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Centripetal Force String Tension

  1. Mar 4, 2014 #1
    1. The problem statement, all variables and given/known data

    Two masses M of the same amount are tied to two stings of length L and 2L. If both masses are swung in unison faster and faster, which string will break first?

    2. Relevant equations

    The formula I've been using is Tension = mv^2/r.

    3. The attempt at a solution

    My understanding is that when the string breaks, tension will be 0. If I plug that into the formula the radius/length of the string will not matter. What am I doing wrong here?
  2. jcsd
  3. Mar 4, 2014 #2


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    Welcome to PF!

    But you are looking for the tension in the string just before it breaks.
  4. Mar 4, 2014 #3

    Andrew Mason

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    Since the two masses have the same rotational speed, to compare the forces you may wish to use: Fc = mω^2r

  5. Mar 4, 2014 #4
    So what variable in this equation changes due to the string length change and how does it affect your force?
  6. Mar 5, 2014 #5
    What is the value ω stand for? I don't think I'm familiar with that equation.

    Because the radius is the only thing that changes in the equation, this is what I'm thinking:

    Fc = mv2/r


    Fc = mv2/2r
    2Fc = mv2/r

    Therefore the longer the string the larger the centripetal force/tension. Am I correct or completely off?
  7. Mar 5, 2014 #6

    Andrew Mason

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    ω is the angular speed in radians per second. There are 2π radians in a circle so if the number of rotations per second is f the angular speed is 2πf. Since T = 1/f you can express ω = 2π/T. Tangential speed v = ωr so mv2/r = mω2r = m4π2r/T2

    Can you see how to use this to compare the centripetal force on each of these two rotating masses?

    You have provided an excellent example of why getting the right answer is not very important when you are a student.

    Are the tangential speeds the same for each rotating mass? Can you compare the tangential speeds? (hint: Can you see why using ω makes this a lot easier?)

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