Analytic Mechanics - kcir.pwr.edu.pl

DEPARTMENT OF CYBERNETICS AND ROBOTICSELECTRONICS FACULTYWROCŁAW UNIVERSITY OF SCIENCE AND TECHNOLOGY

Lecture Notes in Automation and Robotics

Wrocław 2021

Krzysztof TchońRobert Muszyński

Analytic Mechanics

Krzysztof Tchon Robert Muszynski

Analytic Mechanics

Lecture Notes

in Automation and Robotics

Compilation: January 19, 2022

Wroc law 2021

Krzysztof Tchon, Robert Muszynski

Wroc law 2021

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Authors

Krzysztof Tchon

Robert Muszynski

Department of Cybernetics and Robotics,

Electronics Faculty,

Wroc law University of Science and Technology,

Poland

Computer typesetting

Robert Muszynski

Krzysztof Tchon

https://creativecommons.org/licenses/by-sa/4.0/legalcode

https://creativecommons.org/licenses/by-sa/4.0/legalcode

Contents

Nomenclature vii

0 Prelude 1

0.1 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . 1

0.2 Invariants of motion . . . . . . . . . . . . . . . . . . . . . . . 3

0.3 Orbits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

0.4 Trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

0.5 Tasks and exercises . . . . . . . . . . . . . . . . . . . . . . . . 7

0.6 Comments and references . . . . . . . . . . . . . . . . . . . . 7

Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1 Newtonian mechanics: kinematics of motion of a material point 9

1.1 Time, space, and motion . . . . . . . . . . . . . . . . . . . . . 9

1.2 Frenet trihedron . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 Curvature and torsion . . . . . . . . . . . . . . . . . . . . . . 13

1.4 Frenet-Serret equations . . . . . . . . . . . . . . . . . . . . . 15

1.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5.1 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . 17

1.5.2 Planar curve with constant curvature . . . . . . . . . 17

1.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

2 Newtonian mechanics: dynamics of a system of material points 20

2.1 Law of Universal Gravitation . . . . . . . . . . . . . . . . . . 21

2.2 Newton's Laws of Dynamics . . . . . . . . . . . . . . . . . . . 22

2.3 Linear and angular momentum, energy . . . . . . . . . . . . . 23

2.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

2.4.1 Non-uniqueness of solution of the equation of motion . 26

2.4.2 Mathematical pendulum . . . . . . . . . . . . . . . . . 26

iii

Contents iv

2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30

3 Elements of variational calculus 31

3.1 Dierentiation . . . . . . . . . . . . . . . . . . . . . . . . . . 31

3.2 Functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3.3 Extrema of functional . . . . . . . . . . . . . . . . . . . . . . 33

3.4 Conditional extrema . . . . . . . . . . . . . . . . . . . . . . . 35

3.5 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

3.5.1 Gateaux derivative with respect to vector or matrix . 37

3.5.2 Shortest line on a plane . . . . . . . . . . . . . . . . . 37

3.5.3 Curve producing minimal lateral area of a rotational

gure . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3.5.4 Dido's problem . . . . . . . . . . . . . . . . . . . . . . 38

3.5.5 Brockett's integrator . . . . . . . . . . . . . . . . . . . 40

3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

4 Lagrangian mechanics 45

4.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.1.1 Ball and beam . . . . . . . . . . . . . . . . . . . . . . 46

4.1.2 Furuta's pendulum . . . . . . . . . . . . . . . . . . . . 48

4.2 General form of Lagrangian equations of motion . . . . . . . 49

4.3 Geometric interpretation of Lagrangian mechanics . . . . . . 51

4.4 Examples, continuation . . . . . . . . . . . . . . . . . . . . . 53

4.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

5 Hamiltonian mechanics 58

5.1 Legendre transform . . . . . . . . . . . . . . . . . . . . . . . . 58

5.2 Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

5.3 Canonical Hamilton's equations . . . . . . . . . . . . . . . . . 60

5.4 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

5.4.1 Ball and beam . . . . . . . . . . . . . . . . . . . . . . 62

5.4.2 Furuta's pendulum . . . . . . . . . . . . . . . . . . . . 62

5.5 Invariants. Poisson bracket . . . . . . . . . . . . . . . . . . . 63

5.6 Liouville's theorem on invariants . . . . . . . . . . . . . . . . 64

Contents v

5.7 Examples: invariants of motion . . . . . . . . . . . . . . . . . 65

5.8 Liouville's theorem on divergence . . . . . . . . . . . . . . . . 66

5.9 Examples: divergence . . . . . . . . . . . . . . . . . . . . . . 68

5.9.1 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.9.2 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . 69

5.10 Poincare's theorem on recurrence . . . . . . . . . . . . . . . . 69

5.11 Examples: a dynamic system on torus . . . . . . . . . . . . . 70

5.12 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

6 Kinematics and dynamics of rigid body 74

6.1 Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

6.2 Elementary rotations . . . . . . . . . . . . . . . . . . . . . . . 76

6.3 Coordinates for SE(3) . . . . . . . . . . . . . . . . . . . . . . 77

6.4 Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.5 Lagrangian dynamics . . . . . . . . . . . . . . . . . . . . . . . 80

6.6 Euler-Lagrange equations . . . . . . . . . . . . . . . . . . . . 82

6.7 Euler-Newton equations . . . . . . . . . . . . . . . . . . . . . 83

6.8 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

6.9 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86

7 Lagrange top 87

7.1 Euler-Lagrange equations . . . . . . . . . . . . . . . . . . . . 88

7.2 Canonical Hamilton's equations . . . . . . . . . . . . . . . . . 90

7.3 Invariants and quadratures . . . . . . . . . . . . . . . . . . . 90

7.4 Motion of the Lagrange top . . . . . . . . . . . . . . . . . . . 91

7.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

8 Systems with constraints 97

8.1 Conguration constraints . . . . . . . . . . . . . . . . . . . . 97

8.1.1 Spherical pendulum . . . . . . . . . . . . . . . . . . . 98

8.2 Phase constraints . . . . . . . . . . . . . . . . . . . . . . . . . 99

8.2.1 Wheel, skate, and ski . . . . . . . . . . . . . . . . . . . 100

8.2.2 Rolling wheel . . . . . . . . . . . . . . . . . . . . . . . 101

8.2.3 Kinematic car . . . . . . . . . . . . . . . . . . . . . . . 103

Contents vi

8.3 Constraints for rigid body motion . . . . . . . . . . . . . . . 104

8.3.1 Rolling ball . . . . . . . . . . . . . . . . . . . . . . . . 105

8.4 Holonomic and non-holonomic constraints . . . . . . . . . . . 108

8.4.1 Wheel moving without lateral slip . . . . . . . . . . . 109

8.4.2 Non-holonomicity condition . . . . . . . . . . . . . . . 109

8.4.3 Example . . . . . . . . . . . . . . . . . . . . . . . . . . 110

8.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

9 Dynamics of non-holonomic systems 115

9.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117

9.1.1 Chaplygin's skater . . . . . . . . . . . . . . . . . . . . 117

9.1.2 Wheel rolling vertically . . . . . . . . . . . . . . . . . 119

9.1.3 Rolling ball . . . . . . . . . . . . . . . . . . . . . . . . 122

9.1.4 Tilted wheel . . . . . . . . . . . . . . . . . . . . . . . 125

9.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 128


Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129

Index 130

List of Figures 133

List of Theorems 135

The book is typeset with LATEX, the document preparation system, orig-inally written by L. Lamport [Lam94], which is an extension of TEX[Knu86a,Knu86b]. The typeface used for mathematics throughout thisbook, named AMS Euler, is a design by Hermann Zapf [KZ86], com-missioned by the American Mathematical Society. The text is set ina typeface called Concrete Roman and Italic, a special version of Knuth'sComputer Modern family with weights designed to blend with AMS Eu-ler, prepared to typeset [GKP89].

[GKP89] R. L. Graham, D. E. Knuth, and O. Patashnik, ConcreteMathematicsa. Addison-Weslay, Reading, 1989.

[Knu86a] D. E. Knuth, The TEXbook, volume A of Computers andTypesetting. Addison-Wesley, Reading, 1986.

[Knu86b] D. E. Knuth, TEX: The Program, volume B of Computersand Typesetting. Addison-Wesley, Reading, 1986.

[KZ86] D. E. Knuth, and H. Zapf, AMS Euler | A new typeface formathematics. Scholary Publishing, 20:131157, 1986.

[Lam94] L. Lamport, LATEX: A Document Preparation System.Addison-Wesley, Reading, 1994.

Nomenclature

Poisson bracket (63)

A(q) Pfaan matrix (100)

C(q, _q) matrix of Coriolis forces (50)

ckij(q) Christoel's symbols of the 1st kind (50)

Γkij(q) Christoel's symbols of the 2nd kind (50)

div divergence (67)

ϕ(t, x) ow (66)

(ϕ, θ,ψ) Euler angles (78)

H(q,p) Hamiltonian (60)

I(q(·)) action (45)

K curvature (14)

K(q, _q) kinetic energy (45)

L Lie algebra (110)

L(q, _q) Lagrangian (45)

M angular momentum of system (24)

P linear momentum of system (24)

Q(q) inertia matrix (49)

R rotation matrix (74)

R set of real numbers (9)

R3 3-dimensional real space (9)

R(X,α), R(Y,β), R(Z,γ) elementary rotations (77)

S1 unit cirle (70)

S2 unit sphere (53)

SE(3) special Euclidean group (76)

SO(3) special orthogonal group (77)

vii

Nomenclature viii

T torsion (15)

T translation vector (74)

T2 torus (70)

V(q) potential energy (45)

vTQ(q)w Riemannian metric (52)

D derivative of function (31)

X Banach space (31)

X(x) vector eld (66)

Ω matrix rotation velocity (78)

ω vector rotation velocity (79)

Chapter 0

Prelude

These lecture notes serve as a basic teaching aid for a 3-semester under-

graduate course in analytic mechanics for students of control engineering

and robotics at the Electronics Faculty of the Wroc law University of Science

and Technology, Wroc law, Poland. Their objective is to present methods

and tools for dening mathematical descriptions of motion of mechanical

systems encountered in control engineering and robotics. In a sequence of

chapters of the notes we shall show three approaches to mechanics: Newto-

nian, Lagrangian, and Hamiltonian mechanics, and nally give a treatment

of the kinematics and dynamics of systems with constraints.

In order to provide the Reader with an overview of the subject, methods,

and results of analytic mechanics in this Prelude we shall present a deriva-

tion of the 1st Kepler's Law concerned with the motion of Planets around

the Sun. This law predicates that the Planets are moving around the Sun in

elliptic orbits having the Sun in one of their focal points. During our deriva-

tion we shall assume that the Reader is acquainted with the 2nd Newton's

Principle of Dynamics as well as with the Newton's Law of Universal Gravi-

tation. We shall demonstrate the methods used by analytic mechanics: laws

of motion, coordinate changes, quest for invariants of motion, pursuit for

solving the equations of motion by quadratures, determination of orbits and

trajectories of motion.

0.1 Equations of motion

Suppose that the trajectory of motion of a Planet around the Sun lies in

the plane. Both the Planet as well as the Sun will be regarded as material

points. Let attach a coordinate frame at the center of the Sun, and describe

1

Chapter 0. Prelude 2

y

Y

rF

m

x M

θ

X

P

S

Figure 1: Position of the Planet around the Sun

the position of the Planet with respect to the Sun using the Cartesian co-

ordinates (x,y)T , see Figure 1. Let M denote the mas of the Sun, and m

the mass of the Planet.

In order to formulate equations of motion of the Planet we shall invoke

the 2nd Newton's Principle of Dynamics,

ma = F,

where F refers to the force of gravitational attraction exerted on the Planet,

dened by the Law of Universal Gravitation, and a denotes the Planet's

acceleration. Assuming that the position of the Planet at the time instant t

is (x(t),y(t))T we arrive at the following equation of motion in the Cartesian

coordinates∗

m

(x

y

)= −

GMm

x2 + y21√

x2 + y2

(x

y

). (0.1)

To simplify our notations, in what follows we set the coecient GM = 1.

As can be seen, the Cartesian equations look rather complex, so to make

them simpler we shall try to introduce alternative coordinates. The polar

coordinates seem to be a natural candidate for that, i.e.x = r cos θ

y = r sin θ.

Having computed the derivatives and performed necessary mathematical

transformations we have obtained the following equations of motion in polar

∗Following Isaac Newton, in these lecture notes we will use dots above the func-

tions/variables symbols to denote their time derivatives so, the double dots applied

below indicate the second time derivative of the position, i.e. acceleration.


coordinates r− r _θ2 + 1

r2= 0

rθ+ 2_r _θ = 0. (0.2)

Our objective is now to determine the Planet's trajectory (r(t), θ(t))T .

0.2 Invariants of motion

A classical approach to solving equations of motion assumed in analytic

mechanics consists of the determination of so called invariants (or constants)

of motion. The invariants are certain functions of the position and velocity

that remain constant in time on trajectories of motion. As we shall see

later on, given a suitable number and quality of invariants will guarantee

the solvability of the equations of motion by quadratures, i.e. in the closed

form.

To be more specic, let's consider the angular momentum h = mr2 _θ of

the Planet. Its time-derivative along a trajectory of equations (0.2) amounts

to_h = 2mr_r _θ+mr2θ = mr

(rθ+ 2_r _θ

)= 0.

In this way we have proved that the angular momentum h is an invariant

of motion.

Another candidate for an invariant is the total (kinetic plus potential)

energy of the Planet given by

E =1

2m(

_x2 + _y2)−m

r=1

2m_r2 +

1

2h _θ−

m

r,

where the last component stands for the potential energy in the gravitational

eld. The time-derivative of the energy is equal to

_E = m_rr+1

2hθ+

m

r2_r.

Along a trajectory of motion satisfying equations (0.2) there holds

_E = m_r

(r+

1

r2

)+1

2hθ = m_rr _θ2 +

1

2hθ =

1

2

h

r

(rθ+ 2_r _θ

)= 0,

what means that the energy is also an invariant of motion.


0.3 Orbits

As we shall see, these two invariants of motion will allow us to solve the

equations of motion of the Planet. To this objective, instead on the trajec-

tory (r(t), θ(t))T , we temporarily concentrate on nding the orbit r = r(θ)

of motion. More specically, we shall compute the function u(θ) = r−1(θ).

We begin with expressing the energy in terms of u(θ),_r = _(u−1) = −u−2

du

dθ_θ = −r2 _θ

du

dθ_θ =

h

mu2

,

so

E =1

2

h2

m

((du

dθ

)2+ u2

)−mu.

After re-writing the last expression as(du

dθ

)2+ u2 =

2m

h2(E+mu) (0.3)

and dierentiating with respect to time we obtain

_θdu

dθ

(d2u

dθ2+ u−

m2

h2

)= 0.

It follows that, from the mathematical point of view, the Planet is able

to make three kinds of motion: θ = const the rectilinear motion along the

radius r, u = const the circular motion and, nally, the motion described

by the linear dierential equation

d2u

dθ2+ u =

m2

h2. (0.4)

The rst two motions are not supported by observations, so the only possi-

bility is the motion number three.

The corresponding dierential equation is easy to be solved, resulting in

u(θ) = r−1(θ) =m2

h2+ C cos(θ+ θ0)

for certain constants C and θ0. Assuming that θ0 = 0 the other constant C

can be found from the expression (0.3) for the angle θ = π/2. This gives us

u = m2

h2, dudθ = −C, and yields

C =m2

h2

(1+

2Eh2

m3

)1/2.


Having substituted the constant we get the orbit of the Planet

r(1+ ε cos θ) = l, (0.5)

where ε =(1+ 2Eh2

m3

)1/2and l = h2

m2 . The formula (0.5) represents an

ellipse in polar coordinates and constitutes the essence of the 1st Kepler's

Law. The coecient ε is called the ellipse eccentrity and l is the parameter

of the ellipse. Coecients ε and l determine the half-axes: longer a and

shorter b of the ellipse appearing in the well known Cartesian equation of

the ellipse (xa

)2+(yb

)2= 1.

Using the denition of the ellipse (a geometric locus of points with xed

sum of distances from the focal points) it can be shown that

a =l

1− ε2, b =

l√1− ε2

, b = a√1− ε2. (0.6)

Observe that in order to have the elliptic orbit the eccentrity ε must be less

than 1 implying that the energy E of the planet must be negative.

0.4 Trajectory

Relying on two invariants of motion we have been able to discover the orbit

of the Planet that for a given angle θ denes the position r(θ) of the Planet.

However, in some applications of astronomy (like establishing the date of

Easter) we need to know not only the orbit, but also the Planet's trajectory

(r(t), θ(t))T . To recover the trajectory we shall invoke the invariant angular

momentum

h = mr2dθ

dt= const .

Using the equation of the orbit (0.5) we are able to deduce the following

dierential equationh

ml2dt =

dθ

(1+ ε cos θ)2

that, after integration, delivers the function

t(θ) =ml2

h

∫θ0

dϑ

(1+ ε cos ϑ)2. (0.7)


y

Y

r

x

θ

X,X ′

P

S

Y ′

u

ea

b

Figure 2: Ellipse

The inverse function θ(t) = t−1(θ) yields the Planet's trajectory: in polar

coordinates (r(t) =

l

1+ ε cos θ(t), θ(t)

)Tas well as in Cartesian coordinates

x(t) = r(t) cos θ(t)

y(t) = r(t) sin θ(t).

(The remaining portion of material can be skipped.) Below we shall unveil

some details of computation of function (0.7). We use the parametric equa-

tions of the ellipse (see Figure 2). Denoting the parameter by u we get the

following identities x = a cosu− e = r cos θ

y = b sinu = r sin θ, (0.8)

where e is the distance of the focal point of the ellipse from its geometric

center. Now, it is easily shown that

e = a− r(0) = aε,


resulting in x = a(cosu− ε). Referring to Figure 2 and (0.8) we obtain

r =√x2 + y2 = a(1− ε cosu).

Since simultaneously

r =l

1+ ε cos θ,

taking the dierentials of the right hand sides of the two last formulas fol-

lowed by suitable substitutions from (0.8), leads to

ldθ

(1+ ε cos θ)2=a sinu

sin θdu =

la2

b(1− ε cosu)du.

Eventually, the integral we have been searching for is dened by so called

Kepler's equation

t(u) =mab

h(u− ε sinu),

that needs to be solved with respect to the function u(t). The dependence

of the angle θ from the parameter u follows directly from (0.8)

tan θ =b sinu

a(cosu− ε).

0.5 Tasks and exercises

Exercise 0.1 Using the denition of the ellipse derive (0.6).

Exercise 0.2 Relying on the 1st Kepler's Law: The Planet moves around

the Sun in the elliptic orbit, derive the 2nd Law: The polar velocity of the

Planet (the area covered by the radius r in unit time) is constant.

Exercise 0.3 Relying on the 1st and the 2nd Kepler's derive the 3rd Law:

The ratio of squared circulation periods around the Sun of two Planets is

equal to the ratio of cubes of the longer half-axes of their orbits,

T21T22

=a31a32

.

0.6 Comments and references

The Kepler's laws were published in two volumes by Johannes Kepler:

[Kep09] contains the 1st and the 2nd law, while [Kep19] presents the 3rd

law. There exist English translations of these works. Our derivation of the

1st Kepler's Law is based on chapter 2 of the monograph [RK95].


Bibliography

[Kep09] J. Kepler. Astronomia nova. Voegelin, Heidelberg, 1609.

[Kep19] J. Kepler. Harmonices mundi. J. Planck, Linz, 1619.

[RK95] W. Rubinowicz and W. Krolikowski. Mechanika teoretyczna.

PWN, Warszawa, 1995. (in Polish).

Chapter 1

Newtonian mechanics:

kinematics of motion of

a material point

The Newtonian mechanics deals with the motion of a material point (a par-

ticle) or of a collection of material points. The concept of the material point

will be regarded as a primitive, undened concept. The scenery of motion

of a material point is constituted by time and space.

1.1 Time, space, and motion

In mechanics time is identied with the set of real numbers R. The set Ris innite, unbounded, linearly ordered, connected, and continuous; the last

property means that between any two real numbers there is a third number.

The mechanical space is tantamount to the 3-dimensional real space R3.Elements of R3 are triples of real numbers of the form u = (u1,u2,u3)

T ,

v = (v1, v2, v3)T , that are interpreted as coordinates of the material point

with respect to a certain coordinate frame. Elements of space can be added,

u+ v =

u1 + v1u2 + v2u3 + v3

,

and multiplied by real numbers, for any α ∈ R there holds

αu =

αu1αu2αu3

,

9

Chapter 1. Newtonian mechanics: kinematics of. . . 10

what implies that R3 is a vector space over R, and its elements will be called

vectors. Furthermore, two vectors u, v ∈ R3 can also be multiplied by each

other. The result is a number,

(u, v) = uTv =

3∑i=1

uivi ∈ R,

and multiplication is called the inner or scalar product of vectors. Comple-

mentarily, a vector or cross product of a pair of vectors produces a vector

computed as

u× v = det

i j k

u1 u2 u3v1 v2 v3

= i(u2v3 − v2u3) − j(u1v3 − v1u3)+

k(u1v2 − v1u2) ∈ R3,

where i, j, k denote unit vectors in R3. The inner product dened above will

be referred to as Euclidean. It is easily observed that the inner product is

commutative, (u, v) = (v,u) and bilinear, i.e. for two real numbers α1,α2 ∈R and three vectors u, v,w ∈ R3 we have (α1u + α2v,w) = α1(u,w) +

α2(v,w). Dierently to the inner product the cross product of vectors is

antisymmetric, u × v = −v × u (thus irre exive u × u = 0); it is also non-

associative, because (u× v)×w = u× (v×w) + v× (w×u) 6= u× (v×w).By means of the inner product we dene the norm (length) of a vector

||u|| = (u,u)1/2 =

√√√√ 3∑i=1

u2i

referred to as the Euclidean norm. The space R3 equipped with the Eu-

clidean norm will be called the Euclidean space. Let's recall the axioms of

norm: ||u|| = 0 ⇔ u = 0, for any α ∈ R ||αu|| = |α| ||u||, and the triangle

inequality ||u+ v|| 6 ||u||+ ||v||.

Besides the inner and the cross product we often use the mixed product

of vectors. Given three vectors u, v,w ∈ R3, the mixed product is

(u× v,w) = det

u1 u2 u3v1 v2 v3w1 w2 w3

= det

u1 v1 w1u2 v2 w2u3 v3 w3

.

All three kinds of products of vectors from R3 can be given a nice geometric

interpretation:


∠(u, v)

v

u

v− (u, v)u

(u, v)u

Figure 1.1: Inner product

∠(u, v)

v

u

u× v

||v|| sin∠(u, v)

Figure 1.2: Cross product

1. inner product:

(u, v) = ||u|| ||v|| cos∠(u, v)

• if ||u|| = 1 then (u, v) = ||v|| cos∠(u, v) denes a projection of

vector v onto vector u,

• if (u, v) = 0 then either ||u|| = 0 or ||v|| = 0 or cos∠(u, v) = 0; the

last property means that vectors u and v are perpendicular, u⊥v;this interpretation of the inner product is shown in Figure 1.1;

2. cross product:

||u× v|| = ||u|| ||v|| sin∠(u, v)

• length (norm) of the cross product is equal to the area of the

parallelogram spanned by vectors u and v,

• direction of the cross product is perpendicular to the plane spanned

by vectors u and v, and oriented in accordance with the clock-

wise screw; this interpretation of the cross product is displayed

in Figure 1.2;


v

u

u× v

||w|| cos∠(u× v,w)

w

∠(u×v,w

)

Figure 1.3: Mixed product

3. mixed product:

(u× v,w) = ||u× v|| ||w|| cos∠(u× v,w)

is equal to the volume of the parallelepiped spanned by vectors u, v,

w; this is illustrated in Figure 1.3.

Time and space providing the scenery for the motion are often called the

Galilean space-time. Given the scenery, the motion of a material point will

be dened in the following way.

Denition 1.1.1 The motion of a material point is a map of time into

Euclidean space,

c : R −→ R3, c(t) =

c1(t)c2(t)

c3(t)

.

The map c(t) needs to be continuous and be continuously dierentiable

at least up to order 2.

The requirement of continuity is supported by experience; in macro-world

we do not observe discontinuous motions (teleportation is excluded). The

property of dierentiability will allow us to apply in mechanics the tools of

analysis; it may be relaxed to dierentiability almost everywhere (as motion

with impacts). Also, instead of the whole time axis R we may study motions

restricted to a certain nite time interval [t0, t1].


