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A thin ring of radius R is made of material of density ρ and Young’s modulus E. It is spun in its own plane, about an axis through its center, with angular velocity ω. Determine the amount (assumed small) by which its circumference increases.

Step-by-step

Let the tension in the ring be T. Its resolved component acting along the radius towards the centre of rotation is 2T sin(∆θ/2) ≈ T∆θ and this must balance the centripetal force of R∆θAρR{{\omega }^{2}}.
It follows that the longitudinal stress in the ring, T /A, is sρ{{R}^{2}}{{\omega }^{2}} ; the strain ε is {{E}^{-1}} times this. Finally, the increase in circumference, given by 2πRε, is 2πρ{{R}^{3}{{\omega }^{2}}}/E.

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