Physics 01 Courses and Tasks PDF

HASSIBA BENBOUALI UNIVERSITY OF CHLEF FACULTY OF EXACTE SCIENCES AND COMPUTER SCIENCE COMMUN CORE DEPARTEMENT Physics 01 Courses and Tasks Prepared by Dr.HOURIA CHAACHOUA SAMEUT Academic Year 2023-2024 ...

HASSIBA BENBOUALI UNIVERSITY OF CHLEF FACULTY OF EXACTE SCIENCES AND COMPUTER SCIENCE COMMUN CORE DEPARTEMENT Physics 01 Courses and Tasks Prepared by Dr.HOURIA CHAACHOUA SAMEUT Academic Year 2023-2024  Foreword Kinematics is the subdivision of mechanics whose object is the quantitative study of movement of material bodies, regardless of the causes that produce this movement (position, speed, acceleration, etc.), whereas dynamics is the study of the movement of bodies depending on the forces exerted on them. This work is structured in two main parts. The first part will be devoted to the kinematics, preceded by the definition of mathematical tools that make it possible to understand this part as well as the rest. The second part will be the subject of the study of the dynamics followed a chapter on work and energy, which consists of shedding light on the relationships between movements and their causes. This handout represents a course support of mechanics of the material point. It is intended for students of the first semester of the SM track. We want students to find in this support a good working tool that will allow them to fill any gaps that may take place when taking notes while explaining the course or assignments led by their teachers. This handout is just an add-on to the course. He cannot, in any way, exempt the student from his presence in class. 1  Table des matières 1 Mathematics reminder 6 1.1 Error calculations........................ 7 1.1.1 Definitions....................... 7 1.1.1.1 Absolute error............... 7 1.1.1.2 Absolute incertitude............ 7 1.1.1.2.1 Addition and subtraction.... 7 1.1.1.3 Relative uncertainty............. 8 1.1.1.4 Error calculations.............. 8 1.1.1.4.1 Case of a product or a quotient. 9 1.2 Les Vecteurs........................... 9 1.2.1 Définition........................ 9 1.3 Vectors.............................. 10 1.3.1 Definition........................ 10 1.3.2 Orthonormal direct basis................ 11 1.3.2.0.2 Definition............. 11 1.3.3 Coordinates of a point and components of a vector. 11 1.3.4 The Algebra of Vectors................. 12 1.3.4.1 Multiplying a Vector by a Scalar..... 12 1.3.4.2 Adding Vectors.............. 12 1.3.5 Multiplying Vectors.................. 12 1.3.6 Scalar Product (Dot Product)............. 13 2  1.3.7 Vector Product (Cross Product)............ 14 1.3.8 Mixed Product (Scalar triple product )........ 15 1.3.9 Double cross product................. 16 1.3.10 Vector derivation................... 16 1.3.11 coordinates system................... 16 1.3.11.1 Mathematical operators.......... 19 1.3.11.1.1 Gradient operator........ 19 1.3.11.1.2 Divergence operator....... 20 1.3.11.1.3 Rotational operator........ 20 1.4 Corrected exercises....................... 21 2 Kinematics of material point 25 2.1 Kinematics of material point.................. 26 2.1.1 Definition........................ 26 2.1.2 Vector velocity..................... 26 2.1.3 Acceleration vector................... 27 2.1.4 Velocity vector in different coordinate systems... 27 2.1.5 Acceleration vector in different coordinate systems. 28 2.1.6 Components of velocity and acceleration vectors in curvilinear coordinates................. 29 2.1.7 Examples of simple movements............ 32 2.1.7.1 Rectilinear motion............. 32 2.1.7.2 Rectilinear sinusoidal motion....... 33 2.1.7.3 Circular movement............. 33 2.2 Change of reference frame................... 35 2.3 Corrected exercises....................... 37 3 Point material dynamics 48 3.1 Point material dynamics.................... 49 3.1.1 Definition........................ 49 3.1.2 Newton’s laws..................... 49 3.1.2.1 First Principle (Principle of inertia).... 49 3.1.2.2 The second principle (Fundamental Prin- ciple of Dynamics)............. 49 3  3.1.2.3 Third principle (action and reaction)... 50 3.1.3 Base Units and Physical Standards.......... 50 3.1.4 The Algebra of Dimensions.............. 50 3.1.5 Force classification................... 52 3.1.5.1 Real (or external) forces.......... 52 3.1.5.2 Action-at-a-Distance Forces........ 52 3.1.5.3 Contact force................ 54 3.1.6 Quantity of movement and kinetic momentum.... 58 3.1.6.1 Definitions.................. 58 3.1.6.2 The angular momentum theorem..... 58 3.2 Corrected exercises....................... 58 4 Work and Energy 67 4.1 Work and Energy........................ 68 4.1.1 Work and power of a force............... 68 4.1.1.1 Power of a force............... 68 4.1.1.2 Elementary work of a force........ 68 4.1.1.3 Work of a force............... 68 4.1.1.4 Relation between work and power..... 69 4.1.2 Kinetic energy..................... 69 4.1.2.1 Definition.................. 69 4.1.2.2 Kinetic energy theorem........... 69 4.1.3 Conservative forces and potential energy....... 71 4.1.3.1 Definition.................. 71 4.1.3.2 Another definition of a conservative force 71 4.1.3.2.1 Potential energy of a point M in free fall.............. 71 4.1.3.2.2 Potential energy of a charge pla- ced in an electrostatic field... 72 4.1.3.2.3 Potential energy derived by the restoring force or spring force.. 73 4.1.4 Example of a non-conservative force......... 73 4.1.5 Mechanical energy................... 74 4  4.1.5.1 Definition.................. 74 4.1.5.2 Mechanical energy theorem........ 74 4.1.5.2.1 Non conservation of mechani- cal energy............. 74 4.1.5.2.2 conservation of mechanical energy 74 4.1.6 System stability..................... 75 4.1.6.1 Definition of stability............ 75 4.1.6.2 Stability condition............. 75 4.1.7 Corrected exercises................... 76 5  1 Mathematics reminder 6 1.1. ERROR CALCULATIONS CHAPITRE 1. MATHEMATICS REMINDER 1.1 Error calculations 1.1.1 Definitions For any measurable quantity A, it is possible to define : - its measured value a - its exact value a0 which cannot be reached 1.1.1.1 Absolute error It is then defined by : δa = a − a0 This error is the result of several errors (systematic, incidental...) 