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Fictious force: Cylinder on an Accelerating Plank

  1. Feb 24, 2012 #1

    Leb

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    1. The problem statement, all variables and given/known data

    Problem is described in the picture
    Cylinder on an accelerating plank.jpg
    I do not understand how can [itex]\alpha^{'}R=a^{'}[/itex].
    The dimensions do not seem correct. Angular velocity x distance from the origin = tangential velocity, is that correct ? How can this equal acceleration then ?

    3. Attempt to solution

    I think that [itex]\alpha^{'}R=a^{'}[/itex] would hold only if we would consider a unit time. That is:
    [itex]\alpha^{'}R=\frac{d\theta}{dt}R[/itex], which, more by knowing the anticipated result in this case, than by logic, gives [itex]\frac{d\theta}R = a^{'}dt[/itex] which is now dimensionally OK, I think...
     
    Last edited: Feb 24, 2012
  2. jcsd
  3. Feb 24, 2012 #2

    tiny-tim

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    Hi Leb! :smile:
    No, a is (linear) acceleration, and α is angular acceleration

    s = rθ

    v = rω

    a = rα :wink:
     
  4. Feb 24, 2012 #3

    Leb

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    Thanks tiny-tim !
    It was strange to see an alpha instead of omega, but since the author was at times using random notation (such a ro, for distance and w for density...) made me forget about usual notation:)
     
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