Roulement sans glissement d'un cylindre homogène I. Présentation du problème

α
O
X
Z
G
α
θ = ω
I
N
P
UyUz
Ux
T
θ
α
(O, −→ ux,−→ uy,−→ uz)
α
(G, −→ ux,−→ uy,−→ uz)

−→
P=m−→ g=−mg ·sin (α)−→ ux−mg ·cos (α)−→ uz
−→
T=T·−→ ux
−→
N=N·−→ uz
ω=˙ θ−−−→ VI,Rg =˙ θ·−→ uy∧
−→
GI =−R·˙ θ·−→ ux
−−→
VI,R =−−→ VG,R +−−−→ VI,Rg = ˙x·−→ ux−R·˙ θ·−→ ux
˙x=R·˙ θ
Ec(R)=Ec(Rg)+1
2mV 2
G,R
ω=˙ θ
Ec(Rg)=1
2IOy ·˙
θ2=1
4mR2·˙
θ2
Ec(R)=1
4m·˙x2+1 2m·˙x2=3 4m·˙x2
Ec(R)−Eco(R)=W(t)
dEc(R)
dt =p(t)
p(t) = −→ P·
−−→
VG,R +−→ T+−→ N·
−−→
VI,R
p(t)=(−mg ·sin (α)−→ ux−mg ·cos (α)−→ uz)·˙x·−→ ux=−˙x·mg ·sin (α)
dEc(R)
dt =3 4m·d˙x2 dt =3 2m·˙
x·¨x=−˙x·mg ·sin (α)

−−→
aG,R = ¨x·−→ ux=−2
3·g·sin (α)·−→
ux
Ec(R)+Ep(R)=Eco(R)+Epo(R)
Epo(R)= 0
x·sin (α)Ep(R)=mg ·x·sin (α)
3
4m·˙x2+mg ·x·sin (α) = Eco(R)
Ec(R)−Eco(R)=W(t)
3
2m·˙
x·¨x+ ˙x·mg ·sin (α) = 0
−−→
aG,R = ¨x·−→ ux=−2
3·g·sin (α)·−→
ux
−→
P+−→ N+−→ T=m·−−→ aG,R =m·¨x·−→ ux
m·¨x=−mg ·sin (α) + T
−−→
δG,R =−→ GI ∧−→ N+−→ T=−R·−→ uz∧(N·−→ uz+T·−→ ux) = −R·T·−→ uy
¨
θ
−−→
δG,R =−−−→ δG,Rg =IGy ·¨ θ·−→ uy=1
2m·R2·¨
θ·−→ uy=−R·T·−→ uy
T=−1
2m·R·¨
θ

¨x=R·¨ θ
T=−1
2m·¨x
m·¨x=−mg ·sin (α)−1
2m·¨x
−−→
aG,R = ¨x·−→ ux=−2
3·g·sin (α)·−→
ux
T=−1
2m·¨x=1 3·m·g·sin (α)
N=m·g·cos (α)
T < f ·N1
3·m·g·sin (α)< f ·m·g·cos (α)
tan (α)<3·f
IGy =2
5m·R2