Question-47422

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

Commented by ajfour last updated on 09/Nov/18

A v e r t i c a l e l l i p s o i d s h e l l o f

f r i c t i o n l e s s s u r f a c e h o l d s a

f l e x i b l e r o p e r i n g o f m a s s m,

r a d i u s R . F i n d t h e t e n s i o n i n

t h e r o p e .

Answered by MrW3 last updated on 09/Nov/18

e q n . o f e l l i p s e :

\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1

\frac{x}{b^{2}} + \frac{y}{a^{z}} y^{'} = 0

a t x = R :

\frac{R}{b^{2}} + \frac{k}{a^{2}} \tan θ = 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 θ})}^{2} = 1

a^{2} R^{2} \times \frac{1}{\tan^{2} θ} = b^{2} (b^{2} - R^{2})

a^{2} R^{2} \times \frac{1}{\tan^{2} θ} = b^{2} (b^{2} - R^{2})

\tan^{2} θ = \frac{a^{2} R^{2}}{b^{2} (b^{2} - R^{2})}

\Rightarrow \tan θ = \frac{a}{b} \times \frac{\frac{R}{b}}{\sqrt{1 - \frac{R^{2}}{b^{2}}}}

\frac{T}{R} = ρ g \tan θ = \frac{m g}{2 π R} \times \frac{a}{b} \times \frac{\frac{R}{b}}{\sqrt{1 - \frac{R^{2}}{b^{2}}}}

\Rightarrow T = \frac{m g}{2 π} \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

T h a n k y o u S i r .

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