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

Commented by ajfour last updated on 09/Nov/18

$$Averticalellipsoidshellof$$$$frictionlesssurfaceholdsa$$$$flexibleroperingofmassm,$$$$radiusR.Findthetensionin$$$$therope.$$

Answered by MrW3 last updated on 09/Nov/18

$$eqn.ofellipse:$$$$\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$$$$\frac{x}{b^2}+\frac{y}{a^z}{y}^{\prime }=0$$$$atx=R:$$$$\frac{R}{b^2}+\frac{k}{a^2}\tan \theta =0$$$$\frac{R^2}{b^2}+\frac{k^2}{a^2}=1$$$$\Rightarrow \frac{R^2}{b^2}+\frac{1}{a^2}{(-\frac{a^{2}R}{b^2\tan \theta })}^2=1$$$$a^2{R}^2\times \frac{1}{{\tan }^2\theta }=b^2(b^2-R^2)$$$$a^2{R}^2\times \frac{1}{{\tan }^2\theta }=b^2(b^2-R^2)$$$${\tan }^2\theta =\frac{a^2{R}^2}{b^2(b^2-R^2)}$$$$\Rightarrow \tan \theta =\frac{a}{b}\times \frac{\frac{R}{b}}{\sqrt{1-\frac{R^2}{b^2}}}$$$$\frac{T}{R}=\rho g\tan \theta =\frac{mg}{2\pi R}\times \frac{a}{b}\times \frac{\frac{R}{b}}{\sqrt{1-\frac{R^2}{b^2}}}$$$$\Rightarrow T=\frac{mg}{2\pi }\times \frac{a}{b}\times \frac{\frac{R}{b}}{\sqrt{1-\frac{R^2}{b^2}}}$$

Commented by ajfour last updated on 09/Nov/18

$$ThankyouSir.$$

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