Mechanics 1 - OER@AVU - African Virtual University

Transcription

Mechanics 1
Mechanics 1
By Adolphe Ratiarison
African Virtual university
Université Virtuelle Africaine
Universidade Virtual Africana
African Virtual University Notice
This document is published under the conditions of the Creative Commons
http://en.wikipedia.org/wiki/Creative_Commons
Attribution
http://creativecommons.org/licenses/by/2.5/
License (abbreviated “cc-by”), Version 2.5.
African Virtual University Table of contents
I.
Mechanics 1_______________________________________________ 3
II.
Prerequisite Course or Knowledge_ _____________________________ 3
III. Time_____________________________________________________ 3
IV. Materials__________________________________________________ 3
V.
Module Rationale_ __________________________________________ 4
VI. Content___________________________________________________ 4
6.1 Overview ____________________________________________ 4
6.2 Graphic Organizer______________________________________ 6
VII. General Objectives_ _________________________________________ 8
VIII. Specific Learning Objective(s)__________________________________ 9
IX. Learning Activities__________________________________________ 11
X.
Key Concepts_ ____________________________________________ 16
XI. Required Readings_________________________________________ 17
XII. Essentiel Resources_ _______________________________________ 25
XIII. Useful links_______________________________________________ 35
XIV. Learning Activities__________________________________________ 49
XV. Synthesis of the Module____________________________________ 218
XVI. Summative Evaluation______________________________________ 220
XVII.References_ _____________________________________________ 227
XVIII. Grades and results of student’s evaluation_ ____________________ 230
XIX. Main Author of the Module _ ________________________________ 231
XX. File Structure ____________________________________________ 232
African Virtual University I. Mechanics 1
Pr Adolphe RATIARISON
II. Prerequisites / Previous knowledge needed
To follow this module, the students should possess knowledge of the following
concepts:
•
•
•
•
•
•
•
•
•
Space, time, translation, rotation, reference system,
change of coordinates
cartesian coordinates
cylindrical coodinates
spherical coordinates
polar coordinates
vector field paths
velocity fields
differentiation
integration
III. Times
120 hours
IV. Materials
•
•
•
•
Internet access
Computer with CDROM
Adequate software (MS office)
Television
African Virtual University V. Module Rationale
This module is part of a training program for teachers. It helps to lay the basic
skills acquired in the secondary school teaching system.
It addresses the movement of objects that are crucial in the physical universe.
The description of these movements is the essential work of physicists in the
development of science since Galileo and Aristotle.
This module helps the learner to better understand the laws governing movement.
VI. Content
6.1 Brief summary
This module of Mechanics 1 addresses aspects experienced in daily life and in
our environment including:
•
•
•
•
•
Physical quantities and vector operators;
Kinematics of a material point in one dimension and two dimensions:
- Research of parametric equations and trajectories of a moving object
- Calculation of velocity and acceleration vectors in different coordinate
systems
- The composition law of velocities and accelerations
Static solids (forces acting on a system)
The dynamics of material points using Newton’s laws
The concepts of Work, Energy, Power, Mechanical theorem of kinetic
energy and the conservation of mechanical energy.
African Virtual University This module comprises of 4 units :
Unit 1 : (15 hours).
The measurable quantities of physics.
•
•
•
•
classification and measurement, and errors in measurement,
vectors,
scalars,
vector operations.
Kinematics of a material point :
•
•
One-dimensional movement,
Two and three-dimensional movement
•
Statics of solids
•
•
•
•
Composition law of movement
Dynamics of a material point
Work, energy, and mechanical power
Oscillators
African Virtual University 6.2 Graphic organizer
Mechanics 1 Module
Scalars and Vectors
General
physics
Vector
Operations
Uniform movement
Straight movement
Varied uniform movement
Sinusoidal movement
Circular movement
Kinematics of a
material point
Cycloidal movement
Curvilinear
movement
Helical movement
Cylindrical coordinates
Coordinate
systems
Solid
equilibrium
Torques
Center of gravity
Equilibrium
conditions
Spherical coordinates
Dynamics of a
material point
Galilean
reference
Non-Galilean reference
Composition of movements
Newton’s 3 laws
Equilibrium
stability
Kinetic moment
Work-Energy-Power
Oscillators
5
African Virtual University Parametric
equations
Equatio ns pa
ra métr iqu es
x=x=
x(t);x(t);
y=y(t)y=y(t)
f(x,y)=Cte
f(x,y)=Cte
Trajectory
Equatio n d eequation
l a traj ectoire
•
Composante sof
d espeed
l a vit esse
Components
G énér alités sur la
Generalities
cinématique du
Of
point
kinematics
Composante s d e l’ a ccélération
Components of acceleration
Composante s int r ins èques d e l’ a ccélération
Intrinsic components of acceleration
Composante s d e l a vit esse et d e
l’ac céléra tion danof
s diffé r velocity
ents syst èmes deand
Components
coordonn ées
acceleration
systems
in
different
coordinate
Work of a constant force
Work of conservative forces
Work
Energy
Power
Work of non-conservative forces
Kinetic energy theorem
Mechanical energy theorem
•
x ( t ); y( t )
••
••
x ( t ), y ( t )
dv v 2
;
dt
R
African Virtual University VII. General objectives
The student should be able to:
1. learn about
-
-
-
Physical quantities,
Concepts of energy and work
Relation between energy and work
2. learn and apply the following concepts :
-
-
-
-
-
One-dimensional movement
Two and three-dimensional movement
Newton’s 3 laws
Kinetic energy
Working in a group to find solutions to an exercise
3. Students should be able to set, carry out experiments and analyse experimental data to establish relations between physical quantities
African Virtual University VIII. Specific objectives related to learning activities
Units
Specific objectives
Unit 1: (15 H)
Measurable quantities of physics.
- Their classification and nature,
- Sources of error in measurement,
- Vectors,
- Scalars,
- Vector operators
Specific learning objectives
- Define a physical quantity
- Give examples of physical quantities
- Give units of physical quantities
Specific objectives of theoretical knowledge
- Distinguish a vector quantity from a scalar quantity
Unit 2 : (30 H)
Kinematics of a material point :
- One-dimensional movement,
- Two or three-dimensional
movement
Learning objectives-Define the trajectory of a moving object
- Define average velocity
- Recall the components of acceleration vectors in a (o, x, y,
z) coordinate system
- Write the parametric equations of movement.
- Calculate average velocity of a moving object.
- Calculate the instantaneous velocity of a moving object.
- Calculate average acceleration of a moving object.
- Calculate the instantaneous acceleration of a moving
object
- Integrate the instantaneous velocity
- Integrate the instantaneous acceleration
- Trace the trajectory of a moving object
- Caclulate the intrinsic (local) components of the
acceleration
Specific objectives of group work
- Accomplish an exercise with group members
Unit 3 : (30 hours)
- Document forces acting on a body
- Determine equilibrium conditions of a solid :
- in rectangular translation
- in rotation
Statics of solids
African Virtual University 10
Unit 4 : (45 hours)
- The law of movement composition- Dynamics of material points –
- Work, energy, and mechanical power –
- Oscillators
- Newton’s 3 laws
- Apply Newton’s laws to solve problems
- Apply the kinetic energy theory
- Calculate the work done by a constant force
- Calculate the work done by a varying force
- Calculate gravitational potential energy
- Calculate the kinetic energy of a moving object
- Calculate the mechanical energy of a system
- Apply the kinetic energy theorem
- Apply the conservation of mechanical energy theorem to a
system
Specific learning objective (Optional
Be able to use two forms of evaluation
educational material)
Note when evaluations will take place
IX.
Teaching and learning activities
9. Preliminary evaluation
Title of preliminary evaluation: MECHANICS TEST 1
Justification : This test aims to evaluate knowledge that will be essential to
understand this module.
QUESTIONS
1. Movement means the ……………………………of a moving object from
one place to another.
2. We call the trajectory of a moving object the set of points that a moving object
follows when the time t…………..
3. A vector is a straight line :
a. True
b. False
4. Velocity is a scalar :
a. True
b. False
5. Accelerating a car causes the scalar product of its velocity and its acceleration
to be positive:
a. True
b. False
6. We apply a force to :
a. produce movement of an object
b. modify the movement of an object
c. deform an object
Check the correct answer(s)
7. You are in a car approaching a curve with velocity v. When the car
7-1. turns to the left, your body is thrown to the left :
a. true
b. false
7-2. turns to the right, your body is thrown to the left :
a. true
b. false
8. When a driver of a car moving with a speed ,v, on a level road, brakes suddenly,
a. passenger in the car is thrown to a. the front
b. the back
c. the left
d. the right
Check the correct answer(s).
9. Two vectors which are equal in intensity have between them an angle a. R denotes the resultant of two vectors and U the common module of these two vectors.
Associate the correct answers by matching the letter and number
a. a = 90 °
1. R=0
b. a = 0 °
c. a = 180 °
2. R = U 2
3. R=2U
d. a = 45°
4. R = U 3
10. A brick of mass M placed on a smooth table is :
a. subject only to the action of its own weight
b. not exposed to any force
c. exposed only to the reaction of the table
d. exposed to its own weight and the reaction of the table
Choose the correct answer.
11. The product of force and distance is :
a. a vector
b. a scalar
12. Kinetic energy and work have the same units
a. True
b. False
13. Power is the integral of energy
a. True
b. False
14. Gravity is a force
a. True
b. False
15. Mass is a scalar
a. True
b. False
Correct Answers
1. Movement means the displacement of a moving object object from one place
to another. Movement is essentially related to displacement.
2. We call the trajectory of a moving object the set of points that a moving object follows when the time, t varies. Good answer, in fact displacement is
related to time.
3.
a. Read the question carefully before answering
b. Very good. A vector is always oriented and has a measure which is not the
case of a ray.
4.
a. We are talking about a velocity vector, therefore it cannot be a scalar.
b. Very good, velocity is a vector.
5. True. For an accelerated movement, the scalar product of the acceleration
and the velocity is positive
6.
a. Correct. A force can displace objects.
b. Correct. A force can also deviate the movement of an object.
c. Correct. In fact, if we wanted to break an object, we could apply a force to it.
7.
7-1.
a. Watch out, think carefully.
b. Correct. You understand that we are ultimately thrown in the opposite direction.
7-2.
a.
Correct. We are thrown in the opposite direction.
8.
a. Correct. We are thrown in the opposite direction.
b. Incorrect
c. Incorrect
d. Incorrect
9. a2 ; b3 ; c1. are the correct answers, and you know how to calculate the result
of two forces.
For all other answers (a1, a3, b2, b1, c2, c3, d1, d2, d3, d4, a4, b4, c4,) : Think
before answering.
10.
a. Incorrect.
b. Think about it, it is impossible
c. A solid in equilibrium is subjected to at least 2 forces
d. Correct. You are recalling action and reaction.
11.
a. Watch out. Work is a scalar product of two vectors, but is not itself a vector.
b. Correct. The work of a force is a scalar.
12.
a. Correct. Work is a form of energy.
b. Incorrect.
13
a. Watch out. Energy and power do not have the same units.
b. Correct. The integration of energy cannot provide power.
14.
a. Correct. Gravity is indeed a force.
b. Incorrect. Think of the units.
15.
a. Correct. Unlike gravity, mass is a scalar.
b. Reread your lesson.
Comments of students following
Evaluation test in mechanics 1
(100-200 words)
You took the pre-evaluation test.
•
•
•
•
•
If you have correctly answered all question, you have a grade of A +. You
will not have difficulty following this module.
If you answered 75% of the test correctly, you have an A grade. You will
have no difficulty following this module.
If you answered 60% of the test correctly, you have a B grade. You could
very well succeed in this module, but you put in some extra effort.
If you answered between 45% and 50% of the test correctly, you have a
C grade. You must take some supplemental courses.
If you answered under 45% of the test correct, you have the note D. You
should strive to learn everything, while still following the module as it is
taught.
X. Key concepts (glossary)
1. Acceleration: Change in velocity per unit of time
2. Freefall: Movement of an object subject only to its own weight (all resistant
forces are neglected)
3. Kinetic Energy: Energy possessed by a body in motion
4. Potential energy: Energy stored by a body due to its position
5. Force: any means capable of producing a motion, to modify or deform an
object.
6. Movement: Movement of a moving object object from one point to another
7. Mechanical power: work of a force per unit time
8. Referential: Object of reference
9. Mechanical work: Energy provided by a force when its point of application moves.
10. Speed: Change in the position of a moving object object per unit of time
XI. Required readings
UNIT 1 :
Measurable physical quantities.
-
-
-
-
-
Their classification and measure,
Different sources of error of measurement,
Vector quantities,
Scalar quantities,
Vector operators
There are four required readings for Unit 1. They are grouped in Appendix 1.
Reading #1
Complete references :
RATIARISON Adolphe (2006). Grandeurs physiques – Mesures-Incertitudesopérations vectorielles. Madagascar. Université d’Antanarivo.
The first two parts of this document are drawn from the following sites:
http://www.bipm.fr/fr/si/si_brochure/chapter1/1-2.html
http://www.cegep-ste-foy.qc.ca/freesite/index.php?id=3113
http://www.ulb.ac.be/cours/psycho/content/cognum/calcul.html
Summary : The value of a physical quantity is usually expressed as the product of a number by a unit. For a particular quantity, we can use many different
units. Among these units, we distinguish those of the International System (SI)
based on seven base quantities.
The measurement of a physical quantity can be done directly, such as the
length with the meter, the voltage with a voltmeter, or indirectly such as a surface area obtained by the product of the length by width.
Finally, the various operations on vectors are detailed.
Justification:
-
-
Any physicist must know the units of measure because we cannot add
two different sizes without expressing in the same unit.
The vector addition is not only part of the composition of forces, but it is
also vitally important in the composition of movement that we will see
later.
Reading #2
http://tanopah.jo.free.fr/seconde/Vct2.html
Addition, opposing, and subtraction of vectors.
Summary : This course and nearly all elements and programming within,
were designed and made by Jerome ONILLON. It is listed by the Irish tavern.
Addition and subtraction of two vectors are well detailed. It highlights the
properties of vector addition as: commutativity, associativity, existence of neutral elements without forgetting the Chasles relationship
Justification: This completes the reading #1 The parallelogram rule used for
addition and subtraction of vectors is well explained.
Reading #3
http://formation.etud.u-psud.fr/pcsm/physique/outils_nancy/apprendre/ chapitre2/
partie2/Title1res.htm
Vectors. Vector addition.
Summary : Vector addition is an internal composition law and has the following properties :
Associativity •
Commutativity •
Neutral element
•
Symmetric element
Hence, we can talk about subtracting a vector from another, and the Chasles
relationship.
•
The multplication of a vector by a scalar is an external composition law.and
has the following properties:
•
•
•
•
Distributivity with respect to vector addition :
Distributivity with respect to scalar addition :
Associativity :
Neutral element :
These properties are followed by :
The determination of the position of a point M on a segment AB,
•
The linear combination of two vectors.
Justification : Starting from the linear combination of several vectors, we can
define the centroid of several points affected by weights ai
•
Reading #4
http://fr.wikipedia.org/wiki/G%C3%A9om%C3%A9trie_vectorielle
Vector geometry.
Summary : We develop addition and subtraction of vectors, and multipying a
vector by a scalar. Properties of a scalar product, vector product, mixed product and double vector product.
Justification : Allows readers to gain an in-depth understanding of vector
operations.
Unité 2 : Kinematics of a material point
1D, 2D, and 3D movement
There are 3 required readings in Unit 2. They are grouped in Appendix 2.
Reading #5
RATIARISON, A. (2006). Cinématique du point. Mouvement à 1D, 2D ou
3D. Madagascar. Université d’Antananarivo. Cours inédit
Summary : The generality of the kinematic point concerns the definition of
referentials, tracking a moving object in space, the curvilinear abscissa, the
velocity vectors and acceleration vectors.
This manual will then examine the rectilinear uniform motion and uniformly
varied motion.
When considering curvilinear movement, we emphasize the intrinsic components of acceleration, circular motion, cycloidal and spiral motion.
Finally, the different coordinate systems and components of velocity and acceleration vectors in these coordinate systems are considered.
Justification: Before we study the dynamics of a material point, we must have
the kinematics of the point. To do this we need to know the topics listed above.
Reading #6
http://abcsite.free.fr/physique/meca/me_ch3.html
Kinematics of a point
Summary : This reading completes the previous calculations of components
of speed and acceleration vectors in different coordinate systems. Polar coordinates and semi-polar coordinates are also still taught.
In this reading we encounter the so-called hodograph.
The different diagrams are clearly legible.
Justification: This course is easy to read, and can offer significant help to
students.
Reading #7
http://www.chez.com/mecasite/Mecanique/cinematsol.htm
Kinematics of a point.
Summary : This reading reinforces our knowledge of movement, the average
velocity, average acceleration, instantaneous velocity and instantaneous acceleration. The rotational motion and uniform circular motion varied uniformly
are also highly developed.
Justification: In addition to the two previous readings, it completes the course
of the kinematics of the point.
UNIT 3 : Equilibrium of a solid on a horizontal plane
In Unit 3, there are 3 required readings, which are grouped in Appendix 3.
Reading #8
Complete reference :
RATIARISON, A. (2006). Equilibre d’un solide sur un plan – Faculté des
Sciences- Université d’Antananarivo –MADAGASCAR, Cours inédit
Summary : This reading is mainly concerned with the equilibrium of a solid
on a plane. A solid can slide or rotate on a plane if it is not in equilibrium. To
introduce the equilibrium of a solid, we speak of torque, which is a system of
free vectors. This system of free vectors is reduced to the resultant forces and
resultant moment of all forces applied to the system considered. The equilibrium condition is defined by a torque of zero, meaning a general zero result
and zero moment.
Justification: In the module only zero resultant forces have been defined, but
to broaden the knowledge of students we must also mention the zero moment
of the forces applied to the system in question.
Unit 3 :
Reading #9
Statique du solide taken from « http://fr.wikipedia.org/wiki/Statique_du_solide »
A Wikipedia article.
Solid Statics
Summary : The possible movements, sometimes called degrees of freedom
are of two kinds: translations (3 main directions) and rotation (around the three
directions). While the translations may not be caused by forces, rotations are
generated by moments of these forces, or other pairs of force. When the equilibrium point requires that the establishment of 3 algebraic relations (equation
of vector forces in 3 dimensions), while that of the solid demands the consideration of 3 additional equations (moments vector equation). The fundamental
principle of statics can then be considered:
1. the theorem of the resultant (sum of forces is zero).
2. the theorem of the moment (sum of moments is zero).
Justification: The study of equilibrium of a solid always requires the consideration of these 2 theorems, even if in some simple cases of mechanics of
a point, they seem to be resolved with one of the 2 parts. Generally, it is not
possible to treat the two aspects separately (forces and moments): it is actually
a complex 6-dimensional problem.
Unit 3 :
Reading #10
http://www.ac-poitiers.fr/cmrp/cpge/docs/Coursdemodelisationetdestatique.doc
Solid statics
Summary : The mechanical action is anything likely to maintain a body at rest,
create or modify a motion to deform a body, and manifests itself in two forms:
- The translational motion due to the resultant forces applied to the solid
- The rotation due to the resultant moment of these forces
Before stating the Fundamental Principle of Statics (FPS), the author speaks of
the modeling contact activities:
-
-
Contact with a fluid on a solid,
The contact of two solids.
Justification: One of the characteristics of this reading is the mechanical actions applied to a balance:
-
The mechanical action at a distance (gravity, electromagnetic, electrostatic,
...)
- The mechanical action of contact (pressure, contact, ...)
This reading is very beneficial for students.
Unit 4 :
Composition of movements
Dynamics of material pointsWork, energy, and mechanical power –
Oscillators
Reading #11
RATIARISON, A. (2006). Composition de mouvement, Dynamique du point
matériel, Travail – Energie - Puissance, Oscillateurs– Faculté des SciencesUniversité d’Antananarivo –MADAGASCAR, Cours inédit
Topics on dynamics and oscillators were taken from :
http://abcsite.free.fr/index.html
Summary : This unit begins by addressing the law of movement composition
and Newton’s 3 laws, with their practical applications.
It continues by discussing evidence on Coriolis inertial forces.
It highlights the definition and calculation of work produced by the conservative forces and that produced by non-conservative forces.
It establishes the theorem of kinetic energy and the theorem of mechanical
energy.
It concludes with the study of damped harmonic oscillators.
Justification: To get a general idea of absolute motion and relative motion, the
course begins with the generalization of different velocities and accelerations
of the three Newton laws, and theorems of mechanical and kinetic energy on
the basis of the dynamic point.
Reading #12
Papanicola Robert, http://www.sciences-indus-cpge.apinc.org/IMG/pdf/ CIN2_
DERIVATION_VECTORIELLE.pdf
Vectorial derivation.
Summary : This course of vector derivation leads to the composition law of
motion. It therefore complements the course of Ratiarison Adolphe. For the
concept of the composition of three rotations, the author brings the three Euler
angles, namely precession, nutation and proper rotation.
Justification:The three Euler angles are not on the agenda because in practice this concerns the kinematics of sound. It is therefore not essentially that
a student spends considerable time on it. The composition of rotations is well
developed in the course of Ratiarison.
Reading #13
http://abcsite.free.fr/physique/meca/me_ch3.html
Work, energy, and power
Oscillators
Summary : A complete course on the dynamics of a material point. It follows
the site : http://abcsite.free.fr/physique/meca/me_ch3.html that was already cited
in the kinematics portion.
Justification : A course that is easy to read.
Reading #14
DIOUF, S. (2004). L’Evaluation des apprentissages. Sénégal. Université
Cheikh Anta DIOP de Dakar. FASTEF (ex ENS)
Summary: This text is recommended to respond to an optional formal evaluation of educational nature. It contains different parts including:
Evaluation that deals with various issues relating to the assessment
The different forms of evaluation where it is also about the roles and moments
of assessment
Strategies for collecting information. In this section you will find ways to correct the issues relating to objective and subjective correction.
It also includes the steps of building a subject of examination and the characteristics of the evaluation.
Justification: Reading this text enables students to answer questions correctly
during a formative assessment of educational nature. All answers to the assessments are contained in this text.
XII. Essential resources
Resource #1
RATIARISON, A. (2006), Cours de mécanique générale 1
Faculté des Sciences -Université d’Antananarivo - Madagascar
Summary : This mechanics course is taught in the first year of university in
the Faculté des Sciences, at the Ecole Normale and l’Ecole Supérieure Polytechnique at l’Université d’Antananarivo.
It discusses the vector operators, and the kinematic point in a Galilean reference. The laws of composition of movements are dealt with in the kinematics
portion. Central acceleration movements are also developed.
Justification: This is useful for practical exercises from a distance.
Resource #2
Complete reference:
PEREZ, J. P. (1997). Mécanique – Fondements et applications. Université Paul
Sabatier Toulouse – France. Edition MASSON, 120 bd St Germain 75 280
Paris Cedex 06
Summary : It is a comprehensive manual for students from first to third year.
The mechanics of a material point, the dynamics of solids, the movement of
central acceleration, oscillators, analytical mechanics, and fluid mechanics are
treated.
Justification: Students will always need this book during their studies, because apart from the topics of kinematics and mechanics of a solid, there are
also fluid mechanics topics and several applications that are treated..
Resource #3
http ://www.hazelwood.k12.mo.us/grichert/sciweb/applets.htlm
Summary : This site is a compilation of links leading to other internet sites
containing physics-based simulations.
Justification : Students will always have a need for this manual during their
studies, as many physics topics are developed.
Resource #4
Complete reference:
CAZIN, M. (1995), Cours de mécanique générale et industrielle– Gautier
Villars – tome 1, NY 1003 -1995
Summary : A complete course, but slightly difficult to read. However, it
contains many applications and is very useful in further studies.
Justification : The student will find many applications to mechanics. Resource #5
Complete reference:
The Free High School Science Texts: A Textbook for High - School Students
Studying Physics.- FHSST Authors1 - December 9, 2005 -http://savannah.
nongnu.org/projects/fhsst
Summary : Many physics topics are in this manual, including mechanics,
electricity, optics, and electromagnetism.
The student can learn all about physics by consulting this site.
Justification: Students will always find a use for this book, since the physics
topics are very well developed.
Resource #6
Complete reference:
http://www.google.ca/search?client=firefox_a&rls=org.mozilla%3Aen-US%Aofficial&hl=en&q=c3%A9equilibre+d%27un+solide+sur+plan&meta
Summary : This site contains practical exercises of equilibrium of a solid on a
place, subject to many forces.
Justification: Students can do practical exercises on the site.
Resource #7
Complete reference:
http://www.chimix.com/an5/prem5/hotp5/force1htlm
Summary : This site contains practical exercises of equilibrium of a solid on a
place, subject to many forces. Visualisation of free fall and apparent motion.
Justification: Students can do practical exercises on the site.
Resource #8
Complete reference:
http://fr.wikipedia.org/wiki/statique-du-solide
Summary : This page summarizes static equilibrium of a solid.
Justification : Students can do practical exercises on the site.
Resource #9
Complete reference:
http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre2/Title1res.htm
Summary : Developed in this site:
-
-
the vector sum and its properties (associativity, commutativity, identity
element, element symmetrical), the difference of two vectors, the relationship Chasles
the product of a vector by a scalar and its properties (distributivity over
vector addition, distributivity with respect to the addition of scalar, associativity, and the existence of the neutral element).
Applications of vector addition that are found:
-
-
-
-
-
-
The position of a point on a line,
The linear combination of 2 vectors
Coplanar vectors
The centroid of n weighted points
Isobarycentre of 3 points, not aligned (center of gravity of a triangle)
Isobarycentre of 4 points in space (center of gravity of a tetrahedron)
Justification: This document is a valuable resource that can complete the course
and teach that the research of the center of gravity is a vector addition.
Resource #10
Complete reference:
http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre3/partie3
/ Title1res.htm
Summary : This document geometrically represents complex numbers. Addition and subtraction of complex numbers relates to addition and subtraction of
vectors.
Justification: This site demonstrates that addition and subtraction of vectors
has other applications in the field of science.
Resource #11
Complete reference:
http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.op_addition.
aspx
http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.op_methods.
aspx
http://mathexel.site.voila.fr/index.htlm
Summary : These documents execute the addition and subtraction of vectors
on the computer screen.
They show different operations using the « Padding » method.
Justification: The student is, with the aid of an instructor, familiarized with
vector operations.
Resource #12
Complete reference:
http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/lissajou_
j.html
http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/cycloi_j.html
http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/pndhgs_
j.html
Summary: The first document sets out a Lissajous curve corresponding to the
addition of two perpendicular sinusoidal movements of their phase.
A cycloid is the trajectory of a fixed point of a circle when it rolls without slipping on a line. The second document shows how to trace a cycloid as a function of the travel speed of the circle.
The third document highlights the oscillation of a cycloidal pendulum, which
is called the Huygens pendulum.
Justification: The visualization of these phenomena provides a very specific
vision for the student. The equations of Lissajous curves are very complex unless they are reduced to that of the ellipse. Thus, visualization of these curves
is very helpful to students. Similarly, an abstract design and a cycloid cycloidal
pendulum is difficult.
Resource #13
http://electronics.free.fr/school/article.phys3?id_article=9#5
Summary : This site belongs to a young Moroccan born February 5, 1988 who
is passionate about philosophy, mathematics and computing. That’s why he
created this site which is a medium of exchange of knowledge and experience.
This article is a summary of the kinematics of a particle in a Galilean reference. It defines:
-
-
-
-
-
-
-
The position vector
The velocity vector
The acceleration vector
The average acceleration vector
The average velocity
Cartesian coordinates of the acceleration vector
Coordinates of the acceleration vector in the Frenet reference
It develops some specific movements:
- The rectilinear motion (uniform and uniformly varying)
- The circular motion (uniform and uniformly varying)
Justification: We have already developed all of the topics, but it is interesting
to see what others are doing.
Resource #14
Complete reference:
http://www.chez.com/Mecanique/cinematipts.htm
Summary : After defining the kinematic characteristics (velocity and acceleration) the author speaks of:
-
-
-
Uniform rectilinear movements and uniformly varying
Movements of rotation, the normal components and tangential acceleration
Movements of uniform rotation and uniformly varying
Resource #15
http://www.chez.com/Mecanique/dynamiqu.htm
Summary : This site contains :
Fundamental principle of the dynamics of solids under rectilinear translation.
D’Alembert ;s principle ;
Rotational movement with respect to a fixed axis.
Resource #16
http://www.chez.com/Mecanique/energeti.htm
Summary : This site summarizes the mechanical energy of a point. The work
of a force, the work of a couple, the gravitational potential energy, elastic
energy of a spring, kinetic energy of a solid translation, the kinetic energy of
a solid rotation, the average power, the power developed by a force, and the
concept of performance are shown.
Justification : This completes our course.
Resource #17
http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/aquadiff.html
Summary : The purpose of this site is to illustrate the solution of linear differential equations of first and second order frequently encountered in physics.
The solution itself is developed in mathematics courses and will not be detailed
here. However, there is a summary of results and examples.
Illustrations and animations found here include:
- The presentation of the harmonic oscillator.
- The horizontal linear oscillator
- The elastic pendulum
- The weighted pendulum
- The tension of a pendulum wire
- The period of the pendulum weight
- The Botafumeiro
- The cycloidal pendulum
Justification : This is useful for practical exercises from a distance.
Resource # 18
Complete reference
http://www.n-vandewiele.com/TDMeca2.pdf
Summary : Seven corrected exercises concerning the composition of movement and changing reference states.
Justification: Since there are not many exercises concerning the compostion
of movement, the student should familiarize themselves with this topic from
other resources as well.
Resource # 19
http://www.ens-lyon.fr/Infosciences/Climats/Dynam-atmo/Cours-Coriolis
Summary : The purpose of this site is to supplement knowledge of the
Coriolis force. The author tries to introduce the concepts gradually so that the
Coriolis force is understandable with minimal prior knowledge.
•
•
•
This site consists of seven paragraphs:
The first shows the existence of forces of inertia and the Coriolis force.
The second gives us the mathematical expression of the Coriolis force.
•
•
The third shows some manifestations of the Coriolis force.
The fourth examines the movement of air masses moving on the ground
to see how clear the Coriolis force is.
• The fifth gives us additional information.
• The sixth offers a simple experiment to demonstrate the deviation of
trajectories in a rotating frame
• The seventh, a conclusion, examines a planet invented by Antoine de
Saint Exupéry in “Le Petit Prince”
Justification: This site is not only dedicated to the mathematical formulation
of the Coriolis force, but describes the various manifestations of the Coriolis
force in everyday life. It is useful, necessary and even essential that teachers and
students read this site.
Resource #20
http://www.ucd.ma/fs/modules/meca1/um1./modules3/cin2.htm
http://perso.orange.fr/rmchs/physique_05/cours_physique/cours_mecach5_cinematique.pdf
Summary : A course on different coordinate systems and on the laws of composition of velocities and accelerations.
Resource #21
http://www.keepschool.com/cours-fiche-les_systèmes_oscillants
Summary :
This site serves as a supplement to courses on oscillators. It talks about:
-
classification of experimental oscillators (experimental properties and
characteristic properties of oscillators)
- free mechanical oscillators (simple pendulum, horizontal elastic pendulum)
- mechanical oscillators forced (torsion pendulum in forced oscillations,
resonance phenomena)
Justification:
The course of this site is given a simple way. It can help the student understand
oscillators.
Resource #22
Complete reference:
http://www.logitheque.com/fiche.asp?I=18755
Summary : Static is an educational software used in physical science at the
high school level dealing with the static point of solid material subjected to one
or more forces (or forces couples) to show the conditions of equilibrium solids.
STATIC revolves around six themes related to statics
- Equilibrium point;
- Equilibrium of a solid rotation about an axis;
- Equilibrium of a solid subjected to couples of forces;
- Equilibrium of a rod rotating about an axis;
- Static solid on an inclined plane with or without friction forces;
- Static floaters (Archimedes principle).
Justification: The learner will deepen their knowledge on the equilibrium of a
solid rotation about an axis and of static floaters.
Resource #23
http://www.univ-lemans.fr/enseignements/physique/01/statique.htm
Summary : A brief summary of the equilibrium and dynamics of points in a
Galilean reference.
Justification : This module summarizes the different forces that can act on a
point or an object.
Resource #24
http://www.univ-lemans.fr/enseignements/physique/02/meca/couplage3.html
Summary : We have here a simulation of a coupling of 3 mechanic oscillators, neglecting friction. The independent pulsations of the oscillators are w12
= K/M1, w22 = K/M2,. w32 = K/M3.
Each mass is subjected to the restoring force of 2 springs attached to it.
The movement equations are :
If the 3 masses are equal, the solution to the system is :
To treat all cases in the program, this system of coupled differential equations
is solved numerically using the Runge-Kutta order of 4.
By assumption, the initial velocity of the two masses is always zero.
One can see that for any initial conditions the solution is usually a complex
aspect. It is a linear combination of the three proper modes.
It is of the form: Xi = Ai.cos (wpt) + Bi.cos (wqt) + Ci.cos (wrt) (i = 1, 2, 3)
The values of the constants Ai, Bi and Ci are a function of initial conditions.
Justification: The simulation work helps students to understand physical phenomena.
XIII. Useful links
Useful link #1
Title : Parabolic trajectory
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/tir_para_
j.html
Screen capture :
Justification :
We can see the envelope trajectory (parabola of safety) if we vary the direction
and intensity of the initial velocity. The learner can calculate the equation of
the parabola of safety.
Useful link #2
Title :Trajectory and free fall
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/tirchu_
j.html
Screen capture :
Justification :
This animation shows the meeting point of two moving object objects. The
learner can calculate the z component of the meeting point of the projectile and
the other object, and the time at which they will meet.
Useful link #3
Title : Movement of the moon and a solar planet.
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/
terlun_j.html
Screen capture :
Justification :
The animation shows the movement of the moon and a solar planet. The trajectories are presented in two different references.
Useful link #4
Title :Longitudal oscillations of a spring
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/prlong
_j.html
Screen capture :
Justification :
One can observe the sinusoidal signal propagation along the spring and the
displacement of a localized point M of the spring.
Useful link #5
Title : Multiple pendulums
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/pendmu
_j.html
Screen capture :
Justification :
The animation here shows the conservation of mechanical energy. You can
choose the number of pendulums N released without initial velocity and common initial angle.
Useful link # 6
Title :Coupled oscillators
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/osccpl_j.html
Screen capture :
Justification :
We have here the coupling of three springs of respective stiffness k, k0 and K.
You can vary the stiffness k0 in the second spring and we can have the proper
modes of periods T1 and T2.
Useful link #7
Title :Uniform rotational movement
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/manege_
j.html.
Screen capture :
Justification :
The student may vary the period of rotation of the carousel, the pathways of
rays Ra and Rb, and calculate the centrifugal force of inertia applied to A and
B.
Useful link #8
Title :Trajectory of a grenade
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/grenad_j.html
Screen capture :
Justification :
In this site we find another reference that is the centroid reference. We therefore study two different references (the fixed reference and the centroid reference).
Useful link #9
Title : :Oscillation with solid friction
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/frtsol_
j.html
Screen capture
Justification :
