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Tension in String

  1. Nov 15, 2005 #1
    Let's say you have a yo-yo that is constructed from two circular masses connected by a small hub. If the radius of the hub is r and the radius of the circular masses is R, how can I find the tension in the string when the yo-yo is released? Assume each circular mass has mass M and the hub is m.

    I know a bit about angular kinematics and dynamics, but I am having trouble applying that here. I am going to take a guess here and say that the tangential acceleration (the magnitude) is g (9.8 m/s2). The reason I say this is because the yo-yo would have a downward acceleration of g had it not been connected to any string. Because the string is connected, an object on the perimeter of the yo-yo will accelerate in the tangent direction at the same rate as if the yo-yo was in free-fall. Is this reasoning correct? If so, I believe I can make more progess on the problem.

    Thank you.
  2. jcsd
  3. Nov 15, 2005 #2


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    The tension, T, in the string is applying a torque, tau = Tr, where r is the radius of the hub.

    Then tau = I.alpha
    and a = alpha*R

    where a is the tangential acceleration and R is the radius of the circular mass.

    Edit: I is the inertia of the combined masses.

    Edit2: correction to the expression for linear acceln.
    Last edited: Nov 16, 2005
  4. Nov 15, 2005 #3


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    Isn't the tension in the string T = 2Mg + mg ???
  5. Nov 16, 2005 #4
    Am I right about the tangential acceleration though?

    Thanks for the response.
  6. Nov 16, 2005 #5


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    Not absolutely certain, but it doesn't seem right.

    Your argument would apply to every point in the yo-yo, including diametrically opposite points. Your argument would give both those points a downward accln of g, but we know that they would would start to move in opposite dirns.!!

    I'm not absolutely sure about what I said about the Tension in the string - it being equal to 2Mg + mg.
    Somehow or other it feels as though the rotation of the yo-yo would offset some of the string's tension.
    However, at the start, there is no rotation, So there won't be any offset to the tension then, so (I think) the initial tension in the string is T = 2Mg + mg.

    And, if that assumption is true, then you can quite easily show that the initial tangential accln isn't neccesarily g, but depends on I.
    Last edited: Nov 16, 2005
  7. Nov 16, 2005 #6
    Of course points on opposite sides are travelling in opposite directions, but when I say "g," I mean the magnitude of the tangential acceleration (not caring about direction).

    Thanks again Fermat.
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