=

= + =

or 0

1

0n

jk k j

k

a q a=

+ = ( 1, , )j m= (2)

Eq. (2) can be written in matrix form as

01 110

n mm n mA q a

+ = (3)

By comparing these expressions, the elements of the coefficient matrix m n

A

and the column vector 01m

a

are identified to be

j

jk

k

fa

q

( 1, , ; 1, , )j m k n= = and 0

j

j

fa

t

( 1, , )j m= (4)


In this form, the independence of the constraint equations can be verified by checking the rank of the

coefficient matrix A . If the rank of A is “m”, then the equations are independent. If the rank of A is less

than “m”, then the equations are dependent. The rank of A is equal to the number of independent constraint

equations.

Configuration constraints are called holonomic constraints. If they depend explicitly on the time, they are

called rheonomic constraints. If they do not depend explicitly on the time, they are called schleronomic

constraints.

Examples

1. If (as described above) the generalized coordinate set 1 2 3, , , ,G Gq q q x y = is used to describe the

configuration of the single link pendulum, then two configuration constraints can be written as

1 1 2 3 1 3

12

( , , ) sin( ) 0f q q q q L q= − = 2 1 2 3 2 3

12

( , , ) cos( ) 0f q q q q L q= − =

Differentiating these equations with respect to time using the chain rule gives

1

1 3 3 3

2

2 3 3 3

3

1 12 21 12 2

cos( ) 1 0 cos( ) 0

sin( ) 0 1 sin( ) 0

qq Lq q L q

q A qq Lq q L q

q

− −

= = + +

Comparing these results to the general form shown above gives

11 22 1a a= = 12 21 0a a= = 1312

cos( )a L = − 2312

sin( )a L =

0 0ja = ( 1,2)j =

2. If the generalized coordinate set 1 2, ,G Gq q x y= is used to describe the configuration of the single link

pendulum, then there is only one independent configuration constraint. For example,

( )22 2

1 2 1 212

( , ) 0f q q q q L= + − =

Differentiating this equation with respect to time using the chain rule gives

1 1 2 22 2 0q q q q+ = 1 1 2 2 0q q q q + = 1

1 2

2

[ ] 0q

q q A qq

=

So, in this case, 11 1 Ga q x= = , 12 2 Ga q y= = , and 10 0a = .


Motion Constraints for Mechanical Systems

Configuration constraints directly relate some or all the coordinates used to describe the configuration of

a system. The basic form of a configuration constraint is as shown in Eq. (1). In contrast, motion constraints

directly relate some or all of the derivatives of the generalized coordinates as presented in

Eq. (2). If the motion constraint can be integrated, an equivalent configuration constraint in the form of Eq.

(1) can be found. Configuration constraints and motion constraints that can be integrated are called holonomic

constraints. Motion constraints that cannot be integrated are called nonholonomic constraints.

Given that the constraint functions ( 1, , )jf j m= and their derivatives through second order with respect

to all variables are continuous, not only is Eq. (4) true, but all the mixed second partial derivatives must be

equal. That is,

2 2jk j j j

k k k

a f f a

q q q q q q

= = =

, (1, , )k n=

and

2 20j j j ja f f a

q q t t q t

= = =

(1, , )n=

If these equations are satisfied for a given motion constraint, the constraint is integrable and the constraint is

holonomic. Unfortunately, if the above equations are not satisfied, it cannot be immediately concluded that

the constraint is nonholonomic, because it is possible that an integrating factor exists that transforms the

equation into an exact differential. Unfortunately, finding an integrating factor is not always an easy task. In

some (usually simple) cases they are not difficult to find, but in many cases the analyst is left only with a trial

and error process.

Equations of Motion for Systems with Constraints

The forms of Lagrange’s equations and d’Alembert’s principle presented in Units 4 and 5 of this volume

require that a set of independent generalized coordinates be used. The form of Kane’s equations presented in

Unit 5 of this volume allows for a set of dependent generalized coordinates but requires a set of independent

generalized speeds. The use of independent generalized coordinates in Lagrange’s equations or d’Alembert’s

principle or the use of independent generalized speeds in Kane’s equations is not always practical or possible.

In the case of configuration constraints or integrable motion constraints (i.e. holonomic constraints), it is

theoretically possible to eliminate excess generalized coordinates from the analysis by solving the constraint

equations. However, this process may provide equations of motion that are overly complicated and very

tedious to generate. In the case of non-integrable motion constraints (i.e. nonholonomic constraints), it is not

possible to eliminate excess generalized coordinates from the analysis. It is, however, possible to eliminate


the excess generalized speeds. But, as with the elimination of excess generalized coordinates, the process of

eliminating excess generalized speeds may provide equations of motion that are overly complicated and very

tedious to generate.

Lagrange’s Equations and d’Alembert’s Principle for a Set of Dependent Generalized Coordinates

For the reasons mentioned above, it is often advantageous or necessary to use a set of dependent

generalized coordinates ( 1, , )kq k n= to describe the configuration of a mechanical system. If the system

possesses N n m= − degrees of freedom (DOF), then there are m independent constraint equations that can be

written in the form of Eq. (2) or Eq. (3) above. In this case, Lagrange’s equations and d’Alembert’s principle

can be modified using a set of unknown Lagrange multipliers. The specific forms of the equations follow.

Lagrange’s Equations:

1

k

m

q j jk

jk k

d K KF a

dt q q

=

− = +

( 1, , )k n= (5)

or

( )1

k

m

q j jknc

jk k

d L LF a

dt q q

=

− = +

( 1, , )k n= (6)

d’Alembert’s Principle:

( ) ( )1 1 1

B B

i i

i i i i i k

R RN N mG BR R R

i G G B B G q j jk

i i jk k

vm a I H F a

q q

= = =

+ + = +

( 1, , )k n= (7)

These equations are identical to those presented in Units 4 and 5 of this volume except for the summation

term 1

m

j jk

j

a=

that has been added to the right side of the equations. The variables ( 1, )j j m = are a set of

unknown Lagrange multipliers, one for each constraint equation, and the variables

( 1, , ; 1, , )jka j m k n= = are the coefficients from the constraint equations expressed in the form of

Eq. (2). Clearly, the Lagrange multipliers appear in the equations as algebraic unknowns.

The n equations (from Lagrange’s equations or d’Alembert’s principle) and the m constraint equations

form a set of n m+ differential/algebraic equations for the n m+ unknowns – the n generalized coordinates

( 1, , )kq k n= and the m Lagrange multipliers ( 1, , )j j m = . The Lagrange multipliers can be related

to the forces and moments required to maintain the constraints. (See the Addendum at the end of this Unit.) It

is important to note that all terms in the equations be written only in terms of ( 1, , )kq k n= and their

derivatives and no other variables.


