Foreword
Mechanics is an old science, but it acquired its great reputation at the end of the 17th century, due to Newton’s works. A century later, Euler and, above all, Lagrange renewed it and led it towards a formulation not only aesthetically elegant but also capable of applications to other fields of physics. Fifty years later, Hamilton and Jacobi gave their names to very important further contributions. Lastly, at the end of the xix th century, Poincaré took a new step with the introduction of geometry in the analysis of physical problems. During the xx th century, physicists produced new developments from the works of their famous predecessors.
This book is addressed to readers already familiar with the Newtonian approach for mechanics. Several training textbooks – some of them excellent – rely on this approach. On the other hand, it seems that exercises based on Lagrangian and Hamiltonian formulations are rather scarce in the literature; we hope that the present work may help to fill this gap.
In a previous book published in French by EDP, Grenoble-Sciences collection, under the name La mécanique: de la formulation lagrangienne au chaos hamiltonien, we proposed, for undergraduate students, a comprehensible synthesis of all the modern facets of mechanics and their relationships to other domains of physics. This textbook contains a great number of exercises and problems, many of them original, dealing with the theories of Lagrange, Hamilton and Poincaré. We gave only the results or brief hints for solving these problems. Some problems can be considered as difficult, or even disconcerting, and readers encouraged us to provide the solution of those exercises which illustrate all the topics presented in the book. This is the aim of the present work. We retained from the foregoing book most of the problems presented here, very often trying to make them clearer, sometimes trying to find interesting extensions. We also proposed new ones better suited to our pedagogic goal. In the same spirit, others have been withdrawn because we judged them less instructive for physics, even if the mathematical points they dealt with were logical consequences of features treated in the textbook.
Of course, the present work is a natural complement of the course-book; nevertheless, we have tried to make it self-contained and, with this in mind, we add in each chapter a succinct, although clear and complete, summary of each topic. These summaries are adequate to tackle and solve the problems presented afterwards; every concept or notion necessary for obtaining the solution is presented and developed therein. Our aim is to make the reader familiar with the Lagrangian and Hamiltonian approaches, which may be difficult to grasp, to demonstrate the power of this formalism and help to develop skills for managing the techniques essential for this kind of study. The problems are selected with this purpose and they illustrate very often practical physical situations and sometimes aspects of everyday life.
This book is built around eight chapters entitled:
The Lagrangian formulation
Lagrangian systems
Hamilton’s principle (also called the least action principle)
The Hamiltonian formalism
The Hamilton-Jacobi formalism
Integrable systems
Quasi-integrable systems
From order to chaos In each chapter, the reader will find:
A clear, succinct and rather deep summary of all the notions that must be understood, the important points that must be memorized
and the notations and symbols used in the problems.
The statements of the problems which are presented consecutively. These statements are sufficiently detailed so that, with the
help of the lesson summaries, it is unnecessary for the reader to search for other sources of information. Whenever a figure turns out to be essential for a good understanding of the text, it appears in the statement. The progressive difficulty of the problems is symbolized with an increasing number of stars (from 1 to 3) added to the title.
Detailed answers to the problems which are grouped together at the end of the chapter. A number of additional figures are
inserted in the corresponding text in order to exhibit essential points or to avoid lengthy circumlocutions.
At the beginning of the book, a synoptic table gathers, chapter by chapter, the set of all the proposed problems, giving, for each of them, its reference (number, title, page number), its difficulty (1 to 3 stars), as well as the important features or the peculiar aspects treated in this problem.
Two purposes are pursued
To concretize, through simple and often academic examples, notions that seem apparently very abstract.
The methods to find the solution are not necessarily the most elegant or the quickest, but it is important to check, via these simple examples, one’s understanding of these new tools for mechanics. Sometimes the same problem, or the same physical situation, is studied once more in a subsequent chapter with new tools in order to emphasize some novel feature of the method.
To emphasize the power of these new tools in physics applied to fields as miscellaneous as traditional mechanics, optics,
electromagnetism, waves in general, and quantum mechanics. Concerning these fascinating and up to date subjects in physics, we will focus only on the mechanical aspect. However, the reader could satisfy his curiosity, with help of keywords, by looking for further information firstly in a good and complete encyclopedia, then on the web using a engine such as Google (or, for more exotic subjects, Yahoo). He will find exhaustive lectures as well as recent articles.
We strongly recommend that the reader carries out some applications and draws the figures proposed in the detailed solutions. For the drawing or plotting of curves, the authors have used the freewares xfig or xmgrace that can be downloaded from the web. The differential equations have been solved with the help of the Mathematica software package.
Chapter 1 The Lagrangian Formulation
Summary
1.1. Generalized Coordinates
A mechanical system is composed, in fine, of a given number N of elements α, with a mass m α , which can be considered as pointlike and located at position r α . The configuration of this system is specified by the set of the constituent coordinates. However, in most situations, internal constraints (for example in a rigid body the distance between the constituents is independent of the configuration) or external constraints (for example a point subjected to remain on a given surface) impose a number of relationships between the coordinates; in such cases, a smaller set of specifications allows us to characterize the configuration of the system.
The n variables (n ≤ 3N), which unambiguously define the configuration of the system are called generalized coordinates; they are denoted generically as q, for the set (q 1 , q 2 , … , q n ) of the n generalized coordinates q i . In any practical case, generalized coordinates are either lengths, or angles. Generalized coordinates being sufficient to completely describe the configuration, there exist N mathematical relations r α (q, t) (α = 1, … , N), each coordinate position depending only on n variables q i . Sometimes one encounters an explicit time dependence of the constraints, for example when a point moves on a surface which moves itself.
1.2. Lagrange’s Equations
The Lagrangian formulation of mechanics consists in writing Newton’s equations, which depend on N vectorial quantities r α , in terms of n scalar quantities q i (q 1 , q 2 , … , q n ). To begin with, let us consider the case for which the n generalized coordinates are independent; in this case, n is called the number of degrees of freedom for the system. The Lagrangian formalism relies on the kinetic energy T, which is a kinematic quantity defined in terms of velocities 1 v α = r ˙ α = dr α /dt of each element by
If the particle positions are given in terms of generalized coordinates, the kinetic energy is expressed not only in terms of n generalized coordinates q i , but also in terms of n generalized velocities q˙ i = dq i /dt and, possibly, in terms of time: T(q, q,˙ t). From the kinetic energy, one builds n kinematical quantities A i , called generalized accelerations, defined by the following relation:2
(1.1)
Then, Newton’s equations are translated into the Lagrangian formalism through a set of n dynamical equations, called Lagrange’s equations, which are written (1.2)
where \(Q`_i (q, \dot{q},t)\) are dynamical quantities, called generalized forces, which will be defined later.
The Lagrange equations are a set of n coupled differential equations of second order.
Directions for use and precisions
The first task is to obtain the expression of the kinetic energy as a function3 of the generalized velocities, possibly of the generalized coordinates, and
1. As usual in mechanics, a dot above a quantity means its first derivative with respect to time, two dots its second derivative,… : \(\dot{f} = df/dt, \ddot{f} = d^2 f/dt^2\), …
2. With typographical simplicity in view, we will use a simplified notation to define partial derivatives for a function of several variables
frac{partial^2 f(x, y)}{partial x^2}, partial^2_{xy} f(x, y) = frac{partial^2 f(x, y)}{partial x partial y},
3. The choice for generalized coordinates is, a priori, arbitrary. The best starting choice is that which gives the most simple form to the kinetic energy.
(although rarely) of time 4 (see Exercise 1.4). To obtain the kinetic energy, one supposes first that the generalized coordinates depend on time q(t); naturally the derivatives of these functions with respect to time ˙q(t) appear in the expression of the velocities. Thus the kinetic energy is expressed in terms of q and q.˙ Subsequently, these functions are considered as independent. Sometimes, the kinetic energy exhibits only generalized velocities, and sometimes both generalized velocities and coordinates.
Just as an example, let us consider a particle with mass m, moving on a plane: if one locates the particle by the Cartesian coordinates (x, y) the kinetic energy is expressed only as function of generalized velocities since \(T(\dot{x}, \dot{y}) = \frac{1}{2} m( \dot{x}^2 + \dot{y}^2^)\), whereas if one chooses polar coordinates ( ρ , φ ) the same kinetic energy contains, in addition to the generalized velocities, the coordinate ρ since \(T( ρ , \dot{ρ}, \dot{\phi}) = \frac{1}{2} m( \fot{ρ}^2 + ρ^2 \dot{\phi}^2 )\).
Once the expression for the kinetic energy is obtained the rest of the treatment is as follows:
One derives the function \(T(q, \dot{q},t)\) with respect to the generalized
coordinates \(q_i\) to get \(\partial_{q_i} T(q, \dot{q},t)\).
One derives the function \(T(q, \dot{q},t)\) with respect to the generalized
velocities \(\dot{q}_i\) to get \(\partial_{\dot{q}_i} T(q, \dot{q},t)\).
One derives with respect to time the function \(\partial_{\dot{q}_i} T(q, \dot{q},t)\), considering that one handles a function q(t), for which q˙ =
dq(t)/dt and ¨q = d ˙q(t)/dt. The generalized acceleration A i (q, q,˙ ¨q, t) follows from (1.1).
