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French Summer School Phnom Penh 2007

French Summer School Phnom Penh 2007. Mechanics I. ROSSETTO Bruno Institut Universitaire de Technologie Université du Sud-Toulon-Var (France) tél. + 336 08 45 48 54 email: rossetto@univ-tln.fr site: http://rossetto.univ-tln.fr. Mechanics I. Chap. 1 – Coordinates

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French Summer School Phnom Penh 2007

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  1. French Summer School Phnom Penh 2007 Mechanics I ROSSETTO Bruno Institut Universitaire de Technologie Université du Sud-Toulon-Var (France) tél. + 336 08 45 48 54 email: rossetto@univ-tln.fr site: http://rossetto.univ-tln.fr B. Rossetto

  2. Mechanics I Chap. 1 – Coordinates Chap. 2 – Vectors Chap. 3 – Differential operators Chap. 4 – Forces. Equilibrium Chap. 5 – Kinematics. Particle motion Chap. 6 – Relative motion Chap. 7 – System of particles Chap. 8 – Rigid body motion Summary B. Rossetto

  3. Mechanics I M. Alonso and E. J. Finn, Fundamental University Physics, vol. 1 Mechanics, Addison Wesley (1969) C. Kittel, W. D. Knight, M. A. Ruderman, The Berkeley Course on Physics, vol. 1 Mechanics, Mc Graw Hill, (1965) R. W. Feynmann, M. Leighton and M. Sands, The Feynmann Lectures on Physics, vol 1, Mainly Mechanics, Radiation and Heat, Addison Wesley, early 1960s) References B. Rossetto

  4. 1. Coordinates • Cartesian 2-dim. 1 - Origin 0 2 - System of orthogonal axis (0xy) 3 - Unit vectors and y x 0 3-dim. Orientation of the three-dimensional system of coordinates: - screw rule - right hand rule z y x 0 B. Rossetto

  5. 1. Coordinates • Orientation rules z z x x y y B. Rossetto

  6. x y z 1. Coordinates • Orientation rules x y z B. Rossetto

  7. 1. Coordinates • Cylindrical (3-dim.) • Polar (2-dim.) P(r,q,z) y z P(r,q) q q x x 0 q 0 P(r,q) and For both: and 3-dim.: B. Rossetto

  8. 1. Coordinates • Transformations 1 – From polar to cartesian x = r cosq y = r sinq z = z 1 – From cartesian to polar z r z y 0 q r x z = z B. Rossetto

  9. 1. Coordinates • Spherical 1 - Definitions z 2 - Transformations r sin j z j r y 0 q r sin j x B. Rossetto

  10. 1. Coordinates Definition of radian B A q (for the disk : q= 2p radians) r r 0 From this definition: B. Rossetto

  11. 1. Applications (1) 1 – Triangle area from the equation A pb = h If b is the basis and h the height: • Find the equation of the line OA • Use a property of integrals h x 0 b 2 – Surface of a disk from the equation • Find the equation of the circle • Use a property of integrals x -r r 0 B. Rossetto

  12. 1. Applications (1) f(x) = px 1 – Triangle area from the equation pb = h If b is the basis and h the height: Equation: f(x)=px h 2 – Surface of a disk from the equation x 0 b x -r r 0 B. Rossetto

  13. 1. Applications (1) f(x) = px 1 – Triangle area from the equation pb = h If b is the basis and h the height: h 2 – Surface of a disk from the equation x 0 b Let x -r r 0 B. Rossetto

  14. 1. Applications (2) 1 - Length of a circonference Contribution of the angle dq to the length: B A dq r r Total length: sum of contributions: 0 2 - Surface of a disk using polar cordinates The contribution to the area of the sector having r as length and q as angle is the aerea of the triangle having r as basis and rdq as height: r dq r q r 0 dA= Total area : A= B. Rossetto

  15. 1. Applications (2) 1 - Length of a circonference Contribution of the angle dq to the length: B A dq r r Total length : 0 2 - Area of a disk using polar cordinates The contribution to the area of the sector having r as length and q as angle is the aerea of the triangle having r as basis and rdq as height: r dq r q r 0 Total aerea : B. Rossetto

  16. 1. Applications (3) 1 – Surface of a triangle (base b , height: h) f(x)=px Contribution of the infinitesimal surface dy.dx : dA = pb=h h 2 – Surface of the ellipse x b 0 Equation: b dA= A= a 0 B. Rossetto

  17. 1. Applications (3) 1 – Surface of a triangle (base b , height: h) f(x)=px Contribution of the infinitesimal surface dy.dx : dA = dy.dx pb=h Area: h 2 – Surface of the ellipse x b 0 b a 0 B. Rossetto

  18. 1. Applications (4) z • Cylinder area 1 - Double integral: contribution of the element of length r dq, height dz: rdqdz h dA= A= y 0 dz q r rdq x z 2 - Simple integral: contribution of the element of length 2pr height: dz: 2prdz y 0 dA= A= dz x B. Rossetto

  19. 1. Applications (4) z • Cylinder area 1 - Double integral: contribution of the element of length r dq, height dz: dA=rdqdz h y 0 dz q r rdq x z 2 - Simple integral: contribution of the element of length 2pr height: dz: dA=2prdz y 0 dz x B. Rossetto

  20. 1. Applications (5) z • Sphere area using double integral r sin j Contribution of the element lenght : r sinj dq, width: r dj rdj rsinfdq j r dA= y 0 Area : A = sum of contributions q x • Sphere area using symetries (simple integral): contribution of the element: r sin j length: 2p r sinj, width: r dj rdj dA= j r Total area: A= 0 B. Rossetto

  21. 1. Applications (5) z • Sphere area using double integral r sin j Contribution of the element lenght : r sinj dq, width: r dj rdj rsinfdq dA = r2 sinj dq dj j r y 0 q x • Sphere area using symetries (simple integral): contribution of the element: r sin j length: 2p r sinj, width: r dj rdj dA = 2p r2 sinj dj Total area: j r 0 B. Rossetto

  22. 1. Applications (6) • Sphere volume (or mass if homogeneous) z r sin j Contribution of the element length : r sinj dq weidth : r dj height : dr j r y 0 q x r sin j r dj r sin jdq j r 0 B. Rossetto

  23. 1. Applications (6) • Sphere volume (or mass if homogeneous) z r sin j Contribution of the element length : r sinj dq weidth : r dj height : dr j r y 0 q x r sin j r dj r sin jdq j r 0 B. Rossetto

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