As a consequence of our denition, the motion of a material point is

identied with the trajectory of the point in space, For a given motion c(t)

the velocity of motion is dened as the time derivative

_c(t) =

_c1(t)

_c2(t)

_c3(t)

.

1.2 Frenet trihedron

In order to get some insight into the geometry of motion we shall study a

geometric construction known as the Frenet trihedron. To this aim, for a

given velocity vector _c(t) we dene the tangent vector to the trajectory at

the point c(t),

t =_c(t)

|| _c(t)||.

The tangent vector determines at c(t) a plane perpendicular to t, called the

normal plane. Now, denote the point c(t) by P and choose two other points

P1 and P2 lying on the trajectory, from either side of the point P. If not

collinear, these points P1,P,P2 determine a plane in R3. Let P1 and P2approach P. Then the sequence of planes passing through P1,P,P2 tends

in the limit to a plane containing the tangent vector t, called the strictly

tangent plane (to the trajectory c(t)). The intersection of the tangent plane

and the normal plane denes the normal vector n, of unit length, oriented

toward the curvature of the trajectory. The tangent and the normal vectors

are completed by the bi-normal vector b = t × n; these vectors form at

the point c(t) so called Frenet trihedron shown in Figure 1.4. The Frenet

trihedron travels with time along the trajectory of motion of the material

point and allows to characterize its properties.

To pursue our analysis, let's rst recall the formula for the dierential

of length of the trajectory, ds = || _c(t)||dt, implying that if s(0) = 0 then the

length of trajectory traveled from time instant 0 to time instant t is equal

to

s(t) =

∫t0

|| _c(τ)||dτ.

1.3 Curvature and torsion

Let us x on trajectory c(t) a reference point, and consider a point P distant

from it by s. Take the tangent vector t(s) to this trajectory at the point P


b

t

c(t)

n

rectifyingplane

strictly tangentplane

normalplane

tangent

mainnormal

bi-normal

Figure 1.4: Frenet trihedron

and the tangent vector t(s+ ∆s) at a point P ′ shifted with respect to P by

∆s. Let's compute the increment of the tangent vector resulting from the

shift ∆s, and dene the dierential quotient

K(s) = lim∆s→0 t(s+ ∆s) − t(s)

∆s=dt(s)

ds.

The Euclidean norm of this quotient,

K(s) =

∣∣∣∣∣∣∣∣dt(s)ds

∣∣∣∣∣∣∣∣ ,will be referred to as the curvature of the trajectory c(t) at point P,, and

the vectordt(s)

ds= K(s)n(s) = K(s)

is called the vector of curvature. The inverse

R(s) =1

K(s)

of the curvature denes the radius of curvature of the trajectory c(t) at

point P. As a consequence of denition, the curvature denes how much

non-rectilinear the trajectory is. Using the denition we derive the following

expression for the curvature of the trajectory of motion at the point c(t)

distant from the reference point by s =∫t0 || _c(τ)||dτ,

K(s) =

√|| _c||2||c||2 − ( _c, c)2

|| _c||3. (1.1)


Analogously to curvature we introduce the notion of torsion of the tra-

jectory. To this objective, instead of the tangent vector, we look at the

variation along the trajectory of the bi-normal vector

T(s) = lim∆s→0 b(s+ ∆s) − b(s)

∆s=db(s)

ds.

The Euclidean norm of this vector is called the torsion of the trajectory ,

T(s) =

∣∣∣∣∣∣∣∣db(s)ds

∣∣∣∣∣∣∣∣ .By denition, the torsion describes how much non-planar the trajectory of

motion is. Planar trajectories lie in the tangent plane and have zero torsion.

In order to compute the torsion we rst nd the bi-normal vector

b =1

K

1

|| _c||3_c× c,

and then derive the formula

T(s) =1

K2(s)

( _c× c, c(3))

|| _c||6(1.2)

describing the torsion of trajectory c(t) at the point P whose distance from

the reference point is s =∫t0 || _c(τ)||dτ.

1.4 Frenet-Serret equations

As has been observed, at any point of trajectory of the material point one

can dene the frame of three orthogonal unit vectors (t(s), n(s), b(s)) moving

along the trajectory. These vectors are in fact equivalent to the Frenet

trihedron. We want to study the motion of this trihedron. Observe that

the tangent, normal and bi-normal vectors form a coordinate frame in R3,so derivatives of these vectors with respect to parameter s have a unique

representation in this frame. This means that there exist functions αi(s),

βi(s) and γi(s), i = 1, 2, 3, such that

dt

ds= α1t+ α2n+ α3b, (1.3)

dn

ds= β1t+ β2n+ β3b, (1.4)

db

ds= γ1t+ γ2n+ γ3b. (1.5)


Now, our task is to determine this representation. It can be noticed imme-

diately that dtds = Kn, therefore α1 = α3 = 0 and α2 = K. We also know

that vectors t(s), n(s) i b(s)) have length 1, (t, t) = (n, n) = (b, b) = 1, and

are orthogonal, i.e. (t, n) = (t, b) = (n, b) = 0. After dierentiation, the

former set of identities give(dt

ds, t

)=

(dn

ds, n

)=

(db

ds, b

)= 0, (1.6)

while the latter results in(dt

ds, n

)+

(t,dn

ds

)=

(dt

ds, b

)+

(t,db

ds

)=

(dn

ds, b

)+

(n,db

ds

)= 0. (1.7)

By taking the inner products of (1.4) and (1.5), respective by n and b,

and invoking (1.6) we get β2 = γ3 = 0. Next, taking into account that(dtds , n

)= K and the rst of identities (1.7) we nd β1 = K. Analogously,

from the fact that(dtds , b

)= 0 and the second of identities (1.7) there follows

that γ1 = 0. In this way we obtain dbds = γ2n. Furthermore, from the

denition of torsion we deduce∣∣∣∣dbds

∣∣∣∣ = |γ2| = T , i.e. γ2 = ±T. Finally,

the last identity in (1.7) yields β3 = −γ2. Having chosen γ2 = −T we

get β3 = T , and arrive at the following system of equations that is called

Frenet-Serret equations dtds = Kndnds = −Kt+ Tbdbds = −Tn

. (1.8)

We let R(s) = [t(s), n(s), b(s)] denote a matrix whose columns are the

tangent, normal, and bi-normal vectors. It is easy to show that equations

(1.8) can be rendered the following matrix form

dR(s)

ds= R(s)A(s), (1.9)

where matrix A(s) =

[0 −K(s) 0K(s) 0 −T(s)0 T(s) 0

]is a skew-symmetric, AT = −A,

determined by the curvature and torsion. From the theorem on existence

and uniqueness of solution of a dierential equation it follows that the ma-

trix dierential equation (1.9) has a solution R(s) depending on the initial

condition R(0). Selecting the column 1 of the solution matrix R(s) we get


the tangent vector t(s). Assuming that the motion trajectory has been pa-

rameterized by the arc length s, i.e. that c = c(s), we obtain the dierential

equationdc(s)

ds=

_c

|| _c||= t(s).

Its integration results in the trajectory of motion parameterized by s,

c(s) = c(0) +

∫s0

t(τ)dτ. (1.10)

In this way we have demonstrated that the curvature and the torsion deter-

mine the trajectory of motion of the material point modulo the parameter-

ization.

1.5 Examples

1.5.1 Acceleration

In order to explain the role played by the tangent and the normal vectors

we shall describe the vector of acceleration of the material point. Directly

from denition it follows that _c = || _c||t, therefore, the acceleration

c =d|| _c||

dtt+ || _c||

dt

dt.

After substituting into this identity dtdt =

dtdsdsdt = dt

ds || _c||, and making use ofdtds = Kn we get

c =d|| _c||

dtt+ || _c||2Kn,

which means that the acceleration vector lies in the tangent plane.

1.5.2 Planar curve with constant curvature

To exemplify an application of the Frenet-Serret equations we shall deter-

mine a curve with zero torsion and constant curvature K. Consider the

equation (1.9) with constant matrix

A =

0 −K 0

K 0 0

0 0 0

.

It is well known that the solution of this equation has the form R(s) =

R(0) exp (sA), where the matrix exponential is dened as exp (sA) =∑∞i=0

(sA)i

i! ,


and R(0) = [t(0), n(0), b(0)] stands for the initial condition. Suppose that

t(0) = e1, n(0) = e2 i b(0) = 0. From denition we compute

R(s) = R(0)

cosKs − sinKs 0

sinKs cosKs 0

0 0 1

,

that yields

t(s) =

cosKs

sinKs

0

.

Assuming that c(0) = 0, from (1.1) we deducec1(s) =

1K sinKs

c2(s) =1K − 1

K cosKs

c3(s) = 0

.

It is easily found that a planar curve of constant curvature K is the circle

described by

c21 +

(c2 −

1

K

)2=1

K2.

1.6 Exercises

Exercise 1.1 Using the parametric equations compute the curvature and tor-

sion of the circle, the ellipse, the cycloid and the screw line.

Exercise 1.2 Derive the formula (1.1) for curvature and the formula (1.2)

for torsion.

Exercise 1.3 Show that for a planar trajectory c(t) = (x(t),y(t)> the cur-

vature is given by

K(s) =| _xy− _yx|

( _x2 + _y2)3/2.


The essence of the primary concept such as the material point is perfectly

re ected by the denition of the horse provided by the rst Polish En-

cyclopedia authored by rev. Benedykt (Benedictus) Chmielowski [Chm45]:


"A horse, how it looks like, everyone can see". This Encyclopedia and

its author have been referred to at many places by the Polish 2019 Nobel

prize winner, Olga Tokarczuk, in her opus magnum "Ksiegi Jakubowe".

By the way, the title of this Encyclopedia could serve as a template ti-

tle for any ambitious author. It goes like the following: "New Athens or

the Academy, Full of All Sciences, Partitioned into Various Titles like the

Classes, Erected to be Memorized by the Wise, to be Learnt by Idiots, to

be Practiced by Politicians, to Entertain the Melancholics." The particle

dynamics are treated in chapter I of [Gre03]. Additional knowledge on ge-

ometry of curves in R3, including the Frenet-Serret equations, the reader

can nd in the textbook [Goe70].

Bibliography

[Chm45] B. Chmielowski. Nowe Ateny Albo Akademiia Wszelkiej Scyen-

cyi Pe lna, Na Ro_zne Tytu ly Jak Na Classes Podzielona,

Madrym Dla Memoryja lu, Idiotom Dla Nauki, Politykom Dla

Praktyki, Melankolikom Dla Rozrywki Erygowana. P. J. Gol-

czewski, Lwow, 1745. (in Polish).

[Goe70] A. Goetz. Introduction to Dierential Geometry. Addison-

Wesley, Reading, Mass., 1970.

[Gre03] D. T. Greenwood. Advanced Dynamnics. Cambridge University

Press, Cambridge, 2003.

Chapter 2

Newtonian mechanics: dynamics

of a system of material points

Consider a system of n material points in the Euclidean space. By the

motion of this system we mean a continuous and continuously dierentiable

suciently many times function

c : R −→ RN, c(t) =

c1(t)

c2(t)...

cn(t)

,

where ci denotes the position of the point i and N = 3n. In accordance

with the Newtonian mechanics the motion of the system is subject to two

principles: the Principle of Determinism and the Principle of Invariance

(Relativity). The Principle of Determinism states that the motion is de-

ned by the initial position and the initial velocity of points belonging to

the system. A mathematical consequence of this is that the motion can be

described by an ordinary, 2nd order dierential equation,

c = F(c, _c, t), ci = Fi(c, _c, t), i = 1, 2 . . . ,n,

implying that c(t) = ϕ(t, c(0), _c(0)). The function F is called a law of

motion. By default we assume that the equation of motion has a unique

solution. The law of motion in Newtonian mechanics needs to obey the

Principle of Invariance that requires that this law should be:

• invariant with respect to shifts in time,

F(c, _c, t+ s) = F(c, _c, t), s ∈ R,

20

Chapter 2. Newtonian mechanics: dynamics of. . . 21

(in consequence, F does not depend on time explicitly, so c = F(c, _c)),

• invariant with respect to shifts in space,

F(c1 + u, c2 + u, . . . , cn + u, _c) = F(c1, c2, . . . , cn, _c), u ∈ R3,

(specically, F depends on relative positions),

• invariant with respect to uniform motions,

F(c, _c1 + v, _c2 + v, . . . , _cn + v) = F(c, _c1, _c2, . . . , _cn), v ∈ R3,

(F depends on relative velocities)

• invariant with respect to rotations in space,

Fi(Rc1,Rc2, . . . ,Rcn,R _c1,R _c2, . . . ,R _cn

)=

RFi(c1, c2, . . . , cn, _c1, _c2, . . . , _cn

),

for i = 1, 2 . . . ,n and a rotation matrix R, i.e. a 3 × 3 matrix such

that RRT = RTR = I3, detR = 1. Invariance with respect to rotations

implies that the space is isotropic, there are no singular directions in

space.

2.1 Law of Universal Gravitation

As an example of a law of motion satisfying these two Principles we shall

take the Law of Universal Gravitation. Let's choose two material points

with masses mi and mj, and positions, respectively, ci and cj shown in

Figure 2.1. In accordance with the Law of Universal Gravitation the force

exerted by the mass mi on the mass mj is given as

Fij =Gmimj

||ci − cj||2ci − cj

||ci − cj||,

where G stands for the Gravitation constant. In this expression the rst

factor denes the amount of force, while the second its direction; it is easily

observed that the gravitation force is directed toward the mass mi.

Obviously, this force neither depends on time nor on velocity, and de-

pends on the relative position; thus three rst invariances are guaranteed.

In order to check the rotational invariance, let's take a rotation matrix R

and rotate position vectors: ci 7→ Rci and cj 7→ Rcj. By denition of the


Y

Z

mi

X

mj

ci

cj

c j−c i

c i−c j

Figure 2.1: A pair of material points

rotation matrix it follows that ||Rci − Rcj|| = ||R(ci − cj)|| = ||ci − cj|| (after

rotation the length of a vecor remains the same), moreover, we also have

Rci − Rcj = R(ci − cj), therefore

Gmimj

||Rci − Rcj||2Rci − Rcj

||Rci − Rcj||= R

Gmimj

||ci − cj||2ci − cj

||ci − cj||.

2.2 Newton's Laws of Dynamics

It is well known and taught in secondary school that the Newtonian mechacics

relies on the following three laws. The formulation given below is a trans-

lation from Latin.

• Corpus omne perseverare in statu suo quiescendi vel movendi

uniformiter in directum, nisi quatenus illud a viribus impressis

cogitur statuum suum mutare. This means: Every body preserves

its state of rest or of uniform rectilinear motion unless compelled to

change its state by impressed forces.

• Mutationem motus proportinalem esse vi motrici impressae et

eri secundum linear rectam qua vis illa imprimitur. This reads

as: A change of motion (acceleration) is proportional to the impressed

force and directed along a straight line this force is impressed

ci =1

miFi, mic

i = Fi.

• Actioni contrariam semper et aequalem esse reactionem, sive cor-

porum duorum actiones in se mutuo semper esse aequales et in


Y

Z

mi

X

mj

ci

cj

c i−c j

m1

mn

. . .

...

cj − ci

Fjic1

cn

Fzi

Figure 2.2: System of n material points

partes contrarias dirigi. In English: Every action is always accom-

panied by an equal and opposite reaction; i.e. mutual eects on each

other of two bodies have always equal size and opposite direction.

2.3 Linear and angular momentum, energy

When describing the motion of a system of material points within the frame-

work of Newtonian mechanics, it is advantageous to employ concepts of

linear momentum, angular momentum, and energy. In order to introduce

these concepts, we consider a system of n points impressing on each other

gravitational forces, and subject to external forces, going from outside the

system, as shown in Figure 2.2. Given the system, by the 2nd Newton's

Law, the motion of the mass mi satises the equation

mici = Fi = Fwi + Fzi,

where Fwi denotes an internal force, and Fzi refers to an external force (i.e.

a force coming from outside of the system) acting on mi. The provenance

of internal forces is gravitational,

Fwi =∑j6=i

Fji,

where Fji denotes the force impressed by the mass mj on the mass mi. In

accordance with the Law of Universal Gravitation

Fji =Gmjmi

||cj − ci||2cj − ci

||cj − ci||.


In this context, the linear momentum of the mass mi will be dened as

pi = mi _ci,

and the total linear momentum of the system

P =

n∑i=1

pi =

n∑i=1

mi _ci.

Having computed the time-derivative of the total linear momentum we

get the identity

_P =

n∑i=1

mici =

n∑i=1

n∑j=1

Fji +

n∑i=1

Fzi =

n∑i=1

Fzi,

where the last equality results form the 3rd Newton'a Law of Dynamics

(Fji = −Fij). In this way we have come to the following conclusion.

Remark 2.3.1 The speed of change of the total linear momentum of a

system of material points is equal to the sum of external forces acting

on the system. If the sum is 0, the total momentum remains constant.

The last statement expresses the Principle of Conservation of Linear Mo-

mentum.

The next useful concept is that of the angular momentum. The angular

momentum of the mass mi is dened as the cross product

Mi = ci × pi = mici × _ci.

Consequently, the total angular momentum of the system equals

M =

n∑i=1

Mi =

n∑i=1

mici × _ci.

The dierentiation with respect to time yields the identity

_M =

n∑i=1

mi _ci × _ci +

n∑i=1

mici × ci =

n∑i=1

ci ×mici,

where we have used irre exivity of the cross product. Now, from the 2nd

Newton'a Law applied to the mass mi, it follows that

_M =

n∑i=1

ci ×n∑j=1

Fji +

n∑i=1

ci × Fzi.


In the rst component there appear elements of the form

ci × Fji + cj × Fij = ci × Fji − cj × Fji =(ci − cj

)× Fji = 0,

whose vanishing results from the 3rd Newton's Law and collinearity of vec-

tors c ci − cj and Fji (see Figure 2.2). Finally,

_M =

n∑i=1

ci × Fzi,

which can be phrased as the following.

Remark 2.3.2 The speed of change of the total angular momentum of

a system is equal to the sum of torques of external forces acting on

this system. If this sum equals 0, the total angular momentum remains

constant.

The last statement is referred to as the Principle of Conservation of Angular

Momentum.

We are passing to the concept of energy. The kinetic energy of the mass

mi is computed as,

Ki =1

2mi(

_ci, _ci)

,

and the total kinetic energy is equal to

K =

n∑i=i

Ki =1

2

n∑i=1

mi(

_ci, _ci)

.

After time-dierentiation it is straightforward to observe that the speed of

change of the total kinetic energy is tantamount to the power of all (both

internal as well as external) forces acting in the system,

_K =1

2

d

dt

n∑i=1

mi(

_ci, _ci)=

n∑i=1

(_ci,mic

i)=

n∑i=1

(_ci, Fi

).

The total energy of a system consists of the kinetic and the potential

energy

E = K+ V,

where V(c) denotes the potential energy. Given the potential energy, we

call a force Fi a potential force when

Fi = −∂V

∂ci.


The dierentiation of the total energy with respect to time yields the identity

_E = _K+ _V =

n∑i=1

(_ci, Fi

)+

n∑i=1

(∂V

∂ci, _ci)

=

n∑i=1

(_ci, Fi +

∂V

∂ci

).

By inspection of this identity one can observe that if Fi = − ∂V∂ci

, then _E = 0.

This observation constitutes the Principle of Conservation of Energy.

Remark 2.3.3 If all forces acting in a system are potential then the total

energy of the system remains constant.

2.4 Examples

2.4.1 Non-uniqueness of solution of the equation of motion

The example analyzed below we owe the courtesy of Professor Marek Kus

from the Institute of Theoretic Physics of Polish Academy of Sciences.

Consider the potential

V(r) = −2

3r3/2,

and suppose that the motion is caused by the potential force F = −dVdr , i.e.

r = r1/2.

By inspection it is not hard to check that there are two trajectories corre-

sponding to te initial conditions r(0) = 0, _r(0) = 0, namely,

r(t) = 0 and r(t) =1

9 · 16t4.

In this example the motion is not determined by the initial position and

velocity.

2.4.2 Mathematical pendulum

As an example of derivation of the equation of motion using the tools of New-

tonian mechanics we have chosen a well known mechanical device called the

mathematical pendulum presented in Figure 2.3. We also use this exam-

ple in order to illustrate a certain method of analysis and representation of

behavior of a dynamic system, i.e. the phase plane.


X

Y

m

mg

l

mg sinϕ

ϕ

ϕ

g

Figure 2.3: Mathematical pendulum

As a direct consequence of the 2nd Newton's Law of Dynamics the equa-

tion of pendulum motion takes the form

ml2 ϕ = −mgl sinϕ,

where m denotes the mass of the pendulum, l its length, g the gravity

acceleration, and ϕ the swing angle. Assuming for simplicity that gl = 1 we

get the 2nd order ODE

ϕ = − sinϕ.

A standard way of dealing with such an equation consists in transform-

ing the equation into two 1st order ODEs. To this objective we make a

substitution q = ϕ

p = _ϕ,

that results in a pair of dierential equations_q = p

_p = − sinq.

Now, suppose that we are interested temporarily in the dependence of the

form p = p(q) (an orbit). The orbit can be found form a single dierential

equation with separated variables,

pdp = − sinqdq,

whose solution is

p2 = 2 cosq+ C,

where C means a constant of integration. Notice that a solution exists on

condition that C > −2.


Now, to compute the trajectory q(t) we need to integrate the dierential

equation

_q = ±√2 cosq+ C.

Instead, we prefer to exploit so called method of phase space, i.e. to draw

on a plane the plot of dependence p(q) for various choices of integration

constant C. Such a plot is called the phase portrait of the system. It

is easily observed that the phase portrait can be produced by drawing in

(p,q, z) coordinates the surface z = p2− 2 cosq, and then taking its section

by a plane perpendicular to the z-axis and projecting these sections onto the

plane (p,q).This task will be partially realized in what follows, for specic

values of C.

Choosing C = −2 leads to p2 = 2 cosq − 2, which yields a collection

of points q = k2π, p = 0, for k = 0,±1,±2, . . . In turn, if we pick C = 2

then p2 = 2 cosq + 2 = 4 cos2 12q, and the phase curves have the form

p = ±2| cos 12q|. These curves divide the plane into two separate regions:

internal and external, where the pendulum behaves dierently. For this

reason this curve is referred to as the separatrix. Outside the separatrix we

have C > 2, which implies that p2 = 2 cosq + C > 0. The phase curves

are then open curves extending from −∞ up to +∞, lying either above or

below the q-axis. The phase portrait inside the separatrix, for −2 6 C 6 2,looks completely dierent. To be more specic, suppose that both q and p

are close to zero. Expanding the function p2 = 2 cosq + C into the Taylor

series around 0 we obtain p2 ∼= 2(1− 1

2q2)+C, therefore q2 + p2 = C+ 2.

The phase curves inside the separatrix are closed (the closer to 0, the more

these curved resemble the circle).

Summarizing this analysis we discover the following characteristics of

behavior of the mathematical pendulum:

• Equilibrium points q = kπ, p = 0. An Equilibrium point is called

stable if a phase curve initialized near this point stays near this point,

and unstable otherwise. This means that equilibrium points q = k2π,

p = 0 are stable, whereas the points q = (2k+1)π, p = 0 are unstable.

• Inside the separatrix p = ±| cos 12q| the phase curves are closed, i.e.

the pendulum oscillates.

• Outside the separatrix the phase curves are open and look like smoothly

deformed real axis R, i.e. the pendulum rotates.

• The separatrix consists of equilibrium points and deformed open in-

tervals of R. Warning: It is not possible to pass along the separatrix


ϕ

_ϕ

π−π 2π−2π

Figure 2.4: Phase portrait of the pendulum

through any equilibrium point.

It is easily seen that along the phase curves _q = p allowing us to dene

the direction of motion. If p > 0 then the q coordinate increases, conversely,

wherever p < 0 it deceases. A phase portrait of the pendulum demon-

strating all the mentioned properties has been visualized in Figure 2.4.

2.5 Exercises

Exercise 2.1 Compute all trajectories studied in subsection 2.4.1.

Exercise 2.2 Find the trajectory of motion of a system subject to a potential

force with potential

V(r) = r2 +1

2r4,

for initial conditions r(0) = 0, _r(0) = 1. Show that in a nite time the

trajectory escapes to ∞.

Exercise 2.3 Show that every internal force Fwi acting in the system pre-

sented in Figure 2.2 is potential, with the gravitational potential

Vi(c) = −Gmi∑j6=i

mj

||cj − ci||.

Exercise 2.4 Consider a simple electro-mechanical system displayed in Fig-

ure 2.5, consisting of a material point of massm carrying an electric charge q

xed to the lower end of a spring with spring constant k hung vertically over

an electric charge Q attracting q, distant by l from the upper end of the

spring. Ignoring the gravitation derive the equation of motion of m and

examine this motion using the method of phase plane.


