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.

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In the oscillating frame of the pivot,

(1)
\begin{align} g_{\mbox{eff}}=g+\frac{d^2 y}{d t^2}=g-\omega^2a\cos{\omega t} \end{align}

Therefore,

(2)
\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:

(3)
\begin{align} \theta=\bar{\theta}(t)+\theta_1(t) \end{align}

then

(4)
\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

(5)
\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

(6)
\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

(7)
\begin{align} \frac{\omega^2 a^2}{2 g l} > 1 \end{align}