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Tension in a rotating ring

  1. Aug 18, 2012 #1
    Hello people,

    So i found out the tension in a ring rotating with constant angular velocity (in gravity free space)

    Considering a small element of mass dm - tension will provide the centripetal force,
    2Tsin(dθ/2) = dmrω^2
    sindθ ≈ dθ
    dm = m/2πr ds
    ds = rdθ

    T = (mrω^2)/2π


    Now, the other method
    K.E. = K = 1/2 Iω^2 = 1/2 mr^2 ω^2

    If we increase the radius from r to r+dr, then work done by tension
    dW = T d(2πr) = dK
    T = 1/2π dK/dr
    T = (mrω^2)/2π


    Even though i get the same result, i have a doubt whether the second method is correct
    I know that F=-dU/dr , but whether T=dK/ds , i don't know

    Also, i want to know the general approach of calculating tension in situations like electro-magnetic fields, rotation & all.

    Regards
     
    Last edited: Aug 19, 2012
  2. jcsd
  3. Aug 18, 2012 #2

    haruspex

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    Typo: you mean dmrω^2
    I can't think of a justification for that method. Can you describe your reasoning here?
     
  4. Aug 19, 2012 #3
    Yeah dmrω^2 , sorry

    As i said, if radius of the ring is increased by dr, then work done by the tangential force tension will be T*(change in circumference) which will be equal to the change in kinetic energy which in this case is the rotational energy.
     
    Last edited: Aug 19, 2012
  5. Aug 19, 2012 #4

    K^2

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    Yes, the approach is valid. If you have doubts, you can replace the ring by N discrete points with tension T between them and derive dW=T ds for N->Inf.
     
  6. Aug 19, 2012 #5
    Dividing it into N discrete points, why didn't i think of that?

    Anyways, thank you very much.
    I think i understand it now.
     
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