X

m,q

Q

Fk

FQ

l

x

0

k

Figure 2.5: Electro-mechanical system


The Newtonian mechanics is widely treated in chapters 1 and 2 of the mono-

graph [RK95], and also in chapter 1 of the textbook [Tay05]. The Newton's

Laws of Dynamics stated in subsection 2.2 come from the fundamental work

Philosophiae Naturalis Principia Mathematica by Isaac Newton [New10].

Ibidem one also nd a formulation of the Law of Universal Gravitation. Our

exposition of Conservation Principles presented in subsection 2.3 is pat-

terned on chapter 2 of [RK95]. Alternatively, it can be found in chapter 3

of [Tay05].

Bibliography

[New10] I. Newton. The Mathematical Principles of Natural Philosophy.

Gale ECCO, Print Editions, Farmington Hills, MI, 2010.



[Tay05] J. R. Taylor. Classical Mechanics. University Science Books,

Sausalito, CA, 2005.

Chapter 3

Elements of variational calculus

We have seen that the Newtonian mechanics has been founded on the dif-

ferential calculus in the real space Rn. A similar role in the Lagrangian

mechanics is played by the variational calculus concerned with dierenti-

ation of functions dened in a certain innite dimensional Banach space.

We recall that a Banach space is a linear space equipped with a norm, and

complete. Let X denote a Banach space. The linearity of X means that, for

any elements x,y ∈ X and any pair of reals α,β ∈ R the linear combination

αx + βy ∈ X. The next notion, is the norm or the length of an element of

the Banach space dened as a function || · || : X −→ R, such that, for any

x,y ∈ X and α ∈ R

||x|| > 0, ||x|| = 0⇐⇒ x = 0, ||αx|| = |α| ||x||, ||x+ y|| 6 ||x||+ ||y||.

The last dening property of the Banach space is completeness that guar-

antees that any convergent sequence of elements of this space has a limit

belonging to this space; thus the Banach space is closed with respect to

taking limits of sequences.

3.1 Dierentiation

Assume that in a Banach space X there has been dened the function f :

X −→ R. A particular kind of function is the linear function f for which

f(αx + βy) = αf(x) + βf(y). A derivative of function f will be dened in

the following way.

Denition 3.1.1 The derivative Df(x) : X −→ R of function f at a point

31

Chapter 3. Elements of variational calculus 32

x ∈ X is a linear function such that

limv→0 ||f(x+ v) − f(x) −Df(x)v||||v||

= 0.

The derivative dened in such a way is called the Frechet derivative. In com-

putations, instead of the Frechet derivative we use the Gateaux derivative

that is a sort of directional derivative given by the formula

Df(x)v =d

dα

∣∣∣∣α=0

f(x+ αv). (3.1)

The signicance of this last derivative results from the fact that if the

Gateaux derivative exists and is continuous then the Frechet derivatives also

exists and is continuous, and both these derivatives are equal. Obviously,

for any function f : Rn −→ R there holds Df(x) = ∂f(x)∂x ; the superiority of

the new derivative comes out when the space X is innite dimensional. In

the Example subsection we shall compute example Gateaux derivatives.

3.2 Functional

We shall study the space of functions dened on the time interval [t0, t1],

with values in Rn, continuously dierentiable up to a certain order k > 0,

X = x : [t0, t1] −→ Rn.

By denition, elements of this space are curves x(·) = (t, x(t))|t0 6 t 6 t1,where x(t) ∈ Rn. The space X is a Banach space with norm

||x(·)||k = maxt∈[t0,t1]

||x(t)||+ maxt∈[t0,t1]

|| _x(t)||+ . . . + maxt∈[t0,t1]

||x(k)(t)||,

where the norm in Rn is Euclidean, ||v|| =(∑n

i=1 v2i

)1/2.

Traditionally, a function f : X −→ R will be referred to as the functional.

Below we present a couple examples of functionals for n = 2:

1. the length of a curve x(·) in the plane

f1(x(·)) =∫t1t0

√_x21(t) + _x22(t)dt,

2. the area under a curve x(·) in the plane

f2(x(·)) =∫t1t0

x2(t) _x1(t)dt,


x(·)

x2

x1

x(t0) x(t1)ds

2πx2ds = 2πx2√_x21 + _x22dt

Figure 3.1: Rotational gure

3. the lateral area of a gure created by rotation of a curve x(·) around

the x-axis, see Figure 3.1

f3(x(·)) = 2π∫t1t0

x2(t)√

_x21(t) + _x22(t)dt,

4. the mean square value of curvature of a curve x(·) in the plane

f4(x(·)) =1

t1 − t0

∫t1t0

( _x1(t)x2(t) − x1(t) _x2)2

( _x21(t) + _x22(t))3

dt.

It should be noticed that, except for the functional f4, all remaining func-

tional depend on the curve and its rst order derivative; thus they comply

with the form

f(x(·)) =∫t1t0

L(t, x(t), _x(t))dt, (3.2)

where L(t, x, _x) denotes a continuously dierentiable function. Shortly it

will appear that functionals like (3.2) play a pivotal role in the Lagrangian

mechanics.

3.3 Extrema of functional

Similarly as in "ordinary" dierential calculus in Rn, the zeroing of the

Frechet (Gateaux) derivatives establishes a necessary condition for extremum

of a functional. In our case, a search for extrema of functionals f1f4 has


evident practical sense. For this reason, we need to derive a general for-

mula for the derivative of functional (3.2). Specically, we shall compute

the Gateaux derivative.

Let's choose a curve x(·) ∈ X and its increment (a variation) v(·) ∈ X.

Then, the Gateaux derivative is equal to

Df(x(·))v(·) = d

dα

∣∣∣∣α=0

∫t1t0

L(t, x(t) + αv(t), _x(t) + α _v(t))dt =∫t1t0

∂L(t, x, _x)

∂xv(t)dt+

∫t1t0

∂L(t, x, _x)

∂ _x_v(t)dt.

Having applied to the second component the formula for integration by parts∫t1t0

∂L(t, x, _x)

∂ _x_v(t)dt =

(∂L(t, x, _x)

∂ _xv(t)

)∣∣∣∣t1t0

−

∫t1t0

d

dt

∂L(t, x, _x)

∂ _xv(t)dt

and assumed that the variation v(·) vanishes at te ends of the interval of

integration, i.e. v(t0) = v(t1) = 0, we arrive at the following expression for

the derivative of functional (3.2)

Df(x(·)))v(·) =∫t1t0

(∂L(t, x, _x)

∂x−d

dt

∂L(t, x, _x)

∂ _x

)v(t)dt. (3.3)

Now, because (3.3) needs to vanish for any variation v(·), we obtain the fol-

lowing necessary condition for the extremum, known as the Euler-Lagrange

equations.

Theorem 3.3.1 (Euler-Lagrange equations) Suppose that the curve x(·) is

an extremum of the functional (3.2). Then, this curve satises the

Euler-Lagrange equations

∂L(t, x, _x)

∂x−d

dt

∂L(t, x, _x)

∂ _x= 0. (3.4)

Any solution of the Euler-Lagrange equations is a curve that will be called

an extremal of the functional. In this sense (3.4) is a necessary and sucient

condition for the extremal of functional (3.2).

The Euler-Lagrange equation is a nonlinear ODE of order 2,

∂2L(t, x, _x)

∂ _x2x+

∂2L(t, x, _x)

∂x∂ _x_x+

∂2L(t, x, _x)

∂t∂ _x−∂L(t, x, _x)

∂x= 0.

For the reason that the variable x is usually a vector, the Euler-Lagrange

equation is actually a system of ODEs, and in the sequel we shall use equiv-

alently the term "Euler-Lagrange equations".


In a similar way one can deliver an equation for the extremal of a func-

tional containing the second order derivatives, like f4. Assume that

h(x(·)) =∫t1t0

L(t, x, _x, x)dt.

Then we have

Theorem 3.3.2 (Euler-Poisson equations) Extremals of the functional h(x(·))solve the Euler-Poisson equations

∂L(t, x, _x, x)

∂x−d

dt

∂L(t, x, _x, _x)

∂ _x+d2

dt2∂L(t, x, _x, x)

∂x= 0. (3.5)

3.4 Conditional extrema

Similarly as for functions dened on Rn we can study conditional extremum

of a functional. Two kinds of conditional problems will be distinguished:

• Isoperimetric problem: nd extremum of the functional

f(x(·)) =∫t1t0

L(t, x, _x)dt,

on condition that a constraint functional

h(x(·)) =∫t1t0

K(t, x, _x)dt = const .

• Vaconomic problem: nd extremum of the functional

f(x(·)) =∫t1t0

L(t, x, _x)dt,

on condition that, at any time instant, a constraint function

G(t, x, _x) = 0.

The following theorem provides necessary conditions for conditional ex-

trema.

Theorem 3.4.1 (On conditional extremum)


X

Y

x

y

ϕ

_y

_x

v = u1

Figure 3.2: Motion without lateral slip

• The isoperimetric problem: One needs to dene a function

L = L(t, x, _x) + λK(t, x, _x),

λ ∈ R and write the Euler-Lagrange equations (3.4) for L. The

(Lagrange) multiplier λ may be eliminated using the condition

h(x(·)) = const.

• The vaconomic problem: We dene a function

L = L(t, x, _x) + λG(t, x, _x),

where λ = λ(t) denotes a certain function of time, and invoke the

Euler-Lagrange equations (3.4) for L.

The constraints appearing in the vaconomic problem are characteristic

to the description of motion of wheeled mobile robots that excluded wheel

slips or in some control problems. As a simple illustration, consider a wheel,

a skate or a ski moving without a lateral slip, shown as viewed from above

in Figure 3.2. It follows that coordinates needed to describe this motion

include the position of the contact point with the ground and the orientation,

q = (x,y,ϕ)T . In these coordinates, the constraint of no lateral slip means

that projections of the velocities _x, _y at the contact point on the direction

perpendicular to surface of the wheels (i.e. on the wheel's axle) should

cancel each other. But this property requires that at any time instant

G(t,q, _q) = _x sinϕ− _y cosϕ = 0.

The next section includes solutions of example problems for (unconditional

as well conditional) extrema of functionals.


3.5 Examples

3.5.1 Gateaux derivative with respect to vector or matrix

1. f(x) = xTQx, x ∈ Rn, Q a matrix of size n× n. The derivative

Df(x)v =d

dα

∣∣∣∣α=0

(x+ αv)TQ(x+ αv) = vTQx+ xTQv = xT (Q+QT )v.

2. f(X) = trXTX, X a matrix n× n, tr the trace. The derivative

Df(X)V =d

dα

∣∣∣∣α=0

tr(X+ αV)T (X+ αV) = trVTX+ trXTV = 2 trXTV.

3. f(X) = detX, X an n× n matrix. The derivative

Df(X)V =d

dα

∣∣∣∣α=0

det(X+ αV) =

d

dα

∣∣∣∣α=0

det[X1 = αV1,X2 + αV2, . . . ,Xn + αVn] =

det[V1,X2, . . . ,Xn]+det[X1,V2, . . . ,Xn]+ · · ·+det[X1,X2, . . . ,Vn],

where symbols Xi i Vi denote, respectively, the i-th column of matrix

X and V.

3.5.2 Shortest line on a plane

Let x(·) denote a curve in R2, x(t) = (x1(t), x2(t))T . We are looking for the

shortest curve joining two given points, so minimizing the functional f1. In

this case L(t, x, _x) =√

_x21 + _x22. The Euler-Lagrange equations

d

dt

∂L

∂ _x−∂L

∂x= 0

yield ∂L∂ _x1

= _x1L , ∂L∂x1 = 0, ∂L∂ _x2 = _x2

L , ∂L∂x2 = 0, which gives_x1L = C1_x2L = C2

for certain constants C1 and C2. After elimination of time we get dx2dx1= C,

therefore

x2 = Cx1 +D, C,D constants.

Thus, the shortest curve is a segment of the straight line.


3.5.3 Curve producing minimal lateral area of a rotational gure

We consider the functional f3, and we want to nd a curve x(·), x(t) =

(x1(t), x2(t))T , such that the lateral area of the rotational gure be mini-

mum. Having omitted the factor 2π we obtain L(t, x, _x) = x2

√_x21 + _x22. The

calculations based on the Euler-Lagrange equations result in the following:∂L∂ _x1

= x2 _x1√_x21+ _x22

, ∂L∂x1

= 0, ∂L∂ _x2

= x2 _x2√_x21+ _x22

, ∂L∂x2

=√

_x21 + _x22. In consequence,

we arrive at a pair of ODEsx2 _x1√_x21+ _x22

= C

ddt

(x2 _x2√_x21+ _x22

)−√

_x21 + _x22 = 0.

From the former we nd x2√_x21+ _x22

= C_x1

, and substitute into the latter

d

dt

C _x2_x1

=√

_x21 + _x22 =x2 _x1C

,

therefored

dt

dx2dx1

=d2x2

dx21_x1 = a

2x2 _x1,

where a = 1C . Two extremals have been found in this way: the vertical

straight line x1 = const and a curve solving the equation d2x2dx21

= a2x2. For

obvious reasons the rst solution is not acceptable. The second equation is

solved by the curve

x2 = c cosh(ax1 + b), a,b, c constants,

shown in Figure 3.3, resembling the shape of hanging chain, and called

the catenary∗. The gure created by rotation of the catenary is called a

catenoid.

3.5.4 Dido's problem

The Dido's problem consists in nding a closed curve x(t) = (x1(t), x2(t))T

of prescribed perimeter, sweeping the maximum area. This problem is an

example of the isoperimetric problem, with to be extremalized functional

dened by

L(t, x, _x) = x2 _x1

∗in Latin: catena means a chain


x1

x2

catenary

Figure 3.3: Catenary

and constraining functional characterized by

K(t, x, _x) =√

_x21 + _x22.

In order to solve the Dido's problem we invoke Theorem 3.4.1 and dene

a function

L = x2 _x1 + λ√

_x21 + _x22,

where λ ∈ R denotes a Lagrange multiplier. To state the Euler-Lagrange

equations we compute derivatives

∂L∂ _x1

= x2 +λ _x1K ,

∂L∂ _x2

= λ _x2K ,

∂L∂x1

= 0,∂L∂x2

= _x1

that leads to equations x2 +

λ _x1K = C

ddtλ _x2K = _x1 =

ddtx1

.

The former equation implies that

λ _x1K

= C− x2,

while the integration of the latter results in

λ _x2K

= x1 +D,


for certain constants C and D. In order to get rid of the constant λ we

divide these equations sideways and obtain the equation

dx2dx1

=D+ x1C− x2

,

whose solution is a circle

(x1 +D)2 + (x2 − C)2 = R2.

In this way we have proved that the solution of Dido's problem is a circle.

Constants C, D and R can be determined if we know coordinates of a point

on the circle and its length.

3.5.5 Brockett's integrator

The Brockett's integrator is a control system of the form_x1 = u1

_x2 = u2

_x3 = x1u2 − x2u1

,

with state variable x = (x1, x2, x3)T ∈ R3 and control variable u = (u1,u2)

T ∈R2. The control problem of this system consists in determining a con-

trol u(t) that moves the initial system's state x(0) = x0 to the nal state

x(1) = xd within control time T = 1, and simultaneously minimizes the

control energy∫10(u

21(t) + u

22(t))dt.

It is easily noticed that this optimal control problem can be formulated

as a vaconomic problem in with

L(t, x, _x) = _x21 + _x22 and G(t, x, _x) = _x3 − x1 _x2 + x2 _x1.

Now, in accordance with Theorem 3.4.1, we introduce a function

L = _x21 + _x22 + λ( _x3 − x1 _x2 + x2 _x1)

containing an unknown function λ = λ(t). To get the Euler-Lagrange equa-

tion we compute derivatives

∂L∂ _x1

= 2 _x1 + λx2,∂L∂ _x2

= 2 _x2 − λx1,∂L∂ _x3

= λ


and∂L∂x1

= −λ _x2,∂L∂x2

= λ _x1,∂L∂x3

= 0,

that yields λ = const andx1 + λ _x2 = _u1 + λu2 = 0

x2 − λ _x2 = _u2 − λu1 = 0.

Both these ODEs are equivalent to linear, 2nd order equationsu1 + λ

2u1 = 0

u2 + λ2u2 = 0

,

whose solutions have the formu1(t) = A1 cos λt+ B1 sin λt

u2(t) = A2 cos λt+ B2 sin λt.

By solving the vaconomic problem we have demonstrated that the optimal

energetic control of the Brockett's integrator is sinusoidal. Additionally,

having chosen x1(0) = x1(1) = 0 and x2(0) = x2(1) = 0 we discover that

λ = 2kπ, for an integer k.

Eventually, given the controls we compute the trajectory passing through

the point x(0) = 0,x1(t) =

1λ(A1 sin λt− B1 cos λt+ B1)

x2(t) =1λ(A2 sin λt− B2 cos λt+ B2)

x3(t) =1λ2(A1B2 −A2B1)(λt− sin λt)

.

Specically, when A1 = B2 = C and A2 = B1 = 0, we gat a circular motion

in the plane (x1, x2),

x21 +

(x2 −

C

λ

)2=

(C

λ

)2,

C = const, whereas the coordinate x3(t) ∼= t3. The optimal trajectory

(orbit) of the Brockett's integrator is visualized in Figure 3.4.


x1

x3

x2

x1

x3

x2

Figure 3.4: Optimal orbit of Brockett's integrator (a side view and a view

from above)

X

Y

x

yD

H

v

c

ct

a

Figure 3.5: Pursuit problem

3.6 Exercises

Exercise 3.1 (Pursuit problem) A dog runs with constant velocity v and

pursues a hare that runs away with constant velocity c along a straight

line x = a, see Figure 3.5. Provide an equation of the pursuit curve y(x)

assuming that at any point of this curve the direction of dog's velocity is

dened by the position of hare. Solve the pursuit equation and nd the

pursuit curve. Specify a condition of successful pursuit and determine the

time after which the dog will catch up the hare. Hint: observe that at any

time instant

vt =

∫x0

√1+ (y ′u)

2du, y ′x =dy

dx.

Exercise 3.2 (Brachystochrone problem) Find the equation of curve y(x)

along which a material point of mass m slides down in the gravitational


X

Y

x

y

P0

g

P1

m

v

Figure 3.6: Brachystochrone problem

eld between the initial point P0 and the nal point P1, such that the time

of motion is the shortest possible, see Figure 3.6. Hint: Observe that the

velocity of motion can be computed from the identity

1

2mv2 = mgy, and v =

ds

dt.


A basic knowledge of the variation calculus can be found in many text-

books, e.g. in the classic work [GF63], as well as in chapter 6, volume I of

the monograph [Tay05]. Also, it may be instructive to have a look into the

recent book [Kom09]. The term "isoperimetric problem" comes from Greek

words isos equal and perimetron perimeter. A classic example of this

kind of problem is the Dido's problem studied in this chapter. The name of

this problem comes from the name of Dido (Dido means a wanderer), the

ancient princess of Tyre, a Phoenician city-state (now in Lebanon). Dido,

forced to escape from her motherland, ed through the Mediterranean up to

the coast of Numidia (now in Tunisia). There the Numidians oered her a

piece of land near the see coast that can be contained within the skin of an

ox. Dido cut the skin into narrow strips, formed them into a circle, and on

the hill Byrsa contained inside the circle founded the city of Carthage. Dido

was the rst queen of Carthage. A version of the Dido's story was described

by the Latin poet Virgil (Publius Vergilius Maro) in book no IV of the poem

Aeneid [Mar19]. The word "vaconomic" coined the Russian mathematician

V. V. Kozlov to name the "mechanics of variational axiomatic kind". The


control system referred to as the Brockett's integrator was introduced by

Roger W. Brockett, a professor of the Harvard University, author of funda-

mental works on control theory. The brachystochrone problem was stated

in 1696 by z Johann Bernoulli. The name "brachystochone" originates from

Greek words brachystos the shortest and chronos time. The brachys-

tochrone is also called the tautochrone (tautos the same) to describe the

property that the travelling time along the whole tautochrone is the same

as the travelling time needed to get from any point on the tautochrone

to its end point. Our statement of the pursuit problem comes from the

monograph [RK95].

Bibliography

[GF63] I. M. Gelfand and S. W. Fomin. Calculus of Variations. Prentice

Hall, Englewood Clis, NJ, 1963.

[Kom09] L. Komzsik. Applied Calculus of Variations for Engineers. CRC

Press, Boca Raton, 2009.

[Mar19] Publius Vergilius Maro. Aeneis. De Gruyter, Berlin/Boston, 2019.





Chapter 4

Lagrangian mechanics

In this chapter variational calculus will be used as a means for introducing

basic concepts of Lagrangian mechanics. The formulation of Lagrangian me-

chanics relies on the following elements. Suppose that the motion of a system

has been described by generalized coordinates q = (q1,q2, . . . ,qn)T ∈ Rn

and generalized velocities _q = ( _q1, _q2, . . . , _qn)T ∈ Rn. The generalized

coordinates may denote linear or angular positions, and the velocities are

time-derivatives of the positions.

For given q and _q we dene the Lagrangian of the system

L(q, _q) = K(q, _q) − V(q)

equal to the dierence between kinetic and potential energy. We shall as-

sume that the system is moving over a time interval [t0, t1] and is subject

to the variational Principle of the Least Action stating that the system's

trajectory extremalizes the functional

I(q(·)) =∫t1t0

L(q(t), _q(t))dt

called the action functional. In this setting the equations of motion formu-

late the Euler-Lagrange equations

d

dt

∂L

∂ _q−∂L

∂q= F. (4.1)

Symbol F appearing on the right hand side of identity (4.1) refers to non-

potential forces (forces that are not the gradient of a potential energy) ex-

isting in the system, like the forces of friction, traction, control forces, etc.

If non-potential forces are absent, we set F = 0.

45

Chapter 4. Lagrangian mechanics 46

X

Y

m

x

ϕ

gy

π2 +ϕ

Ir

Figure 4.1: Ball and beam

It is easy to notice that equation (4.1) leads to a system of 2nd order

ODEs, of the form∂2L

∂ _q2q+

∂2L

∂q∂ _q_q−

∂L

∂q= F.

This equation contains the Hesse matrix of 2nd order derivatives ∂2L∂ _q2

. If this

matrix is non-singular then the equation can be made explicit with respect

to the acceleration,

q =

(∂2L

∂ _q2

)−1(∂L

∂q−∂2L

∂q∂ _q_q+ F

). (4.2)

The solution of (4.2) for given initial conditions q(t0) and _q(t0) describes

the motion (trajectory) q(t) of the system.

4.1 Examples

In order to illustrate the way of setting Lagrangian equations of motion we

shall study two examples.

4.1.1 Ball and beam

Let the device shown in Figure 4.1 be given, composed of a ball able to

roll down and up along a beam equipped with a rotational joint. Suppose

that the beam has the moment of inertia I, and the ball is represented by a

material point of massm. At any time instant the ball must touch the beam.

As generalized coordinates we choose the rotation angle of the beam ϕ and

the position of the ball r along the beam, so that q = (r,ϕ)T , _q = (_r, _ϕ)T .

The kinetic energy of the system is composed of the rotational kinetic

energy of the beam, and of the kinetic energy of the ball, i.e.

K =1

2I _ϕ2 +

1

2mv2,


where v denotes the ball's velocity. Having denoted the ball's Cartesian

coordinates by x and y, we compute v2 = _x2+ _y2. It follows from Figure 4.1

that x = r cosϕ

y = r sinϕ,

and, after dierentiation,_x = _r cosϕ− r _ϕ sinϕ

_y = _r sinϕ+ r _ϕ cosϕ.

It follows that the squared ball's velocity v2 = _r2 + r2 _ϕ2, and the kinetic

energy of a system consisting of the beam and the ball amounts to

K =1

2(I+mr2) _ϕ2 +

1

2m_r2.

The potential energy of the system comes from the gravitational force acting

on the ball, that can be computed as

V = −m(r, g) = mgr sinϕ,

where bold letters refer to the vector of position of the ball and the vector of

gravitational acceleration, both expressed in the coordinate system (x,y).

Combining so obtained results we get a Lagrangian

L = K− V =1

2(I+mr2) _ϕ2 +

1

2m_r2 −mgr sinϕ.

