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# Time Period of Vertical Circular motion

Note: $$x$$ has been used for the angle with the vertical, measured in anti-clockwise direction.

As shown in the figure, the tangential acceleration $${ a }_{ t }$$ is $$g\sin x$$ . Thus, the angular acceleration will be $$\frac { { a }_{ t } }{ R }$$ , where $$R$$ is the radius of the circle. Writing $\frac { \omega d\omega }{ dx} =-\dfrac { g\sin x }{ R } int$ $\int _{ v/R }^{ \omega }{ \omega \, d\omega } =\dfrac { g\int _{ 0 }^{ x }{ \, sinx\, dx } }{ R } \\ { This\quad gives\\ \omega =\sqrt { \dfrac { 2g(1-\cos x) }{ R } +{ \dfrac { v }{ R } }^{ 2 } } =\dfrac { dx }{ dt } }$

Now, I don't know how to integrate this expression between $$0\quad to\quad 2\pi$$, to calculate the time taken for complete oscillation.

Thanks.

Note by Abhijeet Verma
1 year, 5 months ago

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There is no simple method to evaluate that integral.
Your equation is simply $$\dfrac{d^2 \theta}{d t^2} + \dfrac{g \theta}{R} = 0$$
The solution of this requires an elliptic integral to be solved. · 1 year, 4 months ago