Chapter 8.6, Problem 119RP

To determine

The natural frequency of the inverted pendulum and any other restriction in the solution.

Answer to Problem 119RP

The natural frequency of the inverted pendulum is $\frac{1}{2\pi }\sqrt{\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}}$.

The restriction imposed on the system is $k>\frac{mgl}{2b^2}$.

Explanation of Solution

Draw the schematic diagram.

Write the expression for the kinetic energy of the system $\left(T\right)$.

  $T=\frac{1}{2}m{\left(l\dot{\theta }\right)}^2$        (I)

Here, the mass of the pendulum is $m$, the length of the pendulum is $l$.

Write the expression for the potential energy of the system $\left(V\right)$.

  $V=-mgl\left(1-\cos \theta \right)+\frac{1}{2}k{\left(\delta +b\sin \theta \right)}^2+\frac{1}{2}k{\left(\delta -b\sin \theta \right)}^2$        (II)

Here, the acceleration due to gravity is $g$, the spring stiffness is $k$ and the initial spring compression is $\delta$.

Write the expression for the total energy of the system $\left(E\right)$.

  $E=T+V$        (III)

Write the expression for the natural frequency of the system $\left(f_n\right)$.

  $f_n=\frac{{\omega }_n}{2\pi }$        (IV)

Here, the natural circular frequency is ${\omega }_n$.

Conclusion:

Substitute, Equation (I) and Equation (II) in Equation (III).

  $E=\frac{1}{2}m{\left(l\dot{\theta }\right)}^2-mgl\left(1-\cos \theta \right)+\frac{1}{2}k{\left(\delta +b\sin \theta \right)}^2+\frac{1}{2}k{\left(\delta -b\sin \theta \right)}^2$        (V)

Differentiate Equation (V) and equate to $0$.

  $\begin{array}{l}\frac{dE}{dt}=0\\ \frac{d}{dt}\left[\frac{1}{2}m{\left(l\dot{\theta }\right)}^2-mgl\left(1-\cos \theta \right)+\frac{1}{2}k{\left(\delta +b\sin \theta \right)}^2+\frac{1}{2}k{\left(\delta -b\sin \theta \right)}^2\right]=0\\ m{l}^2\ddot{\theta }-mgl\sin \theta +k\left(\delta +b\sin \theta \right)\left(b\cos \theta \right)+k\left(\delta -b\sin \theta \right)\left(-b\cos \theta \right)=0\end{array}$        (VI)

Rewrite Equation (VI) for small values of $\theta$.

  $\begin{array}{l}m{l}^2\ddot{\theta }-mgl\theta +k\left(\delta +b\theta \right)\left(b\right)+k\left(\delta -b\theta \right)\left(-b\right)=0\\ m{l}^2\ddot{\theta }-mgl\theta +kb\left(\delta +b\theta +b\theta -\delta \right)=0\\ m{l}^2\ddot{\theta }-mgl\theta +2k{b}^2\theta =0\\ \ddot{\theta }-\frac{mgl}{m{l}^2}\theta +\frac{2k{b}^2}{m{l}^2}\theta =0\\ \ddot{\theta }+\left[\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}\right]\theta =0\end{array}$        (VII)

Obtain the value of ${\omega }_n$ from Equation (VII) as:

  $\begin{array}{l}{\omega }_n{}^2=\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}\\ {\omega }_n=\sqrt{\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}}\end{array}$

Substitute, $\sqrt{\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}}$ for ${\omega }_n$ in Equation (IV) .

  $f_n=\frac{1}{2\pi }\sqrt{\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}}$        (VIII)

Thus, the natural frequency of the inverted pendulum is $\frac{1}{2\pi }\sqrt{\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}}$.

From Equation (VIII), we get;

  $\frac{2k{b}^2}{m{l}^2}-\frac{g}{l}>0$

  $\begin{array}{l}\frac{2k{b}^2}{m{l}^2}>\frac{g}{l}\\ k>\frac{mgl}{2b^2}\end{array}$        (IX)

Thus, the restriction imposed on the system is $k>\frac{mgl}{2b^2}$.