L`expérience du pendule de Foucault (1851)

◦
O ~g
g
•O R = 6400 km
P C P C∗
1P C∗ 2
◦

•
•O g = 9,8 m ·s−2
•
•
L= 67 m P M
M
M
P λ = 48◦510=π
2−θ θ
(P, ~ux, ~uy, ~uz)~ux~uy
~uz~ux~uy(x, y, z)
M M z =cste
O P ~ux~uz~uy
(~ux, ~uy, ~uz)~ Ω
M P
M P
M
M~ Tf=Tf
L(−x ~ux−y ~uy+L ~uz)
Tf
M
¨x=−ω2x+α˙y
¨y=−ω2y−α˙x
ω α L g Ωλ
f=x+i y
ω0
f(t) = e−iΩ sin λ t c−e−i ω0t+c+e+i ω0t
f(t)
T0=2π
ω0T0=2π Ω sin λ
T0T0
T0λ=π
2λ= 0
P C P C∗
1P C∗ 2
◦

ρ= 103kg ·m−3p0= 1 bar
x
~v
v H
z= 0 ~g g h0
N
R R0> R
d α
P C P C∗
1P C∗ 2
◦

V
β
V=1
2βc(b2−a2)
∆V∆t
ω v
Dv=∆V
∆t=d h0v
R R0h0v
h0R0ω Dm=ρ Dv= 3,6×103kg ·s−1d= 1,4 m R= 11 m v= 0,60 m ·s−1
p1(z)
p2(z)−−→
gradP =ρ~g
pa=pb=pc=p1(z)
pd=pe=pf=p2(z)
~u
pc=p1(z)pd=p2(z)
~
M=M~ex
p1(z) = pa> pb> pc> pd> pe> pf=p2(z)
H= 3,0 m g= 9,80 m ·s−2Γ
P
P C P C∗
1P C∗ 2
◦

≈0,74 kW
P C P C∗
1P C∗ 2
◦