Tension in a rotating ring

  • Thread starter ShakyAsh
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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
 
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Answers and Replies

  • #2
haruspex
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2Tsin(dθ/2) = dmωr^2
Typo: you mean dmrω^2
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
I can't think of a justification for that method. Can you describe your reasoning here?
 
  • #3
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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.
 
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  • #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.
 
  • #5
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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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