Corrigé de l`examen de mécanique newtonienne du 16

A B
k2
`0~xA~xB
V=k2
2(k~xA−~xBk − `0)2.
A
k~xA(0) −~xB(0)k=`0
B
A
R
x=a(θ+ sin θ)
y=a(1 −cos θ)
yG(0) = 2a

~rG~r M =mA+mB
µ=mAmB
M
~rG=mA~xA+mB~xB
M
~r =~xB−~xA
⇔
~xA=~rG−mB
M~r
~xB=~rG+mA
M~r
.
~vA=˙ ~rA~vB=˙ ~rB~vG=˙ ~rG~ P=
M~vG
E=1
2(mAv2
A+mBv2 B)+V=1
2(Mv2
G+µ˙
~r2)+ k2
2(r−l0)2
E0=1
2µ˙
~r2+k2
2(r−l0)2.
~
L=mA~rA×~vA+mB~rB×~vB=M~rG×
~vG+µ~r ×˙ ~r
~
L0=µ~r ×˙ ~r .
~
L0z
z= 0 L0=r2˙ θ
E0=1
2µ˙r2+L02 2µr2+k2 2(r−l0)2.
r= 0 L0= 0
E0
~vA(0) = ~ 0
˙
~r(0) = ~vB(0) ~vG(0) = mB~vB(0)
M

~xA(0) = ~ 0~xB(0) = l0~ 1x=~r(0) ~ L0˙ ~r(0) =
˙r(0)~ 1x=~vB(0) x
t∗
E0=1
2µv2
B(0) = 1
2µ˙r2(t∗) + k2 2l2
0
|vB(0)| ≥ ωl0ω2=k2
µ
|vB(0)|
|vB(0)|=ωl0
E0=k2l2
0
2=1 2µ˙r2+k2 2(r−l0)2.
¨r=−ω2(r−l0)r(0) = l0|˙r(0)|=ωl0
r=l0±l0sin(ωt).
vB(0) −
t∗=π
2ω
xG(t∗) = rG(0) + vG(0)t∗=mBl0
M(1 −π 2)
+
t∗=3π
2ωxG(t∗) = mBl0 M(1 + 3π 2)

~ω =−ω~ 1b
~
1b=±~ 1z
~vG=vG~ 1t
~
GQ =−R~ 1nQ
~vQ=~vG+~ω ×~ GQ = (vG−Rω)~ 1t=~ 0.
vG=Rω
E=1
2Mv2
G+1
2GIzzω2+MgyG.
GIij =2
5MR2δij
E=1
2Mv2
G+1
5MR2ω2+MgyG=7 10Mv2
G+MgyG.
vG(0) = 0 yG(0) =
2a yQ,min = 0
yG,min =R
E= 2Mga =7
10Mv∗2
G+MgR ⇔v∗ G=r10
7g(2a−R)

~r = (a(θ+ sin θ), a(1 −cos θ))
θ
~rG=~r +R~ 1n
d~r
dθ = (a(1 + cos θ), a sin θ) = 2acos θ 2(cos θ 2,sin θ 2)
~
1t=±(cos θ
2,sin θ 2)~
1n= (−sin θ
2,cos θ 2)
~rG=a(θ+ sin θ)−Rsin θ
2, a(1 −cos θ) + Rcos θ 2
~vG=d~rG
dt =2acos2θ 2−R 2cos θ 2,2acos θ 2sin θ 2−R 2sin θ 2˙
θ
=±2acos θ
2−R 2˙
θ~ 1t
.
E=7
10M2acos θ 2−R 22
˙
θ2+Mg 2asin2θ
2+Rcos θ 2.
θ(0)
yG(0) = 2a= 2asin2θ(0)
2+Rcos θ(0) 2⇒cos θ(0) 2=R 2a.
T=Z2 arccos R
2a
02acos θ
2−R 2
q10
7ME−Mg 2asin2θ 2+Rcos θ 2
dθ