Kane’s Equations for a Set of Dependent Generalized Speeds

For the reasons mentioned above, it may be advantageous to use a set of dependent generalized speeds

( 1, , )ku k n= . If the system possesses N n m= − degrees of freedom, and there are m independent

constraint equations that can be written of the form

0

1

0n

jk k j

k

a u a=

+ = ( 1, , )j m= (8)

then, Kane’s equations can be modified using a set of unknown Lagrange multipliers. In this case, Kane’s

equations can be written as

( ) ( )1 1 1

B B

i i

i i i i i k

R RN N mG BR R R

i G G B B G u j jk

i i jk k

vm a I H F a

u u

= = =

+ + = +

( 1, , )k n= (9)

If the system configuration is described by the generalized coordinates ( 1, , )k cq k n= , then the n Kane's

equations along with the cn kinematic differential equations and the m constraint equations form a set of

cn n m+ + differential and algebraic equations in as many unknowns – the cn generalized coordinates, the

n generalized speeds ( 1, , )ku k n= , and the m Lagrange multipliers ( 1, , )j j m = . As before, the

Lagrange multipliers can be related to the forces and moments required to maintain the constraints. Note that,

it is important that all quantities be written only in terms of ( 1, , )k cq k n= , ( 1, , )ku k n= and their

derivatives and no other variables.

Example 1: Equations of Motion of the Simple Planar Pendulum

The diagram shows a simple pendulum – a point mass (mass, m ) on the

end of a light rod of length L . The X-axis is horizontal, and the Y-axis points

vertically downward. The light rod makes an angle with the vertical, and

the coordinates of the point mass are ( , )x y .

Find: Equations of motion using Lagrange’s equations using

a) one independent generalized coordinate,

b) two dependent generalized coordinates, ( , )x y

Solution:

a) Using as the generalized coordinate, the Lagrangian of the system can be written as

( )2 2 21 12 2

cos( )L K V mv mgy mL mgL = − = − − = +


Using Lagrange's equations in the form of Eq. (6), but with no non-conservative forces and no constraints, the

equation of motion can be written as

( ) ( ) ( )2 2sin sin 0d L L d

mL mgL mL mgLdt dt

− = + = + =

sin( ) 0gL

+ =

b) Using ( ),x y as the dependent generalized coordinates, the Lagrangian of the system can be written as

( )2 2 21 12 2

( )L K V mv mgy m x y mg y= − = − − = + +

The two generalized coordinates can be related using the configuration constraint equation

2 2 2x y L+ =

Differentiating this equation with respect to time and writing the result in the form of Eq. (3) gives

0xx yy+ = or 0x

x yy

=

11a x= 12a y= 10 0a =

Using Lagrange's equations in the form of Eq. (6) with no non-conservative forces and with one constraint

gives

1k

k k

d L La

dt q q

− =

( )1,2k =

Specifically,

( ) 110d L L d

mx mx a xdt x x dt

− = − = = =

0mx x − =

( ) 12

d L L dmy mg my mg a y

dt y y dt

− = − = − = =

0my mg y − − =

Differentiating the constraint equation again gives

2 2 0xx y y x y+ + + =

These last three boxed equations are a complete set of three coupled, second-order, differential/algebraic

equations in three unknowns ( ), ,x y .

Alternatively, the first two of the above equations can be used to eliminate the Lagrange multiplier

from the equations. For example, if the first equation is multiplied by “y” and the second equation by “x” and

the second equation is subtracted from the first, then the first two equations are reduced to a single equation.

In this case, the equations of motion can be written as two coupled, second-order differential equations


2 2

0

0

yx x y g x

xx y y x y

− − =

+ + + =

Note: Clearly the use of a single coordinate is recommended for this simple system, but that will not always

be the case. Both cases are shown here for easy comparison.

Example 2: Equations of Motion of a Planar Slider-Crank Mechanism

The equations of motion of a slider-crank

mechanism can be formulated in various ways. The

system shown to the right consists of a slider-crank

mechanism which has had the slider constraint removed

(or relaxed). The two links and the end-mass are all

connected with simple revolute joints. The two links

have masses ( 1,2)im i = and lengths ( 1,2)i i = , and the

end-mass has mass 3m . The end-mass is assumed to

translate and not rotate.

One approach to finding the equations of motion of

the slider-crank mechanism (shown in the lower figure)

is to first find the equations of motion of the system in

upper figure and then apply the necessary constraints to

form the slider-crank mechanism shown in the lower

figure.

Find: Using Lagrange’s equations with Lagrange multipliers,

a) equations of motion of the unconstrained system

b) equations of motion of the slider-crank mechanism

Assume that both systems shown move in a vertical plane and are driven by a torque T on the shaft at A.

Solution:

a) Kinetic Energy:

The velocities of the mass center and the end of the second link can be written as

2 2 1 2/ 1 1 2 2

R R RG B G Bv v v e r e = + = +

21/ 1 1 2 2R R R

C B C Bv v v e e = + = +

Here, the unit vectors ( 1,2)i

e i = are perpendicular to the first and second links and directed in directions of

increasing ( 1,2)i i = . Using these results, the kinetic energy of the system can be written as

Slider-Crank Mechanism with

Slider Constraint Removed


( ) ( ) ( )

( )2

2

2

1 1 2 2 1

1 2

2 2 2 21 1 1 11 2 3 1 1 2 2 2 32 2 2 2

2 22 2 21 1 1 1

1 1 1 2 1 1 2 2 2 3 1 1 2 22 2 2 2

2 2 2 2 2 2 21 1 1 11 1 1 2 1 1 2 2 2 2 1 2 1 2 2 1 22 2 2 2

213 12

GAG C

G G

G G

K K K K I m v I m v

I m r m e r e I m e e

I m r m m r m r C I

m

−

= + + = + + +

= + + + + + +

= + + + + +

+ 2 2 211 3 2 2 3 1 2 1 2 2 12

m m C −+ +

( ) ( ) ( )1 2

2 2 2 2 2 2 21 11 1 2 1 3 1 1 2 2 3 2 2 2 2 3 2 1 1 2 2 12 2G GK I m r m m I m r m m r m C − = + + + + + + + +

Here, the symbol 2 1C − is used to represent 2 1cos( ) − .

Potential Energy: (using a horizontal datum through point A)

The potential energy function associated with the weight forces can be written as

( ) ( )

( ) ( )

1 2 3 1 1 1 2 1 1 2 2 3 1 1 2 2

1 1 2 1 3 1 1 2 2 3 2 2

V V V V m g r S m g S r S m g S S

m r m m gS m r m gS

= + + = + + + +

= + + + +

In this expression, the symbols 1S and 2S are used to represent 1sin ( ) and 2sin ( ) , respectively.