1.1.1.2 Absolute incertitude Since the absolute error δa is not known, we simply give an upper limit ∆a called the absolute uncertainty such that : |δa| ≤ ∆a ⇒ ∆a > 0 (the absolute uncertainty is always > 0) This means that the absolute uncertainty is the maximum value that the absolute error can reach. 1.1.1.2.1 Addition and subtraction Let is suppose that the quantity we are looking for R is the sum of two measurements A and B R=A+B In this case the uncertainty about the result is ∆R = ∆A + ∆B It is the same for : R = A − B The absolute uncertainty about a sum or a difference is the sum of ab- solute uncertainties of each term. Example A container has mass m = (50 ± 1)g.Filled with water, its mass is : M = (200 ± 1)g. 7 1.1. ERROR CALCULATIONS CHAPITRE 1. MATHEMATICS REMINDER The mass of water it contains is therefore : meau = M − m By applying the rule above : ∆meau = ∆M + ∆m = 1 + 1 = 2g , it follows that : meau = (150 ± 2)g 1.1.1.3 Relative uncertainty ∆a It is defined by the ratio a. 1.1.1.4 Error calculations Let a quantity g = f (x, y, z), sa différentielle totale s’écrit : ∂f ∂f ∂f dg = ∂x dx + ∂y dy + ∂z dz L ’incertitude absolue sur la variable g s’obtient en passant aux variations des variables qui la compose, soit : ∆g = | ∂f ∂f ∂f ∂x |∆x + | ∂y |∆y + | ∂z |∆z Example Either to determine the relative uncertainty ∆ρ ρ ddensity of the substance of à homogeneous cube from the measurement of its mass m and of its edge a Solution If m and a are the approximate values for the mass and the edge of the cube, we can write : masse volume = m a3 ⇒ ρ = ma−3 Deriving volume from mass and edge.This gives : dρ = a−3 dm − 3a−4 mda By approximating small variations d with great variations ∆, it comes : ∆ρ = a−3 ∆m − 3a−4 m∆a ⇒ ∆ρ = 1 1 a3 ∆m − 3 a4 m∆a Hence ∆ρ ∆m m ρ = m + 3 a4 ∆a 8 1.2. VECTORS CHAPITRE 1. MATHEMATICS REMINDER Remarks : We find the same thing by taking Ln(ρ) and differentiating the new equation. Ln :is the natural function. a b x y 1.1.1.4.1 Case of a product or a quotient k (z+t) c We apply the logarithm function to this equation a b x y LnF = Ln(k (z+t) c) LnF = Lnk + Lnxa y b − Ln(z + t)c LnF = Lnk + aLnx + bLny − cLn()z + t We move on to the derivative of this equation, with k constant, dF dy F = a dx dz+dt x + b y − c z+t Finally, we replace the differential elements by the uncertainties on the asso- ciated quantities and transform all the negative signs into positive signs. The result is ∆F ∆y F = a ∆x ∆z+dt x + b y + c z+t The relative uncertainty on a product or a quotient is the sum of re- lative uncertainties of each term.. 1.2 Vectors 1.2.1 Definition A vector is directed line segment. M N , having an origin M and an extremity N. It is completely defined if we specify : 9 1.2. VECTORS CHAPITRE 1. MATHEMATICS REMINDER 1. both its length ( Magnitude) and its di- rection, 2. We can represent a vector graphically by an arrow, showing both its scale length and its direction. 3. A vector is represented graphically by an arrow drawn on a scale as shown in the Figure 4. A vector is denoted with a small arrow −−→ → − − over the symbol like M N , A , → a , etc..... 5. The magnitude of a vector quantity is re- ferred by simple identifier like M N or as the absolute value of the vector as |M N | Unit vector : Each vector can be associated with a unit vector which has the same direction and a magnitute equal to one. The unit vector is obtained by dividing the initial vector by its magnitude : → − k→ − A uk= → − kAk 1.2.2 Orthonormal direct basis → − → − → − 1.2.2.0.2 Definition (B) = ( i , j , k ) is an orthonormal basis if and only if : → − → − → − i ⊥ j ⊥ k (vectors orthogonal to each other) → − → − → − k i k = k j k = k k k = 1 unitary vectors In this course, we will always consider a direct orthonormal basis 1.2.3 Coordinates of a point and components of a vector → − → − → − Let (B) = ( i , j , k ) a direct orthonormal basis associated with a reference frame in space (R) 10 1.2. VECTORS CHAPITRE 1. MATHEMATICS REMINDER A point M is identified by its coordinates in the reference frame (R) :M (x, y, z) −−→ A Vector → − v = OM is defined by its components in the base (B) : → − → − → − → − v =x i +y j +zk 1.2.4 The Algebra of Vectors → − → − → − Let (B) = ( i , j , k ) a direct orthonormal basis and two vectors : → − → − → − → − v 1 = x1 i + y1 j + z1 k → − → − → − → − v 2 = x2 i + y2 j + z2 k Note θ = (→ −v 1, → − v 2 ) the angle where 0 ≤ θ ≤ π 1.2.4.1 Multiplying a Vector by a Scalar → − → − → − ∀µ ∈ t) 2. Instantaneous velocity The instantaneous velocity vector of M with respect to the reference 25 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT frame R at an instant t is obtained by taking the limit ∆ t → 0 in the definition of average velocity −−→0 −−→ −−→ → − V = lim∆ t → 0 OM∆−tOM = d OM dt 2.1.3 Acceleration vector → − − → Let V the velocity of the material point at time t, and V 0 its velocity at time t0. so we define : 1. Average acceleration The average acceleration between the instances t and t0 is defined as : − →0 → − → −0 → − → − a moy = V t0− V = V −V −t 4t 2. Instantaneous acceleration The instantaneous acceleration vector of M with respect to the reference frame R at time t is obtained by taking the limit ∆ t → 0 in the defini- tion of average acceleration → − → −0 →− −−→ → − a = lim∆ t → 0 = dV = V 4t −V = d2 OM dt dt2 2.1.4 Velocity vector in different coordinate systems Note that the Cartesian coordinate basis vectors are fixed, Cartesian coordinates : The velocity vector of point M is obtained by deriving its position vector with respect to time : : → − dx → − dy → − dz → − V = dt i + dt j + dt k The following notation is also used → − → − → − → − V = ẋ i + ẏ j + ż k where the dot on the variable denote the derivative with respect to time. Cylindrical coordinates : To obtain the expression for the velocity vector in cylindrical coordi- nates, we derive the vector position : −−→ → − → − d(ρ → − e ρ +z k) V = dOM dt = dt 26 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT → − → − → − V = + ρ d dte