Proceeding with the previous example and polar coordinates, this series of operations leads to the generalized acceleration:
A ρ = m(¨ρ − ρ φ 2 );
¨φ A φ φ
=
m(
ρ2
2
ρ
˙
ρ
).
From Newton’s equations, the product of mass with acceleration is determined by the forces acting on the system; similarly the link between the generalized accelerations and generalized forces through Lagrange’s equations matches Newton’s equations.
As long as the forces are not specified, the functions q(t) entering the generalized accelerations are arbitrary. Equating generalized accelerations to generalized forces leads to a system of differential equations which are fulfilled only for special functions q(t), which are precisely the solutions of the true physical motion and which are called trajectories. To determine them unambiguously, it is necessary to set the initial values q(0) and ˙q(0).
1.3. Generalized Forces
To define generalized forces, one must first introduce the notion of virtual displacement. Let us imagine that, at a given time, two configurations of the system are described by the coordinates q et q + δq, compatible with the constraints imposed on the system. The quantity δq is called a virtual displacement.
In this displacement, the constituents α are displaced by a quantity δr α and the forces f α acting on them produce a total work
N δW = ∑ f α · δr α . α=1
This last quantity is said to be a virtual work and it can be put under a form expressed in terms of the virtual displacements δq:
n δW = ∑ Q i δq i . i=1
(1.3)
This expression defines the generalized forces 5 Q i (q, q,˙ t). Let us note that a virtual displacement is only compatible with the constraints and can be entirely different from a real displacement of the system which results from the temporal evolution given by Lagrange’s equations (1.2).
Let us emphasize a point. In the Lagrangian formalism, the forces responsible for the constraints are inaccessible, since the generalized coordinates were chosen precisely to get rid of them. Since they are generally uninteresting quantities, this is of little consequence and, in fact, lies at the origin of the elegance of Lagrange’s equations. If, after all, we insist on obtaining the expression of these constraint forces, we have to introduce supplementary generalized coordinates (to get rid of cumbersome constraints) in order to obtain a non vanishing virtual work concerning this type of force (see Problems 1.4 and 1.7).
When the system is at rest, generalized velocities and accelerations vanish; Lagrange’s equations (1.2) then imply a vanishing value for the generalized force. The relation:
Q i = 0 at rest
(1.4)
represents d’Alembert’s principle.
1.4. Lagrange Multipliers
Let us consider now the case where the n generalized coordinates are not independent. It is useful to remind ourselves that these coordinates were introduced with the purpose of taking into account a number of constraints.
The present case thus corresponds to a situation for which the system is subject to additional constraints. Practically, this happens when the constraints are not able to reduce the number of generalized coordinates or when the search for new generalized coordinates turns out to be a too painful procedure.
All the virtual displacements are no longer possible, but compelled to obey new conditions taking into account the supplementary constraints. For simplicity, let us consider only one condition written under a differential form:
n ∑ Λ i δq i = 0. i=1
(1.5)
This equation defines the quantity Λ i , an important ingredient in constrained Lagrange equations. There exists a special very simple kind of constraint, known as holonomic, for which this quantity is the differential of a single function Φ:
n ∑ Λ i δq i = dΦ(q). i=1
The constraint is thus equivalent to the fact that Φ(q) is a constant. This allows us, in principle, to express one generalized coordinate as a function of the n − 1 others; it is enough then to proceed like this in the expressions of the kinetic energy and generalized forces 6 in order to work now with n−1 generalized coordinates. Indeed the system depends on n − 1 rather than n degrees of freedom.
If the constraint is not holonomic, or if elimination is not an easy task, then we keep the original generalized coordinates and introduce Lagrange multipliers. It is not our intention, in this brief summary, to develop
6
In particular, this is the case for rolling without slipping motion in two dimensions. This is no more the case in three dimensions.
Let us stress the fact that a rolling without slipping motion necessarily implies a non vanishing tangential reaction force acting on the rolling surface. Nevertheless in a virtual displacement, this force does not perform work. The deep reason for this is a consequence of the fact that, in this virtual displacement δ φ , the application point of the force follows a cycloid and there is a displacement only of second order in δ φ in the perpendicular direction and of third order in the tangential direction (see Exercise 1.5).
the general theory of Lagrange multipliers. We simply give the form of constrained Lagrange equations when the system is subject to l differential constraints of type (1.5):
l A i (q, q,˙ ¨q, t) = Q i (q, q,˙ t) + ∑ λ k Λ i k (q, t). k=1
(1.6)
The quantities λ k are the Lagrange multipliers. Their values must be determined at the same time as the trajectories q(t) by solving the n differential equations and the l constraints equations imposed on the coordinates. The Lagrange multipliers can be interpreted in terms of reaction forces associated with the constraints (see Exercise 1.8).
Problem Statements
1.1. The Wheel Jack [Solution and Figure p. 24]
This exercise is simply an application of d’Alembert principle
A wheel jack is an articulated machine which is designed to lift up heavy burdens (a coach for example); it is represented in Fig. 1.1. It is composed of two rigid bases (one resting on the ground, the other sustaining the burden with weight P); they form an articulated lozenge with side l, which may be deformed by mean of a threaded stem with a step h, (the axis of the jack changes by a length h for each revolution of the crank) driven by a force F applied on the crank of arm length a.
We assume no friction on the mechanical parts of the jack (fortunately frictional forces do exist and allow the mass to be supported without any effort!).
Examine the constraints imposed on the system and show that it has
only one degree of freedom. Which generalized coordinate seems to you
the most appropriate?
2. Using d’Alembert’s principle, deduce the relationship between the weight to be lifted up and the exerted force, as a function of the characteristics of the jack and of the angle θ between the lozenge side and the threaded stem.
Numerical application: Calculate the ratio of the weight and the force for a jack with crank arm length a = 20 cm, with h = 2 mm, at the beginning of the lifting process when θ = 30 ◦ .
1.2. The Sling [Solution p. 26]
Very simple application of Lagrange’s equations
A rigid stem is maintained fixed at one of its ends O. It turns around O in the horizontal plane with a constant angular speed ω = φ (see Fig. 1.2). A pointlike mass m slips without friction on this stem. It is placed at rest at a point A such that OA = a.
Find the most natural generalized coordinate and assess the real and generalized forces.
Write and solve the Lagrange equation.
1.3. Rope Slipping on a Table [Solution p. 27]
Classical problem for which the Lagrangian formalism is well suited
A part L − l of a rope with length L and with constant linear mass μ is originally at rest on a horizontal table. The rest of the rope, with length l, hangs vertically in a constant gravitational field g.
The rope is placed without any initial velocity. One assumes that the part which hangs over the edge of the table remains always vertical 7 (Fig. 1.3).
Fig. 1.3 Rope slipping on a table
1. In a first study, assume that friction on the table is absent. Find a generalized coordinate and assess the real and generalized forces. Write and solve the Lagrange equation.
2. Assume a solid friction, with a constant friction coefficient f (the dynamical friction coefficient is assumed equal to the static coefficient). What is the minimum length l 0 necessary to induce sliding of the rope. If l > l0 write and solve the Lagrange equation.
1.4. Reaction Force for a Bead on a Hoop
[Solution and Figures p. 28]
Calculation of a reaction force by adding a generalized coordinate
Let us consider a system composed of a pierced bead M, with mass m, which slides without friction on a massless hoop with center O, radius R, which itself rotates around one fixed diameter Oz, parallel to the vertical.
7
Indeed the linear momentum acquired by the horizontal part of the rope when it falls has the consequence that it keeps falling and the rope does not turn at right angles. Moreover a rope is a flexible system which can exhibit transverse deformations and the fall can cause undulations. Of course, all these complications are neglected.
The angle φ , between the plane of the hoop and the fixed vertical plane xOz, varies in time according to a known law imposed by the operator: φ (t). A generalized coordinate is chosen as the angle θ between the direction OM and the vertical direction Oz.
This system is embedded in a constant gravitational field g, acting along the vertical axis.
Give the expression of the kinetic energy in terms of the generalized coordinate.
Give the expression of the generalized force.
Write down the corresponding Lagrange equation.
4. This simple question illustrates the difference between virtual work and real work. We are interested in the reaction force of the hoop on the bead. Introducing a new generalized coordinate which allows virtual work for the component of the reaction force normal to the plane of the hoop, determine this force. Check your result with the Coriolis inertial force.
1.5. Huygens Pendulum
[Solution and Figures p. 31]
Work done by contact forces responsible for a motion without slipping
In a vertical plane xOz, a point M, with mass m, is fixed to a massless hoop, with radius R, which can roll without slipping on a horizontal stem Ox, placed above it. It is well known that the curve followed by M is a cycloid. We choose as generalized coordinate the angle φ , such that R φ is the abscissa of the center C of the hoop. The origin O is taken when M is in its lowest position, CM being then parallel to Oz. The system is subject to a constant gravitational field g directed along the downward vertical.
Write the Lagrange equation relative to the coordinate φ .
2. Make a change of variable and take instead x = sin( φ /2). Show that x varies in time following a harmonic motion with angular frequency ω = √ g/(4R). Deduce that φ evolves periodically with the same angular frequency, independently of its amplitude. This pendulum is said to be isochronous and is known as Huygens pendulum.