To derive the Euler-Lagrange equations we rst compute a number of

derivatives∂L∂ _r = m_r,∂L∂ _ϕ = (I+mr2) _ϕ,∂L∂r = mr _ϕ2 −mg sinϕ,∂L∂ϕ = −mgr cosϕ,

and, under assumption that any non-potential forces are absent, we conclude

the followingddt∂L∂ _r −

∂L∂r = mr−mr _ϕ2 +mg sinϕ = 0

ddt∂L∂ _ϕ − ∂L

∂ϕ = (I+mr2) ϕ+ 2mr_r _ϕ+mgr cosϕ = 0.


Y

Z

m

l ϕ

g

X

z

x

y

θ

a

b

Figure 4.2: Furuta's pendulum

4.1.2 Furuta's pendulum

The Furuta's pendulum consists of a vertical column of length l, able to

rotate, equipped with a perpendicular arm of length a with xed a pendulum

of length b carrying a mass m, see Figure 4.2. The Furuta's pendulum can

be regarded either as a model of a human who can turn around a vertical

axis, with his arm extended straight, holding an object in his hand or as a

model of 2 rotational dof robot. In the following analysis we ignore the mass

of the column and of the arms. As generalized coordinates of the Furuta's

pendulum we shall employ the rotation angle of the column and the swing

angle of the pendulum, q = (θ,ϕ)T .

The kinetic energy is tantamount to the energy of mass m, and equals

K = 12mv

2. Denoting the Cartesian coordinates of mass m as (x,y, z)T we

compute v2 = _x2 + _y2 + _z2. From the gure we deducex = (a+ b sinϕ) cos θ

y = (a+ b sinϕ) sin θ

z = l− b cosϕ

.

After a dierentiation we have_x = −(a+ b sinϕ) _θ sin θ+ b cosϕ _ϕ cos θ

_y = (a+ b sinϕ) _θ cos θ+ b cosϕ _ϕ sin θ

_z = b _ϕ sinϕ

,


that yields a formula for the velocity

v2 = (a+ b sinϕ)2 _θ2 + b2 _ϕ2.

The potential energy of the system is due to gravitational forces acting

on mass m and can be expressed by V = −m(r, g). For the reason that

r = (x,y, z)T , and g = (0, 0,−g)T , we nd V = mgz = mg(l − b cosϕ). By

denition of the Euler-Lagrange equations, a constant term is immaterial, so

we may omit the component mgl. Eventually, we arrive at the Lagrangian

L =1

2m(a+ b sinϕ)2 _θ2 +

1

2mb2 _ϕ2 +mgb cosϕ.

The last stage preceding the derivation of the equations of motion is the

dierentiation

∂L∂ _θ

= m(a+ b sinϕ)2 _θ,∂L∂ _ϕ = mb2 _ϕ,∂L∂θ = 0,∂L∂ϕ = m(a+ b sinϕ)b _θ2 cosϕ−mgb sinϕ.

In conclusion, the Euler-Lagrange equations take the formddt∂L∂ _θ

− 0 = 0⇒ m(a+ b sinϕ)2 _θ = const,ddt∂L∂ _ϕ − ∂L

∂ϕ = mb2 ϕ−m(a+ b sinϕ)b _θ2 cosϕ+mgb sinϕ = 0.

As can be seen, the former equation expresses the conservation of angular

momentum in rotation around the vertical axis.

4.2 General form of Lagrangian equations of motion

It is known that for typical mechanical systems the kinetic energy is de-

scribed by a quadratic form of generalized velocities whose matrix depends

on generalized coordinates. This being so,

K(q, _q) =1

2_qTQ(q) _q =

1

2

n∑i,j=1

Qij(q) _qi _qj.

The matrix Q dening this quadratic form bears the name of inertia matrix

of the system. The inertia matrix is symmetric, QT = Q, and positive

denite, vTQv > 0 for any v 6= 0.


From the form of the kinetic energy there results that the Lagrangian

equals

L(q, _q) =1

2

n∑i,j=1

Qij(q) _qi _qj − V(q). (4.3)

After performing suitable mathematical operations we deduce the following

Euler-Lagrange equations, for k = 1, 2, . . . ,n

n∑i=1

Qki(q)qi +

n∑i,j=1

ckij(q) _qi _qj +∂V(q)

∂qk= Fk. (4.4)

In (4.4) the coecients

ckij(q) =1

2

(∂Qik(q)

∂qj+∂Qkj(q)

∂qi−∂Qij(q)

∂qk

)are known as the Christoel's symbols of the 1st kind, whereas Fk denotes

the force acting on coordinate k. Equations (4.4) can be represented conve-

niently in the matrix form as

Q(q)q+ C(q, _q) _q+D(q) = F, (4.5)

where Q(q) is the inertia matrix, C(q, _q) denotes the matrix of centripetal

and Coriolis forces, D(q) is a vector of potential forces, and vector F collects

non-potential forces. Given the Christoel's symbol of the 1st kind, the

matrix of centripetal and Coriolis forces can be found as

C(q, _q)kj =

n∑i=1

ckij(q) _qi.

Obviously, we have

Dk(q) =∂V(q)

∂qkand F = (F1, F2, . . . , Fn)

T .

Alternatively, the equations of motion (4.4) are sometimes made explicit

with respect to the acceleration,

qk +

n∑i,j=1

Γkij(q) _qi _qj + Dk(q) = Fk, (4.6)

where Γkij(q) stands for the Christoel's symbols of the 2nd kind , and

Dk =

n∑l=1

Q−1kl (q)Dl(q),

Fk =

n∑l=1

Q−1kl (q)Fl.


By denition, the Christoel's symbols of the 2nd kind can be computed

from the identity

Γkij(q) =

n∑l=1

Q−1kl (q)c

lij(q).

Hereabove, Q−1ij (q) denotes the (ij)-th element of the inverse inertia matrix.

It follows that the Christoel's symbols of the 1st kind are symmetric

with respect to a permutation of lower indices, i.e.

ckij(q) = ckji(q).

Also, another important property can be proved, namely, along a trajectory

q(t) there holds_Q(q) = C(q, _q) + CT (q, _q). (4.7)

4.3 Geometric interpretation of Lagrangian mechanics

In this section we shall assume that Lagrangian includes only the kinetic

energy,

L(q, _q) =1

2_qTQ(q) _q. (4.8)

In these circumstances, the Euler-Lagrange equations become

Q(q)q+ C(q, _q) _q = 0. (4.9)

Now, we shall show that along a trajectory q(t) this Lagrangian is constant,

dL(q(t), _q(t))

dt= 0.

To this aim, we compute

dL(q(t), _q(t))

dt= _qTQ(q)q+

1

2_qT _Q(q) _q.

Having invoked the equations of motion, and then taking into account the

property (4.7) the right hand side of this identity gets equal to

1

2_qT _Q(q) _q− _qTC(q, _q) _q =

1

2_qT(

_Q(q) − 2C(q, _q))

_q =

1

2_qT(CT (q, _q) − C(q, _q)

)_q = 0,

where the last identity results from the fact that a quadratic form whose

matrix is skew symmetric must be zero.


Let's stay with assumption (4.8) and the Euler-Lagrange equations (4.9).

A consequence of Lagrangian mechanics is that the trajectory satisfying

these equations is an extremal of the action functional,

2I =

∫t1t0

_qTQ(q) _qdt =

∫t1t0

L(q, _q)dt, (4.10)

because extremals of the both functionals I i 2I coincide. Now, let us con-

sider the functional

J =

∫t1t0

√_qTQ(q) _qdt =

∫t1t0

√L(q, _q)dt, (4.11)

and corresponding Euler-Lagrange equations

d

dt

∂√L

∂ _q−∂√L

∂q=d

dt

1

2√L

∂L

∂ _q−

1

2√L

∂L

∂q.

Since the Lagrangian L remains constant in time, after dierentiation we

obtain1

2√L

(d

dt

∂L

∂ _q−∂L

∂q

)= 0

which is true because q(t) extremalizes functional I. These arguments can

be summarized in the following way

Remark 4.3.1 Extremals of the action functional are are extremals of

the J.

There is a remarkable analogy between functional J and the length of the

trajectory q(t), dened by∫t1t0

√_qT _qdt, and containing the Euclidean inner

product of velocities. Relying on this analogy one can recognize that the

functional J also denes a length of this trajectory, the dierence being only

that the inner product of velocities _q needs to be computed with the help of

the inertia matrix Q(q), therefore, at each point q the way of multiplying

velocities is dierent. The inner product

(v,w)Q = vTQ(q)w

is called the Riemannian metric in the space of generalized coordinates. An

extremal of the length functional is known as the geodesic (line). In this

context, Remark 4.3.1 leads to the following conclusion.

Theorem 4.3.1 (On geodesics of Riemannian metric) For Lagrangian of the

form (4.8) the system moves along the geodesic of the Riemannian met-

ric determined by the inertia matrix.

Theorem 4.3.1 provides a geometric interpretation of Lagrangian mechanics.


Y

Z

X

y

x

z

ϕ

Pu1

u2

θ

Figure 4.3: Sphere S2

4.4 Examples, continuation

In order to facilitate the Reader getting acquainted with the concept of

Riemannian metric we shall compute this metric for the sphere S2 vizualized

in Figure 4.3. To this objective, we introduce spherical coordinates F :

R2 −→ R3, F(ϕ, θ) = (x,y, z)T , wherex = sin θ cosϕ

y = sin θ sinϕ

z = cos θ

.

The Jacobian DF(ϕ, θ) acts on tangent vectors to the sphere at the point

of coordinates (ϕ, θ). Consider images of two tangent vectors u1 and u2shown in the gure, so let

v1 = DF(ϕ, θ)u1, i v2 = DF(ϕ, θ)u2.

Vectors v1 and v2 belong to the Euclidean space R3, bearing the Euclidean

inner product (v1, v2) = vT1v2. Therefore, the Riemannian metric on S2

should be dened in such a way that

(u1,u2)Q = uT1Qu2 = vT1v2.

Now, we compute

vT1v2 = uT1 (DF(ϕ, θ))TDF(ϕ, θ)︸︷︷︸

Q

u2,


X

Y

m

x

ϕg

M

l

Figure 4.4: Inverted pendulum

concluding that the Riemannian metric on the sphere is dened by the

matrix

Q(ϕ, θ) =

[sin2 θ 0

0 1

].

4.5 Exercises

Exercise 4.1 Prove the property (4.7).

Exercise 4.2 Using the tools of Lagrangian mechanics obtain equations of

motion of the inverted pendulum shown in Figure 4.4. The inverted pen-

dulum consists of a point mass M moving along the X-axis and carrying a

pendulum of length l and mass m.

Exercise 4.3 Using the tools of Lagrangian mechanics derive equations of

motion of a leg of a jumping robot displayed in Figure 4.5 during its contact

with the ground. Assume that the leg consists of a mass m mounted to a

spring of length l. The potential energy of the spring is equal to 12kl

2. Also

assume that during the contact phase the contact point of the leg with the

ground does not change its position.

Exercise 4.4 Provide Lagrangian equations of motion of a planar 2R robotic

manipulator shown in Figure 4.6. Ignore masses of the arms and let the

object of manipulation have mass m.

Exercise 4.5 Derive Lagrangian equations of motion of a planar space robot

shown in Figure 4.7. The robot is composed of a free oating base of massM

and moment of inertia I, and a rotating arm with variable length l holding

a point mass m. Choose as generalized coordinates the orientation of the


X

m

ϕ

l

Y

g

Figure 4.5: Jumping robot's leg

Y

g

X

l1

ϕ1

ϕ2

m

l2

Figure 4.6: 2R robotic manipulator

base θ, the extension of the arm l, and the rotation angle of the arm ϕ.

Ignore the gravitational forces and the mass of the arm.

Exercise 4.6 Obtain Lagrangian equations of motion of a planar Ballbot

robot presented in Figure 4.8. The robot resembles the inverted pendulum

(body) mounted on a wheel that is able to roll along the X-axis. The length

and the mass of the pundulum are, respectively, equal to l andm. The wheel

has radius r, mass M, and moment of inertia I. Assume that the rotation

angle of the wheel with respect to the body equals θ, and the orientation of

the body is described by the angle ϕ. Assume that the wheel rolls without

sliding, which means that the path travelled by the wheel x = r(ϕ+ θ).

Exercise 4.7 By means of the formalism of Lagrangian mechanics derive

equations of motion of the elastic spherical pendulum, see Figure 4.9. The

arm of the pendulum is telescopic, of variable length l and the spring con-

stant k. The tip of the pendulum holds a mass m, and can take positions

on the sphere of radius l. In the derivation ignore the mass of the arm and

use the formula Vs =12kl

2 for the potential energy of the spring.


X

Y

m

ϕ

l

x

y θ

M, I

Figure 4.7: Space robot

X

m

ϕ l

x

θ

M, I

r

Y

g

Figure 4.8: Ballbot


Aditional information on Lagrangian mechanics can be found in subchapter

14 of chapter 2 of the monograph [RK95], in chapter II of the book [Gre03] or

in chapter 7 of the textbook [Tay05]. The Principle of the Least Action, often

attributed to Hamilton, belongs to the variation principles of mechanics,

see chapter 3 of [Gut71]. A theory and example derivations of the Euler-

Lagrange equations are contained in chapter 4 of this reference.

Bibliography




Y

Z

X

ϕ

θ m

l

g

Figure 4.9: Elastic spherical pendulum

[Gut71] R. Gutowski. Mechanika analityczna. PWN, Warszawa, 1971. (in

Polish).





Chapter 5

Hamiltonian mechanics

Alternative to Lagrangian way of formulation of equations of motion is pro-

vided by Hamiltonian mechanics. Conceptually, we shall derive fundamental

concepts of Hamiltonian mechanics from the Lagrangian mechanics. Sup-

pose then that a mechanical system has been described in terms of gener-

alized coordinates and velocities, there has been dened a Lagrangian and

stated the Euler-Lagrange equations of motion. Our objective will be to

introduce corresponding concepts of Hamiltonian mechanics, culminating

in Hamiltonian equations of motion. To this aim we need to employ the

Legendre transform of functions.

5.1 Legendre transform

To a given function f : R −→ R such that y = f(v) we associate a new

function F(p) dened in the following way

F(p) = maxv

(pv− f(v)).

Function F(p) will be called the Legendre transform of f(v). Under as-

sumption that the original function f(v) is continuously dierentiable, the

maximum condition yields

d

dv(pv− f(v)) = 0,

which means that p =df(v)dv . Having computed from this identity the locus

of maximum v = v(p) we nd out that

F(p) = pv(p) − f(v(p)).

58

Chapter 5. Hamiltonian mechanics 59

v

f(v)

F(p)

pv

Figure 5.1: Legendre transform

The idea of Legendre transform has been illustrated in Figure5.1. It follows

that there exist functions for which this transform cannot be dened: for

example, the Legendre transform is well dened for either f(v) = v2 or

f(v) = v4, although it does not exist for cubic function f(v) = v3.

The denition of Legendre transform generalizes in a natural way to

functions depending on a vector argument, f : Rn −→ R. In such a case we

choose p ∈ Rn and obtain

F(p) = maxv

(pTv− f(v)

).

The maximum locus v(p) solves the equation

p =∂f(v)

∂v,

i.e.

F(p) = pTv(p) − f(v(p)).

5.2 Hamiltonian

As a starting point we take a Lagrangian L(q, v) = K(q, v)−V(q), q, v ∈ Rn,

where v stands for velocity _q. The fundamental concept of Hamiltonian

mechanics is the Hamiltonian that will be introduced below.

Denition 5.2.1 The Hamiltonian of a system characterized by Lagrangian

L(q, v) is the Legendre transform of Lagrangian with respect to velocity

v, with q treated as a parameter,

H(q,p) = maxv

(pTv− L(q, v)

).


The maximum locus v(q,p) results from identity

p =∂L(q, v)

∂v,

so, in consequence, the Hamiltonian equals

H(q,p) = pTv(q,p) − L(q, v(q,p)). (5.1)

Variable p is referred to a generalized momentum of the system. Under

assumption that the kinetic energy is given as the quadratic form K(q, v) =12vTQ(q)v we get

p =∂L(q, v)

∂v= Q(q)v,

therefore

v(q,p) = Q−1(q)p.

Finally, a substitution to (5.1) delivers the following expression for Hamil-

tonian

H(q,p) =1

2pTQ−1(q)p+ V(q). (5.2)

It turns out that the Hamiltonian is equal to the total energy of the system,

i.e. the sum of kinetic energy (expressed in terms of momenta) and the

potential energy.

5.3 Canonical Hamilton's equations

As has been shown in the previous chapter, Lagrangian mechanic relies on

the concept of generalized coordinates and velocities

q, _q ∈ Rn,

Lagrangian

L(q, _q) = K(q, _q) − V(q),

and the Euler-Lagrange equations

d

dt

∂L(q, _q)

∂ _q−∂L(q. _q)

∂q= F.

Analogous categories of Hamiltonian mechanics are generalized coordinates

and momenta

q,p ∈ Rn, p =∂L(q, v)

∂v,


and Hamiltonian

H(q,p) = pTv(q,p) − L(q, v(q,p)).

In order to dene Hamiltonian equations of motion we compute partial

derivatives of Hamiltonian,

∂H

∂qi= pT

∂v

∂qi−∂L

∂qi−

(∂L

∂v

)T∂v

∂qi.

Now, using the denition of momentum p = ∂L∂v , we get

∂H

∂qi= −

∂L

∂qi.

Also, from the Euler-Lagrange equations, there follows that

∂L

∂qi= _pi − Fi,

so

_pi = −∂H

∂qi+ Fi.

Continuing in this way, we nd

∂H

∂pi= vi + p

T ∂v

∂pi−

(∂L

∂v

)T∂v

∂pi.

The denition of momentum yields

∂H

∂pi= vi,

which means that

_qi =∂H

∂pi.

The equations we have derived for _qi and _pi are called canonical Hamilton's

equations of motion. In vector notation they can be given the following form_q =

∂H(q,p)

∂p

_p = −∂H(q,p)

∂q+ F

, (5.3)

where F refers to non-potential forces acting in the system. If F is absent, the

canonical equations allows us to prove that the Hamiltonian is an invariant

(a constant of motion).

Theorem 5.3.1 (On invariance of Hamiltonian) Hamiltonian is constant on

trajectories of canonical Hamilton's equations,

dH(q(t),p(t))

dt=

(∂H

∂q

)T_q+

(∂H

∂p

)T_p = − _pT _q+ _qT _p = 0.


5.4 Examples

5.4.1 Ball and beam

We shall derive Hamiltonian equations of motion for the ball and beam

system studied in the previous chapter by means of the tools of Lagrangian

mechanics. We have had q = (r,ϕ)T , p = (p1,p2)T , and Lagrangian

L = K− V =1

2

(I+mr2

)_ϕ2 +

1

2m_r2 −mgr sinϕ.

To dene the Hamiltonian we shall use the inertia matrix contained in the

kinetic energy.

Q =

[m 0

0 I+mr2

].

By virtue of (5.2), the inverse matrix

Q−1 =

[1m 0

0 1I+mr2

],

denes the Hamiltonian,

H(q,p) =1

2

(p21m

+p22

I+mr2

)+mgr sinϕ.

The canonical Hamilton's equations related to this Hamiltonian become_r = ∂H

∂p1= p1m

_ϕ = ∂H∂p2

= p2I+mr2

_p1 = −∂H∂r =mrp22

(I+mr2)2−mg sinϕ

_p2 = −∂H∂ϕ = −mgr cosϕ

.

5.4.2 Furuta's pendulum

The position of the Furuta's pendulum has been described by two angles

q = (θ,ϕ)T . The corresponding vector of momenta is p = (p1,p2)T . In the

previous chapter we have computed the Lagrangian as

L =1

2m(a+ b sinϕ)2 _θ2 +

1

2mb2 _ϕ2 +mgb cosϕ.

Form the form of kinetic energy we deduce the inertia matrix

Q =

[m(a+ b sinϕ)2 0

0 mb2

].


Its inverse,

Q−1 =

[1

m(a+b sinϕ)20

0 1mb2

]allows us to dene Hamiltonian

H(q,p) =1

2

(p21

m(a+ b sinϕ)2+

p22mb2

)−mgb cosϕ.

In conclusion, the canonical Hamilton's equations of motion of the Furuta's

pendulum take the following form

_θ = ∂H∂p1

= p1m(a+b sinϕ)2

_ϕ = ∂H∂p2

= p2mb2

_p1 = −∂H∂θ = 0

_p2 = −∂H∂ϕ =bp21 cosϕ

m(a+b sinϕ)3−mgb sinϕ

.

5.5 Invariants. Poisson bracket

Consider a mechanical system characterized by generalized coordinates q ∈Rn and momenta p ∈ Rn, with Hamiltonian H(q,p). A function F(q,p) will

be called an invariant of motion, a constant of motion or a rst integral of

the system if F(q(t),p(t)) = const along solutions q(t),p(t) of the canonical

Hamilton's equations, i.e.

dF(q(t),p(t))

dt= 0.

In Theorem 5.3.1 we have established that the Hamiltonian is an example

invariant. In order to check that a given function F(q,p) is an invariant, we

simply dierentiate

dF(q(t),p(t))

dt=

(∂F

∂q

)T_q+

(∂F

∂p

)T_p =

(∂F

∂q

)T∂H

∂p−

(∂F

∂p

)T∂H

∂q= F,H.

The expression appearing on the right hand side of this identity is called

the Poisson bracket of function F and Hamiltonian H. It follows that for F

to be an invariant the Poisson bracket must vanish,

F,H = 0.

The Poisson bracket can be dened for any (suciently smooth) func-

tions F1(q,p) i F2(q,p),

F1, F2 =

(∂F1∂q

)T∂F2∂p

−

(∂F1∂p

)T∂F2∂q

. (5.4)


The Poisson bracket may be regarded as a sort of product of functions,

assigning to a pair of functions F1 and F2 another function F1, F2. This

product has the following properties:

1. F, F = 0 irre exivity,

2. F2, F1 = −F1, F2 antisymmetry,

3. F1, F2, F3+ F2, F3, F1+ F3, F1, F2 = 0 Jacobi identity.

It is worth to note that by the Jacobi identity the Poisson bracket is non-

associative, resembling the cross product of vectors in R3.The signicance of Poisson bracket for Hamiltonian mechanics consists

in the fact that by means of this bracket one can generate new invariants.

This is described formally in the following.

Theorem 5.5.1 (On Poisson bracket) Suppose that F1 i F2 are invariants.

Then, the Poisson bracket F1, F2 is also an invariant.

5.6 Liouville's theorem on invariants

In chapter 0 we have signalized the role played by invariants of motion

in solving the equations of motion: we have shown that the invariance of

angular momentum and energy of the Planet results in computing explicitly

the Planet's orbit as predicted by the 1st Kepler's law. In reference to a

solution of a dierential equation obtained analytically (symbolically, in

closed form), as has been in the case of the equations of motion of the

Planet around the Sun, we shall use the term solvability by quadratures.

Obviously, the solvabiity by quadratures is an extremely desired property.

Sucient conditions for that are provided by the following

Theorem 5.6.1 (Liouville's on invariants) Let the system of canonical Hamil-

ton's equations _q = ∂H

∂p

_p = −∂H∂q,

q,p ∈ Rn, have n invariants F1 = H, F2, . . . , Fn that are independent

and in involution. Then

1. trajectory (q(t),p(t)) of the system lies on the n-dimensional man-

ifold

Mα =(q,p) ∈ R2n

∣∣ F1(q,p)=α1, F2(q,p)=α2, . . . , Fn(q,p)=αn

,


and

2. canonical Hamilton's equations can be solved by quadratures.

In this theorem α = (α1,α2, . . . ,αn)T denotes a vector of constants. For

a given initial condition (q0,p0), we set αi = Fi(q0,p0), i = 1, 2, . . . ,n.

The independence of invariants is understood as independence of functions

F1, F2, . . . , Fn, which means that for any (q,p) ∈Mα

rank

∂F1∂q1

. . . ∂F1∂qn

∂F1∂p1

. . . ∂F1∂pn

......

∂Fn∂q1

. . . ∂Fn∂qn

∂Fn∂p1

. . . ∂Fn∂pn

(q,p) = n.

Finally, the involution of invariants is tantamount to the requirement that

for any i, j = 1, 2, . . . ,n along trajectories of the canonical Hamilton's equa-

tions the Poisson bracket Fi, Fj = 0. Thus, the invariants dealt with in

Theorem 5.6.1 form a maximal system of invariants in the sense that it is

not possible to generate out of them any new invariants. We recall that a

k-dimensional (dierentiable) manifold in Rn is a subset of Rn, on which

n− k independent (and suciently smooth) functions vanish.

5.7 Examples: invariants of motion

Let's come back once again to the Furuta's pendulum. We have n = 2.