ρ + dz dρ → − dt e ρ dt k → − → − → − Knowing that k a fixed vector, its derivative is zero ddtk = 0 The vector being mobile → − eρ its derivative is generally not zero. In fact implicitly on t through its dependence on angle ϕ.So d→ − eρ d→ −eρ dϕ dt = dϕ dt → − → − → − Using the expression for the vector → − eρ in the base ( i , j , k ) we obtain : d→ − → − → − eρ = d(cos ϕ i +sin ϕ j ) = − sin ϕ → − → − i + cos ϕ j = → − eϕ dϕ dϕ The derivative with respect to time is then given by : d→ − eρ dϕ → − dt = dt eϕ For the velocity vector, we obtain : → − → − V = ρ̇→ − eρ + ρ ϕ̇→ − eϕ + ż k Spherical coordinates : The velocity vector is obtained by deriving the position vector : −−→ → − OM = dr → d − d→ − er V = dt dt er + r dt er = ∂ → → − − er dθ + ∂ → − er dϕ Avec ddt ∂ θ dt ∂ϕ. dt Ainsi que ∂→− er → − ∂θ = eθ. → − ∂ er → − ∂ ϕ = sin θ eϕ le vecteur vitesse s’écrit : → − V = ṙ→ − er + rθ̇→ − eθ + rϕ̇ sin θ→ − eϕ 2.1.5 Acceleration vector in different coordinate systems Cartesian coordinates : Using the velocity vector expression in cartesian coordinates, we have : → − → − → − → − a = d(ẋ i +ẏ j +ż k ) dt Since the Cartesian coordinate basis vectors are fixed, we only need to derive the components of the velocity vector, which gives donne : → − → − → − → − a = ẍ i + ÿ j + z̈ k where the two points on a variable mean the second derivative of the variable with respect to to time. Cylindrical coordinates : We use the expression for the velocity vector in cylindrical coordinates : 27 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT → − → − d(ρ̇→− eρ +ρϕ̇→ − eϕ +ż k ) a = dt → − d→ − e d→ − eϕ dż → − a (= dt → d ρ̇ − eρ + ρ̇ dtρ + dρdt ϕ̇→ −e ϕ + ρϕ̇ dt + dt k d→− e Avec dtρ = ϕ̇→ − eϕ et → − d eϕ → − deϕ dϕ → − → − d(− sin ϕ i +cos ϕ j ) dt =. dϕ dt = dϕ = −ϕ̇→ −eρ Substituting the above expression for the acceleration we obtain : → − a = → − (ρ̈ − ρϕ̇2 )→ − eρ + (ρϕ̈ + 2ρ̇ϕ̇)→ − eϕ + z̈ k Spherical coordinates : We use the velocity vector’s expression in spherical coordinates : → − d(ṙ→ − er +rθ̇→ − eθ +rϕ̇ sin θ→ − eϕ ) a = dt → − a = (r̈ − rθ̇2 − rϕ̇2 sin θ2 )→ − er + (2ṙθ̇ + rθ̈ − rϕ̇2 sin θ cos θ)→ − eθ + (2ṙϕ̇ sin θ + 2rθ̇ϕ̇ cos θ + rϕ̈ sin θ)→ − eϕ 2.1.6 Components of velocity and acceleration vectors in curvilinear coordinates Let (C) a trajectory described by a material point M relative to the frame → − → − →− R(O, i , j , k ) and P an arbitrary point considered as an origin on(C). The length of the arc (P M ) is called the curvilinear abscissa s (P M = s). 28 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT Let M 0 the position of M at time (t + ∆t) −−→ −−→ −−→ −−→ −−−→ OM (t + ∆t) − OM (t) = OM 0 − OM = M M 0 The speed vector is : → − −−→ −−→ V = dOMdt = dOM ds ds dt −−→ −−→0 −−→ dOM OM −OM with ds = lim ∆s→0 ∆s = −−−→0 lim∆s→0 M∆s M If ∆t → 0, ∆s → 0 et M 0 very close to −−−→ −−−→ M. So kM M 0 k = ∆s et M M 0 is tangent to the trajectory at point M and therefore −−−→0 −−−→0 → − →− lim∆s→0 M∆s M = lim MM ∆s→0 kM M 0 k = T ( T is therefore a unit vector tangent to the trajectory at the point M ).So, The instantaneous velocity vector can there- fore be written as → − ds → − V = dt T 29 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT Acceleration vector : The derivative with respect to time of the velocity vec- tor gives the acceleration vector, which 2 →− → − is written as follows : → − a = ddt2s T + ds dT dt dt → − → − Or ddtT = ddθT dt dθ and → − → − dT dθ = N is the normal vector directed towards the centre of curvature of the trajectory of point M. also we have, ∆s = Rs θ où ds = Rc dθ. So dθ 1 ds v dt = Rc dt = Rc → − → − Therefore, ddtT = Rvc N The instantaneous acceleration vector of point M is written as : → − dv → − v2 → − → − → − a = dt T + Rc N = aT T + aN N → − - The direction defined by the vector N is the principal normal at M to the trajectory.. → − → − → − - The unit vector B = T ∧ N is called the binormal vector. → − → − → − - The reference frame (M, T , N , B ) is the reference of Frenet-Serret. Calculation of the curvature’s radius → − → → − → − 2 →− → − → − → − ṡ2 → − ṡ3 → − V ∧− a = ṡ T ∧ [s̈ T + Rṡ c N ] = ṡ T ∧ s̈ T + ṡ T ∧ Rc N = Rc B → − → → − − ṡ3 ṡ3 k V k3 ⇒ k V ∧ a k = Rc kBk = Rc = Rc So the radius of curvature is calculated by : → − k V k3 Rc = → − → k V ∧−ak Remarks → − → − -→ − aV >0⇒− a→ T and V have the same direction. The motion is there- fore accelerated 30 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT → − → − -→ − aV 0 or decelerated (→ − a V < 0). ẍ = a0 ⇒ x(t) = 21 a0 t2 + V0 t + x0 v(M ) = a0 t + V0 avec (x(t = 0) = x0 ) et (V (t = 0) = V0 ). 31 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT 2.1.7.2 Rectilinear sinusoidal motion Rectilinear sinusoidal motion if the position vector in the reference frame → − → − →− R(O, i , j , k ) can be written as : −−→ → − → − OM = x(t) i = Xm cos(ωt + φ) i With Xm : the amplitude of movement (m) ω : pulse (rad.s−1 ) φ : phase at origin(rad). So −−→ → − d OM (M/R) → − V (M/R) = dt = −Xm ω sin(ωt + φ) i → − → − γ = d V (M/R) et → − dt = −Xm ω 2 cos(ωt + φ) i −−→ We can see that : → − γ (M/R) = −ω 2 OM −−→ → − d2 OM (M/R) and as γ (M/R) = ( dt2 ) −−→ Rectilinear sinusoidal motion is governed by the differential equation d OM 2 dt2 + −−→ ω 2 OM = 0 → − Projected according to i , this differential equation is written as : ẍ + ω 2 x = 0 2.1.7.3 Circular movement This is a movement such that the trajectory is a circle of constant radius. polar coordinates are well adapted to the study of the motion of a point des- cribing a circular trajectory. In the case of circular motion, the constant radius (ρ = R) 32 2.1. KINEMATICS OF MATERIAL POINT CHAPITRE 2. KINEMATICS OF MATERIAL POINT In this case the position vector can be written as : −−→ OM = R→ − eρ. The velocity vector : → − −−→ (dR→ − eρ ) Rd(→ − eρ ) → − V (M ) = dOM dt = dt = dt = Rϕ̇→ − eϕ ⇒ k V (M )k = Rϕ̇ The acceleration vector : → − d(Rϕ̇→ − eϕ ) → − → − V = γ (M ) = ddt = Rϕ̈→ − e ϕ ) = Rϕ̈→ eϕ + Rϕ̇ d( dt − eϕ − Rϕ̇2 → − eρ ⇒ k→ − γ (R)k = q dt R ϕ̈ + ϕ̇4 → − dk V (M )k → − ϕ̇) → − → − − γ→ T = ( dt ) T = d(R dt T = Rϕ̈ T − γ→ → − − → 2→− 2→ − N = γ (M ) − γT = −Rϕ̇ eρ = Rϕ̇ N → − → − → − with N = B ∧ T = → − ez ∧ →− eϕ = −→ − eρ Remarques - ϕ̇ : Angular velocity → − -→ − ω = ϕ̇ k : Angular velocity vector. → − → − −−→ - V (M ) = Rϕ̇→ − eϕ = Rϕ̇ k ∧ → − eρ = → − ω ∧ OM. - ω̈ : angular acceleration. - Si ϕ̇ = ω0 = cte ⇒ the movement is uniformly circular (ϕ(t) = ω0 t + ϕ0 ) - Si ϕ̈ = γ0 = cte ⇒ the motion is circular and uniformly varied. 