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A spinning loop

  1. Aug 18, 2009 #1
    1. The problem statement, all variables and given/known data
    I never thought I would have this kind of elementary problem
    consider a string closed-loop spinning around an axis, and its shape is a circle, I wanted to find the centripital force at each point. (uniform density is assumed). I have problem expressing the mass


    2. Relevant equations



    3. The attempt at a solution
    I apply F=ma , and F=m f(angle), the acceleration is obviously a function of the angle. But how do I write the mass at a point of a continuous string. The formula F=m f(angle) works for 1000, or 10000 mass points spinning around an axis, since the mass is given for each mass point, but what about a string, its continuous. But if I write dm (differential form), it would distort the formula.......
     
  2. jcsd
  3. Aug 18, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    If you have a continuous string, you can pick a point on the loop and consider a piece of arc length ds at that point. Then if the mass of the loop is M and its radius R, the mass of your piece is

    dm = Mds/2πR

    Once you have this, the centripetal force on that element is

    dF = dm ω2r

    where ω is he instantaneous angular speed and r the distance of your element of mass dm to the axis of rotation.

    Is this what you had in mind?
     
  4. Aug 18, 2009 #3
    yea, but I feel weird having to integrate dF. The formula works for finite masses, but as the number goes infinite, it fails. But the formula for continuous rope should have the form similar to that of finite. that's why I feel weird to integrate dF.
     
  5. Aug 18, 2009 #4

    tms

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    Why do you want to integrate? The question, as you posted it, asks for the force at each point, not the total force.
    If you do integrate, the mass will be a constant; you will integrate over [tex]ds[/tex].
     
  6. Aug 18, 2009 #5
    thats exaclty the problem I am having. how to modify that formula for the continuous situation
     
  7. Aug 20, 2009 #6

    tms

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    You have already been given the main part of the answer.
     
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