Furuta pendulum

Fig. 1: Schematic of the single rotary inverted pendulum system.

Consider the rotational inverted pendulum mounted to a DC motor as shown in Fig. 1. The DC motor is used to apply a torque ${\displaystyle \tau _{1}}$ to Arm 1. The link between Arm 1 and Arm 2 is not actuated but free to rotate. The two arms have lengths ${\displaystyle L_{1}}$ and ${\displaystyle L_{2}}$. The arms have masses ${\displaystyle m_{1}}$ and ${\displaystyle m_{2}}$ which are located at ${\displaystyle l_{1}}$ and ${\displaystyle l_{2}}$ respectively, which are the lengths from the point of rotation of the arm to its center of mass. The arms have inertia tensors ${\displaystyle {\boldsymbol {J}}_{1}}$ and ${\displaystyle {\boldsymbol {J}}_{2}}$ (about the centre of mass of the arms respectively). Each rotational joint is viscously damped with damping coefficients ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$, where ${\displaystyle b_{1}}$ is the damping provided by the motor bearings and ${\displaystyle b_{2}}$ is the damping arising from the pin coupling between Arm 1 and Arm 2.

A right hand coordinate system has been used to define the inputs, states and the Cartesian coordinate systems 1 and 2. The coordinate axes of Arm 1 and Arm 2 are the principal axes such that the inertia tensors are diagonal.

The angular rotation of Arm 1, ${\displaystyle \theta _{1}}$, is measured in the horizontal plane where a counter-clockwise direction (when viewed from above) is positive. The angular rotation of Arm 2, ${\displaystyle \theta _{2}}$, is measured in the vertical plane where a counter-clockwise direction (when viewed from the front) is positive. When the Arm is hanging down in the stable equilibrium position ${\displaystyle \theta _{2}=0}$ .

The torque the servo-motor applies to Arm 1, ${\displaystyle \tau _{1}}$, is positive in a counter-clockwise direction (when viewed from above). A disturbance torque, ${\displaystyle \tau _{2}}$, is experienced by Arm 2, where a counter-clockwise direction (when viewed from the front) is positive.

Before deriving the dynamics of the system a number of assumptions must be made. These are:

• The motor shaft and Arm 1 are assumed to be rigidly coupled and infinitely stiff.
• Arm 2 is assumed to be infinitely stiff.
• The coordinate axes of Arm1 and Arm 2 are the principal axes such that the inertia tensors are diagonal.
• The motor rotor inertia is assumed to be negligible. However, this term may be easily added to the moment of inertia of Arm 1.
• Only viscous damping is considered. All other forms of damping (such as Coulomb) have been neglected, however it is a simple exercise to add this to the final governing DE.

The non-linear equations of motion are given by[15]

${\displaystyle {\ddot {\theta }}_{1}\left(J_{1zz}+m_{1}l_{1}^{2}+m_{2}L_{1}^{2}+(J_{2yy}+m_{2}l_{2}^{2})\sin ^{2}(\theta _{2})+J_{2xx}\cos ^{2}(\theta _{2})\right)+{\ddot {\theta }}_{2}m_{2}L_{1}l_{2}\cos(\theta _{2})-m_{2}L_{1}l_{2}\sin(\theta _{2}){\dot {\theta }}_{2}^{2}+{\dot {\theta }}_{1}{\dot {\theta }}_{2}\sin(2\theta _{2})(m_{2}l_{2}^{2}+J_{2yy}-J_{2xx})+b_{1}{\dot {\theta }}_{1}=\tau _{1}}$

and

${\displaystyle {\ddot {\theta }}_{1}m_{2}L_{1}l_{2}\cos(\theta _{2})+{\ddot {\theta }}_{2}(m_{2}l_{2}^{2}+J_{2zz})+1/2{\dot {\theta }}_{1}^{2}\sin(2\theta _{2})(-m_{2}l_{2}^{2}-J_{2yy}+J_{2xx})+b_{2}{\dot {\theta }}_{2}+gm_{2}l_{2}\sin(\theta _{2})=\tau _{2}}$

Most Furuta pendulums tend to have long slender arms, such that the moment of inertia along the axis of the arms is negligible. In addition, most arms have rotational symmetry such that the moments of inertia in two of the principal axes are equal. Thus, the inertia tensors may be approximated as follows:

${\displaystyle {\boldsymbol {J}}_{1}=diag[J_{1xx},J_{1yy},J_{1zz}]=diag[0,J_{1},J_{1}]}$

${\displaystyle {\boldsymbol {J}}_{2}=diag[J_{2xx},J_{2yy},J_{2zz}]=diag[0,J_{2},J_{2}]}$

Further simplifications are obtained by making the following substitutions. The total moment of inertia of Arm 1 about the pivot point (using the parallel axis theorem) is ${\displaystyle {\hat {J_{1}}}=J_{1}+m_{1}l_{1}^{2}}$. The total moment of inertia of Arm 2 about its pivot point is ${\displaystyle {\hat {J_{2}}}=J_{2}+m_{2}l_{2}^{2}}$. Finally, define the total moment of inertia the motor rotor experiences when the pendulum (Arm 2) is in its equilibrium position (hanging vertically down), ${\displaystyle {\hat {J_{0}}}={\hat {J}}_{1}+m_{2}L_{1}^{2}=J_{1}+m_{1}l_{1}^{2}+m_{2}L_{1}^{2}}$.

Substituting the previous definitions into the governing DEs gives the more compact form

${\displaystyle {\ddot {\theta }}_{1}\left({\hat {J_{0}}}+{\hat {J_{2}}}\sin ^{2}(\theta _{2})\right)+{\ddot {\theta }}_{2}m_{2}L_{1}l_{2}\cos(\theta _{2})-m_{2}L_{1}l_{2}\sin(\theta _{2}){\dot {\theta }}_{2}^{2}+{\dot {\theta }}_{1}{\dot {\theta }}_{2}\sin(2\theta _{2}){\hat {J_{2}}}+b_{1}{\dot {\theta }}_{1}=\tau _{1}}$

and

${\displaystyle {\ddot {\theta }}_{1}m_{2}L_{1}l_{2}\cos(\theta _{2})+{\ddot {\theta }}_{2}{\hat {J_{2}}}-1/2{\dot {\theta }}_{1}^{2}\sin(2\theta _{2}){\hat {J_{2}}}+b_{2}{\dot {\theta }}_{2}+gm_{2}l_{2}\sin(\theta _{2})=\tau _{2}}$