Consider a simple electromagnetic suspension system shown in Figure P1.14.

Figure P1.14 

The electromagnetic force \(f_{m}\) is given by

\[ f_{m}=\alpha \frac{I^{2}}{h^{2}} \]

where \(I\) and \(h\) are the coil current and the air gap, respectively. The constant \(\alpha=\mu_{0} N^{2} A_{p}\) where \(\mu_{0}, N\), and \(A_{p}\) are the air permeability, the number of coil turns, and the face area per single pole of the magnet, respectively. Let \(h_{0}\) be the desired air gap. Then, the current \(I_{0}\) is calculated from the following static equilibrium condition:

\[ \alpha \frac{I_{0}^{2}}{h_{0}^{2}}=m g \]

Let \(x(t)\) be the dynamic displacement of the mass with respect to the static equilibrium position. Derive the differential equation of motion and determine the stability of system. Show that the dynamic characteristic of this system is equivalent to that of an inverted pendulum.

Transcribed Image Text:

I ho fm x(t) mg