By changing the parameters b, x0 and T0, we visualize the range of equilibrium and location of extremum. The student can then write the equation of
motion.
Useful link #10
Title :Trajectory of the sun in the galaxy
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/galaxi_
j.html
Screen capture:
Justification :
The trajectory of the sun is not perfectly circular because it describes a circle
of radius R in the plane Oxy galactic center O, but it varies more along the axis
Oz, perpendicular on both sides of this plane.
Useful link #11
Title : The Big Wheel
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/gdroue_
j.html
Screen capture :
Justification:
Study of circular movement in a vertical plane.
Useful link #12
Title :Epicycle of Ptoémée
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/epiclc_j.html
Screen capture :
Justification :
The Earth T and another planet P demonstrate a circular motion around the
sun. It shows the relative motion of P relative to T.
Useful link #13
Title :Trajectory of a dog running after its master
URL : http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/chien_
j.html
Screen capture :
Justification :
By varying the velocity of the dog and its handler, it displays the path of the
dog and the student can find the equation of the trajectory based on these velocities.
Useful link #14
Title :Trajectory of 4 flies
http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/4mouche_j.html
Screen capture :
Justification :
Before finding the distance traveled by each fly, the student tries to find the
equation of the trajectory of each fly.
XIV. Learning activities
Learning activity 1
Activity title •
•
•
Vector quantities
Addition and subtraction of vectors
Vector operations
The learner must be able to:
•
•
•
•
•
•
•
•
•
•
Properly represent a physical quantity
Recall some units of physical quantities
Add vectors
Subtract two vectors
Perform scalar products
Recall the physical meaning of the scalar product
Perform vector products
Recall the physical meaning of vector products
Calculate the double vector product
Calculate the mixed product
Activity summary
The main aim of this module is the dynamics of material points subjected to
various forces that can be represented by vectors. Thus, this activity is to familiarize oneself with the common vector operators. As the movement of particles
can be movements of translation and / or rotation, we can not confine ourselves
to only addition and subtraction of vectors, but we must also investigate other
operations of vectors, such as the vector product.
Key concepts
General result: vector sum of several vectors
Scalar product: Operation vector giving the projection of a vector on the support another to a constant.
Vector product: Operation vector whose norm is equal to the surface generated by the two vectors and the vector product is directly perpendicular
to both vectors.
Mixed product: Operation vector containing both the scalar product and vector
product, scalar representing the volume generated by the 3 vectors.
Double vector product: Twice the vector product.
Appropriate readings
(IN APPENDIX 1)
1° RATIARISON, A. (2006). Grandeurs physiques – Mesures-Incertitudesopérations vectorielles.Madagascar. Université d’Antanarivo
2° ht tp://tanopah.jo.free.fr/seconde/Vct2.html
Addition and subtraction of vectors
3° http://formation.etud.u-psud.fr/pcsm/physique/outils_nancy/apprendre/
chapitre2/partie2/Title1res.htm
Vectors. Addition of vectors.
4° http://fr.wikipedia.org/wiki/G%C3%A9om%C3%A9trie_vectorielle
Vector geometry.
Wikipedia article.
Appropriate resources
ANSERMET J.-P. (Version 2004-2005), La mécanique rationnelle – Formation
de base des Sciences et des ingénieurs – Institut de Physique des nanostructures- Ecole Polytechnique Fédérale de Lausanne de Lausanne– PHB
– Ecublens, 1015 Lausanne
TIPLER, P. A. (1995) , Physics for Scientists and Engineers – New York, NY
1003. Worth Publishers
PEREZ, J. P. (1997), Mécanique: Fondements et applications, MASSON
The Free High School Sciences: A Textbook for High School Students Studing
physics – FHSSt Authors- December 9, 2005 from http://savannah.nongnu.
org/projects/fhsst
http://www.google.ca/search?client=firefox_a&rls=org.mozilla%3Aen-US%Aofficial&hl=en&q=c3%A9equilibre+d%27un+solide+sur+plan&meta
Useful links
http://www.infoline.ru/g23/5495/Physics/English/waves.htlm
http://www.infoline.ru/g23/5495/index.htlm
http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre2/Title1res.htm
http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre3/partie3 /Title1res.htm
http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.
op_addition.aspx
http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.
op_methods.aspx
http://mathexel.site.voila.fr/index.htlm
Detailed activity description
In this activity, there are twenty independent questions on vector calculus.
They can check whether students master the various operations on vectors.
The students will discuss on an online chat to have the same understanding of
the different parts of the course: Physical - Measurement-Uncertainty-vector
operations. For each exercise, students will be organized in groups for collaborative work.
These 20 questions and corrected exercises include:
-
-
-
-
-
-
Units of physical quantities.
The practical meaning of vector operations,
Addition and subtraction vector treated graphically and analytically;
The vector product;
Double vector product;
The combination product.
Evaluation of Activity 1
Twenty questions and exercises
Exercise 1
What does following physical quantity signify ? :
M=5,25 ± 0,02 Kg
Exercise 2
Write the following expression correctly :
D= 15,83379 ± 0,173 m
Exercise 3
Consider m as a physical quantity defined by the following equality :
m=m1-m2-m3
Write Dm as a function of ∆m1, ∆m2 et ∆m3
Exercise 4
What does the relative uncertainty
∆a
.represent.
a
Give the precision of the measure if we have: m= 25,4 ± 0,2 Kg
Exercise 5
How do we add two vectors :
-
-
If they are parallel
If they are not parallel
Exercise 6
What is the scalar product of a vector with any unit vector?
Exercise 7
What does the modulus of the vector product of two vectors represent?
Exercise 8
When is the vector product equal to zero ?
Exercise 9
What does the mixed product of three vectors represent geometrically ?
Exercise 10
The mixed product of three vectors is invariant under permutation of these
(
three vectors. Write the mixed product in different forms V1 , V2 , V3
)
Exercise 11. Vector division.
Consider the equality : a ∧ x = b (1) where a and b are 2 given vectors
and x the unknown vector. Our goal is to solve this equation in the following
scenarios :
a) What is the solution x if a = 0 and b = 0
b) What is the solution x si a ≠ 0 and b = 0
c) What is the solution x if a = 0 and b ≠ 0
d) We suppose a ≠ 0 and b ≠ 0 and a is not perpendicular to b
e) We suppose the general case: a ≠ 0 and b ≠ 0 and a perpendicular to b
•
We suppose that the equation in question has a particular solution x0 . We
can thus write a ∧ x 0 = b (2). By subtracting term by term in equations (1)
and (2), write the general form of the solution x of equation (1) knowing
a particular solution x0
•
We are looking for a particular solution x0 of equation (2). For this we
multiply vectorally the left of equation (2). What is the particularity of
x0 to achieve it in an easier fashion. Write x0 .
•
Write the solution x of equation (1).
•
We represent x by the vector OM . What is the complete M portion of
the solution of the equation a ∧ OM = b .
Exercise 12
Consider 2 vectors V1 and V2 , of respective modulus 5 m/s and 3 m/s ,
(
that have an angle of a=30° between then. Calculate V1 + V2
(
)
2
)
Find the modulus of the vector sum V1 + V2 . Trace the vector sum.
Exercise 13
A boat is crossing a river at a speed of 6 km / h. The velocity of the incoming flow perpendicular to the boat is 3 km / h (these velocities are measured
from the reference ground, say, by an observer located on a shoreline). In what
direction the boat is headed?
Exercise 14
Prove that the diagonals of a parallelogram intersect in the middle.
Exercise 15
Prove that the diagonals of a rhombus are perpendicular.
Exercise 16
Determine a unit vector perpendicular to the plane formed by the vectors
A = 2i − 6 j − 3k and B = 4i + 3 j − k
Exercise 17
Show that the sine law holds in a triangle.
.
Exercise 18
Consider two vectors
ur ur 2 ur ur 2 ur 2 ur
A ∧ B + A.B = A B
( )
. Show that
2
Exercise 19
How is it that a person moving along the x axis with velocity U must tilt their
umbrella to protect most of the rain that falls parallel to the y-axis with velocity V.
Exercise 20
Consider 3 non-aligned points and a point O as an origin
We let OA = a , , OB = b,.
r
r r
r r
OC = c
r
Show that : a ∧ b + b ∧ c + c ∧ a is a perpendicular vector in the plane ABC.
Corrected exercises
Evaluation of activity 1
1°) M=5,25 ± 0,02 Kg signifies that 5,23 Kg < M < 5,27 Kg
2°) A more correct way of writing it would be D= 15,8 ± 0,2 m
3°) m=m1-m2-m3 ⇒∆m = ∆m1+ ∆m2 + ∆m3
4°) The relative uncertainty
∆a
represents the precision of the measurement.
a
If we have : m= 25,4 ± 0,2 Kg, the precision of the measurement of m is
∆m 0,2
=
= 0,8%
m 25,4
5) If the two vectors are parallel, we draw the second vector on the line of action
of the first vector.
If the two vectors are not parallel, we use the parallelogram rule.
6°) The scalar product of a vector with a given unit vector is an orthogonal projection of the first vector on the unit vector.
7°) The modulus of the vector product of two vectors is the area of the parallelogram generated by the two vectors.
8°) The vector product is zero only if :
- one of the vectors is zero
- the two vectors are parallel.
9°) The mixed product of three vectors represents the volume of the parallelepiped
defined by the three vectors.
ur
(
ur
u
ur
u
) (
ur ur
u ur
u
)
ur
u
(
ur
u
ur
ur
u
)
(
ur
ur
u
10°) V1 • V 2 ∧ V3 = V1 ,V 2 ,V3 = V 2 • V3 ∧ V1 = V3 • V1 ∧ V 2
)
11°)
Vector division
r
r
r
r
r
r
r
r
a) a ∧ x = b ➙ 0 = 0 ➙ indetermination : all vectors x of the vectorial
space is a solution to (1)
r
r
b) a ∧ x = 0 ➙ x = la , the vectors x parallel to the vector a are solutions to (1)
c) Equation (1) is written 0 = b , which is impossible : There is no vector x
satisfying equation (1).
r
r
r
r
r
d) no solution since a ∧ x = b implies that a ⊥ b
r
r
r
e) a ∧ x = b (1)
r ur
u r
a ∧ x0 = b (2)
r r ur
u
r
r ur
u
r
(1)-(2) ➙ a ∧ (x − x0 ) = 0 ➙ x = x0 + la
•
r r ur
u
r r
r ur
u r
ur
u r r
a ∧ a ∧ x0 = a ∧ b ➙ a.x0 a − a 2 x0 = a ∧ b . . If x0 is perpendicular
r r
ur
u r ur
u
a∧b
to a , we can easily have x0 = 0 ➙ x0 = − 2
a
r r
r
r
a∧b
•
x = − 2 + la
a
(
•
)
( )
b
( )
−
a ∧b
a2
H
M
a
The group of points M, at the extremities of vectors x belong to the segment
r r
uuur
a∧b
(∆) passing by the point H, such tat OH = − 2 and (∆) is parallel to a .
a
Exercise 12
(V
1
+ V2
)=V
2
2
2
V1 + V2 =
2
2
+ V2 + 2V1 .V2 = V1 + V2 + 2 V1 .V2 cos α
1
2
2
V1 + V2 + 2 V1 . V2 cos 30o =
25 + 9 + 2.5.3.
=7.745
V1 +V2
V2
V1
Exercise 13
V 1 Vitesse de la
pirogue
V2: Vitesse du
courant
Rives
The boat is deviated by an angle b of its original direction such that
tan β =
V2 3 1
= =
V1 6 2
3
2
Exercise 14
Call ABCD the parallelogram, P the intersection point of AC and BD ,
a = AB = DC , b = AD = BC .
We have :
Or
uuur uuur uuur
r r
BD = CD − CB = −a + b
uuur
r r
BP = x(−a + b)
uuur uuur uuur r r
AC = AB + BC = a + b
uuur
r r
AP = y(a + b)
uuur uuur uuur r r
AB = AP + PB a + b
r
r r
r r
a = y(a + b) − x(−a + b)
r
r
r
a = (x + y)a + ( y − x)b
For the equality to be true, the following system must remain true :
x + y = 1 and y – x =0 ➙x = ½
and y=1/2 .
P is thus the midpoint of BP and AC.
Exercise 15
Call OPQR the rhombus. PR and OQ the two diagonals.
P
Q
O
R
Form the scalar product OQ.RP
OQ.RP = (OR + RQ).(RQ + QP)
2
2
2
OQ.RP = OR.2 cos(RO, RQ) + OR � OR + RQ cos(QR , QP) The an-
gles
are supplementary, as
they have opposite cosines.
Thus
are perpendicular.
Exercise 16
A ∧ B is a perpendicular vector in the place formed by vectors A and B
2
4
15
A ∧ B = − 6 ∧ 3 = − 10
− 3 −1
30
The unit vector parallel to à A ∧ B is
u =
A∧B
A∧B
u=
15i − 10 j − 30 k
(15 ) + ( − 10 ) + ( 30 )
=
3
2
6
i − j− k
7
7
7
Exercise 17
a , b, c represents the 3 sides of triangle ABC and α, β, γ the three angles.
a+b+c=0
If we multiply this vector sum by a or by b or by c , we will always obtain a
zero vector.
Ainsi :
r
r
r
r r
r
r
r r
r
r
r r
a ∧ a + b + c =b ∧ a + b + c =c ∧ a + b + c =0
(
) (
) (
r
r
r r
r
r r
r
r
r
a ∧ ( b + c) = b ∧ ( a + c) = c ∧ ( a + b ) = 0
r
r r r r r r r r r r r
r
a ∧ b+ a ∧ c = b ∧ a + b∧ c = c ∧ a + c ∧ b = 0
And thus :
r
r r r r r
a ∧ b = b∧ c = c ∧ a
This vector equality implies the equality of the moduluses :
)
r
r
r r
r r
a ∧ b = b∧ c = c ∧ a
absinγ =bcsinα=casinβ
By dividing this double equality by the product abc, we have :
sinγ sinα sin β
=
=
c
a
b
Exercise 18
ur ur 2 ur ur
A ∧ B + A.B
ur ur
A∧ B
ur ur
A∧ B
2
2
( )
ur ur
+ ( A.B )
ur ur
+ ( A.B )
2
2
2
ur r ⎞ 2 ⎛ ur ur
ur r ⎞ 2
⎛ ur ur
= ⎜ A B sin( A,b)⎟ . + ⎜ A B cos( A,b)⎟
⎝
⎠
⎝
⎠
ur 2 ur 2
ur r
ur r
= A B sin 2 ( A,b). + cos 2 ( A,b)
{
ur 2 ur
= A B
}
2
Exercise 19
V’
is the velocity of the raindrops with respect to (R’).
uur ur ur
V / = V −U
U is the driving velocity
The person must tilt their umbrella by an angle a, such that tan α=U/ V.
Exercise 20
OA = a , , OB = b,. OC = c .
r r r
r r r
a ∧ b + b ∧ c + c ∧ a is a perpendicular vector to the plane ABC.
M is a point on plane ABC.
Assume OM = r .
r
r
r
r
(( )
r
r
)
The mixed product (r − a ). b − a ∧ (a − c) = 0
r r r r r r r r
(r − a ). b ∧ c − b ∧ a − a ∧ c = 0
(
)
r r r r r r r r
(r − a ).( a ∧ b + b ∧ c + c ∧ a ) = 0
From this last equation, we deduce that :
(
r r r r r r
r r
a ∧ b + b ∧ c + c ∧ a is perpendicular to (r − a ) .
)
(
r
r
r
r
r
r
)
Thus a ∧ b + b ∧ c + c ∧ a is perpendicular to the plane ABC.,
Learning activities
The students must do all the exercises. They are organized in groups for collaborative work. Each group completes the proposed exercises and designates
a leader of the group who will report for each group. The professor gives a
deadline for each exercise, at which time each group will send an attached file
of their reports to the professor of the course.
Teacher’s guide
The professor will correct the group reports, and will place the corrections in a
workspace accessible to students. The corrections are accompanied by adequate feedback. The scores for each group are assigned to group members and
will count for 20% of the final evaluation of the module.
Learning activity 2
Activity title
1D, 2D and 3D Kinematics of a Point
The student must be able to:
•
•
•
•
•
•
•
•
Locate the position of a moving object in a given reference system
Establish the parametric equations of a moving object
Calculate the coordinates of a vector in a given reference system
Determine the velocity of a moving object
Calculate the coordinates of the acceleration vector of a moving object
Calculate the acceleration of a moving object
Calculate intrinsic components of acceleration
Find the equations of the trajectory of a moving object
Activity summary
This activity includes:
•
•
•
identifying the position,
finding the path,
calculate speed and acceleration of a moving object in a Cartesian, spherical, cylindrical or polar coordinate system, and the intrinsic components
of acceleration for curved movements.
Key concepts
•
•
•
•
•
Ray vector: Vector indicating the position of the moving object at time t
Peak: Maximum height reached by a projectile
Range: Maximum distance reached by the projectile from the point of
shooting
Details of a moving object: the ray vector components over time.
Cartesian coordinates [ x(t), y(t), z(t)]:
Cylindrical coordinates [ ρ (t), θ (t), z(t)]
Spherical coordinates [ ρ (t), θ (t), ϕ(t)]
Polar coordinates [ ρ (t), θ (t)])
Trajectory: curve described by the moving object when the time t varies.
•
•
Binormal: vector directly perpendicular to the unit vector of the tangent
and the unit vector of the normal to the trajectory.
Intrinsic components of acceleration: Components of the acceleration
vector on the tangent and normal to the trajectory.
(IN APPENDIX 2)
1°) RATIARISON, A. (2006). Cinématique du point. Madagascar. Université
d’Antanarivo. Cours inédit
2° http://abcsite.free.fr/physique/meca/me_ch3.html
3°) http://www.chez.com/mecasite/Mecanique/cinematsol.htm
- http://www.infoline.ru/g23/5495/Physics/English/waves.htlm
- http://www.infoline.ru/g23/5495/index.htlm
Useful links
•
•
•
•
•
•
•
•
•
•
•
http://perso.orange.fr/rmchs/physique_05/cours_physique/cours_
mecach5_cinematique.pdf
Calculs vectoriels
Composition de forces
Dérivation vectorielle
PEREZ, J. P. (1997), Professeur de Physique – Université Paul Sabatier
Toulouse – France. - Mécanique – Fondements et applications , Edition
MASSON, 120 bd St Germain 75 280 Paris Cedex 06
M. CAZIN (1995), Cours de mécanique générale et industrielle– gautier
viullars – tome 1, NY 1003 -1995
http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/
chien_j.html
4mouche_j.html
lissajou_j.html
cycloi_j.html
•
•
•
•
•
•
pndhgs_j.html
The activity has two parts :
Part I: Detailed study of free fall.
This section serves not only as a complement to the course, but also during
revision of what we have already seen in secondary classes. In this section, the
student will have the opportunity to see:
- Free fall without friction force,
- Free fall with friction force proportional to the velocity of fall
- Free fall on a slope.
The student will become familiar with the equation of projectile trajectories by
varying the firing angle, keeping the other parameters constant, and doing the
same by varying the initial velocity. He or she will review methods for integrating equations of first or second order. By downloading the website below
he or she will have fun checking the calculations of the peak, range (depending
on the angle of the shot) and the initial velocity of the projectile.
Part Two: A series of 9 corrected exercises, dealing mostly with the calculation of parametric equations, velocity and acceleration of a moving object in a
coordinate system. In some cases, one must determine the equation of a trajectory. The student is already familiar with integration methods he or she saw in
the first part of Unit 2.
Evaluation
First portion of activity 2 evaluation :
Detailed study of free fall
http://e.m.c.2.fr_chute-libre.htm\e.m.c.2.free.fr\pj00wdel.htm
A – Studied situation:
See the virtual animation on the site
http://e.m.c.2.fr_chute-libre.htm\e.m.c.2.free.fr\pj00wdel.htm
We shoot with initial velocity an uncharged object of constant mass m in a
region small enough that the intensity of gravity can be regarded as constant.
We choose the point O as the starting point of the moving object M. We choose
a plane xOz containing
and in the following fashion:
The x axis is horizontal in the direction of initial movement.
The z axis is vertical and directed upwards.
We choose a y-axis perpendicular to the 2 previous, thus horizontal, and such
as the trihedral Ox, Oy, Oz is direct or that:
We have
as the oriented angle
B - System: The projectile of constant mass m, uncharged
C - Referentiel: Assumed Galilean.
D – Balance of forces:
a- Gravity
, vertical, directed downward, in the following manner:
b - Air resistance neglected (this approximation is the least justified). It may
be important, it is partly why airplanes fly at 30 000 feet. It depends on the
velocity: F = - k V (laminar regime), F = - k = V2 or F - k V3 or more complex
in the turbulent regimes. The last km / h of velocity achieved by the TGV, for
example, are very expensive.
Skydiving : a form of free fall.
Beginning of fall : 3 600 m. End of fall : 1 500 m. Duration : 45 s.
Competition between 2 forces: the weight, being constant, and the air resistance, increasing with velocity and is opposed to weight. It is therefore
expected, after an extended period of time, to have an equilibrium between
the 2 forces, resulting in a zero resistance, a zero acceleration and therefore a
constant velocity. During this jump, we travel after about 8 seconds, at 55 m / s
or 200 km / h.
c – Buoyancy of air:
neglected since
The air has a density of about 1.3 kg/m3 while the bodies that we are studying
the movement of have densities of about 1000 kg/m3 or more.
E – Fundamental relation of dynamics - Acceleration:
If the chosen system was “Earth + projectile”, the weight is not an external
force.
The mass is simplified because it is the same mass that the projectile weight
and inertia to the movement express.
We obtain the following vector differential equation:
The acceleration is independent of time. It is equal to - g (with the orientation
of Oz upward).
It depends (as does g), altitude and latitude of the land.
It also depends on the mass of the attractor (moon, sun, planet ...) and the altitude and latitude.
By integrating, we must not forget that the constants can be determined by the
initial conditions imposed by the situation:
F - First integration - The velocity:
There are two constants in this movement:
-
-
Vy the velocity along the axis Oy initially zero, remains zero, this means
that the projectile remains in the plane xOz. Now we can ignore what
happens outside the plane xOz.
The velocity vx (horizontal component) along the axis Ox is always equal
to v0 cos α. In effect, the force (weight) has a zero component along Ox
and Oy since there can not be any change in velocity along these 2 axes.
However, the velocity vz (vertical component) depends on time t. (The
weight has a nonzero vertical component).
vz is positive if
is directed upward.
vz vanished at the moment
THe projectile elevates since Oz
vz is negative
diminishes with time.
The projectile rises.
The altitude of the projectile
Behaviour of the velocity and its components vx and vz.
The angle which the direction of the velocity with the horizontal axis Ox makes at time 0, but in general, the generic instant t:
if
Soon, we will better understand the particular significance of this moment.
This is the moment when the projectile from an altitude z0 (eventually 0) passes through the same altitude z0 (eventually 0).
The normal of the velocity also depends on time:
if
, time of launch.
, the instant when the projectile has the same
altitude as the launch altitude.
G - Second integration – Time law of movement – The parametric equation
of trajectory:
Recall the solution
At the instant t = 0:
Where the time law of movement is:
Uniform rectilinear motion of the orthogonal projection of the projectile on the
axis Ox.
The flat trajectory is contained in the plane xOz.
Parabolic behavior of the border (or altitude) z.
This law is also called the time parametric equation of the trajectory (the parameter being the time) because by giving t, we can calculate x and z and find
the corresponding set of points where the projectile passes.
H – Cartesian equation z ( x ) of the trajectory:
Recall the parametric equation:
Eliminate the time t in line 3 by using equation in line 1:
1 gives :
which, inserted in 3, gives the cartesian equation of the trajectory:
It is a concave parabola oriented downward ( coefficient of x2 < 0 and axis Oz
upward ).
The quotient is in meters: the first term of the second expression is homogeneous in length.
Experiment at home :
Webcam - 25 images per second - Pose 1/250 s – Rule of 30 cm.
I – Calculation of the peak h:
1°) Utilizing tS :
and
In
thus m.
Is a maximum for = 90°.
2°) Utilisation of the derivative dz / dx:
We express the abscissa xS of the summit of the parabola.
or
Thus:
Maximum for
Then we insert in
J – Calculation of the scope:
Let C be the point where the projectile struck the ground. The scope is OC. C
is the intersection of the trajectory of the projectile and the line z = 0 is
thus
Maximum for
.
K - Variant 1: The projectile from a non-zero altitude z0. We must restart
from the initial condition of z.
All the above (acceleration, velocity, x (t) and y (t)) remain unchanged. Only
change z (t). At time t = 0, z is not equal to 0 but is equal to z0; C3 constant is
then equal to z0 where the new time law is:
The moment of transition in S is the same. The peak is increased by z0.
It is more difficult to determine the scope. We must solve the quadratic equation:
with
And here,
L - Variant 2 – Sloping terrain:
The terrain ( sloping up or down ) can be represented by the segment z =
z’0 ± a . x or z = z’0 ± x . tan
There is not much extra difficulty, except to solve the equation
To obtain the abcissa of the point of impact.
First portion of activity 2 evaluation :
Series of exercises
1. a, k and ω are constants. We give the acceleration vector of a point on a
moving object M
(
a = ak 2 6ωt x − 48ωt y + 12z
)
Determine the velocity vector V ( M ) and the position vector M 0 M . M0
is the initial position of the moving object M and we note V0 as the initial
velocity.
(
)
, x , y , z , consider a point
2. With respect to a fixed orthonormal reference R O
on a moving object M whose trajectory is given as a function of time by the
following parametric equations:
⎧ x = 4t 2
t3
⎪⎪
y
=
4
(
t
−
)
⎨
3
⎪
⎪⎩ z = 3t + t 3
a) Determine the velocity vector of the moving object M
b) Determine the velocity algebraically
c) Prove that the tangent to the trajectory is a constant angle with the axis
(O, z )
3. Compared to an orthonormal reference, a point M is animated by a movement
defined in cylindrical coordinates by:
⎧r = 1 + cos θ
⎪
⎨ θ = ωt
⎪ z = sin θ
⎩
a) Calculate the components in cylindrical coordinates of the velocity and
acceleration vectors
b) Let m be the orthogonal projection of M in the xy plane. Write the polar
equation of m.
4. A point M describes the curve of parametric equations:
⎧x=t
⎪
2
⎨y = t
⎪z = t 3
⎩
a) Determine the unit vectors of the Frenet reference at t = 1.
b) Write the equations of the tangent, normal and binormal at a point M of
the trajectory.
5. Consider a point describing a curve (C). In a fixed, direct orthonormal
reference R at time t, the components of point M are:
⎧ x = 4 cos t
⎪
⎨ y = −4 sin t
⎪ z = 3t + 5
⎩
a) Determine the unit vectors of the Frenet trihedral
b) Determine the curvature and torsion of the curve (C) M.
6. If V , a and S are respectively the velocity, acceleration and jerk-acceleration (derivative of acceleration), a point M describes a curve (C), show
that the curvature and torsion can be written as:
ur r
V ∧a
= ur 3
V
and
et
ur r ur
V ,a ,S
= − ur r 2
V ∧a
(
)
7. In a direct orthonormal R, the coordinates of a moving object M are at every moment t:
a and w are positive constants ( a, is in the dimension of length and w is the
inverse of the time t).
⎧ x = a ( 2e
⎪
⎨ y = 2a (e
⎪z = 0
⎩
− ω t
−e
− 2 ω t
)
− ω t
−e
− 2 ω t
)
a) Write the cartesian equation of the trajectory of the point M
b) Write the velocity vector of the point M
c) Draw the trajectory for t>0
d) Determine the acceleration vector a and the tangential acceleration
aT
e) Determine the path taken by the moving object and the area swept by
OM for t varying from 0 to .
8. A canon is fired at a boat from the top of a cliff. The height of the cliff is in
and the distance of the boat from the foot of a cliff is D.
a) What angle α must be drawn such that the initial velocity of the projectile
is as low as possible?
b) What will be the velocity v0?
c) What is the velocity of the projectile at the impact on the boat?
d) When does this impact occur?
a. A material element is launched from a fixed point O downstream at an
inclined plane of angle a with respect to the horizontal plane. If the launch
velocity makes an angle of g with the inclined plane and the projectile
falls at a point P on ; determinethe distance OP.
b. the angle g as a function of a such that OP is a maximum, and calculate
this maximum.
Correction of exercises of evaluation of activity 2
Exercise 1.
dV
= a ➙ d V = adt ➙ V =
dt
a)
V = ak
(
∫ adt ➙
)
∫ 6 ω t x − 48 ω t y + 12 z dt ➙
(
)
V = ak 2 3ω 2 t 2 x − 16ω 2 t 3 y + 12t z + V0
dM 0 M
=V
dt
b )
➙
d M 0 M = Vdt
➙
M0M =
(
)
∫ Vdt
➙
M 0 M = ak 2 ∫ 3ω 2 t 2 x − 16ω 2 t 3 y + 12t z + V0 dt
(
M 0 M = ak 2 ω 2 t 3 x − 4ω 2 t 4 y + 6t 2 z + V0 t
)
Exercise 2.
1°) Velocity vector of M
8t
d OM V = 4(1 − t 2 )
V=
➙
dt
3(1 + t 2 )
2°) Algebraic velocity of M
s is the curvilinear abcissa
V=
ds
dt
We have :
⎛ ds ⎞ ⎛ dx ⎞ ⎛ dy ⎞ ⎛ dz ⎞
⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 64t + 16(1 − t ) + 9(1 + t )
⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠
⎛ ds ⎞
⎜ ⎟ = 25t + 50t + 25 = 25(1 + t )
⎝ dt ⎠
⎛ ds ⎞ ⎛ dx ⎞ ⎛ dy ⎞ ⎛ dz ⎞
⎜ ⎟ = ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 64t + 16(1 − t ) + 9(1 + t )
⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠
⎛ ds ⎞
⎜ ⎟ = 25t + 50t + 25 = 25(1 + t )
⎝ dt ⎠
By orienting the trajectory in the direction of movement, we have :
V=
ds
= 5 1+ t2
dt
(
)
3°) Angle of the tangent to the trajectory and the axis Oz
dz
dz dt 3(1 + t 2 ) 3
Cos(Oz, V) =
=
=
=
ds ds 5(1 + t 2 ) 5
dt
The cosine of this angle is constant so the angle between the tangent to the
trajectory and the axis Oz is constant.
Exercise 3.
a) The cyclindrical coordinates :( r, q, z).
The coordinates of the velocity vector in cylindrical coordinates are (equation
71 of the course):
dr
= −ω sinθ
dt
•
uuuuuuur
dθ
VR ( M ) = r θ = r
= (1 + cosθ)ω
dt
•
dz
z=
= ω cosθ
dt
•
r=
The coordinates of the acceleration vector are (equation 72 of the course) :
⎛ • • •2 ⎞
2
2
2
⎜⎝ r − r θ ⎟⎠ = −ω cosθ − (1 + cosθ)ω = −ω (1 + 2cosϑ )
uuuuuuur
aR ( M ) =
• •
⎛ ••
⎞
2
r
θ
+
+2
r
θ
⎜⎝
⎟⎠ = −2ω sinθ
••
z = −ω 2 sinθ
Rc
uuuuuuur
aR ( M ) =
Rc
−ω 2 (1 + 2cosϑ )
−2ω 2 sinθ
−ω 2 sinθ
b) The polar equation of m, orthogonal projection of M in the place xOy.
The parametric equations of the point m are:
r=1+cosθ and θ = ωt
The polar equation of m is r=1+cosθ which is a cardioid.
Exercise 4.
Determination of unit vectors of the Frenet reference.
(
)
, i , j , k , the vector position is written as :
In the referentiel O
OM = t i + t 2 j + t 3 k
d OM
dt
= i + 2t j + 3t k
2
d 2 OM
= 2 j + 6t k
dt 2
(M, e ,e
T
N
)
, eB is the Frenet trihedral
The vector e T is tangent to the trajectory
uuuur r
r
r r r
r
ur
u 1 dOM i + 2t j + 3t 2 k i + 2 j + 3k
eT =
=
=
V dt
14
9t 4 + 4t 2 + 1
If we orient the trajectory in the direction of the movement
V = 12 + ( 2t) 2 + ( 3t 2 ) 2 = 9t 4 + 4t 2 +1
The vector e T is thus :
eT =
1 d OM
i + 2t j + 3t 2 k
i + 2 j + 3k
=
=
4
2
V dt
14
9t + 4t + 1
The binormal e B is directed as follows:
ur uuuuur uuuur
B = V(M) ∧ a(M) =| 1 ∧ | 0 =| 6
2 2 −6
3 6 2
r r
r
r r r
ur
u 6i − 6 j + 2k 3i − 3 j + k
eB =
=
76
19
The vector N of the normal at t=1 is :
ur ur ur
N = B ∧ T =| 6 ∧ | 1 =| −22
−6 2 −16
2
3 18
r
r
r
r r
r
ur
u −22i − 16 j +18k −11i − 8 j + 9k
eN =
=
1064
266
For t=1, x=1, y=1 and z=1
The equation of the tangent is :
x − 1
=
1
y − 1
2
z − 1
=
3
The equation of the principal normal is :
x − 1
=
− 11
y − 1
− 8
z − 1
=
9
The equation of the binormal is :
x − 1
=
3
y − 1
− 3
z − 1
=
1
Exercise 5.
The vectors of the Frenet trihedral are :
eT =
d OM
1 de T
, eN =
, eB = eT ∧ e N
ds
R ds
x
OM
=
y
=
t
− 4 sin
=
z =
eT =
4 cos
t
3 t + 5
dOM dOM dt
=
x
ds
dt
ds
2
2
2
2
⎛ ds ⎞ = ⎛ dx ⎞ + ⎛ dy ⎞ + ⎛ dz ⎞
⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟
⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠ ⎝ dt ⎠
= 16 sin 2 t + 16 cos 2 t + 9
2
⎛ ds ⎞ = 25
⎜ ⎟
⎝ dt ⎠
⇒
Calculation of
dt 1
=
ds 5
− 4 sin t
d OM
= − 4 cos t ➙
, e
dt
3
=
1
5
− 4 sin t
− 4 cos t
3
With
ur
u
2
deT
⎛ 4⎞
R=
= ⎜ ⎟
⎝ 25 ⎠
ds
ur
u 25 −4
ur
u
, eN =
|
cost ⇒ eN =| −cost
4 25
sint
4
Where
sint
0
25
0
ur
u ur
u ur
u 1
1
eB = eT ∧ eN = | −4sint ∧ | −cost = | −3sint
5
5
−4cost sint
−3cost
3
0
−4
Calculation of the torsion
de
ds
de
ds
τ =
= τe
=
=
de
dt
dt
ds
de
=
ds
=
1 de
5
=
dt
− cos t
3
sin t
25
0
3
25
Exercise 6
The velocity vector V is the derivative of the ray vector OM
V=
d OM d OM ds
=
= Ve T
dt
ds dt
The acceleration a is the derivative of the velocity
a =
dV
dt
=
dV
dt
eT + V
d e T ds
ds
dt
=
dV
dt
2
e T + V ρe N
ρ=
1
is called the curve of the trajectory
R
By calculating the vector product
⎛ dV
⎞
V ∧ a = Ve T ∧ ⎜
e T + V 2 ρe N ⎟ = V 3 ρe B
⎝ dt
⎠
By taking the modulus, we have :
ρ =
V∧a
V
3
The jerk-acceleration is the derivative of the acceleration.
S=
S =
S=
S=
S=
d e ds
da d 2 V
dV d e T ds
dV
= 2 eT +
.
+ 2V
ρe N + ρV 2 N
dt
dt ds dt
dt
ds dt
dt
da
d V
=
dt
e
+ V
dt
dV
dt
.
e
+ 2V
ρ
dV
dt
ρe
+ ρV
de
ds
da d 2 V
dV e N
dV
d
=
eT + V
.
+ 2V
ρe N + ρV 3
eB ∧ eT
2
dt
dt ρ
dt
ds
dt
(
da
=
dt
d V
dt
e +V
)
de
de
dV e
dV
.
+ 2V
ρe + ρV
∧ e + ρV e ∧
dt ρ
dt
ds
ds
e
d 2V
dV e N
dV
eT + V
.
+ 2V
ρe N + ρV 3 τe N ∧ e T + ρV 3 e B ∧ N
2
dt ρ
dt
ρ
dt
( )
Form the mixed product S V, a
(
S,V ,a
)
= − ρ τV
Where
τ=-
(V, a , S)
V∧a
2
Exercise 7
a) Cartesian equation of the trajectory :
⎧ x = a ( 2e
⎪
⎨ y = 2a (e
⎪z = 0
⎩
− ω t
−e
− 2 ω t
)
− ω t
−e
− 2 ω t
)
− ωt
By letting f = e
, this system is equivalent to :
⎧ x = a ( 2f − f )
⎪
⎨ y = 2a ( f − f )
⎪z = 0
⎩
⎧
⎪
⎪⎪
➙⎨
⎪
⎪
⎩⎪
x
= ( 2f − f 2 )
a
y
x
y
= (f − f 2 ) ⇒ −
=f
2a
a 2a
⎧ x = a ( 2f − f )
⎪
⎨ y = 2a ( f − f )
⎪z = 0
⎩
⎧
⎪
⎪⎪
➙⎨
⎪
⎪
⎪⎩
x
= (2f − f 2 )
a
y
x y
= 2(f − f 2 ) ⇒ − = f 2
a
a a
2
2
2
2
x
y
−
=f
a 2a
x y
− =f2
a a
2
⎛x y ⎞
⎛x y⎞
➙⎜ −
⎟ = ⎜ − ⎟➙ (2x − y )
⎝ a 2a ⎠
⎝a a⎠
The trajectory is the branch of a parabola.
b) Vector velocity
•
V=
x = −2aω(f − f 2 )
•
y = −2aω(f − 2f 2 )
2
= 4a ( x − y )
b) Draw the trajectory for t>0
We study the variation of the curve path as we did in the tracing of parametric
curves.
The table of variation is below.
T
0
∞
F
1
1/2
0
•
x
0
-
0
X
a
3a/ 4
0
•
y
2aω
+
0
-
0
Y
a/2
0
0
y
M
x
I(t=0)
3a/4
0 (à t=∞)
d) The acceleration vector a
••
r x = 2aω 2 ( f − 2 f 2 )
a = ••
y = 2aω 2 ( f − 4 f 2 )
The tangential acceleration a T
ur 2
V = 4a 2ω 2 f 2 2 − 6 f + 5 f 2
(
)
By orienting the trajectory in the direction of the movement
V = 2aω f
(2 − 6 f + 5 f )
2
The modulus of the tangential acceleration is :
dV
d
aT =
= ⎡⎢ 2aω f 2 − 6 f + 5 f 2 ⎤⎥ = aω 2 f
⎦
dt dt ⎣
⎡ −2 + 9 f − 10 f 2 ⎤
⎢
⎥
⎢⎣ 2 − 6 f + 5 f 2 ⎥⎦
e) Path traveled by the moving object and the area swept by OM for t varying from 0 à ∞.
The arc IM =
∫ vdt
= 2a
∫ωf
2 − 6f + 5f
dt
To calculate this primitive, is written in a conventional manner:
2
2
⎡⎛
3⎞ ⎛1⎞ ⎤ 1
5f 2 − 6f + 2 = 5⎢⎜ f − ⎟ + ⎜ ⎟ ⎥ = ch 2 θ
5 ⎠ ⎝ 5 ⎠ ⎥⎦ 5
⎢⎣⎝
3 1
f − = shθ;
5 5
1
df = chθdθ
5
ch 2 θdθ
5f 2 − 6f + 2.df
5 5
1
(θ + sh θch θ ) , with θ as a function of f and f as a
The primitive is :
10
5
function of t.
shθ = 5f − 3;
chθ =
1 + (5f − 3) 2
θ = Ln[5f − 3 +
(5f − 3) 2 + 1 ]
(
)
For t=0, f0=1, θ 0 = Ln 2 + 5 , shθ0=2, chθ 0 =
For t=∞, f∞=0, θ
∩
The arc IO =
= Ln
( 10
)
− 3, shθ =-3,
∞
5
chθ ∞ = 10
⎤
a ⎡
2+ 5
+ 2 5 + 3 10 ⎥
⎢Ln
⎥⎦
5 5 ⎣⎢
10 − 3
Exercise 8
a) Expression of α
P designates the position of the boat, a the launch angle
We are looking for the equation of the projectile’s trajectory.
Following the Oy axis, the movement is uniform with the equation :
y = V0 cosα.t
Following the Oz axis, the movement is uniformly varied with a constant acceleration –g, of equation : z = −
The trajectory equation is :
1 2
gt + V0 sinα.t
2
2
1
y
z=− g 2
+ V0 tanα.y
2 V0 cos 2 α
The boat that is at a distance D from the cliff is hit by the projectile if and only
if
y=D and z= -h
1
D2
−h = − g 2
+ tanα D (1)
2 V0 cos 2 α
z
O
y
h
D
P
Take V0 from this expression and write that V0 is a minimum, however is it
best to write :
f =
−gD 2
2V02
Which gives :
−h =
f (α )
+ D tanα
cos 2 α
By deriving this expression with respect to a, we have :
0=
1` df (α )
D
2 f sinα
+
+
2
2
cos α dα
cos α
cos 2 α
And write that
df (α )
= 0 , which gives
dα
V02 = gD tanα (2)
0 = D + 2 f tanα or also
Place equation (2) in equation (1).
Thus :
1
D2
−h = − g
+ tanα D
2 gD tanα cos 2 α
After simplification, we find : tan 2α =
Or
tan 2α =
⎛ D⎞
D
1
➙ α = Arc tan ⎜ ⎟
h
2
⎝ h⎠
D
2 tanα
2 tanα
h
=
⇒ 1 − tan 2 α =
⇒ 1 − tan 2 α − 2 tanα = 0
2
D
h 1 − tan α
D
h
We now have a second degree equation in X=tana which has a positive solution :
X = h2 + D 2 − h = tanα
b) Calculation of V0.
The velocity V0 is such that V 2 = gD tanα = gD
0
(
)
h2 + D 2 − h
c) velocity of the projectile at the moment of impact on the boat
The movement is uniformly varied and we can write :
V 2 − V02 = 2gh ⇒ V 2 = g ⎡⎢ h2 + D 2 + h⎤⎥
⎣
⎦
d) Time of impact of the projectile on the boat :
Starting with the equation y = V0 cosα.t , on a : D = V0 cosα.t ➙
t=
D
V0 cosα
We raised to the power of 2 and we have :
t2 =
D2
2D
2 h2 + D 2
=
=
g
V02 cos 2 α gsin 2α
Exercise 9
Trajectory equation is :
z=−
g
y2
+ y tan(γ − α )
2 V02 cos 2 (γ − α )
W are looking for coodinates of P under the form : yp= rcosa et zp= rsina
Substitute P in the trajectory equation and we obtain :
2V02 cos 2 (γ − α )
2V02 cos(γ − α )sinγ
⎡⎣ tanα + tan(γ − α ) ⎤⎦ =
r = OP =
g
cosα
g
cos 2 α
For r to be a maximum, the following expression must represent a maximum :
sinγ cos(γ − α )
The maxmum value of this expression corresponds to that of g. that has a zero
derivative with respect to g.
Thus :
− sin(γ − α )sinγ + cos(γ − α )cosγ = 0 ⇒ cos(2γ − α ) = 0
2γ − α =
π
π α
➙γ = +
2
4 2
In these conditions rm ax imum
V02
=
g(1 − sinα )
Learning activities
Teacher’s guide
Learning activity 3
Activity title: Equilibrium of solids
-
become familiar with the concepts of the equilibrium of a solid
distinguish translational motion from rotational motion
recall the result of a moment and the result of several forces.
calculate the position of center of gravity of a solid or a group of solids
Activity summary Statics is the study of conditions for which the bodies are immovable objects,
relative to a reference R, Galilean or not, related to the observer. These conditions relate particularly to the distribution of forces on the body at rest.
For this study, we introduced the notion of torsor which is reduced to the general resultant and resultant moment of all forces applied to a solid. The equilibrium of a solid subjected to one or more forces results in a zero torsor (general
zero resultant and zero moment).
Key concepts •
•
•
•
•
General Resultant = the vector sum of a system of vectors (forces)
Time = the resulting vector sum of moments of a system of vectors
(forces)
Torsor = system of free vectors (forces) which reduces to the general
resultant and resultant moment
Reaction: force opposed to an action
Center of gravity = Point which where the entire mass of a system of
material points appears to be concentrated.
Appropriate readings (IN APPENDIX 3)
1°) RATIARISON, A. (2006). Equilibre des solides sur un plan. Madagascar.
Université d’Antanarivo. Cours inédit
2°) http://fr.wikipedia.org/wiki/Statique_du_solide
Solid statics.
3°) http://www.ac-poitiers.fr/cmrp/cpge/docs/Coursdemodelisationetdestatique.
doc
Solid statics.
Appropriate resources The Free High School Sciences: A Textbook for High School Students Studing
physics – FHSSt Authors- December 9, 2005
ANSERMET, J.-P. ( 2004-2005), La mécanique rationnelle – Formation de
base des Sciences et des ingénieurs – Institut de Physique des nanostructures- Ecole Polytechnique Fédérale de Lausanne de Lausanne– PHB
– Ecublens, 1015 Lausanne
TIPLER, P. A. (1995). Physics for Scientists and Engineers –– Worth Publishers. New York, NY 1003
PEREZ, J.P. (1997). Mécanique : Fondements et applications. MASSON
http://www.google.ca/search?client=firefox_a&rls=org.mozilla%3AenUS%Aofficial-&hl=en&q=c3%A9equilibre+d%27un+solide+sur+plan&meta
Useful links
http://savannah.nongnu.org/projects/fhsst
This activity has two parts:
• The first part includes an additional course on the forces acting on a solid.
Forces can, depending on their intensity:
-
-
-
-
Maintain a strong balance
Develop a strong translational motion,
Develop a solid rotation around a fixed axis.
Deform an object.
To supplement the knowledge of students on these facts, it seems useful and
necessary to make a point of being on these points. This add-on courses:
-
-
-
-
-
The weight of a body,
The reaction of a support,
The forces of solid-solid friction,
The equilibrium of a solid on an inclined plane with rough contact surfaces,
The study of the deformation of a spring under different forces.
• a second part consisting of eight exerciss that relate to a solid balanced on a
support. For each exercise, the corresponding scheme is given. In most exercises, it is required to find the equilibrium conditions of the solid or of the
system.
Evaluation
Part 1 : Course compliment – Forces acting on a solid
macroscopic forces on a solid
resultant of microscopic forces
distribution over a surface or volume
The weight
of a body
Definition : We call the weight of a point object, situated in a given point M,
the foce opposing the force exerted by a thread that holds this point object at
rest relative to the solid Earth, taken as reference.
In this reference system, the weight
form:
of the point object can be put in the
= m or is, by definition, the vector field of terrestrial gravity at the
point M considered. Remark : For an object of finite dimensions, the assembly should be under vacuum in order to overcome the buoyancy. Note that
each microscopic particle (atom, molecule, ion, etc..) of the object is subjected
to the gravitational pull of the Earth, represented by the vector force. . The
sum
of the forces dispersed in the volume of the solid body is
.
Can we confuse the weight of an object and the gravitational attractive force
that the Earth applies to an object ?