From the equations of motion derived in subsection 5.4.2 we infer that there

exist two invariants: Hamiltonian H and momentum p1, so the number of

invariants satises the premises of Theorem 5.6.1. In order to check their

independence we build a matrix[∂H∂θ

∂H∂ϕ

∂H∂p1

∂H∂p2

∂p1∂θ

∂p1∂ϕ

∂p1∂p1

∂p1∂p2

].

Now, from the canonical Hamilton's equations and the mutual independence

of coordinates and momenta this matrix takes the form

rank

[0 − _p2 _θ _ϕ

0 0 1 0

]= 2.

It follows that the invariants are independent provided that either _ϕ 6= 0 or

_p2 6= 0. Equivalently, since p2 = mb2 _ϕ, the invariants are independent if

either _ϕ 6= 0 or ϕ 6= 0. This means that the pendulum's arm needs to be


swinging, and if it stops then its acceleration must be non-zero. One can say

that in "natural" motions of the Furuta's pendulum the invariants fulll the

condition of independence. Next, by the properties of the Poisson bracket,

for checking the condition of involution it suces to compute H,p1,

H,p1 =

(∂H

∂q

)T∂p1∂p

−

(∂H

∂p

)T∂p1∂q

.

Because ∂H∂q = (0, ∗)T , the asterisk refers to an element whose value is not

important, moreover ∂p1∂p = (1, 0)T and ∂p1∂q = (0, 0)T , we conclude that

H,p1 = 0.

In this way we have demonstrated that the Hamiltonian equations of motion

of the Furuta's pendulum have two invariants that are independent and

in involution. This being so, from Theorem 5.6.1 we deduce that these

equations of motion are solvable by quadratures.

5.8 Liouville's theorem on divergence

The system of canonical Hamilton's equations (5.3) without non-potential

forces can be regarded as a system of ODE or as a dynamic system

_x = X(x), x = (q,p)T , X(x) =

(∂H(q,p)∂p

−∂H(q,p)∂q

)

operating in R2n. Such a system is called Hamiltonian. Set 2n = s. Func-

tion X(x) is named the vector eld. Given an initial state x of a dynamic

system, its trajectory x(t) = ϕ(t, x). Function

ϕ : R× Rs −→ Rs

is referred to as the ow of dynamic system. By denition, the ow deter-

mines the state of dynamic system at the time instant t, initialized from the

state x at time 0. With xed t, the ow denes a function

ϕt : Rs −→ Rs, ϕt(x) = ϕ(t, x).

This function has the following properties:

1. ϕ0(x) = x identity property,

2. ϕt+s(x) = ϕt(ϕs(x)) semi-group property.


ϕt(D)Dϕt(x)

Figure 5.2: Flow of dynamic system

A direct consequence of these two is that ϕ−1t (x) = ϕ−t(x).

Consider a set of initial states D ⊂ Rs of a dynamic system with ow

ϕ(t, x), and compute its image ϕt(D) at time t, see Figure 5.2. The way

the set D is transformed by function ϕt (this describes how the ow "is

owing") provides essential information on the behavior of the dynamic

system. Specically, let Vt denotes the volume of image ϕt(D),

Vt = vol(ϕt(D)).

By denition, V0 = vol(D). We distinguish the following three relationships

between Vt and V0:

1. Vt < V0 the ow is contracting,

2. Vt > V0 the ow is expanding,

3. Vt = V0 the ow is volume-preserving (isochoric).

It is clear that these cases do not exhaust all possibilities; it may happen

that at a certain time instant the volume grows up, and at another decreases.

Undoubtedly, establishing one of the three properties of the ow would be

instructive.

To this objective we need the concept of divergence of a vector eld,

namely, given vector eld X(x), its divergence equals

divX(x) = tr∂X(x)

∂x,

where trM denotes the trace (sum of elements along the main diagonal) of

a matrix M. Now, using the rules of changing coordinates in the integral


calculus it is not hard to prove that

dV(t)

dt=

∫ϕt(D)

divX(x)dx. (5.5)

This identity yields directly the following

Theorem 5.8.1 (Liouville's on divergence) For a dynamic system dened

by the vector eld X(x), with ow ϕt(x), if the divergence

divX(x) = 0,

then the ow is isochoric. The ow of a Hamiltonian system is iso-

choric.

Checking that the vector eld of a Hamiltonian system has zero divergence

is immediate, namely,

∂X(x)

∂x=

∂2H(q,p)∂q∂p

∂2H(q,p)∂p2

−∂2H(q,p)∂q2

−∂2H(q,p)∂p∂q

,

therefore

divX(x) = tr[∂2H(q,p)∂q∂p

]− tr

[∂2H(q,p)∂p∂q

]= 0

by virtue of the identity of mixed partial derivatives.

5.9 Examples: divergence

For illustration of the concept of divergence we shall look at two examples.

5.9.1 Example 1

As shown in subsection 2.4.2, the equation of motion of a pendulum

x = − sin x

can be represented as a dynamic system_x1 = x2

_x2 = − sin x1,

with vector eld X(x) = (x2,− sin x1)T . We compute the derivative

DX(x) =

[0 1

− cos x1 0

],

and nd that divX(q) = trDX(x) = 0. This means that the ow of the

pendulum is volume-preserving.


D

ϕt ′(x ′)

U

x x ′

Figure 5.3: Theorem on recurrence

5.9.2 Example 2

Next, we consider the dynamic system_x1 = x1 + x

22

_x2 = x31

,

containing vector eld X(x) = (x1 + x22, x

31)T . Its divergence

divX(x) = tr

[1 2x23x21 0

]= 1,

therefore, the ow is expanding.

5.10 Poincare's theorem on recurrence

A consequence of the Liouville's theorem on divergence is the following result

that provides an important characterization of trajectories of a Hamiltonian

system.

Theorem 5.10.1 (Poincare's on recurrence) Suppose that the ow ϕt(x) of

a dynamic system _x = X(x), x ∈ Rs, has zero divergence, divX(x) = 0.

Let D ⊂ Rs be invariant with respect to the ow (ϕt(D) ⊂ D), and

bounded (vol(D) <∞). Then, in any neighborhood U of the point x ∈ Dthere exists a point x ′ ∈ U such that at a certain time instant t ′ there

holds ϕt ′(x′) ∈ U or, equivalently, ϕt ′(U) ∩U 6= ∅, see Figure 5.3.

For the reason that the divergence of a Hamiltonian vector led is zero,

Theorem 5.10.1 applies to Hamiltonian systems dened on a bounded subset

of the phase space of positions and momenta.


α1α2

Figure 5.4: Torus T2

X (α1)

Y

(α2)

1

3

2

1 2 3 4 50

Figure 5.5: Dening the torus T2

5.11 Examples: a dynamic system on torus

We think that an appealing illustration of the Poincare's theorem on re-

currence is provided by the following dynamic system dened on the torus

T2 = S1 × S1. As displayed in Figure 5.4, as coordinates describing the

position of a point on torus one can take two angles (α1,α2). The torus T2

is created by identifying in the plane R2 all points of integer coordinates,

see Figure 5.5. Figuratively speaking, we rst "roll up" the plane R2 in

the vertical direction, to make all horizontal lines of integer ordinates co-

incide, and then "roll up" the obtained innite cylinder horizontally to get

the coincidence of all lines of integers abscissas. By design, the torus is a

bounded subset of R3; its area being equal to 4π2. Now, let us dene on T2

a dynamic system _α1 = a

_α2 = b,


where a ∈ R and b ∈ R are integers. Obviously, X(α1,α2) = (a,b)>,

and divX = 0. Also, the system's trajectories stay on the torus, which

means that the torus is an invariant set. It follows that all conditions of the

Poincare's theorem on recurrence are fullled. For zero initial condition,

α1(0) = α2(0) = 0, the orbits of the system satisfy

dα2dα1

=b

a,

i.e. they are straight lines α2 = baα1, as shown in Figure 5.5. It should be

noticed that a position of such line on the torus depends on the ratio ba .

If this ratio is a rational number then the line α2 = baα1 passes through a

point in the plane of integer coordinates; this produces a closed orbit on the

torus. Every point of this orbit returns innitely many times. The situation

is completely dierent for irrational ratio ba , as the line α2 = b

aα1 will not

pass through any point of the plane of integer coordinates. The orbit on

torus is winding up continuously. This example helps to imagine which kind

of trajectories/orbits may appear after solving the Hamiltonian equations

of motion.

5.12 Exercises

Exercise 5.1 Compute the Legendre transform of function f(v) = v2 and

f(v) = v4.

Exercise 5.2 Derive Hamiltonian equations of motion of a Planet aroud the

Sun. Hint: consult chapter 0.

Exercise 5.3 Obtain Hamiltonian equations of motion of the inverted pen-

dulum and the leg of a jumping robot included as exercises into the previous

chapter.

Exercise 5.4 Provide Hamiltonian equations of motion of the sperical pen-

dulum shown in Figure 5.6. The tip of this pendulum takes positions on

the sphere S2.

Exercise 5.5 Derive Hamiltonian equations of motion of the space robot

displayed in Figure 4.5. Ignore the motion of the robot's base and use three

generalized coordinates q = (φ, θ, l)T .

Exercise 5.6 Derive Hamiltonian equations of motion for the robot Ballbot

described in the previous chapter, in Exercise 4.6.


ϕ

θ

Y

Z

X

m

l

g

Figure 5.6: Spherical pendulum

Exercise 5.7 Prove Theorem 5.5.1. Hint: use the Jacobi identity.

Exercise 5.8 Prove the identity (5.5). Hint: observe that

dV(t)

dt=dV(t+ s)

ds

∣∣∣∣s=0

,

and consult chapter 3 of [Arn78].


Foundations of Hamiltonian mechanics have been exposed in chapter 3 of

the monograph [RK95], in chapter II of [Gre03], as well as in chapter 13 of

[Tay05]. The Legendre transform and a derivation of the canonical Hamil-

ton's equations can be found in chapter 7 of [Gut71]. Instructive addi-

tional information on Hamiltonian mechanics is contained in the monograph

[Arn78]. In the same reference there is a complete statement and a proof

of the Liouville's theorem on invariants; sometimes this theorem is referred

to as the Liouville'-Arnold's theorem. A proof of the Liouville's theorem

on divergence has been provided in chapter 3 of [Arn78]. This chapter also

includes a simple proof of the Poincare's theorem on recurrence, and the

example presented in section 5.11.


Bibliography

[Arn78] V. I. Arnold. Mathematical Methods of Classical Mechanics.

Springer, Berlin, 1978.



[Gut71] R. Gutowski. Mechanika analityczna. PWN, Warszawa, 1971. (in

Polish).





Chapter 6

Kinematics and dynamics of

rigid body

6.1 Motion

A rigid body will be dened as a compact (closed and bounded) subset of

the Euclidean space, B ⊂ R3. We recall that in a closed set every convergent

sequence of elements has a limit belonging to this set, and bounded means

that B can be packed into a ball of nite volume. The denition of the rigid

body includes also a requirement that during its displacement lengths of

vectors and angles between vectors within the rigid body remain unchanged;

with reference to Figure 6.1 this means that the (Euclidean) norm of vectors

||u−w|| and ||v−w|| as well as the angle between vectors u−w i v−w will

not change.

Suppose that a coordinate frame (XS, YS,ZS) has been xed, called

the space frame. To rigid body B we attach another coordinate frame

(XB, YB,ZB), that will be referred to as the body frame. Both these frames

have been displayed in Figure 6.2. The displacement of the rigid body will

be described as a transformation of the frame (XS, YS,ZS) into the frame

(XB, YB,ZB) realized by a 4× 4 matrix

A =

[R T

0T 1

]. (6.1)

This matrix is composed of a rotation matrix R of dimension 3×3, obeying

the orthogonality condition RRT = RTR = I3 (I3 unit 3 × 3 matrix) and

such that detR = +1, and a translation vector T ∈ R3.Given a point P of the body B whose position with respect to the body

frame is described by a vector r ∈ R3, its uniform coordinates are dened

74

Chapter 6. Kinematics and dynamics of rigid body 75

Y

Z

X

B

w

u

v

∠(u−v,w−v)

Figure 6.1: Rigid body

YS

ZSP

XS

B ZB

YB

XB

r

A=

[ R T

0T

1

]

Figure 6.2: Displacement of rigid body

as a pair (r, 1) ∈ R4. The transformation matrix A allows one to determine

uniform coordinates of the point P relative to the space frame as(s

1

)=

[R T

0T 1

](r

1

)=

(Rr+ T

1

).

For the uniform coordinates of origin of the body frame in the body frame

are (0, 1), it is easily seen that the origin's coordinates with respect to the

space frame are equal to (T , 1), therefore, vector T denes the position of

origin of the body frame in the space frame.

Now, assume that T = 0, which means that the origins of both the frames

coincide, and let r1, r2 and r3 denote columns of a rotation matrix R. It

follows that images of the versors (unit vectors) e1, e2, e3 of the body frame


are precisely equal to s1 = Re1 = r1, s2 = Re2 = r2 and s3 = Re3 = r3.

In this way we have come to the conclusion that matrix R characterizes

the rigid body orientation, while vector T describes its position. The set of

matrices (6.1) describing displacements of the rigid body denes the special

Euclidean group SE(3). If

A1 =

[R1 T10T 1

], A2 =

[R2 T20T 1

]are two elements of SE(3) then the the group operation is tantamount to

the block matrix multiplication,

A1A2 =

[R1R2 R1T2 + T10T 1

].

SE(3) is a matrix group, with unit element I4 and inverse element

A−1 =

[RT −RTT

0T 1

].

By analogy to the motion of the material point, the motion of the rigid

body will be dened as a transformation of time into the special Euclidean

group,

c : R −→ SE(3), c(t) =

[R(t) T(t)

0T 1

]that needs to be continuous and continuously dierentiable up to at least

order 2. We shall say that the scenery of the rigid body motion is constituted

by time R and the special Euclidean group SE(3). A description of rigid

body motion requires establishing at every time instant the position and

the orientation of the body frame with respect to the space frame. We can

think of the description of motion of an airplane relative to an airport as an

example.

6.2 Elementary rotations

Suppose that the space and the body frames coincide, and that the latter

rotates with respect to the former. By elementary rotations we mean rota-

tions around the axes XS, YS, and ZS. Our objective is to dene rotation

matrices R(X,α), R(Y,β), and R(Z,γ) describing elementary rotation, re-

spectively, around the axis X by an angle α, around the axis Y by β, and

around the axis Z by an angle γ. As an example, we shall derive the form


YS

ZS

XS

α

e1

αe2

e3

Figure 6.3: Elemenetary rotation around the X axis

of matrix R(X,α), see Figure 6.3, the remaining elementary rotations can

be found analogously.

Let's take into account the versors of the body frame. By denition,

the columns of the rotation matrix are images of these versors in the space

frame. Denote them by r1, r2, r3. Obviously, after the rotation around the

X axis the versor e1 does not change, so r1 = e1. Next, space coordinates

of the versor e2 equal (0, cosα, sinα)T , and the coordinates of versor e3 are

(0,− sinα, cosα)T . In this way we have obtained the identity

R(X,α) =

1 0 0

0 cosα − sinα

0 sinα cosα

.

Two remaining elementary rotations are described by the matrices

R(Y,β) =

cosβ 0 sinβ

0 1 0

−sinβ 0 cosβ

, R(Z,γ) =

cosγ − sinγ 0

sinγ cosγ 0

0 0 1

.

6.3 Coordinates for SE(3)

Let a matrix

A =

[R T

0T 1

]∈ SE(3)

be given, describing position and orientation of the rigid body. The rota-

tion matrix R contains 9 elements that satisfy 6 independent orthogonality

conditions, RRT = I3. In consequence, rotation matrices form in R9 a 3-

dimensional submanifold. We shall denote it by SO(3) and call the special


orthogonal group. The translation vector T ∈ R3 can be arbitrary, therefore

the whole group SE(3) is a 6-dimensional manifold contained in R12.For computational reasons it is convenient to characterize point of a

manifold by means of coordinates; in the case of the group SE(3) we need

6 coordinates. In what follows we shall focus on one example coordinate

system. For the vector of translation T = (T1, T2, T3)T we shall use Cartesian

coordinates that are identical with elements of this vector. For the rotation

matrix we have chosen so-called Euler ZYZ angles, that can be computed

for a given rotation matrix R from the identity

R = R(Z,ϕ)R(Y, θ)R(Z,ψ) = E(ϕ, θ,ψ) =[cosϕ cos θ cosψ− sinϕ sinψ − cosϕ cos θ sinψ− sinϕ cosψ cosϕ sin θ

sinϕ cos θ cosψ+ cosϕ sinψ − sinϕ cos θ sinψ+ cosϕ cosψ sinϕ sin θ

− sin θ cosψ sin θ sinψ cos θ

].

The Euler angles form a local coordinate system; notice that if θ = 0 or

θ = ±π then it is not possible to nd uniquely the remaining two angles.

Global coordinates for SO(3) do not exist, therefore they do not exists for

SE(3) either.

6.4 Velocity

In contrast to the velocity of motion of a material point, dening a concept

of velocity of rigid body is not obvious. In what follows we shall conne

exclusively to changes of the rigid body orientation, and assume that

c(t) = R(t).

Since R(t) is orthogonal, at any time instant there holds

R(t)RT (t) = RT (t)R(t) = I3,

resulting in

_R(t)RT (t) + R(t) _RT (t) = _RT (t)R(t) + RT (t) _R(t) = 0. (6.2)

In consequence, we get two matrix rotation velocities of the rigid body:

1. ΩS = _RRT velocity in space (space velocity),

2. ΩB = RT _R velocity in body (body velocity).


By denition, it follows that

_R = ΩSR = RΩB,

which means that

ΩS = RΩBRT .

By a substitution to the identity (6.2) we discover that the matrix rotation

velocities fulll the condition

Ω+ΩT = 0,

i.e. the matrices ΩS and ΩB are skew symmetric. From linear algebra we

know that any skew symmetric 3 × 3 matrix is fully determined by its 3

entries. We shall associate with the vector ω = (ω1,ω2,ω3)T a matrix

Ω = [ω] =

0 −ω3 ω2ω3 0 −ω1−ω2 ω1 0

.

The positioning of components of vector ω within Ω guarantees that

Ωv = ω× v,

where v ∈ R3, and × stands for the cross product of vectors. Applying

this to both the matrix rotation velocities we introduce a vector rotation

velocity ωS in space, and a vector rotation velocity ωB in body, so that

ΩS = [ωS], ΩB = [ωB].

We have deduced the vector rotation velocities from properties of the

rotation matrix. The velocity ωB resides in the body frame, whereas ωSbelongs to the space frame. The two velocities are related to each other as

ωS = RωB.

The space velocity denes the physical rotation velocity of the body with

respect to the space frame. Unlike this, the body velocity is not a velocity

of the body with respect to the body frame, because the body relative to

this frame does not move. This being so, the velocity ωB can be inter-

preted as the space velocity ωS transferred to (observed from) the body

frame. Notwithstanding their physical sense, both these velocities are for-

mally equivalent. We shall see that, when dening equations of motion of


YS

ZS

dm

XS

B ZB

YB

XB

r

A =

[R T

0T 1

]

Figure 6.4: Motion of element of mass dm

the rigid body, the body rotation velocity is ever more convenient than the

space velocity.

When using coordinates it is useful to recognize the following relation

between the body rotation velocity and the time derivatives of Euler angles

e = (ϕ, θ,ψ)

ET (ϕ, θ,ψ) _E(ϕ, θ,ψ) = [ωB], (6.3)

where ωB

ωB =ME _e =

− sin θ cosψ sinψ 0

sin θ sinψ cosψ 0

cos θ 0 1

_ϕ_θ_ψ

.

6.5 Lagrangian dynamics

Let rigid body B equipped with body frame (XB, YB,ZB) move with respect

to space frame (XS, YS,ZS), so that the transformation of frames is described

by the matrix

A =

[R T

0T 1

].

Choose an element of mass dm (a material point) of the body B, whose

position in body frame is given by a vector r ∈ R3, see Figure 6.4. In order

to derive equations of motion of the body B we shall exploit the Lagrangian

formalism. Let v denote the velocity of element of mass dm relative to the

space frame. The kinetic energy of mass dm,

dK =1

2dmvTv =

1

2dm tr(vvT ),


where we have used the property that, for a number of α ∈ R the trace

trα = α and that for two matrices A and B tr(AB) = tr(BA). The position

s of mass dm in space frame s = Rr+ T , therefore the velocity

v = _s = _Rr+ _T ,

for r does not change in time. Using the form of velocity we get

dK =1

2dm tr

((_Rr+ _T

)(rT _RT + _TT

)).

A further expansion yields

dK =1

2dm tr( _RrrT _RT ) +

1

2dm tr( _Rr _TT ) +

1

2dm tr( _TrT _RT ) +

1

2dm tr( _T _TT ).

Finally, taking once again into account the property tr(AB) = tr(BA), and

the fact that trA = trAT and trα = α, we deduce the following

dK =1

2dm tr

(_RrrT _RT

)+ dm _TT _Rr+

1

2dm|| _T ||2.

After these preparations, the kinetic energy of the whole body B is an

integral of dK over all points of the body,

K =

∫B

dK =1

2tr( _R

∫B

rrTdm _RT ) + _TT _R

∫B

rdm+1

2mB|| _T ||

2,

where mB denotes the mass of the body. The integral that has appeared

in the rst component of the sum is called the pseudoinertia matrix of B.

With notation r = (x,y, z)T , we have

JB =

∫B

rrTdm =

∫B x2dm

∫B xydm

∫B xzdm∫

B yxdm∫B y

2dm∫B yzdm∫

B zxdm∫B zydm

∫B z2dm

. (6.4)

Also, observe that in the second component there have appeared coordinates

of the center of mass of B,

rB =1

mB

∫B

rdm,

expressed in the body frame. Based on these observations we can rewrite

the kinetic energy in the form

K =1

2tr(

_RJB _RT)+mB _TT _RrB +

1

2mB|| _T ||

2.


As we know, the time derivative of the rotation matrix can be expressed

in terms of either the matrix rotation velocity in the space or in the body,_R = ΩSR or _R = RΩB. We select the latter possibility that leads to the

following

tr(

_RJB _RT)= tr

(RΩBJBΩ

TBRT)= − tr

(JBΩ

2B

).

Hereabout we have invoked for the third time the property of trace tr(AB) =

tr(BA) and the skew symmetry of matrix ΩB. We also compute

mB _TT _RrB = mB _TTRΩBrB = mB _TTR(ωB × rB).

All these mathematical developments result in the kinetic energy

K = −1

2tr(JBΩ

2B) +mB _TTR(ωB × rB) +

1

2mB|| _T ||

2.

In order to compute the potential energy of body B we assume that its

total mass is concentrated in the center of mass, so z

gV = −mB(g, sB) = −mB(g,RrB + T).

Symbol sB refers to the position of the center of mass in space frame, and

g denotes the vector of gravitational acceleration expressed in this frame.

Combining our results we arrive at the Lagrangian

L = K−V = −1

2tr(JBΩ

2B

)+mB _TTR(ωB×rB)+

1

2mB|| _T ||

2+mB(g,RrB+T).

The expression for Lagrangian gets even more transparent after introducing

the body inertia matrix

IB =

JB22 + JB33 −JB12 −JB13−JB21 JB11 + JB33 −JB23−JB31 −JB32 JB11 + JB22

.

With the help of the inertia matrix the Lagrangian can be written as follows

L =1

2ωTBIBωB +mB _TTR(ωB × rB) +

1

2mB|| _T ||

2 +mB(g,RrB + T). (6.5)

6.6 Euler-Lagrange equations

A derivation of the Euler-Lagrange equations of motion for the rigid body

needs to be preceded by expressing the Lagrangian (6.5) in coordinates. To


this aim we shall use the Cartesian coordinates for the position and the

Euler angles ZYZ e = (ϕ, θ,ψ) for the orientation. We assume that the

body frame has been attached at the center of mass of the body, so rB = 0.

Having taken into account the formula (6.3) we obtain

L =1

2_eTMT

EIBME _e+1

2mB( _T21 + _T22 + _T23 ) +mB(g, T).

Let assume additionally that the inertia matrix is diagonal, IB = diagIB1,

IB2, IB3, and that the vector of gravitational acceleration is directed along

the ZS axis, i.e. g = −ge3. With these notations the coordinate expression

of the Lagrangian is the following

L =1

2(IB1 sin2 θ cos2ψ+ IB2 sin2 θ sin2ψ+ IB3 cos2 θ) _ϕ2+

1

2(IB1 sin2ψ+ IB2 cos2ψ) _θ2 +

1

2IB3 _ψ2 − (IB1 − IB2) sin θ sinψ cosψ _ϕ _θ+

IB3 cos θ _ϕ _ψ+1

2mB( _T21 + _T22 + _T23 ) −mBgT3. (6.6)

Given Lagrangian (6.6), the equations of motion of the rigid body can be

derived in a standard way.