1.6. Cylinder Rolling on a Moving Tray
[Solution and Figure p. 33]
Work performed by contact forces responsible for a motion without slipping
A homogeneous cylinder, with radius R, mass M and moment of inertia I around its axis, rolls without slipping on a horizontal tray. We impose a translational motion on the tray, perpendicular to the axis of the cylinder, with a given time law a(t). This situation represents for instance the motion of a bottle in the boot of a car.
As generalized coordinate, one can choose the angle θ that specifies an
arbitrary point of the cylinder along the horizontal direction. Show that
the position X of the center of the cylinder in the Galilean frame is
linked to θ by an holonomic constraint which is to be determined. What
is the corresponding generalized acceleration? Solve the corresponding
Lagrange equation and give the real acceleration of the cylinder.
Repeat this question choosing now as the generalized coordinate the position X of the center of the cylinder.
1.7. Motion of a Badly Balanced Cylinder
[Solution and Figure p. 35]
Application of Koenig’s theorem; holonomic forces
An inhomogeneous cylinder (center C), with radius R and mass M, has its center of mass G at a distance a from its axis. The mass density is constant along a straight line parallel to the axis. This property implies that one of the principal axes for the cylinder is also parallel to its axis. We denote by I the moment of inertia of the cylinder with respect to the straight line parallel to the axis which passes through G. We study the motion without slipping of the cylinder subject to a constant vertical gravitational field g; the cylinder rolls on a fixed horizontal plane, the plane of its circular section being always fixed (the instantaneous rotation vector ω is always parallel to the axis).
Equation of motion 1. To define the cylinder configuration, let us take as the single generalized coordinate the angle θ between the downward vertical and the direction CG. Taking into account the constraint for rolling without slipping, write the corresponding Lagrange equation.
2. After multiplying both sides of this equation by the angular velocity ˙ θ, express the conservation of energy E (we speak of a constant of the
motion). To get an idea of this type of motion, plot the angular velocity as ˙ a function of the angle, for several different values of the energy: θ(θ, E).
Vertical reaction force
Explain why the cylinder can exhibit singular behaviour if it rolls too quickly. Using a second generalized coordinate, which breaks the contact with the plane, determine the vertical component F v (θ, E) of the reaction force to the plane. It is naturally supposed that this force is weakest when the center of mass is in its highest position. Deduce the maximum energy of the system.
Horizontal reaction force
Rolling without slipping is possible only because the plane exerts a horizontal reaction force to the cylinder. But we know that the ratio between the horizontal and vertical components of the reaction force cannot exceed the friction coefficient f. To obtain this horizontal reaction force F h (θ, E), we must consider two generalized coordinates in order to break the constraint of rolling without slipping. Study graphically, as a function of the energy, the conditions that must be fulfilled to achieve rolling without slipping.
1.8. Free Axle on a Inclined Plane
[Solution and Figures p. 39]
To understand how to use Lagrange multipliers
A massless axle CC ′ maintains two identical wheels, of centers C and C ′ and radius R, in planes normal to it and separated by a distance L = CC ′ . These wheels, for which the axle is a symmetry axis, have a mass m, and the three moments of inertia are I 1 = I 2 = I (in the plane of the wheel) and I 3 (along CC ′ ).
The mechanical system consists of the set of the axle and the two wheels (see Fig. 1.4). We study the rolling without slipping of this system on a inclined plane making an angle α with the horizontal plane. For rigidly locked wheels, the motion is identical to that of a cylinder, that is a uniformly accelerated motion.
The aim of this problem is to study the motion when the wheels roll independently of each other.
One chooses a system of perpendicular axes in the inclined plane: horizontal XX ′ , and Y Y ′ along the direction of steepest upward slope. The center O of the axle is characterized by its coordinates X et Y in this frame with an arbitrary origin A. The direction C ′ C makes an angle θ with the horizontal line XX ′ . We denote by φ and φ ′ the angles which mark the positions of reference points on the circumference of the wheels with respect to the line normal to the inclined plane. Thus, the system is described in terms of 5 generalized coordinates (X, Y, θ, φ , φ ′ ).
α
Fig. 1.4 Axle with independent wheels rolling without slipping on a inclined plane
1. There exist four scalar relationships concerning the constraints of rolling without slipping for each of the wheels (two per wheel). In fact, two of them are identical. Give the three independent constraint relationships and show that one of them is holonomic whereas the other two are not.
Introducing three Lagrange multipliers strained Lagrange equations.
λ 1 , λ 2 , λ 3 , write the five con-
Interpret the three Lagrange multipliers in terms of contact forces.
4. To solve the eight equations (five Lagrange equations plus three constraint equations), it is judicious to change variables by defining σ = ( φ + φ ′ )/2
and δ = ( φ − φ ′ ).
Rewrite the Lagrange equations in terms of these new variables. According to the initial conditions, study the various types of behavior for the axle. In particular, give the equations of the motion if, initially, the axle center is located at A and sets off down the slope with a speed V 0 , the axle itself being horizontal and having an initial angular velocity ˙ θ(0) = ω.
In this framework, calculate the Lagrange multipliers λ i which represent the reaction forces.
1.9. The Turn Indicator
[Solution and Figure p. 43]
Mechanics in the “clouds”
In the absence of any visual reference, the pilot of an aircraft would ignore whether he is turning or not, without a small gyroscope (10 cm or so), refereed to as “turn indicator” or “needle”. Such a gyroscope of center O is presented in Fig. 1.5. An axis X ′ OX, parallel and firmly attached to the fuselage of the aircraft, is assumed to remain horizontal during the turn. A frame, with normal OZ, is free to oscillate around X ′ X. The mechanical system under study is the inertia flywheel of the gyroscope which is a cylinder with symmetry axis Y ′ OY . The axes of inertia are OX, OY , OZ, which form a direct orthogonal trihedron OXY Z, and the corresponding moments of inertia are respectively I X = I Z and I Y = I.
z
z ′
ω
Z
α
θ
Y ′
O
Y
Fig. 1.5 Gyroscope inside a plane. Only the axis Y ′ Y of the gyroscope, the true vertical Oz and the apparent vertical Oz ′ are represented
A small electric motor maintains the flywheel rotation around the axis Y ′ Y and imposes a constant angular velocity Ω on it. The apparent vertical for the aircraft – namely the normal to the wing plane– is denoted Oz ′ . A small spiral spring acts to force the axes Oz ′ and OZ to coincide with a restoring torque C = −kθ, where θ is the angle between the Oz ′ and OZ axes. A needle measures the angle θ.
The aircraft turns, with constant angular velocity ω (see Fig. 1.5; the effective rotation axis is located out of the plane, but this does not matter for the reasoning), around the true vertical Oz (parallel to gravitational acceleration g ). The angle between Oz and Oz ′ is denoted α and it is assumed to be constant throughout the turn.
One chooses as generalized coordinate the angle θ which is the only freedom left to the gyroscope.
In a first study, the apparent vertical is supposed to coincide with the true vertical: α = 0.
Write down the kinetic energy of the flywheel.
In the following, we assume the condition ω ≪ Ω, which is always satisfied in practical circumstances.
3. We are interested in the equilibrium solution (θ = const) (obtained in practice with a small pneumatic shock absorber). Give ω as a function of θ; the second order terms ω 2 are neglected. The pilot reads the angle θ and deduces the value ω.
As a matter of fact, the real situation is a little more complicated. Exactly as does a cyclist, the aircraft banks during the turn, and this corresponds to an angle α = 0. The pilot maintains the inclination and the velocity V ̸ of the plane during the turn.
Give α as a function of V , ω and g. The simplest method is to consider a static problem in the frame of the plane.
2. How is the relationship between ω and θ modified if the (obligatory) inclination of the plane is correctly taken into account. What relation should exist between V , g and the characteristics of the instrument in order to achieve maximum sensitivity (to give the biggest value of θ for a given ω). It is legitimate to employ the approximation α = ∼ tan(α).
1.10. An Experiment to Measure the Rotational Velocity of the Earth
[Solution p. 46]
An alternative to the Foucault pendulum, realized by A.H. Compton
Imagine yourself sitting on a seat of a carousel turning with constant angular velocity ω. You now take hold of the axis of a disc which can rotate without friction with an angular velocity φ . Initially, the axis is maintained in the vertical direction and the disc is motionless in the frame of the carousel. Now pivot the axis into the horizontal plane. In so doing, you feel a reaction
due to the axis and, to your great surprise, the disc starts turning spontaneously around its axis. Pursue the change of orientation of the axis until its complete reversal along the vertical. You will notice that the rotation velocity increases.
It is easy to do the experiment, sitting on a turning stool, with a bike wheel grasped in your hands.
Being located at the pole and considering the Earth as the carousel, this simple experiment directly establishes the earth’s rotation, using a much less cumbersome set up than Foucault’s pendulum. It was proposed and realized by A.H. Compton (Phys. Rev. 5, February 1915, 109).
Determine the angular velocity of the disc φ around its axis for each
angle θ(t) between its axis and the rotation axis of the carousel.
What is the prediction of the calculation if the experiment is realized not at the pole but at a place located at latitude λ ?
1.11. Generalized Inertial Forces
[Solution p. 48]
How to use the Lagrangian formalism in a non Galilean frame?
