Furuta pendulum

Equations of motion

Despite the great deal of attention the system has received, very few publications successfully derive (or use) the full dynamics. Many authors have only considered the rotational inertia of the pendulum for a single principal axis (or neglected it altogether). In other words, the inertia tensor only has a single non-zero element (or none), and the remaining two diagonal terms are zero. It is possible to find a pendulum system where the moment of inertia in one of the three principal axes is approximately zero, but not two.

A few authors have considered slender symmetric pendulums where the moments of inertia for two of the principal axes are equal and the remaining moment of inertia is zero. Of the dozens of publications surveyed for this wiki only a single conference paper and journal paper were found to include all three principal inertial terms of the pendulum. Both papers used a Lagrangian formulation but each contained minor errors (presumably typographical).

The equations of motion presented here are an extract from a paper on the Furuta pendulum dynamics derived at the University of Adelaide.

Definitions

Consider the rotational inverted pendulum mounted to a DC motor as shown in Fig. 1. The DC motor is used to apply a torque τ 1 to Arm 1. The link between Arm 1 and Arm 2 is not actuated but free to rotate. The two arms have lengths L 1 and L 2 . The arms have masses m 1 and m 2 which are located at l 1 and 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 J 1 and J 2 (about the centre of mass of the arm). Each rotational joint is viscously damped with damping coefficients b 1 and b 2 , where b 1 is the damping provided by the motor bearings and 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, θ 1 , is measured in the horizontal plane where a counter-clockwise direction (when viewed from above) is positive. The angular rotation of Arm 2, θ 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 θ 2 = 0 .

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

Assumptions

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

Non-linear Equations of Motion

The non-linear equations of motion are given by

θ ¨ 1 ( J 1 z z + m 1 l 1 2 + m 2 L 1 2 + ( J 2 y y + m 2 l 2 2 ) sin 2 ⁡ ( θ 2 ) + J 2 x x cos 2 ⁡ ( θ 2 ) ) + θ ¨ 2 m 2 L 1 l 2 cos ⁡ ( θ 2 ) − m 2 L 1 l 2 sin ⁡ ( θ 2 ) θ ˙ 2 2 + θ ˙ 1 θ ˙ 2 sin ⁡ ( 2 θ 2 ) ( m 2 l 2 2 + J 2 y y − J 2 x x ) + b 1 θ ˙ 1 = τ 1

and

θ ¨ 1 m 2 L 1 l 2 cos ⁡ ( θ 2 ) + θ ¨ 2 ( m 2 l 2 2 + J 2 z z ) + 1 / 2 θ ˙ 1 2 sin ⁡ ( 2 θ 2 ) ( − m 2 l 2 2 − J 2 y y + J 2 x x ) + b 2 θ ˙ 2 + g m 2 l 2 sin ⁡ ( θ 2 ) = τ 2

Simplifications

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:

J 1 = d i a g [ J 1 x x , J 1 y y , J 1 z z ] = d i a g [ 0 , J 1 , J 1 ]

J 2 = d i a g [ J 2 x x , J 2 y y , J 2 z z ] = d i a g [ 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 J 1 ^ = J 1 + m 1 l 1 2 . The total moment of inertia of Arm 2 about its pivot point is 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), J 0 ^ = 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

θ ¨ 1 ( J 0 ^ + J 2 ^ sin 2 ⁡ ( θ 2 ) ) + θ ¨ 2 m 2 L 1 l 2 cos ⁡ ( θ 2 ) − m 2 L 1 l 2 sin ⁡ ( θ 2 ) θ ˙ 2 2 + θ ˙ 1 θ ˙ 2 sin ⁡ ( 2 θ 2 ) J 2 ^ + b 1 θ ˙ 1 = τ 1

and

θ ¨ 1 m 2 L 1 l 2 cos ⁡ ( θ 2 ) + θ ¨ 2 J 2 ^ − 1 / 2 θ ˙ 1 2 sin ⁡ ( 2 θ 2 ) J 2 ^ + b 2 θ ˙ 2 + g m 2 l 2 sin ⁡ ( θ 2 ) = τ 2