g-l-sin-0-Exact-form-May-include-elliptic-integral-

Question Number 127974 by Dwaipayan Shikari last updated on 03/Jan/21

\overset{∙ ∙}{θ} + \frac{g}{l} s i n θ = 0

E x a c t f o r m (M a y i n c l u d e e l l i p t i c i n t e g r a l)

Commented by Dwaipayan Shikari last updated on 03/Jan/21

M y t r y

\overset{∙ ∙}{θ} \overset{∙}{θ} + \overset{∙}{θ} \frac{g}{l} s i n θ = 0 \Rightarrow \frac{1}{2} . \frac{d}{d t} {(\overset{∙}{θ})}^{2} - \frac{g}{l} . \frac{d}{d t} (c o s θ) = 0

\Rightarrow {(\overset{∙}{θ})}^{2} - \frac{2 g}{l} c o s (θ) = C

\Rightarrow {(\overset{∙}{θ})}^{2} + \frac{4 g}{l} {s i n}^{2} \frac{θ}{2} = C + \frac{2 g}{l} \Rightarrow \overset{∙}{θ} = \sqrt{C + \frac{2 g}{l} - \frac{4 g}{l} {s i n}^{2} \frac{θ}{2}}

\Rightarrow \int \frac{d θ}{\sqrt{C + \frac{2 g}{l}} (\sqrt{1 - \frac{4 g}{C . l + 2 g} {s i n}^{2} \frac{θ}{2}})} = \int d t

\Rightarrow 2 \int \frac{d ζ}{\sqrt{C + \frac{2 g}{l}} (\sqrt{1 - {(\sqrt{\frac{4 g}{C . l + 2 g}})}^{2} {s i n}^{2} ζ})} = t + Φ

\Rightarrow 2 \frac{1}{\sqrt{C + \frac{2 g}{l}}} F (\frac{θ}{2} ∣ \sqrt{\frac{4 g}{C . l + 2 g}}) = t + Φ

\Rightarrow F (\frac{θ}{2} ∣ \sqrt{\frac{4 g}{C . l + 2 g}}) = \frac{1}{2} (t + Φ) \sqrt{C + \frac{2 g}{l}}

Answered by mr W last updated on 03/Jan/21

\overset{∙}{θ} \frac{d \overset{∙}{θ}}{d θ} = - \frac{g}{l} \sin θ

{(\overset{∙}{θ})}^{2} = \frac{2 g}{l} (\cos θ - \cos θ_{0})

\overset{∙}{θ} = \frac{d θ}{d t} = \sqrt{\frac{2 g (\cos θ - \cos θ_{0})}{l}}

\frac{d θ}{\sqrt{\cos θ - \cos θ_{0}}} = \sqrt{\frac{2 g}{l}} d t

\Rightarrow t = \sqrt{\frac{l}{2 g}} \int_{0}^{θ} \frac{d θ}{\sqrt{\cos θ - \cos θ_{0}}}

\Rightarrow t = \sqrt{\frac{2 l}{g}} \times \frac{F (\frac{θ}{2} ∣ \frac{2}{1 - \cos θ_{0}})}{\sqrt{1 - \cos θ_{0}}}

Commented by Dwaipayan Shikari last updated on 03/Jan/21

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