corrigé - Physique
TD13 : Solide en rotation autour d’un axe fixe – corrigé
Exercice 1
J2 J4 J1 J3
J4> J2
J4> J2> J3> J1
J2> J3 sent bien
Exercice 2
L∆=J∆Ω
M dm ∆
dσ∆=dmr2Ω M ∆ v=rΩ
L∆=X
solide
dσ∆=X
solide
dmr2Ω = r2ΩX
solide
dm =mr2Ω = J∆Ω
J∆=mr2
~e∆
Exercice 3
H O ∆ (D)
~
F ∆
Γ∆=|| ~
OP ∧~
F|| =||(~
OH +~
HP )∧~
F|| =|| ~
OH ∧~
F||
~
HP ∧~
F= 0 P
(D)
(D)
P
~
F
∆
H
Exercice 4
Γ = −rF (π/4) = −rF
√2
Γ = 0
Γ = F r
Γ = F r cf
Exercice 5
FT
µsN Fl=µsN
Ml=rvFl1
Ml rtFl2=Ml=rvFl1
Fl2=rv
rt
Fl1
rt> rv
Exercice 6
∆ σ∆=mr2ω
L∆= 8mr2ω
∆ J∆= 8mr2
˙ω=ωf
T J∆˙ω= Γ Γ = J∆
ωf
T
Ec=1
2J∆ω2 ω(t) = ωft
T T
Ec=J∆ω2
f
2T2t2
dEc
dt =P P
P(t) = J∆ω2
ft
T2
t=T P=J∆ω2
f
T
m'
4g=r ω2
f ωf=q4g
r T= P=J∆ω2
f
T=8mr ×4g
T P'
Exercice 7
m
~
P=m~g ~
T
~
T=−m~g
h(t) = −rθ(t)
m h(t)
Oz m¨
h(t) = T−mg
J∆˙ω=M∆(~
T) = rT.
T=m(g−r¨
θ)
J∆˙ω=mr(g−r˙ω) ˙ω
α= ˙ω=mrg
J∆+mr2
a=−rα =−mr2g
J∆+mr2 |a|=g1
1 + J∆/mr2< g
g
a'. − α'. −
m Ec=−mgh
~e∆
r
θ(t)
M
m
z
h
0
~
Tc
~
Tm
~
P
||~
Tm|| =||~
Tc|| =T
Exercice 8
S ∆ S
∆ J∆ S Γ θ
Γ = −Cθ
t= 0 θ0
S JƬ
θ= Γ = −Cθ
¨
θ+C
J∆
θ= 0
θ(t) = A(ω0t+ϕ) ω0=qC
J∆
ϕ= 0 A=θ0 θ(t) = θ0(ω0t)
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