In establishing formula (1.1), there is no hypothesis concerning the choice of the physical frame. The kinetic energy and the acceleration that come out are those relative to this peculiar frame. If this frame is not Galilean, one has to take into account inertial forces and equate m α a α to f α (v) + f α (i) , the sum of the true force acting on the particle α and the corresponding inertial force.
It is important to recall that the inertial force is itself the sum of a driving force due to the acceleration of the origin, of a Coriolis force (depending on the velocity v α in the given frame) and of a centrifugal force:
f (i) α = −m α a (e) − 2m α ω × v α − m α (dω/dt) × r α − m α ω × (ω × r α ).
In this case, the formalism leads to Lagrange equations containing additional generalized forces Q i → Q i (v) + Q i (i) .
If the given frame moves translationally with an acceleration a (e) (which
can depend on time) with respect to the Galilean frame, show that the
generalized inertial force is simply:
Q i = −Ma (e) ∂ q i R cm ,
where M is the total mass of the system and R cm is the center of mass coordinate. As an application, write Lagrange’s equations for a pendulum
of length l in a constant gravitational field, whose point of suspension is subject to an imposed arbitrary vertical motion h(t).
2. If the given frame rotates uniformly with a constant instantaneous rotation vector ω with respect to the Galilean frame, show that there exists a generalized Coriolis force
d [∂q ˙ i Q (cor) i = ∂ q i (ω · L) − (ω · L)] dt
and a generalized centrifugal force
Q (cent) i = ∂ q i T,
where
L ∑ α
=
mα
rα
×
v
α
is the angular momentum of the system about a point of the axis in the chosen frame and
1 = m α × r α )2 (ω T 2 ∑ α
is the driving kinetic energy of the system (energy of the coincident points).
Hints: it is expedient to introduce the mixed product [a, b, c] = a · (b × c) and its invariance properties under even permutations and change of sign under odd permutations. The following vectorial calculus formulae may also be useful.
a × (b × c) = (a · c) b − (a · b) c;
(a × b) · (c × d) = (a · c) (b · d) − (a · d) (b · c) .
Problem Solutions
1.1. The Wheel Jack [Statement and Figure p. 14]
1. Let ABCD denote the lozenge of the jack, the apex A lying beneath the weight, B at the crank and C on the ground; let O be the center of the lozenge, in the middle of the threaded stem BD (see Fig. 1.6).
A priori the configuration of the system is given by α, the angle between the crank and the vertical, and by the form of the lozenge, that is by the values of DB and AC. A first holonomic constraint is due to the invariance of the length, l, of the side of the lozenge: OA 2 + OB 2 = l 2 .
One has a second holonomic constraint due to the threaded stem, which gives a relation between α and DB (when α varies by 2π, DB varies by h). Finally only the angle α is needed to describe the configuration of the system; it has one degree of freedom.
A
l
l
a
D
O
C
B
Fig. 1.6 Lozenge ABCD representing schematically the wheel-jack
Let us make a virtual displacement δα such that the stem length DB increases by an amount δx = h δα/(2π) and the length OB = OD by δOB = δx/2. The length OA decreases, in order to fulfill the relation
OA 2 + OB 2 = l 2 .
Using these conditions, it is easy to see that δCA = 2δOA = −δx/ tan θ. Thus the variation in the altitude of the weight is given as a function of the virtual displacement by: δz = δCA = −h δα/(2π tan θ).
The forces concerned are – the weight P acting at A, which produces an amount of work:
δW P = −P δz = Ph δα/(2π tan θ);
– the reaction force of the ground acting at C which remains at rest; thus the work due to this force vanishes;
– the force F acting on the crank, the virtual work of which is given by δW F = −Fa δα (if one wishes to maintain equilibrium, the force must be opposed to the direct rotation considered previously).
The total virtual work is the sum of all these contributions, namely
Ph Fa = Qα δW = δα. δα [ 2π tan θ ]
One deduces the generalized force Q α = Ph/(2π tan θ) − Fa.
At equilibrium, d’Alembert’s principle imposes a null generalized force and leads to the required expression:
P 2π a tan θ = . F h
To insure a ratio as large as possible (this is precisely the justification of the jack principle), one must thus choose a large crank arm and/or a small step h for the screw.
Numerical application: With a = 20 cm, h = 0.2 cm and tan θ = 0.577, one finds P/F ∼ 363; these values allow us to maintain a 16,000 N coach with a = force of only F = 11 N (remember that P = 16, 000/4 in this special case).
1.2. The Sling [Statement and Figure p. 15]
The system Oxyz is Galilean; the axis Oz is vertical and its unit vector k is directed upwards. In the plane xOy, it is natural to specify the position of the mass by its distance to O: OM = ρ . The angle φ between Ox and OM is proportional to time, φ = ωt, since the angular velocity is kept constant φ = ω. Note that φ is not a coordinate, since it is an externally imposed function.
The forces are the weight, along Oz, and the reaction force of the mass on
the stem, perpendicular to the stem since we have a frictionless contact.
None of these forces performs work during the virtual displacement δ ρ
along OM. The virtual work thus vanishes and the resulting generalized
force is null:
Q ρ = 0.
The expression for the kinetic energy is easy to obtain; it is the usual expression in polar coordinates
1 T = + ω 2 ρ 2 ) . m 2 ( ρ
From this, one obtains, with (1.1), the generalized acceleration A ρ = m ( ¨ρ − ω 2 ρ ) . Lastly, the Lagrange equation A ρ = Q ρ = 0 provides the differential equation ¨ρ − ω 2 ρ = 0. The general solution is well known: ρ (t) = A cosh(ωt)+B sinh(ωt). The integration constants are determined from the initial conditions ρ ˙ (0) = 0 and ρ (0) = a. One finds A = a, B = 0. The solution is thus given by:
ρ (t) = a cosh(ωt).
Note: The equation ¨ρ − ω 2 ρ = 0 represents the fundamental principle of dynamics in a rotating frame (acceleration = centrifugal force). In this case, a classical treatment is even simpler.
1.3. Rope Slipping on a Table
[Statement and Figure p. 16]
1. Let M be the total mass of the rope and μ = L/M its linear mass. One can choose as generalized coordinate the length x which hangs vertically. Since the rope is not elastic, all the points α of the rope have the same velocity v α = x,˙ ∀α. The kinetic energy of the rope is deduced:
1 1 1 T = ∑ mα v α 2 = = . M x L x 2 α 2 2 μ
The acceleration follows from (1.1): A = μ L¨x.
Concerning the real external forces, one distinguishes the weight of the rope and the reaction force of the table (perpendicular to the table since there is no friction); both are vertical. Let us make a virtual displacement δx. The work produced by the weight and by the reaction force on
the horizontal part of the table vanishes because the displacement is perpendicular to them. There remains the work of the hanging portion μ gx of the weight. This work is equal to δW = μ gx δx = Q δx; the expression of the generalized force follows: Q = μ gx.
The Lagrange equation A = Q leads to μ L¨x = μ gx, or ¨x − ω 2 x = 0 with ω = √ g/L. The solution of this differential equation is x(t) = A cosh(ωt) + B sinh(ωt). The integration constants are determined from the initial conditions ˙x(0) = 0, x(0) = l. They imply A = l, B = 0, hence the solution:
g/L . x(t) = l cosh t ( √ )
2. In this case, the reaction force R has both a vertical component R v and a horizontal one R h . At equilibrium, the part on the table is subject to the weight P , to the reaction force R and to the rope tension T due to the hanging part. One must have P + R + T = 0. Projection of this equality on the vertical axis gives P = R v = μ (L − l)g. Projection on the horizontal axis gives R h = T. On the other hand, the tension is also equal to the weight of the hanging part (in order to insure equilibrium):
T = μ l g = R h .
This reasoning “à la Newton” is simpler to understand. The condition of static solid friction imposes R h ≤ f R v or l ≤ f(L − l). It follows that there exists a critical length l 0 for equilibrium: l ≤ l 0 . This minimum length necessary for the motion of the rope is thus:
f l0 = L. 1+f
The kinetic energy takes the same form as before and hence A = μ L¨x. In contrast to the previous case, the horizontal reaction produces work (the tension is an internal force that does no work) and its virtual work is −R h δx (the force acts against the motion). The total virtual work is thus δW = ( μ gx − R h ) δx. On the other hand, for a dynamical friction action, one has: R h = fR v = fP = f μ g(L − x). The generalized force is derived as: Q = μ g [(1 + f)x − fL].
The Lagrange equation A = Q implies
g g ¨x = or ¨x = ) = ω d 2 (x − l 0 ), [(1 + f)x fL] (1 + f)(x l0 L L
where we introduced a new dynamical angular frequency in the presence of friction ω d = √ g(1 + f)/L. The solution of the resulting differential equation, with the correct initial conditions, is given by an expression of the form:
g(1 + f) x(t) = l 0 + (l − l 0 ) cosh t . L √ ( )
which is valid for a time less than the time required for the rope to fall.
One should think about the fact, which may seem paradoxical, that, in the presence of friction, the variation in time for the hanging length is greater than the corresponding rate without friction: ω d > ω.
