DM 3 - Correction - Site de la PSI du lycée Paul Eluard

x y z
r θ φ
x=rsin θcos φ
y=rsin θsin φ
z=rcos θ
−→
v
t=µ−→
g−−−→ Pt
−→
v=−→ 0
−−→ Pt=µ−→
g
−→
g=−g−→ uzOx Oy
∂Pt
∂x =∂Pt ∂y = 0
Ptz Oz
Pt
z=−µ g
µ z
Pt(z) = −µgz +
−→
v= 0 µ
t= 0
f−→−−→f=−→ 0
φ
−→
v=−−→ φ−→−→ v=−→−−→ φ
−→−→
v=−→ 0

u
−→
vu=−−→ φu=−−→ (uz) = u−−→ z
−→
vu=u−→ ez
z=rcos θ φu=u r cos θ
−→
vu=−−→ φu=−−→ (u r cos θ) = u∂(rcos θ)
∂r
−→
er+1
r
∂(rcos θ)
∂θ
−→
eθ+∂(rcos θ)
∂z
−→
ez
−→
vu=ucos θ−→ er−sin θ−→ eθ
−→
vs=−−→ φs=∂φs
∂r
−→
er+1
r
∂φs
∂θ
−→
eθ+∂φs
∂z
−→
ez
φu=u r +b
r2cos θ
−→
vs=u−2b
r3cos θ−→
er−u+b
r3sin θ−→
eθ
−→
vs
−→
eθθ
−→
vs=2
ru−2b r3cos θ+∂ ∂r u−2b r3cos θ−cos θ rsin θu+b r3sin θ−1 r
∂
∂θ u+b r3sin θ
−→
vs= cos θ2u
r−4b r4+0+6b r4−u r−b r4−u r−b r4= cos θu r2−1−1+b r4−4+6−1−1
−→
vs= 0
vr
−→
vsvr=u−2b
r3cos θ vr= 0
u=2b
r3r=a=2b u1/3
b=u a3
2
b−→ vs
−→
vs=u1−a3
r3cos θ−→
er−1 + a3
2r3sin θ−→
eθ

u= 1
r=a−→ vs=−3
2sin θ−→
eθr= 2a−→ vs=7
8cos θ−→
er−17
16 sin θ−→
eθ
θ
−→
v θ
r=a r = 2a
−→
v
−→−→
v=−→ 0
−→
v·−−→−→ v=−−→ v2
2
ρ−−→ v2
2=−−−→ P
ρ
−−→ ρv2
2+P= 0
ρv2
2+P=C

r−→ vs
−→
vs(∞) = ucos θ−→ er−sin θ−→ eθ=u−→ ez
rlimr→∞ P(r) = 0
C=ρu2
2
r=a−→ vs=−3
2usin θ−→
eθ
P=C−ρ
23 2usin θ2
=ρ u2
21−9 4sin2θ
θ
P(r=a, −θ, φ) = P(r=a, θ, φ)
m V
m
−→
u
t=m−→
g−6πηa−→ u−ρ V −→ g
V=4
3πa3m=ρbV
−→
u
t+9η 2ρba2
−→
u=1−ρ
ρb−→
g
(Ox) (Oy)
−→
u(0) = −→ 0ux=uy= 0 −→ u=u−→ ez
Oz
u(t) = A e−t/τ −1−ρ
ρbτ g τ =2ρba2 9η
u(0) = 0 A=1−ρ
ρbτ g
−→
u(t) = −1−ρ
ρbτ g1−e−t/τ −→
ez
−→
u∞
−→
u∞=−1−ρ
ρbτ g−→
ez

a= 1 µ
u∞=1−ρair
ρeau 2ρeaua2 9ηg=1−1,3 1032.103(10−6)2 9×1,8.10−5×9,8≈2,8 9×1,8.10−4≈1,2.10−4−1
a3a2
6u∞≈120 −1
−1
−1
a= 10−6
Re=ρair a v
ηair
=1,3×10−6×1,2.10−4
1,8.10−5≈1,6 1,8.10−5≈10−5
a103v106
109Re≈104
Re < 1
F
3au
4r= 1
−→
v=−2 cos θ−→ er+ sin θ−→ eθ
d
a−→
v−→
−v
−→
F=−6πηa(−−→ v) = 6πηa−→ v
−→
v−→ v