Lagrangian:

( ) ( ) ( )

( ) ( )1 2

2 2 2 2 2 2 21 11 1 2 1 3 1 1 2 2 3 2 2 2 2 3 2 1 1 2 2 12 2

1 1 2 1 3 1 1 2 2 3 2 2

G G

L K V

I m r m m I m r m m r m C

m r m m gS m r m gS

−

= −

= + + + + + + + +

− + + − +

Partial Derivatives of the Lagrangian:

( ) ( )1

2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1

1

G

LI m r m m m r m C

−

= + + + + +

( ) ( )2

2 22 2 3 2 1 1 2 1 2 2 3 2 2

2

G

Lm r m C I m r m

−

= + + + +

( ) ( )2 2 3 2 1 1 2 2 1 1 1 2 1 3 1 1

1

Lm r m S m r m m gC

−

= + − + +

( ) ( )2 2 3 2 1 1 2 2 1 2 2 3 2 2

2

Lm r m S m r m gC

−

= − + − +

Time Derivatives:

( ) ( ) ( ) ( )1

2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2 1 2 1

1

G

d LI m r m m m r m C m r m S

dt

− −

= + + + + + − + −


( ) ( ) ( ) ( )2

2 22 2 3 2 1 1 2 1 2 2 3 2 1 1 2 1 2 1 2 2 3 2 2

2

G

d Lm r m C m r m S I m r m

dt

− −

= + − + − + + +

Lagrange’s Equations: (unconstrained system)

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( )

1

1

1 1

2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2 1 2 1

2 2 3 2 1 1 2 2 1 1 1 2 1 3 1 1

2 2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2

G

G

d L L

dt

I m r m m m r m C m r m S

m r m S m r m m gC

I m r m m m r m C m r m S

− −

−

− −

−

= + + + + + − + −

− + + + +

= + + + + + − +

( )

1

2 2 3 2 1 1 2 2 1m r m S −+ + ( )2 2 3 2 1 1 2 2 1m r m S −− + ( )1 1 2 1 3 1 1m r m m gC+ + +

( ) ( ) ( )

( )1

1

2 2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2 1

1 1 2 1 3 1 1

GI m r m m m r m C m r m S

m r m m gC F

− −+ + + + + − +

+ + + =

( ) ( ) ( ) ( )( ) ( )

( ) ( ) ( )

2

2

2 2

2 22 2 3 2 1 1 2 1 2 2 3 2 1 1 2 1 2 1 2 2 3 2 2

2 2 3 2 1 1 2 2 1 2 2 3 2 2

2 2 22 2 3 2 1 1 2 1 2 2 3 2 2 2 2 3 2 1 1 2 1

2 2 3

G

G

d L L

dt

m r m C m r m S I m r m

m r m S m r m gC

m r m C I m r m m r m S

m r m

− −

−

− −

−

= + − + − + + +

+ + + +

= + + + + + +

− +( )2 1 1 2 2 1S − ( )2 2 3 2 1 1 2 2 1m r m S −+ + ( )2 2 3 2 2m r m gC+ +

( ) ( ) ( )

( )2

2

2 2 22 2 3 2 1 1 2 1 2 2 3 2 2 2 2 3 2 1 1 2 1

2 2 3 2 2

Gm r m C I m r m m r m S

m r m gC F

− −+ + + + + +

+ + =

Generalized Forces:

The contributions of the driving torque to the equations of motion can be written as

1 1 1F T k T k k T = = =

2 1 2 0 0F T k T k = = =

Summary:

The two equations of motion of the unconstrained system can be written as


( ) ( ) ( )

( )1

2 2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2 1

1 1 2 1 3 1 1


m r m m gC T

− −+ + + + + − +

+ + + =

( ) ( ) ( ) ( )2

2 2 22 2 3 2 1 1 2 1 2 2 3 2 2 2 2 3 2 1 1 2 1 2 2 3 2 2 0Gm r m C I m r m m r m S m r m gC − −+ + + + + + + + =

b) To convert the unconstrained system into a simple slider-crank mechanism with zero offset (as shown),

mass 3m must be constrained to move along a horizontal line passing through point A. The constraint equation

associated with this restriction of motion is

1 1 2 2 0S S+ =

This constraint is converted to standard form by differentiating the expression with respect to time.

1 1 1 2 2 2 0C C + = or 1 1

1 1 2 2 11 12

2 2

0C C a a

= =

Lagrange’s Equations: (for the constrained system)

1 ( 1,2)i i

i i

d L LF a i

dt

− = + =

Adding the Lagrange multiplier terms to the right side of the equations of motion of part (a) gives the equations

of motion for the slider crank mechanism.

( ) ( ) ( )

( ) ( )1

2 2 2 21 1 2 1 3 1 1 2 2 3 2 1 2 2 1 2 2 3 2 1 2 2 1

1 1 2 1 3 1 1 1 1


m r m m gC T C

− −+ + + + + − +

+ + + = +

( ) ( ) ( )

( ) ( )2

2 2 22 2 3 2 1 1 2 1 2 2 3 2 2 2 2 3 2 1 1 2 1

2 2 3 2 2 2 2

Gm r m C I m r m m r m S

m r m gC C

− −+ + + + + +

+ + =

These two equations are now supplemented with the twice differentiated constraint equation.

2 2

1 1 1 2 2 2 1 1 1 2 2 2 0C C S S + − − =

These three equations represent a set of three second-order, differential/algebraic equations in three

unknowns – 1 , 2 , and .


Notes:

1. The use of a Lagrange multiplier in this problem allows construction of the equations of motion in

terms of two variables instead of just one. This is a much easier alternative to developing the

Lagrangian in terms of a single variable, say 1 , and then differentiating to find a single differential

equation of motion (as would be required by Lagrange’s equations without a Lagrange multiplier). This

clearly demonstrates the advantage of using Lagrange multipliers.

2. There are many ways of formulating the equations for complex systems using Lagrange multipliers.

For example, consider developing the equations of motion of the slider crank mechanism using the

following two systems.

First, develop the equations of motion of the individual systems. System 1 is a single degree-of-freedom

system with one equation of motion for the angle 1 . System 2 is a two degree-of-freedom system with

two equations of motion for the coordinate x and the angle 2 . To form the slider-crank mechanism,

the points labeled B in each system must coincide. This gives rise to two constraint equations and two

Lagrange multipliers.

1 1 2 2 0C C x− − = and 1 1 2 2 0S S− =

Modifying the original equations of motion using the constraint equations and their associated

Lagrange multipliers gives three equations of motion in five unknowns – 1 , 2 , x, 1 , and 2 .