1.4. Reaction Force for a Bead on a Hoop
[Statement p. 16]
1. It is possible to begin with Cartesian coordinates expressed in terms of R, θ, φ , but it is as simple to deal directly with spherical coordinates since the proposed variables are precisely this type of coordinate. Let us denote as usual the unit vector u r (along OM), u θ (along the motion on the circle) and u φ (along the normal to the hoop plane). The expression for the bead velocity is given by v = R( ˙ θu θ + φ sin θu φ ). This is simply the velocity expressed with spherical coordinates when the bead is
constrained to move on a circle (R ˙ = 0). From the velocity, the kinetic energy is expressed as
1 T = θ ˙ 2 + φ (t) 2 sin 2 θ mR2 . 2 ( )
In this particular case, it would be incorrect to consider φ as a generalized coordinate; it is simply an externally imposed function. We are faced with a constraint (the hoop) which varies with time. As a consequence, the kinetic energy depends explicitly on time through the function φ (t) (see Fig. 1.7).
z
θ
M
O
y
φ
x
X
Fig. 1.7 Bead M slipping without rubbing on a hoop with an imposed external rotation
2. Let us give the bead a virtual displacement δθ; it moves physically with δr = R δθ u θ . Since we have a contact without friction, the reaction force is always perpendicular to the hoop and does no work. The only force which produces work is the weight
P = mg(sin θ u θ − cos θ u r ).
The corresponding virtual work is
δW = P · δr = mgR sin θ δθ.
Identifying this expression to Q θ δθ, one obtains the generalized force:
Q θ = mgR sin θ.
From the kinetic energy and using (1.1), one deduces the acceleration:
¨θ A θ = mR 2 ( − φ 2 cos θ sin θ).
The corresponding Lagrange equation A θ = Q θ leads, after simplification, to the differential equation:8
¨θ g = φ (t) 2 cos θ sin θ. sin + θ R
It is easy to check this result with the help of the fundamental principle of dynamics using the momentum of the weight and of the centrifugal force; this gives the time derivative of the angular momentum (see Fig. 1.8).
θ
mR sin θ φ 2
mg
Fig. 1.8 Weight and centrifugal force acting on the bead
The system has only one degree of freedom and it is meaningless to calculate an acceleration A φ , because φ is not a generalized coordinate.
4. We are concerned with the component f of the reaction force along the normal u φ to the plane of the hoop. 9 If one wishes to calculate it, one must introduce generalized coordinates which give a non null work for this force during the virtual displacement. In order to do this, let us introduce, in addition to θ, the angle φ between the plane of the hoop and the plane xOz. It coincides with the angle corresponding to the imposed rotation but, now, instead of considering it as a given function, it must be considered as a full generalized coordinate. Let δ φ be a virtual displacement of the bead. It moves with δr = R sin θ · δ φ u φ . The virtual work is δW = f · δr = f R sin θ δ φ = Q φ δ φ . Hence the expression for the generalized force is Q φ = f R sin θ.
8
9
The sign in front of the gravitational restoring term may seem strange. It follows from our choice concerning the definition of the angle θ (from the vertical axis directed upwards).
We could be interested as well by the component in the hoop plane, but along the radial direction.
Using the kinetic energy and (1.1), the Lagrange equation A φ = Qφ explicitly gives
¨φ mR 2 sin 2 θ + 2mR 2 φ θ sin θ cos θ = fR sin θ.
After simplification, one arrives at the required expression:
¨φ f(t) = 2m R φ (t) θ ˙ cos θ + m R (t) sin θ.
In contrast to its virtual work which vanishes, this force produces work during a real displacement of the bead.
Once more, one can check this result classically. In the rotating frame, there exist two terms in the component of the inertial force perpendicular to the hoop plane: the usual Coriolis force and a contribution due to the variation of the angular velocity.
1.5. The Huygens Pendulum
[Statement and Figure p. 17]
Remarks concerning the cycloid
Let us consider an arbitrary point P on a circle (not depicted) such that when the center C lies on the vertical through O, the angle CP with the upward vertical is α. When the circle has rolled by an angle φ , the contact point I is horizontally displaced by R φ (rolling without slipping). The coordinates for P are easily obtained:
(R( φ − sin( φ + α)), R(cos( φ + α) − 1)) .
z
I
O
x
φ
C
M
m g
Fig. 1.9 The rope of the Huygens pendulum oscillates between two cycloids. Thus, its length decreases with the deviation from the
vertical
The trajectory followed by the point P is a cycloid which exhibits cusps when φ = −α modulo 2π.
For the point M under consideration, α = π and its coordinates are
(R( φ + sin φ ), −R(1 + cos φ )), see Fig. 1.9.
The kinetic energy of this point can be calculated at once:
1 mv2 T = = mR 2 φ 2 (1 + cos φ ) 2
and the generalized acceleration follows from (1.1):
¨φ A φ = mR 2 (1 + cos φ ) − 2 sin φ 2 . [ φ ]
Let us make a virtual displacement δ φ and study the various forces. – First, the weight: the corresponding work is expressed as
δW = P · δOM = −mgR sin φ δ φ ,
which provides us with the generalized force Q φ = −mgR sin φ .
– Secondly, the contact force exerted at the point of contact I between the circle and the axis Ox. To first order in δ φ the arbitrary point P is displaced by
(Rδ φ (1 − cos( φ + α)), −Rδ φ sin( φ + α)) .
For the given point I, α = − φ , and, to first order, this displacement vanishes. It is a cusp for which both velocity components are null. The contact force does not furnish virtual work and the corresponding generalized force is null.
The Lagrange equation (1.2) A φ = Q φ leads, with the definition
1 g/R, ω = 2 √
to the following differential equation:
¨φ 2 (1 + cos φ ) − φ 2 sin φ = −4ω 2 sin φ .
2. Let us transform first the expression for the Lagrange equation with the help of well known trigonometric formulae in terms of φ /2; after sim¨φ plification, we are left with the equation: 2 cos( φ /2) − φ 2 sin( φ /2) = −4ω 2 sin( φ /2). Now, let us switch to the variable x = sin( φ /2). This last equation is then transformed into the much simpler differential equation:
¨x + ω 2 x = 0.
The solution is x(t) = x 0 sin ωt = sin( φ (t, x 0 )/2). Let T = 2π/ω; then, it is easily seen that φ (t+T, x 0 ) = φ (t, x 0 ), independently of the amplitude x 0 . In other words, we have to deal with a synchronous pendulum with a period T given by:
R T = 4π . √ g
Now let us consider a simple pendulum whose string is attached at one end to a fixed point on the Oz axis, with a length 4R and which is constrained by two cycloids symmetric with respect to Oz, in such a way that the free string length decreases with the oscillation amplitude. It is possible to show that the pendulum bob follows a cycloid similar to that studied in this problem. Indeed, a very good isochronism can be obtained by attaching the string to a flexible blade.
1.6. Cylinder Rolling on a Moving Tray
[Statement p. 18]
Let Oxyz represent a Galilean frame, C the center of the cylinder with abscissa X, H the contact point of the cylinder on the tray (see Fig. 1.10). Considering this point as belonging to the tray, its velocity is a˙ (imposed by the operator); considering this point as belonging to the cylinder, its ˙ velocity is X ˙ + R θ. The non slipping rolling condition imposes equality ˙ for both velocities a˙ = X ˙ + R θ. This expression gives a link between the generalized coordinate X and the generalized coordinate θ. The constraint is holonomic.
θ
C
H
a(t)
X
Fig. 1.10 Cylinder rolling without slipping on a tray driven with a motion a(t). X denotes the coordinate of the center of the cylinder and θ its rotation angle
Now consider the cylinder; its kinetic energy is given by T = 1 2 M X ˙ 2 + 1 2 I θ 2 . The system is described by one degree of freedom.
Let us choose first the θ coordinate. Taking into account the previous relation, the kinetic energy can be recast as
1 ˙ 1 T = θ ˙ + θ 2 . M a R I 2 ( ) 2 2
Using (1.1), the value for the acceleration is easily obtained:
¨θ = + A θ ( I MR 2 ) − MR¨a(t).
The only force to be considered for the virtual work is the weight (the force necessary for the rolling has already be taken into account through the relation between the X and θ coordinates (see Problem 1.5)). For a virtual displacement δθ, the center of mass altitude does not vary and the work performed by the weight is null. One deduces a vanishing generalized force: Q θ = 0. The Lagrange equation leads to:
¨θ MR = ¨a, I + MR2
¨θ which, coupled with the already quoted relation = (¨a − ¨X)/R, allows us to find the acceleration of the center of the cylinder:
I ¨X = ¨a(t). I + MR2
Let us now choose X as the coordinate. The expression for the kinetic energy is at present:
1 1 I 2 ˙ T = X ˙ 2 + (˙a − X) 2 . M 2 2R
One obtains the corresponding acceleration as:
I I A X = M + ¨X − ¨a. ( R 2 ) R2
For a virtual displacement δX, the virtual work furnished by the weight is still null, with the consequence of a vanishing generalized force Q X = 0. In this case, the Lagrange equation leads to
I I M + ¨X − ¨a = 0, ( R 2 ) R2
or, in other words:
I ¨X = ¨a(t). I + MR2
One finds the same result, as required.
1.7. Motion of a Badly Balanced Cylinder
[Statement p. 18]
The Galilean frame Oxyz is depicted in Fig. 1.11 and the angle θ is defined positively in the trigonometric sense.
z
G
a
C
θ
O
R
X
I
x
Fig. 1.11 Cylinder rolling without slipping on a horizontal plane. The center of gravity G is out of true by a distance a
The velocity component for point G can be calculated very easily:
v G = (X ˙ + a θ ˙ cos θ, a θ ˙ sin θ).
