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Playing with a yoyo (Angular momentum problem)

  1. Mar 9, 2008 #1
    [SOLVED] Playing with a yoyo (Angular momentum problem)

    1. The problem statement, all variables and given/known data
    This problem is from Gregory. This is only the first part of the problem, since I understand how the solution to the first part implies the second part.

    A uniform circular cylinder (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 Z(t) at time t. Find the acceleration of the yo-yo.

    a: the radius of the yo-yo
    m: the mass of the yo-yo
    I: moment of inertia of the yo-yo
    v: velocity of the center of mass of the yo-yo
    [tex]\omega[/tex]: angular velocity of the yo-yo
    g: acceleration of gravity
    Z(t): the displacement of the support
    T: the tension force the string exerts on the yo-yo
    [tex]\tau[/tex]: torque of the yo-yo

    2. Relevant equations
    Since the yo-yo is a cylinder, I = ma^2/2
    [tex]\tau[/tex] = T*a = I [tex]\omega[/tex]'
    m v' = -mg + T + mZ''(t)
    v = -a[tex]\omega[/tex]

    3. The attempt at a solution
    Work a little substitution magic:
    m v' = -mg + I [tex]\omega[/tex]'/a + mZ''(t)
    m v' = -mg - ma^2/2 v'/a^2 + mZ''(t)
    m v' = -mg - m v'/2 + mZ''(t)
    3/2m v' = -mg + mZ''(t)
    v' = -2/3g + 2/3Z''(t)

    The solution given in the book is v' = -2/3g + 1/3Z''(t). So, somewhere along the line I missed something with Z''(t). Can anybody guess where I went wrong?
     
  2. jcsd
  3. Mar 10, 2008 #2
    Nevermind, I got it. Turns out it was omega = (v'-z')/a. Sometimes just typing it out seems to help, eh?
     
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