Summer School 2007

B. Rossetto 1

French Summer School Phnom Penh 2007

Mechanics I

ROSSETTO Bruno Institut Universitaire de TechnologieUniversité du Sud-Toulon-Var (France)

tél. + 336 08 45 48 54 email: rossetto@univ-tln.fr site: http://rossetto.univ-tln.fr

Summer School 2007

B. Rossetto 2

Mechanics I Summary

Chap. 1 – CoordinatesChap. 2 – VectorsChap. 3 – Differential operatorsChap. 4 – Forces. EquilibriumChap. 5 – Kinematics. Particle motionChap. 6 – Relative motionChap. 7 – System of particlesChap. 8 – Rigid body motion

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Mechanics I

References

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)

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1. Coordinates Cartesian

x

y

0 22222222222222xu

22222222222222yu

2-dim.

1 - Origin 0 2 - System of orthogonal axis (0xy) 3 - Unit vectors andxu

22222222222222yu

22222222222222

x

y

022222222222222xu

22222222222222yu

zu22222222222222

z3-dim.

Orientation of the three-dimensional system of coordinates:

- screw rule - right hand rule

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1. Coordinates Orientation rules

x

y

z

y

z

x

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1. Coordinates Orientation rules

x

y

z

y

x

z

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1. Coordinates

x

y

0 22222222222222xu

u22222222222222

Polar (2-dim.)

ru22222222222222

22222222222222yu

Cylindrical (3-dim.)

P(r,)

x0

zu22222222222222z

ru22222222222222

P(r,,z)

P(r,)r 0

0 2

P(r, ,z): r 0, 0 2 , z - , +

r cos sin x yu u u222222222222222222222222222222222222222222

sin cos θ x yu u u222222222222222222222222222222222222222222

andFor both:

u22222222222222 zu

22222222222222

x r cos sin u u u222222222222222222222222222222222222222222

y r sin cos u u u222222222222222222222222222222222222222222

and

z zu u2222222222222222222222222222

3-dim.:

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1. Coordinates Transformations

ru22222222222222

r

x

y

z

0u

22222222222222zu

22222222222222

r

z

1 – From polar to cartesian x = r cos y = r sin z = z

1 – From cartesian to polar

2 2r = x y

y = Arctg and sign of x or y

x

z = z

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1. Coordinates Spherical

ru22222222222222

r

r sin

x

y

z

0

u22222222222222

u22222222222222

r sin

z

P(r, , ) : r 0, 0 2 , 0

x r sin cos

y r sin sin

z r cos

2 - Transformations

2 2 2

2 2

r = x y z

y = Arctan and sign of x or y

x

x +y = Arctan , 0

z

1 - Definitions

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1. Coordinates

System of coordinates

Differential line elements along the coordinate axis of the system

Cartesian (x, y, z) dx, dy, dz

Cylindrical (r, , z) dr, r d, dz (cf applications 2, 4)

Spherical (r, , ) dr, r sind, r d(cf applications 5, 6)

Definition of radianAB

(rad)r

(for the disk : = 2 radians)

AB : length of the oriented arc of circonferencer

0r

A

B AB r

From this definition:

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1. Applications (1)

1 – Triangle area from the equation

b

pb = h

x0

h

rx

0-r

2 – Surface of a disk from the equation

If b is the basis and h the height:

- Find the equation of the line OA- Use a property of integrals

A

- Find the equation of the circle- Use a property of integrals

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1. Applications (1)

1 – Triangle area from the equation

b

f(x) = px

pb = h

x0

h

rx

0-r

2 2f(x) = ± r - x

2 – Surface of a disk from the equation

If b is the basis and h the height:

Equation: f(x)=px

2 2Equation : f(x) = ± r - x

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1. Applications (1)

b b2

0 0

1 1A = f(x)dx = px dx = pb = hb

2 2

1 – Triangle area from the equation

b

f(x) = px

pb = h

x0

h

rx

0-r

2 2f(x) = ± r - x

2 – Surface of a disk from the equation

r2 2

0

A = 4 r - x dxr 2

0

x= 4 r 1 - dx

r

x dxsinθ = , cos θ dθ =

r rLet

22 2

0

A = 4r cos θ dθ

2

2 2

0

= 2r 1 + cos 2θ dθ = r

If b is the basis and h the height:

