**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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThere 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.

Log in to reply