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.

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

Thanks.

#Mechanics

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Comments

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.

Ameya Daigavane - 5 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$