insa toulouse 1a mecanique du point examen mai 2010 correction
TRANSCRIPT
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8/9/2019 INSA Toulouse 1A Mecanique Du Point Examen Mai 2010 Correction
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WF1tour
F
WF1tour
F0 2
WF1tour =
2
0F .d =
2
0 b sin(b.).R.d = [ R. cos(b.) ]20 = R. (1 cos(2.b))
b = 1, 2, 3... WF1tour
Fi.e. rot
F= 0
)
b h m M [0, /2]
= ddt
t = 0 = 0
O,ex, ey, ez O, er, e, ek ere
e e ez t = 0R = N+ F N
F e
er e ek
P = m.g.ez = m.g. (cos.er sin. ek)R = Fe + N.ek
aM/ a/ac er e ek
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OM= r.er aM/ =d2
dt2
OM= d2rdt2 .er
a/ =d2 OO
dt2
=0
+
d/dt
OM
=0
+/
/ OM
a/ = .ez (.ez r.er)
a/ = r.2.ez (ez er)
a/ = r.2.ez (sin( ).e)
a/ = r.2. sin.ez e sin( ) = sin
a/ = r.2. sin.e = r.2. sin. (cos.ek + sin.er)
ac = 2./ vM/ = 2..drdt . (ez er)
ac = 2..drdt. sin( ).e = 2..
drdt. sin.e
F Fe Fc = m.aM/ Fm.a/ m.ac = m.aM/(m.g. cos.er m.g. sin. ek) + (Fe + N.ek) + m.r.2. sin. cos.ek + m.r.2 sin
2 .er
2.m..drdt . sin.e
= m.d
2rdt2 .er
er, e, ek
m.g. cos + m.r.2 sin2 = m.d2rdt2
F 2.m.drdt .. sin = 0m.g. sin + N+ m.r.2. sin. cos= 0
r(t)
r(t)
=g cos
2. sin2 [cosh(.t. sin) 1]
d2rdt2
2. sin2 .r = g. cos
r = . sin
d2rdt2
2.r = 0
r(t) = A.e.t + B.e
.t
r(t) = Cte
Cte = g. cos2
r(t) = A.e.t + B.e
.t g. cos2
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t = 0 r = 0 0 = A + B g. cos2
t = 0 v = 0 0 = AB v(t) = drdt = A..e
.t B. .e.t
0=A + B g. cos20= AB
A = B = g. cos22
r(t) = g. cos2 .e
.t+e.t
2
g. cos2
r(t) = g. cos2 . cosh(.t) g. cos2
r(t) = g. cos2 sin2
. [cosh( sin.t) 1]
F
F = 2.m.drdt.. sin
F = 2.m. sin. g. cos2 sin2
. sin. sinh( sin.t)F = 2.m.g. cos sinh( sin.t).e
er
N= m.g. sinm.r.2 cos sinN= m.g. sinm.2 cos sin. g. cos2 sin2 . [cosh( sin.t) 1]
N= m.g. sin2
sin m.g cos2
sin[cosh( sin.t) 1]
N= m.g. sin2 m.g. cos2 . cosh( sin.t)+m.g. cos2
sin
N= m.gm.g. cos2 . cosh( sin.t)sin
N= m.gsin
.
1 cos2 . cosh( sin.t)
cosh(x) = ex+ex2x = 0 [0, /2[ N
t N = 0N
N 0
= /2 N= m.g ek ez
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