Combining these equations with the differentiated constraint equations gives five equations in five

unknowns.

3. Lagrange multipliers clearly make it easier to formulate the equations of motion. The cost of using

them comes from having to solve a set of differential/algebraic equations instead of a smaller set of

ordinary differential equations. Thus, the solution process is generally more detailed.

System 1

System 2


Example 3: Three-Dimensional Rotating Frame and Bar

The system shown has two degrees of freedom and consists

of two bodies, the frame F and the bar B. As F rotates about the

fixed vertical direction, B rotates relative to the horizontal arm

of F. The orientation of F is given by the angle ( ) = , and

the orientation of B is given by the angle ( ) = . The bar

has mass m and length . The frame is assumed to be light. The

motor torque ( )M t is applied to F by the ground, and the

motor torque ( )M t is applied to B by F.

Find:

Using Kane’s equations, find the equations of motion of the system. Use the body-fixed angular velocity

components of the bar as a set of dependent generalized speeds.

Solution:

The angular velocity of the bar can be written as

( )2 2 1 3 1 2 3 1 1 2 2 3 3R

B e k e S e C e S e e C e e e e = + = + − + = − + + + +

From this expression, the angular velocity components 1 and 3 are not independent. The two can be related

with the following constraint equation.

1 3 0C S + = or

1 1

2 11 12 13 2

3 3

0 0C S a a a

=

Given this set-up, Kane’s equations with a single Lagrange multiplier may be written as

( ) ( ) 1k

RRBR R RG

G G B B G k

k k

vm a I H F a

+ + = + ( 1,2,3)k =

Kinematics:

The velocity of the mass center of the bar can be written as

( ) ( )/ 2 1 3 2 1 1 3 3 2R R

G F G Ov r k d n S e C e d n e e d n = = = − + = +

( )3 1 1 3R

Gv d e e = − +

By direct differentiation, the acceleration of the mass center of the bar can be written as

Reference Frames


( ) ( ) ( )1 2 3

3 1 1 3 3 1 1 3 3 1 1 3 1 2 3

3 10

R RG B

e n e

a d e e d e e d e e d

= − + + − + = − + +

−

( ) ( ) ( )2 23 1 2 1 1 3 2 1 2 3 3

RGa d e d e d e = − + − + + +

Partial Angular Velocities and Partial Velocities:

The three partial angular velocities can be written as

RB

k

k

e

=

( )1,2,3k =

The three partial velocities can be written as

3

1

RGv

d e

=

2

0

RGv

=

1

3

RGv

d e

= −

Terms on the left side of Kane’s equations:

( ) ( ) ( )( ) ( )2 2 23 1 2 1 1 3 2 1 2 3 3 3 1 2 3

1

RR G

G

vm a m d e d e d e d e md

= − + − + + + = +

2

0R

R GG

vm a

=

( ) ( ) ( )( ) ( ) ( )2 2 23 1 2 1 1 3 2 1 2 3 3 1 3 1 2

3

RR G

G

vm a m d e d e d e d e m d

= − + − + + + − = −

( )11

2 2 1 1 1 2 2 2 3 3 3

3 3

0 0

0 0

0 0

R RG B G B

I

I I I I e I n I e

I

= + +

( )R

BRG B k k

k

I I

=

( )1,2,3k =

( )11

2 2 1 1 1 2 2 2 3 3 3

3 3

0 0

0 0

0 0

RG G B G

I

H I I H I e I n I e

I

= = + +


( ) ( ) ( )

1 2 3

1 2 3 3 2 2 3 1 1 3 1 3 2 2 1 1 2 3

1 1 2 2 3 3

RB G

e n e

H I I e I I n I I e

I I I

= = − + − + −

( ) ( )3 2 2 3

1

RBR

B GH I I

= −

( ) ( )1 3 1 3

2

RBR

B GH I I

= −

( ) ( )2 1 1 2

3

RBR

B GH I I

= −

Generalized Forces:

The three generalized forces associated with the driving torques can be written as

( )1 2 1 3 2 1

1 1

R RB B

F M k M n M S e C e M n e S M

= + = − + + = −

( )2 2 1 3 2 2

2 2

R RB B

F M k M n M S e C e M n n M

= + = − + + =

( )3 2 1 3 2 3

3 3

R RB B

F M k M n M S e C e M n e C M

= + = − + + =

Note here that the torque M is transferred directly to the bar B because the frame F is light.

Kane’s Equations of Motion:

Substituting the intermediate results into Kane’s equations gives

1:k = ( ) ( )21 2 3 1 1 3 2 2 3md I I I S M C + + + − = − +

( ) ( )2 21 1 3 2 2 3I md I md I S M C + + + − = − +

2 :k = ( )2 2 1 3 1 3I I I M + − =

3:k = ( ) ( )23 1 2 3 3 2 1 1 2md I I I C M S − + + − = +

( ) ( )2 23 3 2 1 1 2I md I I md C M S + + − − = +


Note:

If the first of these equations is multiplied by S− and the third by C and the resulting equations added

together, the Lagrange multiplier can be eliminated from the equations. Further, if the angular velocity

component 3 is written in terms of 1 , then the two resulting equations (along with a set of kinematic

equations relating the angular velocity components 1 and 2 to and ) can be solved without regard to

the Lagrange multiplier. Following this process and treating the bar as slender, the above equations can be

shown to be identical to those found in previous Units using Lagrange’s equations and d’Alembert’s principle.

Example 4: Double Pendulum

The system shown is a three-dimensional double

pendulum or arm. The first link is connected to ground and

the second link is connected to the first with ball and socket

joints at O and A. The orientation of each link is defined

relative to the ground using a 3-1-3 body-fixed rotation

sequence. The lengths of the links are 1 and 2 . The links

are assumed to be slender bars with mass centers at their

midpoints. Assume the 2N direction is vertical.

Reference frames:

1 2 3: , ,R N N N (fixed frame)

1 2 3: , , ( 1,2)ii i iL n n n i = (fixed in the two links)

Find:

Using Kane’s equations with Lagrange multipliers, find the equations of motion describing the free motion

of the double pendulum under the action of gravity. Use the twelve dependent generalized speeds

1 2 3 4 5 6 7 8 9 10 11 12 11 12 13 21 22 23 11 12 13 21 22 23, , , , , , , , , , , , , , , , , , , , , ,u u u u u u u u u u u u v v v v v v =

The first six generalized speeds are the body-fixed angular velocity components of the two links, and the last

six generalized speeds are the inertial components of the velocities of the mass centers of links.

