Physics Help Forum inverted nonlinear pendulum, show unstable.

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 Sep 26th 2008, 03:28 PM #1 Junior Member   Join Date: May 2008 Posts: 12 inverted nonlinear pendulum, show unstable. Hey there. I have a pendulum quandary. Suppose we have a nonlinear pendulum. How can we show that the inverted position is unstable?. And what is the exponential behavior of the angle in the neighborhood of this unstable equilibrium position?. I know the inverted position is at $\displaystyle {\theta}={\pi}$. If even one initial condition causes the solution to tend away from equilibrium then it is unstable. But I do not know how to show that, much less explain the exponential behavior. I am thinking we can write the angle in terms of sine and cosine. linearized stability analysis shows that if $\displaystyle f'(x_{E})<0$, then it is unstable. The displacement from equilibrium will grow exponentially for most initial conditions. It would appear I need to take the derivative of some function using pi and show it is < 0. $\displaystyle sin(\pi)=0, \;\ cos(\pi)=-1$ Does anyone have any thoughts?. Thank you.

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