Question-203636

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Question Number 203636 by ajfour last updated on 24/Jan/24

Commented by ajfour last updated on 24/Jan/24

$${Take}\:{string}\:{length}\:\boldsymbol{{s}}\:{for}\:{question}\:\left(\mathrm{1}\right). \\ $$

Answered by mr W last updated on 24/Jan/24

Commented by mr W last updated on 27/Jan/24

F_x =((mgR)/( (√(s^2 −R^2 )))) T=((mgs)/( (√(s^2 −R^2 ))))=((mg)/( (√(1−(R^2 /s^2 ))))) θ_k =((2πk)/n), k=1...n−1 ϕ=(π/2)−(θ_k /2)=(π/2)−((kπ)/n) r_k =2R sin (θ_k /2)=2R sin ((kπ)/n) F_k =((cq^2 )/r_k ^2 )=((cq^2 )/(4R^2 sin^2 ((kπ)/n))) F_(k,x) =F_k cos ϕ=((cq^2 )/(4R^2 sin ((kπ)/n))) F_x =Σ_(k=1) ^(n−1) F_(k,x) =((cq^2 )/(4R^2 ))Σ_(k=1) ^(n−1) (1/(sin ((kπ)/n)))=((mgR)/( (√(s^2 −R^2 )))) (R^3 /( (√(s^2 −R^2 ))))=((cq^2 )/(4mg))Σ_(k=1) ^(n−1) (1/(sin ((kπ)/n)))=λ^2 with λ=(√(((cq^2 )/(4mg))Σ_(k=1) ^(n−1) (1/(sin ((kπ)/n))))) R^6 +λ^4 R^2 −λ^4 s^2 =0 R^2 =((λ^4 ((√((s^4 /4)+(λ^4 /(27))))+(s^2 /2))))^(1/3) −((λ^4 ((√((s^4 /4)+(λ^4 /(27))))−(s^2 /2))))^(1/3) special case: n=3 Σ_(k=1) ^(n−1) (1/(sin ((kπ)/n)))=(1/(sin (π/3)))+(1/(sin ((2π)/3)))=(4/( (√3))) λ=(√(((cq^2 )/(4mg))×(4/( (√3)))))=(μ/( (√(√3)))) with μ=(√((cq^2 )/(mg))) R^2 =(((μ^4 /3)((√((s^4 /4)+(μ^4 /(81))))+(s^2 /2))))^(1/3) −(((μ^4 /3)((√((s^4 /4)+(μ^4 /(81))))−(s^2 /2))))^(1/3) (T/(mg))=(s/( (√(s^2 +(((μ^4 /3)((√((s^4 /4)+(μ^4 /(81))))−(s^2 /2))))^(1/3) −(((μ^4 /3)((√((s^4 /4)+(μ^4 /(81))))+(s^2 /2))))^(1/3) ))))

$${F}_{{x}} =\frac{{mgR}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$${T}=\frac{{mgs}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }}=\frac{{mg}}{\:\sqrt{\mathrm{1}−\frac{{R}^{\mathrm{2}} }{{s}^{\mathrm{2}} }}} \\ $$$$\theta_{{k}} =\frac{\mathrm{2}\pi{k}}{{n}},\:{k}=\mathrm{1}…{n}−\mathrm{1} \\ $$$$\varphi=\frac{\pi}{\mathrm{2}}−\frac{\theta_{{k}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{k}\pi}{{n}} \\ $$$${r}_{{k}} =\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta_{{k}} }{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{{k}\pi}{{n}} \\ $$$${F}_{{k}} =\frac{{cq}^{\mathrm{2}} }{{r}_{{k}} ^{\mathrm{2}} }=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\frac{{k}\pi}{{n}}} \\ $$$${F}_{{k},{x}} ={F}_{{k}} \mathrm{cos}\:\varphi=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{sin}\:\frac{{k}\pi}{{n}}} \\ $$$${F}_{{x}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{F}_{{k},{x}} =\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\frac{{mgR}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$$\frac{{R}^{\mathrm{3}} }{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }}=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\lambda^{\mathrm{2}} \\ $$$${with}\:\lambda=\sqrt{\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}} \\ $$$${R}^{\mathrm{6}} +\lambda^{\mathrm{4}} {R}^{\mathrm{2}} −\lambda^{\mathrm{4}} {s}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\lambda^{\mathrm{4}} \left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\lambda^{\mathrm{4}} }{\mathrm{27}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\lambda^{\mathrm{4}} \left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\lambda^{\mathrm{4}} }{\mathrm{27}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$ \\ $$$${special}\:{case}:\:{n}=\mathrm{3} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$$\lambda=\sqrt{\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}×\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}}=\frac{\mu}{\:\sqrt{\sqrt{\mathrm{3}}}}\:{with}\:\mu=\sqrt{\frac{{cq}^{\mathrm{2}} }{{mg}}} \\ $$$${R}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$\frac{{T}}{{mg}}=\frac{{s}}{\:\sqrt{{s}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}}} \\ $$

Commented by ajfour last updated on 24/Jan/24