A thin uniform elastic rope forms a circular ring of natural radius R. The material has density ρ and Young's modulus Y. The ring rotates about its centre in its own plane with angular speed ω. Assuming small strain, what is the fractional increase in the circumference of the ring?
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Question

A thin uniform elastic rope forms a circular ring of natural radius R. The material has density ρ and Young's modulus Y. The ring rotates about its centre in its own plane with angular speed ω. Assuming small strain, what is the fractional increase in the circumference of the ring?

Options

G1 Physics · Accepted Answer

Consider a small ring element subtending angle dθ. If T is the tension, the inward resultant of the two end tensions is approximately T dθ. Mass of the element is dm = λR dθ, where λ is mass per unit length. The required centripetal force is dm ω²R = λω²R² dθ. Therefore, T dθ = λω²R² dθ. Hence, T = λω²R². Since λ = ρA, stress in the rope is σ = T/A = ρω²R². Therefore circumferential strain is ε = σ/Y = [ρω²R²] / Y. The fractional increase in circumference equals this strain. Correct option: B.