1

1 1 111 1 12 2 13 3

RL n n n = + +

2

2 2 221 1 22 2 23 3

RL n n n = + +

11 1 12 2 13 31

RGv v N v N v N= + + 21 1 22 2 23 32

RGv v N v N v N= + +


Solution:

Transformation Matrices and Angular Velocities: (previous results)

The transformation matrices and the body-fixed components of the angular velocities of the links

( 1,2)iL i = are as given in Unit 5 of Volume I.

1 3 1 2 3 1 3 1 2 3 2 3

1 3 1 2 3 1 3 1 2 3 2 3

1 2 1 2 2

i i i i i i i i i i i i

i i i i i i i i i i i i

i i i i i

i

C C S C S S C C C S S S

R C S S C C S S C C C S C

S S C S C

− +

= − − − +

−

1 1 2 3 2 3

2 1 2 3 2 3

3 3 1 2

i i i i i i

i i i i i i

i i i i

S S C

S C S

C

= +

= −

= +

( 1,2)i =

Angular Momentum: (previous results)

The inertia matrices of the links about their mass centers can written as

11

22

33

0 0

0 0

0 0

i

iG

i

i Li

I

I I

I

=

( 1,2)i =

Here, because the links are assumed to be slender bars, inertias about the 1in and 3

in directions are identical

and the inertias about the 2in directions are small. Using these inertia matrices, the angular momenta of the

links about their mass centers can be written as

11 1

22 2 11 1 1 22 2 2 33 3 3

333

0 0

0 0

0 0

i

i

R i i i i i i iG G L i G i i i

ii

i i i i

I

H I I H I n I n I n

I

= → = + +

( 1,2)i =

Finally, the cross product of the angular velocity with the angular momentum for each of the links can be

written as

( ) ( ) ( )

1 2 3

1 2 3

11 1 22 2 33 3

33 22 2 3 1 11 33 1 3 2 22 11 1 2 3

i i i

RL G i i i

i i ii i i

i i i i i i i i ii i i i i i

i i

n n n

H

I I I

I I n I I n I I n

=

= − + − + −

( 1,2)i =

Constraint Equations:

The generalized speeds associated with translational and rotational motion can be related by first finding

the fixed-frame components of the position vectors of the mass centers of the links. Then, these components

can be differentiated to relate the generalized speeds.


The fixed-frame position vector components of mass center 1G can be written in matrix form as

1

1 1 1 1 1

1

1 12 2

0 0

1

0 0

T T

x

y R R

z

= − = −

Using results from Addendum 2 of Unit 4 of this volume, this result can be differentiated.

11 13 12

12 1 1 1 1 1 1 1 13 11

13 12 11

1 1 12 2 2

00 0 0

1 1 0 1

0 0 00

T T T

v

v R R R

v

−

= − = − = − −

−

11 13

12 1 1

1113

12

0

0 0

0

T

v

v R

v

−

+ =

This second term on the left side of this expression can be expanded to give

1 1 1 111 13 31 11 31 1113 11

1 1 1 11 1 1 12 13 32 11 1 32 12 12

1 1 1 111 1313 13 33 11 33 13

1 1 12 2 2

0

0 0

0

T

R R R R

R R R R R

R R R R

− + − −

= − + = −

− + −

Substituting the expanded results into the previous equation gives

1 131 1111 11

1 112 1 32 12 12

1 113 1333 13

12

0 0

0 0

00

R Rv

v R R

v R R

− + − =

−

The fixed-frame position vector components of mass center 2G can be written in matrix form as follows.

2

2 1 1 2 2 1 1 2 2

2

1 12 2

0 0 0 0

1 1

0 0 0 0

T T T T

x

y R R R R

z

= − + − = − −

Again, using results from Addendum 2 of Unit 4 of this volume, this result can be differentiated as follows.

21

22 1 1 2 2 1 1 1 2 2 2

23

1 12 2

0 0 0 0

1 1 1 1

0 0 0 0

T T T T

v

v R R R R

v

= − − = − −


13 12 23 22

1 1 13 11 2 2 23 21

12 11 22 21

12

0 00 0

0 1 0 1

0 00 0

T TR R

− −

= − − − − − −

21 13 23

22 1 1 2 2

11 2123

12

0

0 0 0

0

T T

v

v R R

v

− −

+ + =

The second and third terms on the left side of this equation can be expanded to give

1 1 2 211 13 31 11 11 23 31 2113 23

1 1 2 21 1 2 2 1 12 13 32 11 2 12 23 32 21

1 1 2 211 21

13 13 33 11 13 23 33 21

1 12 2

0 0T T

R R R R

R R R R R R

R R R R

− + − +− −

+ = − + + − +

− + − +

1 1 2 231 11 31 1113 23 11 21

1 1 2 21 1 2 2 1 32 12 12 2 32 12 22

1 1 2 211 21 13 2333 13 33 13

1 12 2

0 0

0 0 0 0

0 0

T T

R R R R

R R R R R R

R R R R

− − − −

+ = − + −

− −

Substituting the expanded results into the previous result gives

1 1 2 231 11 31 1121 11 21

1 1 2 222 1 32 12 12 2 32 12 22

1 1 2 223 13 2333 13 33 13

12

0 0 0

0 0 0

00 0

R R R Rv

v R R R R

v R R R R

− − + − + − =

− −

Combining the constraint equations and rearranging into the standard form for constraint equations gives

the following.

12 1 6 16 120A u

=

The coefficient matrix A is shown here partitioned into a set of 3 3 matrices.

1 131 11

1 11 32 12

1 133 13

1 1 2 231 11 31 11

1 1 2 21 32 12 2 32 12

1 1 2 233 13 33 13

6 12

12

12

0 0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0 0

0 0 0 0 0 1 0 0 00

0 0 0

0 0

0 0

R R

R R

R RA

R R R R

R R R R

R R R R

− − −

= − − − −

− −

0 0 1 0 0

0 0 0 0 1 0

0 0 0 0 0 1


The vector of generalized speeds is shown partitioned into a set of 3 1 vectors.