33 2.2. CHANGE OF REFERENCE FRAMECHAPITRE 2. KINEMATICS OF MATERIAL POINT 2.2 Change of reference frame Let’s study the motion of a particle M relative to a fixed reference frame R, called absolute reference. It is sometimes useful to introduce a second reference frame R0 , known as the relative frame, compared with in which the movement of M is simple to study. R(O, X, Y, Z) an absolute reference (fixed reference point). R0 (O0 , X 0 , Y 0 , Z 0 ) a relative reference (a reference that moves relative to R). R0 can be animated by a movement of translation or rotation relative to R. The rotation of R0 with respect to R occurs with an angular velocity ω(R0 /R) such that : In the R reference , → −0 → − di dt |R =→ − ω (R0 /R) ∧ i0 → −0 → − dj dt |R =→ − ω (R0 /R) ∧ j 0 → −0 → − dk dt |R =→ − ω (R0 /R) ∧ k 0 In the R0 reference , → −0 → −0 → −0 di dj dk → − dt |R = dt |R = dt |R = 0 0 0 0 Derivation in a moving reference : → − Lets A an unusual vector. In the R frame, this vector is written as : → − → − → − → − A =x i +y j +zk. → − In the frame R0 the vector A is written as, → − → − → − → − A = x0 i0 + y 0 j 0 + z 0 k 0. → − → − → −0 → − → −0 → − dA →− → − → − 0 i0 +x0 d i | +y 0 j 0 +y 0 d j | +z 0 k 0 +z 0 dk0 | dt |R = ẋ i + ẏ j + ż k = x dt R dt R dt R which can also be written as, → − → − → − → − → − → − dA 0 0 dt |R = x i + y 0 j 0 + z 0 k 0 + x0 → − ω (R0 /R) ∧ i0 + y 0 → − ω (R0 /R) ∧ j 0 + → − z0→ − ω (R0 /R) ∧ k 0. → − → − → − dA A| 0 +→ = ddt − ω (R0 /R) ∧ A dt |R R 34 2.2. CHANGE OF REFERENCE FRAMECHAPITRE 2. KINEMATICS OF MATERIAL POINT – Velocity composition Let R(O, XY Z) an absolute reference and R0 (O0 , x0 , Y 0 , Z 0 ) relative reference. the position vectors of the particle M in the reference frames R and R0 are, respectively : −−→ → −−→ → − OM = − r et O0 M = r0. We can write, −−→ −−→0 −− → OM = OO + O0 M. So the absolute velocity of point M is, −−→ −−→0 −−0→ −−→0 − → → − dOM dOO dO M dOO Va (M ) = V (M/R) = dt |R = dt |R + dt |R = dt |R + −− 0→ −− dO M 0 → − 0 0→ dt |R + ω (R /R) ∧ O M... −−0→ → − → − où V (M/R0 ) = V r (M ) = dOdtM |R0 denotes the relative velocity point M. And, −−→0 −−→ → − → − Ve (M ) = dOO dt |R + ω (R 0 /R) ∧ O 0 M. is the training velocity of M. Training speed of M is the absolute speed of the (imaginary) point which coincides with M at time t and assumed to be fixed in t R0. We can also note the training speed of M as follows, → − −−→ Ve (M ) = dOMdt |R (M f ixed inR’). So we have, − → → − → − Va (M ) = Vr (M ) + Ve (M ). ∗ Acceleration composition The absolute acceleration of point M is, 35 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT −−→ − → → − dVa γa (M ) = → − γ (M/R) = d OM 2 dt 2 |R = dt |R → − → − → − −−→0 −−→ → − d(Vr (M )+Ve (M )) dVr (M ) d dOO → − 0 /R) V O 0 M | ] γa (M ) = |R = |R + [ |R + ω (R R → − − dt → dt → − dt dt dVr (M ) dt |R = dVrdt(M ) |R0 + →−ω (R0 /R) ∧ Vr = → − γ r (M ) + → − ω (R0 /R) ∧ → − V r. −−→ → − 0 −−→ − 0 −−0→ d → − 0 /R) ∧ O 0 M )| = d ω (R /R) ∧ O 0 M + → ∧ dO M dt ( ω (R R dt ω (R /R) dt |R The absolute acceleration can therefore be − written −→ as : → − 0 → − 0 /R) → − γa (M ) = → − γr (M ) + 2→ − ω (R0 /R) ∧ Vr (M ) + d2 OO dt2 |R + d ω (R dt ∧ −− → − 0 −−→ O0 M + →ω (R /R) ∧ (→− ω (R0 /R) ∧ O0 M ). Of−− which → 0 → − 0 /R) −−→0 → −−→ d2 OO dt2 |R + d ω (R dt ∧ OM + − ω (R 0 /R) ∧ (→ − ω (R 0 /R) ∧ O 0 M ) = → − γe (M ) denotes the training acceleration, and → − 2→ − ω (R0 /R) ∧ Vr (M ) = − γ→ C (M ). is the Coriolis or complementary acceleration. We then write. → − γa (M ) = → − γr (M ) + → − γe (M ) + → − γc (M ). · Special cases When the reference R0 is in translation with respect to R, → − → − ω (R0 /R) = → − o ⇒→ − γc (M ) = 0 Therefore : → − γa (M ) = → − γr (M ) + → − γe (M ) 2.3 Corrected exercises Task 01 : → − → − → − Let R(0, i , j , k ) be a direct orthonormal reference. Consider a ma- terial point M which describes an orthonormal motion in the plane → − → − (O, i , j ) along the path shown in Figure 1. The equation of this trajec- tory is given in polar coordinates by : ρ = 21 ρ0 (1 + cos ϕ). 36 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT Where ρ0 is a given length, 0 ≤ ϕ ≤ π and ϕ̇ > 0. 1. Show that the velocity of M in R can be written in the basis (→ − e ρ, → − e ϕ) in the form : : → − V (M/R) = ρ0 ϕ̇ cos( ϕ2 )[− sin( ϕ2 )→ − e ρ + cos( ϕ2 )→ − e ϕ] → − → − 2. Determine k V (M/R)k the magnitude of the velocity. V (M/R). → − 3. Deduce T the unit vector tangent to the trajectory. d→ − 4. Show that the angle (→ −e ϕ, T ) = ϕ. 2 → − 5. Represent graphically the vector T. 6. Determine γT and γN the tangential and normal acceleration vectors respectively.. 7. Deduce ρc the radius of curvature of the trajectory and the unit vec- → − tor N. Solution : → − →− → − Let R(0, i , j , k ) be a direct orthonormal reference. Consider a material point M which describes an orthonormal motion in the → − → − plane (O, i , j ) along the path shown in Figure 1. The equation of this trajectory is given in polar coordinates by : −−→ ρ = 12 ρ0 (1+cos ϕ). 1- We have : OM = ρ→ −e ρ = ρ = 12 ρ0 (1+cos ϕ)→ − eρ → − V (M/R) = − 12 ρ0 ϕ̇ sin ϕ→ − e ρ + 12 ρ0 ϕ̇(1 + cos ϕ)→ − eϕ Similarly, we have 37 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT sin ϕ = sin( ϕ2 + ϕ2 ) = 2 sin( ϕ2 ) cos( ϕ2 ). cos ϕ = cos( ϕ2 + ϕ2 ) = 2 cos2 ( ϕ2 ) − 1 ⇒ 1 + cos ϕ = 2 cos2 ( ϕ2 ) ⇒ → − V (M/R) = −ρ0 ϕ̇ sin( ϕ2 ) cos( ϕ2 )→ − e ρ + ρ0 ϕ̇ cos2 ( ϕ2 )→ − eϕ → − Hence : V (M/R) = ρ0 ϕ̇ cos( 2 )[− sin( 2 )−e ρ + cos( ϕ2 )→ ϕ ϕ → −e ϕ] → − 2-k V (M/R)k = kρ0 ϕ̇ cos( )[− sin( )−e ρ + cos( )→ ϕ ϕ → 2 2 ϕ −e ϕ ]k 2 =kρ0 ϕ̇ cos( ϕ2 )k k[− sin( ϕ2 )→ − e ρ + cos( ϕ2 )→ − e ϕ ]k = ρ0 ϕ̇ cos( ϕ2 ) → − → − V (M/R) 3-We know that T = → − k V (M/R)k → − So : T = −sin( 2 ) e ρ + cos( ϕ2 )→ ϕ →− − eϕ → −→ 4-We have : T −e ϕ = cos( ϕ2 ). d→ − So (→ −e ϕ , T ) = ϕ2. 