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Lagrange - yo-yo with moving support

  1. May 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A uniform circular cylinder of mass `m' (a yo-yo) has a light inextensible string wrapped around it so that it does not slip. The free end of the string is fastened to a support and the yo-yo moves in a vertical straight line with the straight part of the string also vertical. At the same time the support is made to move vertically having upward displacement [itex]Z(t)[/itex] at time `t'. Find the upward acceleration of the yo-yo.

    2. Relevant equations

    Define `a' as the radius of the cylinder.
    Define theta as the rotation angle of the yo-yo.

    [tex]V = -mg(a\theta - Z)[/tex]

    [tex]T = 3/4 m v^2 = 3/4 m (a \dot{\theta} - \dot{Z})^2[/tex]

    [tex]L = T-V[/tex]

    Use, [tex]\frac{d}{dt} ( \frac{dL}{d\dot{\theta}} ) - \frac{dL}{d\theta} = 0[/tex]

    3. The attempt at a solution

    [tex]\frac{d}{dt} ( \frac{dL}{d\dot{\theta}} ) = 3/2m(a\ddot{\theta} - \ddot{Z})[/tex]

    [tex]\frac{dL}{d\theta} = mga[/tex]

    So solving gives me,
    [tex] \ddot{\theta} = \frac{2/3g +\ddot{Z}}{a} [/tex]
    (downwards angular acceleration.)

    Therefore upwards acceleration of the yo-yo is,

    [tex] \ddot{z} = -a \ddot{\theta} = \ddot{Z}-2/3g[/tex].

    But I'm missing a factor of 1/3 in front of [tex]\ddot{Z}[/tex] or,

    [tex]\ddot{z} = 1/3(\ddot{Z}-2g)[/tex].
     
    Last edited: May 26, 2009
  2. jcsd
  3. May 26, 2009 #2

    diazona

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    Homework Helper

    It seems that you didn't correctly take into account the rotation of the yo-yo... the kinetic energy of the system consists of two parts, rotation and translation. The latter just arises from the change in height of the yo-yo, with [tex]v = a\dot\theta - \dot{Z}[/tex]. But there is also the rotation of the yo-yo around its axis, specified by [tex]\omega = \dot\theta[/tex] and the moment of inertia [tex]I = \frac{1}{2}ma^2[/tex]. When you put all that together, the kinetic energy comes out to be slightly different than what you wrote.
     
  4. May 26, 2009 #3
    So [tex]L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)][/tex] ?

    [tex]\frac{d}{dt}\frac{dL}{d\dot{\theta}} = ma(a\ddot{\theta}-\ddot{Z})+1/2ma^2\dot{\theta}[/tex](*)
    [tex]\frac{dL}{d\theta} = mga[/tex](**)

    (*)-(**)=0

    I'm getting [tex]\ddot{\theta} = \frac{2}{3a} (\ddot{Z} - g)[/tex].

    Then the linear acceleration of the yo-yo relative to the support is,

    [tex]\ddot{z} = a \ddot{\theta} = 2/3(\ddot{Z}+g)[/tex]

    So relative to the ground I'd have acceleration,

    [tex]\ddot{z}' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})[/tex]

    So the upwards acceleration would be,

    [tex]- \ddot{z} ' = 1/3(\ddot{Z}-2g)[/tex]

    Does this look ok?
     
    Last edited: May 26, 2009
  5. May 26, 2009 #4

    diazona

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    Homework Helper

    Just a couple of typos:
    [tex]L = T-V = [1/2m(a\dot{\theta}-\dot{Z})^2 + 1/4ma^2\dot{\theta}^2] - [-mg(a\theta-Z)][/tex]
    and
    [tex]\ddot{z}\,' = 2/3(\ddot{Z}+g) - \ddot{Z} = 1/3(2g-\ddot{Z})[/tex]
    Other than that it looks reasonable. Note that the upwards acceleration will be [tex]+\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases upwards (which is what it looks like you did), or [tex]-\ddot{z}\,'[/tex] if [tex]z'[/tex] is defined such that it increases downwards.
     
  6. May 26, 2009 #5
    Thanks, fixed them. I just assigned the variable `z' randomly (yeah, sloppy hehe) but thanks for the reminder.

    If I wanted to now find total energy at a time `t' should I use the [tex]\ddot{\theta}[/tex] I just found and integrate twice over t'=0..t then sub into T+V as a have defined above? Looks like a lot of algebra heh.
     
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