6.7 Euler-Newton equations

As an alternative to the Lagrangian formulation, in this subsection we shall

describe the rigid body motion in terms of the Euler-Newton equations.

Let the space and the body frames be given, and assume that the body

frame is placed in the body center of mass. As usual, the body motion is

characterized by a rotation matrix R and a position vector T . Let IB denote

the inertia matrix of the body relative to the body frame, and let vB and

ωB be the linear and angular velocities in body frame. In body frame we

compute the momentum and the angular momentum,

pB = mBvB,

MB = IBωB,

and transform them to the space frame,

pS = RpB = mBRvB,

MS = RMB = RIBωB.


If there are no external forces/torques acting on the body, the Conserva-

tion Principles of Momentum and of Angular Momentum hold in the space

frame, therefore we have

_pS = mB _RvB +mBR _vB = 0,

_MS = _RIBωB + RIB _ωB = 0.

Now, a substitution _R = RΩB = R[ωB] results in

mBRΩBvB +mBR _vB = R(mBωB × vB +mB _vB) = 0

RΩBIBωB + RIB _ωB = R(ωB × (IBωB) + IB _ωB) = 0.

After multiplying both these identities from the left by the RT we end up

with the Euler-Newton equations of motion_vB = vB ×ωBIB _ωB = (IBωB)×ωB

. (6.7)

If forces FB or torques τB act in the body frame, they should be added

to the right hand side of equations(6.7). Having solved the Euler-Newton

equations with respect to velocities vB and ωB, from identity _R = R[ωB] we

nd the body orientation matrix R. Then, we compute the space velocity

vS = RvB. Eventually, the position of the rigid body is found from equation_T = vS.

6.8 Examples

The Euler-Newton equations will be illustrated by a derivation of so called

Euler's equations of attitude of rigid body. We assume that the inertia

matrix of the body is diagonal,

I = diagI1, I2, I3.

From the second of equations (6.7) it follows that

I _ω = (Iω)×ω.

Now, taking into account the form of the inertia matrix we obtain the Euler's

equations for the body angular velocity ω = (ω1,ω2,ω3)T

_ω1 =I2−I3I1

ω2ω3

_ω2 =I3−I1I2

ω1ω3

_ω3 =I1−I2I3

ω1ω2

.


6.9 Exercises

Exercise 6.1 Dene the group multiplication, the neutral element, and the

inverse element in special orthogonal group SO(3) and in special Euclidean

group SE(3).

Exercise 6.2 Derive formulas for elementary rotations around axes Y and Z.

Exercise 6.3 Using the property Ωv = [ω]v = ω× v show that

R[ω]RT = [Rω].

Exercise 6.4 Let a pesudoinertia matrix JB be given, dened by (6.4). As-

suming that rB denotes position of the body center of mass in the body

frame, and vector T ∈ R3 and matrix R ∈ SO(3) represent position and

orientation of the body frame with respect to the space frame show that

the pseudoinertia matrix in space frame satises the generalized Steiner's

formula

JS = RJBRT +mB(RrBT

T + TrTBRT ) +mB||T ||

2.

Exercise 6.5 Derive formula (6.5).

Exercise 6.6 Write the Euler-Lagrange equations for rigid body motion cor-

responding to Lagrangian (6.6).

Exercise 6.7 Prove that for I1 = I2 the norm ||ω|| of angular velocity satis-

fying the Euler's equations is constant. Solve these equations, and obtain

ω1(t) and ω2(t).


The rigid body dynamics is studied in chapter III of [Gre03]. Additional

information on the rigid body motion can be found in chapter 6 of the book

[Arn78], including a derivation of the Euler's equations. The rotational mo-

tion of rigid body is the subject of chapter 10 of [Tay05]. An application of

Lagrangian mechanics to the modeling of robot dynamics has been described

in monograph [TMD+00] and in lecture notes [TM18].


Bibliography



[Gre03] D. T. Greenwood. Advanced Dynamnics. Cambridge Univer-

sity Press, Cambridge, 2003.



[TM18] K. Tchon and R. Muszynski. Robotyka. Wroc law University of

Science and Technology, project AZON, 2018. (in Polish).

[TMD+00] K. Tchon, A. Mazur, I. Duleba, R. Hossa, and R. Muszynski.

Manipulators and Mobile Robots: Models, Motion Planning,

Control. Akademicka Ocyna Wydawnicza PLJ, Warsaw, 2000.

(Wroc law University of Science and Technology, project AZON,

2018, in Polish).

Chapter 7

Lagrange top

The Lagrange top is a symmetric rigid body of conical shape shown in Fig-

ure 7.1. We shall assume that the top spins around its axis, and the position

of its contact point with a horizontal plane (the ground) is xed. Our task

consists in deriving equations of motion of the top using the Lagrangian

and the Hamiltonian formalisms, and then performing an analysis of this

motion. To this objective we dene the space frame (XS, YS,ZS) and the

body frame (XB, YB,ZB) attached to the top in the way presented in the

gure. Let the top's mass equal m, and its moments of inertia relative to

axes of the body frame amount to I1, I2, and I3. Due to the top's symme-

try, the moments relative to axes XB and YB are the same, I1 = I2. It is

assumed that the center of mass of the top is located on the ZB axis, at the

distance r from the origin of the body frame. It should be noticed that the

body frame has not been placed at the center of mass of the top, but at the

xed contact point of the top with the ground. For this reason the position

vector T = 0.

To describe the top's orientation it is convenient to use the angles (α,β,γ)

shown in the gure. These angles dene a sequence of elementary rotations

that should be accomplished in order that the space frame coincide with

the body frame: we begin with a rotation around the ZS axis until the XSaxis gets perpendicular to the plane ZS,ZB, then make a turn around the

instantaneous axis X ′S resulting in the ZS axis coinciding with the axis ZB,

and end up with a rotation around the axis ZS = ZB after which the X ′S axis

overlaps XB. This being so, the orientation coordinates q = (α,β,γ)T be-

come generalized coordinates of the Lagrange top. It turns out that the ZYZ

Euler angles (ϕ, θ,ψ) introduced in the previous chapter can be expressed

87

Chapter 7. Lagrange top 88

YS

ZS, _α

XS

X ′S,_β

α

γ

βZB

, _γ,ω3

XB,ω1

YB,ω2

mr

⊥ ZB

g Y ′S

Figure 7.1: Lagrange top

in terms of the generalized coordinates asϕ = α− π/2

θ = β

ψ = γ+ π/2

.

7.1 Euler-Lagrange equations

In the previous chapter we have derived a formula for Lagrangian of the

rigid body whose orientation is dened by the Euler angles ZYZ. Although

in the case of the Lagrange top the origin of the body frame does not lie at

the center of mass of the body, nevertheless the second component in the

formula (6.5)

L =1

2ωTBIBωB +m _TTR(ωB × rB) +

1

2mB|| _T ||

2 +mB(g,RrB + T)

vanishes because the position vector T = 0, which implies _T = 0. Also the

third component vanishes. Taking this into account we can invoke the (6.6),


with added potential energy in accordance with (6.5),

L =1

2

(IB1 sin2 θ cos2ψ+ IB2 sin2 θ sin2ψ+ IB3 cos2 θ

)_ϕ2+

1

2

(IB1 sin2ψ+ IB2 cos2ψ

)_θ2 +

1

2IB3 _ψ2−

(IB1 − IB2) sin θ sinψ cosψ _ϕ _θ+ IB3 cos θ _ϕ _ψ+mB(g,RrB).

The matrix R appearing in the last component of the sum is the matrix

of Euler angles E(ϕ, θ,ψ) dealt with in section 6.3. Since rB = (0, 0, r)T ,

mB = m, as well as g = (0, 0,−g)T , we compute

mB(g,RrB) = −mgr cos θ.

Next, due to the identity of moments of inertia I1 = I2, we get

L =1

2I1

(_θ2 + _ϕ2 sin2 θ

)+1

2I3( _ψ+ _ϕ cos θ)2 −mgr cos θ.

Finally, after insertion of angles α,β,γ, the Lagrangian of the Lagrange top

takes the following form

L =1

2I1

(_β2 + _α2 sin2 β

)+1

2I3( _γ+ _α cosβ)2 −mgr cosβ.

Under assumption that any non-potential forces are absent, the Euler-

Lagrange equations of motion have the form

d

dt

∂L

∂ _q−∂L

∂q= 0.

We nd derivatives

∂L∂ _α = I1 _α sin2 β+ I3( _γ+ _α cosβ) cosβ,∂L∂α = 0,∂L∂ _β

= I1 _β,∂L∂β = I1 _α2 sinβ cosβ− I3( _γ+ _α cosβ) _α sinβ+mgr sinβ,∂L∂ _γ = I3( _γ+ _α cosβ),∂L∂γ = 0,

and arrive at the following equations of motion for the Lagrange top(I1 sin2 β+ I3 cos2 β

)_α+ I3 _γ cosβ = const

I1β+ (I3 − I1) _α2 sinβ cosβ+ I3 _α _γ sinβ−mgr sinβ = 0

I3( _γ+ _α cosβ) = const

. (7.1)


7.2 Canonical Hamilton's equations

Before dening the Hamiltonian we rst need to compute the inertia matrix

of the kinetic energy of the top,

Q(q) =

I1 sin2 β+ I3 cos2 β 0 I3 cosβ

0 I1 0

I3 cosβ 0 I3

.

Its inverse matrix,

Q−1(q) =1

I21I3 sin 2β

I1I3 0 −I1I3 cosβ

0 I1I3 sin2 β 0

−I1I3 cosβ 0 I1(I1 sin2 β+ I3 cos2 β)

.

Let now p = (p1,p2.p3)T denote the vector of generalized momenta.

The Hamiltonian

H(q,p) =1

2pTQ−1(q)p+ V(q),

therefore, for the Lagrange top we get

H(q,p) =1

2

p21I1 sin2 β

+1

2

p22I1

+1

2p23I1 sin2 β+ I3 cos2 β

I1I3 sin2 β−

p1p3 cosβ

I1 sin2 β+mgr cosβ.

By denition, the canonical Hamilton's equations related with this Hamil-

tonian are the following

_α = ∂H∂p1

= p1−p3 cosβ

I1 sin2β

_β = ∂H∂p2

= p2I1

_γ = ∂H∂p3

=p3(I1 sin

2β+I3 cos2β)−p1I3 cosβ

I1I3 sin2β

_p1 = −∂H∂α = 0

_p2 = −∂H∂β = −p21 cosβ+p

23 cosβ+p1p3(1+cos2β)

I1 sin3β

+mgr sinβ

_p3 = −∂H∂γ = 0

. (7.2)

7.3 Invariants and quadratures

As we know, the Hamiltonian is always an invariant of the Hamilton's canon-

ical equations; furthermore, it follows from equations number 4 and 6 of (7.2)


that momenta p1 and p3 are also invariants of motion. Thus, the number

of invariants is exactly as big as required by the Liouville's theorem on in-

variants. We only need to check that these invariants are independent and

in involution. To this objective we rst dene a matrix∂H∂α

∂H∂β

∂H∂γ

∂H∂p1

∂H∂p2

∂H∂p3

∂p1∂α

∂p1∂β

∂p1∂γ

∂p1∂p1

∂p1∂p2

∂p1∂p3

∂p3∂α

∂p3∂β

∂p3∂γ

∂p3∂p1

∂p3∂p2

∂p3∂p3

=

0 − _p2 = −I1β 0 _α _β _γ

0 0 0 1 0 0

0 0 0 0 0 1

.

It is easily seen that the invariants are independent on condition that veloc-

ity or acceleration of changing the angle β are non-zero. Next, we compute

the Poisson brackets of invariants,

H,p1 =(∂H∂q

)T∂p1∂p −

(∂H∂p

)T∂p1∂q = −∂H∂α = 0,

H,p3 =(∂H∂q

)T∂p3∂p −

(∂H∂p

)T∂p3∂q = −∂H∂γ = 0,

p1,p3 =(∂p1∂q

)T∂p3∂p −

(∂p1∂p

)T∂p3∂q = 0.

Summarizing, we have demonstrated that the three invariants of motion are

independent and in involution. As a result, equations (7.2) can be solved

by quadratures.

7.4 Motion of the Lagrange top

Making use of invariants, we shall analyze the motion of the Lagrange top.

For, during the motion only the top's orientation changes, its motion consists

of three rotations named, respectively, precession (α algle), nutation (β

angle) and spinning (γ angle). As a staring point of our analysis we shall

take the Hamiltonia

H(q,p) =1

2

p21I1 sin2 β

+1

2

p22I1

+1

2p23I1 sin2 β+ I3 cos2 β

I1I3 sin2 β+

p1p3 cosβ

I1 sin2 β+mgr cosβ.

Obviously, the Hamiltonian is well dened for nutation angles β 6= 0,π. The

component number 3 of the Hamiltonian can be represented as a sum of two

terms,1

2

p23I3

+1

2

p23 cos2 β

I1 sin2 β.


Using the fact that H, p1 and p3 are invariants of motion, the Hamiltonian

can be re-written in the following form

E = H−1

2

p23I3

=1

2

(p1 − p3 cosβ)2

I1 sin2 β+1

2I1 _β2 +mgr cosβ = const .

Observe that only two variables, β and _β have appeared in this formula.

An introduction of a new variable u = cosβ, such that _u = − sinβ _β yields

_u2 = sin2 β _β2 = (1− u2) _β2.

On the other hand we have

_β2 =2E

I1−

(pn1− p3u)2

I21(1− u2)

−2mgru

I1.

The last two identities result in a dierential equation for the variable u,

namely

_u2 = (1− u2)2(E−mgru)

I1−

(p1 − p3u)2

I21=

(a− bu)(1− u2) − (c− du)2 = f(u), (7.3)

where

a =2E

I1, b =

2mgr

I1, c =

p1I1

, d =p3I1

.

Equations number 1 and 3 of (7.2) allow us to express by means of the

variable u the angular velocities of the Lagrange top,_α = c−du

1−u2

_γ =(d−e)u2−cu+e

1−u2

,

with the notation e = p3I3

.

The function f(u), dened in the equation (7.3), is a polynomial of

order 3; suppose that its graph looks like in Figure 7.2. Because _u2 > 0, we

are interested in the portion of the graph of f(u) corresponding to abscissae

u1 and u2. If ui = cosβi, i = 1, 2, it means that the angle β2 6 β 6 β1,

so the inclination of the top's axis ZB must t in this range.

We have shown that the speed of change of the precession angle α

amounts to

_α =c− du

1− u2=

L(u)

1− u2.


f(u)

u1u1 u20

Figure 7.2: Function f(u)

L(u)

u1u1 u20 cd

Figure 7.3: Function L(u)


Y

Z

X

precession

nutation

spinning

Figure 7.4: L(u) < 0

Y

Z

X

precession

nutation

spinning

Figure 7.5: L(u) > 0

The denominator of this ratio is positive all the time, therefore the direction

of the angle α changes depends on the numerator L(u). Feasible positions

of the straight line L(u) are shown in Figure 7.3. In the case when in the

interval [u1,u2] there holds L < 0, the speed of change of α is negative,

and the ZB axis of the top rotates left. If in this interval L > 0, the top's

axis rotates right. When in the interval [u1,u2] the sign of L changes, the

direction of rotation of the top's axis changes. These three situations are

displayed in Figures 7.4, 7.5 and 7.6.

The speed of change of the angle γ has been expressed as

_γ =(d− e)u2 − cu+ e

1− u2=

L(u)

1− u2.


Y

Z

X

precession

nutation

spinning

Figure 7.6: L(u) changes sign

L(u)

u1u1 u20

Figure 7.7: Function L(u)

The numerator L(u) of this formula is a quadratic function of the variable

u whose possible shapes for I1 > I3 and p3 > 0 are shown in Figure 7.7.

It can be observed that, analogously as for the precession angle α, in the

interval [u1,u2] the function L can be negative, positive or change its sign.

Correspondingly, the top spins around the ZB axis left, right or changes its

direction of spinning.

7.5 Exercises

Exercise 7.1 Assuming that the body frame has been placed at the center

of mass, derive the Euler-Newton equations of attitude of the Lagrange top.



The material concerned with the Lagrange top presented in this chapter is

based on chapter 6 of monograph [Arn78]. Various aspects of motion of the

top are also studied in chapter IV, paragraph 8 of the book [RK95]. The

term "nutation" originates from Latin word nutare, that means to waive.

The term "precession" comes from Latin precedere, i.e. go before, precede.

Bibliography





Chapter 8

Systems with constraints

Until now we have assumed that the motion of systems under study is not

subject to any constraints, and their generalized coordinates as well veloci-

ties take values in the whole space Rn. However, the motion of real system

obeys some constraints, and our task in this chapter consists in deriving

equations of motion for such systems. In doing this we shall be guided by

the Lagrangian formalism. Suppose therefore that a system is given, de-

scribed by coordinates iq ∈ Rn and velocities _q ∈ Rn, with Lagrangian

L(q, _q). Two kinds of constraints will be imposed on the motion of the

system: conguration constraints and phase constraints.

8.1 Conguration constraints

The conguration constraints refer solely to the system's coordinates. We

shall assume that they are represented by a number l of independent equa-

tions

F(q) = (F1(q), F2(q), . . . , Fl(q)) = 0, (8.1)

where l 6 n. Functions Fi(q) should have continuous derivatives of appro-

priate order, and their independence means that

rankDF(q) = l.

By denition, a system subject to conguration constraints moves within a

conguration manifold

MF = q ∈ Rn| F(q) = 0

contained in Rn. The dimension of the manifold MF equals m = n − l.

In order to formulate the equations of motion we need to dene on the

97

Chapter 8. Systems with constraints 98

X

y

xY

Z

z

ϕ

θ

m

r

Figure 8.1: Conguration constraints

conguration manifold a coordinate system ~q ∈ Rm, then to express the

Lagrangian in new coordinates, and nally write the corresponding Euler-

Lagrange equations. It should be noticed that usually the coordinates on

a manifold are local (the topology of MF is dierent than the topology of

a linear space Rm), therefore the equations of motion will hold only on

this region of the manifold where the coordinates are well dened. On

the manifold SO(3) an example of a coordinate systems is dened by the

Euler angles introduced in the previous chapter. As an illustration of the

conguration constraints we shall exploit the following example.

8.1.1 Spherical pendulum

We study the motion of a material point of mass m in the Euclidean

space R3, see Figure 8.1. Let q = (x,y, z)T ∈ R3 denote the position of

the point, and _q = ( _x, _y, _z)T its velocity. We compute the kinetic energy of

the point K = 12m( _x2+ _y2+ _z2), its potential energy V = mgz, and get the

Lagrangian

L =1

2m(

_x2 + _y2 + _z2)−mgz.

Now suppose that the position of the point is constrained according to

F(q) = x2 + y2 + z2 − r2 = 0,

which means that the mass m is moving over the surface of a sphere, and

forms a spherical pendulum. Indeed, the conguration manifold

MF =(x,y, z) ∈ R3

∣∣ x2 + y2 + z2 − r2 = 0


is a sphere of radius r. The dimension of this manifold m = 2. The inde-

pendence of constraints means that at any point q ∈ MF the dierential

DF(q) = 2(x,y, z)T 6= 0.In agreement with the procedure outlined above we dene spherical co-

ordinates ~q = (θ,ϕ), such thatx = r sin θ cosϕ

y = r sin θ sinϕ

z = r cos θ,

compute velocities_x = r _θ cos θ cosϕ− r _ϕ sin θ sinϕ

_y = r _θ cos θ sinϕ+ r _ϕ sin θ cosϕ

z = −r _θ sin θ,

and express the Lagrangian by means of coordinates ~q and velocities _~q,

~L =1

2m(r2 _θ2 + r2 sin2 θ _ϕ2

)−mgr cos θ.

Next, we compute derivatives

∂~L∂ _θ

= mr2 _θ,

∂~L∂θ = mr2 sin θ cos θ _ϕ2 +mgr sin θ,

∂~L∂ _ϕ = mr2 sin2 θ _ϕ,

∂~L∂ϕ = 0,

and arrive at the Euler-Lagrange equations of motion of the spherical pen-

dulum mr2θ−mr2 sin θ cos θ _ϕ2 −mgr sin θ = 0

mr2 sin2 θ _ϕ = const.

8.2 Phase constraints

The other kind of constraints imposed on the motion is concerned simul-

taneously with coordinates and velocities. Such constraints will be called

phase constraints. We shall concentrate on the phase constraints in the

Pfaan form, i.e. linear with respect to velocity,

A(q) _q = 0, (8.2)


X

Y

x

y

ϕ

_y

_x

v = u1

Figure 8.2: Wheel moving without lateral slip

where A(q) denotes a matrix of dimension l × n, l 6 n, of full rank,

rankA(q) = l, called the Pfaan matrix. The formula (8.2) states that

at a xed position q admissible system's velocities belong to the null space

of matrix A(q), i.e. _q ∈ KerA(q), that forms an m = n − l-dimensional

linear subspace of the velocity space Rn. Under assumption that the vectors

(more precisely: vector elds) g1(q),g2(q), . . . ,gm(q) span this subspace,

so that A(q)gi(q) = 0, the Pfaan constraints can be represented as a

control system

_q = G(q)u =

m∑i=1

gi(q)ui. (8.3)

System (8.3) is referred to as a control system associated with the phase

constraints or, simply, an associate system. Despite that at any position

the admissible system's velocities are restricted to a certain subspace, the

reachable positions of the system do not have to be restricted. The question

if a system subject to constraints (8.2) is able to reach any position q ∈ Rn

is tantamount to the question of controllability of system (8.3).

The Pfaan phase constraints are met when analyzing the motion of

wheeled mobile robots that move without slips of the wheels. Now, we shall

study a couple of examples of such systems in order to derive corresponding

Pfaan constraints (8.2).

8.2.1 Wheel, skate, and ski

Figure 8.2 show a top view of a wheel moving over the plane XY. Since

we ignore the rotation angle of the wheel, this gure may also represent a

view of a skate or of a ski. We assume that the lateral slip of the wheel

is not permitted. As generalized coordinates q = (x,y,ϕ)T we choose the


position of the contact point of the wheel with the ground and the wheel's

orientation. Let velocity at the contact point has components _x and _y.

Then, the no-lateral slip condition means that the velocity in the direction

perpendicular to the wheel (along the wheel's axle) must be zero. From the

gure it follows that this condition can be written as

_x sinϕ− _y cosϕ = 0

i.e.

A(q) _q = 0,

with Pfaan matrix

A(q) =[sinϕ − cosϕ 0

].

In our case n = 3 and l = 1, so the null space of matrix A(q) is 2-

dimensional, m = 2. As generators of the null space we can take vector elds

g1(q) = (cosϕ, sinϕ, 0)T and g2(q) = (0, 0, 1)T . The associated system

(8.3) is of the form _x = u1 cosϕ

_y = u1 sinϕ

_ϕ = u2

.

It is easy to see that the control u1 can be interpreted as a linear velocity

of the wheel, while the control u2 is a velocity of turning the wheel.

8.2.2 Rolling wheel

Consider a wheel that is rolling over a plane, and let θ denote its angle of

rotation. To the top view from Figure 8.2 we add a side view of the wheel,

Figure8.3, presenting the contact point of the wheel with the ground. The

vector of coordinates has 4 components, q = (x,y,ϕ, θ)T . The rolling of the

wheel consists in the motion without both the lateral and the longitudinal

slip. The condition of no lateral slip is the same as before,

_x sinϕ− _y cosϕ = 0.

The longitudinal slip occurs in two cases: either the displacement velocity

of the wheel is less than its rolling velocity (spinning) or the displacement

velocity exceeds the rolling velocity (blocking). In Figure 8.3 the former

velocity has been denoted by v, the latter by r _θ. A lack of the longitudinal

slip implies the identity

v = r _θ.


θ

r _θ v

r

P

Figure 8.3: Rolling wheel

Now, from Figure 8.2 it follows that_x = v cosϕ

_y = v sinϕ.

Having multiplied the rst equality by cosϕ, while the second by sinϕ, and

added sideways we get

v = _x cosϕ+ _y sinϕ,

so the condition of no longitudinal slip takes the form

_x cosϕ+ _y sinϕ− r _θ = 0.

In conclusion, the Pfaan matrix of the rolling wheel

A(q) =

[sinϕ − cosϕ 0 0

cosϕ sinϕ 0 −r

].

