Time Period of Vertical Circular motion

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

As shown in the figure, the tangential acceleration at{ a }_{ t } is gsinxg\sin x . Thus, the angular acceleration will be atR\frac { { a }_{ t } }{ R } , where RR is the radius of the circle. Writing ωdωdx=gsinxRint\frac { \omega d\omega }{ dx} =-\dfrac { g\sin x }{ R } int v/Rωωdω=g0xsinxdxRThisgivesω=2g(1cosx)R+vR2=dxdt\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 0to2π0\quad to\quad 2\pi , to calculate the time taken for complete oscillation.

So, please help by proceeding from here or if there is any other method to calculate the time period, please mention.


Note by Abhijeet Verma
5 years, 4 months ago

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There is no simple method to evaluate that integral.
Your equation is simply d2θdt2+gθR=0 \dfrac{d^2 \theta}{d t^2} + \dfrac{g \theta}{R} = 0
The solution of this requires an elliptic integral to be solved.

Ameya Daigavane - 5 years, 4 months ago

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