INSA Toulouse 1A Mecanique Du Point Examen Mai 2009 Correction
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Transcript of INSA Toulouse 1A Mecanique Du Point Examen Mai 2009 Correction
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8/9/2019 INSA Toulouse 1A Mecanique Du Point Examen Mai 2009 Correction
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ere
nd
a
a ab cos sin tan
m1 m2
v1 v2
v1 v2 v1 v2
v
1 v
2 v1 v2 v1 v2
v1 =2v2 + (k 1).v1
k + 1v2 =
2.k.v1 + (1 k).v2k + 1
k = m1/m2
m1 m2 v1 v2
v1 = v1 v2 = 0
v
1 = v2
v
2 = v1
v1 = v1 v2 = 2v1
tL
t
0 < t < tL
t > tL g = 10m.s2
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{O,Ox,Oy} {ex, ey, ez} v = v0.ex {O, Ox, Oy}
m = 160g
ddt
(t) = .t
= 10 rad.s2
{e, e, ez} O
M
t = 0
= 0
tL
L = 225o
= 0.6m
L = 1.5 m
vL [Ox) t = tL
tL L tL
tL d/dt = .t (t) = .t2/2+Cte
t = 0
= 0
Cte = 0
tL =
2L
L L =
54 rd tL =
2L
=
25104 0.9s
ex ey {e, e, ez}
ex = cos .e sin .e ey = sin .e cos .e
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OM
v
a
{e, e , ez}
OM = OO + OM
OM = x.ex + L.ey + .e
OM = x. (cos e sin .e) + L. ( sin .e cos .e) + .eOM = (x. cos L. sin + ) .e (x. sin + L. cos ) .e
v =d OM
dt=
d
dt(x.ex + L.ey + .e) =
dx
dt.ex + .
dedt
= v0. (cos .e sin .e) + . .t.e = v0. cos .e + (..t v0. sin ) .e
a =dv
dt=
d
dt(v0.ex + ..t.e) =
d
dt(..t.e) = .2.t2.e + ..e
{ex ey, ez}
tL v0 = 7 m.s1
||v|| =
(v0. cos )2 + (..t v0. sin )2
||v
||= v20 + ..t. (..t
2.v0 sin )
||v|| =
v20 + ..t.
..t 2.v0 sin
.t2
2
e = cos .ex sin .ey e = sin .ex cos .ey
e e v
v = (v0 ..t. sin ) ex ..t. cos .ry
tan =vyvx
t = tL
tan =..tL. cos L
(v0 ..tL. sin L) =..tL. cos
.t2L2
v0 ..tL. sin.t2
L
2
= 19.26o
P = m.g.ey
T
T
Tn = Tn.e
F = F.e
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t = tL
a = ..t.e .2.t2.e
F = P + Tn + F = m.g.ey Tn.e + F.e
m.g.ey Tn.e + F.e = m.
..t.e .2.t2.e
m.g. ( sin .e cos .e) Tn.e + F.e = m.
. .t.e .2.t2.e
(m.g. sin Tn) .e + (m.g. cos + F) .e = m...e m..2.t2.e
e m.g. sin Tn = m..2.t2
Tn = m.g. sin + m..2.t2
tL | Tn| N = 0.16 10 sin(5/4) + 0.16 0.6 102 0.92 6.4 N
tL = 19.3o
y0 = 1.92 m vL = 11.4m.s1
fair = .v = 0.05 tL
ex
ey
vx(t) = vL. cos .exp
.tm
vy
(t) = vL
. sin +g.m
.exp.t
m
g.m
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P = m.g.ez fair = .v = .(vx.ex + vy.ey) a = dvdt = d(vxe+vy .ey)dt =dvxdt .ex +
dvydt .ey
(1) m.dvxdt
= .vx
(2) m.dvydt
= .vy mg
vx = C.exp.tm
vy = D.exp.tm
m.g
vx(t=0) = vL. cos vy(t=0) = vL. sin C = vL. cos D = vL. sin + m.g/
vx(t) = vL. cos .exp
.tm
vy(t) =
vL. sin +
g.m
.exp
.tm
g.m
x(t)
y(t)
x(t) = m
.vL. cos .exp
.tm
+ C
y(t) =
m.vL. sin
+
g.m2
2
.exp
.tm
g.m.t
+ D
t = 0
x = 0
y = y0
x(t) =m
.vL. cos .
1 exp
.tm
y(t) =
m.vL. sin
+
g.m2
2
.
1 exp
.tm
g.m.t
+ y0
vf
vf = 35.45 km.h1
Efc Eic = W( F) = Wfair + WP
WP =fi
P .dl =fi
m.g.ey.(dxex + dyey) =0y0
m.g.dy = m.g.y0
Efc Eic = 12 .m.v2f 12 .m.v2L
Wfair = Ec WP = 12 .m.(v2f v2L) m.g.y0
vL = 11.4 m.s1
vf = 10.12 m.s1
m = 0.160Kg
g = 10m.s2
y0 = 1.92 m Wfair = 5.28 J