INSA Toulouse 1A Mecanique Du Point Examen Mai 2008 Correction

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  • 8/9/2019 INSA Toulouse 1A Mecanique Du Point Examen Mai 2008 Correction

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    ere

    nd

    L = OM m.v

    OM = 0. e v = 0.. e L = m.20.. ez

    M =

    i

    OM Fi OM = 0. e F1 = P = m.g.ey F2 = T = T. e

    M = 0. e (m.g.ey) + 0. e (T. e) =0

    = m.g.0. ( e ey) = m.g.0. sin( ). ez = m.g.0. sin(). ez

    M = ddtL

    m.g.0. sin(). ez = ddt

    m.20.. ez

    = m.20.. ez +

    g0

    sin = 0

    = g0 [] = [L].[T]2

    [L]1/2 = [T]1

    ex = cos . ex + sin . ey ey = sin . ex + cos . ey

    OM = R2 cos . ex + R2 sin . ey v(M/) = R2.. sin . ex + R2.. cos . ey

    a(M/) = R2.2. cos . ex R2.2. sin . ey

    e = e. ez = . ez

    v(M/) = R1.

    . sin . ex + R1.. cos . ey

    ddt ( OO)

    + R2.. cos .

    ey R2.

    . sin .

    ex

    e OM

    +R2.. sin .

    ex + R2.

    . cos .

    ey

    v(M/)

    v(M/) = R1.. sin . ex + R1.. cos . ey 2R2.. sin . ex + 2R2.. cos . eyv(M/) =

    R1.. sin 2R2.. sin2

    . ex +

    R1.. cos + 2R2.. cos2

    . ey

    a(M/) = R1 .2. cos . ex R1.

    2 . sin . ey

    d2

    dt2

    OO

    + 0

    d edt

    OM

    +

    R2 .2. cos . ex R2.

    2 . sin . ey

    e( eOM )

    +

    2.R2.2. sin . ey 2.R2 .

    2. cos . ex

    2 ev(M/)

    +

    R2 .2. sin . ey R2 .

    2 . cos . ex

    a(M/)

    a(M/) = R1.2. cos . ex R1.2. sin . ey 4R2.2. cos . ex 4R2.2. sin . ey

    a(M/) = R1.2. cos . ex R1.2. sin . ey 4R2.2. cos2. ex 4R2.2. sin2. ey

    a(M/) =R1.2. cos 4R2.2. cos2

    . ex +

    R1.2. sin 4R2.2. sin2

    . ey

    OM = OO + OM = R1 cos . ex + R1 sin . ey OO

    + R2 cos . ex + R2 sin . ey

    OM

    OM = R1 cos . ex + R1 sin . ey + R2 cos2. ex + R2 sin2. ey

    OM = (R1 cos + R2 cos2) . ex + (R1 sin + R2 sin2) . ey

    ddt

    OM

    =

    R1.. sin 2R2.. sin2

    . ex +

    R1.. cos + 2R2.. cos2

    . ey

    d2

    dt2

    OM

    =

    R1.2. cos 4R2.2. cos2

    . ex +

    R1.2. sin 4R2.2. sin2

    . ey

  • 8/9/2019 INSA Toulouse 1A Mecanique Du Point Examen Mai 2008 Correction

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    P2 = m.g.ey UP = m.g.y + Cte R2 = +m.g.ey

    N = N. ex

    F = k.x.ex U1 = 12 .k.x2

    f = f. v||v|| = f. ex f Uf = f.x + Cte

    || vf1 || = 0 || vf2 || = vi1

    Eif Eic =

    WF =xM0

    F . d +xM0

    f . d =xM0

    dU1 +xM0

    f ex.dx. exEfc Eic = 12k.x2M f.xM Efc = 0 x = xM

    1

    2

    m. vi

    12

    =

    1

    2

    k.x2

    M f.xM

    12k.x

    2M + f.xM 12m.(vi1)2 = 0

    g

    V = .R2. (H z)

    M = ..V

    A = ..V.g = ..g..R2. (H z) A = . (H z) . ez = ..g..R2

    P = m.g.ez A = . (H z) . ez

    a = z. ez

    F = m.a

    P + A = m.a m.g.ez + . (H z) . ez = m.z. ez z

    z + m .z =.H

    m g

    z(t) =

    C1.ei

    m .t + C2.e

    i

    m .t

    zp = Cte

    zp = H

    m.g

    z(t) = C1.ei

    m .t + C2.e

    i

    m .t + H m.g

    z(t = 0) = z0 z(t = 0) = 0 C1 + C2 + H m.g = z0 C1 C2 = 0 C1 = C2 =

    z0H+mg

    2

    z(t) =

    z0 H + mg

    cos

    m .t

    + H m.g

    =

    m =

    ..g..R2

    m =

    10001.6103.140.052

    31.4=

    10001.6103.140.0025

    3.1410=

    1.6 2.5 = 4 = 2 s1 T = 2. = s