Strictly speaking, we should write :
-
=
+
+
is the weight of the object.
-
is the gravitational attractive force of high intensity that the Earth
applies to an object.
-
is a force of low intensity due to the attraction of terrestrial bodies
other t than the Earth (moon, sun, etc.) on this object.
-
is a force of low intensity due to the rotation of the Earth acting on
the object.
- In the problems studied in Unit 1, we can neglect
and
.
We confuse the weight of the object and the gravitational attraction exerted by
earth on this object.
We write :
=
Remark : In certain problems studied in further courses, we cannot neglect
and
. Actually :
is responsible for the tides.
Force
explains that the direction of the plumb line does not pass ex
Force
actly through the center of the Earth.
The weight
of an object is characterised by :
-
Its point of application: the center of gravity of the object coincides with
the center of inertia.
- The direction of the plumb line, almost confused with the vertical.
- Its orientation: downward.
- its Intensity P = mg (1) where m is the mass of the object (in kg) and
g, the intensity of the gravity vector (in N / kg).
The value of g varies slightly with latitude and heavily with altitude. In France,
at sea level, g = 9,81 N/kg.
The reaction of a support
Consider a solid S at rest on a horizontal table.
Figure 1
Reference to be studied : the solid Earth.
System to be studied : the solid S.
The solid S is subjected to 2 forces :
-
the weight
-
the force
(essentially the gravitational action of the Earth on the solid S)
(vertical action of the table on the solid S)
The existance of the force is governed by the principle of inertia, studied in
high school. For a terrestrial observer, every body perseveres in its state of rest
or uniform rectilinear motion, if the forces exerted on him compensate each
other.
Here, for a terrestrial observer, the solid S is at rest. The sum of forces acting
on it must be zero.
To compensate for the vertical weight , directed downward, the table must
exert a vertical contact force , directed upward such that :
(2)
Remark : The force represents the sum of the forces distributed over the
contact area between the table and the solid S. These forces are due to electromagnetic interactions between atoms of the table and the solid.
Solid-solid frictional forces.
Place a solid S in a slightly inclined place. If the contact is sufficiently rough,
solid S is at rest relative to the reference ground. It does not slip on the inclined
plane.
Figure 2
According to the principle of inertia (for a terrestrial observer, every body
perseveres in its state of rest or uniform rectilinear motion, if the forces exerted compensate for the observer’s forces) the weight vertical, directed
downward, is again compensated by a force , vertical, directed upward such
that:
(3)
This contact force exerted by the inclined plane on the solid S peut être can
be broken down into two components :
, the normal action of the inclined plane on the solid, perpendicular to the
plane, that prevents the solid from penetrating the support.
, tangential action of the inclined plane on the solid, on the tangent parallel to the line of greatest slope on the plane, which opposes movement of the
solid. This force
models the frictional forces when we are dealing with a
rough surface.
We can write
=
+
(4)
(3) which describes the equili- Use relation (4) in relation
brium of the solid S with a terrestrial reference. We obtain :
-
+
+
=
from equation (4)
Remark : according to the principle of inertia, the relation +
+
=
in equation (4), which is equivalent to the relation
in equation (3),
is still satisfied if the solid slides along the portion of the inclined plane with
the greatest slope.
Absence of friction: Without friction, the forces on the solid reduce to
.
Under the action of and
equilibrium of the solid is impossible (the sum
+
is different that the zero vector ). The solid, placed without velocity on the inclined plane, assumes a movement of rectilinear motion with with
an increasing velocity following the line of greatest slope on the plane.
Figure 3
2- Exemples d’effets produits par des forces s’exercant sur un solide
-
-
-
-
Forces can maintain a solid in equilibrium (see example 1 and example
2).
Forces can put an object into translational movement (see example 3).
Forces can make a solid rotate around a fixed axis. Consider, for example,
a door ajar and an immovable object. A force applied on the gate will
generally revolve around the fixed axis (unless the force is parallel to the
axis if the force meets the shaft).
Forces may have other effects. We will study a little more detail of their
work on a spring.
Exercise : Deformation of a spring.
Consider a spring wound with non-touching turns with lenth Lo = 15 cm and
negligible mass. It hangs at one end to a fixed support. When we attach a solid
S (mass marked m) at its other end, its length is L.
We vary m and note the different lengths L at equilibrium:
m (in g)
L (in cm)
0
15,0
50
15,5
100
20,0
150
22,5
200
25,0
a- Draw the force diagram. Represent the two forces acting on the solid and
the spring.
b- Construct the graph representing the common value T of the two forces
acting on the spring according to the spring extension.
c- Show that we can write T = k (L - Lo). Determine the value of k, called
the coefficient of stiffness of the spring.
We will assume g = 10 N/kg.
Solution
FORCE DIAGRAM :
Figure 4
Now represent the forces acting on the spring of negligible mass. Figure 5
b- Construct the graph representing the common value T of the two forces acting on the spring according to the spring extension.
In the table given, we add two lines giving the value of the force T and the
value of the extension x. It uses the international system of units.
M in Kg
L in cm
T=mg=10.m in N
X=L-L0 =L-0,150 in m
0
0,150
0
0
0,050
0,175
0,500
0,025
0,100
0,200
1,000
0,050
0,150
0,225
1,500
0,075
0,200
0,250
2,000
0,100
Figure 6
c-Show that we can write T = k (L - Lo)
The graph above shows that the points are on a line through the origin
The force T is a linear function of the elongation x = L - Lo.
We can write T = K (L - Lo).
The tension T of the spring is proportional to its elongation.
The coefficient K is called the coefficient of stiffness of the spring. Its value is
:
K = T / (L - Lo) , K = 2,00 / 0,100 thus : K = 20 N / m
Remarks :
Under vector form, we can write :
=K
(figure 5).
The relation associated with the equilibirum of a solid S in a terrestrial reference system is :
+
m
=
+K
(figure 4)
=
or : m
-K
=
Second Portion of Activity 3 Evaluation :
Series of Exercises
Exercise 1.
Ra
L3 = 3 m
Rb
L = 10 m
L1 = 1 m
D
A
L2=1 m
F
C
B
E
P
W
A plank of 12 m weighing P1 = 90 N is held by two supports A and B. A is 2
m from B, in the orientation of vector BA , we place a weight of P2 = 360 N.
Find the forces exerted by A and B.
Exercise 2
A ladder of length L = 5m and weight of intensity P = 60 N rests on a rough
horizontal floor AB and a vertical wall BC. Assume that the wall exerts no
frictional force on the ladder. The foot of the ladder on level ground is at z = 3
m from the foot of the wall.
F1
K
C
m
5
y
F2
P
Fn
/
Horizontal floor
A
B
FS
x
z=3m
1° What are the forces in place that allow equilibrium of the ladder ?
2° Draw these forces on the diagram.
3° Find the distance BC = y,
4° The center of gravity of the ladder is G, which is projected orthogonally
on D on the horizontal surface. Find the distance AJ=x.
5° The scale is in static equilibrium under the effect of several non-parallel
forces means that these forces are under concurrent support.
What is the minimum coefficient of static friction necessary between the ladder
and the ground so that the ladder does not slip.
Exercise 3
A ladder AB of mass m and length l, is leaning against a wall of height
h (see figure), the contacts A and B being void of friction, maintains
the scale tilted at an angle α relative to the vertical through a horizontal
wire OA.
Calculate the reaction supports in A and D and the wire tension.
B
D
g
C
h
O
A
Exercise 4
We want to hang a picture, height h on a smooth wall, so that the nail attachment
F is on the wall at the same level as the highest point on the wall (see figure).
1. Show that the point of attachment to the rear of the table should be placed
at a fixed distance from the lowest point of the table.
2. Assess, based according to the wire length and h, the angle of the table
with the vertical. Deduce that l must be between two values that we express
F
T
g
C
A
Exercise 5
We wish to make efforts to link the console shown in the figure. The bar AB of
length l is horizontal, and the bar of length
4l / 3 makes an angle of 30o with the vertical.
The horizontal bar supports a uniformly distributed load value P.
We neglect the weight of the bars.
Calculate the efforts in A, D and E.
A
D
30°
E
P
B
g
Exercise 6
A moving object ABC, shaped like an isosceles triangle (AB = AC = h)
is installed on a pipe of diameter 2r as shown in the figure. The factor of
static friction between the console and the pipe is mS.
Neglect the weight of the console in front of load P placed on AC.
Calculate the distance to the x axis of the pipe to which the burden can
be supported without sliding.
x
A’
A
C
h
h
g
P
B’
B
Exercise 7
A ladder consists of two simple scales, AO and BO of the same length, same
weight Mg, articulated without friction at the common vertex O.
The angle is 2a at the top of the two ladders and m, the friction factor with the
ground. A man of weight mg, climbs the ladder AO to point H at a distance, x,
from the top O.
Show that if the angle a increases, it is the ladder BO that will slide first.
Discuss the influence of x in the special case where m = M.
Exercise 8.
D
H
E
30°
C
4a
2b
c
F
A
2a
B
x
Three rods AD, DB and CH of negligible mass lie in the same vertical planeas
shown.
The assembly is subjected to an outside vertical force F1 .
Find the reaction on each joint, so that the system is in equilibrium. (For reasons of simplicity of the problem, it is required that the reaction C is normal to
the rod DB).
Exercise 9
Determine the center of mass (center of gravity) of a quarter of a homogeneous
disk.
Exercise10
Determine the center of mass (center of gravity) of half a homogeneous disk of
radius R and mass M.
Exercise11.
Determine the center of mass (center of gravity) of a homogeneous half-sphere
of radius R and mass M.
Correction of Activity 3 exercises :
Exercise 1
This system is in equilibrium if the torsor of the effect in A or in B is zero.
➙The general resultant of forces applied to the system is zero:
Ra + Rb + W + P = 0 ➙Ra +Rb-W-P=0➙Ra+Rb=W+P
➙The resultant moment in A is zero :
-W.AC - P.AC + Rb.AB=0
Rb =
W.AC + P.AC
= and Ra=W+P-Rb=
AB
Exercise 2
The forces acting on the plate are: its weight P , the horizontal reaction of
the wall F1 and the floor reaction F2 . These three forces must be concurrent
and let K be the point of intersection of the supports of these three forces. The
components of the force F2 along the vertical and the horizontal are respectively Fn and Fs .
The equilibrium condition is no torsor:
ur
u ur
u ur r
following the horizontal
⎧ Fs − F1 = 0 ⇒ Fs = F1
F1 + F2 + P = 0 ⇒ ⎨
⎩ Fn − P = 0 ⇒ Fn = P = 60 N following the vertical
The moment of these forces on O is zero :
F1.BC – P. x =0 ➙ F1 =
Px 60 x1,5
=
= 22,5 N
BC
4
If m is the coefficient of friction of the floor, the floor does not slide if :
Fs ≤ µFn ==> m =
F s 22,5
=
= 0,395
Fn
60
F1
K
C
m
5
y
F2
P
Fn
/
Horizontal floor
A
B
FS
x
z=3m
Exercise 3
T, R A , R D are respectively the tension of the wire, the reaction in A and
the reaction in D.
The ladder being an immovable object :
mg + T + R A + R D = 0 and
uuur
ur uuur uur r
AC ∧ mg + AD ∧ R D = 0
By projecting following the axes, we have :
-T+RDcosα=0
-mg+ RA + RD .sinα=0
mg(l/2)sinα=0
mg(l/2)sinα-(RDh)/cosα=0
Thus: R D =
mgl
sin(2α ),T = R D cosα,etR A = mg − R D sinα
4h
Exercise 4
1. For the torsor associated to the three forces mg , the reaction R in O and
thread tension T to be zero, they must to be coplanar and concurrent. It follows
that the points F, A and H are aligned. The triangle AFO and AHC are homothetic. Therefore, if Ao = a, we have:
h
− a CH 1
h
2
=
= d’où a =
a
OF 2
3
2. Since l 2 = h2 cos 2 θ + a 2 − 2ahcos 2 θ ➙ cos 2 θ =
0 ≤ cos 2 θ ≤ 1 , we find that :
h
2h
≤l ≤
3
3
3l 2 1
− et
h2 3
Exercise 5
The balance of the rod AB is achieved if the resultant force is zero and also the
resultant moment of the force at A is zero.
Let (XA, YA) be the components of the reaction A
(XD, YD) those of reaction D
(XE, YE) those of reaction E
XA+XD = 0 ; YA+ YD – P = 0 and
P
l
2l
− YD
=0
2
3
Similarly, the equilibrium of DE results in writing that the sum and moment of
the forces are zero in D:
XE-XD = 0 ; YE - YD = 0 and
(−YE + X E 3)
2l
=0
3
Thus :
YD = YE =
3
3
P
P ; X D = X E = − X A = P
; YA =
4
4
4
Exercise 6
Under the action of the load, the contacts appear in B and point A ‘, symmetrical with respect to the axis of the pipe.
Translate the equilibrium conditions by canceling the torsor of external actions
in A ‘.
Let (XA ‘YA’) be the components of the reaction in A ‘,
(XB, YB) those of the reaction in D
We have : XA’+XB = 0 ; YA’+ YB – P = 0 and
(x+r)P-hXB – 2rYB =0
Since A ‘is based on the pipe, XA’ <0, XB> 0, IS ‘> 0, YB> 0 because these
components are opposed to the emerging shift downward.
Solving this system gives :
P = YA ' + YB = −ms X A ' + ms X B = 2ms X B
x+r =
x=
h + 2r ms
2ms
h
2ms
Exercise 7
Translate the equilibrium of the system by canceling the torsor of the forces
exerted on O and the torsor of the forces exerted on the stem of O. Thus:
and
ur
ur
ur
ur r
uuur ur
uuur ur
uuur
ur r
R A + R B + 2 M g + mg = 0 et OA Λ R A + OB Λ R B + OH Λ mg = 0
ur
ur
ur
ur r
uuur ur
uuur ur
uuur
ur r
R
et
OA
Λ
R
A + R B + 2 M g + mg = 0
A + OB Λ R B + OH Λ mg = 0
uuur
And for OB :
and
ur
ur
ur r
uuur
ur uuur
ur r
R B + R a → b + M g = 0 et OG 2 Λ M g + OB Λ M g = 0
ur
ur
ur r
uuur
ur uuur
ur r
R B + R a → b + M g = 0 et OG 2 Λ M g + OB Λ M g = 0
r r
By projecting in the base (e x , e y ) , we obtain :
X A + X B = 0, YA + YB − (2 M + m)g = 0 and :
X A l sinα
X B −xsinα
−l sinα
0
0
−l cosα Λ YA + −l cosα Λ YB + −xcosα Λ −mg = 0
0
0
0
0
0
0
0
It results that :
−l YA sinα + l X A cosα + l YB sinα + l X B cosα + mgxsinα = 0 such
soit that
l(YB − YA ) + mgx
sinα + l X B cosα + mgxsinα = 0 soit l(YB − YA ) + mgx = 0
Since
X A + X B = 0 . We deduce that :
YB = (2 M + m)
g
gx
g
x
−m
= Mg + m (1 − ) et YA = (2 M + m)g − YA
2
2 l
2
l
uuur
Cancel the moments of the actions that exert themselves in O on OB :
XB
0
l sinα
0
−l cos(α / 2) Λ − Mg + −l cosα Λ YB = 0
0
0
0
0
0
l sin(α / 2)
By letting u = m(1 − x / l ) / M , we obtain :
X A = − X B = tanα(−
and
YB = Mg +
Mg
Mg tanα
+ YB ) =
2
2
mg
x
u
(1 − ) = Mg(1 + )
2
l
2
⎡
m⎛
1−
⎢1 +
M ⎜⎝
⎣
x ⎞ ⎤ Mg tanα
(1 + u)
⎥=
l ⎟⎠ ⎦
2
The discussion is facilitated by using a Cartesian plane where the abscissa and
the ordinate of the forces are 1 (Fig. S23.5). In the case where m = M , we
have, in reduced coordinates :
XA
tanα
= −xB =
(1 + u)
Mg
2
Y
u
yB = B = (1 + )
Mg
2
Y
u
y A = A = 3 − yB = 2 −
Mg
2
xA =
For equilibrium to be possible, we need the segment | x A / m s | to cut the segments representing
Ya and Yb in the interval [0,1] for u : Several cases present themselves:
: No equilibrium possible
: Points A and B will slide
: Point B slides before A
: No sliding
Exercise 8
Here we have a problem with 5 joints: the joints are at points A, B, C, D and E.
In the system of coordinates on the diagram the various components of forces
applied to the system are:
F1
0
F1
, R1
X1
Y1
, R2
X2
Y2
, R3
X3
Y3
, R4
X4
Y4
, R5
X5
Y5
R 1 the reaction in A
R 2 the reaction in B
R 3 the reaction in C
R 4 the reaction in D
R 5 the reaction in E
The immobility of HC gives the following equations :
X5 X3 0
0
ur
u ur
u ur
u r
⎧⎪ X 5 + X 3 = 0
➙ F1 + R5 + R3 = 0 ⇒ − F1 + Y5 + Y3 = 0 ⇒ ⎨
− F + Y + Y3 = 0
0 ⎪⎩ 1 5
0
0
0
(1)
(2)
➙ Sum of the moments with respect to E is zero
F1.HE+Y3.EC=0
F1.c +Y3.a =0 (3) ➙ Y3 = −
(3)
F1c
a
(4)
0
R 3 cos 30
R 3 = R 3 sin 30 0 =
0
X 3 = R3
5
R3 =
X5 = −
3
2
R3
2
0
3
3
c 3
= − X 5 or X 3 = R3
= − F1
2
2
a
x = −R
(4)➙
R3
3
3
2 (5)
−2F1 c
a , in modulus R 3 = 2F1 c
a
F1c 3
Fc 3
,in modulus X 5 = 1
a
a
(6)
(2) and (4) give
⎛ F c⎞
⎛
c⎞
Y5 = F1 − Y3 = F1 − ⎜ − 1 ⎟ = F 1⎜ 1 + ⎟
a⎠
⎝
⎝ a ⎠
Immobility of the stem AD
uur uur uur
r
R4 + R5 + R1 = 0
(7)
X4
X5
X1
0
Y4 + Y5 + Y1 = 0
0
0
0
0
moment
⎧ X 4 = X1
⎪
⇒ ⎨ X 4 + X 5 + X1 = 0
⎪Y + Y + Y = 0
5
1
⎩ 4
(8) and (9)
Immobility of the stem DB
uur uur uur
r
R4 + R3 + R2 = 0
X4
X3
−X2
0
Y4 + Y3 + −Y2 = 0
0
0
0
0
moment with respect to point B
uuur uur uuur ur
u r
BD ∧ R4 + BC ∧ R3 = 0
(8)
(9)
(10)
X 4 −2a sin 30 X 3
−4a sin 30
0
+4a cos30 ∧ Y4 + +2a cos30 ∧ Y3 = 0
0
0
0
0
0
−4a sin 30Y4 − 4a cos30 X 4 − 2a sin 30Y3 − 2a cos30 X 3 = 0
We have the relation
Y4 + Y3 − Y2 = 0
cF1 cF1
cF
−
− Y2 = 0 ⇒ Y2 = − 1
2a
a
2a
3c
F =0
2a 1
2c
3c
⇒ Y4 +
F1 −
F =0
2a
2a 1
Y4 + R3 −
⇒ Y4 −
Then
Y4 + Y3 − Y2 = 0
cF1 cF1
cF
−
− Y2 = 0 ⇒ Y2 = − 1
2a
a
2a
c
F =0
2a 1
ou bien Y4 =
c
F
2a 1
X4 + X3 − X2 = 0
and
c 3F1 cF1 3
c 3F1
+
− X2 = 0 ⇒ X2 = −
2a
a
2a
−
From (10)
Y1 = − Y4 − Y5 = 0
cF1
cF
− F1 − 1
2a
a
⎛
3c ⎞
Y1 = − F1 ⎜ 1 + ⎟
2a ⎠
⎝
Y1 = −
Exercise 9
uuur
The center of gravity is defined by M OG =
∫
uuuur
OM dm
(S )
Consider a base sector rdq , of radius r. The angle q is counted beginning at the
axis Ox.
Center of gravity
du
64748
⎛2
⎞
r
cosθ
⎜⎝ 3
⎟⎠
lecentr e de gr avite
xG =
1
M
π
2
∫σ
0
sec teur
Surface
⎛ 1 2 ⎞ 4r
⎜⎝ 2 r dθ ⎟⎠ = 3π
1424
3
sur facedu sec teur
Center of gravity
du
64748
⎛2
⎞
⎜⎝ 3 r sinθ ⎟⎠
lecentr e de gr avite
1
M
yG =
π
2
∫σ
0
sec teur
Surface
⎛ 1 2 ⎞ 4r
⎜⎝ 2 r dθ ⎟⎠ = 3π
1424
3
sur facedu sec teur
Exercise 10
dy
By reason of symmetry, the center of gravity is found on the axis Oy .
dm = σ dS = 2σ dyr
r = R 2 − y2
R
MyG = ∫ 2σ ydy R 2 − y2
0
We let
0
R 2 − y2 = u and we integrate in terms of U going from R to 0
yG = ∫ −2σu2 du =
R
4R
3π
Exercise 11
R
R
0
0
R
MzG = ∫ zρdV = ∫ zρπr dz = ∫ zρ(R 2 − z2 )dz =
zG =
2
0
ρπ R 4
4
3R
8
Learning activities
The students should read a complementary course on forces being applied to
solids. The students must do all the exercises. They are organized in groups
for collaborative work. Each group completes the proposed exercises and
designates a leader of the group who will report for each group. The professor
gives a deadline for each exercise, at which time each group will send an attached file of their reports to the professor of the course.
Teacher’s guide
Learning activity 4
Activity title
•
•
•
•
Composition of movementsDynamics of a material point–
Work, energy power –
Oscillators
•
•
•
•
•
•
Calculate the kinematic characteristics of a moving object (velocity and
acceleration)
Calculate the kinetic energy
Calculate the work of a force
Apply energy theorems
Use the laws of composition of movements
Activity summary
This actvities allow students to accomplish the following
• Master calculations of velocity and acceleration
• Describe the various forces applied to a given system,
• Learn to manipulate vector operators and mathematical tools
Then, knowing the Newton’s three laws, work, the theorem of kinetic energy, we
can easily write and solve differential systems governing the laws of motion.
Key concepts
•
•
•
•
•
•
•
•
Work = Energy granted by a force to move a body from one place to
another.
Absolute Movement = movement against a fixed reference or absolute
Relative motion = movement against a moving object reference
Training Movement = movement that a moving object would have, if it
was fixed relative to the reference of the moving object.
Energy = kinetic energy of moving masses
Pendulum = all suspended bodies
Oscillator = any body that oscillates around an equilibrium position
Period = time taken by an oscillator to complete a full oscillation
(Toutes en Annexe 4)
1°) RATIARISON, A. (2006). Référentiels, Dynamique du point matériel,
Travail-Puissance-Energie.-fFaculté des Sciences -Oscillateurs. Université
d’Antanarivo. Cours inédit.
2°) Papanicola Robert, http://www.sciences-indus-cpge.apinc.org/IMG/
pdf/ CIN2_DERIVATION_VECTORIELLE.pdf
Vectorial derivation
3°) http://abcsite.free.fr/physique/meca/me_ch3.html
Work, energy, power
Oscillators
4°) RATIARISON, A. (2006). Grandeurs physiques, Opérations vectoriellles.
Madagascar. Université d’Antanarivo. Cours inédit.
*
*
*
*
*
*
*
*
*
*
*
CAZIN, M. (1972). Cours de mécanique générale et industriielle- Gauthier
Villars- Tome 1
PEREZ, J. P. (1995), Mécanique Fondement et applications –MASSONTIPLER, P. A. – Worth Publishers (1995) , Physics for Scientists and
Engineers – New York, NY 1003 –
* The Free High School Sciences: A Textbook for High School Students
Studing physics – FHSSt Authors- December 9, 2005 from http://savannah.nongnu.org/projects/fhsst
ANSERMET, J. P. (Version 2004-2005), La mécanique rationnelle
– Formation de base des Sciences et des ingénieurs – Institut de Physique des nanostructures- Ecole Polytechnique Fédérale de Lausanne de
Lausanne– PHB – Ecublens, 1015 Lausanne.
CLAUDE, S. B. Exercises de mécanique from http://www.univ-nantes.
fr
MEYER, D. Exercises de mécanique. from http://www.chimix.com/mines/dyna1.htlm
SAINT BLANQUET Claude, Exercises de mécanique, http://www.univnantes.fr
D. MEYER, Exercises de mécanique, http://www.chimix.com/mines/
dyna1.htlm
Useful links
•
•
•
•
•
•
•
•
Vectorial calculations
Kinematics
Vectorial derivation
Integration of functions of several variables
Vectorial integration
•
•
•
http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/aquadiff.
html
•
•
•
•
•
•
•
•
http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/epiclc.
html
http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/prlong.
htm
http://www.univ-lemans.fr/enseignements/physique/02/meca/couplage3.html
http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/froflu.
htm
This course is divided into two main parts:
Part I: A supplementary course with basic knowledge and general knowledge.
In addition to this course, there are four exercises showing us physical phenomena seen in everyday life such as:
-
-
-
-
The field of terrestrial gravity,
The deviation towards the East when you drop an object without initial
velocity,
The deviation of the Foucault pendulum,
The tides.
The students can download the sites below to see animations.
Part II: Four exercises, related to the laws of composition of velocity and acceleration, application of the fundamental principle of dynamics in a fixed reference
or a reference relative to that of the theorem of kinetic energy and damped and
undamped mechanical oscillators.
These exercises are required. They will be solved individually by each student
who will produce their own report to be sent by email to the professor.
Evaluation
First portion of activity 4 evaluation
Course complement
Table of Contents
Contact the author
Phenomena of terrestrial dynamics
Illustrations and animations of Geneviève Tulloue
Rotation of the Earth Coriolis Effect Plan
1. The field of terrestrial
gravity 2. Deviation to the East 3. Foucault pendulum 4. The tides
Pendulum on a turning surface Foucault pendulum
Depressions Tides (influence of the sun) Tides (influence of the Moon and the sun) Roche’s limit The seasons
CABRI
In a reference linked to earth’s surface, the fundamental equation of dynamics
can be written:
a r (M ) is the relative acceleration of M
m is the mass of M
uuuuuuur
KM r
G T ( M ) = − 2 T z is the universal field of attraction at a distance r RT ,
r
radius of the Earth
ur
Ω is the velocity of rotation of the Earth
uuuuuuur uuuuuur
m⎡G A ( M ) − G A (C ) ⎤ is the universal attraction of planetary bodies other
⎣
⎦
than the Earth
Fa is the force applied on the object M
The field of terrestrial gravity
and the gravity field
The weight of a body are defined by the relation :
where radius of the Earth
is the universal field of attracton at a distance
We call the direction of the vector as downward.
At ground level, and , also as a first approximation we often confuse
and .
explains the variations of the terrestrial gravity
field. Often, we assume
.
Deviation towards the East
if we neglect the gravitation of planets other than
Earth, and if there are no forces acting on the object.
By projecting the Oxyz reference, we obtain:
By neglecting the term I
At the instant above the ground
in
front of , we drop the object without initial velocity from a height
.
By integration, and by neglecting z in front of and
in front of ,
we find and .
This phenomenon is known under the name of deviation to the East.
is equal to 27 mm for et . This effect is very weak and is often neglected.
There, in the Northern Hemisphere (South), a deviation from the South (North)
even more negligible.
The Foucault pendulum The simple pendulum of length is only sli-
ghtly sloped with respect to the vertical.
; ; (
)
; ; The cord of the simple pendulum is inextensible and the mass m is subject to
its weight and tension
due to the cord.
The proper rotation vector of the Earth is : also ,
In the last equation, the term we neglect the Coriolis force), the term
is negligible (it is not written when
also (pendulum always slightly
away from vertical). This allows us to write .
The equations become To solve this system can be decoupled by derivation, or better, using the complex variable which leads to By integrating, we obtain :
This solution shows a turning movement in the plane Oxy to the pulsation
(period where is the sidereal period of Earth). The
rotation is counterclockwise in the Northern Hemisphere and clockwise in the
Southern Hemisphere.
This phenomenon has been studied by Foucault in the last century with a pendulum of 67 m in length suspended in the dome of the Pantheon.
Note: There are other phenomena showing the rotation of the Earth itself, the
most famous are the swirling drain of a sink, and preferential erosion of one
bank of a river ‘s flow towards the north or south.
4. The tides Considering the huge mass of water of an ocean, it is no longer possible to
ignore the term which is responsible for the tides.
To start with, we will only consider the influence of the moon.
At the two points Z and N, is directed outward from
the Earth and is opposed to Earth’s gravity while staying in the same direction.
On a point M, has a horizontal component [maximum if Z
(or N) is 45 ° CM] producing a deviation from the vertical.
It is the same for the Sun.
The maximum deviations are 0.017 «for the action of the moon, of 0.008» for
the Sun.
These deviations, although very low, are highlighted by the variation of the
surface of the ocean: it changes every minute so as to be constantly perpendicular to the vertical.
The effects of the other stars are negligible, so that in practice only the effects
due to the Moon and Sun are considered. These effects are added when the
Sun, Earth and Moon are aligned: the tides of spring are compensated when the
directions Sun-Earth-Moon and Earth are perpendicular: it is the tides of neap
tide. Finally, during the equinoxes of spring and autumn, the effect due to the
sun is at its highest: it is the equinox tides.
The rotation of the Earth itself explains the existence of two high tides and two
low tides each day.
Second portion of activity 4 evaluation
Series of exercises
Exercise 1.
A ring of mass m, dimensions negligible, slips and moves on a circular helix
axis Oz and whose parametric equations are: x = r cos q, y = r sin q, z = h q.
The applied forces are the weight, a frictional force of constant intensity f, collinear with the velocity but opposite in direction, and the reaction of the track,
normal to the movement at every moment.
In a terrestrial Galilean reference :
1. Express the Cartesian components of the friction force against f, r, q, and a,
the angle between the velocity and the horizontal plane. For the following
we shall assume that this angle is constant and tan a is expressed in terms of
h and r.
2. Calculate the work of the friction force when moving between points B (q p
= 4) and A (q = 0).
3. Expressed in terms of given data the weight and reaction of the track between
B and A.
4. Deduce the velocity of the ring at point A given that its initial velocity was
zero at B. Discuss.
Exercise II
- A solid (S) mass m = 5 kg is moving on rails ABC located in a vertical plane.
AB = 4.0 m; BD is the arc of a circle of radius R = 10 m. (S) is initially an immovable object in A. We exert, between A and B, on (S), a force F parallel to
AB and constant. The solid goes to D and then goes back. H = 3 m g = 9.8 m
s-2. The friction is negligible.
1. Express and calculate the velocity of (S) B.
2. Express and calculate the value of F.
3. Express and calculate the velocity of (S) in C. (H = 1.5 m). Show that the
velocity C is similar going and returning.
4. Determine the action R of the support at point C.
5. At point D can the solid be balanced?
6. Compare the length of journeys AB and BA.
II-friction is no longer neglected. The value of friction f is constant. The solid
stops as it returns to B.
1. 2. Express and calculate f and F.
Compare the forward and return journeys:
- The values of the velocity at any point of the arc BD
- Travel time BC and CB.
III We exert on the solid (S) a weak force F ‘; it reaches D, then stops, returning to a height h’ = 0.5 m. Justify this behavior of the solid.
Exercise 3
We are interested in some properties of the oscillators in one dimension, ie
whose evolution over time can be analyzed by a function x (t). Called harmonic oscillator any system whose function x (t) is the corresponding solution
of the differential equation:
w0 rad/s,is the pulsation of the oscillator
The elastic pendulum :
Consider a solid (S) mass m, center of mass G. When S is an immovable object
in the laboratory reference, G is in O, origin of the horizontal axis x’x. The
solid S ‘s only movement possible is rectilinear translation along the axis x’x.
S is submitted to a single force, the tension T of an elastic spring of stiffness
constant k, of negligible mass. There is no friction and the weight of the solid
is compensated by the reaction of the support. The position of S is marked by
the abscissa of G.
1. Prove that we have made a harmonic oscillator whose evolution over time
is governed by the equation:
Explain the function x (t) if the initial conditions of movement are at t = 0, x
(0) = A, positive amplitude and v (0) = 0.
2. We call phase space (in this case, plan of phases) the plan of coordinates (x,
dx / dt). What is, for the harmonic oscillator, the trajectory of P characteristic
of the state of the system in the phase space?
3. From which potential energy function does the tension T derive itself from?
Retrieve the above equation using conservation of mechanical energy.
Exercise 4
1. The solid S previous stated, comparable to a material point, is a moving
object without friction along the x axis. It is subjected to a force F collinear
with the axis x’x and deriving from a potential energy function Ep (x) that
does not depend explicitly on time
-
-
Can the two trajectories of point P in the phase space, defined in the previous question, be cut?
What conditions Ep (x) and its derivatives with respect to x have to verify
so that the movement of M is in the vicinity of an equilibrium position
M0, abscissa x0 comparable to that of a harmonic oscillator.
2. The force field is performed as follows: M is the object moving without friction on the axis x’x through a spring of stiffness k and void length l0. O is
the orthogonal projection of A on the x-axis, and we let OA = l which can be
larger or smaller than l0.
-
-
-
Express the potential energy Ep(x) as a function of x and the parameters
k, l and l0.
Are there equilibrium positions? What are the types of the function
Ep(x)?
In the vicinity of these positions can one speak of a harmonic oscillator?
1. Various physical systems can be modeled by a solid S similar to a material
point of mass m moving along one axis, subjected to a tension of a spring
and a viscous friction force, ie opposite to the velocity (- kx ‘), k is a positive
constant.
2. Establish the differential equation satisfied by x (t). Specify the type of solutions depending on the value of the coefficient k. -
-
Why would using a Foucault current braking device be better than the
braking force of a pallet in a liquid?
What about the cases where l <0?
Exercise 5
Weighted pendulum :
It consists of a solid of mass m and center of gravity G, a moving object
without friction around a horizontal axis D perpendicular to the plane of the
figure. The moment of inertia of the solid relative to this axis is J.
1. Establish the differential equation verified by q (t). Show that if q is small,
the pendulum weight can be treated as a harmonic oscillator of angular
frequency ω″0 = mga / J.
2. Consider briefly the case of a simple pendulum for which the solid consists
of a particle of mass m suspended from a taut, of massless length l.
Exercise 6
The potential energy of interaction between the atoms of a diatomic molecule
is given by the expression of the Morse potential: E(r) =A(1-exp-a(r-r0) )².
r : variable distance between atoms, a: positive constant, r0 and A: positive
parameters which will determine the physical meaning.
1. What is the expression of molecular interaction forces. Determine r, for
there balance? This equilibrium is stable?
2. Find these results using the potential energy.
3. Calculate the dissociation energy of the molecule.
4. The potential energy of interaction between 2 nucleons is given by the
Yukawa potential: E (r) =- B r0 / r exp-r/r0. Deduce the interaction of
nuclear forces.
5. Represent the previous forces and compare them.
Exercise 7
Consider a point particle of mass M orbiting at a distance r = OM from the
center of a spherical body that is subject to an attractive force:
weight of the particle is negligible.
The
1. Express its potential energy as a function of K and r.
2. The trajectory of the particle is circular with center O, show that the
movement is uniform, and calculate the kinetic energy of the particle.
3. Calculate the mechanical energy.
4. It causes a relative decrease of 10-4 of mechanical energy. What happens
to the velocity and radius of the trajectory?
5. The initial distance is OM = r0. What are the minimum energy and velocity
needed to provide for the rescue of the attraction of the spherical body.
Exercise 8
A particle moves in the force field
Following the trajectory defined by the parametric equations, in the SI system
of units:
x=3t ; y=2t2 ; z=t-2
1. Calculate the power received by the particle at time t.
2. What is the position of the particle when the power is minimized?
3. Calculate the work done by the force field between times t1 = 0s and t2 =
2s?
4. What is the work if the particle is compelled to move in a straight line from
its position at time t1 = 0 to its position at time t2 = 2s? Conclusions?
Exercise 9
1. The suspension of a car of mass M = 600kg, is shown by a spring of stiffness k. We note that the wheels, which we neglect the mass, leave the
ground when the car is raised to a height h = 30cm. Determine:
2. The stiffness k of spring.
3. The equation of vertical motion, and the period of vertical oscillations of
the empty car.
4. What is the period with 4 passengers totaling a mass m = 300kg.
5. We add to the previous suspension a damper which creates a frictional force proportional to the vertical velocity f =-bv.
At rest, the system damping is critical. Write the equation of vertical
motion. B. Determine
6. When the car has 4 passengers, what are
- The equation of vertical motion.
- The pseudo-period T ‘, and compare it to the natural period of
damped oscillation.
Assume g = 10m.s-2. Exercise 10
An elastic pendulum of mass m = 0.1 kg, stiffness = 20 N / m oscillates without solid friction along a rod that is at a constant angle with the vertical downward.
1) a) Establish the dynamic differential equation of the motion of A along the
fixed rod. Find the position around which the movement takes place and
calculate the oscillation period of the movement.
b) We note that the oscillations are damped according to an exponential
law. To what type of force do we attribute this depreciation? After 30
oscillations, the amplitude is divided by three. Calculate the logarithmic
decrement and the quality factor of the pendulum.
2) The rod is rotated uniformly around the stem (the vertical) with an angular
velocity. Express, in the rotational frame, the kinetic energy of the pendulum,
the gravitational potential energy, elastic potential energy and the centrifugal
potential energy. Establish, in the absence of friction, the differential equation of motion. Determine the equilibrium position and period of oscillations
knowing that r sinθ o = 0,9ω o
Exercise 11
Consider two springs the same length L and different stiffness k1 and k2.
1) The springs are placed vertically in parallel. The upper end is fixed and
the other carries a weight of mass m. Find the expression of the pulsation
of the oscillation formed. Conclude.
2) Answer the same question when both springs are placed in series. Conclude.
Exercise 12
Throughout the exercise, we take g = 10 m / s2. We neglect the friction. It uses
a spring of negligible mass, wound with non-touching joints.
1 Preliminary study of the spring
To determine the stiffness k of a spring, we hang one end to a fixed support.
When we hang a mass m = 200 marked g at its other end, the spring stretches
by 10.0 cm.
a) Verify that the spring rate is 20.0 N / m.
b) b) Using the theorem of the center of inertia, show that the stiffness can
also be expressed in kg / s2.
In which quantity does
express itself ?
2- Study of an elastic oscillator
a) We now fix the studied spring as shown in Figure 2. The spring is horizontal, one end is fixed. At its other end is a solid (S) mass m = 200 g. This
solid can move without friction along a horizontal axis Ox. At equilibrium, the center G of the solid coincides with the origin 0 of the reference.
- Establish the differential equation that governs the movement of G.
Verify that, if we correctly choose To, the function
, of period To, is the solution to the previous differential equation.
- numerically, calculate the value of the period To.
-
b We compress the spring to the left. Point G occupies the position GO such
that OGo = - 0.15 m.
At time t = 0, we drop the solid without initial velocity. Determine the amplitude XM and phase F of movement and the expression of the velocity v
(t) of the solid. Determine the value of the maximum velocity.
c) Identify and express the mechanical energy of the undamped oscillator. Calculate
its value at t = 0. (We take the potential energy of the initial spring when x = 0).
By accepting and using the conservation of this mechanical energy, find the
maximum value of the velocity of the solid.
Exercise 13
The numerical values necessary for the solution are given at the end of the
exercise.
To model the spring suspension system of car, one student suggests using a
spring of spring constant k (value specified by the supplier).
A – Study of free oscillations
The student uses a system of data acquisition schematically shown in Figure 1.
Two electrodes A and B, immovable objects, immersed in the solution S, are
attached to positive and negative terminals of a voltage generator. A metal rod,
covered with insulation along its entire length, is attached to the mass m. Its
end E, slightly stripped of its insulation, follows exactly the movement of the
mass m.
Measuring the voltage between point E and the 0 V terminal of the generator
can detect the position of E (the measuring device is not shown in the diagram). Thus, it is possible to know the position of mass m during the oscillations.
After setting the parameters of the acquisition software, the student moves the
mass m down by 1 cm, and leaves the system oscillate freely.
The triggering of the acquisition is by moving to the position of equilibrium.
The curve obtained is given in figure a.
• A-1 Discuss the shape of the curve. What type of oscillations is it?
• A-2 Determine graphically the T0 period of oscillations of mass m hanging
from the spring.
• A-3 Is this value consistent with the theoretical value
•
A-4 We add to the extremity E of the stem a horizontal disc of negligible volume and mass. Frictional forces are now involved.
Draw the shape of the curve obtained after a new acquisition.
B – Study of forced oscillations
Students now connects the upper end of the spring to an eccentric driven by a
motor (Figure 2) and made several recordings for different velocity of rotation
of the engine measured by the rotation frequency f in Hertz. At the end of the
rod is always horizontal disk volume and negligible mass.
The student records the amplitude of each curve.
f (Hz)
1,5 2,0 2,5 2,8 3,1 3,2 3,3 3,6 4,0 4,5
xmax (cm) 0,4 0,6 1,0 1,5 2,1 2,3 2,0 1,5 1,0 0,7
• B-1 What name do you give the engine fitted to the eccentric and the
system (spring + mass)?
• B-2 Build a plot of xmax as a function of the frequency f. What phenomenon
do we observe at f = 3,2 Hz ?
Compare the frequency fR of oscillations at a resonance corresponding
to the frequency f0 of free oscillations of the undamped system (spring +
