Etudes et évaluation de processus océaniques par des hiérarchies ...

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Exercise 32:
The moment of inertia is L = Iω = mr 2 /2ω = mρV/(4π)ω/H where we used, that the volume
V of a cylinder is given by V = 2πr 2 H. As the mass and the volume are constant during the
stretching of flattening process we obtain that conservation of angular momentum implies that
ω/H is constant.
Exercise 38:
The (kinetic) energy is given by: E = α(U 2 + V 2 ), where the constant α = Aρ/(2H) is the
horizontal surface area times the density divided by twice the layer thickness. ∂ t E = α(∂ t U 2 +
∂ t V 2 ) = α2(U∂ t U+V ∂ t V ) using eqs. (6.13) and (6.13) we obtain ∂ t E = α(fV U+Uτ x −fUV ) =
0, as the velocity is perpendicular to the forcing, that is, U = 0.
tel-00545911, version 1 - 13 Dec 2010
Exercise 41:
Summing the first and the third line of the matrix equation gives U 1 + U 2 = 0, summing the
second and the fourth line of the matrix equation gives f(V 1 + V 2 ) = τ x . Eliminating U 2 and
V 2 in the first and third equation we obtain:
˜rU 1 − fV 1 = τ x
fU 1 + ˜rV 1 = (r/f)τ x .
with ˜r = r(H 1 + H 2 )/(H 1 H 2 ) solving this equations give:
U 1 =
rτ x
f 2 + ˜r 2 1
H 1
V 1 = − rτ x
f 2 + ˜r 2(f r +
U 2 = − rτ x
f 2 + ˜r 2 1
V 2 =
H 1
rτ x ˜r
f 2 + ˜r 2 fH 1
˜r
fH 2
)