INSA Toulouse 1A Mecanique Du Point Examen Mai 2010
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Transcript of INSA Toulouse 1A Mecanique Du Point Examen Mai 2010
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8/9/2019 INSA Toulouse 1A Mecanique Du Point Examen Mai 2010
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ere
nd
a
p Ec
M
m
R
O
{ e e ez } (t)
F = b sin(b.).e + T.e b b = 1, 2, 3... T
F
e t = 0
= 0
v =
2Rm (1 cos(b.))
d2dt2 =
vR2
dvd
v =
2Rm (1 cos(b.))
T
v =
2Rm
(1 cos(b.))
WF1tour
F
WF1tour
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b
h
m
M
[0, /2]
= ddt
t = 0
= 0
O,ex, ey, ez O, er, e, ek er
e
e e ez t = 0
R =
N+
F
N
F
e
er e ek
aM/ a/ ac er e ek
m.g. cos + m.r.2 sin2 = m.d2r
dt2
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]
F
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