mass).
• B-3 (s) What change (s) would we observe if we used a more viscous solution S.
C – Suspension of a moving object
The suspension system of a moving object includes springs and shock absorbers.
The moving object is a mechanical system oscillating at a frequency f0.
Some desert tracks have an aspect of “corrugated iron”: they have a pattern of
bumps, with a spacing of L (a few tens of centimeters).
For a velocity VR, the vehicle undergoes large amplitude oscillations that dangerously diminish its handling.
C-1 Explain this phenomenon.
C-2 Express the velocity VR as a function of f0 and L.
Calculate this velocity in km.h - 1 with f0 = 5,0 Hz and L = 80 cm.
Given :
k = 40 N.m -1
m = 100 g
π= 3,14
Exercise 14
Throughout the exercise, we take g = 10 m / s2. We neglect the friction. It uses
a spring of negligible mass, wound with non-touching joints.
1 Preliminary study of the spring
To determine the stiffness k of a spring, we hang one end to a fixed support.
When we hang a mass m = 200 marked g at its other end, the spring stretches
by 10.0 cm.
a) Verify that the spring rate is 20.0 N / m.
b) Using the theorem of the center of inertia, show that the stiffness can also
be expressed in kg / s2.
In which quantity does
express itself ?
a) We now fix the studied spring as shown in Figure 2. The spring is horizontal, one end is fixed. At its other end a solid (S) mass m = 200 g. This
solid can move without friction along a horizontal axis Ox. At equilibrium, the center G of the solid coincides with the origin 0 of the reference.
- Establish the differential equation that governs the movement of G.
-
, of period To, is the solution to the previous differential equation.
-
numurically calculate the value of the period To.
b) We compress the spring to the left. Point G occupies the position GO such
that OGo = - 0.15 m.
At time t = 0, we drop the solid without initial velocity. Determine the amplitude XM and phase F of movement and the expression of the velocity v
(t) of the solid. Determine the value of the maximum velocity.
c) Identify and express the mechanical energy of the undamped oscillator.
Calculate its value at t = 0. (We take the potential energy of the initial
spring when x = 0).
By accepting and using the conservation of this mechanical energy, find the
maximum value of the velocity of the solid.
Exercise 15
A spring of stiffness K is horizontal, one end is fixed and at its other end a
solid mass m. This solid can move along a horizontal axis Ox. O is the position
of center of inertia of the body in equilibrium (see Figure).
There is friction present. They are reduced to a force = - h, where it denotes the
instantaneous velocity of the solid. The coefficient h is positive.
•
1 - Establish the characteristic differential equation of motion of the solid.
• 2 - What is the nature of this movement? What is the nature of x (t) depending on the value of damping coefficient A = h / m.?
• 3 - The energy of the oscillator.
a) Give the expression of the mechanical energy of the solid-spring system.
b) Establish the relationship between the derivative of the mechanical
energy versus time and power of the friction force.
c) Comment on this relationship in terms of energy transfers.
• 4 - Using an interface connected to a computer, there was a voltage portional to x u (t).
The computer is programmed so that 1 volt corresponds 1 cm.
From the graph above:
a) Determine the initial conditions imposed on this oscillator
b) Calculate the pseudo-period.
c) Determine the mechanical energy E m of the oscillator each time it
is passed by a negative extremum of x (limited to the first four).
What can be said of the value (Em) i / (Em) i + 1? Data: k = 10 N / m.
d) Determine the work of the frictional force between the passages of the 1st
and 4th negative extremum.
Exercise 16
A point moving object of mass m is located relative to a Galilean frame
(
)
R O, e x , e y , e z . We call R’ the reference of the center O and the base
ur ur ur
er ,eθ ,ez with OM = r e r + ze z
(
)
ez
M
ey
r
ex
e
H
H
er
(a)We take R ‘as a relative reference and a projection reference of the resultant
vectors. Calculate by composition the movement V R ( M )
and find them again via a direct calculation.
and a R ( M ) ,
(b)To continue the problem, we suppose z=0 ; the point M is a moving object
ur
ur
on the plane xoy. In addition, its weight P = −mgez , is subjected to 2 forces
F1 and F2 , defined by :
• ur
ur
u
ur
F1 = mr ⎛ 2θ er + eθ ⎞
⎝
⎠
t.
and
ur
u
ur ur
u
F2 = mlez ∧ VR (M ) , l is a scalar function of
a) From the fundamental relation of dynamics, give the scalar equations of
motion of M.
b) Write the theorem of kinetic energy and show that there exists a first
integral of energy.
(c) We assume l=0 the point M is a frictionless moving object on a circle of
center O with a radius that reacts with a bonding force L . We suppose
at the instant t=0, the point M is an immovable object with q=0. With the
help of the first energy integral, express L as a function of only q.
Find the differential expression of movement M from the theorem of angular
momentum at a point.
Exercise 17
A material point M moving object of mass m is located relative to a Galilean
(
)
reference frame R O, e x , e y , e z by the spherical coordinates (r, q, f). We call
(
ur ur ur
)
R’ the center reference O and of basis er ,eθ ,eφ , with OM = r e r .
i)
We take R ‘as a relative reference and projection reference on the resulting
vectors. Calculate the composition by motion V R ( M )
Confirm them by direct calculation
and a R ( M ) .
ii) ° The point M describes the frictionless meridian circle with center O and
(
uur uuuur
)
radius a plan Oxz. A position is identified by the angle θ = Oz,OM .
With respect to R. We consider the coordinates of the following points :
A(0,0,-a), B(0,0,a), C(a,0,0)
Assume that M is dropped without initial velocity at point B. During its move-
ur
ur
u
ment, it is subject to its own weight P = −mgez , , to the reaction L of the
support, to the force F = k AM , k a positive constant.
a) Explain the basic law of dynamics and derive the differential equation of
motion and expression of L.
b) Write the theorem of kinetic energy and derive a first integral of motion,
then the expression L as a function of q ; this will give the differential
equation of movement.
c) Given in part C, the expression of the linear velocity v of M, and that of
the reaction L.
d) Show that there exists a value of q for which L = 0. Numerical application
mg
.
a
k=
Exercise 18
(
We consider a fixed reference R 0 O, x 0 , y 0 , z 0
) and a relative reference
ur
u r
ur
u ur
R 1 O, x , y, z 0 , such that x0 , x = y0 , y = ψ (t) . Consider a point M defined
(
)
( ) ( )
( )
in the plane (Oyz0) by its polar coordinates ρ,θ ) on which we associate part of
the unit vector v of OM , and the unit vector w , deduced from v by an angular
rotation of
π
in the plane (Oyz0).
2
(
)
The vectors x, v and w form the reference R O, x , v, w , of origin O.
r and q are functions of the time t.
In all of the problems, the results are expressed in a reference frame
(
)
R O, x , v, w .
(a) We designate v R 0 (M ) the velocity of M in its movement with respect to
(
)
R 0 O, x 0 , y 0 , z 0 .
Determine v R (M ) , v R 1 (M ) and v R 0 (M )
uuuuuuuuuuuur
(b) We designate vR 0 ( M ∈ R1 ) the trailing velocity of M in its movement
(
) (
with respect to R 0 O, x 0 , y 0 , z 0 , R 1 O, x , y, z 0
reference.
uuuuuuuuuuuur
uuuuuuuuuuuur
) being the relative
uuuuuuuuuuuur
Prove the relation : vR 0 ( M ∈ R1 ) = vR 1 ( M ∈ R ) + vR 0 ( M ∈ R1 )
uuuuuuur
uuuuuuur
uuuuuur
(c) D e t e r m i n e t h e a c c e l e r a t i o n s : a R ( M ) , a R 1 ( M ) , ac ( M )
uuuuuuuuuuuuur
a R 0 ( M ∈ R1 )
,
(d) (i) For what conditions is the relative motion of M the central acceleration
of center O?
(ii) For what conditions other than the previous, is the absolute motion of M is also the central acceleration of center O?
(e) Assume three true propositions:
-
r= a = Cte
-
The relative motion of M is not central acceleration of center O
-
The absolute motion of M is central acceleration of center O.
•
•
Find ψ and θ as a function of q only.
Correction of Activity 4 exercises
Exercise 1
In the local reference (u, t, k) the components of the frictional force are : (0 ; f
cos a ; f sin a )
Components of velocity, derived from the vector position with respect to time :
(-r sinq q ' ; r cosq q ' ; hq ' )
Or in the local reference (u, t, k) :
(0, rq ' ; hq ' )
tan a = hq ' / (rq ') = h / r.
Work of the frictional force :
Elementary displacement :
dx = -r sinθ dθ ; dy = r cosθ dθ ; dz = h dθ ;
friction (-f cosα sinθ ; f cosα cosθ ; f sinα)
scalar products between friction vectors and displacement :
-f r cosα sin²θ dθ + f r cosα cos²θ + f h sinα dθ .
To obtain the work on the displacement of B to A, integrate between 4π and 0,
a and r are constants
While noticing that sin²q = ½(1-cos(2q)) and that cos²q = ½(1+cos(2q))
W = -4π f [r cosα + h sinα].
Work of weight of B in A :
Elementary work during elementary displacement hdq :
dW = mghdθ
To obtain the work on the displacement of B to A, integrate between 4π and 0
W = 4π mgh .
The work of the normal reaction is zero ( perpendicular force to displacement)
velocity in A :
Write the kinetic energy theorem between B and A : (in B the velocity is zero)
½mv²A = 4π mgh -4π f [r cosα + h sinα].
v²A =8π [gh - f /m[r cosα + h sinα]].
This is possible when friction is not important :
mgh>4π f [r cosα + h sinα].
Exercise 2
Kinetic energy theorem :
From B to D : perpendicular to the velocity does not work. The working
weight is resistant (upward) and is WP= - mgH
The kinetic energy in D is zero (stop) ; the kinetic energy in B is : ½mv² ( v :
velocity in B)
The change in kinetic energy is equal to the sum of the work of forces applied
to the solid: 0
-½mv² = -mgH
v²= 2gH = 2*9,8 * 3 = 58,8 ; v = 7,7 m/s.
From A to B : les forces R et P perpendiculaires à la velocity, ne travaillent pas.
Le travail de la force F vaut WF=F AB.
The kinetic energy at A is zero (stop); kinetic energy at B is ½ mv ² (v: velocity
in B)
to the solid:
½mv² -0= F AB
F= mv²/(2AB) = 5*7,7² / 8 = 36,7 N.
velocity from (S) to C :
From B to C : R, perpendicular to the velocity does not work. The working
weight is resistant (up) and is: WP= - mgh
The kinetic energy in C is ½ mv ² C; kinetic energy at B is ½ mv ² (v: velocity
in B)
to the solid : ½mv²C -½mv² = -mgh
v²C = v² -2gh = 58,8 -2*9,8*1,5 =29,4 ; vC = 5,4 m/s.
The expression of this velocity vC indicates that it only depends on the velocity
at B and the altitude h, whatever the direction of travel.
or we can apply the theorem of kinetic energy between the two passages of
solid M located on the arc BD going and returning:
R does not work and the work of the weight is zero (at the same point the
altitude difference is zero). So the kinetic energy, the value of the velocity does
not change the outward and return journeys.
Action R of the support at point C :
Write Newton’s second law on an axis normal to the path and headed toward O
:
R is inward pointing to O and its value is: R= m[v²C/OC + g cosq]
cos q = (OB-h) / OB = (10 -1,5) / 10 = 0,85 ; q = 31,8°.
R= 5(29,4/10+9,8 * 0,85)= 56,3 N.
At point D the solid can not be in equilibrium: the vector sum of forces is not
zero. Duration of the journeys AB and BA:
AB : write Newton’s second law : F=ma such that a = F/m = constane ; initial
zero velocity :
AB=½at² = ½F/mt² and t² = 2AB m/F = 8*5/36,7 =117,4 ; t = 1,1 s.
BA : pseudo-isolated solid, thus uniform rectilinear motion . AB = vt
t = AB/v = 4/7,7=0,52 s.
Friction is no longer neglected. The value of friction f is constant. The
solid stops on its return along B.
Express and calculate f and F : kinetic energy theorem :
From D to B : R, perpendicular to the velocity does not work. The work of
weight is
downhill and is : WP= mgH
The work of f (resistance) is : Wf= -f *length of the arc of the circle DB = -f
OC q (q in radians )
cos q = (OB-H) / OB = (10 -3) / 10 = 0,7 ; q = 0,795 rad.
The kinetic energy is zero in D (stop); kinetic energy at B is zero (stop)
to the solid:
0-0= mgH - f *OCq where f = mgH /(OC q) = 5*9,8*3/(10*0,795)= 18,5 N.
On the entire trip: R, perpendicular to the velocity does not work. The weight
does not work
(identical elevation at the departure A and arrival B)
The work of f (resistance) is : Wf= -f (AB+2OC q ) ; the work of F is WF=F AB
The kinetic energy of A is zero ( stop) ; the kinetic energy in B is zero (stop)
0-0 = -f (AB+2OC q ) + F AB and F= f (AB+2OC q ) /AB = 18,5(4+20*0,795)
/ 4 = 92 N.
Compare the outward and return:
- The values of the velocity at any point of the arc BD
apply the theorem of kinetic energy between the two passages of solid M located on the arc BD outward and then return:
R does not work and the work of the weight is zero (at the same point the
altitude difference is zero). The work of friction is negative, thus the velocity decreases: the return velocity is smaller than the outward velocity.
- Travel time BC and CB:
The velocity is lower on the return journey, so the length of return is greater
than the outward.
________________________________________
The solid stops while returning on the arc when the vector sum of forces is
zero.
The angle q 1 is then: cos q 1 = (OB-h ') / OB = (10-0,5) / 10 = 0.95; q 1 =
0.318 rad.
On the descent the work of weight is active. The driving force is
mg sin q 1 = 5 * 9.8 * 0.312 = 15.3 N, a value insufficient to overcome friction
(18.5 N).
However the driving force at D is: mg sin q = 5 * 9.8 * 0.714 = 35 N a value
sufficient to overcome the friction: the moving object does not remain at D.
Exercise 3
The system studied is the solid S. The reference is a laboratory, supposed as
Galilean.
Applying the theorem of the center of inertia, the only force exerted on S is the
spring tension.
with w02= k/m
funtion solution to the differential equation : x(t) = A cos (ω0 t + ϕ)
at t=0 : A =Acosϕ where cosϕ = 1 and ϕ = 0
x'(t) = -Aω0sin(ω0 t)
at t=0 x’(0) = 0
x(t) = A cos (ω0 t ).
Plan of phases:
x’(t) = -Aω0sin((ω0 t) soit x’² =A²ω²0sin²(ω0 t)--> sin²(ω0 t) = x’² /(A²ω²0)
x² = A² cos²(ω0 t)-->cos²(ω0 t) =x² / A²
or cos²(ω0 t) +sin²(ω0 t) =1
path whose equation is given above is, in terms of phases, an ellipse of semimajor axis A and semi minor axis Aω0.
Energy :
The tension derives from a potential energy, there exists a function such that
Ep:
the origin of the elastic potential energy is chosen at the equilibrium position
(G in O).
In the absence of friction there is conservation of mechanical energy.
E =½mv² +½kx²
Derived with respect to time ( the derivative of u² being 2 uu’)
½ m 2 v v’ + ½ k 2 x x’ = 0
with v’ = x” at x’ = v simplifying with v, we find : m x” + k x=0
Exercise 4
The physical system has a single degree of freedom. Write the conservation of
mechanical energy and then derive: we obtain a differential equation of second
order of the form x ‘= f (x, x’).
This equation admits a unique solution for given initial conditions.
Develop Ep (x) near the equilibrium position x0:
Ep(x) = Ep(x0) + (x-x0) (dEp / dx) x0 + ½(x-x0)² (d²Ep / dx²) x0 + ...
F = - dEp / dx
x0 is an equilibrium position if F(x0) =0 and (dEp / dx) x0 = 0
derive Ep(x) to study the stability of equilibrium :
F= - dEp/dx =-dEp / dx) x0- (x-x0) (d²Ep / dx²) x0= - (x-x0) (d²Ep / dx²) x0
if the second derivative is positive, then F is of opposite sign to x-x0; the material point is subjected to a force that brings it to its equilibrium position x0. The
equilibrium is called “stable”.
However, if the second derivative is negative, it tends to point away from the
position of equilibrium and the equilibrium is called “unstable”.
Potential energy :
Ep(x)= ½ kDl²= ½k[ (l²+x²)½-l0]²
The derivative with respect to x :
dEp/dx= kx[1- l0(l²+x²)-½ ]
si l >l0, the only equilibrium position is x=0
si l <l0, there are three equilibrium positions : x=0; x = (l0²-l²)½; x = -(l0²-l²)½
the shape of Ep is represented below:
Stability of equilibrium :
for x=0 : d²Ep/dx = k(1- l0/l) so if l>l0 the equilibrium is stable. If l<l0 the equilibrium is unstable.
for x = (l0²-l²)½ ou x = -(l0²-l²)½ , d²Ep/dx =k((1- l²0/l²) positive; the equilibrium
positions are stable.
Near the stable equilibrium positions, we have a harmonic oscillator.
The theorem of center of inertia is written : R (action of the support)
The solutions of this differential equation depend on the sign of the discriminant of the associated characteristic equation: ∆ = l ²-4km
∆ <0 low friction, pseudo-periodic regime
∆ = 0, the critical regime
∆> 0, aperiodic regime
the viscous friction force is not strictly proportional to the velocity. The braking force by Foucault current is proportional to the velocity.
l <0: The oscillator receives energy. The amplitude of oscillations increases.
Exercise 5
The system studied is the pendulum, a laboratory frame assumed Galilean.
The forces acting on the system are the weight and response of the axis. The
moment of the response of the axis with respect to O is zero.
The moment of the weight relative to O is GL-mg sin q.
Applying the theorem of angular momentum relative to a fixed axis:
q jd ² / dt ² =-sin q mgOG.
if the angle is small: sin q is near q rad.
Jq" + mgOG q =0
q" + mgOG / J q =0
it is the differential equation of a harmonic oscillator of angular frequency [mg
OG / J]½.
For a simple pendulum, J= ml² and the pulsation is w0² = g / l.
Exercise 6
Laboratory reference supposed Galilean
The molecular interaction force is conservative and derives from the potential
E(r)
Fm(r) = -dE / dr = -2Aa exp(-a(r-r0)) (1-exp(-a(r-r0))).
There is an equilibrium when the force is zero, meaning when r = r0.
r0 represents the average distance between two atoms forming the molecule.
Is the equilibrium stable ?
-2Aa exp(-a(r-r0)) is negative.
if r>r0, -a(r-r0)<0 and (1-exp(-a(r-r0))) is positive : Fm attractive.
if r<r0, -a(r-r0)>0 and (1-exp(-a(r-r0))) is negative : Fm repulsive.
Fm is a force towards the equilibrium position : this is stable.
Does the potential energy pass by an extremum ?
We derive the potential energy :
dE / dr = 2Aa exp(-a(r-r0)) (1-exp(-a(r-r0)))
this derivative is zero for :1-exp(-a(r-r0))=0 and r = r0
The potential energy passes by a minimum for r= r0, which corresponds to
stable equilibrium.
Dissociation energy
work done by an outside operator for separating the two atoms constituting the
molecule
dWop = -Fm dr
A represents the dissociation energy of the molecule.
The nuclear interaction forces are conservative and are derived from a potential energy.
Fn (r) = -dE / dr
distances between the molecules are about 10-10m; distances in the nucleus are
approximately 10-15m. In the short and medium range, nuclear forces are attractive, intense when the nucleons are close.
The molecular forces are attractive or repulsive.
Exercise 7
We choose a reference system linked to a spherical body.
The force depends only on the variable r. This force derives from a potential
energy if there is a function E such that f(r) = - dE/dr.
or dE = -f(r) dr : it suffices to research a primitive of f(r)
E=-K/r + Cte
When the distance r becomes very large, the force and potential energy no
longer exist, the constant is set equal to zero as r tends to infinity.
The velocity has a constant standard: the movement is uniform
The kinetic energy is : Ec =½mv²=K/(2r).
The mechanical energy is the sum of the kinetic and potential energy:
E= -K/r+K/(2r) =-½ K/r
Logarithmic derivation of the expression of energy
dE/E = dr /r = -10-4.
Logarithmic derivation of the expression of velocity : 2 dv/v = -dr/r
dv/v = -½ dr/r = + 0,5 10-4 the velocity increases.
pulingl the particle attraction of the spherical body amounts to provide
energy until its total energy becomes zero.
Energy from the operator : Eop = ½K/r0
Kinetic energy needed
½mv² =½K/r0
where v²= K/(mr0)
Exercise 8
power :
z-x = t-2 -3t = -2-2t ;
2z²-x = 2(t-2)²-3t = 2t²-11t + 8
Force vector (25t²/3 ; -2-2t ; 2t²-11t + 8 )
Velocity vector : derived from the position vector with respect to time
( 3 ; 4t ; 1)
Power: scalar product of force vector and velocity vector
P= 12,5 *2t² + (-2-2t)*4t + 2t²-11t + 8
P= 19t² -19t +8
power passes through an extreme value (minimum or maximum) when its
derivative with respect to time is zero
and 38t-19 =0 where t= 0,5 s
à t < 0,5 s, the derivative is negative, thus the power diminishes, and passes by
a minimum at t=0,5 s and diminishes for t>0,5 s.
Position of the particle at t=0,5 s :
x= 3*0,5 = 1,5 ; y= 2*0,5² = 0,5 ; z= 0,5-2 = -1,5.
Work between t = 0 and t = 2s :
if the transfer takes place along a line segment AB, the work is equal to:
Cooridnates of A ( initial position at t=0) : (0 ; 0 ; -2)
Coordinates of B (final position at t=2) : (6 ; 8 ; 0)
Displacement vector : (6 ; 8 ; 2 )
Force vector (25t²/3 ; -2-2t ; 2t²-11t + 8 )
work = scalar product of the displacement vector with the force vector :
50 t²+8(-2-2t)+2*(2t²-11t + 8 ) = 50 t2-38 t
at t= 2 s, this work is equal to : 140 J
work depends on the path followed, the two work values are different between
t=0 et t=2 s.
The force is not conservative.
Exercise 9
Spring stiffness :
Weight of the vehicle : 600*10 = 6000 N
Height : h=0,3 m
Stiffness k= 6000 / 0,3 = 2
104 N/m.
Period :
the upper end of the spring is subject to vehicle weight and spring tension.
the fundamental relationship of dynamics can be written: mz”= mg -k(l-l0)
the origin is chosen at the equilibrium position : l = lequi+ z
k(l-l0 )= k(lequi-l0 + z ) = mg + kz
thus : mz»= -kz or z»
+ k/m z =0.
solutions of this differential equation are of the form z = A cos (0t + w j) and
the movement is sinusoidal of pulsation w0 = square root (k / m) = (2 104 / 600)
0,5
= 5,77 rad /s.
the period is : T0= 2p / w0 =6,28 /5,77 = 1,088 s.
with 4 passengers the pulsation becomes : (2 104 / 900) 0,5 = 4,714 rad /s. and
period : 1,33 s.
critical regime damping:
the differential equation above is written as : z» -b/m z’ + k/m z =0
characteristic equation : r² -b/m r + ω0 ² =0
discriminant :∆ =(b/m)²-4ω0 ² ;
critical regime ∆=0 where b²= 4km and b=2 square root (20000*600)= 6928
kg /s.
Pseudoperiodic damping regime :
∆ =(b/m)²-4ω0 ² = 4[(b/(2m))²-4ω0 ² ]
the discriminant of the characteristic equation is negative if m is equal to
900kg instead of 600kg
the movement is sinusoidal damping of pseudo pulsation w such that :
ω² = ω0 ² -(b/(2m))² = 4,714²-(6928/1800)² =22,22-14,81 = 7,41
ω = 2,72 rad/s and the pseudo period T ' is 6,28/2,72 = 2,3 s
Exercise 10
(i) Compared to the Galilean reference system linked to the stem, the fundamental
relationship of dynamics can be written :
(i) Compared to the Galilean reference system linked to the stem, the fundamental
relationship of dynamics can be written :
ur ur
ur ur
mΓ = T + mg + R (I)
uuuur
r
ur
r
OM = r er ⇒ Γ = r&&er
ur
r
r
T = −kxer = −k(r − l o )er
rr
mg.er = mgcosθ o
ur r
R .er = 0
( frictionless)
The projection following the stem (I) gives :
m&&
r = −k(r − l o ) + mgcosθ o
Equilibrium position re /
(1)
&r& = 0
⇒ −k(re − l o ) + mgcosθ o = 0
(2)
mgcosθ
+ lo
k
⇒ re =
(3)
(1) − (2) ⇒ m&&
r = −k(r − l o ) + mgcosθ o + k(re − l o ) − mgcosθ o
⇒ m&&
r = −k(r − re )
⇒ R&& = r&&
We have :
mR&& = −kR ⇔ R&& + ω o2 R = 0
(4)
Where :
ω o2 =
k
2π
⇒ To =
= 0,44 s
m
ωo
(5)
b) friction is due to air resistance, therefore, the viscous friction force that opposes movement of the pendulum is proportional to the velocity:
ur
f = −α R&
(6)
Alors :
α
k
mR&& = −kR − α R& ⇒ R&& + R& + x = 0
m
m
Where
α 1
k
= et ω o2 =
m τe
m
⇒ R&& +
R&
+ ω o2 R = 0
τe
(7)
The resolution involves the characteristic equation:
R = est ⇒ R& = sest et R&& = s2 est
s
1
⇒ s2 + + ω o2 = 0 ⇒ s = 2 − 4ω o2
τe
τe
In the case of weakly damped oscillations, was:
s<0⇒
So :
s=−
1
− 4ω o2 < 0 ⇒ 1 < 4ω o2τ e2 ⇒ 2ω oτ e > 1
2
τe
∆
1
± j
2τ e
2
où
∆ = ω a = ∆ = 4ω o2 −
(8)
1
1
= ω o2 (4 − 2 2 )
2
τe
ω oτ e
∆
1 12
= ω o (1 −
)
2
4ω o2τ e
ωa =
⇒∆=−
1
± jω a
2τ e
(−
⇒ R = C 1e
−
R=e
1
2τ e
t
1
2
+ jω a ) t
(C 1e
jω a t
⇒ R = R m exp(−
(−
+ C2e
1
2
− jω a ) t
− jω a t
+ C2e
)
t
)cos(ω a t + φ),
2τ e
ω a = ω o (1 −
1
)
4ω o2τ e
Exponential decrease of elongations :
R
t
O
Figure – 1 : the shape of the curve in the case of an oscillatory damped pseudo
2π
1 − 12
- pulse and period Ta =
= To (1 −
)
ωa
4ω o2τ e2
If at t=0, we have R=0 and R`=vo
& = vo = R m (−ω a sinφ −
x(0) = 0 = R m cosφ and x(0)
We have
1
cosφ)
2τ e
t
1
R& = R m exp(−
)[−ω a sin(ω a t + φ) −
cos(ω a t + φ)]
2τ
2τ
e
e
However
φ=
Thus
π
2
et
Rm = −
vo
ω a ⇒ R = vo exp(− t )sinω t
a
ωa
2τ e
Logarithmic decrement δ
At Α(t) = R m exp(−
t
) the pseudo – sinusoid is at a maximum at times t C
2τ e
cos(ω a t + φ) = ±1 = cos(kπ )
⇒ tC =
kπ − φ kTa φTa
=
−
ωa
2
2π
Α(tC ) = Α(k) = R m exp[−
⇒ Α(k) = R m exp(
(kTa / 2) − (φTa / 2π )
2τ e
φTa
kT
)exp( a )
4πτ e
4τ e
The logarithmic decrement of the movement is the quantity :
⇒δ =
Ta
2τ e
The logarithmic decrement characterizes the decay of elongation in each period due to depreciation of the oscillatory motion. It is deduced by measuring
the number therefore stretching moments separated by n periods:
⎛ R ⎞ 1
x(t)
1
1
= exp(nδ ) ⇒ δ = [Logx(t) − Log[x(t + nT a )] i.e : Log ⎜ o ⎟ =
Log3
x(t + nT a )
n
n
⎝ R S o ⎠ 50
2π
⎧
= 0, 44 s
⎪T a ≈ T o =
ωo
⎨
⎪δ = 0, 022
T
T
⎩
Et δ = a ⇒ τ e = a
2τ e
2δ
So τ e = 10 s : duration of relaxation.
As energy is the duration after which the amplitude is represented by
1
exp( ) ≈ 1,5 . Since the amplitude is zero after several values of τ e , we need
2
for τ e to characterise the life of damped oscillations. This is why we call it the
duration of relaxation of energy.
Quality factor :
It is a dimensionless number Q that characterizes the damping of an oscillator. It is defined by Q = ω oτ e
Here, Q = ω oτ e =
(thus linked to the duration of the relaxation).
τe
10
=
= 22,7
To 0,44
(ii) ) In a rotating reference (linked to the stem) ;
We have E C =
1 2
1
m&r ; E P P = mgz = mgr cosθ o ; E P E = k(r − l o )2
2
2
Centrifugal potential energy :
During the rotational movement, the point M is subjected to the centrifugal
ur
ur
ur
ur
ur uuur
force of inertia : F = −mΓ e ; où Γ e = ΩΛ(ΩΛOA)
À
ur
uur
r
r uuur
uuur uuur uuur r
uuur r uuur r uuu
R = cst Γ e = Ω2 ez Λ (ez ΛOA) and OA = OH + HA; ez Λ OA = e12
z ΛOH + ez ΛHA
4
4
3
=0
ur
r
Ω = −Ωez
ur
ur ur ur
uuur
uuur uuur
uuur
1
dW (F ie ) = F ie .d A = F ie .d(HA) = mΩ2 HA.d(HA) = d[ mΩ2 (HA)2 ]
2
ur
ur ur ur
uuur
uuur uuur
uuur
1
⇒ W (F ie ) = F ie .d A = F ie .d(HA) = mΩ2 HA.d(HA) = d[ mΩ2 (HA)2 ]
2
ur
uuur
ur
1
2
2
te
⇒ W (F ie ) = mΩ (HA) + C ⇒ E P = −W (F ie )
2
uuur
1
⇒ E P = − mΩ2 (HA)2 + C te et C te = 0 ⇒ H = A
C
2
uuur
1
1
⇒ E P = − mΩ2 (HA)2 = − mΩ2 r 2 sin 2 θ o
C
2
2
b) the oscillator system – ground isolated, the reference rotates
⇒ E R ( A) = C te = E C ( A) + E P P + E P e + E P C
R
1
1
1
m&r 2 + mgr cosθ o + k2(r − l o )2 − mΩ2 r 2 sin 2 θ o = E R ( A) = C te
2
2
2
By describing with respect to t , we have :
1
1
1
m&r&&
r + mg&r cosθ o + k(r − l o ).&r − mΩ2 r r& sin 2 θ o = 0
2
2
2
2
⇒ m&&
r + mgcosθ o + k(r − l o ) − mΩ (sinθ o ).r = 0
⇒
k
k
− Ω2 sin 2 )r = l o − gcosθ o
m
m
k
⇒ r&& + (ω o2 − Ω2 sin 2 )r = l o − gcosθ o
m
⇒ r&& + (
c) Equilibrium position :
ω o l o − gcosθ o
lo −
2
Ω = C ⇒ Ω e=
te
ω o 2 − Ω2 sin 2 θ o
=
gcosθ o
ω o2
2
⎛ Ω⎞
1 − ⎜ ⎟ sin 2 θ o
⎝ ωo ⎠
And the pulsation of the oscillating movement is :
ω 0ʹ = ω o2 − Ω2 sin 2 θ o = 0,19s ⇒ Toʹ =
To
2π
=
≈ 1s
ω oʹ 0,436
Exercise 11
1°) RFD
ur ur ur
ur
mΓ = T 1 + T 2 + mg
⇒ m&&
x = −k1 (x − l1 ) − k2 (x − l 2 ) + mg
The two springs have the same extension x
Pulsation ω o2 =
k1 + k2 k
= ; où k = k1 + k2
m
m
2°)
ur ur
ur
ur
⎧T 1 + T 2 + mg = mΓ
ur
ur r
⎪⎪ ur
⎨T o1 + T o 2 + mg = 0
r
⎪ ur ur
T
1 +T 2 = 0
⎩⎪
−k1 (x1 − l o1 ) − k2 (x2 − l o 2 ) + mg = m&&
x
−k1 (x1e − l o1 ) − k2 (x2 e − l o 2 ) + mg = 0
−k1 (x1 − x1e ) − k2 (x2 − x2 e ) = m&&
x
x = x1 + x2
2
ur
r
r
r
r
& z = r&er + c ez ⇒ v2 = r& 2 + c
V R ( M ) = r&er + rθe
r
r2
dv2 2
d
dt
( r& ) = ( r& 2 )
dr
dt
dr
d 2
1
&r . = 2&&
( r& ) = 2r&&
r
dr
r&
1 d 2
c2
(v ) = (&&
r − 3)
2 dr
r
Rotating reference :
uuuur
r
OM = xe x
r ur r
⎧⎪ R = (O, x o , y , z o )
o
o
r ur r
⎨
⎪⎩ R = (O, x, y, z o )
x = −k2 (x − xI − l ) + mg
⎪⎧ m&&
⎨
⎪⎩0 = k2 (x − xI − l ) − k1 (xI − l )
0 = k2 x − k2 l − k2 xI − k1 xI + k1l
⇒ xI =
k2 x + k1l − k2 l
k1 + k2
m&&
x = −k2 (x − l −
k2 x + k1l − k2 l
) + mg
k1 + k2
m&&
x = −k2 (x − l ) + k2 xI + mg
m&&
x = −k2 (x − l ) + mg +
m&&
x=
(k22 − k1 k2 − k1 2 )x (k2 k1 + k22 + k1 k2 − k2 )l
+
+ mg
k1 + k2
k1 + k2
m&&
x=−
0=−
k2
(k x + k1l − k2 l )
k1 + k2 2
k1 k2
2k k
x + 1 2 l + mg
k1 + k2
k1 + k2
k1 k2
k1 + k2
⇒ m&&
x+
xe +
k1 k2
k1 + k2
2k1 k2
k1 + k2
l + mg
(x − xe ) = 0
X = x − xe ⇒ mX&& + kX = 0
où k =
k1 k2
1 1 1
⇔ = +
k1 + k2
k k1 k2
Exercise 12
1- Preliminary study of the spring
a) Check that the coefficient of stiffness of the spring is equal to k = 20,0 N.m - 1.
Figure 1
Galilean reference : the solid Earth.
System studied : the solid attached to the free end of the spring.
The equilibrium condition is :
+
m
=
(1)
+k
=
By projecting on Oz :
m
k=m
-k
=0
/
k = (0,200 x 10) / 0,100 = 2 / 0,100
k = 20 N / m (2)
b) the uni N is equivalent to kg . m / s 2. Thus :
k = 20 N / m = 20 kg . m / s 2 m = 20 kg / s 2 = 20 kg . s - 2
- The coefficient of the spring stiffness is expressed in kg / s 2.
- the quantity m / k is expressed in kg / kg . s - 2 and in s2.
- the quantity
is expressed in s.
a) differential equation of movement.
Figure 2
Studied system : the solid attached to the free end of the spring.
Applied forces : the soli dis subjected to three forces :
The weight
The force
(the action of the Earth on the solid).
exerted by the spring on the solid as
=k
=-k
(3) .
the force exerted by the support on the solid (it is perpendicular to the surfaces in contact since friction is neglected)
-
Let us apply Newton’s second law (theorem of the center of inertia):
In a Galilean reference frame, the sum of external forces applied to a solid
is the product of the mass of the solid by the acceleration of its center of
inertia :
Here, we write :
+
+
=m
(4) with
=-k
Where, by projecting on the unit vector
and, by consequence, Fx = - k x
:
0+0-kx=m
m
+kx=0
The equation
(5) is the differential equation of movement of the soli
It is a differential equation of second order with constant coefficients.
-
(6) is the solution of the differential equation:
(5)
(6) with respect to time :
Derive (7)
Derive
(8)
- Forming :
(9)
-
this expression (9) will be zero, according to
condition :
(11)
= 0 (10), with the
(6) is the solution to the differential equation of movemen
the condition:
(11)
- Calculate the numerical value of the period T0 = 2 p
g = 0,200 kg and k = 20 N / m
with m = 200
T0 = 0,628 s (12)
b) Study of a particular case Initital conditions : at t = 0 s, we have x (o) = - 0,15 m and v (0) = 0 m/s
- transfer these values into the expressions of the position and the velocity :
x = X M cos (w0 t + j) (6) and
(9)
Thus, at t = 0 s :
- 0,15 = X M cos (ϕ)
0 = - (2 π / To) X M sin (ϕ)
And :
X M = - 0,15 / cos (ϕ)
0 = - sin (ϕ)
There appears to be 2 solutions :
ϕ1 = 0 (modulo 2 π)with X1M = - 0,15 m
ϕ2 = π (modulo 2 π)with X2 M = 0,15 m
However, these solutions are in fact equal, since cos cos (a + π) = - cos (a) :
x1 = - 0,15 cos (10 t) = 0,15 cos (10 t + π) = x2
we retain :
The position x of the center of inertia G of the solid is, at each instant, given by
:
x = 0,15 cos (10 t + π) (13)
by deriving x with respect to time, we obtain the velocity of the solid in rectilinear translation :
v = - 1,5 sin (10 t + π) (14)
the velocity thus varies between - 1,5 m/s and + 1,5 m/s which is the maximum
value.
Remark : by deriving with respect to time, we obtain the acceleration of the
solid.
a = - 15 cos (10 t + π)
c) Energetic study of the undamped oscillator.
- Recall that x = X M cos (10 t + π ) (13)andt v = - 10 X M sin ( 10 t + π ) (14)
- The solid-spring system has, in the Galilean terrestrial reference, mechanical energy:
Em = EP + EC =
Em =
k x2 +
m v2
k Xm2 cos2 (10 t + p) +
m (- 10)2 Xm2 sin2 (10 t + p)
But 100 = k / m (4)
Em =
k Xm2 cos2 (10 t + π) +
m k/m Xm2 sin2 (10 t + π)
Em =
k Xm2 [ cos2 (10 t + π) + sin2 (10 t + π) ]
We know that [ cos2 (10 t + π) + sin2 (10 t + π) ] = 1
Finally :
Em =
k Xm2 (13)
This energy remains constant as time elapses (note that we have neglected any
friction).
With k = 20 N/m et Xm = - 0,15 m, we obtain :
Em = 0,225 J (14)
- We have seen that in the absence of friction, the mechanical energy of the
spring-solid system remained constant. In passing through x = 0 the potential
energy of spring
mum. Thus :
k x2 is zero, thus the kinetic energy of the system is a maxi-
m v2max = 0,225 and
0,1 v2max = 0,225
v2max = 2,25 m2 / s2
The velocity v varies between extreme values - 1.5 m / s and +1.5 m / s which
is maximum value already found in question 2-b.
Exercise 13
A – Study of free oscillations
A-1 Comment on the shape of the curve and specify the type of oscillations
observed.
The amplitude of oscillations is constant. The oscillations, free and undamped,
are periodic.
A-2 Graphically determine the oscillation period T0 of mass m hanging spring.
The duration of two oscillations is around 0.63 s. The natural period T0 of free,
undamped oscillations is:
T0 = 0,315 s
A-3 Comparison with the theoretical value
.
This gives m = 100 g = 0,100 kg and k = 40 N / m. We obtain :
T0 = 0,314 s
The calculation confirms the value obtained experimentally. We retain :
T0 = 0,32 s
A-4 We add to the extremity E of the stem a horizontal disc of negligible
volume and mass. Frictional forces are now involved.
The mass of the disk and hence its weight is negligible.
The buoyant force exerted by water on the disc is also negligible, since its
volume is negligible.
However, the oscillator is now damped by viscous friction.
In the presence of moderate friction (small hard surface, placed in a low viscosity liquid like water, salt) the amplitude of oscillations gradually decreases
(pseudo-periodic regime). The pseudo period T1 is close to the natural period
T0.
Let us draw the shape of the curve obtained after a new acquisition:
B – Study of forced oscillations B-1 The engine fitted to the eccentric is the exciter. The system (spring + mass)
is the resonator.
It requires that the excitation frequency f of the resonator frequency f0 = 1 / T0.
B-2 The curve giving the variation of the amplitude xmax of the oscillations of
the resonator as a function of frequency f which is imposed by the exciter, is
called resonance curve.
f (Hz)
1,5 2,0 2,5 2,8 3,1 3,2 3,3 3,6 4,0 4,5
xmax (cm) 0,4 0,6 1,0 1,5 2,1 2,3 2,0 1,5 1,0 0,7
For a frequency of excitation fR = 3.2 Hz we observe that the amplitude of oscillations of the resonator is maximized. This is the phenomenon of amplitude
resonance. It occurs for a frequency of excitatory fR = 3.2 Hz near the frequency f0 = 1 / T0 = 3.18 Hz of the resonator.
B-3 If we used a solution S ‘over the viscous friction force increase, the resonance becomes blurred.
There would be more resonance if the damping became very important.
C – Suspension of a moving object
C-1 For a velocity VR, the vehicle undergoes large amplitude oscillations that
dangerously diminish its handling. Let us explain this phenomenon.
The suspension of a moving object is comparable to a mechanical system oscillating at a frequency f0. It acts as a resonator, however, it is damped.
With each bump (exciter), the system receives a nearly vertical stimulus. If
these pulses are periodic and if the frequency f is close to f0, thus the oscillations of the car, though muted, can achieve a large amplitude. The “handling”
of the car may be compromised.
C-2 Express the velocity VR as a function of f0 and L.
This phenomenon occurs if the time between two passes over a hump, and
L / VR is equal to the natural period T0 = 1 / f 0 of the oscillating mechanical
system:
L / VR = 1/ f0
VR = L . f0
With f0 = 5,0 Hz et L = 80 cm = 0,80 m, we calculate :
VR = L . f0 = 0,80 5,0 = 4,0 m / s = 14400 m / h
VR = 14,4 km / h
Exercise 14
1- Preliminary study of the spring
a) Check that the coefficient of stiffness of the spring is equal to k = 20,0 N.m
-1
.
Figure 1
System studied : the solid attached to the free end of the spring.
The equilibrium condition is :
+
m
=
(1)
+k
=
By projecting on Oz :
m
-k
k=m
=0
/
k = (0,200 x 10) / 0,100 = 2 / 0,100
k = 20 N / m (2)
b) the uni N is equivalent to kg . m / s 2. Thus :
k = 20 N / m = 20 kg . m / s 2 m = 20 kg / s 2 = 20 kg . s - 2
- The coefficient of the spring stiffness is expressed in kg / s 2.
- the quantity m / k is expressed in kg / kg . s - 2 and in s2.
- the quantity
is expressed in s.
a) differential equation of movement.
Figure 2
Studied system : the solid attached to the free end of the spring.
Applied forces : the solid disc is subjected to three forces :
The weight
The force
(the action of the Earth on the solid).
exerted by the spring on the solid as
=k
=-k
(3) .
the force exerted by the support on the solid (it is perpendicular to the surfaces in contact since friction is neglected)
- Let us apply Newton’s second law (theorem of the center of inertia):
In a Galilean reference frame, the sum of external forces applied to a solid is the
product of the mass of the solid by the acceleration of its center of inertia :
Here, we write :
+
+
=m
(4) with
=-k
Where, by projecting on the unit vector
and, by consequence, Fx = - k x
:
0+0-kx=m
m
+kx=0
The equation
(5) is the differential equation of movement of the soli
It is a differential equation of second order with constant coefficients.
- Verify that, if we correctly choose To, the function
(6) is the solution of the differential equation:
Derive Derive
(5)
(7)
(8)
- Forming :
- this expression (9) will be zero, according to
tion :
(9)
= 0 (10), with the condi-
(11)
(6) is the solution to the differential
= 0 (5), with the condition:
equation of movement
(11)
- Calculate the numerical value of the period T0 = 2 p
0,200 kg and k = 20 N / m
with m = 200 g =
T0 = 0,628 s (12)
b) Study of a particular case Initital conditions : at t = 0 s, we have x (o) = - 0,15 m and v (0) = 0 m/s
- transfer these values into the expressions of the position and the velocity :
x = X M cos (w0 t + j) (6) and
(9)
Thus, at t = 0 s :
- 0,15 = X M cos (ϕ)
0 = - (2 π / To) X M sin (ϕ)
And :
X M = - 0,15 / cos (ϕ)
0 = - sin (ϕ)
There appears to be 2 solutions :
ϕ1 = 0 (modulo 2 π) with X1M = - 0,15 m
ϕ2 = π (modulo 2 π) with X2 M = 0,15 m
However, these solutions are in fact equal, since cos (a + π) = - cos (a) :
x1 = - 0,15 cos (10 t) = 0,15 cos (10 t + π) = x2
we retain :
The position x of the center of inertia G of the solid is, at each instant, given by
:
x = 0,15 cos (10 t + π) (13)
by deriving x with respect to time, we obtain the velocity of the solid in rectilinear translation :
v = - 1,5 sin (10 t + π) (14)
the velocity thus varies between - 1,5 m/s and + 1,5 m/s which is the maximum
value.
Remark : by deriving with respect to time, we obtain the acceleration of the
solid.
a = - 15 cos (10 t + π)
c) Energetic study of the undamped oscillator.
- Recall that x = X M cos (10 t + π ) (13) andt v = - 10 X M sin ( 10 t +
π ) (14)
- The solid-spring system has, in the Galilean terrestrial reference, mechanical energy:
Em = EP + EC =
Em =
k x2 +
m v2
k Xm2 cos2 (10 t + π) +
m (- 10)2 Xm2 sin2 (10 t + π)
But 100 = k / m (4)
Em =
k Xm2 cos2 (10 t + π) +
m k/m Xm2 sin2 (10 t + π)
Em =
k Xm2 [ cos2 (10 t + π) + sin2 (10 t + π) ]
We know that[ cos2 (10 t + π) + sin2 (10 t + π) ] = 1
Finally :
Em =
k Xm2 (13)
This energy remains constant as time elapses (note that we have neglected any
friction).
With k = 20 N/m et Xm = - 0,15 m, we obtain :
Em = 0,225 J (14)
- We have seen that in the absence of friction, the mechanical energy of the
spring-solid system remained constant. In passing through x = 0 the potential
energy of spring
mum. Thus :
k x2 is zero, thus the kinetic energy of the system is a maxi-
m v2max = 0,225 and
0,1 v2max = 0,225
v2max = 2,25 m2 / s2
The velocity v varies between extreme values - 1.5 m / s and +1.5 m / s which
is maximum value already found in question 2-b.
Exercise 15
1- Establish the differential equation of movement of the solid.
Use the figure below :
Associated orthonormal reference : O,
System studied : the solid of mass m.
The solid disc subjected to 4 forces :
-
: weight (due to gravity)
-
: normal action of the stem on the moving object
-
: action of the spring
-
: friction of the fluid on the moving object
Apply Newton`s second law ( lesson 11) :
In a Galilean reference frame, the sum of external forces applied to a solid is
equal to the product of the mass m of the solid by the acceleration
of its
center of inertia :
Here, the law is written :
+
+
+
Note that
+
=m
=K
-K
(1)
=-K
-h
=m
(2) and that
m
+h
:
=m
+ K x = 0 (4)
+ (h / m)
+A
(3)
(1 bis)
Project (1 bis) on the axis of the unit vector
0+0-Kx-h
=-h
+ (K / m) x = 0 (5)
+ (K / m) x = 0 (6) with :
A = h / m (7) (A is the damping coefficient).
Equation (5) is the differential equation of movment of the solid of mass m. It
is a second order differential equation with constant coefficients.
In this equation (5) figure the term A
force = - h .
which corresponds to the frictional
2- Nature of the movement of the solid body.
If A could be zero, then the movement would be periodic.
Since there is friction, the movement cannot be periodic.
