A ring of radius r made of wire of density ρ is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring as shown in the figure. Determine the angular velocity (in$$ $$rad/s) of ring at which the ring breaks. The Wire breaks at tensile stress σ. Ignore gravity. Take σ$$/$$ρ $$=$$ $$4$$ and r $$=$$ $$1$$ m.

Question

A ring of radius r made of wire of density ρ is rotated about a stationary vertical axis passing through its centre and perpendicular to the plane of the ring as shown in the figure. Determine the angular velocity (in$$ $$rad/s) of ring at which the ring breaks. The Wire breaks at tensile stress σ. Ignore gravity. Take σ$$/$$ρ $$=$$ $$4$$ and r $$=$$ $$1$$ m.

Infinity Learn · Accepted Answer

$$2Tsin$$⁡Δθ$$2=dm$$ω$$2r2T$$Δθ$$2=$$ρ×A×$$r2$$ω$$2$$∆θ σ$$=TA=$$ρ$$r2$$ω$$2$$∴ ω$$=1r$$σρ$$=2$$ $$rad/s$$

Elastic behaviour of solids

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$\begin{array}{l} 2 Tsin \frac{Δθ}{2} = {dmω}^{2} r \\ 2 T (\frac{Δθ}{2}) = ρ \times A \times r^{2} ω^{2} ∆ θ \\ σ = \frac{T}{A} = {ρr}^{2} ω^{2} \\ ∴ ω = \frac{1}{r} \sqrt{\frac{σ}{ρ}} = 2 rad / s \end{array}$

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Elastic behaviour of solids

Remember concepts with our Masterclasses.

2Tsin⁡Δθ2=dmω2r2TΔθ2=ρ×A×r2ω2∆θ σ=TA=ρr2ω2∴ ω=1rσρ=2 rad/s

Ready to Test Your Skills?

free study material

$\begin{array}{l} 2 Tsin \frac{Δθ}{2} = {dmω}^{2} r \\ 2 T (\frac{Δθ}{2}) = ρ \times A \times r^{2} ω^{2} ∆ θ \\ σ = \frac{T}{A} = {ρr}^{2} ω^{2} \\ ∴ ω = \frac{1}{r} \sqrt{\frac{σ}{ρ}} = 2 rad / s \end{array}$