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
$${F}_{{x}} =\frac{{mgR}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$${T}=\frac{{mgs}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }}=\frac{{mg}}{\:\sqrt{\mathrm{1}−\frac{{R}^{\mathrm{2}} }{{s}^{\mathrm{2}} }}} \\ $$$$\theta_{{k}} =\frac{\mathrm{2}\pi{k}}{{n}},\:{k}=\mathrm{1}...{n}−\mathrm{1} \\ $$$$\varphi=\frac{\pi}{\mathrm{2}}−\frac{\theta_{{k}} }{\mathrm{2}}=\frac{\pi}{\mathrm{2}}−\frac{{k}\pi}{{n}} \\ $$$${r}_{{k}} =\mathrm{2}{R}\:\mathrm{sin}\:\frac{\theta_{{k}} }{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{{k}\pi}{{n}} \\ $$$${F}_{{k}} =\frac{{cq}^{\mathrm{2}} }{{r}_{{k}} ^{\mathrm{2}} }=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\frac{{k}\pi}{{n}}} \\ $$$${F}_{{k},{x}} ={F}_{{k}} \mathrm{cos}\:\varphi=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} \:\mathrm{sin}\:\frac{{k}\pi}{{n}}} \\ $$$${F}_{{x}} =\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}{F}_{{k},{x}} =\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{R}^{\mathrm{2}} }\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\frac{{mgR}}{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }} \\ $$$$\frac{{R}^{\mathrm{3}} }{\:\sqrt{{s}^{\mathrm{2}} −{R}^{\mathrm{2}} }}=\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\lambda^{\mathrm{2}} \\ $$$${with}\:\lambda=\sqrt{\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}} \\ $$$${R}^{\mathrm{6}} +\lambda^{\mathrm{4}} {R}^{\mathrm{2}} −\lambda^{\mathrm{4}} {s}^{\mathrm{2}} =\mathrm{0} \\ $$$${R}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\lambda^{\mathrm{4}} \left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\lambda^{\mathrm{4}} }{\mathrm{27}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\lambda^{\mathrm{4}} \left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\lambda^{\mathrm{4}} }{\mathrm{27}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$ \\ $$$${special}\:{case}:\:{n}=\mathrm{3} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{sin}\:\frac{{k}\pi}{{n}}}=\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\pi}{\mathrm{3}}}+\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\mathrm{2}\pi}{\mathrm{3}}}=\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$$\lambda=\sqrt{\frac{{cq}^{\mathrm{2}} }{\mathrm{4}{mg}}×\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}}=\frac{\mu}{\:\sqrt{\sqrt{\mathrm{3}}}}\:{with}\:\mu=\sqrt{\frac{{cq}^{\mathrm{2}} }{{mg}}} \\ $$$${R}^{\mathrm{2}} =\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)} \\ $$$$\frac{{T}}{{mg}}=\frac{{s}}{\:\sqrt{{s}^{\mathrm{2}} +\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}−\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}−\sqrt[{\mathrm{3}}]{\frac{\mu^{\mathrm{4}} }{\mathrm{3}}\left(\sqrt{\frac{{s}^{\mathrm{4}} }{\mathrm{4}}+\frac{\mu^{\mathrm{4}} }{\mathrm{81}}}+\frac{{s}^{\mathrm{2}} }{\mathrm{2}}\right)}}} \\ $$
Commented by ajfour last updated on 24/Jan/24
$${utterly}\:{amazing}\:{Sir}. \\ $$
Commented by mr W last updated on 25/Jan/24
$${thanks}\:{sir}!\:{the}\:{question}\:\mathrm{2}\:{is}\:{much} \\ $$$${harder}.\:{have}\:{you}\:{tried}? \\ $$
Commented by ajfour last updated on 25/Jan/24
Thats why i havnt tried, but i m starting to think, few hours...
Commented by mr W last updated on 27/Jan/24
$${solution}\:{to}\:{question}\:\mathrm{2}\: \\ $$$${see}\:{Q}\mathrm{203726} \\ $$