We have n = 4, l = 2, m = 2. The control vector elds of system (8.3) can

be chosen as g1(q) = (r cosϕ, r sinϕ, 0, 1)T and g2(q) = (0, 0, 1, 0)T . The

associated system representing the rolling wheel is then the following_x = u1r cosϕ

_y = u1r sinϕ

_ϕ = u2_θ = u1

.

Observe that the controls appearing in this system have the meaning of

angular rolling velocity u1 and turning velocity u2 of the wheel.


x

y

θ

ϕ

η

ξ

ϕ

X

Y

_ξ

_η

l

_x

_y

Figure 8.4: Kinematic car

8.2.3 Kinematic car

A top view of the kinematic car is shown in Figure 8.4. The front axle of the

car is turnable, and the car's length is equal to l. The vector of coordinates

q = (x,y,ϕ, θ)T includes the position of the center of the rear axle, the car's

orientation, and the turning angle of the front axle (or of the driving wheel),

te velocity vector _q = ( _x, _y, _ϕ, _θ)T . It is assumed that the car is moving

without the lateral slip of both the rear as well the front wheels. With this

assumption the two rear wheels and the two front wheels can be identied

with each other, leading to a bicycle model of the kinematic car. By rigidity

of axles, if a lateral slip exists, it is the same at the contact points of wheels

with the ground and at any other point of the wheel's axle; for this reason

it suces to eliminate the slip of the centers of the rear and the front axle.

The no-slip condition for the rear axle takes the form we already know,

_x sinϕ− _y cosϕ = 0.

Denoting by (ξ,η) the coordinates of the center of front axle we get an

analogous condition

_ξ sin(ϕ+ θ) − _η cos(ϕ+ θ) = 0.

Now, the ony thing left is to express time derivatives of coordinates ξ and

η by the velocities _q. From the gure it follows thatξ = x+ l cosϕ

η = y+ l sinϕ,


therefore _ξ = _x− l _ϕ sinϕ

_η = _y+ l _ϕ cosϕ.

A substitution to the condition for lateral no-slip yields

_x sin(ϕ+ θ) − _y cos(ϕ+ θ) − l _ϕ cos θ = 0.

Summarizing, the Pfaan matrix for the kinematic car takes the following

form

A(q) =

[sinϕ − cosϕ 0 0

sin(ϕ+ θ) − cos(ϕ+ θ) −l cos θ 0

].

It is easy to check that generators of the null space of matrix A(q) can

be chosen as g1(q) = (l cosϕ cos θ, l sinϕ cos θ, sinθ, 0)T , and g2(q) = (0, 0,

0, 1)T . They dene an associated control system representing the kinematics

of the kinematic car moving without lateral slips_x = u1l cosϕ cos θ

_y = u1l sinϕ cos θ

_ϕ = u1 sin θ

_θ = u2

.

Controls u1, u2 appearing hereabout have clear physical interpretation. The

control u1 is a scaled linear velocity of the center of front axle, which can

be understood as the front drive of the car∗. Complementarily, the control

u2 is the turn speed of the steering wheel.

8.3 Constraints for rigid body motion

In the examples studied above we have been able to derive the phase con-

straints in a direct and intuitive way. These constraints are a special case of

the phase constraints for the rigid body motion that will be provided in this

section. Suppose that a rigid body B is moving relative to the space frame

(XS, YS,ZS), and let (XB, YB,ZB) denote the body frame, see Figure 8.5. It

will be assumed that the body's orientation is characterized by a rotation

matrix R ∈ SO(3), and its position by a vector T ∈ R3. Let p ∈ R3 denote

the position of the contact P of the body with the ground with respect to the

∗Alternatively, picking the vector eld g1(q) in the form g1(q) =(cosϕ, sinϕ, sinθ

l cosθ, 0)T

we obtain a model of kinematics of the kinematic car, where u1will refer to the linear velocity of the center of rear axle (a rear drive car).


YS

ZS

XS

P

B ZB

YB

XBT

s

s− Tωs

_T

ωs × (T − s)

Figure 8.5: Phase constraints

body frame. Relying on the analysis performed in chapter 6, the coordinates

s of the contact point in the space frame are computed as

s = Rp+ T .

The phase constraints come from a requirement that the velocity of the

contact point relative to the space frame be zero. This means that

_s = _Rp+ _T = 0.

Invoking the denition of the matrix and vector rotation velocity in space,_R = ΩSR = [ωs]R, taking into account the identity ΩSRp = ωS × Rp, and

then substituting Rp = s − T we arrive at a general formula for the phase

constraints of the rigid body

ωS × (s− T) + _T = 0. (8.4)

Hereabout ωS refers to the rotation velocity in space, and s, T denote,

respectively, coordinates of the contact point and of the origin of the body

frame with respect to the space frame.

8.3.1 Rolling ball

For illustration of usability of the formula (8.4) we shall now dene phase

constraints for a ball rolling over a plane as shown in Figure 8.6. Let us x


P

YB

ZB

XB

r

ϕ

YP

XP

θ

ZP

ψ

YS

ZS

XS

P

XPYP

y

x

Figure 8.6: Rolling ball

a space frame (XS, YS,ZS), a body frame (XB, YB,ZB) attached to the ball,

and an additional coordinate frame (XP, YP,ZP) attached at the contact

point of the ball with the ground. The ball has radius r. The ball's motion

will be described in terms of coordinates q = (x,y,α,β,γ)T , denoting,

respectively, the position (x,y) (Cartesian coordinates) of the contact point

in space frame, the position (α,β) (spherical coordinates) of the contact

point in body frame, and the orientation of the ball dened by the angle

between axes XP and XS, see the gure. Intuitively, by rolling of a ball we

understand a motion free of slips along the meridian and the parallel as well

as free of pure spinning, i.e. the spinning not related to a ball displacement.

The phase constraints of rolling will be derived from the (8.4). We

start from determining position and orientation of the ball. Obviously, the

position of the origin of body frame T = (x,y, r)T . The position of the

contact point in space frame s = (x,y, 0)T . In order to nd the velocity ωSwe invoke the identity ΩS = _RRT , where R denotes the orientation matrix

of the ball, composed of elementary rotations of the space frame that make

this frame to overlap the body frame. Instead of matrix R it will be easier

to obtain the transpose matrix RT , i.e. a sequence of elementary rotations

transforming the body frame into the space frame. From Figure 8.6 we

deduce

RT = R(Z,α)R(Y,β)R(X,π)R(Z,−γ),

so that, due to the property R(axis, -angle) = RT (axis, angle) and RT (X,π) =

R(X,π), the orientation matrix

R = R(Z,γ)R(X,π)R(Y,−β)R(Z,−α).


Now, we need to compute the matrix rotation velocity in space

ΩS = _RRT = ( _R(Z,γ)R(X,π)R(Y,−β)R(Z,−α)+

R(Z,γ)R(X,π) _R(Y,−β)R(Z,−α)+

R(Z,γ)R(X,π)R(Y,−β) _R(Z,−α))R(Z,α)R(Y,β)R(X,π)R(Z,−γ).

Relying on the description of elementary rotations provided in chapter 6,

as well as the identity RΩRT = R[ω]R = [Rω] (compare Exercise 6.3), we

derive that the vector rotation velocity of the ball in space

ωS =

_α cosγ sinβ− _β sinγ

_α sinγ sinβ+ _β cosγ

_α cosβ+ _γ

.

After necessary substitutions to (8.4) we nd the following conditions of no

slip along the meridian and the parallel_x− r _α sinγ sinβ− r _β cosγ = 0

_y+ r _α cosγ sinβ− r _β sinγ = 0.

In order to eliminate pure spinning we request that the third component of

the space velocity ωs needs to be 0,

_α cosβ+ _γ = 0.

A combination of the all three rolling constraints results in the Pfaan

matrix

A(q) =

1 0 −r sinγ sinβ −r cosγ 0

0 1 r cosγ sinβ −r sinγ 0

0 0 cosβ 0 1

.

The vector elds annihilated by the Pfaan matrix are equal to g1(q) =

(r sinγ sinβ,−r cosγ sinβ, 1, 0,− cosβ)T , g2(q) = (r cosγ, r sinγ, 0, 1, 0)T ,

and the associated control system representing the kinematics of the rolling

ball assumes the form

_x = u1r sinγ sinβ+ u2r cosγ

_y = −u1r cosγ sinβ+ u2r sinγ

_α = u1_β = u2

_γ = −u1 cosβ

.


8.4 Holonomic and non-holonomic constraints

It turns out that the Pfaan phase constraints may take a specic form,

like e.g.

q = (x,y, z)T , A(q) =[x y z

].

In consequence, we get

A(q) _q = x _x+ y _y+ z _z = 0,

or, equivalently,

x2 + y2 + z2 = const .

As we see, the last constraint is a conguration constraint. If satised then

the system is moving over a 2-dimensional sphere. This example demon-

strates that there are phase constraints that can be reduced to conguration

constraints by integration. Such constraints will be called holonomic. The

phase constraints that are not integrable are referred to as non-holonomic.

The property of being holonomic can be characterized in the following

way. Consider phase constraints described by the identity A(q) _q = 0. It

is clear that these constraints will remain unchanged after a multiplication

from the left by any non-singular matrix M(q) of dimension l × l; their

equivalent form being M(q)A(q) _q = 0. Suppose that we are able to select

the matrix M(q) in such a way that

M(q)A(q) = DF(q)

for a certain function F : Rn −→ Rl continuously dierentiable to at least

order 2. If this holds, we get

M(q)A(q) _q = DF(q) _q =d

dtF(q(t)) = 0,

along the trajectoryq(t). But this means that along the trajectories the

function F(q) = const, i.e. the constraints are holonomic. In this way we

have arrived at a denition of holonomic constraints, namely we say that

the phase constraints A(q) _q = 0 are holonomic if there exists a non-singular

matrixM(q) and a function F(q), such thatM(q)A(q) = DF(q). In general,

deciding that constraints are holonomic directly from the denition may not

be easy. In the next subsection we study a simple example where this has

been possible.


8.4.1 Wheel moving without lateral slip

As we have already shown, a wheel moving without the lateral slip is de-

scribed by coordinates q = (x,y,ϕ)T and subject to phase constraints de-

ned by the matrix A(q) =[sinϕ − cosϕ 0

]. We ask a question if these

constraints are possibly holonomic. Assume temporarily the answer is ar-

mative. If so, there exists a non-singular matrix of dimension 1 × 1 (i.e. a

non-zero function m(q) 6= 0), and a function F(q) such that

m(q)A(q) =

(∂F(q)

∂x,∂F(q)

∂y,∂F(q)

∂ϕ

).

Taking into account the form of matrix A(q) we derive from this condition

three equations m(q) sinϕ =

∂F(q)∂x

m(q) cosϕ = −∂F(q)∂y

0 =∂F(q)∂ϕ

.

Now, since the function F(q) has continuous partial derivatives of order 2,

the mixed derivatives are equal, therefore

∂2F(q)

∂x∂ϕ= 0, a tak_ze

∂2F(q)

∂y∂ϕ= 0.

We have obtained a system of linear equations[sinϕ cosϕ

cosϕ − sinϕ

](∂m(q)∂ϕ

m(q)

)= 0,

whose solution is ∂m(q)∂ϕ = 0 andm(q) = 0. This contradicts our assumption

thatm(q) 6= 0, i.e. the phase constraints imposed on the motion of the wheel

must be non-holonomic.

8.4.2 Non-holonomicity condition

Given an associated control system

_q = G(q)u =

m∑i=1

gi(q)ui,

a condition for Pfa constraints being non-holonomic can be expressed

by means of certain dierential operations involving control vector elds

g1(q),g2(q), . . . ,gm(q). To be more specic, for a pair of vector elds


gi(q) and gj(q) we introduce a kind of multiplication operation, called the

Lie bracket, dened in the following way

[gi,gj](q) = Dgj(q)gi(q) −Dgi(q)gj(q).

Properties of the Lie bracket resemble those of the Poisson bracket presented

in the chapter on Hamiltonian mechanics. This means that

• [gi,gi] = 0 irre exivity

• [gj,gi] = −[gi,gj] antisymmetry,

• [gi, [gj,gk]] + [gj, [gk,gi]] + [gk, [gi,gj]] = 0 Jacobi identity.

For the associated system we dene a Lie algebra L of the system as the

minimal linear space over reals R that contains the control vector elds gi,

and is closed with respect to the operation of Lie bracket. In this context

we state the following

Theorem 8.4.1 (On non-holonomic constraints) If at any point q ∈ Rn the

rank of the Lie algebra of associated system is equal to n,

rankL(q) = n,

then the phase constraints are non-holonomic.

The rank of a Lie algebra at a point q equals the dimension of the vector

space spanned at this point by vector elds belonging to the algebra. Al-

though the Lie algebra is usually innite-dimensional, in order to check the

rank condition it often suces to take the control vector elds, and compute

a few Lie brackets.

8.4.3 Example

By a simple example we shall demonstrate how to use the condition for

non-holonomicity. Consider the rolling wheel studied in subsection 8.2.2.

We recall that the vector of coordinates q = (x,y,ϕ, θ)T , and the control

vector elds of the associated system are

g1(q) =

r cosϕ

r sinϕ

0

1

, g2(q) =

0

0

1

0

.


Let's compute the Lie bracket of these vector elds

g1,g2](q) = Dg2(q)g1(q) −Dg1(q)g2(q) = −Dg1(q)g2(q) =

∂g1(q)

∂ϕ= r(sinϕ,− cosϕ, 0, 0)T .

The vector space spanned by three vector elds g1, g2, g12 = [g1,g2] has

dimension equal to the rank of the matrix

rank

r cosϕ 0 r sinϕ

r sinϕ 0 −r cosϕ

0 1 0

1 0 0

= 3.

By denition, the Lie algebra of the associated system (the associated Lie al-

gebra) also contains vector elds g112(q) = [g1,g12] and g212(q) = [g2,g12]

computed below

g112(q) = Dg12(q)g1(q) −Dg1(q)g12(q) = r sinϕ∂g12(q)

∂x+

r sinϕ∂g12(q)

∂y+∂g12(q)

∂θ− r sinϕ

∂g1(q)

∂x+ r cosϕ

∂g1(q)

∂y= 0,

and

g212(q) = Dg12(q)g2(q) −Dg2(q)g12(q) =

∂g12(q)

∂ϕ= (r cosϕ, r sinϕ, 0, 0)T .

The dimension of the vector space spanned by g1, g2, g12 and g212 is equal

to the rank of the matrix

rank

r cosϕ 0 r sinϕ r cosϕ

r sinϕ 0 −r cosϕ r sinϕ

0 1 0 0

1 0 0 0

= 4.

Since the rank of the associated Lie algebra is 4, the phase constraints are

non-holonomic. In the similar way one can decide holonomicity or non-

holonomicity of phase constraints for other robotic systems.

8.5 Exercises

Exercise 8.1 Using formula (8.4) derive phase constraints for the rolling

wheel.


θ1rK1

θ2rK2

x

yϕ

X

Y

K1

K2

l

l

Figure 8.7: Wheeled mobile robot

θ0

X

Y

x

y

ϕ0

l0

ϕ1l1

Figure 8.8: Car with trailer

Exercise 8.2 Complete details in the derivation of the rotation velocity in

space of the rolling ball.

Exercise 8.3 Describe phase constraints for the motion of a 2-wheel mobile

robot shown in Figure 8.7. Exclude both the lateral and the longitudinal

slip of the wheels. The length of the axle equals 2l.

Exercise 8.4 Derive a formula for phase constraints for a car pulling a trailer,

see Figure 8.8.

Exercise 8.5 Derive a formula for phase constraints for a re truck presented

in Figure 8.9, moving without the lateral slip of the wheels. Notice that both

the front as well the rear wheels are steerable.

Exercise 8.6 Check non-holonomicity of phase constraints for selected mo-

bile robots.


θ0

X

Y

x

yϕ0

l0

ϕ1l1θ1

Figure 8.9: Fire truck


Systems with constraints are studied in chapter 2 of the book [RK95], and

also in chapter 1 of [Gut71], where the terms: geometric and kinematic

constraints are employed. The viewpoint dominating in robotics has been

presented in chapter 11 of the monograph [TMD+00], in the Lecture notes

[TM18], and in chapter 7 of the monograph [MZS94]. From the last reference

we have borrowed exercise 8.5. Control of non-holonomic robotic systems

is the central thread of the monograph [Maz09]. The term "holonomic" was

coined by Hertz, and comes from Greek words holos, meaning a whole, and

nomos a law.

Bibliography

[Gut71] R. Gutowski. Mechanika analityczna. PWN, Warszawa, 1971.

(in Polish).

[Maz09] A. Mazur. Sterowanie oparte na modelu dla nieholo-

nomicznych manipulatorow mobilnych. Ocyna Wydawnicza

PWr, Wroc law, 2009. (in Polish).

[MZS94] R. Murray, Li Zexiang, and S. Sastry. A Mathematical Intro-

duction to Robotic Manipulation. CRC Press, Boca Raton,

1994.




[TM18] K. Tchon and R. Muszynski. Mathematical Methods of

Robotics and Automation. Wroc law University of Science and

Technology, project AZON, 2018.





2018, in Polish).

Chapter 9

Dynamics of non-holonomic

systems

In the previous chapter we have distinguished conguration and phase con-

straints of motion, and the latter divided into holonomic and non-holonomic.

We have also explained how to derive equations of motion for systems sub-

ject to conguration constraints. Now, we shall deal with te dynamics of

non-holonomic systems.

Let such a system be given, described by generalized coordinates q ∈ Rn,

and velocities _q ∈ Rn, with Lagrangian L(q, _q), subject to l 6 n non-

holonomic phase constraints in the form of Pfa, A(q) _q = 0. We shall

assume that these constraints result from an action of certain non-potential

forces F that will be called traction forces. A procedure of deriving the

equations of motion of this system is the following:

1. Write the Euler-Lagrange equations in the form

d

dt

∂L

∂ _q−∂L

∂q= Q(q)q+ P(q, _q) = F,

where P(q, _q) stands for the centripetal, Coriolis and potential, and F

denotes the traction forces.

2. Determine the traction forces in accordance with the d'Alembert's

principle: the traction forces do no work along admissible displace-

ments. By admissible displacements we mean displacements obeying

the phase constraints A(q) _q = 0. The d'Alembert's principle yields

A(q) _q = 0→ (F, _q) = FT _q = 0.

115

Chapter 9. Dynamics of non-holonomic systems 116

Geometrically, the d'Alembert's principle implies that the traction

forces need to be perpendicular to the velocity, similarly as the rows

of matrix A(q). It follows that there exists a vector λ ∈ Rl (dependent

on q i _q) such that

FT = λTA(q), i.e. F = AT (q)λ.

3. State the dynamics equations in the form

Q(q)q+ P(q, _q) = AT (q)λ. (9.1)

4. Using a representation of the phase constraints in the form of the

associated control system

_q = G(q)η,

η ∈ Rm=n−l, A(q)G(q) = 0, eliminate the vector λ by a left multipli-

cation of identity (9.1) by the matrix GT (q),

GT (q)Q(q)q+GT (q)P(q, _q) = 0.

5. Compute acceleration as

q = G(q) _η+ _G(q)η,

where _G(q) denotes the time derivative of matrix G(q) along the tra-

jectory q(t), and substitute to the obtained equation,

GT (q)Q(q)G(q) _η+GT (q)Q(q) _G(q)η+GT (q)P(q, _q) = 0.

Due to the independence of phase constraints, the matrix G(q) of

dimension n ×m has full rank m. The inertia matrix Q(q) is non-

singular, therefore the m×m matrix GTQG is invertible.

6. Using the inverse of matrix GTQG obtain the following equations of

motion of a system subject to non-holonomic phase constraints_q = G(q)η

_η = −(GT (q)Q(q)G(q))−1GT (q)(Q(q) _G(q)η+ P(q, _q)). (9.2)


α

X

Z Y

α y

Sϕ

x

Figure 9.1: Chaplygin's skater

Remark 9.1 Instead of eliminating the vector λ from equation (9.1), we

can compute this vector, and obtain the traction forces. To this objec-

tive we multiply both sides of this equation by the matrix A(q),

A(q)Q(q)q+A(q)P(q, _q) = A(q)AT (q)λ,

and then exploit the invertibility of matrix AAT to get

F = AT (q)λ = AT (q)(A(q)AT (q)

)−1A(q)(Q(q)q+ P(q, _q)). (9.3)

The identity (9.3) allows one to check e.g. whether the friction forces

acting on the system are sucient to enforce fulllment of the non-

holonomic constraints.

The formation of equations of motion for non-holonomic systems will be

analyzed on a couple of examples.

9.1 Examples

9.1.1 Chaplygin's skater

Consider a skater (a skier) going down a slope of inclination angle α, pre-

sented in Figure 9.1. As generalized coordinates q = (x,y,ϕ)T of the skater

S we take its position and orientation relative the the coordinate frame (X, Y)

lying on the slope. Suppose that the skater starts from the point of coordi-

nates (0, 0, 0)T , with zero initial velocity, and with a certain angular velocity

ω, and is sliding without the lateral slip. The corresponding Pfaan matrix

is the following

A(q) =[sinϕ − cosϕ 0

].


As we have already shown in section 8, the phase constraints imposed on the

motion of the skater are non-holonomic. The control matrix of associated

system

G(q) =

cosϕ 0

sinϕ 0

0 1

has also been computed in the previous chapter.

For simplicity we shall assume that the mass and the moment of inertia

of the skater have been chosen in such a way that the skater's kinetic energy

K = 12( _x2+ _y2+ _ϕ2), and dimensions of the slope give the skater's potential

energy V = −x. With these assumptions the Lagrangian of the Chaplygin's

skater

L = K− V =1

2

(_x2 + _y2 + _ϕ2

)+ x.

The dynamics equations of the skater will be obtained in accordance

with the procedure provided above. We begin with equation (9.1)x− 1 = λ sinϕ

y = −λ cosϕ

ϕ = 0

.

In order to eliminate the multiplier λ we multiply the rst equation by cosϕ,

the second by sinϕ, and add sidewise,

x cosϕ− cosϕ+ y sinϕ = 0. (9.4)

Next, from equation _q = G(q)η we nd_x = η1 cosϕ

_y = η1 sinϕ

_ϕ = η2

.

Since the angular acceleration of the skater is zero, we get η2 = ω =

const, therefore the skater's orientation

ϕ(t) = ωt.

The linear accelerations are equal tox = _η1 cosϕ− η1 _ϕ sinϕ

y = _η1 sinϕ+ η1 _ϕ cosϕ,


y

xforward backward forward backward

Figure 9.2: Path of Chaplygin's skater

which, after a substitution into (9.4), yields

_η1 = cosϕ = cosωt.

The zero initial velocity of the skater means that η1(0) = 0, so

η1(t) =1

ωsinωt.

An integration of the equations for linear velocity_x(t) = 1

2ω sin 2ωt

_y = 12ω(1− cos 2ωt)

at zero initial conditions results in the skater's trajectoryx(t) = 1

4ω2(1− cos 2ωt)

y(t) = 14ω2

(2ωt− sin 2ωt).

We have obtained parametric equations of the cycloid evolving in the direc-

tion of the Y axis of the coordinate frame placed on the slope, see Figure 9.2.

It turns out that the Chaplygin's skater having a non-zero initial angular

velocity will not slide down the slope, but is moving along the upper edge

of the slope going forward and backward along a cycloid.

9.1.2 Wheel rolling vertically

Now we are going to derive equations of motion of a wheel of radius r

rolling over a horizontal plane (XS, YS), situated vertically to the plane, see

Figure 9.3. We make an assumption that the wheel is thin and uniform.

The body frame will be placed in the center of the wheel, in such a way that

its XB axis lies in the wheel's plane, and the YB coincides with the rotation

axis of the wheel. As coordinates q = (x,y,α,β)T we choose the position


x

y

ZS

g

P

YS

XS

α

β

m

r

XB

YB

Figure 9.3: Vertical wheel

of the contact point of the wheel with the ground, and the orientation and

rotation angles of the wheel.

For the reason that during its motion the wheel touches the ground at

any time, its potential energy remains constant; we can assume that V = 0.

This yields the Lagrangian to be equal to the kinetic energy of the wheel,

L = K =1

2m(

_x2 + _y2)+1

2I2 _β2 +

1

2I3 _α2,

where m denotes the mass of the wheel, I2 = 12mr

2 its moment of inertia

relative to the YB axis, and I3 =14mr

2 is the moment of inertia of the wheel

with respect to the ZB axis.

In the previous chapter we have found that the phase constraints corre-

sponding to the rolling are dened by the Pfaan matrix

A(q) =

[sinα − cosα 0 0

cosα sinα 0 −r

].