11 12 13 21 22 23 11 12 13 21 22 2312 1

Tu v v v v v v

=

Kane’s Equation of Motion:

( ) ( )2 2 6

1 1 1k

i i

i i i i i

R RG LR R R

i G G L L G u j jk

i i jk k

vm a I H F a

u u

= = =

+ + = +

( 1, ,9)k =

Terms associated with translational motion:

1

11

zero

0

RGR

G

k

vm a

u

=

( 1, ,6)k = and ( 10,11,12)k =

1

11 1 11

7

RGR

G

vm a m v

u

=

1

11 1 12

8

RGR

G

vm a m v

u

=

1

11 1 13

9

RGR

G

vm a m v

u

=

2

22

zero

0

RGR

G

k

vm a

u

=

( 1, ,9)k =

2

22 2 21

10

RGR

G

vm a m v

u

=

2

22 2 22

11

RGR

G

vm a m v

u

=

2

22 2 23

12

RGR

G

vm a m v

u

=

Terms associated with rotational motion:

( ) ( ) ( )1

1 1 1 1

1 1 111 11 33 22 12 13

1

RLR R

G L L GI H I I Iu

+ = + −

( ) ( ) ( )1

1 1 1 1

1 1 122 12 11 33 11 13

2

RLR R

G L L GI H I I Iu

+ = + −

( ) ( ) ( )1

1 1 1 1

1 1 133 13 22 11 11 12

3

RLR R

G L L GI H I I Iu

+ = + −

( ) ( ) 1

1 1 1 10

RLR R

G L L G

k

I Hu

+ =

( 4, 12)k =

( ) ( ) 2

2 2 2 20

RLR R

G L L G

k

I Hu

+ =

( 1,2,3)k = and ( 7, 12)k =


( ) ( ) ( )2

2 2 2 2

2 2 211 21 33 22 22 23

4

RLR R

G L L GI H I I Iu

+ = + −

( ) ( ) ( )2

2 2 2 2

2 2 222 22 11 33 21 23

5

RLR R

G L L GI H I I Iu

+ = + −

( ) ( ) ( )2

2 2 2 2

2 2 233 23 22 11 21 22

6

RLR R

G L L GI H I I Iu

+ = + −

Generalized Forces:

The only two active forces are the weight forces that act at the mass centers of the two links. The

generalized forces associated with these two forces can be written as

1 2

1 2k

R RG G

k k

u

v vF W W

u u

= +

( 1, 12)k =

Specifically,

1 2

1 2

zero zero

k

R RG G

k k

u

v vF W W

u u

= +

( 1, ,6)k =

( ) 1

1

7 1 2

7

0

RG

N

u

vF m g N

u

= − =

( ) 1

2

8 1 2 1

8

RG

N

u

vF m g N m g

u

= − = −

( ) 1

3

9 1 2

7

0

RG

N

u

vF m g N

u

= − =

( ) 2

1

10 2 2

10

0

RG

N

u

vF m g N

u

= − =

( ) 2

2

11 2 2 2

11

RG

N

u

vF m g N m g

u

= − = −

( ) 2

3

12 2 2

12

0

RG

N

u

vF m g N

u

= − =


Substituting into Kane’s equations of motion:

Equation for 1 11u = :

( ) ( ) ( ) ( )

( ) ( ) ( )

1 1 1 1 1 111 11 33 22 12 13 1 31 1 1 32 2 1 33 3

1 1 11 31 4 1 32 5 1 33 6

1 1 12 2 2

I I I R R R

R R R

+ − = + +

+ + + (10)


( )1 1 122 12 11 33 11 13 0I I I + − = (11)


( ) ( ) ( ) ( )

( ) ( ) ( )

1 1 1 1 1 133 13 22 11 11 12 1 11 1 1 12 2 1 13 3

1 1 11 11 4 1 12 5 1 13 6

1 1 12 2 2

I I I R R R

R R R

+ − = − − −

− − − (12)


( ) ( ) ( ) ( )2 2 2 2 2 211 21 33 22 22 23 2 31 4 2 32 5 2 33 6

1 1 12 2 2

I I I R R R + − = + + (13)


( )2 2 222 22 11 33 21 23 0I I I + − = (14)


( ) ( ) ( ) ( )2 2 2 2 2 233 23 22 11 21 22 2 11 4 2 12 5 2 13 6

1 1 12 2 2

I I I R R R + − = − − − (15)

Equation for 7 1u x= : 1 11 1m v = (16)

Equation for 8 1u y= : 1 12 1 2m v m g = − + (17)

Equation for 9 1u z= : 1 13 3m v = (18)

Equation for 10 2u x= : 2 21 4m v = (19)

Equation for 11 2u y= : 2 22 2 5m v m g = − + (20)

Equation for 12 2u z= : 2 23 6m v = (21)

Differentiated Constraint Equations:

As determined above, the constraint equations relating the inertial mass-center velocities components and

the body-fixed angular velocity components are

11 13

12 1 1

1113

12

0T

v

v R

v

= −

21 13 23

22 1 1 2 2

11 2123

12

0 0T T

v

v R R

v

= + − −


These equations can be differentiated again to form a set of six, second-order differential equations.

Differentiating the first set of equations gives

11 13 13 13 13

12 1 1 1 1 1 1 1 1 1

11 11 11 1113

13

1 1 1 1

11

1 1 1 12 2 2 2

1 12 2

0 0 0 0

0

0

T T T T

T T

v

v R R R R

v

R R

= + = + − − − −

= + −

13 12 13

13 11

1112 11

11 1213

2 21 1 1 1 13 11

11 12 13

1 12 2

0 0

0

0T T

R R

−

− −−

−

= + + − −

13 11 1211

2 212 1 1 11 13

13 11 12 13

12

0

0

0

T

v

v R

v

−

− + =

− −

(22)

Differentiating the second set of equations gives

21 13 23 13 23

22 1 1 2 2 1 1 2 2

11 21 11 2123

13 23

1 1 2 2

11 21

1 12 2

12

0 0 0 0

0 0

T T T T

T T

v

v R R R R

v

R R

= + + + − − − −

= + − −

13 23

1 1 1 2 2 2

11 21

12

0 0T T

R R

+ + − −

13 23

1 1 2 2

11 21

13 12 13 23 22 23

1 1 13 11 2 2 23 21

11 2112 11 22 21

12

12

0 0

0 0

0 0 0 0

0 0

T T

T T

R R

R R

= + − −

− −

+ − + − − −− −

13 11 12 23 21 2221

2 2 2 222 1 1 11 13 2 2 21 23

23 11 12 13 21 22 23

12

0

0

0

T T

v

v R R

v

− −

− + − + =

− − − −

(23)


Eqs. (10) through (23) form a set of eighteen, first-order, differential/algebraic equations in as many

unknowns – 11 12 13 21 22 23 11 12 13 21 22 23 1 2 3 4 5 6, , , , , , , , , , , , , , , , ,v v v v v v .

It can be shown that Eqs. (10) through (23) are equivalent to those found in Example 6 of Unit 5. In that

example, Kane’s equations were used to generate the equations of motion for a set of six independent

generalized speeds – the body-fixed angular velocity components. To show the equations are equivalent, first

solve Eqs. (16) through (21) for the Lagrange multipliers in terms of the six velocity components. Then

substitute those results into the right sides of Eqs. (10), (12), (13), and (15). Combining those terms into a

matrix form, the differentiated constraint equations (22) and (23) can be used to eliminate the velocity

components in favor of the angular velocity components. The resulting equations are identical to those found

in Unit 5.