5-See figure1. → − → − dk V (M/R)k → − d(ρ0 ϕ̇ cos( ϕ2 )) → − 6- → − γ T = γT T = dt T = dt T = [ρ0 ϕ̈ cos( ϕ2 ) − 1 2 ϕ → − 2 ρ0 ϕ̇ sin( 2 )] T → − → − → − γN = γN N = k V (M/R)k ddtT = ρ0 ϕ̇ cos( ϕ̇2 ) dt d [−sin( ϕ2 )→ − e ρ +cos( ϕ2 )→ − e ϕ] = ρ0 ϕ̇ cos( ϕ )[− ϕ̇ cos( ϕ )→ 2 2 2 − e ρ − ϕ̇ sin( ϕ )→ 2 − e ϕ − ϕ̇ sin( ϕ )→ 2 − e ϕ − ϕ̇ cos( ϕ )→ 2 − e ρ] 2 ϕ 3 ϕ →− 3 ϕ→ − = ρ0 ϕ̇ cos( 2 )[− 2 ϕ̇ cos( 2 ) e ρ − 2 ϕ̇ sin( 2 e ϕ ]) = 32 ρ0 ϕ̇2 cos( ϕ2 )[− cos( ϕ2 )→ − e ρ − sin( ϕ2 )→ − e ϕ] → − ρ2 ϕ̇2 cos2 ( ϕ ) 7- We know that :γN = k V (M/R)k 2 ρc = 2 ρc 2 and like : γN = kγN k = 23 ρ0 ϕ̇2 cos( ϕ2 ) So : ρ20 ϕ̇2 cos2 ( ϕ2 ) ρc = 23 ρ0 ϕ̇2 cos( ϕ2 ) 2 ρ cos( ϕ ) ⇒ ρc = 0 3 2. → − → − And as :γN = γN N then N = [− cos( ϕ2 )→ − e ρ − sin( ϕ2 )→ − e ϕ ]. 38 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT Task 02 : The velocity diagram for a moving body A animated by rectilinear mo- tion on an axis Ox is given Figure 2, knowing that at time t = 0s, x = 0m , v = 0m/s 1. Plot the acceleration versus time diagram. 2. Plot the space diagram for the time interval [0, 7s]. 3. Determine the position of the mobile at time t = 10s. 4. What is the distance travelled between the two instants t = 0 and t = 10s 5. Specify the nature of the movement in each phase 6. Determine the equations of velocity v(t) and motion x(t) for each phase 7. Plot the vectors : position, velocity and acceleration at time t = 8s on the trajectory. Solution : 1. The acceleration versus time diagram. Acceleration is measured by the slope of the curve (tan) v2 −v1 a= t2 −t1 3−0 – t ∈ [0, 3]s; a1 = 3−0 = 1m/s2 39 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT – t ∈ [3, 6]s a2 = −1m/s2 – t ∈ [6, 7]s a3 = 0m/s2 – t ∈ [7, 10]s a4 = −1m/s2 2. The space diagram for the time interval [0, 7s]. 3. Position of the mobile at time t = 10s X(10) = [x(10) − x(0)] + [x(7) − x(0)] x(10) = −4.5 + 9 = 4.5m 4. Distance travelled between the two instants t = 0 and t = 10s d = X(10) = |[x(10) − x(0)]| + |[x(7) − x(0)]| 5. The nature of the movement in each phase – [0, 3s] : v.a > 0 uniform movement – [3, 6s] : a.v < 0 decelerate movement – [6, 7s] : a.v = 0 at rest – [7, 10s] : a.v > 0 accelerate movement 6. the equations of velocity v(t) and motion x(t) for each phase 40 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT – [0, 3s] a1 = 1m/s2 dv1 (t) R v1 (t) Rt a1 = dt ⇒ dv1 = a1 dt ⇒ 0 dv(t) = t0 =0 dt v1 (t) = t dx(t) v1 (t) = dt ⇒ R x1 (t) R t=0 x(0)=0 dx(t) = t v1 (t)dt ⇒ x1 (t) = 12 t2 – [3, 6s] a2 = −1m/s2 R v2 (t) a2 = dvdt2 (t) Rt ⇒ dv2 = a2 dt ⇒ v(3) dv(t) = t=3 a2 dt v2 (t) − 3 = −1(t − 3) ⇒ v3 (t) = −t + 6 dx(t) v2 (t) = dt ⇒ R x2 (t) Rt x(3)=4.5 dx(t) = t=3 v2 (t)dt Rt 2 x2 (t) − 4.5 = t=3 (−t + 6)dt ⇒ x2 (t) = − t2 + 6t − 9 – [6, 7s] a3 = 0m/s2 , v = 0, x3 (t) = 9 m – [7, 10s] a4 = −1m/s2 41 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT dv4 (t) R v4 (t) Rt a4 = dt ⇒ dv4 = a4 dt ⇒ v(7)=0 dv(t) = t=7 a4 dt v4 (t) = −t + 7 dx(t) v4 (t) = dt ⇒ R x4 (t) Rt x(7)=9 dx(t) = t=7 (−t + 7)dt 2 x4 (t) = − t2 + 7t − 15.5 7. Piloting of vectors : position, velocity and acceleration Task 03 : The coordinates of a material point M in a direct orthonormal reference → − → − →− frame R(O, i , j , k ) are given as a function of time by : x(t) = t − 1 , y(t) = −t2 + 1 et Z(t) = 0. 1. Determine the equation of the trajectory of M. 2. Determine the velocity and acceleration vectors of M. 3. Calculate the tangential and normal accelerations of M. 4. Deduce the radius of curvature Rc of the trajectory as a function of time Solution : 1. x(t) + 1 = t ⇒ y = −x2 − 2x ⇒ the trajectory is a parabola. −−→ → − dx → − dy → − dZ → − → − → − 2. V (M/R) = dOM dt = dt i + dt j + dt k = i − 2t j → − −−→ → − γ (M/R) = d V (M/R) 2 → − = d2 OM 2 2 = dd2xt + dd2yt + dd2Zt = −2 j dt dt2 → − dk V (M/R)k → − 3.Tangential acceleration : γT = dt T → − → − V (M/R) = √ 1 → − 2t → − T = → − 1+4t2 i − √1+4t 2 j k V (M/R)k 42 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT → − √ dk V (M/R)k d( 1+4t2 ) 4t dt = dt = √1+4t2 → − 4t → − 8t2 → − γ T = 1+4t2 i − 1+4t2 j. 4t → − Normal acceleration : → − γN = → − γ (M/R) − → − γT ⇒ → − γ N = − 1+4t 2 i − 2 →− 1+4t2 j → − 3 k V (M/R)k3 (1+4t2 ) 2 4.Radius of curvature : Rc = → − V→ − = 2 k V (M/R) γ (M/R)k 3 2 2 ⇒ Rc = (1+4t 2 ). → − 2→ − → − 2 → − → − or γ N = k V (M/R)k else → − Rc N γ N k = k V (M/R)k ⇒ k→ − Rc k N k, k N k = 1 ⇒ → − 3 Rc = k V (M/R)k 2 (1+4t2 ) (1+4t2 ) 2 → − =q = k γ Nk ( 4t 2 )2 +( 1+4t 2 2 1+4t2 2) Task 04 : Consider a moving point M whose cylindrical coordinates at each ins- tant are ρ(t) = a0 t2 + ρ0 , ϕ(t) = ωt − ϕ0 and Z(t) = −V t, with ρ0 = 1m, a0 = 1m.s−2 , ω = 3rad.s−1 , ϕ0 = 2 rad and V = 2m.s−1. 1. Give in the cylindrical base (→ − e ρ, → − e ϕ, → − e Z ), the velocity and accele- ration vectors of M. 2. Calculate the magnitude of the velocity vector of M at time t = 1s. 3. Calculate the magnitude of the acceleration vector of M at the initial instant (t = 0s). Solution : → − −−→ → − d(ρ→ − e ρ +Z k ) = dρ → − d→ − eρ dZ → − 1. V (M/R) = dOM dt = dt dt e ρ + ρ dt + dt k → − → − ⇒ V (M/R) = 2aa t→ − e ρ + (a0 t2 + ρ0 )ω → − eϕ−V k → − q ⇒ k V (M/R)k = (2a0 t)2 + ((a0 t2 + ρ0 )ω)2 + V 2 → − γ (M/R) = d V (M/R) → − dt = 2a0 → − e ρ + 2a0 ωt→ −e ϕ − (a0 t2 + ρ0 )ω 2 → − e ρ. ⇒ γ(M/R) = [2a0 − (a0 t2 + ρ0 ω 2 )]2 + (4a0 ωt)2 2. The magnitude of the velocity vector of M at time t = 1s. → − q k V (M/R)k = (2a0 )2 + ((a0 + ρ0 )ω)2 + V 2 = 6.63m/s. 3.The magnitude of the acceleration vector of M at the initial instant (t = 0s) k→ − q γ (M/R)k = (2a0 − ρω 2 )2 = 7ms−2 43 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT Task 05 : A ball h is dropped from a building of height M with no initial velo- → − →− → − city.In a reference frame R(O, i , j , k ) linked to this building it falls vertically according to a uniformly accelerated movement of accelera- → − tion → −γ = −g k (See the figure below). −−→ 1. Determine the position vector O0 M of the ball in a reference frame → − → − → − ntiel R0 (O0 , i0 , j 0 , k 0 ) related to a car moving in a uniform rectili- → − near motion with speed → −u = u i and passing through the vertical fall at the moment