Fall 2007 12 Mech

The pivot of an inverted simple pendulum is rapidly oscillated vertically with amplitude a and frequency $\omega$ (see diagram).


Find a condition on $\omega$ such that $\theta$ = 0 is a point of stable equilibrium.

Hint: Separate the equation of motion into "fast" and "slow" parts. Eliminate the former. The remaining equation for the slow part determines whether the system is stable.




In the oscillating frame of the pivot,

\begin{align} g_{\mbox{eff}}=g+\frac{d^2 y}{d t^2}=g-\omega^2a\cos{\omega t} \end{align}


\begin{align} \ddot{\theta}=\frac{g}{l}\left(1-\frac{\omega^2 a}{g}\cos{\omega t} \right)\theta \end{align}

Let us decompose $\theta$ into a "slow" $\bar{\theta}$ and a "fast" $\theta_1$ parts:

\begin{align} \theta=\bar{\theta}(t)+\theta_1(t) \end{align}


\begin{align} \frac{d^2\theta_1}{dt^2}+\frac{d^2\bar{\theta}}{dt^2}=\frac{g}{l}\left(1-\frac{\omega^2a}{g}\cos{\omega t}\right)(\bar{\theta}+\theta_1) \end{align}

The "fast" equation is

\begin{align} \frac{d^2\omega_1}{dt^2}=-\frac{g}{l}\frac{\omega^2a}{g}\cos{(\omega t)}\bar{\theta}\;\;\therefore\;\;\theta_1=\frac{a}{l}\bar{\theta}\cos{\omega t} \end{align}

The "slow" equation is

\begin{align} \frac{d^2\bar{\theta}}{dt^2}=\frac{g}{l}\bar{\theta}-\frac{g}{l}\left<\theta_1 \cos{\omega t} \right>\frac{\omega^2 a}{g}=\frac{g}{l}\left(1-\frac{\omega^2 a}{g}\frac{a}{2l} \right)\bar{\theta} \end{align}

The stability is achieved if

\begin{align} \frac{\omega^2 a^2}{2 g l} > 1 \end{align}
Add a New Comment
or Sign in as Wikidot user
(will not be published)
- +
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-Share Alike 2.5 License.