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In the given question with constant angular velocity. It asks us to find longitudinal stress at each of the positions. Now I'm not even sure what longitudinal stress is but here is what I tried as A is the farthest it will have the highest Radius and hence highest centripetal force. Therefore component of tension should be the highest at A. But the answer is given is that it's minimum at A. Can someone tell me where am I going wrong?

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    $\begingroup$ Exact duplicate: physics.stackexchange.com/questions/659283/… $\endgroup$
    – ACB
    Aug 15, 2021 at 16:47
  • $\begingroup$ I would guess that the longitudinal stress is the tension in the hoop which is tangential to the hoop at each point. $\endgroup$
    – R.W. Bird
    Aug 15, 2021 at 17:05

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Hint: It depends on how much mass it outside the 'radius of rotation'.

For example at B and C, there is tension there, as a centripetal force is needed to make the half circle (green) between B and C rotate. You would need to integrate $\delta m \times \omega^2r$ for each mass element in the green section, where $r$ is the distance of the mass element from $O$.

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At A there is no mass outside that radius, so there can be low tension as a centripetal force isn't required.

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  • $\begingroup$ Now according to me for A we would have tension along the ring in direction of AB and AC and the centripetal force on A is dm w^2 2R which is provided by the component of tension, where am I going wrong? $\endgroup$ Aug 15, 2021 at 18:13
  • $\begingroup$ @ BlackThunder there are two things reducing the tension at A 1) there is no mass outside the 2R radius that the tension has to 'hold in', whereas at B and C the tension there has to hold in the green section. Also 2) the direction of any tension at B and C all causes 'longitudinal stress' but near A the tension the centripetal force would be at right angles to the hoop. $\endgroup$ Aug 15, 2021 at 18:20
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I would think that the tensions at points B and C should combine to produce the centripetal acceleration of the center of mass for the right side of the hoop. I currently have no idea how you would find the tensions at O and A.

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