The non slipping rolling condition imposes the constraint
X ˙ + R θ ˙ = 0.
This holonomic constraint allows us to retain the angle θ as the unique coordinate (because X = −Rθ). Applying Koenig’s theorem, one obtains the total kinetic energy of the cylinder as the sum of the translational energy for the center of mass 1 2 Mv G 2 and the rotational energy in the ˙ center of mass frame which is simply 1 2 I θ 2 . Explicitly:
1 T(θ, θ) ˙ = θ ˙ 2 [ I + M(R 2 + a 2 − 2aR cos θ) ] . 2
With the help of formula (1.1), the acceleration is derived
¨θ A θ = [ I + M(R 2 + a 2 − 2aR cos θ) ] + MaR θ ˙ 2 sin θ.
The generalized force must now be calculated; the weight P is the only force that performs work during a virtual displacement δθ:
δW = P · δz G = −Mga sin θ δθ,
which is identified with the expression δW = Q θ δθ, in order to give the generalized force Q θ = −Mga sin θ. The Lagrange equation follows from (1.2):
¨θ ˙ 2 θ [ I + M(R 2 + a 2 − 2aR cos θ) ] + MaR sin θ = −Mga sin θ. (1.7)
Multiplying by θ, ˙ the Lagrange equation, A θ − Q θ = 0, can be recast ˙ in the form dE(θ, ˙ θ)/dt = 0 where E(θ, θ) is the energy function, which remains constant at the value E.
1 E = θ ˙ 2 [ I + M(R 2 + a 2 − 2aR cos θ) ] − Mga cos θ. 2
From this last equation, it is easy to deduce the velocity in terms of the coordinate
2(E + Mga cos θ) . θ ˙ = ± √ I + M(R 2 + a 2 − 2aR cos θ)
(1.8)
It is useful to discuss the problem of sign and distinguish several regimes which depend on the sign of the quantity E + Mga cos θ.
– If E < −Mga, the sign of the numerator under the square root is always negative and Equation (1.8) cannot be satisfied. No motion is possible.
˙ – If E > Mga, the sign is always positive and θ maintains a constant sign, which depends on the initial conditions. The cylinder always rolls in the same direction, the angular velocity being comprised between two extreme values.
˙ – If −Mga < E < Mga, the numerator of θ in (1.8) vanishes for two values of the angle: θ = ±θ 0 , with cos θ 0 = | E | /(Mga). The cylinder moves by oscillating between these two values where the velocity vanishes and then changes sign.
The curve which corresponds to the value E = Mga discriminating the last two regimes is called a separatrix.
All these regimes are illustrated in the upper part of Fig. 1.12.
2. To obtain the vertical component F v of the reaction force, it is necessary that it does work to which end one must introduce another coordinate which allows such work. Thus the ordinate of the center C is no longer considered to be a constant R but rather a new coordinate q, subject to a virtual variation. In contrast, the cylinder radius is still R and the constraint relation remains unchanged. The Cartesian coordinates for
point G are changed to x G = X + a sin θ, z G = q − a cos θ.
We recalculate the total kinetic energy as:
37
1 1 T(q,˙ θ, θ) ˙ = + 2aq˙ θ ˙ sin θ θ ˙ 2 [ I + M(R 2 + a 2 − 2aR cos θ) ] . M q + 2 [ ] 2
˙ θ
4 3 2 1 0 −1 −2 −3 −4
−π
−4 3π
−2 π
−4 π
0
π
4
π
2
3π
π
4
F h /Fv
2
1
0
−1
θ
θ
˙ Fig. 1.12 Upper part: phase portrait θ(θ, E) for several values of the energy E. The outermost curve corresponds to maximum energy just before the cylinder takes off
Lower part: ratio between the tangential and normal components of the reaction force. This ratio must be less than the f coefficient.
The parameters are: M = R = g = 1, a = 0, 3, I = 0, 4.
The acceleration relative to q can now be obtained:
¨θ ˙ 2 A q = M ¨q + a sin θ + a θ cos θ . [ ]
For a virtual displacement δq, the weight produces an amount of work −Mgδq and work corresponding to the reaction force an amount F v δq (it acts so as to oppose the weight). Hence, we obtain the generalized force Q q = (F v − Mg). The Lagrange equation, A q = Q q , in which we substitute q = R (¨q = 0) finally provides the reaction force:
¨θ d 2 cos θ F v (θ, E) = M sin θ + θ ˙ 2 cos θ) 2 . g + a( = M g a [ ] [ dt ]
The energy dependence for the F v component is obtained using the rela˙ ¨θ tion θ(θ, E) given by (1.8) and the expression (θ, E) from (1.7). Moreover the condition F v > 0 must be satisfied in order to keep contact with the ground; this is an effect of a centrifugal force which is too strong. The outer curve of the upper part of the Fig. 1.12 corresponds to an energy ˜E responsible for the critical situation F v (π, ˜E) = 0. For a greater value of the energy, the cylinder no longer stays on the ground and all the previous equations are meaningless.
3. We now investigate the horizontal component F h of the reaction force. In order to make it perform work, we have to consider the X coordinate as an independent coordinate no longer connected to θ by a constraint relationship. The Cartesian coordinates of point G are, in this case,
x G = X + a sin θ, z G = R − a cos θ. The total kinetic energy becomes:
1 1 T(X, ˙ θ, θ) ˙ = X ˙ 2 + 2aX ˙ θ ˙ cos θ θ ˙ 2 [ I + Ma 2 ] . M + 2 [ ] 2
The acceleration relative to the X coordinate is deduced from (1.1)
A X = M ¨X + a d( θ ˙ cos θ)/dt . [ ]
During a virtual displacement δX, the only work comes from the F h force: δW = F h δX (F h as given here includes its sign which can be positive or negative); the value of the generalized force is derived at once: Q X = F h . ˙ The Lagrange equation A X = Q X , in which one inserts X ˙ = −R θ, gives the expression for the component of the reaction force:
d . F h (θ, E) = M ˙ θ(a cos θ − R) dt [ ]
The ratio between the tangential and normal components of the force is plotted in the lower part of Fig. 1.12. For a given friction coefficient f, characteristic of the materials, the energy must be such as to allow this ratio to be less than < f.
1.8. Free Axle on a Inclined Plane
[Statement p. 19]
Let us use the convention for axes proposed in the statement. The natural frame is defined by the inclined plane, with origin A, with horizontal axis AX, axis AY in the direction of steepest slope, and axis AZ normal to the plane. The coordinates of the center of the axle, O, are denoted X and Y . Let K be the unit vector normal to the plane, u the unit vector along OC and v the unit vector of the plane perpendicular to OC. Of course, one has OC = (L/2)u. Starting with AC = AO + OC the velocity for point ˙ C follows: V C = V O + (L/2) θ v. In the following discussion, to obtain a quantity relative to wheel C ′ , it is enough to change L → −L and φ → φ ′ in the corresponding quantity relative to the wheel C (see the configuration in Fig. 1.13).
C
v
θ
u
Y
O
K
C’
g sin α
A
X
Fig. 1.13 Position of the axle and the two wheels on the inclined plane. The axle center O is specified by its two coordinates X, Y and the axle direction by the angle θ made with the horizontal
1. The axle being massless, only the wheels contribute to the kinetic energy. The instantaneous rotation vector for wheel C is ω = θ ˙ K + φ u. The kinetic energy for this wheel is T C = 1 2 mV C 2 +T C (r) . The rotational energy T C (r) is calculated using ω, the moments of inertia of the wheel and the translational kinetic energy using the expression for the velocity given previously. Thus, one obtains:
1 1 T C = + (L 2 /4) θ ˙ 2 + L θ ˙ v · V O θ ˙ 2 + I 3 φ 2 m + I VO 2 2 [ ] 2 [ ]
with a corresponding expression for the other wheel. With V O 2 = X ˙ 2 +Y 2 , the kinetic energy of the system, which is the sum of the kinetic energy of the two wheels, is calculated as:
1 1 + I + mL2 I3 . T = m X ˙ 2 + Y ˙ 2 θ ˙ 2 + φ 2 + φ ( ) ( 4 ) 2 ( )
From this expression and from definition (1.1), one deduces the accelerations
A X = 2m¨X,
A Y = 2m¨Y ,
¨θ 1 Aθ =2 I+ , mL2 ( 4 )
¨φ Aφ =I 3 ,
¨φ ′ A φ ′ =I 3 .
We may now express the conditions for rolling without slipping.
Let H be the contact point of the wheel C with the plane; the required conditions impose V H = 0. This last velocity is calculated from that of C and from the instantaneous rotation vector: V H = V C +ω×CH. The non slipping rolling condition provides two scalar conditions. The same thing is applied to wheel C ′ . Among these four conditions, two of them are identical (those referred to (1.9)). Finally, one has three constraint equations:
X ˙ cos θ + Y ˙ sin θ = 0;
(1.9)
1 Y ˙ cos θ − X ˙ sin θ + θ ˙ + R φ = 0; L 2
(1.10)
1 Y ˙ cos θ − X ˙ sin θ − θ ˙ + R φ = 0 L 2
(1.11)
After simple elimination, these conditions can be recast in the simpler form:
2X ˙ − R φ + φ sin = 0; θ ( )
(1.12)
2Y ˙ + R φ + φ cos = 0; θ ( )
(1.13)
L θ ˙ + R φ − φ = 0. ( )
(1.14)
The conditions (1.14) (holonomic) and (1.12), (1.13) (non holonomic) are the relations required for a non slipping rolling motion. We are faced with 5 generalized coordinates X, Y, θ, φ , φ ′ and 3 differential conditions of type (1.5):
5 ∑ Λ i (k) δq i = 0.