If A is weak, then the movement of the solid is pseudo-periodic.
If A is strong, the movement of the soli dis aperiodic.
Remark There is a damping called «critical» that ensures a return to the position of equilibrium faster than aperiodic or pseudo-periodic.
3- Study the mechanical energy Em of the system (spring + solid).
a) The mechanical energy of the system (spring + solid) is :
Em = Epotentiel of the spring + Ekinetic of the mass
Em =
k x2 +
m v2 =
( k x2 + m
) (8)
b) Let us establish the relationship between the derivative of the mechanical
energy versus time and power of the friction force.
Drive Em with respect to time :
d Em / dt = k x.
+m
Use the relation m
.
+h
(10) d Em / dt = ( - h
=(kx+m
)
(9)
+ K x = 0 (4)
)
= fx vx =
.
= P (power developed from friction).
c) Comment on this relationshio in terms of energy transfer
Power developed from friction is P = d Em / dt = - h
tive.
, which is always nega-
The system (spring and solid) loses mechanical energy. This energy is transformed to heat. 4- Graphical representation.
a) Determine the initial conditions imposed on the oscillator
At time t = 0 s, the graph given by the computer can show u (o) = 0 V. We
deduce that at the same time the solid mass m passes through the point O of an
abscissa of zero.
What is its velocity V (o)?
We know that the velocity is Vx = dx / dt. Its value is given by the coefficient
of the tangent to the curve giving x in terms of t.
The graph gives du / dt = - 6.6 V / s at time t = 0 s (check after drawing the
tangent at the origin).
The scale is such that 1 volt corresponds to 1 cm, it follows that Vx (0) = - 6.6
cm / s = - 0.066 m / s.
At time t = 0 s, we have :
- initial abscissa x(0) = 0 m
- initial velocity V(0) = - 0,066 m/s (11)
b)
Calculate the pseudo-period T.
The graph shows that 5 oscillations last 10 s. Thus :
5 T = 10
The pseudo period is T = 2,0 s (12)
c) The mechanical energy of the system is Em =
k x2 +
m v2
At each extremum i, the velocity is zero. The mechanical energy is :
(Em) i =
k x2i = 5 x2i (13)
(Em) i / (Em) i + 1 is quasi constant.
d) Determine the work and frictional force
and 4th negative extremum.
between the passages of the 1st
Passing by the 1st negative extremum, the mechanical energy is :
Em1 = 0,0054 J (14)
Passing by the 4th negative extremum, the mechanical energy is :
Em4 = 0,00018 J (15)
In the presence of friction, the mechanical energy of the solid-spring system
varies. Here, it decreases.
It turns gradually into heat energy which heats the system and its surroundings.
The variation of the mechanical energy of the solid-spring system is equal to
the work of the frictional force :
Em4 - Em1 = W14 (
W14 (
) ) = Em4 - Em1 = 0,00018 - 0,0054
(16)
W14 (
) = - 0,0052 J = - 5,2 x 10 - 3 J (17)
The work of the friction force is negative because it is resistant work.
Note: Sometimes, a friction force provides a positive working engine. This is
the case, for example, when a treadmill turns on, carrying an object. The work
of the frictional force exerted by the belt on the bag is positive.
Exercise 16
c
R
uur uur ur
u
(O , e x , e y, e z )
R ' (O
a)
uuur
VR
uuuuuur
et
OM
ur
u ur
u ur
u
, e r , e θ, e z )
ur
u
ur
u
= r .e r + z .e z
uuuuuuur
• ur
u
Ω R '/ R = θ e z
uuuur
uuur
( M )= VR' ( M ) + VR ( M ∈R ' )
=
d
dt/ R
r
( r.e +
• ur
• ur
uuur
u • ur
u
u
VR (M) = r e r + z e z + r θ eθ
r
•
r
)
+
θ
z.e z
r
∧ ( r.e +
R '/ R
r
r
z.e z )
ur
ur
Γ a (M) = Γ R ' (M)
b)
ur
Γ R (M)
=
d
dt/ R
2°)
(
+
ur
ur
Γ e(M) + Γ c(M)
r
r
r
• r
• r
• r
r + • r ) + ••
∧(
+
)+
∧(
+
)
r .e r z .e z θ R '/ R.e z r.e r z.e z
2.θ .e z r .e r z .e z
r
r
r
ur
u
•
+
∧ [ e ∧ ( r.e + z.e
.e
z
r
z
(θ)2 z
•
uur
eθ
ey
r
F1
uur
er
r
F2
r
θ
uur
ex
r
mg
⎡ 2mrθ ⎤
⎢
⎥
F1 = ⎢ mr ⎥
⎢ 0 ⎥
⎣
⎦
uur
à
ur
P
a)
uur
F 2 = ml
r ur
e z ∧ V R (M)
r
r
r
= − mg e y = − mg ( sin θe r + cosθeθ )
ur
P
ur
+F +
1
ur
ur
=
m
F2
Γ R (M)
)]
•
••
• •
− mg sin θ + 2mrθ − mlr θ = m[ r − r (θ)2 ]
•
••
(1)
••
− mg cosθ + mr + ml r = m[ 2 r θ + r θ ]
• •• • •
⎧
2
⎪⎪− g sin θ + 2rθ − lr θ = r − r (θ)
⇒⎨
•
••
••
⎪
⎪⎩− g cosθ + r + l r = 2 r θ + r θ
ur
ur
u
(2)
(3)
(4)
ur
u
b) dEc (M) = dW(P ) + dW(F1 ) + dW(F2 )
ur uuuur ur
u uuur
ur
u uuur
1
2
d[ m V R (M)]= d(P.OM ) + F1.V R (M).dt + F2 .V R (M).dt
14
4244
3
2
0
•r
ur
u uuur
r
r •r
F1.V R (M) = (2mrθe r + mreθ )(r e r + r θ eθ )
•
•
= 2m r rθ + mr 2 θ = d(mr 2θ)
ur uuuur
P.OM = − mgr sin θ
•
•
1
d [ m( r 2 + r 2 θ2 ) ]= d ( mr 2θ − mgr sin θ)
2
•
2
•
2 2
⇒ r + r θ = 2r 2θ − 2gr sin θ + cte
(5)
(cte = cons tan te)
The relation (5) is the first integral of energy
3)
ur ur
u ur
uuur
P + F1 + L = m Γ R (M)
ur
ur
u
où L = Le r
•
⎧
L
⎪⎪ 2aθ − g sin θ +
= − a θ2
m
⇒⎨
••
⎪
⎪⎩ a − g cosθ = a θ
(6)
(7)
••
(7) ⇒ a θ + g cosθ = a
(8)
(8) : Is the dynamic differential equation of M
ur
ur
u
ur
dEc (M) = dW(P ) + dW(F1 ) + dW(L)
123
0
•
2 2
⇒ a θ = 2a 2θ − 2ga sin θ + cte
(9)
à t = 0 ⇒ θ = 0 ,et la relation (9) nousdonne : cte = 0
•
2
⇒ a θ = 2aθ − 2g sin θ
(10)
(6) et (10) ⇒ L = ( − 2aθ + 2g sin θ )m + mg sin θ − 2maθ
b) ⇒ L = 3mg sin θ − 4maθ
(11)
4)
d uur uuur ur
L = M o ( ∑ F)
dt / Ro o
• ur
uur uuuur
uuur
ur
u
u
où L o = OM ∧ m VR (M) = a e r ∧ ma θ eθ
• ur
uur
u
L o = ma 2 θ e z
uuur ur
uuuur
ur ur
u ur
M o ( ∑ F) = OM ∧ ( P + F1 + L )
ur
u
ur
u
ur
u
ur
u
ur
u
ur
u
= a e r ∧ [Le r + 2maθe r + maeθ − mg sin θe r − mg cosθeθ ]
ur
u
ur
u
= a e r ∧ [maθ − mg cosθ]eθ
ur
u
= ma(a − g cosθ)e z
•• ur
u
ur
u
⇒ ma 2 θ e z = ma(a − g cosθ)e z
••
⇒ a θ + g cosθ = a
Exercise 17
(8)
uur uur ur
u
(
,
,
,
R O e x e y e z)
ur
u ur
u ur
u
(
,
,
,
R ' O e r e θ e φ)
uur
uur uur ur
u
ur
u ur
u ur
u
(φ,ez )
(
,
,
,
)
⎯
⎯⎯
→
(
,
,
,
R O ex e y ez
R 1 O eρ eφ e z)
uuuuuuuur
Ω R1 / R
R 1(O
uur
ur
u ur
u ur
u
ur
u ur
u ur
u
(θ,eφ )
, e ρ, e φ, e z ) ⎯⎯⎯→ R '(O , e r , e θ, e φ)
uuuuuuuur
Ω R '/ R1
uuuuuuur
• ur
•
u
ur
u
ur
u
= φ e z = φ(cosθe r − sin θe θ )
• ur
u
= θeφ
uuuuuuuur
uuuuuuur
• ur
u
• ur
u
Ω R '/ R = Ω R '/ R1 + Ω R1/ R = θ e φ + φ e z
uuuuuuur
Ω R '/ R
• ur
u •
ur
u
ur
u
= θ e φ + φ(cosθe r − sin θe θ )
(1) ) By composition de movement
a) vector velocity
uuur
uuur
uuur
VR (M) = VR ' (M) + VR (M ∈R ')
ur
u
uuuuuur uuuur
d
=
(re ) + Ω R '/ R ∧ OM
dt / R ' r
•r
• r
•
r
r
r
= r e r + r θ(eφ ∧ e r ) − r φ sin θ(eθ ∧ e r )
•r
•r
•
uuur
r
VR (M) = r e r + r θ eθ + r φ sin θeφ
b) acceleration vector
uuur
uuur
uur
uur
Γ R (M) = Γ R ' (M) + Γ c (M) + Γ e (M)
uuur
Γ R ' (M) =
•• ur
u
d uuur
VR ' (M) = r e r
dt / R '
uur
uuuuuur uuur
Γ c (M) = 2Ω R '/ R ∧ VR ' (M)
• • r
•
•
r
r r
r
r
= 2 r[θ(eφ ∧ e r ) + φ cosθ(e r ∧ e r ) − φ sin θ(eθ ∧ e r )
••r
••
r
= 2 r θ eθ + 2 r φ sin θeφ
uur
Γe(M) =
uuuur uuuuuur
uuuuuur uuuur
d uuuuuur
(Ω R '/ R ) ∧ OM + Ω R '/ R ∧ (Ω R '/ R ∧ OM )
dt / R
•• ur
••
••
u
ur
u
ur
u ur
u uuuuuur
uuuuuur uuuur
= θ eφ ∧ re r +[− φ sin θ − φ θ cosθ] (eθ ∧ e r ) + Ω R '/ R ∧ (Ω R '/ R ∧ OM )
•
φ cosθ
•
•
rθ
2
•
2
−r φ sin θ − r θ
0
•
uuuuuur
uuuuuur uuuur
avec Ω R '/ R ∧ (Ω R '/ R ∧ OM ) = − φ sin θ ∧
θ
•
2
•
2
= −r φ sin θcosθ
•
r φ sin θ
••
r θ φ cosθ
The vectors are expressed in the reference R’
where
•
• ur
•
••
••
••
uur
u
ur
u
ur
u
2
2
2
Γe(M) = − r(φ sin θ + θ )e r + r( θ − φ2 sin θcosθ)eθ + r[φ sin θ + 2φ θ cosθ]eφ
•
•
⎛
⎞
••
⎜
⎟
r − r(φ2 sin 2 θ + θ2 )
⎜
⎟
•
••
••
uuur
⎜
⎟
⇒ Γ R (M) = ⎜ r( θ − φ2 sin θcosθ) + 2 r θ ⎟
⎜ ••
⎟
••
••
⎜ r[φ sin θ + 2φ θ cosθ] + 2 r φ sin θ⎟
⎜
⎟
⎝
⎠
Components of the acceleration vector in the reference R’
By direct calculation
Velocity vector
uuur
VR (M) =
uuuuuur uuuur
d uuuur
(OM ) + Ω R '/ R ∧ OM
dt / R '
•r
• r
•
•
r
r r
r
r
= r e r + r θ(eφ ∧ e r ) + r φ cosθ(e r ∧ e r ) − r φ sin θ(eθ ∧ e r )
•r
•r
•
uuur
r
VR (M) = r e r + r θ eθ + r φ sin θeφ
Acceleration vector
uuur
Γ R (M) =
uuuuuur uuur
d uuur
d uuur
VR (M) =
VR (M) + Ω R '/ R ∧ VR (M)
dt / R
dt / R '
•
r
=
d
dt / R '
•
•
r
•
rθ
+ − φ sin θ ∧
r φ sin θ
θ
•
•
φ cosθ
•
•
rθ
•
r φ sin θ
•
•
⎛
⎞
••
⎜
⎟
r − r(φ2 sin 2 θ + θ2 )
⎜
⎟
•
••
••
uuur
⎜
⎟
⇒ Γ R (M) = ⎜ r( θ − φ2 sin θcosθ) + 2 r θ ⎟
⎜ ••
⎟
••
••
⎜ r[φ sin θ + 2φ θ cosθ] + 2 r φ sin θ⎟
⎜
⎟
⎝
⎠
2)
ur
ur
u
uuuur ur
u
L = Le r
OM = ae r
ur
r
ur
u
r
r
P = mg = − mge z = − mg ( cosθe r − sin θeθ )
ur
uuuur
uuur uuuur
r
r
F = − kAM = − k(AO + OM ) = − k ( ae z + ae r )
ur
r
r
F = − k a[(1 + cosθ)e r − sin θeθ ]
a)
ur ur ur
uuur
L + P + F = mΓ R (M)
•
⎧
⎪ L − mg cosθ − k a(1 + cosθ) = − ma θ2
⇒⎨
••
⎪
mg
sin
θ
+
ka
sin
θ
=
ma
θ
⎩
••
⎛g k⎞
(2) ⇒ θ = ⎜ + ⎟ sin θ
⎝ a m⎠
(1)
(2)
(3)
Equation (3) is the dynamic differential equation of movement
expression of L
•
2
(1) ⇒ L = (ka + mg)cosθ + k a − ma θ
(4)
b) kinetic energy theorem
uuur
r
ur
dE c (M) = dW(mg) + dW(F) + dW(L)
1
a2 • 2
2
Ec (M) = m V R (m) = m (θ)
2
r 2 ur uuuur
W(mg ) = mg .OM = − mga cosθ
ur ur uuur
dW(L) = L.dM = 0
ur ur uuur
dW( F) = F.VR (M).dt
•r
•
r
r
r
= − k a[(1 + cosθ)e r − sin θeθ ].[a θ eθ + a φ sin θeφ ].dt
•
= ka 2 sin θθ dt = ka 2 sin θdθ = d(−ka 2 cosθ)
d[m
a2 • 2
(θ) ] = d[−(ka 2 + mga )cosθ]
2
a2 • 2
⇒ m (θ) = − (ka 2 + mga )cosθ + cte
2
•
à t = 0⇒ θ = 0 et θ = 0 ⇒ cte = ka 2 + mga
⇒m
a2 • 2
(θ) = (ka 2 + mga )(1 − cosθ )
2
Where the first energy integral is :
•
ma 2 (θ)2 = 2 (ka 2 + mga )(1 − cosθ )
(4) and (5)
expression of L
(5)
⇒ L = (ka + mg)cosθ + k a − 2(ka + mg)(1 − cosθ )
L = k a + (ka + mg)(3cosθ − 2)
(6)
Differential equation of movement
• ••
•
(5) ⇒ ma.2.θ θ = 2(ka + mg)(sin θ)θ
•
••
et θ ≠ 0 ⇒ θ =
ka + mg
sin θ
ma
We find equation (3)
c) at point C, we have
θ=
π
2
et
•
v=
uuur
v c = a θc et (5)⇒
⇒ vc =
•
V R (M) = a θ
2
[(ka 2 + mga)(1 − cosθ)]
ma
2
ka 2 + mga
m
and
L = ka + (ka + mg)(−2) = − 2mg − ka
ur
ur
u
L = − (ka + 2mg)e r
d)
L = 0 alors (6)⇒3cosθ − 2 = −
1
ka
cosθ = (2 −
)
3
ka + mg
ka
ka + mg
1
ka
(2 −
) ≤1
3
ka + mg
2ka − ka + 2mg
− 3≤ (
) ≤3
ka + mg
ka + 2mg
− 3≤ (
) ≤3
ka + mg
ka + 2mg
≤3
ka + mg
L = 0 existe ⇔ − 1≤
d 'où
ka + 2mg ≤3(ka + mg)
2ka + mg ≥0 toujours vraie
1
1 1
ka = mg ⇒ cosθ = (2 − ) =
3
2 2
⇒θ = ±
π
⇒
3
θ=
π
3
Exercise 18
uur uur ur
u
u ) ⎫⎪
R 0 ( O, uxuor , uyuor ,zur
R 0 ( O, xro , ryour
u,z oo ) ⎫⎬⎪
u ) ⎬⎪
R1 ( O, xr , yr ,zur
R1 ( O, x , y,z oo ) ⎭⎪⎭
r r ur
ur )
R ( O, xr , vr ,w
R ( O, x , v ,w )
R est un repère de
R
Rr isun
a projection
uuuu
rrepère de
est
uOM
uuur = ρvr
OM = ρv
}}
uuuuuuuur • ur
u
• z
uuuuuuuur = ψ
ur
u
Ω
1 / R0 = ψ z o
ΩR
R /R
o
1
0
uuuuuur • r
uuuuuur = θ• xr
Ω
/ R1
ΩR
R / R1 = θ x
projection
projection
1°)
uuur
d uuuur
d r •r
a) VR (M) =
OM =
ρv = ρ v
dt / R
dt / R
uuuur
b) VR (M) =
1
d
dt / R
1
uuuur
d uuuur uuuuuur uuuur
OM =
OM + Ω R / R ∧ OM
1
dt / R
r •r
r
= ρ v + θ x ∧ ρv
•r
• ur
= ρ v + ρθ w
•
2°)
c)
uuuur
VR (M) =
o
uuuur
d uuuur uuuuuur uuuur
OM =
OM + Ω R / Ro ∧ OM
dt / R
d
dt / R
o
r •r • r
r
= ρ v + (θ x ∧ ψ z ) ∧ ρv
•r
• ur
•
r
= ρ v + ρθ w + ρψ sin θ(−x )
•
• ur
r •r
= − ρψ sin θ x + ρ v + ρθ w
•
uuuur
uuuuuur uuuur uuuuuur uuuuuur uuuur
VR (M ∈R) = Ω R / Ro ∧ OM = (Ω R / Ro + Ω R / R ) ∧ OM
o
uuuuuur uuuur uuuuuur uuuur 1
= Ω R / Ro ∧ OM + Ω R / R ∧ OM
uuuur
uuuu1r
= VR (M ∈R) + VR (M ∈R1 )
1
o
3°)
•• r
uuur
d uuur
a) Γ R (M) =
VR (M) = ρ v
dt / R
uuuur
b) Γ R (M) =
1
d
dt / R
1
uuuur
uuuuuur uuuur
d uuuur
VR (M) =
VR (M) + Ω R / R ∧ VR (M)
1
1
1
1
dt / R
• ur
•r
•r
• ur
d •r
=
( ρ v + ρθ w ) + θ x ∧ ( ρ v + ρθ w )
dt / R
• r
•• r
••
•• ur • • ur
uuuur
Γ R (M) = ρ v + (ρ θ + ρ θ ) w + θ ρ w − ρθ2 v
1
• r
••
••
•• ur
uuuur
2
Γ R (M) = ( ρ − ρθ ) v + (2ρ θ + ρ θ ) w
1
c)
• ur
uur
uuuuuuur uuuur
u • r • • ur
Γ c (M) = 2Ω R1/ Ro ∧ VR (M) = 2 ψ z o ∧ (ρ v + ρ θ w )
1
• •
r
r
= −2 ψ ρ cosθx + 2 ψ ρθ sin θx
• r •
•
uur
Γ c (M) = 2 ψ x (ρθ sin θ − ρ cosθ )
• •
uuuur
uuuuuuur uuuur
d uuuuuuur uuuur uuuuuuur
Γ R (M ∈R1 ) = (
Ω R1/ Ro ) ∧ OM = Ω R1/ Ro ) ∧ (Ω R1/ Ro ∧ OM )
o
dt / R
o
•• ur
u
• ur
r • ur
u
u
r
= ψ z o ∧ ρv + ψ z o ∧ (ψ z o ∧ ρv )
• ur
••
r
u
r
= − ψ ρcosθx − ρ ψ 2 z o ∧ cosθx
•
r
r
= − ψ ρcosθx − ρ ψ 2 cosθy
••
•
••
uuuur
r
r
ur
Γ R (M ∈R1 ) = − ψ ρcosθx − ρ ψ 2 cosθ(cosθv − sin w )
o
•
•
••
uuuur
r
r
ur
2
2
Γ R (M ∈R1 ) = − ψ ρcosθx − ρ ψ cos θv + ρ ψ 2 cosθsin θw
o
4°)
a)
••
••
•
uuuur
uuuur r
d
Γ R (M) ∧ OM = 0 ⇒ ρ θ + 2ρ θ = 0 = ( ρ2 θ)
1
dt
•
⇒ ρ2 θ = cte
b)
uuuur
uuuur r
Γ R (M) ∧ OM = 0
o
(1)
(2)
uuuur
uuuur
uur
uur
Γ R (M) = Γ R (M) + Γ c (M) + Γ e (M)
o
1
••
•
•
•
−ρψ cosθ + 2 ψ (ρθ sin θ − ρ cosθ )
•
2
•
2
••
2
−ρψ cos θ + ρ − ρθ
=
•
2
••
••
−ρψ cosθsin θ + ρ θ + 2ρ θ
Relative components of the reference R
••
• •
• •
⎧
⎪ρ( − ψ cosθ + 2 ψ θ sin θ) − 2ρ ψ cosθ = 0
(2) ⇒ ⎨
•
••
••
⎪
2
5°)
ρ(ψ
cosθsin
θ
+
θ
)
+
2ρ
θ = 0
⎩
•
(3)
(4)
••
ρ=a ⇒ρ= ρ =0
••
• •
(3) ⇒ − ψ cosθ + 2 ψ θ sin θ = 0
•
2
(5)
••
(4) ⇒ ψ cosθsin θ + θ = 0
(5) ≡
•
d •
(ψ cos2 θ) = 0 ⇒ ψ cos2 θ = c1
dt
•
⇒
2
c sin θ = 0
θ+ 1
cos3 θ
• ••
(7)
2
c1
2
⇒ ψ =
••
(6)
cos4 θ
(8)
2
••
c sin θ dθ = 0
⇔ θ dθ + 1
cos3 θ
•
•
•
•
et θ θ dt = θ d θ ⇒θ d θ −
2
c1d(cosθ) = 0
cos3 θ
•
1
d( θ2 −
2
•
2
⇒θ +
•
2
⇒ θ
2
c1 cos
−2
−2
θ
•
2
1
c1 ) = 0
) = d(θ2 +
2
cos2 θ
2
c1
cos2 θ
= c2
2
= c2 −
c1
cos2 θ
(9)
Learning activities
Teacher’s guide
will count for 20% of the final evaluation of the module
Optional evaluation of educational nature
Activity title : Forms of activity evaluation
Time of learning
4 hours
Being able to:
• Name two forms of assessment
• Recognize the evaluation moments
• Identify the roles of evaluation
Appropriate reading
DIOUF, S. (2004). L’évaluation des apprentissages. Sénégal. Faculté des
Sciences et Technologies de l’Education et de la Formation (FASTEF). Université Cheikh Anta DIOP (UCAD) de Dakar
Justification for the appropriate reading
The text on the assessment of learning gives readers information on the issue
of assessment, evaluation forms and their role in correcting the issues objectively including the short questions, different types of feedback, etc. .
Activity summary
This activity serves to identify forms of assessment from the time of evaluation, and to clarify the roles of assessment and evaluation times.
Evaluation
A newly assigned teacher in a school receives their schedule.
In the classroom and immediately after the presentations, there is a written test
lasting one hour (1 hour).
The following week, they begin their classes and after two sessions of the
mechanics of one and two dimensions, there must be a test on this part of the
course.
1) In what form of assessment can we have the first written test?
2) In what form of assessment can we have the second written test?
3) At what time do we have the summative evaluation?
4) What role does each of the first two evaluations play?
Learning activities
Key answers
1) The first written evaluation should be before the learning. It is a preliminary diagnostic evaluation .
2) The second evaluation will be written during the learning. It is a formal
evaluation.
3) The summative evaluation should be at the end of the learning.
4) Role of the diagnostic evaluation : identify strengths and weaknesses in
students so that they will know what they have to review.
5) The predictive evaluation should give an indication of chances of success
or failure of the students.
The formal evaluation is to help students improve their knowledge as they are
learning.
Self-evaluation
This assessment allows students to be aware of the importance of the moments
of evaluation and the role they can play. This will allow them to be better
equipped and prepare accordingly.
Teacher’s Guide
This evaluation is optional. It is not mandatory. Only those ones who want will
complete it. The professor will correct their productions, but the grades will
not be taken into account in the final assessment.
XV. Synthesis of the module
The MODULE « mechanics 1 » includes prerequisites, objectives, general
learning and specific learning objectives. It is divided into four units of learning. Each unit is accompanied by readings comprised of links and resources
that are valuable aids to the students
Unit 1
Unit 1 includes:
-
-
Four required readings,
Twenty questions and exrcices.
This will facilitate the reception of knowledge for the student.
We started learning the international system of units (SI), based on seven basic
units. Then we continued by physical measurements, measurement errors and
uncertainties. For conclusion, we discussed the particular vector operations of
addition and subtraction of vectors.
Unit 2 Unit 2 includes:
- Three required readings,
- Completion of a course that examines the movements of free fall
- And a series of nine exercises.
In most cases, we seek the kinematic characteristics of a moving object: trajectory, velocity, acceleration, time equations .... For complete knowledge of
students, we worked on different systems coordinates (spherical coordinates,
cylindrical ...).
Unit 3 Unit 3 includes
-
-
-
Three required readings,
Completion of a course dealing with the reaction supports, the forces of
solid-solid friction and the study of the deformation of a spring.
A series of eight exercises based on the application of zero torsor
Unit 4 :
Unit 4 includes:
-
-
-
Three required readings related to the derivation of vector dynamics,
points in a Galilean and non-Galilean reference, work and oscillators.
Additional courses formed by small exercises treating physical phenomena
seen in everyday life
Eighteen exercises and problems solved.
XVI. Summative evaluation
- An instant velocity vector has the hollowing components : Vx , Vy, Vz ;
its normal is equal to :
V = Vx +Vy+Vz
V = Vx + Vy + Vz
V = Vx2 + Vy2+Vz2
V = V(M) = Vx2 + Vy2 + Vz2
Circle the correct answer
- A moving object M is on a horizontal route. Its position M at instants t1,
t2, t3 has abscissas of:
X1 = 2t +20 ; y1 = 4t
X2 = 2t ; y2 = 4t
X3 = 2t – 4 ; y3 = 8- 4t
-
-
-
-
Uniform rectilinear motion
Uniformly accelerated rectilinear motion
Uniformly decelerated rectilinear motion
Rectilinear sinusoidal motion
Circle the correct answer.
3. Complete the following sentences :
a. A movement is helical if and only if its trajectory is given by …. .
b. A helical movement is uniform if its angular velocity is…..
4. The moment of a vector with respect to an axis is a vector when the moment of a vector with respect to a point is a scalar.
a. True
b. False
Circle the correct answer
5. The moment of a vector with respect to an axis is a vector when the moment of a vector with respect to a point is a scalar.
a. True
b. False
6. A solid mass m in equilibrium on a smooth horizontal plane is subjected
to:
a. b. c. d. e. its weight only
the reaction of the plane only
the reaction of the plane acting in the opposite direction, and its weight
the reaction of the plane acting in the same direction, and its weight
neither its weight or the reaction of the plane
7. The equilibrium of a body is more stable when:
a. b. c. d. its weight is large
its weight is small
its center of gravity is low
its center of gravity is high
8. Match the following two columns
Forces at a distance
2. Frictional forces
Contact forces
3. Electrostatic forces
4. Electromagnetic forces
5. Weight of a body
9. A force is said to be conservative if its work :
a. depends on the path followed
b. is independent of the path followed
10. Match the following two columns :
1. Pure oscillation
2. Damped oscillation
3. Forced oscillation
a.
b.
c.
d.
energy input
exchange of energy
energy dissipation
conservation of energy
11. When a solid is in equilibrium, the resultant forces applied to the solid
is zero. This condition is:
a. necessary
b. sufficient
Choose the correct answer
12. Two forces can turn a solid. True or False
13. Equilibrium of a solid placed on an inclined plane is possible if :
a. rough
b. smooth
Check the correct answer.
14. Complete the following sentence :
The mechanical energy is conserved when, occasionally, the only forces working derive from ______energy.
15. An advantage to having conservation of mechanical energy is :
a. a first degree integral
b. a second degree equation
16. For a solid body, which of the following physical quantities varies with
altitude ?
a. its mass
b. its weight
17. Is the moment of a force, with respect to a point, a vector because it
results from a product of type:
a. vectorial of two vectors
b. scalar of two vectors
18. The moment of a force with respect to a segment is a scalar.
a. True
b. False
19. A body left at the top of the vertical mast of a boat that is animated by a
rectilinear uniform motion compared to a terrestrial reference, falls at
the foot of the mast. Why?
20. So that kinetic energy is always non-negative, why can the mechanical
energy of a system be negative?
21. A stationary magnetic field does not increase the norm of the velocity of
a charged particle. Why is it so widely used in particle accelerators ?
22. Why is the mechanical energy of a system not conserved in all cases ?
23. When the balance of forces acting on a vehicle or a person walking along
a horizontal path, one is led to conclude that it is the force of friction
which allows the movement. Is this paradoxical? Why?
Answers Keys
1.
a. Think carefully before answering
b. Have you thought about the units of velocity vector before your choice?
c. The unit velocity in the International System (SI) is not the (m / s) 2
d. Good answer. It must indeed be the square root of the sum of the squares
of the coordinates.
2.
a.
Very good. Indeed the velocity is constant, the motion is rectilinear uniform.
b. Did you calculate the velocity of the moving object before choosing?
c. Reread the question and see if the velocity depends on time.
d. For sinusoidal motion, the horizontal axis is a sinusoidal function of time
(x = Asin (wt + Φ)), which is not the case here.
3. a. by a propeller. Good answer, you have learned your lesson
b. is constant. Good answer, your lesson is learned
4.
a. Carefully reread the question
b. Good answer, you do well to distinguish the two moments
5. a. Good answer, indeed the moment of a vector with respect to an axis is a
scalar, not a vector
b. Reread the question before answering.
6.
a. Attention, there should be more than a force for equilibrium of a solid on
a plane.
b. Only one force cannot maintain a solid in equilibrium on a plane.
c. Very good. You see it is the principle of action and reaction.
d. A body can be in equilibrium on a plane under the action of two forces
having the same direction and same orientation.
e. Every body in equilibrium on a plane is submitted to at least two forces
7
a. Attention. a stable equilibrium does not depend solely on body weight
b. Think again, a stable equilibrium can not depend solely on force.
c. Very good, actually more the center of gravity of a body is, the more the
equilibrium is stable.
d. Reread the question and you can tell that between two people of the same
weight and size , which is most stable if they are standing.
8.1b ; 1c ; 1d ; 2a 9
a. Have you really learned your lesson before replying?
b. Bravo. You have learned your lesson. The work of a conservative force
is actually independent of the path.
10. 1d ; 2c ; 3a. Bravo.
11
a. Very good, it is necessary, however it does not suffice by itself
b. Incorrect, it does not suffice by itself
12. True. Good answer, you know the applications of groups of forces
13.
a. Bravo. If the slope is rough, there is a coefficient of friction generated by
a frictional force that keeps the solid equilibrium.
b. A solid on an inclined plane has drag and can not therefore remain in
equilibrium on a smooth surface.
14. Potential energy. Bravo.
15.
a.
Very good. In fact, the advantage is to have a first integral that the solution
is easier than the equations from the fundamental principle of dynamics,
which are of second order.
b. The equations of the second level are not easy to solve and are thus a
disadvantage
16.
a. Attention, the mass of a body does not vary
b. Bravo. The weight of a body varies with altitude.
17.
a. Good answer. The moment of a vector from a point is actually a vector.
b. Attention, you are confusing it with the moment of a vector with respect
to a segment
18.
a. Good answer. This is indeed the moment of a vector with respect to a line
which is a scalar.
b. Proofread well and think before responding.
19. The reference R ʹ linked to the boat is also Galilean. With respect to R ʹ , the
laws of mechanics and thus those of free fall are written in the same way as
in R . Very good, you are mastering your course.
20. The energies Ek and EP are defined at a close constant additive. Conventionallyt, Ek = 0 for the body of an immovable object, so that EP = 0 for bodies
infinitely far from each other. It results that EP is negative if the interaction is
positive ; thus En can be negative, contrary to Ek . Very good answer,
21. The magnetic field can bring in, through its properties of deviation, any charged
particle in a region where there is an electric field, which can only increase
the norm of the velocity of the particle. Very good answer.
22. The mechanical energy is not conserved in the general case, because of external and internal forces that do not derive a potential energy; thus this is
not a conservative quantity. In contrast, the total energy, total kinetic energy
of the macroscopic potential energy of external forces and of the internal
energy, is conservative: it is a statement of the first law of thermodynamics.
Very good.
23. To advance, we must rely on the ground. The friction force is an essential
intermediary. Very good. Without the frictional forces we would be unable
to walk properly.
XVII. References
Readings
MAGER, R,F. (1975). Comment définir les objectifs pédagogiques. Bordas.
Paris
RATIARISON, A. (2006). Grandeurs physiques – Mesures-incertitudesOpérations vectorielles. Madagascar. Université d’Antananarivo. Cours
inédit.
CAZIN, M. (1972). Mécanique générale et industriielle Tome 1- Gauthier
villarsPEREZ–MASSON J.-P. (1997) Mécanique Fondement et applications
Paul, A.T.(1995). Physics for Scientists and Engineers – Worth Publishers
– New York, NY 1003
The Free High School Sciences : A Textbook for High School Students Studing
physics – FHSSt Authors- December 9, 2005 from http://savannah.nongnu.
org/projects/fhsst (consulte le 15 sptembre 2006)
ANSERMET, J.P. ( 2004-2005). La mécanique rationnelle – Formation de
base des Sciences et des ingénieurs. Institut de Physique des nanostructures- Ecole Polytechnique Fédérale de Lausanne – PHB – Ecublens,
1015 Lausanne.
DIOUF, S. (2004). L’évaluation des apprentissages. UCAD de Dakar.
FASTEF(ex ENS). Cours inédit
DE KETELE, J. M. (1986). L’évaluation : approche descriptive ou prescriptive ? De Boeck Université
DE LANDSHEERE , G (1984). Evaluation continue et examens. Précis de
docimologie. Editions Labor. Education 2000.
DAWOUD. M. (1995). Elaboration d’un examen de rendement scolaire. Techniques et procédures. Editions Nouvelles.
ODILE, et VESLIN, J. (1992). Corriger des copies. Evaluer for former.
DEPOVER , C et al .(1994) La conception des logiciels éducatifs (Title provisoire). Inédit
Links and resources
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
29.
30.
31.
32.
ABC SITE
http://www.hazelwood.k12.mo.us/grichert/sciweb/applets.htlm.
http://jas.eng.buffalo.edu
http://lectureonline.cl.msu.edu/∼mmp/kap5/work/work.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5a.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5b.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5c.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5d.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5e.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5c.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5d.htlm
http://www.hazelwood.k12.mo.su/∼grischert/explore/dswmedia/freefall.htlm
http://www.google.ca/search?client=firefox_a&rls=org.mozilla%3AenUS%Aofficial-&hl=en&q=c3%A9equilibre+d%27un+solide+sur+plan
&meta
http://e.m.c.2.fr_chute-libre.htm
http://e.m.c.2.free.fr\pj00wd3l.html
http://www.hazelwood.k12.mo.us/grichert/sciweb/applets.htlm.
http://jas.eng.buffalo.edu
http://lectureonline.cl.msu.edu/∼mmp/kap5/work/work.htlm
http://www.glenkrook.k12.il.us/gbsci/phys/Class/1DKin/U1L5a.htlm
http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre2/
titre1res.htm
33. http://formation.edu-psud.fr/pcsm/physique/outils_nancy/apprendre/Chapitre3/
partie3 / titre1res.htm
34. http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.op_addition.aspx
35. http://msch2.microsoft.com/fr-fr/library/system.windows.forms.paddings.
op_methods.aspx
36. http://mathexel.site.voila.fr/index.htlm
37. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/chien_
j.html
38. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/4mouche_
j.html
39. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/lissajou_j.html
40. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/hyp_ep
_j.html
41. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/plttrn
_j.html
42. http://www.sciences.univ-nantes.fr/physique/perso/corticol/bibliohtml/pndhgs_j.html
43.
44.
45.
46.
47.
48.
49.
50.
51.
http://www.sciences.univ-nantes.fr/physique/perso/gtulloue/aquadiff.
html
52.
53.
54. http://www.univ-lemans.fr/enseignements/physique/01/statique.htm
55. http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/prlong.
htm
56. http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/prlong.
html
57. http://www.sciences.univ-nantes.fr/ physique/perso/cortial/bibliohtm/froflu.
html
XVIII. Grades and results of student’s evaluation
Name of Excel file : Evaluation - Students
Mechanics 1
Professor : Adolphe RATIARISON
Student evaluation spreadsheet
Academic year
Mechanics 1:
Professor:
Last
Name
RATIARISON Adolphe
First
Name
A.1
/20
0
A.2
/20
0
A.3
/20
0
A.4
/20
0
Summ.
Eval.
/20
0
Total
/100
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Average
/20
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Result
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
Bad
XIX. Author of the module
Pr Adolphe RATIARISON
Département de Physique
Faculté des Sciences
Université d’Antananarivo,
Madagascar
Tel 261 32 04 266 58
:E_mail : [email protected]
Thesis professor in states of physics, Adolphe RATIARISON teaches :
-
-
-
In first cycle, general mechanics I et II,
In second cycle, heat transfer and turbulent boundary layer
In third cycle, atmospheric physics, physical oceanography
His research in centered on :
-
-
-
-
-
-
The prediction of trajectories of tropical cyclones
The prediction of rain and no rain
Forecasting hail
Siltation of harbors and estuaries
Hydrology
Medical Imaging
XX. Files structure
Name of module (in a WORD file) :
Kinematics of a point
Equilibrium of solids on a plane
Physics quantities-meausurements- Uncertainties- Vector operations
References-Dynamics of a material point-Work- Power, energy, oscillators
Name of other files (WORD, PDF, PPT, etc.) of the module
Evaluation of activities
MÉCANIQUE
Lectures Obligatoires
Source: Wikipedia.org
1
Table des matières
Mouvement de rotation .............................................................................................................................. 5
Définition ................................................................................................................................................ 5
Cinématique dans l'espace ................................................................................................................ 8
Dynamique et énergétique ..................................................................................................................... 9
Centre instantané de rotation .................................................................................................................. 10
Définition .............................................................................................................................................. 10
Exemple des danseuses de cancan ...................................................................................................... 11
Justification .......................................................................................................................................... 11
Utilisation du CIR dans un problème de cinématique plane ........................................................... 13
Orbite ......................................................................................................................................................... 14
Éléments orbitaux ................................................................................................................................ 14
Périodes ................................................................................................................................................. 17
Relations entre les anomalies et les rayons ........................................................................................ 17
Barycentre (physique) .............................................................................................................................. 18
Historique ............................................................................................................................................. 18
Développement mathématique ............................................................................................................ 19
Développements physiques .................................................................................................................. 20
Centre d'inertie ................................................................................................................................ 21
Centre de gravité .............................................................................................................................. 21
Méthode graphique .......................................................................................................................... 22
Astronomie ............................................................................................................................................ 22
Localisation du centre de gravité d'une plaque à deux dimensions ................................................ 23
Système masse-ressort .............................................................................................................................. 23
Oscillations rectilignes d'une masse soumise à l'action d'un ressort ............................................... 23
Amélioration ......................................................................................................................................... 25
Autre amélioration ............................................................................................................................... 25
Moment de force (mécanique) ................................................................................................................. 25
Translation d'une force ....................................................................................................................... 26
Moment par rapport à un point ...................................................................................................... 26
Moment par rapport à un axe ......................................................................................................... 28
Couple de forces ............................................................................................................................... 29
2
Théorème de Varignon .................................................................................................................... 30
En dynamique ....................................................................................................................................... 30
Moment d'inertie....................................................................................................................................... 31
Approche empirique ............................................................................................................................ 31
Détermination du moment d'inertie ................................................................................................... 31
Moments d'inertie particuliers ........................................................................................................... 32
La boule ............................................................................................................................................. 32
La barre ............................................................................................................................................ 33
Le cylindre plein ............................................................................................................................... 33
Le cylindre creux .............................................................................................................................. 33
Théorème de transport (ou Théorème d'Huygens ou Théorème de Steiner) ................................. 34
Énergie cinétique....................................................................................................................................... 34
Historique ............................................................................................................................................. 34
Conventions .......................................................................................................................................... 34
Définitions ............................................................................................................................................. 35
Cas d'un point matériel ................................................................................................................... 35
Cas d'un système de points .............................................................................................................. 35
Unité ...................................................................................................................................................... 36
Théorème de König .............................................................................................................................. 36
Enoncé ............................................................................................................................................... 36
Application à un solide .................................................................................................................... 36
En mécanique relativiste ..................................................................................................................... 37
Théorème de l’énergie cinétique ......................................................................................................... 38
Énoncé ............................................................................................................................................... 38
Démonstration .................................................................................................................................. 39
Théorème de la puissance cinétique ................................................................................................... 39
L’énergie thermique en tant qu’énergie cinétique ............................................................................ 40
Énergie mécanique .................................................................................................................................... 40
]Expression ........................................................................................................................................... 40
Solide ponctuel ................................................................................................................................. 40
Solide étendu non déformable ......................................................................................................... 41
Solide déformable ............................................................................................................................. 41
3
Théorème de l'énergie mécanique ...................................................................................................... 41
Conservation ......................................................................................................................................... 42
4
Mouvement de rotation
Sphère en rotation autour d'un de ses diamètres
La rotation est l'un des deux mouvements simples fondamentaux des solides, avec la translation
rectiligne. En génie mécanique, il correspond au mouvement d'une pièce en liaison pivot par
rapport à une autre.
La notion de mouvement circulaire est une notion de cinématique du point : on décrit la position
d'un point dans le plan. La rotation est une notion de cinématique du solide : on décrit
l'orientation d'un solide dans l'espace.
L'étude du mouvement de rotation est la base de la méthode du centre instantané de rotation
(CIR).
Définition []
Un solide est en rotation si la trajectoire de tous ses points sont des cercles dont le centre est une
une même droite ; cette droite est appelée « axe de rotation », et habituellement notée Δ.
En cinématique dans le plan, les trajectoires des points sont des cercles concentriques, le centre
commun de ces cercle est appelé « centre de rotation » et habituellement noté O.
Définitions []
Définition de l'orientation et de la vitesse angulaire
5
L'orientation du solide est repérée par un angle habituellement noté θ (voir Angles d'Euler). En
cinématique plane, cet angle peut être défini comme l'angle entre


