I- Pendule de FOUCAULT

◦
~
fG~
fie
g
g
L P
M m ~ T
OM m ~g
M
M M
θ
Ep=m g z =−m g L cos θ
Ec=1
2m v2=1 2m L2˙
θ2
~
T
dEm
dt =m g L ˙
θsin θ+m L2˙ θ¨ θ= 0
sin θ≈θ¨ θ+ω2θ= 0 ω=pg
L=2π T
T= 2πqL
g= 16 s
P
P C P C∗
1P C∗ 2◦

(~ux, ~uy, ~uz)~ Ω
~
Ω = Ω (sin θ ~ux+ cos θ ~uz) = Ω (cos λ ~ux+ sin λ ~uz)
~
fiC =−2m~ Ω∧~v =−2mΩ (cos λ ~ux+ sin λ ~uz)∧( ˙x ~ux+ ˙y ~uy)
M P
~
fiC = 2 mΩ (sin λ˙y ~ux−sin λ˙x ~uy−cos λ˙y ~uz)
~
fiCv =−2mΩ cos λ˙y ~uz
k~ fiCv k ≈ 2mΩv≈2mΩLrg
L= 2 mΩpg L
k~ fiCv k
m g ≈2 Ω sL g=4π 86400r67 10 = 3,8×10−41
sin λ > 0
M
~
Tf−−→ MP = (−x ~ux−y ~uy−z ~uz)|x| |y|  L
−z≈L
~
Tf=Tf
L(−x ~ux−y ~uy+L ~uz)
Tf
M
P C P C∗
1P C∗ 2◦

•m ~g =−m g ~uz
•~ Tf=Tf
L(−x ~ux−y ~uy+L ~uz)
•~ fiCh = 2 mΩ (sin λ˙y ~ux−sin λ˙x ~uy)
M
m(¨x ~ux+ ¨y ~uy) = −m g ~uz+Tf
L(−x ~ux−y ~uy+L ~uz)+2mΩ (sin λ˙y ~ux−sin λ˙x ~uy−cos λ˙y ~uz)
~uz
m g =Tf
(m¨x=−Tf
Lx+ 2 mΩ sin λ˙y
m¨y=−Tf
Ly−2mΩ sin λ˙x
m g =Tfm
¨x=−ω2x+α˙y
¨y=−ω2y−α˙x
ω=pg
Lα= 2 Ω sin λ
f=x+i y
¨
f= ¨x+i¨y=−ω2x+α˙y−i ω2y−i α ˙x=−ω2(x+i y)−i α ˙ (x+i y)
f¨ f+i α ˙ f+ω2f= 0 x y
r2+i α r +ω2= 0
∆ = −α2+ 4 ω2<0
r=−i α ±i√α2+ 4 ω2
2
f(t) = e−iΩ sin λ t c−e−i ω0t+c+e+i ω0t
ω0=√α2+4 ω2
2=pΩ2sin2λ+ω2=qΩ2sin2λ+g L
•T0=2π
√Ω2sin2λ+g
L
=2π
q(2π
86400 )2sin2(48◦510)+ 9,8 67 ≈2π
9,8
67
=T T0= 16 s
•T0=2π
Ω sin λ=2π
q(2π
86400 )2sin2(48◦510)=86400
sin(48◦510)= 11,5×103s 32 h
λ=π
2T0= 24 h
λ= 0 T0→ ∞
β∆V=1
2βd(R02−R2)
∆t β ω =β
∆t
∆V=Dv∆t=1
2βd(R02−R2) = d h0v∆t
ω=β
∆t=2d h0v d(R02−R2)
P C P C∗
1P C∗ 2◦

ω=2h0v
R02−R2
vm=R+R0
2ω
vm
ω=R+R0 2=(R02−R2) 2h0
=(R0−R) (R0+R)
2h0
h0=R0−R
Dv=∆V
∆t=d h0v h0=Dv d v =Dm ρdv =3,6×103 103×1,4×0,60 h0= 4,3 m
h0=R0−R R0= 4,3 + 11 R0= 15 m
ω=2×4,3×0,60
152−112ω== 50 rmmrad ·s−1
dP
dz =−ρ g
p1(z) = p0−ρ g z p2(z) = p0−ρ g (z+H)
−−→
d2F= (p1(z)−p2(z)) dx dz ~u =ρ g H dx dz ~u
~u
~
F=ρ g H d (R0−R)~u
~
F
R+R0
2α≈0
~
M=ρ g H d (R0−R)R+R0
2~ex
~
Mk= (pk−pk−1)d(R0−R)R+R0
2~ex
X
k
~
Mk= (p1−p2)d(R0−R)R+R0
2~ex=ρ g H d (R0−R)R+R0 2~ex=~
M
Γ = 103×9,80 ×3,0×1,4×15,32−11,02
2= 2,3×106N·m
P
P= Γ ω=ρ g H d R02−R2
2
2h0v
(R02−R2)=ρ g H d h0v
P= 103×9,80 ×3,0×1,4×4,3×0,60 = 1,1×105W
P0= 150 ×740 = 110 kW
Γ0=P
ω≈150 ×740
2π×3000
60
= 350 N ·m
P C P C∗
1P C∗ 2◦

L O
M m ~ T
OM m ~g θ
Ep=m g z =−m g L cos θ
Ec=1
2m v2=1 2m L2˙
θ2
~
T
dEm
dt =m g L ˙
θsin θ+m L2˙ θ¨ θ= 0
sin θ≈θ¨ θ+ω2
0θ= 0 ω0=pg
L
ω0=pg
L=2π T0
T=T0
2=πsL g=πr0,90 9,81 = 0,95 s
Ep=m g z =−m g L
2cos θ
Ec=1
2JΩ2=1 6m L2˙
θ2
~
T
dEm
dt =m g L 2˙
θsin θ+mL2
3˙
θ¨ θ= 0
sin θ≈θ¨ θ+ω2
0θ= 0 ω0=q3g
2L
T=T0
2=πs2L 3g=πr2×0,90 3×9,81 = 0,78 s
•L
•a
•h
L a h
v a T v =a
T
M
P C P C∗
1P C∗ 2◦