"I.
KINEMATICS: TRANSFORMATION OF COORDINATE SYSTEMS
Let ei and ei (i = 1, . . .3) two orthonormal basis sets defining two coordinate systems. We allow ei to be time dependent, but ei is fixed in time: ˙ ei = dei = 0. dt
A position vector r can be represented in both coordinate systems as
3 3
r=
i=1
ri ei =
j=1
rj ej ,
the components in the two representations will be denoted as r = (r1 , r2 , r3 ) The components are not independent
3 3
r = (r1 , r2 , r3 )
ri =
j=1
rj ei · ej =
j=1
Cij rj ,
where the basis transformation matrix is Cij = ei · ej and ej =
i=1 3
Cij ei
r = Cr
Due to the ortonormality of the basis C −1 = C T detC = ±1.
The time derivative of the ej basis vectors can be expressed as a linear combination of the ej basis vectors:
3
˙ ej =
i=1
Fij ei
where ˙ Fij = ei · ej By calculating the time derivative of ei · ej = δij 1
one obtains ˙ ˙ 0 = ei · ej + ei · ej = Fij + Fji which means that Fij is antisymmetric Fij = −Fji . A 3x3 antisymmetric matrix has three independent elements (the diagonal must be zero and only the lower or upper triangle is filled with independent elements), so we can write it in the form: F = 0 −ω3 ω3 −ω2 ω2 0 −ω1 . ω1 0
The time derivative of the position vector in the two coordinate systems is represented by the components v = (v1 , v2 , v3 ) In the ei system
3
v = (v1 , v2 , v3 )
˙ v=r=
i=1
˙ ri ei that is v = r ˙
In the ei system
3 3
˙ r=
i=1
ri ei + ˙
i=1
˙ ri ei that is v = r + F r . ˙
The last expression can be written in a more elegant form: v =r +ω×r, ˙ ˙ Using this equation, one can easily calculate v a = v + ω × v = r + 2ω × r + ω × (ω × r ) + ω × r ˙ ¨ ˙ ˙ ω = (ω1 , ω2 , ω3 ).
A.
Example: Cartesian coordinate system
ex = (1, 0, 0) ey = (0, 1, 0) ez = (0, 0, 1) 2
B.
Example: Polar coordinate system
Basis vectors: er eφ = = cosφex + sinφey −sinφex + cosφey
ez = ez
r = rer Basis transformation
r = (r, 0, 0)
cosφ −sinφ 0 C = sinφ cosφ 0 . 0 01 ˙0 0 −φ ˙ F = φ 0 0 . 0 00 ˙ ω = (0, 0, φ) ˙˙ v = (r, rφ, 0) ˙ ¨ a = (¨ − rφ2 , 2rφ + rφ, 0) r ˙˙
C.
Example: Spherical polar coordinate system
Basis vectors: er = sinθcosφex + sinθsinφey + cosθez eθ = cosθcosφex + cosθsinφey − sinθez eφ = −sinφex + cosφey
r = rer
r = (r, 0, 0)
3
basis transformation sinθcosφ cosθcosφ −sinφ C = sinθsinφ cosθsinφ cosφ . cosθ −sinθ 0 ˙ ˙ ˙ ω = (φcosθ, −φsinθ, θ) v = (r, rθ, rφsinθ) ˙˙˙ ˙ ˙ ¨ ˙ ¨ ˙˙ a = (¨ − rθ2 − rφ2 sin2 θ, 2rθ + rθ − rφ2 sinθcosθ, 2rφsinθ + rφsinθ + 2rφθcosθ) r ˙˙ ˙˙
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II.
DYNAMICS: BASIC PRINCIPLES A. Newton’s laws
(i) In the inertial frame, every body remains at rest or in a uniform motion unless acted on by a force F. F = 0 ⇒ p = constant (ii) Application of a force alters the momentum ˙ F=p (iii) To each action there is an equal and opposite reaction F12 = −F21 (iv) Superposition of forces F=
i
Fi
B.
Conservation laws
Linear momentum conservation F = 0 ⇒ p = constant Angular momentum conservation L=r×p ˙ ˙ ˙ L=r×p+r×p=r×F=Γ (Γ is the torque). Beware! L = (r − r0 ) × p, where r0 is the origin of the reference frame. L depends on the choice of the coordinate system. Work: W1→2 =
1 2
F(r) · dr 5
the integral is along a line: r = r(σ)
2
W1→2 =
1
F(r) ·
dr dσ dσ
Conservative forces: F(r) = − U (r), (where U is called potential). Alternative definitions: × F(r) = 0
2
F(r) · dr = 0
1
for all closed paths
Work in case of conservative forces:
2 2
W1→2 = −
1
U (r) · dr = −
1
dU (r) = U1 − U2
Kinetic energy 1 T = mv2 2 Time derivative of the kinetic energy ˙ ˙ T = mvv = Fv Energy conservation: By integrating this equation between times t1 and t2 1 1 2 2 W1→2 = mv1 − mv2 2 2 v1 = v(t1 ) v2 = v(t2 ).
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