of release.Derive the equation of the trajectory → − → − → − of the ball in R0 (O0 , i , j , k.) −−→ 2. Determine the position vector O0 M of the ball in the same frame → − → − → − of reference R0 (O0 , i0 , j 0 , k 0 ) if we assume that, at the moment of release and starting from the vertical of the fall a uniformly acce- → − lerated rectilinear movement of acceleration γ = a i. Deduce the → − → − → − equation for the trajectory of the ball in R0 (O0 , i0 , j 0 , k 0 ). Solution : → − → − → − The position vector of the ball in the reference frame R(O, i , j , k ) is : −−→ → − OM = z k (x = y = 0). → − →− → − The ball is in free fall with no initial velocity in R(O, i , j , k ) with → − acceleration → −γ = −g k d2 z ⇒ γz = dt2 = −g ⇒ z = − 21 gt2 + V0 t + z0 −−→ → − At = 0, V0 = 0, z0 = h ⇒ z = − 21 gt2 + h ⇒ OM = (− 21 gt2 + h) k → − → − → − The position vector of the ball in the reference frame R0 (O0 , i0 , j 0 , k 0 ) is : 44 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT −− → → − → − → − O0 M = x0 i0 + y 0 j 0 + z 0 k 0 → − → − → − → − −−→ i = i0 , k = k 0 et y 0 = 0 ( M moves in the plane (xoz))⇒ O0 M = → − → − x0 i + z 0 k −−→ −−→0 −− → −−→ −−→ −−→ OM = OO + O0 M ⇒ O0 M = OM − O0 O. → − → − → − 1. 1st case : R0 (O0 , i0 , j 0 , k 0 ) is in uniform rectilinear motion of speed → − → − u = u i and passing through the vertical fall at the moment of release : −−→0 → − →− →− → − The point O0 is fixed in R0 (O0 , i0 , j 0 , k 0 ) ⇒ V (M/R) = dOO dt = → − dxO0 → − ui = dt i ⇒ xO0 = ut + c −−→ A t = 0 ⇒ O = O0 ⇒ xO0 = 0 ⇒ c = 0 ⇒ xO0 = ut ⇒ OO0 = → − −−→ → − → − −−→ −−→ → − → − ut i ⇒ O0 M = x0 i0 + z 0 k 0 = OM − OO0 = (− 12 gt2 + h) k 0 − ut i0 ⇒ 02 x0 = −ut et z 0 = − 21 gt2 + h ⇒ z 0 = − 21 g ux2 +h ⇒ la the trajectory of → − → − → − the ball in R0 (O0 , i0 , j 0 , k 0 ). → − → − → − 2. 2ed case : R0 (O0 , i0 , j 0 , k 0 ) is in uniformly rectilinear motion accele- → − ration → − γ = a i and passing through the vertical fall at the moment of release : −−→0 → − → − → − The point O0 is fixed in R0 (O0 , i0 , j 0 , k 0 ) ⇒ → − γ (O0 /R) = d OO 2 dt2 = → − 2 → − a i = d dtx2O0 i ⇒ x0 = 12 at2 + c1 t + c2. At t = 0 ⇒ O = O0 ⇒ xO0 = 0, ẋO0 = 0 ⇒ c1 = c2 = 0 ⇒ xO0 = −−→0 1 2 →− 1 2 2 at ⇒ OO = 2 at i0 ⇒ x0 = − 12 at2 et z 0 = − 12 gt2 + h ⇒ z 0 = ag x0 + h. → − → − → − ⇒ the trajectory of the ball in R0 (O0 , i0 , j 0 , k 0 ) is a straight line with equation z 0 = ag x0 + h Task 06 : In the xOy a straight line Ox0 rotates about Oz with a constant angular velocity A moving body M moves along the straight line Ox0 according to the law : r = a sin θ avec θ = ωt et a = cte. 1. Determine at time t as a function of a and ω the relative speed and the driving speed of M by their projections in the moving reference 45 2.3. CORRECTED EXERCISES CHAPITRE 2. KINEMATICS OF MATERIAL POINT frame x0 Oy 0.Deduce the absolute velocity expressed in this same projection basis, and show that the modulus of is constant. 2. Determine at time t as a function of a and ω the relative accelera- tion, training acceleration of M by their projections in the moving reference frame x0 Oy 0 and complementary acceleration. Deduce the absolute acceleration expressed in this same projection basis, and show that the modulus of is constant. Solution : → − −−→ → −0 1. V r = dOM dt |R = aω cos ωt i 0 → − → − V −−→ → − V e = ω k 0 OM = aω sin ωt j 0. → − → − → − → − → − V a = V r + V e = aω cos ωt i0 + aω sin ωt j 0 The unit vectors are written as follows → −0 → − → − i = cos ωt i + sin ωt j et → −0 → − → − j = − sin ωt i + cos ωt j If we replace these expressions in the absolute speed : → − → − → − → − → − V a = aω cos ωt(cos ωt i + sin ωt j ) + ω sin ωt(− sin ωt i + cos ωt j ) → − → − → − V a = aωω(cos ωt cos ωt−sin ωt sin ωt) i +(cos ωt sin ωt+sin ωt cos ωt) j −−→ → −0 2. → − γ r = d OM 2 2 dt2 |R = −aω sin ωt i. 0 → − → − γ e = −aω 2 sin ωt i0 → − → − γ c = 2aω 2 cos ωt j 0 → − → − → − γ +→ r − γ +→e − γ = −2aω 2 sin ωt i0 + 2aω 2 cos ωt j 0 c 46  3 Point material dynamics 47 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS 3.1 Point material dynamics 3.1.1 Definition Dynamics is the study of the movement of bodies as a function of the forces exerted on them. All the knowledge of so-called classical mechanics : dynamics, statics, hydrodynamics, resistance of materials, etc... are based on three prin- ciples (or laws) set out and published by Isaac Newton in 1687. A principle is a statement that is accepted without demonstration. 3.1.2 Newton’s laws 3.1.2.1 First Principle (Principle of inertia) Definition of inertia : is the tendency of an object to resist changes in its state of motion, i.e. to resist a change in speed or orientation. Galilean reference frame : A galilean or inertial frame of reference is a frame of reference in which the trajectory of a particle undergoing no external action is straight. Principle statement : In a Galilean reference frame, an isolated mate- rial point (which is not subject to any force) is either at rest or in uniform rectilinear motion. 3.1.2.2 The second principle (Fundamental Principle of Dynamics) Principle statement : In a Galilean reference frame (R), the resul- tant of the external forces applied to a material point is proportional to its acceleration vector, the proportionality coefficient being its mass m. P→ − F ext = m→ − γ (M ) The acceleration of a particle is directly proportional to the force applied to it and inversely proportional to its mass. Kg m Force units in the system IS : N (N ewton) = s2 48 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS 3.1.2.3 Third principle (action and reaction) Third principle (action and reaction) When a body A acts on a body → − B with a force F A→B then body B acts on body A in a reciprocal way. → − ,with a force F B→A , which has the same direction and the same ma- → − → − gnitude,but in the opposite sense : F A→B = − F B→A For every action there is an equal and opposite reaction. 