The vectors Λ (k) possess the following components:
Λ (1) = (2, 0, 0, −R sin θ, −R sin θ);
Λ (2) = (0, 2, 0, R cos θ, R cos θ);
Λ (3) = (0, 0, L, R, −R)
2. The application points for the reaction forces due to the ground are not displaced during a virtual displacement, because of the non slipping rolling condition (see the reasoning of Problem 1.5); these reaction forces do not imply generalized forces. The only non vanishing virtual work comes from the weight; it is calculated from the displacement of the center of mass O. We easily get δW = −2mg δz O = −2mg sin α δY . The only non vanishing generalized force is thus Q Y = −2mg sin α.
Introducing three Lagrange multipliers λ 1 , λ 2 , λ 3 , the constrained Lagrange equations (1.6) are written
m¨X = λ 1 ;
(1.15)
m¨Y = −mg sin α + λ 2 ;
(1.16)
¨θ 1 2 I + = Lλ 3 ; mL2 ( 4 )
(1.17)
¨φ I 3 = −λ 1 R sin θ + λ 2 R cos θ + Rλ 3 ; ¨φ ′ I 3 = −λ 1 R sin θ + λ 2 R cos θ − Rλ 3 .
(1.18)
(1.19)
The interpretation of the Lagrange multipliers
The right hand side of Equation (1.6), multiplied by δq i and summed over i, gives, in the case of only one constraint:
∑ Q i δq i + ∑ λΛ i δq i .
The last term, which is null because of the constraint, takes the form of a virtual work, product of the force responsible for the constraint by the displacement of the application point. In our problem, the expression corresponding to the multiplier λ 1 is λ 1 (2δX − R (δ φ + δ φ ′ ) sin θ). It produces the work performed by the sum of the horizontal components of the reaction force λ 1 to cancel the horizontal displacement δX −Rδ φ sin θ of the first wheel and δX − Rδ φ ′ sin θ of the second wheel with respect to the plane.
λ 2 is interpreted as the sum of the components of reaction forces along OY , and, lastly, λ 3 is interpreted as the difference of the components along v which acts against the axle rotation.10
4. Let us make the proposed change of variables. After some rearrangements θ ˙ + R δ ˙ = 0; and from (1.18–1.19) λ3 based on ¨δ (1.14), one obtains L = (I 3 /2R) . Using (1.17), one arrives at
¨θ 1 2I + + (I 3 L 2 /2R 2 ) = 0, mL2 [ 2 ]
¨θ ¨δ or = 0, then = 0. The axle spins with a constant angular velocity ω and, with a convenient choice of the time origin, θ = ωt. It follows that:
δ − δ 0 = −(Lω/R) t and λ 3 = 0.
Let us derive the two non-holonomic constraints (1.10), (1.11). Using the proposed variables, after some algebra, one obtains
¨σ ( I 3 + mR 2 ) = mgR cos α cos(ωt),
which can be integrated to give
4Γ 2 V0 t , σ − σ 0 = − cos(ωt) Rω ( R )
where Γ = 4 1 ( mgR 2 sin α ) / ( I 3 + mR 2 ) .
Other results are obtained with no particular difficulty. Let us summarize the solution of the problem
θ(t) = ωt;
Lω t; δ(t) − δ 0 = R
4Γ2 V0 mgR2 sin α t , σ(t) − σ 0 = − cos(ωt) − with Γ = Rω R 4 (I 3 + mR 2 )
V0 Γ X(t) = (2ωt − sin(2ωt)) ; cos(ωt) 1 + ω [ ωV 0 ]
V0 Γ Y (t) = (cos(2ωt) − 1) , sin(ωt) + ω [ ωV 0 ]
δ 0 and σ 0 are two integration constants which fix the initial values of the angles φ and φ ′ .
10
In order to find each reaction force separately, a relation is missing. In fact, very much as in a hyperstatic system (a table with four legs or more on the ground) it is impossible, without further information, to obtain the distribution of the reaction forces (reaction force on each leg).
If we move in a frame which drifts horizontally with constant speed 2Γ/ω, we recognize a periodic trajectory which passes through the four partic-
ular points (0, 0), (−2, 0), (−1, 1 − 2Γ/(ωV 0 )), (−1, −1 − 2Γ/(ωV 0 )).
In the inclined plane, the trajectory of the axle center is plotted in Fig. 1.14 in units of V 0 /ω and for several values of the Γ parameter. The line of steepest slope is directed downwards while the horizontal is from left to right.
Fig. 1.14 Trajectories of the axle center for several values of parameter Γ in the inclined plane
Finally, the reaction forces are obtained quite easily
λ 1 (t) = m (−ωV 0 cos(ωt) + 4Γ sin(2ωt));
λ 2 (t) = m (−ωV 0 sin(ωt) − 4Γ cos(2ωt) + g sin α);
λ 3 (t) = 0.
1.9. The Turn Indicator [Statement p. 21]
We will refer to Fig. 1.15 for the axis and frame conventions. The flywheel ˆ rotates around Y with a constant angular velocity Ω. The frame rotates ˙ ˆ with respect to the plane XOz with the instantaneous rotation vector θ X. Finally, the plane XOz itself rotates with respect to the Earth’s frame of reference (assumed to be Galilean) with the instantaneous rotation vector ωˆ z . The instantaneous rotation vector of the flywheel with respect to the ˆ Galilean frame is thus ω = ωˆ z + θ ˙ X ˆ + ΩY . This vector is projected onto the axes (XY Z), which are the principle axes of the flywheel, to obtain the components: ω X , ω Y , ω Z .
Fig. 1.15 Gyroscope in rotation around the axis Y ′ Y of its frame with an imposed constant velocity Ω. The frame can oscillate also around an axis X ′ X locked on a tray rotating at constant velocity ω. The only degree of freedom is the angle θ between the normal to the frame Z ′ Z and the rotation axis Ox of the tray
The rotational kinetic energy is equal to
1 T rot = ω X 2 + I Y ω Y 2 + I Z ω Z 2 ) . IX 2 (
To this energy one must, in principle, add the center of mass kinetic energy of the flywheel. Since this energy is independent of θ, it is of no consequence for our study. Performing the calculations with the previously obtained components of ω one obtains:
˙ 1 1 T = I X θ 2 + ω 2 cos 2 θ . + I (Ω sin ω θ)2 2 ( ) 2
2. The only dynamical variable is θ and the system has only one degree of freedom, since ω and Ω are imposed variables. The acceleration is calculated from (1.1):
¨θ 1 A θ = I X + IωΩ cos θ + (I X − I) sin(2θ). ω2 2
The forces acting on the flywheel are the weight and the efforts exerted on the axis by the frame and the restoring force of the spring. For a
virtual displacement δθ, the center of mass altitude does not vary and the work of the weight vanishes. The axis X ′ X does not change its direction and the forces that maintain it perform no work. Lastly, the restoring force performs an amount of work δW = −C δθ = −kθ δθ; this implies a generalized force Q θ = −kθ. The Lagrange equation A θ = Qθ leads to the differential equation:
¨θ 1 I X + IωΩ cos θ + (I X − I) sin(2θ) = −kθ. ω2 2
¨θ 3. At equilibrium, one has θ = const ⇒ = 0. If the terms in ω 2 are neglected with respect to ωΩ, the previous equation gives the required relation:
kθ . ω =IΩ cos θ
4. The lift (component of the air reaction force perpendicular to the relative velocity) is perpendicular to the wings and is thus directed along the apparent vertical. In the frame of the aeroplane, this lift balances the weight (vertical) and the centrifugal force (horizontal). A simple drawing shows immediately that tan(α) = centrifugal force/weight, or (R being the radius of the circle corresponding to the turn):
V2 ωV = . tan α = Rg g
5. The restoring torque is exerted between the apparent vertical and the normal to the frame; the angle between these two directions is θ. However the angle between the instantaneous rotation vector ω and the true vertical is now θ + α; this is precisely the angle which appears in the expression of the kinetic energy whose value is presently
1 1 T = θ ˙ 2 + ω 2 cos 2 (θ + α) . IX + I (Ω sin(θ + ω α))2 2 ( ) 2
The rest of the calculation is similar to that quoted in questions 2 and 3. We arrive at the following result:
kθ . ω =IΩ cos(θ + α)
To obtain a maximum sensitivity, one requires that a small speed variation leads to a large variation of the reading; this happens when cos(θ+α) is maximum. In the vicinity of θ ∼ −α, the relation ω(θ) becomes linear = and
∼ kα ∼ k tan α | ω | = IΩ = IΩ ,
or, with the result of the last question, ω = kωV/(IΩg), that is:
ΩI V = . k g
Consequently, the rotational direction and the speed of the engine must be correctly chosen in order that the flywheel plane is as close as possible to the true vertical; this is obviously not the situation depicted in the drawing of the statement!
