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Question Number 47422 by ajfour last updated on 09/Nov/18

Commented by ajfour last updated on 09/Nov/18

A vertical ellipsoid shell of frictionless surface holds a flexible rope ring of mass m, radius R. Find the tension in the rope.

$${A}\:{vertical}\:{ellipsoid}\:{shell}\:{of} \\ $$$${frictionless}\:{surface}\:{holds}\:{a} \\ $$$${flexible}\:{rope}\:{ring}\:{of}\:{mass}\:{m}, \\ $$$${radius}\:{R}.\:{Find}\:{the}\:{tension}\:{in}\: \\ $$$${the}\:{rope}. \\ $$

Answered by MrW3 last updated on 09/Nov/18

eqn. of ellipse: (x^2 /b^2 )+(y^2 /a^2 )=1 (x/b^2 )+(y/a^z )y′=0 at x=R: (R/b^2 )+(k/a^2 )tan θ=0 (R^2 /b^2 )+(k^2 /a^2 )=1 ⇒(R^2 /b^2 )+(1/a^2 )(−((a^2 R)/(b^2 tan θ)))^2 =1 a^2 R^2 ×(1/(tan^2 θ))=b^2 (b^2 −R^2 ) a^2 R^2 ×(1/(tan^2 θ))=b^2 (b^2 −R^2 ) tan^2 θ=((a^2 R^2 )/(b^2 (b^2 −R^2 ))) ⇒tan θ=(a/b)×((R/b)/(√(1−(R^2 /b^2 )))) (T/R)=ρg tan θ=((mg)/(2πR))×(a/b)×((R/b)/(√(1−(R^2 /b^2 )))) ⇒T=((mg)/(2π))×(a/b)×((R/b)/(√(1−(R^2 /b^2 ))))

$${eqn}.\:{of}\:{ellipse}: \\ $$$$\frac{{x}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}}{{b}^{\mathrm{2}} }+\frac{{y}}{{a}^{{z}} }{y}'=\mathrm{0} \\ $$$${at}\:{x}={R}: \\ $$$$\frac{{R}}{{b}^{\mathrm{2}} }+\frac{{k}}{{a}^{\mathrm{2}} }\mathrm{tan}\:\theta=\mathrm{0} \\ $$$$\frac{{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{{k}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\Rightarrow\frac{{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\left(−\frac{{a}^{\mathrm{2}} {R}}{{b}^{\mathrm{2}} \mathrm{tan}\:\theta}\right)^{\mathrm{2}} =\mathrm{1} \\ $$$${a}^{\mathrm{2}} {R}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\theta}={b}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{R}^{\mathrm{2}} \right) \\ $$$${a}^{\mathrm{2}} {R}^{\mathrm{2}} ×\frac{\mathrm{1}}{\mathrm{tan}^{\mathrm{2}} \:\theta}={b}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{R}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}^{\mathrm{2}} \:\theta=\frac{{a}^{\mathrm{2}} {R}^{\mathrm{2}} }{{b}^{\mathrm{2}} \left({b}^{\mathrm{2}} −{R}^{\mathrm{2}} \right)} \\ $$$$\Rightarrow\mathrm{tan}\:\theta=\frac{{a}}{{b}}×\frac{\frac{{R}}{{b}}}{\sqrt{\mathrm{1}−\frac{{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\frac{{T}}{{R}}=\rho{g}\:\mathrm{tan}\:\theta=\frac{{mg}}{\mathrm{2}\pi{R}}×\frac{{a}}{{b}}×\frac{\frac{{R}}{{b}}}{\sqrt{\mathrm{1}−\frac{{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$$$\Rightarrow{T}=\frac{{mg}}{\mathrm{2}\pi}×\frac{{a}}{{b}}×\frac{\frac{{R}}{{b}}}{\sqrt{\mathrm{1}−\frac{{R}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}} \\ $$

Commented by ajfour last updated on 09/Nov/18

Thank you Sir.

$${Thank}\:{you}\:{Sir}. \\ $$

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