3.1.3 Base Units and Physical Standards The concepts of length, time, and mass are fundamental to every branch of physics. The units of length, time, and mass are known as the base units of physics, and are defined by a set of physical standards that are augmented by descriptions of the procedures for employing them. Base units are not mere matters of practical convenience : because they embody the underlying concepts, they are foundations of physical science. This section presents a brief description of the base units and the systems of units derived from them that are universally used in physics. The SI base units of length, mass, and time are the meter, kilogram, and second. A related system, the CGS system (for centimeter, gram, second), differs from SI only in scaling factors.CGS units appear in older databases and are sometimes used in chemical and bio- logical research.Yet another system of units, the English system, is used for non-scientific measurements in Britain and North America, although Britain also uses the SI system. English units are related to SI units by legally agreed scaling factors ; for example, the inch is legally defined as 2.54cm. The table lists some principal units in the SI, CGS, and English systems. 3.1.4 The Algebra of Dimensions Equations in physics are not meaningful unless they are dimensionally consistent. In this context, the term dimension refers to the type of physical quantity (ul- timately expressed in units of mass, length, and time), in contrast to usage in mathematics, where dimension refers to the number of coordinates needed 49 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS to specify a point. A useful check on a calculation is to see whether the units agree on both sides of the final result. If they don’t, there is evidently an error somewhere.in mechanics the physical quantities such as velocity and force are measured in units constructed from the base units of mass, length, and time.Regardless of the system of units we use, in Newtonian mechanics every quantity depends on mass, length, and time in a unique way. For example, the units of velocity are m/s in the SI system and cm/s in the CGS system, but both have dimensions length/time. In analyzing the consistency of units in an equation, the dimension of mass is abbreviated M , the dimension of length is L, and the dimension of time is T. James Clerk Maxwell, who de- veloped the theory of electromagnetism, was the first to use the convenient notation of square brackets to stand for the dimensions of a quantity. [mass] = M , [length] = L [time] = T. The dimensions of quantities in mechanics can always be expressed in terms of powers of M , L, and T. For instance, [velocity] = LT −1 , [force] = M LT −2. Units must agree on both sides of an equation and this is only possible if the underlying dimensions also agree.Note that M , L, and T are independent quantities ; we cannot express mass in terms of time, or length in terms of mass. Consequently, for an equation to be valid, the powers of 50 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS M , L, and T must separately agree no matter what system of units we choose to employ. 3.1.5 Force classification 3.1.5.1 Real (or external) forces There are two types of real force : 3.1.5.2 Action-at-a-Distance Forces It is a force exerted without the help of a material support. Exemples Electrostatic force : force exerted by one charge on another → − F = krq2q́ → − ur → − F :electrostatic force exerted on q 0 by q , with K = 9.109 (SI) Coulomb’s constant 51 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS r : Distance between q 0 and q. → − ur :Unit vector. Electromagnetic force : Electromagnetic force : → − is a force exerted on a point of charge q , of velocity V placed in an → − → − → − → − → − V→− electrostatic field E and a magnetic field B : F = q( E + V B ) Gravitational force It is an interaction force at a distance. It is exerted between two masses. It was formulated by Newton in 1650. A mass M at point O is interacting with another mass m at point P , such that the distance between O and P is equal to r, is written as : → − → − F = G mM r2 u op G = 6.6710−11 (SI) gravitational constant. This force can be written as → − F = −m→ −g (P ) g = G M2 → → − − u op r This last expression is called the gravitational field created by m at any point P in space. It has the same size as a acceleration. The Earth’s gravitational field at a point P in space outside the Earth. this field has the expression : g = G (RTM+r)2 → → − − u op – M M Earth’s mass : M = 5.981024 kg – RT radius of the Earth = 6.37106 m – r Represents the altitude of point P relative to the earth’s surface. This field at the earth’s surface will have a value of g0 = 9.83 52 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS 3.1.5.3 Contact force Are types of forces in which the two interacting objects are physically contac- ting each other. Exemple 1. Normal Force : A body of mass m is placed on a solid horizontal support, for example, a table (Figure 1).Its weight P = m g is applied to its centre of gravity G and normal to the surface of the table. By virtue of the principle of action and reaction, the table exerts an equal and opposite force to → − p : → − R N = −→ − p → − its magnitude k R N k = k→ −p k = mg For a perfectly smooth curvilinear support, its reaction on the object is normal to the tangent to the support at the point of contact I (figure 2). 2. Friction force These forces occur between material bodies in motion relative to each other. There are two types of friction : solid friction and viscous friction. a-Solid friction : Two solid bodies A and B with a certain roughness are in contact. Body → − B (the support) exerts a force on body A R (called reaction)composed → − of a normal reaction R N (at the contact surface) and a tangential reaction → − called R T friction force that opposes movement. 