une direction de référence passant par O, en général l'axe (Ox), et
une droite passant par O et par un point A donné du solide distinct de O.
La vitesse de rotation ω est définie par
.
l'accélération angulaire α est définie par
soit également
.
À l'instar du mouvement de translation et du mouvement circulaire, on distingue le mouvement
de rotation uniforme et le mouvement de rotation uniformément varié.
Mouvement de rotation uniforme []
Dans le cas du mouvement de rotation uniforme, on a une accélération angulaire nulle
α=0
donc la vitesse de rotation est constante
ω = ω0
et l'angle croît de manière linéaire
θ = θ0 + ω0×t
où θ0 est l'orientation à l'instant initial. Ce mouvement idéal est en général utilisé pour décrire la
partie centrale d'un mouvement (vitesse angulaire stable).
Mouvement de rotation uniformément varié []
Dans le cas du mouvement de rotation uniforme, on a une accélération angulaire constante
6
α = α0
donc la vitesse de rotation varie de manière uniforme
ω = ω0 + α0×t
où ω0 est la vitesse à l'instant initial, et l'angle croît de manière quadratique
θ = θ0 + ω0×t + 1/2×α0×t2
où θ0 est l'orientation à l'instant initial. Ce mouvement idéal est en général utilisé pour décrire le
début et la fin d'un mouvement (mise en route ou arrêt).
Mouvement des points []
Triangle des vitesses dans le cas d'une barre en rotation
Triangle des vitesses dans le de points situés sur des axes différents
Chaque point M de l'objet a une trajectoire circulaire, donc décrit un cercle de centre O et de
rayon R = OM. Le vecteur vitesse instantané est tangent au cercle, donc perpendiculaire au rayon
[OM]. Sa norme vaut
v = ω×R.
7
Les équations horaires du point dans le cas des mouvements uniforme est décrit dans l'article
Mouvement circulaire uniforme. Dans le cas général, elles sont décrites dans l'article Mouvement
circulaire non uniforme.
Graphiquement, si l'on considère les vecteurs vitesse des points appartenant à une même droite
passant par O, leurs extrémités sont sur une droite passant par O (en raison de la proportionnalité
en R) ; la figure ainsi formée est appelée « triangle des vitesses ».
Cela permet une résolution graphique de problèmes cinématiques : si l'on connaît la vitesse d'un
point du solide — par exemple point en contact avec un actionneur (extrémité de tige d'un vérin,
dent d'engrenage), on peut déterminer le vecteur vitesse de tous les points du solide :