Vectors spanning the null space of matrix A(q) are taken as columns of the

control matrix

G(q) =

r cosα 0

r sinα 0

0 1

1 0

of the associated control system.


This being so, the equation (9.1) for the rolling wheel becomes the fol-

lowing m _x = λ1 sinα+ λ2 cosα

my = −λ1 cosα+ λ2 sinα

I3 _α = 0

I2 _β = −λ2r

.

Having multiplied these equations by the matrix GT we get two equationsmrx cosα+mry sinα+ I2β = 0

I3α = 0. (9.5)

From the second of them it follows that _α = const = ω, that, accounting

for the zero initial orientation of the wheel, α(0) = 0, results in

α(t) = ωt.

The associated system is dened by the matrix G(q), and takes the form_x = η1r cosα

_y = η1r sinα

_α = η2_β = η1

.

Knowing that _α = ω, from the third equation we nd η2 = ω. The remain-

ing equations allow one to compute the accelerationsx = _η1r cosα− η1r _α sinα

y = _η1r sinα+ η1r _α cosα

β = _η1

.

Insertion of these accelerations to the rst of equations (9.5) yields an iden-

tity

(mr2 + I2) _η1 = 0⇒ η1 = const = η,

from which it follows directly that for β(0) = 0 the trajectory

β(t) = ηt.

Trajectories of position coordinates of the contact point can be found

relying on equalities _x = ηr cosωt

_y = ηr sinωt.


y

x

sgn ηω

= +1ω< 0

ω > 0

ω = 0

a)

y

x

ω> 0

ω < 0

ω = 0

b)

sgn ηω

= −1

η < 0 η > 0

Figure 9.4: Wheel's paths: a) η < 0, b) η > 0

After integration with zero initial conditions x(0) = y(0) = 0 we getx(t) = ηr

ω sinωt

y(t) = ηrω (1− cosωt)

.

In conclusion, the paths of the contact point appear to be circles

x2 +(y−

ηr

ω

)2=(ηrω

)2,

with center at the points(0, ηrω

)and radius ηrω , displayed in Figure 9.4.

9.1.3 Rolling ball

As the next example of dening equations of motion for non-holonomic sys-

tems we shall study a ball able to roll over a plane, presented in Figure 9.5.

It will be assumed that the ball has mass m and the moments of inertia

I1 = I2 = I3 = I. Generalized coordinates assigned to the ball will be the

same as in subsection 8.3.1, q = (x,y,α,β,γ)T , however, the orientation an-

gles have been denoted by α, β, and γ to distinguish them from the Euler

angles. The Paan matrix

A(q) =

1 0 −r sinβ sinγ −r cosγ 0

0 1 r sinβ cosγ −r sinγ 0

0 0 cosβ 0 1

,


P

γ

YB

ZB

XB

r

α

YP

XP

β

YS

ZS

XS

ZP

P

XPYP

m

y

x

g

Figure 9.5: Rolling ball

whereas the control matrix of the associated system

G(q) =

r sinβ sinγ r cosγ

−r sinβ cosγ r sinγ

1 0

0 1

− cosβ 0

.

Taking into account the form of matrix R describing the ball's orienta-

tion, found in subsection 8.3.1, one can show that

R = R(Z,γ)R(X,π)R(Y,−β)R(Z,−α) = E(ϕ, θ,ψ)

for Euler angles e = (π+γ,π−β,−α)T . Observe that the vector T refers to

the position of the center of mass of the ball in the space frame, therefore

T = (x,y, r)T . Having substituted the Euler angles e and the coordinates T

to the Lagrangian (6.6) we obtain

L =1

2I(

_α2 + _β2 + _γ2)+ I _α _γ cosβ+

1

2

(_x2 + _y2

)−mgr.

The last component of this Lagrangian describes the potential energy of the

ball that is constant, so plays no role in the Euler-Lagrange equation and

can be ignored.

A starting point for the derivation of equations of motion for the ball is


the formula (9.1). In the case of the ball it species to

x = λ1

y = λ2

Iα+ Iγ cosβ− I _β _γ sinβ = −λ1r sinβ sinγ+

λ2r sinβ cosγ+ λ3 cosβ

Iβ+ I _α _γ sinβ = −λ1r cosγ− λ2r sinγ

Iγ+ Iα cosβ− I _α _β sinβ = λ3

.

In order to eliminate the multipliers λ we multiply these equations from the

left by the matrix GT arriving at the following identitiesmrx cosγ+mry sinγ+ Iβ+ I _α _γ sinβ = 0

mrx sinβ sinγ−mry sinβ cosγ+ Iα sin2 β−

I _β _γ sinβ+ I _α _β sinβ cosβ = 0

.

The next step consists in computing the accelerations from equation of

the associated control system _q = G(q)η,

_x = η1r sinβ sinγ+ η2r cosγ

_y = −η1r sinβ cosγ+ η2r sinγ

_α = η1_β = η2

_γ = −η1 cosβ

.

From these equations we deduce

x = _η1r sinβ sinγ+ η1r _γ sinβ cosγ+

η1r _β cosβ sinγ+ _η2r cosγ− η2r _γ sinγ

y = − _η1r sinβ cosγ+ η1r _γ sinβ sinγ−

η1r _β cosβ cosγ+ _η2r sinγ− η2r _γ cosγ

α = _η1β = _η2

γ = − _η1 cosβ+ η1 _β sinβ

.

A combination of the two last identities containing accelerations followed by

suitable mathematical transformations leads to the nal dynamics equations(mr2 + I

)_η1 sin2 β+ 2

(mr2 + I

)η1η2 sinβ cosβ = 0(

mr2 + I)

_η2 −(mr2 + I

)η21 sinβ cosβ = 0

.


ZS

α

g

x

_β

m

r

YB

γ

β

Pξ

β

C

β

_α

ZB

_γ

XB

YS

XS

yη

Figure 9.6: Tilted wheel

In conclusion, complete equations of motion of the rolling ball can be given

the following form

_x = η1r sinβ sinγ+ η2r cosγ

_y = −η1r sinβ cosγ+ η2r sinγ

_α = η1

_γ = −η1 cosβ

_β = η2

_η1 sin2 β+ 2η1η2 sinβ cosβ = 0

_η2 − η21 sinβ cosβ = 0

.

An inspection reveals that the last three equations of motion are indepen-

dent of the remaining ones.

9.1.4 Tilted wheel

Our last example of derivation of equations of motion for non-holonomic

systems is concerned with a tilted wheel. Dierently than in subsection 9.1.2

we shall now allow the wheel to tilt with respect to the horizontal plane,

see Figure 9.6. Although this example is a natural generalization of the

vertical wheel studied in subsection 9.1.2, we shall see that the derivation

of equations of motion is technically quite complex. We let the wheel have


radius r, mass m, and its moments of inertia with respect to the body frame

equal I1 = I2 = 14mr

2 and I3 = 12mr

2. Denote by q = (x,y,α,β,γ)T the

vector of coordinates of the wheel composed of x,y the coordinates of the

contact point of the wheel with the ground, α the wheel's orientation, β

its tilt angle, and γ the turning angle of the wheel. The rolling of the

wheel is subordinated to non-holonomic constraints dened by the Pfaan

matrix

A(q) =

[sinα − cosα 0 0 0

cosα sinα 0 0 −r

].

The associated control system has the control matrix

G(q) =

r cosα 0 0

r sinα 0 0

0 1 0

0 0 1

1 0 0

.

It follows from the Figure that the Euler angles ZYZ characterizing the

orientation of the tilted wheel are equal to

ϕ = α−π

2, θ = β−

π

2, ψ = γ.

Also, one deduces from the Figure that the position of the center of mass of

the wheel in the space frame

T =

x+ r sinα sinβ

y− r cosα sinβ

r cosβ

,

yielding the velocities_T1 = _x+ r _α cosα sinβ+ r _β sinα cosβ

_T2 = _y+ r _α sinα sinβ− r _β cosα cosβ

_T3 = −r _β sinβ

.

Using this data we compute the Lagrangian for the tilted wheel from the

formula (6.6),

L =1

2I1

(_α2 cos2 β+ _β2

)+1

2I3 ( _α sinβ+ _γ)2+

1

2m ( _x+ r _α cosα sinβ)2 +

1

2m ( _y+ r _α sinα sinβ)2+

1

2mr2 _β2 +mr _β ( _x sinα− _y cosα) cosβ−mgr cosβ.


A derivation of the equations of motion starts from the formula (9.1),

mx+mrα cosα sinβ+mrβ sinα cosβ+

2mr _α _β cosα cosβ−mr _α2 sinα sinβ−

mr(

_α2 + _β2)

sinα sinβ = λ1 sinα+ λ2 cosα

my+mrα sinα sinβ−mrβ cosα cosβ+

2mr _α _β sinα cosβ+mr _α2 sinα sinβ−

mr(

_α2 + _β2)

cosα sinβ = −λ1 cosα+ λ2 sinα

mr (x cosα+ y sinα) sinβ+(I1 cos2 β+

(I3 +mr

2)

sin2 β)

α+

I3γ sinβ+ 2(I3 − I1 +mr

2)

_α _β sinα cosα+ I2 _β _γ cosβ = 0

mr (x sinα− y cosα) cosβ+(I1 +mr

2)

β−(I3 − I1 +mr

2)

_α2 sinβ cosβ− I3 _α _γ cosβ−mgr sinβ = 0

I3 (α sinβ+ γ) + I3 _α _β cosβ = −λ2r

.

To eliminate the multiplier λ we multiply these identities by the matrix GT

to get

mrx cosα+mry sinα+(I3 +mr

2)

α sinβ+

I3γ+(I3 + 2mr

2)

_α _β cosβ = 0

mrx cosα sinβ+mry sinα cosβ+(mr2 sin2 β+ I1 cos2 β+ I3 sin2 β

)α+ I3γ sinβ+

2(mr2 + I3 − I1

)_α _β sinβ cosβ− I3 _β _γ cosβ = 0

mr (x sinα− y cosα) cosβ+(I1 +mr

2)

β−(mr2 + I3 − I1

)_α2 sinβ cosβ− I3 _α _γ cosβ+mgr sinβ = 0

.

The associated control system takes the following form

_x = η1r cosα

_y = η1r sinα

_α = η2_β = η3

_γ = η1

.


From this system we nd the accelerations

x = _η1r cosα− η1r _α sinα

y = _η1r sinα+ η1r _α cosα

α = _η2β = _η3

γ = _η1

and insert them into the former equations. Eventually, the equations of

motion of the tilted wheel assume the form

_x = η1r cosα

_y = η1r sinα

_α = η2_β = η3

_γ = η1(mr2 + I3

)_η1 +

(mr2 + I3

)_η2 sinβ+

(I3 + 2mr

2)η2η3 cosβ = 0(

mr2 + I3)

_η1 sinβ+((mr2 + I3

)sin2 β+ I1 cos2 β

)_η2+

2(mr2 + I3 − I1

)η2η3 sinβ cosβ− I3η1η3 cosβ = 0(

mr2 + I1)

_η3 −mr2η1η2 cosβ−(

mr2 + I3 − I1)η22 sinβ cosβ− I3η1η2 cosβ+mgr sinβ = 0

.

9.2 Exercises

Exercise 9.1 Show that for the zero tilt, β = 0, the Lagrangian for the tilted

wheel coincides with the Lagrangian for the vertical wheel.


A classical reference concernd with dynamics of non-holonomic systems is

the book [NF72]. Alternatively, we recommend chapter 6 of the monograph

[Gut71]. An exhaustive treatment of models of the robots dynamics can be

found in the monograph [TMD+00]. Robot control algorithms based on the

models of their dynamics are discussed in [Maz09].


Bibliography

[Gut71] R. Gutowski. Mechanika analityczna. PWN, Warszawa, 1971.

(in Polish).

[Maz09] A. Mazur. Sterowanie oparte na modelu dla nieholo-

nomicznych manipulatorow mobilnych. Ocyna Wydawnicza

PWr, Wroc law, 2009. (in Polish).

[NF72] J. I. Neimark and N. A. Fufaev. Dynamics of Nonholonomic

Systems. AMS, Translations of Mathematical Monographs,

Providence, RI, 1972.





2018, in Polish).

Index

A

acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

angular momentum. . . . . . . . . . . . . . . . . . . . . 24

B

ball and beam

Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . 62

Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . 46

bi-normal vector . . . . . . . . . . . . . . . . . . . . . . . . 13

Brockett's integrator. . . . . . . . . . . . . . . . . . . .40

C

Chaplygin's skater . . . . . . . . . . . . . . . . . . . . . 117

Christoel's symbols

1st kind . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2nd kind . . . . . . . . . . . . . . . . . . . . . . . . . . 50

constant of motion . . . . . . . . . . . . . . . . . . . 3, 63

constraints

conguration . . . . . . . . . . . . . . . . . . . . . . 97

holonomic . . . . . . . . . . . . . . . . . . . . . . . . 108

non-holonomic. . . . . . . . . . . . . . . . . . . .108

phase . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

Pfaan . . . . . . . . . . . . . . . . . . . . . . . . . 99

curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

D

derivative

Gateaux . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

Frechet . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

dynamic system . . . . . . . . . . . . . . . . . . . . . . . . 66

ow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

on torus . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

E

elementary rotations . . . . . . . . . . . . . . . . . . . . 76

energy

kinetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . .25

potential . . . . . . . . . . . . . . . . . . . . . . . . . . 25

equations

Euler-Lagrange . . . . . . . . . . . . . . . . 34, 45

Euler-Poisson . . . . . . . . . . . . . . . . . . . . . . 35

Frenet-Serret . . . . . . . . . . . . . . . . . . . . . . 16

Hamilton's canonical. . . . . . . . . . . . . . .61

of motion

ball and beam . . . . . . . . . . . . . . . 47, 62

Furuta's pendulum . . . . . . . . . . 49, 63

Lagrange top . . . . . . . . . . . . . . . . 89, 90

mathematical pendulum. . . . . . . . .26

non-holonomic system. . . . . . . . . .116

non-uniqueness . . . . . . . . . . . . . . . . . .26

Planet around Sun . . . . . . . . . . . . . . . 1

rolling ball . . . . . . . . . . . . . . . . . . . . . 125

spherical pendulum . . . . . . . . . . . . . 99

tilted wheel . . . . . . . . . . . . . . . . . . . . 128

vertical wheel . . . . . . . . . . . . . . . . . . 121

Euclidean

norm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

product . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

Euler angles . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

F

rst integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

forces

centripetal . . . . . . . . . . . . . . . . . . . . . . . . . 50

control . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Coriolis . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

non-potential . . . . . . . . . . . . . . . . . . 45, 61

traction . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

Frenet trihedron . . . . . . . . . . . . . . . . . . . . . . . . 13

functional . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

action . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

extremum . . . . . . . . . . . . . . . . . . . . . . . . . 33

conditional . . . . . . . . . . . . . . . . . . . . . . 35

G

geodesic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

H

Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

130

Index 131

hamiltonian system. . . . . . . . . . . . . . . . . . . . . 66

I

invariant . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3, 63

K

Kepler's equation . . . . . . . . . . . . . . . . . . . . . . . . 7

kinematic car . . . . . . . . . . . . . . . . . . . . . . . . . .103

L

Lagrange top . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Hamiltonian . . . . . . . . . . . . . . . . . . . . . . . 90

invariants . . . . . . . . . . . . . . . . . . . . . . . . . .90

Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . 89

nutation . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

precession . . . . . . . . . . . . . . . . . . . . . . . . . 91

spinning . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

Lagrangian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

of rigid body . . . . . . . . . . . . . . . . . . . . . . 82

LATEX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vi

law of universal gravitation . . . . . . . . . . . . . 21

Legendre transform . . . . . . . . . . . . . . . . . . . . . 58

length

of trajectory . . . . . . . . . . . . . . . . . . . . . . . 13

of vector . . . . . . . . . . . . . . . . . . . . . . . . . . .10

linear momentum. . . . . . . . . . . . . . . . . . . . . . .24

M

manifold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .65

conguration . . . . . . . . . . . . . . . . . . . . . . 97

material point . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

matrix

inertia . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

orthogonal . . . . . . . . . . . . . . . . . . . . . . . . . 74

Pfaan. . . . . . . . . . . . . . . . . . . . . . . . . . .100

rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

mixed product. . . . . . . . . . . . . . . . . . . . . . . . . .10

motion

of Lagrange top. . . . . . . . . . . . . . . . . . . .91

of material point. . . . . . . . . . . . . . . . . . .12

of rigid body . . . . . . . . . . . . . . . . . . . . . . 74

of system of material points . . . . . . . 20

N

Newton's laws of dynamics . . . . . . . . . . . . . 22

norm

Euclidean. . . . . . . . . . . . . . . . . . . . . . . . . .10

of vector . . . . . . . . . . . . . . . . . . . . . . . . . . .10

normal vector . . . . . . . . . . . . . . . . . . . . . . . . . . 13

O

orbit. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .4

P

pendulum

Furuta. . . . . . . . . . . . . . . . . . . . . . . . . . . . .48

Hamiltonian. . . . . . . . . . . . . . . . . . . . .62

invariants . . . . . . . . . . . . . . . . . . . . . . . 65

Lagrangian . . . . . . . . . . . . . . . . . . . . . . 48

mathematical . . . . . . . . . . . . . . . . . . . . . . 26

divergence . . . . . . . . . . . . . . . . . . . . . . 68

spherical. . . . . . . . . . . . . . . . . . . . . . . . . . .98

conguration manifold . . . . . . . . . . 98

Lagrangian . . . . . . . . . . . . . . . . . . . . . . 98

phase portrait . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Poisson bracket . . . . . . . . . . . . . . . . . . . . . . . . . 63

principle

conservation of angular momentum 25

conservation of energy . . . . . . . . . . . . . 26

conservation of linear momentum . . 24

d'Alembert's . . . . . . . . . . . . . . . . . . . . . 115

of determinism . . . . . . . . . . . . . . . . . . . . 20

of invariance . . . . . . . . . . . . . . . . . . . . . . . 20

problem

Dido's . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

isoperimetric . . . . . . . . . . . . . . . . . . . . . . 35

vaconomic . . . . . . . . . . . . . . . . . . . . . . . . . 35

R

Riemannian metric . . . . . . . . . . . . . . . . . . . . . 52

rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

body frame . . . . . . . . . . . . . . . . . . . . . . . . 74

displacement . . . . . . . . . . . . . . . . . . . . . . 74

equations

Euler-Lagrange . . . . . . . . . . . . . . . . . .83

Euler-Newton . . . . . . . . . . . . . . . . . . . 84

phase constraints . . . . . . . . . . . . . . . . . 105

space frame. . . . . . . . . . . . . . . . . . . . . . . .74

uniform coordinates . . . . . . . . . . . . . . . 75

rolling ball . . . . . . . . . . . . . . . . . . . . . . . 105, 122

Lagrangian . . . . . . . . . . . . . . . . . . . . . . . 123

rotation group . . . . . . . . . . . . . . . . . . . . . . . . . . 78

S

scalar product . . . . . . . . . . . . . . . . . . . . . . . . . . 10

skate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

ski . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

special Euclidean group. . . . . . . . . . . . . . . . .76

spherical coordinates . . . . . . . . . . . . . . . . . . . 53

Index 132

T

tangent vector . . . . . . . . . . . . . . . . . . . . . . . . . . 13

time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

torsion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

torus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

trajectory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4, 5

length . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

translation vector . . . . . . . . . . . . . . . . . . . . . . .74

V

vector

eld . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66

divergence . . . . . . . . . . . . . . . . . . . . . . 67

product . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

velocity

of material point. . . . . . . . . . . . . . . . . . .13

of rigid body

in body . . . . . . . . . . . . . . . . . . . . . . . . . 78

in space . . . . . . . . . . . . . . . . . . . . . . . . . 78

W

wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100

rolling . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

condition for non-holonomicity . 110

tilted . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

Lagrangian. . . . . . . . . . . . . . . . . . . . .126

vertical . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

Lagrangian. . . . . . . . . . . . . . . . . . . . .120

List of Figures

1 Position of the Planet around the Sun . . . . . . . . . . . . . 2

2 Ellipse . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

1.1 Inner product . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.2 Cross product . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

1.3 Mixed product . . . . . . . . . . . . . . . . . . . . . . . . . . 12

1.4 Frenet trihedron . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.1 A pair of material points . . . . . . . . . . . . . . . . . . . . . 22

2.2 System of n material points . . . . . . . . . . . . . . . . . . . 23

2.3 Mathematical pendulum . . . . . . . . . . . . . . . . . . . . . 27

2.4 Phase portrait of the pendulum . . . . . . . . . . . . . . . . . 29

2.5 Electro-mechanical system . . . . . . . . . . . . . . . . . . . . 30

3.1 Rotational gure . . . . . . . . . . . . . . . . . . . . . . . . . 33

3.2 Motion without lateral slip . . . . . . . . . . . . . . . . . . . 36

3.3 Catenary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

3.4 Optimal orbit of Brockett's integrator (a side view and a view

from above) . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.5 Pursuit problem . . . . . . . . . . . . . . . . . . . . . . . . . 42

3.6 Brachystochrone problem . . . . . . . . . . . . . . . . . . . . 43

4.1 Ball and beam . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.2 Furuta's pendulum . . . . . . . . . . . . . . . . . . . . . . . . 48

4.3 Sphere S2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.4 Inverted pendulum . . . . . . . . . . . . . . . . . . . . . . . . 54

4.5 Jumping robot's leg . . . . . . . . . . . . . . . . . . . . . . . 55

4.6 2R robotic manipulator . . . . . . . . . . . . . . . . . . . . . 55

4.7 Space robot . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.8 Ballbot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

133

List of Figures 134

4.9 Elastic spherical pendulum . . . . . . . . . . . . . . . . . . . 57

5.1 Legendre transform . . . . . . . . . . . . . . . . . . . . . . . . 59

5.2 Flow of dynamic system . . . . . . . . . . . . . . . . . . . . . 67

5.3 Theorem on recurrence . . . . . . . . . . . . . . . . . . . . . . 69

5.4 Torus T2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

5.5 Dening the torus T2 . . . . . . . . . . . . . . . . . . . . . . 70

5.6 Spherical pendulum . . . . . . . . . . . . . . . . . . . . . . . 72

6.1 Rigid body . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6.2 Displacement of rigid body . . . . . . . . . . . . . . . . . . . 75

6.3 Elemenetary rotation around the X axis . . . . . . . . . . . . 77

6.4 Motion of element of mass dm . . . . . . . . . . . . . . . . . 80

7.1 Lagrange top . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

7.2 Function f(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

7.3 Function L(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

7.4 L(u) < 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.5 L(u) > 0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.6 L(u) changes sign . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.7 Function L(u) . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

8.1 Conguration constraints . . . . . . . . . . . . . . . . . . . . 98

8.2 Wheel moving without lateral slip . . . . . . . . . . . . . . . 100

8.3 Rolling wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.4 Kinematic car . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

8.5 Phase constraints . . . . . . . . . . . . . . . . . . . . . . . . . 105

8.6 Rolling ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

8.7 Wheeled mobile robot . . . . . . . . . . . . . . . . . . . . . . 112

8.8 Car with trailer . . . . . . . . . . . . . . . . . . . . . . . . . . 112

8.9 Fire truck . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

9.1 Chaplygin's skater . . . . . . . . . . . . . . . . . . . . . . . . 117

9.2 Path of Chaplygin's skater . . . . . . . . . . . . . . . . . . . . 119

9.3 Vertical wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

9.4 Wheel's paths: a) η < 0, b) η > 0 . . . . . . . . . . . . . . . . 122

9.5 Rolling ball . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123

9.6 Tilted wheel . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

List of Theorems

3.3.1 Theorem (Euler-Lagrange equations) . . . . . . . . . . . . . 34

3.3.2 Theorem (Euler-Poisson equations) . . . . . . . . . . . . . . 35

3.4.1 Theorem (On conditional extremum) . . . . . . . . . . . . . 35

4.3.1 Theorem (On geodesics of Riemannian metric) . . . . . . . 52

5.3.1 Theorem (On invariance of Hamiltonian) . . . . . . . . . . . 61

5.5.1 Theorem (On Poisson bracket) . . . . . . . . . . . . . . . . 64

5.6.1 Theorem (Liouville's on invariants) . . . . . . . . . . . . . . 64

5.8.1 Theorem (Liouville's on divergence) . . . . . . . . . . . . . . 68

5.10.1 Theorem (Poincare's on recurrence) . . . . . . . . . . . . . . 69

8.4.1 Theorem (On non-holonomic constraints) . . . . . . . . . . 110

135

Documents

Analytic Mechanics - kcir.pwr.edu.pl