53 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS Experience shows that the magnitude of the normal reaction force is → − proportional to the magnitude of the normal reaction force R N → − → − (Coulomb’s Law) : k R N k ∝ k R T k The coefficient → − of friction is defined as follows : µ = tgϕ = k→R T k with : ϕ = (→− → − R N , R ). − kR Nk Remarks : µ has no dimension (without unit). → − → − → − The total reaction of the solid support on the object is : R = R T + R N → − N.B. : In the following, we will refer to the friction force as f and the → − normal reaction as N. There are two types of solid friction : Static solid friction : To make A slide over B , the force applied to A must have a certain value below which it remains stationary. The limiting friction force which balances the driving force and prevents A from moving is then at its maximum. This is called the static friction force to which corresponds the coefficient of static friction noted µs and we have : → − → − → − → − k f kmax = k f s k = µs k N k with µs = tgϕmax = k f→ kmax −. kN k Solid kinetic friction : As soon as body A begins to slide on support B , the frictional force → − → − decreases until it reaches a value of k f c k = µc k N k called the kine- 54 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS tic friction force. µc being the coefficient of kinetic friction, which is therefore less than µs :µc < µs The kinetic friction force is opposite to the direction of motion and is tangent to the trajectory : → − − → → − V f c = −µc k N k → −. kV k b-Viscous friction : When the movement of a solid body occurs in a liquid (e.g. water) or a gas (e.g. air), the friction is called viscous → − friction.In the case where the velocity V of the object is relatively low, the expression for the viscous friction force opposing the motion of the solid is : → − → − f = −kη V.Where k : is a constant that depends on the shape of the object (Unit : m) η :is the coefficient of viscosity of the medium (unit : N.m − 2.s or P a.s) Example : for a spherical object in a medium of viscosity, the friction force experienced by the object in contact with the medium is given by is given by Stockes’ law : → − → − f = −6π R η V. → − → − Note : The viscous friction force is often denoted f = −µ V , avec µ coefficient of viscous friction (Unit S.I. : kg.s−1 ). 3. La force de rappel (ou élastique) d’un ressort (loi de Hooke) A spring is characterised by its initial length l0 and its stiffness constant k k (unit :N/m) which depends on the nature of the material. When sub- jected to elongation, it tends to return to its initial length provided that 55 3.1. POINT MATERIAL DYNAMICS CHAPITRE 3. POINT MATERIAL DYNAMICS this elongation is not too great (to avoid irreversible deformation).The spring is placed on a horizontal plane without friction, with one end fixed„and at the other is attached an object M of mass m.It is stretched by an elongation a, its length is l = l0 + a , then release the M. It begins to oscillate about point O, the position of the mass at the initial instant (immobility).The force exerted by the spring on the mass is proportio- nal to the instantaneous elongation → − X = l − l0 , with : k F k = k|l − l0 | Figure (a) :Initial state. The spring has a length lo. Figure (b) : The spring has a length l > lo.It is stretched and acts on the mass so as to regain its equilibrium length by exerting a restoring force → − → − F from right to left, opposite to the vector i → − −−→ → − F = −k OM = −kX i with X > 0 Figure (c) :The spring has a length l < lo.It is compressed and acts on the mass so as to regain its length at equilibrium by exerting a restoring → − → − force F from left to right, in the same direction as the vector i : we → − −−→ → − also have F = −k OM = −kX i but with X < 0 (x, the abscissa of M , is an algebraic quantity ). Therefore, in all configurations, the spring tension, or return force is → − −−→ → − written as : F = −k OM = −kX i 56 3.2. CORRECTED EXERCISES CHAPITRE 3. POINT MATERIAL DYNAMICS 3.1.6 Quantity of movement and kinetic momentum 3.1.6.1 Definitions → − Assume a material point M with mass m and velocity vector V (M ) in a reference frame R(O, xyz). → − → − The momentum of M in frame R is : P (M ) = m V (M ) The angular momentum of M relative to the fixed point O is : → − −−→ →− −−→ → − σ 0 (M ) = OM ∧ P (M ) = OM ∧ m V (M ) The angular momentum of point M with respect to a straight line (D), passing through O and having unit vector → − u , is given by the scalar, → − MD ( P ) = → − σ 0 (M ).→ − u The dynamic moment of a particle M at a fixed point O , in a reference frame R, is by definition : → − −−→ δ 0 (M ) = OM ∧ m→ − γ 3.1.6.2 The angular momentum theorem −−→ → − d→ − σ 0 (M ) | → − → − −−→ → − dt R = d(OM ∧m dt V (M )) | R = V (M ) ∧ m V (M ) + OM ∧ m→ − γ → − d→ − σ (M/R) −−→ P So : δ 0 (M )|R = 0 dt |R = OM ∧ Fext 3.2 Corrected exercises Exercice 01 Consider a simple pendulum consisting of a point object M M of mass m attached to an inextensible wire of length l l and negligible mass.It moves in the vertical plane (xOy) of the fixed reference frame R(o, xyz). The pendulum is moved away from its equilibrium position by an angle θ (θ = 0) and release without initial speed.Frictional forces are assumed to be non-existent. The assembly is located in the gravity field g considered uniform. 57 3.2. CORRECTED EXERCISES CHAPITRE 3. POINT MATERIAL DYNAMICS → − 1. Express the forces applied to point M in the base (→ − e r, → − e θ , k ). → − 2. Calculate V (M/R) and →−γ (M/R) respectively the vectors velocity and acceleration of M on R. 3. Applying the PFD in the Galilean reference frame R : a-Establish the differential equation of motion in the case of small oscillations. b- Solve this differential equation. 4. Establish the expression for the wire voltage T. 5. Find the differential equation of motion by applying the kinetic energy theorem.. Solution : 1- The forces applied at point M are : → − → − - Its weight P with P = m→ − g = mg cos θ→ − e r − mg sin θ→ − eθ → − → − - Wire tension T with : T = −T → − er 2- The PDF in this Galilean reference frame is as follows : P→ − → − → − F ext = m→ − γ (M/R) ⇒ m→ −γ (M/R) = P + T ⇒ −mlθ̇2 →−e r + mlθ̈→ − e θ = (mg cos θ − T )→ − e r − mg sin θ→ − e θ. 3- a-The projection of PFD onto → −e gives : θ mlθ̈ = −mg sin θ ⇒ mlθ̈ + mg sin θ = 0 58 3.2. CORRECTED EXERCISES CHAPITRE 3. POINT MATERIAL DYNAMICS θ̈ + gl sin θ = 0 For small oscillations, sin θ ∼ = θ. θ very small Finally, we find : θ̈ + gl θ = 0 q g b- ω0 = l. 4-The PFD projection on → − e r gives : −mlθ̇2 = mgcosθ − T ⇒ mg