Exercises:

6.1 In exercises in Units 4 and 5 of this volume, equations of motion were

formulated for this system using a single generalized coordinate .

a) Using Lagrange’s equations or d’Alembert’s principle, formulate the

equations of motion of the system using the set of dependent generalized

coordinates ( , ,x y ). There are five differential equations in all. The

equations will contain five variables: ( , , )x y and two Lagrange

multipliers 1 2( , ) . b) By eliminating 1 ,

2 , x , and y from the

equations of motion, show that the five equations found in part (a) are

equivalent to the single equation found in Exercises 4.3 and 5.1.

Answers:

Three differential/algebraic equations of motion Two differentiated constraint equations

( ) ( ) ( )

1

2

211 212

14

1 14 2

( ) ( ) ( )p p

mx cx k x

m m y m m g F t

m C S

+ + =

+ + + = −

= − +

6.2 In exercises in Units 4 and 5 of this volume, equations of

motion were formulated for this system using two

generalized coordinates and x . a) Using Lagrange’s

equations or d’Alembert’s principle, formulate the

equations of motion of the system using the set of

dependent generalized coordinates ( , , , )G Gx x y . Here,

( , )G Gx y represent the X and Y coordinates of G relative

to A. There are six differential equations in all.

( ) ( )

( ) ( )

2

2

0

0

x C S

y S C

− + =

+ + =


The equations will contain six variables: ( , , , )G Gx x y and two Lagrange multipliers 1 2( , ) . b) By

eliminating Gx , Gy , 1 , and 2 from the equations of motion, show the six equations found in part (a)

are equivalent to the two equations of motion found in Exercises 4.4 and 5.2.

Answers:

Four differential/algebraic equations of motion Two differentiated constraint equations

( ) ( ) ( )

1 2 2

2 2 1

2 2 2

212 1 212

1 12 2

( ) ( )G

G

G

m m x m x cx k x F t

m x m x

m y m g

m C S

+ + + + =

+ =

− =

= − +

6.3 Using Lagrange’s equations or d’Alembert’s principle, formulate

the equations of motion of a slider-crank mechanism based on the

two systems shown. Use 1 2( , , )x as three dependent

generalized coordinates. There is a total of five differential

equations for the five variables: three coordinates 1 2( , , )x and

two Lagrange multipliers 1 2( , ) .

Answers:

Equations of motion

( ) 1

21 1 1 1 1 1 1 21 1 1

( ) ( )GI m r m g r C T C S + + = + −

( ) 2 2

22 2 2 2 2 2 2 2 1 2 22 2 2

( ) ( )GI m r m r S x m g r C C S + − + = − +

( ) 22 3 2 2 2 2 2 2 22 2

m m x m r S m r C + − − = −

Constraint equations

1 2 1 2

2 21 1 2 2 1 1 2 2 0C C S S − − + =

2 2

2 21 1 2 2 1 1 2 21 1

0S S x C C − + − − + =

System 2

( ) ( )

( ) ( )

2

2

1 12 2

1 12 2

0

0

G

G

x C S

y S C

− + =

+ + =


6.4 Find the differential equations of motion of the two degree-of-

freedom system shown using Lagrange’s equations with the set

of four dependent generalized coordinates – angles and ,

and disk-fixed coordinates Gy and Gz of G relative to O. The

system consists of a disk D of mass dm and radius R, and a

uniform slender bar B of mass m and length . The angle

describes the rotation of the disk about the Z axis, and the angle

describes the rotation of the bar B about the axisX . A

linear, torsional spring-damper is located between B and D at

pin P. The torsional spring has stiffness k and is unstretched

when 0 = . The torsional damper has coefficient c. A motor

torque M is applied to the disk about the Z axis, and a motor

torque M is applied to B about the axisX .

Answers:

Equations of Motion

( )2 2 22 2

12 12 6

2d G G G

m mm R my my y S S C M + + + + =

( ) 21 2

2 21 1

12 12 2 2m m

S C c k M C S − + + = − −

2

1G Gm y m y − = 2Gm z mg + =

Constraint equations

( ) ( ) 21 12 2Gy C S = − ( ) ( ) 21 1

2 2Gz S C = +

6.5 Spinning Top – Using d’Alembert’s Principle, find the equations of

motion of the three degree-of freedom spinning top shown in the

diagram. Assume the moments of inertia of the top about the 1e and

2e directions are 1 2I I I= = , and the moment of inertia about the 3e

direction is 3I . Also, assume point O is fixed and acts like a ball-and-

socket joint. Use the dependent set of four Euler parameters as

generalized coordinates to define the orientation of the top.

1 2 3 44 1

Tq

=

The unit vector set 1 2 3: ( , , )T e e e is fixed-in and rotates with the top.


Answers:

d’Alembert’s Principle

( ) ( ) ( ) ( )

( ) ( )

2 2 24 1 3 2 2 3 3 3 3 1 3

2 2 24 3 2 3 1 3 4 1

2 2 2 2

2 4

I mL I mL I I mL I

I I mL mgL

+ − + + − + −

+ − − = + +

( ) ( ) ( ) ( )

( ) ( )

2 2 23 1 4 2 1 3 3 4 3 1 3

2 2 23 3 2 3 2 4 3 2

2 2 2 2

2 4

I mL I mL I I mL I

I I mL mgL

+ + + − + + −

+ − − = + +

( ) ( ) ( ) ( )

( ) ( )

2 2 22 1 1 2 4 3 3 1 3 1 3

2 2 22 3 2 3 3 1 2 3

2 2 2 2

2 4

I mL I mL I I mL I

I mL I mgL

− + + + + + + −

+ + − = − + +

( ) ( ) ( ) ( )

( ) ( )

2 2 21 1 2 2 3 3 3 2 3 1 3

2 2 21 3 2 3 4 1 2 4

2 2 2 2

2 4

I mL I mL I I mL I

I mL I mgL

− + − + − − + −

+ + − = − + +

Kinematical differential equations (which includes the constraint equation):

1 4 3 214 3 2 11

2 3 4 123 4 1 2

2

2 1 4 33 2 1 43

31 2 3 44 1 2 3

1 12 2

0

− − − − = = − − − − − − − −

6.6 Given the equations of motion of the spinning top of Exercise 6.5, show that the value of the Lagrange

multiplier is 0 = . Hint: This can be done by multiplying the first equation by 1 , the second by 2 ,

the third by 3 , and the fourth by 4 , and then adding the equations. A similar process can be used to

show the equations of motion of Exercise 6.5 are equivalent to those found in Exercise 5.6 of this volume.

