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Rotation of rigid bodies: yo-yo

  1. Apr 14, 2006 #1
    consider a yo-yo consisting of two cylinders of radius R1, (combined) mass M1, glued to another smaller cylinder of radius R2, mass M2. find the final velocity after falling a height h using newton's second law. assume the string is vertical.

    [​IMG]

    the inner radius R2 is R0, the outer radius R1 is R in this picture

    so, i started off by applying f = ma to M2..

    Fnet, y = (M1 + M2) - T = (M1 + M2)(a)cm

    tnet, ext = T(R1) = I(alpha), alpha = (a)cm/R1

    i calculated the moment of inertial to be the sum of the moments of inertia of the cylinders..

    I = 1/2*(M1*R1^2 + M2*R2^2)

    solving for (a)cm, i get a very nasty formula..

    a = (M1 + M2)g / ( ( I/(2R1^2)) + (M1 + M2))

    my original plan was to integrate this wrt time to get a velocity.
    but this formula doesn't involve anything that changes with time.. so how do i obtain the final velocity? am i even on the right track?
     
    Last edited: Apr 14, 2006
  2. jcsd
  3. Apr 14, 2006 #2

    Fermat

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    Homework Helper

    It looks ok.
    As you noticed, the expression for a doesn't involve anything that changes wrt time. In other words, a is constant!

    If you had an object that was falling under gravity, what would its velocity be after having fallen a distance h?

    So then, what if g = a?
     
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