leur direction est perpendiculaire au rayon en ce point ;
la norme de la vitesse de tous les points situés sur un même cercle de centre O est
identique ;
si l'on « rabat » les points sur une même droite passant par O, les vecteurs forment le
triangle des vitesse.
Par « rabattre le point B sur la droite », on entend trouver le point B' de la droite situé sur le
même cercle de centre O.
Cinématique dans l'espace []
Position et vecteur vitesse de rotation
Dans le cas de la cinématique dans l'espace, on prend un axe de référence normal à l'axe de
rotation et le coupant en O, et un point A du solide situé dans le plan normal à l'axe de rotation et
passant par O.
Le vecteur vitesse de rotation



est le vecteur
ayant pour direction l'axe de rotation ;
dont le sens est déterminé par la règle conventionnelle d'orientation : règle de la main
droite, sens de vissage ;
dont la norme est la dérivée de la position par rapport au temps.
8
Le vecteur accélération angulaire
est la dérivée vectorielle de
:
Si O est un point de l'axe de rotation et A un point quelconque du solide, le vecteur vitesse
A est obtenu par
en
.
Le vecteur vitesse angulaire est la résultante du torseur cinématique. Le vecteur vitesse en A
est le moment de ce torseur en ce point de réduction.
Dynamique et énergétique []
On peut appliquer la dynamique du point à chaque élément de matière du solide. En intégrant sur
la totalité du solide, on trouve les résultats suivants :


l'inertie en rotation, ou inertie à la rotation, par rapport à l'axe Δ est exprimée par le
moment d'inertie JΔ ;
l'accélération angulaire est reliée aux couples extérieurs Cext et aux moments des forces
extérieures par rapport à l'axe
par le principe fondamental de la dynamique :
ou, sous forme vectorielle
.
Article détaillé : Dynamique de rotation.
Par ailleurs, l'énergie cinétique en rotation Ec s'exprime par
et le théorème de l'énergie cinétique énonce que la variation de l'énergie cinétique est égale à la
somme des travaux des couples et moments internes et externes. Le travail d'un couple C
constant entre deux positions θ1 et θ2 s'écrit
Wθ1→θ2(C) = C⋅(θ2 - θ1),
le paramètre (θ2 - θ1) étant l'amplitude du mouvement. Si le couple varie, on définit alors le
travail élémentaire pour une petite rotation d'un angle dθ
9
dWC = C⋅dθ
et
.
La puissance P du couple se définit par
PC = C⋅ω.
Sous forme vectorielle, la puissance devient
.
Centre instantané de rotation
Le centre instantané de rotation (CIR) est un terme utilisé en mécanique classique et plus
particulièrement en cinématique pour désigner le point autour duquel tourne un solide à un
instant donné par rapport à un repère de référence.
Définition []
À l'instant t, I est le centre instantané de rotation du solide S dans le repère R défini par les axes
Ox et Oy.
10
Lorsqu'un solide isolé au sens mécanique du terme, se déplace suivant une trajectoire comprise
dans un plan, le CIR se définit comme le point où le vecteur vitesse est nul.
Le CIR se situe sur la perpendiculaire à chaque vecteur vitesse du solide isolé passant par le
point d'application de ce dernier.
Lorsque le solide isolé se déplace uniquement en translation dans un plan, le CIR est projeté à
l'infini.
Le torseur cinématique réduit au CIR est :
Exemple des danseuses de cancan []
Danseuses de French cancan vues de haut
L'illustration représente des danseuses de cancan vues de dessus. Si on considère que l'ensemble
des cinq danseuses est un solide isolé au sens mécanique du terme, on peut dire que le centre
instantané de rotation est la danseuse centrale, puisqu'elle n'a pas de vitesse relative
contrairement à ses compagnes qui ont une vitesse proportionnelle à leur éloignement du centre.
Justification []
11
Sur un court instant, le mouvement d'une bielle (bas) dans un système manivelle-bielle-piston est
équivalent à une rotation autour du CIR (haut)
Considérons une pièce ayant un mouvement plan quelconque, par exemple le mouvement d'une
bielle. Si l'on prend une photographie, on a un flou en raison du mouvement : les points
« filent », et les segments de droite générés par les points sont une image des vecteurs vitesse.
Si la bielle était en rotation autour de son CIR, on obtiendrait une photo semblable, avec le même
flou. Sur un très court instant — le temps de pose de la photographie —, les deux mouvements
sont équivalents.
De manière plus rigoureuse : le torseur cinématique d'un solide en mouvement plan dans le plan
(Oxy ) réduit à un point quelconque A s'écrit :
c'est un glisseur puisque
et
sont orthogonaux. Il existe donc un point B tel que
. d'après la propriété d'équiprojectivité, on a
.
12
Si l'on note (X, Y, 0) les composantes de
, on a alors
si ωz n'est pas nul, alors le point B existe et est unique ; il est appelé centre instantané de rotation.
Utilisation du CIR dans un problème de cinématique plane []
Considérons un mouvement plan qui n'est pas une mouvement de translation. Durant un court
instant, tout se passe comme si le solide était en mouvement de rotation autour de son CIR. On
peut alors appliquer les relations établies dans le cas des mouvements de rotation, et en
particulier la notion de triangle des vitesses. Cela permet de déterminer le vecteur vitesse en un
point quelconque du solide, à condition de connaître :


le vecteur vitesse en un point ;
la position du CIR.
La méthode est une alternative à la méthode de l'équiprojectivité.
Application à une voiture dans un virage
Prenons l'exemple d'une voiture en virage, dont on connaît la direction, le sens, le point
d'application et l'intensité (5 m/s) du vecteur vitesse de la roue avant. On connaît également la
direction, le point d'application et le sens de la roue arrière. Les points A et B sont les centres des
roues et respectivement les points d'application de leur vecteur vitesse.
L'objectif est de déterminer l'intensité du vecteur vitesse de la roue arrière.
Résolution graphique grâce au CIR :
13
1.
2.
3.
4.
On choisit une échelle des vitesses, par exemple 10 mm pour 1 m/s ;
On place le vecteur vitesse de la roue avant au point A ;
On trace (en rouge) la direction du vecteur vitesse de la roue arrière au point B ;
Le CIR se situe sur une droite passant par le point d'application des vecteurs vitesse et
perpendiculaire à ces derniers : on trace donc les traits verts, et on déduit le CIR ;
5. On mesure le segment [CIR B] et on reporte la mesure sur le segment [CIR A] trait bleu ;
6. On trace une droite passant par le CIR et par l'extrémité du vecteur vitesse associé au
point A ;
7. On trace un segment perpendiculaire à [CIR A] passant par le mesure reportée sur [CIR
A] et coupant le segment passant par CIR et par l'extrémité de
;
8. On mesure ce dernier segment et en fonction de l'échelle on trouve l'intensité du vecteur
vitesse
.
Orbite
Orbite circulaire de deux corps de masse différentes autour de leur barycentre (croix rouge).
En mécanique céleste, une orbite est la trajectoire que décrit dans l'espace un corps autour d'un
autre corps sous l'effet de la gravitation.
L'exemple classique est celui du système solaire où la Terre, les autres planètes, les astéroïdes et
les comètes sont en orbite autour du Soleil. De même, des planètes possèdent des satellites
naturels en orbite. De nos jours, beaucoup de satellites artificiels sont en orbite autour de la
Terre.
Les trois lois de Kepler permettent de déterminer par le calcul le mouvement orbital.
Éléments orbitaux []
14
Orbite elliptique
Une orbite elliptique peut se définir dans l'espace selon six paramètres permettant de calculer très
précisément la trajectoire complète. Deux de ces paramètres (excentricité et demi-grand axe)
définissent la trajectoire dans un plan, trois autres (inclinaison, longitude du nœud ascendant et
argument du péricentre) définissent l'orientation du plan dans l'espace et le dernier (instant de
passage au péricentre) définit la position de l'objet. Voici la description plus détaillée de ces
paramètres :


Demi-grand axe a : la moitié de la distance qui sépare le péricentre de l'apocentre (le
plus grand diamètre de l'ellipse). Ce paramètre définit la taille absolue de l'orbite. Il n'a de
sens en réalité que dans le cas d'une trajectoire elliptique ou circulaire (le demi-grand-axe
est infini dans le cas d'une parabole ou d'une hyperbole)
Excentricité e : une ellipse est le lieu des points dont la somme des distances à deux
points fixes, les foyers (S et S' sur le diagramme), est constante. L'excentricité mesure le
décalage des foyers par rapport au centre de l'ellipse (C sur le diagramme); c'est le
rapport de la distance centre-foyer au demi-grand-axe. Le type de trajectoire dépend de
l'excentricité :
o e = 0 : trajectoire circulaire
o 0 < e < 1 : trajectoire elliptique
o e = 1 : trajectoire parabolique
o e > 1 : trajectoire hyperbolique
15




Inclinaison i : l'inclinaison (entre 0 et 180 degrés) est l'angle que fait le plan orbital avec
un plan de référence. Ce dernier étant en général le plan de l'écliptique dans le cas
d'orbites planétaires (plan contenant la trajectoire de la Terre; en noir dans la figure 1).
L'inclinaison est l'angle orange dans la figure 1.
Longitude du nœud ascendant ☊ : il s'agit de l'angle entre la direction du point vernal
et la ligne des nœuds, dans le plan de l'écliptique. La direction du point vernal (en noir
dans la figure 1) est la droite contenant le Soleil et le point vernal (point de repère
astronomique correspondant à la position du Soleil au moment de l'équinoxe du
printemps). La ligne des nœuds (en vert dans la figure 1) est la droite à laquelle
appartiennent les nœuds ascendant (le point de l'orbite où l'objet passe du côté nord de
l'écliptique) et descendant (le point de l'orbite où l'objet passe du côté sud de l'écliptique).
Argument du périhélie ω: il s'agit de l'angle formé par la ligne des nœuds et la direction
du périhélie (la droite à laquelle appartiennent le Soleil et le périhélie de la trajectoire de
l'objet), dans le plan orbital. Il est en bleu dans la figure 1. La longitude du périhélie est
la somme de la longitude du nœud ascendant et de l'argument du périhélie.
Instant τ de passage au périhélie : La position de l'objet sur son orbite à un instant
donné est nécessaire pour pouvoir la prédire pour tout autre instant. Il y a deux façons de
donner ce paramètre. La première consiste à spécifier l'instant du passage au périhélie. La
16
seconde consiste à spécifier l'anomalie moyenne M (en rouge dans la figure 1) de l'objet
pour un instant conventionnel (l'époque de l'orbite). L'anomalie moyenne n'est pas un
angle physique, mais spécifie la fraction de la surface de l'orbite balayée par la ligne
joignant le foyer à l'objet depuis son dernier passage au périhélie, exprimée sous forme
angulaire. Par exemple, si la ligne joignant le foyer à l'objet a parcouru le quart de la
surface de l'orbite, l'anomalie moyenne est
° = 90°. La longitude moyenne
de l'objet est la somme de la longitude du périhélie et de l'anomalie moyenne.
Périodes []
Lorsqu'on parle de la période d'un objet, il s'agit en général de sa période sidérale, mais il y a
plusieurs périodes possibles :





