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You swing one yo-yo around your head in a horizontal circle. Then you swing another yo-yo with twice the mass of the first one, but you don't change the length of the string or the period. How do the tensions in the strings differ?

  • Physics -

    m v^2/r
    v the same
    r the same
    twice m

  • Physics -

    Wait a minute, this horizontal circle does not mean the string is horizontal.
    r will not be constant in the real case.

  • Physics -

    Say string at angle A from vertical.
    Then r of circle = R sin A if R is string length
    Inward force on mass by string = T sin A
    T sin A = m v^2/r = m v^2/(R sin A)
    T sin^2 A = m v^2/R

    Vertical force on mass by string = T cos A
    T cos A = m g

    period = p = 2 pi r/v = 2 pi R sin A/v
    v = 2 pi R sin A/p
    v^2 = 4 pi^2 R^2 sin^2A/p^2


    T sin^2 A = m v^2/R = m * 4 pi^2 R^2 sin^2A/p^2
    T = m 4 pi^2 R^2/p^2
    interesting, if p is the same when you double m you double the tension
    I suppose that makes sense in the limit of very slow and hanging just about straight down, the tension is mostly m g and has to be double for twice the mass.

  • Physics -

    And we already did the limit for very fast, string horizontal also gives twice the tension for twice the mass.
    Cute problem !

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