1.10. An Experiment to Measure the Rotational Velocity of the Earth
[Statement p. 22]
We still refer to the Fig. 1.15 of Problem 1.9 since we deal with the same system, a gyroscope, but employed in a different context.
With respect to its frame OXY Z, the disc rotates around the axis OY with angular velocity Ω which, following the statement notation, is simply ˙ φ . The frame itself rotates with angular velocity θ around the axis OX. Lastly, the carousel rotates around axis Oz with angular velocity ω with respect to a Galilean frame.
As a consequence, the instantaneous rotation vector is Ω = ωˆ z + θ ˙ ˆ X+ φ Y ,
which can be rewritten in terms of its components in the frame of the
inertial axes of the disc
ˆ Ω = Ω X X ˆ + Ω Y Y ˆ + Ω Z Z.
Simple projection leads to the expression Ω =
ˆ ω cos θ Z.
1
θ ˙ X ˆ + (ω sin θ + φ
)Y ˆ +
The kinetic energy, written first as T = recast as
2
[ IΩX 2
IΩ Y 2 + I Z ΩZ 2
] , is
1 T = θ ˙ 2 + ω 2 sin 2 θ φ + ω cos θ I Y . I + 2 [ ( ) ( ) 2 ]
The rotation ω and the pivoting motion θ(t) are imposed externally. The only coordinate describing the system is thus φ . As a result, the acceleration is
d cos + . Aφ =I Y φ ω θ dt ( )
Let us perform a virtual displacement which consists, at a given time, of a small rotation δ φ for the disc while maintaining the frame fixed.
The weight performs no work since the center of mass is kept fixed; the forces that lock the frame do not furnish work since the frame does not move. Finally, if we suppose a frictionless rotation of the disc around the axis OY , the reaction force on the rotational axis also does not produce work. In summary, the virtual work of external forces vanishes and the generalized force Q φ is null.
Be careful: In a real displacement, the momentum of the forces exerted on the frame is not null.
The Lagrange equation A φ = Q φ = 0 leads to the interesting conclusion φ +ω cos θ = const. At the initial time t = 0, one has θ = 0 and φ = 0 (the disc is at rest with its axis in the vertical position), whence ω = const. We are led to the desired expression:
φ (t) = ω (1 − cos θ(t)) .
When the axis is horizontal (θ = π/2) one obtains
complete turn (θ = π) φ = 2ω.
φ
= ω and after a
2. It is always possible to choose the carousel axes OXY z with the axis Oz along the true vertical, the axis OX in the southerly direction and axis OY in the easterly direction. The instantaneous rotation vector of ˆ the carousel is now given by (ω sin λ) zˆ − (ω cos λ) X. The rest of the treatment is completely similar to that of the previous question. The instantaneous rotation vector for the disc is therefore written:
Ω = (ω sin λ)ˆ z + ( θ ˙ − ω cos λ) X ˆ + φ Y .
With respect to the previous study, it is sufficient to replace ω by ω sin λ and θ ˙ by θ ˙ − ω cos λ. We then arrive at the equation φ + ω sin λ cos θ = const, which, using the initial conditions, leads to the final expression:
φ (t) = ω sin λ (1 − cos θ(t)) .
After a complete turn, the angular velocity of the disc is: φ = 2ω sin λ. There is no effect at all at the equator, whereas the effect is maximum at the pole.
Remarks:
– In contrast to some other problems, the Lagrangian formalism is much more convenient here than a treatment “à la Newton”.
– Mistakes are not reserved to beginners. A.H Compton, Nobel prize, who imagined this experiment, found a result which was half the exact value.
1.11. Generalized Inertial Forces
[Statement p. 23]
The previously derived formula
d A i = T) − ∂ q i T = ∑ m α a α · (∂ q i r α ) (∂q ˙ i dt α
is valid in any frame. On the other hand, Newton’s formula m α a α = f α (v) , in terms of the true force, is valid only in a Galilean frame. If we want to work in the frame under consideration, we must replace this last equation (e) by m α a α = f α (v) + f α (e) + f (cor) α + f (cent) α where f α = −m α a α (e) is the driven inertial force of the origin, f (cor) α = −m α a (cor) α is the inertial Coriolis force and f (cent) α = −m α a (cent) α is the inertial centrifugal force. Substituting these values in the expression A i and performing virtual displacements, the virtual work can be expressed in the form
∑ A i δq i = δW (v) + δW (e) + δW (cor) + δW (cent) . i
The work
δW (u) ∑ i
=
Qi (u)
δqi
is that due to the generalized force of type u:
Q i (u) = ∑ f α (u) · (∂ q i r α ). α
Translation case
In case of pure translation, the instantaneous rotation vector is null ω = 0. This means that the Coriolis and centrifugal forces are also null; there remains only the driven inertial force. Moreover, the corresponding (e) acceleration is independent of the point mass: a α = a (e) ; ∀α. The generalized force is derived as:
Q i (e) = −a (e) · ∑ m α (∂ q i r α ), α
or ∂q i ∑ [ ( α )]
Qi (e)
=
−a(e)
·
mα
rα
.
Introducing the center of mass coordinate
1 R cm = ∑ m α r α , M α
one easily arrives at the desired formula:
∂Rcm Q i (e) = −Ma (e) · . ∂qi
Application to the pendulum:
Let us direct the vertical downwards and let us specify the direction of the pendulum, with mass m and length l, by the angle θ. The point of suspension, A, is subject to a variation OA = h(t). In the system of ˙ reference of the pendulum, the kinetic energy is simply T = 1 2 ml 2 θ 2 and the generalized force due to the weight is Q (v) = −mgl sin θ. However, one must add to this force the driven inertial force. Since a (e) = (0, ¨h) and (∂R cm /∂θ)= (l cos θ, −l sin θ), the inertial force is easily derived as Q (e) = ml¨h sin θ. The Lagrange equation in the pendulum frame is written, after simplification:
¨θ ¨h(t) g + sin θ = 0. l
Uniform rotation case
In the case of a uniform rotation around a fixed point taken as origin, the driven acceleration of the origin vanishes and the instantaneous rotation vector ω is constant, with the consequence dω/dt = 0. Students often forget the term dω/dt in generalized forces which vanishes only in the case of a uniform rotation; this is often a good approximation (the Earth’s rotation around its axis, or the revolution of the Earth around the sun), but not a general situation. With our hypothesis, we are faced with two inertial forces:
– the Coriolis force: f (cor) α = −2m α ω × v α where v α is the velocity of the point α in the considered frame (relative velocity);
– the centrifugal force: f (cent) α = −m α ω × (ω × r α ).
They give rise to two generalized forces Q (cor) i and Q (cent) i .
We consider first the Coriolis force
Q (cor) i = −2 ∑ m α (ω × v α ) · (∂ q i r α ) = −2 ∑ m α [ω, v α , ∂ q i r α ] , α α
using the notation [ ] for the mixed product.
Let us introduce, in our system of reference, the angular momentum with respect to an arbitrary point O chosen on the axis: L= ∑ α r α × m α v α , which, with the known relation v α = dr α /dt = ∑ i q˙ i (∂ q i r α ) + ∂ t r α , allows us to write
ω · L = ∑ m α q˙ i [ω, r α , ∂ q i r α ] + ∑ m α [ω, r α , ∂ t r α ] . α,i α,i
Then
∂ q ˙ i ∑ α
(ω
·
=
mα
[ω,
rα
,
∂q i
rα
].
We take the total derivative with respect to time, in which we put dr α /dt
= v α and d(∂ q i r α )/dt = ∂ q i v α , to find
d [∂q ˙ i (ω · L)] = m α [ω, v α , ∂ q i r α ] + m α [ω, r α , ∂ q i v α ] . dt ∑ ∑ α α
Moreover, one has
∂ q i (ω · L) = ∑ m α [ω, ∂ q i r α , v α ] + ∑ m α [ω, r α , ∂ q i v α ] . α α
It then suffices to take the difference between these two last equations to find:
· L) d ∂(ω ∂(ω · L) Q (cor) i = − . ∂ q i dt ( ∂ q ˙ i )
Note that it is not necessary to suppose a uniform rotation; if a term ω ˙ is present in the Coriolis force, it appears in the term d[∂ q ˙ i (ω · L)]/dt and the previous formula is still valid.
Let us consider now the centrifugal force
Q (cent) i = − ∑ m α [ω × (ω × r α )] · (∂ q i r α ). α
A well known formula in vector analysis gives ω × (ω × r α ) = (ω · r α )ω −ω 2 r α and allows us to write:
Q (cent) i = − ∑ m α [(ω · r α ) (ω · (∂ q i r α )) − ω 2 r α · (∂ q i r α )]. α
On the other hand, the driving rotational kinetic energy (energy of the coincident points) is
1 T = ∑ m α (ω × r α ) 2 . 2 α
After derivation, one obtains
∂ q i T = ∑ m α (ω × r α ) · (ω × ∂ q i r α ) . α
Finally, let us use the vectorial property
(ω × r α ) · (ω × ∂ q i r α ) = ω 2 (r α · (∂ q i r α )) − (ω · r α ) (ω · (∂ q i r α )) .
Then:
∂T
Q (cent) i ∑ ∂qi α