Période sidérale : Temps qui s'écoule entre deux passages de l'objet devant une étoile
distante. C'est la période « absolue » au sens newtonien du terme.
Période anomalistique : temps qui s'écoule entre deux passages de l'objet à son périastre.
Selon que ce dernier est en précession ou en récession, cette période sera plus courte ou
longue que la sidérale.
Période draconitique : temps qui s'écoule entre deux passages de l'objet à son nœud
ascendant ou descendant. Elle dépendra donc des précessions des deux plans impliqués
(l'orbite de l'objet et le plan de référence, généralement l'écliptique).
Période tropique : temps qui s'écoule entre deux passages de l'objet à l'ascension droite
zéro. À cause de la précession des équinoxes, cette période est légèrement et
systématiquement plus courte que la sidérale.
Période synodique : temps qui s'écoule entre deux moments où l'objet prend le même
aspect (conjonction, quadrature, opposition, etc.). Par exemple, la période synodique de
Mars est le temps séparant deux oppositions de Mars par rapport à la Terre; comme les
deux planètes sont en mouvement, leur vitesses angulaires relatives se soustraient, et la
période synodique de Mars s'avère être 779,964 d (1,135 années martiennes).
Relations entre les anomalies et les rayons []
Dans ce qui suit, e est l'excentricité, T l'anomalie vraie, E l'anomalie excentrique et M l'anomalie
moyenne.
Le rayon r de l'ellipse (mesuré depuis un foyer) est donné par :
Les relations suivantes existent entre les anomalies :
17
ou encore
Une application fréquente consiste à trouver E à partir de M. Il suffit alors d'itérer l'expression :
Si on utilise une valeur initiale E0 = π, la convergence est garantie, et est toujours très rapide (dix
chiffres significatifs en quatre itérations).
Barycentre (physique)
En physique et en mécanique, le barycentre (ou centre de masse) d’un solide est le centre des
poids.
La notion est également utilisée en astronomie
En mécanique du solide, on parle spécifiquement de moment : moment d'inertie, moment
cinétique.
Historique []
Le barycentre de barus (poids) et centre est initialement le centre des poids. C'est donc une
notion physique et mécanique. Le premier à avoir étudié le barycentre en tant que centre des
poids (ce qu'on appelle de nos jours le centre de gravité) est le mathématicien et physicien
Archimède. Il est un des premiers à comprendre et expliciter le principe des moments, le principe
des leviers et le principe du barycentre. Il écrit dans son traité Sur le centre de gravité de surface
plane:
« Tout corps pesant a un centre de gravité bien défini en lequel tout le poids du corps peut être
considéré comme concentré. »
Son principe des moments et des leviers lui permet de construire assez simplement le barycentre
O de deux points de masses m1 et m2 différentes.
18
Pour que la balance soit en équilibre, il faut que les moments
et
soient
égaux. Si par exemple la masse m1 est 4 fois plus importante que la masse m2, il faudra que la
longueur OA soit 4 fois plus petite que la longueur OB. Cette condition se traduit de nos jours par
l'égalité vectorielle
Il est le premier à avoir cherché des centres de gravité de surface comme des demi-disques, des
paraboles. Il procède par approximations successives et a pu prouver que la recherche d'un centre
de gravité utilise des méthodes analogues à celle du calcul d'aire. Son travail est prolongé par
celui de Paul Guldin (1635/1640) dans son traité Centrobaryca et celui de Leibniz à qui l'on doit
la fonction vectorielle de Leibniz.
La notion de centre d'inertie G pour un système non solide est une notion dégagée par Christiaan
Huygens (1654), lors de l'établissement de sa théorie des chocs : même s'il sait que P = P0, il n'est
pas évident pour lui que G ira à vitesse constante. En particulier au moment de la percussion, où
des forces quasi-infinies entrent en jeu, avec éventuellement bris de la cible, G n'en continue pas
moins imperturbé son mouvement : cela paraît mirifique à Huygens, qui ne connaît pas encore le
calcul différentiel. C'est alors qu'il énonce le principe de mécanique :
« Le barycentre d'un système matériel se meut comme si toute la masse du système y était
transportée, les forces extérieures du système agissant toutes sur ce barycentre. »
On peut remarquer le glissement subtil entre barycentre, centre des poids (= centre de gravité)
comme le voyait Archimède et barycentre, centre des masses (= centre d'inertie).
Développement mathématique []
Article détaillé : barycentre (géométrie affine).
Les mathématiques généralisent la construction d'Archimède du point d'équilibre de deux points
affectés de deux masses positives progressivement à des ensembles plus complexes. Les
coefficients peuvent être négatifs : Le barycentre des points A et B affectés des masses a et b (a +
b non nul) est l'unique point G tel que
.
Les coordonnées de G sont alors
19
Le nombre de points peut passer à trois points, quatre points et se généraliser à n points. Si la
somme des masses ai est non nulle, le barycentre du système
que
est le point G tel
.
Les coordonnées sont données par les formules, pour j variant de 1 à la dimension de l'espace
C'est sous cette forme qu'il devient un outil puissant en géométrie affine.
Le nombre de points peut même devenir infini, permettant de trouver le barycentre d'une courbe
ou d'une surface.
Si l'ensemble constitue un domaine D continu, à chaque point M du domaine on affecte une
densité g(M) où g est une fonction continue (un champ scalaire). Le barycentre est alors le point
G tel que
dans l'espace ou
dans le plan .
Si les points M ont pour coordonnées (x1;x2,x3) la fonction de densité s'écrit g(x1,x2,x3) et les
coordonnées de G s'écrivent
Si l'on se ramène à une dimension, ou bien si l'on considère chaque coordonnée séparément, on
retrouve la formule de la moyenne pondérée :
Développements physiques []
20
Centre d'inertie []
En mécanique, le centre d'inertie d'un corps correspond au barycentre des particules qui
composent le corps en question ; chaque particule étant pondérée par sa masse propre. C'est donc
le point par rapport auquel la masse est uniformément répartie.
Dans le cas d'un corps continu , on emploie comme fonction de pondération la masse
volumique ρ du corps. Dans ce cas, la position du centre d'inertie G est défini par la relation
suivante (O étant un point quelconque de l'espace) :
ou
Le centre d'inertie ne dépend donc que de la masse volumique et de la forme du corps. C'est une
caractéristique intrinsèque.
Une propriété étonnante du centre d'inertie est que son mouvement est parfaitement déterminé
par les lois du mouvement, quoi qu'il arrive à ses composants aussi longtemps que ceux-ci ne
subissent pas eux-mêmes de force nouvelle. Ainsi par exemple si un obus éclate en vol, le centre
d'inertie de ses fragments continue à suivre imperturbablement une parabole comme si de rien
n'était (aux effets de résistance de l'air près) avant, pendant et après l'explosion. Attention : ceci
ne s'applique évidemment pas à un obus balistique ou un astéroïde, précisément parce que la
force sur chaque éclat d'obus varie.
Centre de gravité []
Article détaillé : Centre de gravité.
Le centre de gravité d'un corps correspond au barycentre des particules qui composent le corps
en question ; chaque particule étant pondérée par son poids propre.
La position du centre de gravité Gg est défini par la relation suivante (
gravité au point M):
étant le champ de
Le centre de gravité est fondamentalement lié au champ de gravité dans lequel le corps est
plongé. Il n'existe pas forcément !
Très souvent en mécanique, la dimension des corps étant faible devant la rotondité de la terre, on
considère un champ de gravité uniforme. Sous cette hypothèse, le centre de gravité et le centre
d'inertie sont confondus.
21
Méthode graphique []
Dans le cas d'un assemblage composé de pièces dont on connaît la masse et le centre de gravité,
on peut déterminer le centre de gravité de l'ensemble avec la méthode du dynamique et du
funiculaire :
1. on détermine la résultante des différents poids avec un premier funiculaire, la pièce étant
dans une position donnée ;
2. on effectue un second funiculaire en considérant les « poids horizontaux », ce qui revient
à tourner la pièce d'un quart de tour.
Astronomie []
Animation impliquant 2 corps de faible différence de masse. Le barycentre se trouve à l'extérieur
du corps principal comme dans le cas du couple Pluton/Charon.
On parle de barycentre en ce qui concerne le couple formé par un corps stellaire possédant un
satellite. Le barycentre est le point autour duquel l'objet secondaire gravite. Si la plupart des
couples connus possède leur barycentre à l'intérieur de l'objet principal, il existe des exceptions
notables :




Le cas du couple Pluton/Charon : la différence de masse entre ces deux corps est
relativement faible, le barycentre se trouve donc à l'extérieur de Pluton. Pour certains
astronomes, plutôt que de parler de planètes et de satellites, il conviendrait dans ce cas
précis de retenir la notion de « planète double ».
Plusieurs astéroïdes reproduisent le cas de figure ci-dessus.
Le barycentre du couple Jupiter/Soleil se trouve à l'extérieur de ce dernier à environ un
rayon solaire de distance.
On retrouve aussi cette particularité chez certaines étoiles doubles
22
Localisation du centre de gravité d'une plaque à deux
dimensions []
Cette méthode est utile lorsque l'on souhaite trouver le centre de gravité d'un objet plan dont la
forme est complexe et dont on ne connait pas les dimensions exactes.
Étape 2: Suspendre la plaque
en un point proche d'un
Étape 1: Une plaque de forme sommet et attendre la position
arbitraire.
d'équilibre. À l'aide d'un fil à
plomb, tracer la verticale
passant par ce point.
Étape 3: Suspendre la plaque
en un autre point et tracer une
seconde verticale. Le centre de
gravité est à l'intersection des
deux droites.
Système masse-ressort
Un système masse-ressort est un système mécanique à un degré de liberté. Il est constitué par
une masse accrochée à un ressort contrainte de se déplacer dans une seule direction. Son
mouvement est dû à trois forces :



une force de rappel FR,
une force d'amortissement FA,
une force extérieure FE.
Le système masse-ressort est un sujet d'étude simple dans le cadre des oscillateurs harmoniques.
Oscillations rectilignes d'une masse soumise à l'action d'un
ressort []
Mouvement horizontal
23
Oscillation verticale
On peut mettre en oscillation une masse soumise à l'action d'un ressort. Ces oscillations peuvent
être, suivant les cas, des oscillations verticales ou des oscillations horizontales (en utilisant un
dispositif permettant de minimiser les frottements sur le support).
Dans les deux cas, les oscillations sont harmoniques : la fonction du temps [x(t)] de la position
de la masse de part et d'autre de la position d'équilibre (statique) est une fonction sinus. Dans le
cas de l'oscillateur vertical, l'effet de la pesanteur n'introduit qu'une translation de la position
d'équilibre statique. La relation déduite de l'application du théorème du centre d'inertie peut
s'écrire :
, avec
ω0 est appelée pulsation propre de l'oscillateur harmonique. Les solutions de l'équation
différentielle sont de la forme
oscillateur harmonique.
, ce qui est caractéristique d'un
La période est indépendante de l’amplitude (isochronisme des oscillations) : elle ne dépend que
de l'inertie du système (masse m) et de la caractéristique de la force de rappel (constante de
raideur k du ressort) :
24
Remarque : cet oscillateur est soumis à la conservation de l'énergie mécanique : celle-ci est de la
forme
En dérivant membre à membre l'équation par rapport au temps on retrouve l'équation
différentielle précédente.
Amélioration []
Ce qui précède est valable si la masse du ressort est négligeable par rapport à celle de la masse
qui oscille. L'expérience montre que la période est plus proche de :
où
le tiers de la masse du ressort ;
= la masse suspendue au ressort ;
= la constante élastique ou raideur du ressort.
Autre amélioration []
Ceci est de nouveau une approximation. Une étude complète se trouve dans les liens externes.
Chercher : « Étude de la période d'oscillation d'un ressort ».
On montre que la période correcte d'oscillation est :
où est défini par la relation :
= la masse du ressort ;
= la masse suspendue au ressort ;
= la constante élastique ou raideur du ressort.
Une manière de calculer
est d'itérer :
en commençant par :
Moment de force (mécanique)
25
Le moment de force est l'aptitude d'une force à faire tourner un système mécanique autour d'un
point donné, qu'on nomme pivot.
Translation d'une force []
Basculera, basculera pas ?
Soit une planche en équilibre au bord d'un muret. Pour la déséquilibrer, on peut poser une charge
sur la partie en porte-à-faux, au-dessus du vide. La capacité de cette charge à faire basculer la
planche n'est pas la même suivant qu'elle est posée près du muret ou au bout de la planche. De
même, on peut au même endroit, placer une charge plus lourde et constater une différence de
bascule.
Le « pouvoir de basculement » dépend donc de l'intensité de la force, mais également de la
position relative du point d'application de la force, et du point de rotation réel ou virtuel
considéré.
On intègre ces trois composantes du problème par le modèle de moment d'une force, qui est
l'aptitude d'une force à faire tourner un système mécanique autour d'un point donné, qu'on
nommera pivot.
Moment par rapport à un point []
26
Définition vectorielle
Expression vectorielle []
Le moment d'une force s'exerçant au point A par rapport au pivot P, est le pseudovecteur :
.
où
désigne le produit vectoriel.
Remarque sur la notation
Il existe plusieurs variantes de notation des moments de force ; certaines (comme sur
l'image ci-contre) comportent des parenthèses autour du vecteur, parfois autour de
l'ensemble. D'autres ajoutent même à la notation l'élément agissant et l'élément subissant
l'action. Une notation plus compacte consiste à nommer la force par la même lettre que
celle désignant le point d'application, ce qui rend plus rapide l'identification des cas de
nullité de moments.
Ce pseudovecteur est à la fois orthogonal à et au bipoint
et finalement normal au plan dans
lequel se déroule la rotation que peut provoquer la force, et son sens donne le sens de rotation (la
rotation est positive dans le plan orienté par
).
Si d est la distance orthogonale du pivot P à la droite d'action, c’est-à-dire PH, alors sa norme
vaut :
.
27
La longueur d est appelée bras de levier. Dans le cas bidimensionnel, il est fréquent de
considérer la norme du moment comme le moment lui-même, celui-ci ne comportant qu'une
composante non nulle.
Les composantes et la norme d'un moment de force sont exprimées en newton-mètre (Nm), dans
le système international d'unités et leurs dimensions sont ML2T − 2.
Cas de nullité du moment []
Puisqu'il s'agit ensuite d'établir la somme nulle des moments, on peut naturellement s'intéresser
aux cas de nullité individuelle des moments de force ; de par les propriétés du produit vectoriel :



la force est nulle ;
le bipoint
est . La force est donc appliquée en P.
et
sont colinéaires ; alors la droite d'action passe par P, ce qui inclut aussi le cas
précédent.
Formule de Varignon []
Lorsqu'on connaît le moment d'une force en un point, il est possible de le recalculer en n'importe
quel point de l'espace. Cette opération est inévitable lorsqu'on manipule les torseurs d'actions
mécaniques. Cela revient à poser une rallonge au levier AP. On montre alors la relation suivante :
.
On peut vérifier alors :
.
En réalité une force est modélisée par un vecteur (représentant la force) et son point
d'application. Il est possible de représenter cette action mécanique par le couple de vecteurs force
et moment en un point, qui sont les éléments de réduction du torseur d'action mécanique. La
relation d'équilibre liée au principe fondamental de la statique devient une somme de torseurs ;
en pratique, on effectuera parallèlement la somme des forces, et la somme des moments tous
exprimés au même point, d'où l'intérêt de la formule de transport de moments.
Moment par rapport à un axe []
Lorsqu'un solide est animé d'un mouvement de rotation effectif autour d'un axe (cas d'une roue
guidée par un palier) il est intéressant de ne considérer que la part utile du moment d'une force.
On définit le moment de la force par rapport à l'axe (Δ) par
,
28
où est un vecteur unitaire de (Δ), P est un point quelconque de (Δ) et où les crochets dénotent
le produit mixte.
En résumé il s'agit de la composante suivant du moment de calculé en P. De ce fait il s'agit
d'un nombre scalaire : «
» est une opération de projection sur l'axe . Sur le plan mécanique,
c'est la seule composante (dans le cas d'une liaison parfaite au pivot) susceptible de fournir (ou
consommer) une puissance. Le « reste » du moment sera subi par le palier. Cette partie
complémentaire intéressera le technologue qui prendra en compte ces valeurs pour le
dimensionnement du palier.
Le moment par rapport à l'axe est nul si


le moment par rapport au point est nul (cas général précédent).
la force est dans la direction de l'axe considéré.
Couple de forces []
Si on considère deux forces opposées appliquée en A et
appliquée en B, points distincts d'un
même système, il est évident que leur somme est nulle. Qu'en est-il de la somme de leur moment
en un point P de l'espace ?
.
On remarque que le résultat est indépendant du point de pivot P considéré. Cette quantité est
appelée couple. Il n'est pas besoin de préciser le point de rotation. Les deux forces constituent
alors un couple de forces.
Outre les autres cas évidents, le couple est nul lorsque les deux forces ont la même droite
d'action. Le couple augmente avec l'intensité commune des forces, mais aussi avec l'éloignement
des points. Il est maximal lorsque
et sont orthogonaux.
Cas général []
En réalité le couple n'existe pas intrinsèquement. Il est toujours associé à un ensemble de forces
s'annulant vectoriellement mais dont les moments s'ajoutent sans s'annuler. C'est par exemple le
résultat de l'action du vent sur une éolienne, ou l'action des forces électromagnétiques sur l'induit
d'un moteur électrique.
On ne doit donc pas faire le raccourci « somme des moments = moment de la somme ». Cela
n'est vrai que pour un ensemble de forces appliquées au même point. Cela montre enfin qu'une
29
action mécanique n'est pas représentable par un seul vecteur force. La considération du point
d'application est primordiale.
Théorème de Varignon []
Le moment en P de la résultante de plusieurs forces
somme des moments en P de ces différentes forces :
concourantes en A est égal à la
,
avec
.
En effet :
En dynamique []
En mécanique dynamique, on peut montrer que le moment des forces est la dérivée du moment
cinétique par rapport au temps :
Ceci est l'équivalent du principe fondamental de la dynamique (deuxième loi de Newton) en
rotation.
On peut aussi montrer que si



est le vecteur vitesse angulaire, c'est-à-dire le vecteur
colinéaire à l'axe de rotation Δ,
dont la norme est la vitesse angulaire
et orienté de façon que l'orientation positive d'un plan normal correspond au sens de
rotation, alors :
où JΔ est le moment d'inertie du solide par rapport à l'axe de rotation Δ.
30
Moment d'inertie
Lequel de ces mouvements est le plus difficile ?
Le moment d'inertie quantifie la résistance d'un corps soumis à une mise en rotation (ou plus
généralement à une accélération angulaire), et a pour grandeur physique [M.L²]. C'est l'analogue
de la masse inertielle qui, elle, mesure la résistance d'un corps soumis à une accélération linéaire.
Cette appellation est aussi utilisée en mécanique des matériaux pour déterminer la contrainte
dans une poutre soumise à flexion. Il s'agit alors d'une notion physique différente, encore appelée
moment quadratique, qui a pour grandeur physique [L4].
Approche empirique []
Lorsque l'on prend un balai en main au milieu du manche et qu'on le fait tourner comme sur la
figure ci-contre. Il est plus aisé de le faire tourner autour de l'axe du manche (1), qu'autour de
l'axe transversal indiqué (2).
Cela est dû au fait que dans le deuxième cas, la matière constituant le balai se trouve plus
éloignée de l'axe de rotation. Comme pour un solide en rotation, la vitesse linéaire d'un point
croît en proportion avec cet éloignement, il est nécessaire de communiquer une plus grande
énergie cinétique aux points éloignés. D'où la plus grande résistance du balai à tourner autour
d'un axe transversal qu'autour de l'axe du manche.
Détermination du moment d'inertie []
Soit un objet physique composé de plusieurs points solidaires i de masse mi. Cet objet tourne
autour d'un axe Δ, à la vitesse angulaire ω. La distance de i à Δ est ri.
31
Le calcul de l'énergie cinétique de cet objet donne :
On définit alors le moment d'inertie JΔ par rapport à l'axe Δ par :
Par extension dans un solide considéré comme ensemble continu de points matériels x affectés
d'une masse volumique ρ, le moment d'inertie s'écrit :
où



d(x,Δ) est la distance entre le point x et l'axe Δ ;
dV est un volume élémentaire autour de x ;
dm est la masse de ce volume élémentaire
Cette définition peut également prendre une forme vectorielle :
où


O est un point sur l'axe Δ
est un vecteur unitaire de l'axe Δ
Il découle de la définition du moment d'inertie que plus la masse d'un solide est répartie loin de
l'axe de rotation, plus son moment d'inertie est important. Ainsi, le patineur sur glace rapproche
les bras de son corps lors d'une pirouette. Cela a pour effet de diminuer son moment d'inertie, ce
qui, par conservation du moment cinétique, implique une plus grande vitesse de rotation.
Moments d'inertie particuliers []
Pour les exemples suivants, nous considérerons des solides de densité uniforme ρ et de masse M.
La boule []
32
Pour une boule de rayon R et de centre O, les moments d'inertie au centre de la boule par rapport
aux trois axes sont égaux :
(avec
)
La barre []
Dans le cas d'une barre de section négligeable et de longueur L, le moment d'inertie selon un axe
perpendiculaire à la barre est, en son centre :
(avec M = ρL)
ρ est ici une densité linéique.
Le cylindre plein []
Dans le cas d'un cylindre de rayon R et de hauteur h, le moment d'inertie selon l'axe du cylindre
est :
(avec M = ρπR2h)
Le cylindre creux []
Dans le cas d'un cylindre creux de rayons intérieur R1 et extérieur R2, et de hauteur h, le moment
d'inertie selon l'axe du cylindre est :
(avec
)
33
Théorème de transport (ou Théorème d'Huygens ou
Théorème de Steiner) []
Soit l'axe Δ passant par le centre de masse de l'objet, et un axe Δ' parallèle à Δ et distant de d. En
calculant comme précédemment le moment d'inertie, on retrouve la relation établie par
Christiaan Huygens connue sous le nom de théorème de transport[1] ou théorème de Huygens
ou théorème de Steiner ou théorème des axes parallèles qui donne le moment d'inertie JΔ' en
fonction de JΔ :
À l'énergie cinétique de rotation propre d'un corps, s'ajoute celle de « translation » circulaire du
centre de masse auquel on a affecté la masse totale du solide.
Une conséquence immédiate du théorème de Huygens est qu'il est moins coûteux (en énergie) de
faire tourner un corps autour d'un axe passant par le centre de masse.
Énergie cinétique
L'énergie cinétique est l’énergie que possède un corps du fait de son mouvement réel. L’énergie
cinétique d’un corps est égale au travail nécessaire pour faire passer le dit corps du repos à son
mouvement de translation ou de rotation.
Historique []
Article détaillé : vis viva.
Gottfried Leibniz, s'opposant ainsi à Descartes, qui estimait que la quantité de mouvement se
conservait toujours, développa l'idée de la « force vive » (vis viva), à laquelle il attribuait la
valeur mv2. La force vive est donc le double de l'énergie cinétique.
« Il y a longtemps déjà que j’ai corrigé la doctrine de la conservation de la quantité de
mouvement, et que j’ai posé à sa place quelque chose d’absolu, justement la chose qu’il faut, la
force (vive) absolue… On peut prouver, par raison et par expérience, que c’est la force vive qui
se conserve… » [1]
Conventions []
L'énergie cinétique est généralement notée Ec ou Ek, l'indice c faisant référence au mot
« cinétique » et l'indice k à son équivalent anglais, « kinetic ».
34
Définitions []

Ec formule de l'énergie cinétique selon la masse et la vitesse :
où m est la masse, et v la vitesse. Exemple: 1/2 x 45 kg x (8.3 m/s)2 = 1550,025 joules
Cas d'un point matériel []
Dans le domaine de validité de la mécanique newtonienne, la notion d'énergie cinétique peut être
facilement mise en évidence pour un point matériel, corps considéré comme ponctuel de masse
m constante.
En effet, la relation fondamentale de la dynamique s'écrit :
, avec
somme des forces appliquées au point matériel de masse m (y
compris les "forces d'inertie" dans le cas d'un référentiel non galiléen).
En prenant le produit scalaire, membre à membre, par la vitesse
, or
du corps, il vient :
, il vient ainsi :
.
On met en évidence dans le membre de gauche la quantité
appelée énergie
cinétique du point matériel, dont la variation est égale à la somme des puissances
des
forces appliquées au corps (théorème de l'énergie cinétique, forme "instantanée").
On peut obtenir une expression plus générale en considérant que l'on a donc
, puisque
infinitésimale de la quantité de mouvement du corps,
. En introduisant la variation
, il vient au final l'expression :
.
Cas d'un système de points []
35
Dans le cas d'un corps que l'on ne peut considérer ponctuel, il est possible de l'assimiler à un
système (d'une infinité) de points matériels Mi de masses mi avec
corps.
masse totale du
L'énergie cinétique Ec du système de points peut être alors simplement définie comme la somme
des énergies cinétiques associées aux points matériels constituant le système :
, (1). Cette expression est générale et ne préjuge pas de la nature
du système, déformable ou pas.
Remarque : en considérant la limite des milieux continus on a
étant un point courant du système (S).
,M
Unité []
L'unité légale est le joule. Les calculs s'effectuent avec les masses en kg et les vitesses en
.
Théorème de König []
L'expression (1) n'est guère utilisable directement, bien que générale. Il est possible de la réécrire
sous une autre forme, dont l'interprétation physique est plus aisée.
Enoncé []
Ce théorème se démontre en faisant intervenir le référentiel barycentrique (R*) lié au centre
d'inertie G du système, et en mouvement de translation par rapport au référentiel d'étude (R). Il
s'écrit:
.
L'énergie cinétique d'un système est alors la somme de deux termes: l'énergie cinétique du centre
de masse de (S) affectée de sa masse totale M,
dans (R*),
, et l'énergie cinétique propre du système
.
Application à un solide []
36
Un solide est un système de points tels que les distances entre deux points quelconques de (S)
sont constantes. Il s'agit d'une idéalisation d'un solide réel, considéré comme absolument rigide.
Cas général : axe instantané de rotation []
Dans ce cas, le mouvement du solide peut être décomposé en un mouvement de son centre de
masse G dans (R) et un mouvement de rotation autour d'un axe instantané (Δ) dans le référentiel
barycentrique (R*).
Plus précisément, pour un solide on peut écrire le champ des vitesses dans le référentiel
barycentrique (R*) sous la forme
, étant le vecteur rotation instantané du
solide dans (R*) [ou (R), puisque les deux référentiels sont en translation]. Son énergie cinétique
propre
s'exprime alors
,
puisque
au moment cinétique propre
, moment cinétique du solide par rapport à G, égal
(voir théorèmes de König).
D'après le théorème de König, l’énergie cinétique totale d’un solide s'écrit donc ainsi:
,
que l'on peut considérer comme la somme d’une énergie cinétique "de translation" et d’une
énergie cinétique de rotation
, aussi appelée énergie cinétique angulaire.
Cas de la rotation autour d'un axe fixe []
Si, de surcroît, il y a rotation autour d'un axe (Δ) fixe dans (R), le moment cinétique par rapport à
G du solide s'écrit
, où IΔ est le moment d'inertie du solide par rapport à l'axe de
rotation (Δ). Son énergie cinétique de rotation se mettra ainsi sous la forme:
.
En mécanique relativiste []
Dans la théorie de la relativité d’Einstein (utilisée principalement pour les vitesses proches de la
vitesse de la lumière, mais valable pour toutes vitesses), l’énergie cinétique est :
37
Ek = mc2(γ − 1) = γmc2 − mc2
Avec :






; (le facteur relativiste)
Ek : l’énergie cinétique du corps (dans le référentiel considéré) ;
v : la vitesse du corps (dans le référentiel considéré) ;
m : sa masse au repos (dans son référentiel) ;
c : la vitesse de la lumière dans le vide (dans TOUT référentiel) ;
γmc2 : l’énergie totale du corps (dans le référentiel considéré) ;
mc2 est l’énergie au repos (90 pétajoules par kilogramme) exprimée en unités
conventionnelles.
La théorie de la relativité affirme que l’énergie cinétique d’un objet (ayant une masse « au
repos[2]» non nulle) tend vers l’infini quand sa vitesse s’approche de la vitesse de la lumière et
que, par conséquent, il est impossible d’accélérer un objet jusqu’à cette vitesse.
On peut montrer que le rapport de l’énergie cinétique relativiste sur l’énergie cinétique
newtonienne tend vers 1 quand la vitesse v tend vers 0, i.e.,
Ce résultat peut être obtenu par un développement limité au premier ordre du rapport. Le terme
de second ordre est 0,375 mv4/c4, c’est-à-dire que, pour une vitesse de 10 km/s il vaut 0,04 J/kg,
et que, pour une vitesse de 100 km/s il vaut 40 J/kg, etc.
Quand la gravité est faible et que l’objet se déplace à des vitesses très inférieures à la vitesse de
la lumière (c’est le cas de la plupart des phénomènes observés sur Terre), la formule de la
mécanique newtonienne est une excellente approximation de l’énergie cinétique relativiste.
Théorème de l’énergie cinétique []
Ce théorème, valable uniquement dans le cadre de la mécanique newtonienne, permet de lier
l’énergie cinétique d’un système au travail des forces auxquelles celui-ci est soumis.
Énoncé []
38
Dans un référentiel galiléen, pour un corps ponctuel de masse m constante parcourant un chemin
reliant un point A à un point B, la variation d’énergie cinétique est égale à la somme W des
travaux des forces extérieures et intérieures qui s’exercent sur le solide en question :
où EkA et EkB sont respectivement l’énergie cinétique du solide aux points A et B.
Démonstration []
D’après la 2e loi de Newton, l’accélération du centre de gravité est liée aux forces qui s’exercent
sur le solide par la relation suivante :
Pendant une durée dt, le solide se déplace de
déduit le travail élémentaire des forces :
où
est la vitesse du solide. On en
Si le solide parcourt un chemin d’un point A à un point B, alors le travail total s’obtient en
faisant une intégrale le long du chemin :
étant une différentielle totale, l’intégrale ne dépend pas du chemin suivi entre A et B et
peut donc être obtenue explicitement :
Théorème de la puissance cinétique []
Dans un référentiel galiléen, la puissance des forces s'appliquant au point M est égale à la dérivée
par rapport au temps de l'énergie cinétique.
39
L’énergie thermique en tant qu’énergie cinétique []
L’énergie thermique est une forme d’énergie due à l’énergie cinétique totale des molécules et des
atomes qui forment la matière. La relation entre la chaleur, la température et l’énergie cinétique
des atomes et des molécules est l’objet de la mécanique statistique et de la thermodynamique.
De nature quantique, l’énergie thermique se transforme en énergie électromagnétique par le
phénomène de rayonnement du corps noir.
La chaleur, qui représente un échange d’énergie thermique, est aussi analogue à un travail dans
le sens où elle représente une variation de l’énergie interne du système. L’énergie représentée par
la chaleur fait directement référence à l’énergie associée à l’agitation moléculaire. La
conservation de la chaleur et de l’énergie mécanique est l’objet du premier principe de la
thermodynamique.
Énergie mécanique
L'énergie mécanique est une quantité utilisée en mécanique classique pour désigner l'énergie
d'un système emmagasinée sous forme d'énergie cinétique et d'énergie potentielle mécanique.
C'est une quantité conservée en l'absence de frottement ou de choc et s'avère pour cela pratique à
utiliser.
L'énergie mécanique n'est pas un invariant galiléen et dépend donc du référentiel choisi.
]Expression []
L'énergie mécanique s'exprime généralement :
Em = Ec + Ep
où :



Em est l'énergie mécanique
Ec est l'énergie cinétique // formule: 1/2mv² (m: masse en kg, v2: vitesse en mètre par
seconde au carré) exemple: 1/2 X 45 kg X (8,3 m/s)2 = 1550,025 J
Ep est l'énergie potentielle ou l'énergie de position // formule de l'énergie potentielle de
pesanteur : M x G x H (m : masse en kilogramme, g : accélération de la pesanteur sur
Terre (9.8 newton par kg), H: différence d'altitude en mètre) exemple: 0,5 kg X 9,8 N/kg
X 0,3 m = 1,47 Joules)
Solide ponctuel []
40
Pour un solide ponctuel M l'énergie potentielle mécanique est donnée par sa position et l'énergie
cinétique par sa vitesse. On a donc
où :



m est la masse du solide
v est la vitesse du centre de gravité ;
V est le potentiel au niveau du point M
Solide étendu non déformable []
Pour un solide indéformable non ponctuel, il convient d'ajouter l'énergie cinétique de rotation.
L'énergie potentielle est donnée, dans le cas d'un potentiel gravitationnel, par la position du
centre de gravité G.
où, toutes notations égales par ailleurs



J est le moment d'inertie du solide par rapport à son axe de rotation ;
Ω est sa vitesse angulaire de rotation.
V est le potentiel gravitationnel dans lequel se déplace la masse.
Solide déformable []
Pour un solide déformable, interviennent des termes de déformation (tension, torsion,
contraction) tant dans l'énergie cinétique que l'énergie potentielle mécanique.
Théorème de l'énergie mécanique []
En dérivant l'expression de l'énergie mécanique on obtient :
dEm = dEc + dEp
Or d'après le Théorème de l'énergie cinétique, on a :
dEc = δW(Fc) + δW(Fnc)
avec δW(Fc) le travail des forces conservatives et δW(Fnc) le travail des forces non
conservatives.
et on a aussi : dEp = − δW(Fc).
D'où le résultat :
dEm = δW(Fnc)
41
On a ainsi le théorème de la puissance mécanique, la dérivée de l'énergie mécanique est
égale à la puissance des forces non conservatives :
Ainsi si toutes les forces sont conservatives, l'énergie mécanique se conserve.
Conservation []
L'énergie mécanique d'un système soumis à des forces conservatives, c'est-à-dire dérivant d'un
potentiel, est conservée.
42

Mechanics 1 - OER@AVU - African Virtual University

Transcription

Documents pareils

Symposium L`actualité de la recherche en droit franco

Appel à publication d`articles juridiques - African Law e

Visiting professors - Faculté de Droit

Un patient virtuel dialogant

Table ronde - Centre de recherche en immigration, ethnicité et

Utiliser Virtual magnyfying glass

Canadian Society for History and Philosophy of Mathematics

Lecteur de CD - SDE Distribution SDE Distribution