6.7 Spinning Top – Using d’Alembert’s Principle, find the equations of

motion of the three degree-of freedom spinning top shown in the

diagram. Assume the moments of inertia of the top about the 1e and 2e

directions are 1 2I I I= = , and the moment of inertia about the 3e

direction is 3I . Also, assume point O is fixed and acts like a ball in

socket joint. Use the following dependent set of generalized coordinates

– the base-frame coordinates of G to define its location relative to O,

and Euler parameters to define the orientation of the top. That is,

1 2 3 1 2 3 47 1

Tq x x x

=


The unit vector set 1 2 3: ( , , )T e e e is fixed-in and rotates with the top.

Answers:

Position coordinates:

1 1m x = 2 2m x = 3 3m x m g = − +

Euler parameters:

( ) ( )

( ) ( ) ( )

4 1 3 2 3 3 2 3 1 3 2 3 3

3 1 4 2 1 3 1 4

2 2 2

2 2 2

I I I I I I I

L L L

+ − − + − +

= − + + +

( ) ( )

( ) ( ) ( )

3 1 3 2 3 4 2 3 1 3 1 3 3

4 1 3 2 2 3 2 4

2 2 2

2 2 2

I I I I I I I

L L L

+ − + + − −

= − − + +

( ) ( )

( ) ( ) ( )

2 1 3 2 3 1 2 3 1 3 4 3 3

1 1 2 2 3 3 3 4

2 2 2

2 2 2

I I I I I I I

L L L

− + − + + − +

= − − − +

( ) ( )

( ) ( ) ( )

1 1 3 2 3 2 2 3 1 3 3 3 3

2 1 1 2 4 3 4 4

2 2 2

2 2 2

I I I I I I I

L L L

− + − − + − −

= − + − +

Kinematical Differential Equations and Constraint Equations:

1 4 3 214 3 2 11

2 3 4 123 4 1 2

2

2 1 4 33 2 1 43

31 2 3 44 1 2 3

1 12 2

0

− − − − = = − − − − − − − −

( )

( )

( )

1 3 1 4 2 1 3 2 4

2 4 1 3 2 2 3 1 4

3 1 1 2 2 3 3 4 4

20

2 0

02

x L

x L

x L

− + + +

− − + + − =

+ + − −


References:

1. L. Meirovitch, Methods of Analytical Dynamics, McGraw-Hill, 1970.

2. M.L. Boas, Mathematical Methods in the Physical Sciences, John Wiley & Sons, Inc., 1966

3. T.R. Kane, P.W. Likins, and D.A. Levinson, Spacecraft Dynamics, McGraw-Hill, 1983

4. T.R. Kane and D.A. Levinson, Dynamics: Theory and Application, McGraw-Hill, 1985

5. R.L. Huston, Multibody Dynamics, Butterworth-Heinemann, 1990

6. H. Baruh, Analytical Dynamics, McGraw-Hill, 1999

7. H. Josephs and R.L. Huston, Dynamics of Mechanical Systems, CRC Press, 2002

8. R.C. Hibbeler, Engineering Mechanics: Dynamics, 13th Ed., Pearson Prentice Hall, 2013

9. J.L. Meriam and L.G. Craig, Engineering Mechanics: Dynamics, 3rd Ed, 1992

10. F.P. Beer and E.R. Johnston, Jr. Vector Mechanics for Engineers: Dynamics, 4th Ed, 1984


Addendum – The Constraint Relaxation Method: Meaning of Lagrange Multipliers

Previously, it was noted that if a dynamic system is described using generalized coordinates

( 1, , )kq k n= , and if the system is subjected to a set of independent constraint equations of the form

0

1

0n

jk k j

k

a q a=

+ = ( 1, , )j m=

then, Lagrange’s equations of motion can be written using Lagrange multipliers as

1

k

m

q j jk

jk k

d K KF a

dt q q

=

− = +

( 1, , )k n=

or

( )1

k

m

q j jknc

jk k

d L LF a

dt q q

=

− = +

( 1, , )k n=

Here, K is the kinetic energy of the system, kqF is the generalized force associated with the generalized

coordinate kq , L is the Lagrangian of the system, V is the potential energy function for the conservative

forces and torques, ( )kq

ncF is the generalized force associated with kq for the non-conservative forces and

torques, only, j is the Lagrange multiplier associated with the thj constraint equation, and

( 1, , ; 1, , )jka j m k n= = are the coefficients from the constraint equations.

Alternatively, some or all the constraints can be relaxed (or removed) and replaced with unknown force

and/or torque components that are required to maintain the constraints. Then, formulate the equations of

motion including the effects of the constraint forces. Using this approach, the two forms of Lagrange’s

equations can be written as

( )constraintsk kq q

k k

d K KF F

dt q q

− = +

( 1, , )k n=

and

( ) ( )constraintsk kq q

nck k

d L LF F

dt q q

− = +

( 1, , )k n=

Comparing Lagrange’s equations from the two approaches, it is clear that

( )constraints

1k

m

q j jk

j

F a=

= ( 1, , )k n=


It is clear from this last result that the Lagrange multipliers are directly related to the forces and/or torques

required to maintain the constraints.

Example: The Simple Pendulum of Example 1

Consider the simple pendulum of Example 1 with coordinates ( ),x y to

describe the position of the mass m . Then relax the length constraint of

the pendulum and replace the light connecting rod with a force T which is

directed from the mass to the external support. Using this approach,

Lagrange's equations can be written in the form

( ) ( ) ( )constraint constraint

zero

k k kq q qnc

k k

d L LF F F

dt q q

− = + =

Where the Lagrangian is ( )2 212

L m x y mgy= + + , and the contributions of the constraint force on the right

sides of the equations are

( ) ( ) ( ) ( )( ) ( ) ( )constraintxF T v x T x L i y L j x i y j x T x L= = − − + = −

( ) ( ) ( ) ( )( ) ( ) ( )constraintyF T v y T x L i y L j x i y j y T y L= = − − + = −

Substituting into Lagrange's equations and supplementing with the twice differentiated constraint equation

gives the following equations of motion

( )

( )2 2

0

0

0

xL

yL

mx T

m y mg T

xx y y x y

+ =

− + =

+ + + =

In Example 1 the equations of motion for this system were written using Lagrange multipliers as follows.

2 2

0

0

0

mx x

my mg y

xx yy x y

− =

− − =

+ + + =

Comparing these results, T L = − . Hence, the Lagrange multiplier is related to the force per unit length in

the light connecting rod.

an introduction to three-dimensional, rigid body dynamics ......unit 6 lagrange’s equations,...

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