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1. Applications (2)

0

r d

r

r

2 - Surface of a disk using polar cordinates The contribution to the area of the sector having r as length and as angle is the aerea of the triangle having r as basis and rd as height:

r0

rd

AB

dl AB rd

1 - Length of a circonference Contribution of the angle d to the length:

Total length: sum of contributions:

Total area : A=

dA=

2

0

L dl

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1. Applications (2)

22 2

0

1A = r dθ = r

2

0

r d

r

r

21dA = r dθ

2

2 - Area of a disk using polar cordinates The contribution to the area of the sector having r as length and as angle is the aerea of the triangle having r as basis and rd as height:

Total aerea :

r0

rd

AB AB rd

1 - Length of a circonference Contribution of the angle d to the length:

Total length :2

0

L = rdθ = 2 r

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1. Applications (3)

1 – Surface of a triangle (base b , height: h)

b

f(x)=px

pb=h

x0

h

2 – Surface of the ellipse

0a

b2 2

2 2x y

+ = 1a b

dA=

A=

Contribution of the infinitesimal surface dy.dx :

dA =

Equation:

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1. Applications (3)

1 – Surface of a triangle (base b , height: h)

b

f(x)=px

pb=h

x0

h

pxb b2

0 0 0

1 1A = dy dx = pxdx = pb = hb

2 2

2 – Surface of the ellipse

x dxLet sinθ = , cosθdθ = . We find : ab

a a

2xb 1-

2a aa 2

20 y=0 0

xdxdy = 4 dy dx = 4b 1- dx

a

0

a

b

2 2

2 2x y

+ = 1a b

Area:

Contribution of the infinitesimal surface dy.dx : dA = dy.dx

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1. Applications (4)

x

y

z Cylinder area

0r

x

y

z

0dz

h

rd

1 - Double integral: contribution of theelement of length r d height dz: rddz

2 - Simple integral: contribution of the element of length 2r height: dz:2rdz

dzdA=

A=

dA=

A=

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1. Applications (4)

x

y

z Cylinder area

h 2

0 0

A r dθ dz =2 rh

h

0

A = 2 r dz = 2 rh

0r

x

y

z

0dz

h

rd

1 - Double integral: contribution of theelement of length r d height dz:dA=rddz

2 - Simple integral: contribution of the element of length 2r height: dz:dA=2rdz

dz

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1. Applications (5)

r sin

x

y

z

rd

r

length: 2 r sin, width: r d

Total area: A=

Sphere area using symetries (simple integral): contribution of the element:

Sphere area using double integral

Contribution of the elementlenght : r sin d, width: r d

r sin

0

0

r

rd

rsind

Area : A = sum of contributions

dA=

dA=

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1. Applications (5)

r sin

x

y

z

rd

r

length: 2 r sin, width: r d dA = 2 r2 sin d

Total area: 2 2

0

A = 2 r sinφ dφ = 4 r

Sphere area using symetries (simple integral): contribution of the element:

Sphere area using double integral

Contribution of the elementlenght : r sin d, width: r d

22 2 2

0 0 0

A = r d sin d = 2 r sin d = 4 r

r sin

0

0

r

rd

rsind

dA = r2 sin d d

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1. Applications (6)

r

r sin

x

y

z

r d

r sin d

r

Sphere volume (or mass if homogeneous)

Contribution of the element length : r sin d weidth : r d height : dr

r sin

0

0

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1. Applications (6)

r

r sin

x

y

z

r d

r sin d

r

Sphere volume (or mass if homogeneous)Contribution of the element length : r sin d weidth : r d height : dr

r 22

0 0 0

V r sinφ dφ dθ dr

r2 3

0

4V = 4 r dr = r

3

r sin

0

0

2dV=r sinφ dφ dθ