Inverted pendulum

A schematic drawing of the inverted pendulum on a cart. The rod is considered massless. The mass of the cart and the pointmass at the end of the rod are denoted by M and m. The rod has a length l.

Balancing cart, a simple robotics solution

An inverted pendulum is a pendulum which has its mass above its pivot point. It is often implemented with the pivot point mounted on a cart that can move horizontally and may be called a cart and pole. Whereas a normal pendulum is stable when hanging downwards, an inverted pendulum is inherently unstable, and must be actively balanced in order to remain upright, either by applying a torque at the pivot point or by moving the pivot point horizontally as part of a feedback system. A simple demonstration is achieved by balancing an upturned broomstick on the end of one's finger.

Overview

The inverted pendulum is a classic problem in dynamics and control theory and is widely used as a benchmark for testing control algorithms (PID controllers, neural networks, fuzzy control, genetic algorithms, etc.). Variations on this problem include multiple links, allowing the motion of the cart to be commanded while maintaining the pendulum, and balancing the cart-pendulum system on a see-saw. The inverted pendulum is related to rocket or missile guidance, where the center of gravity is located behind the center of drag causing aerodynamic instability. The understanding of a similar problem can be shown by simple robotics in the form of a balancing cart. Balancing an upturned broomstick on the end of one's finger is a simple demonstration, and the problem is solved in the technology of the Segway PT, a self-balancing transportation device.

Another way that an inverted pendulum may be stabilized, without any feedback or control mechanism, is by oscillating the support rapidly up and down. If the oscillation is sufficiently strong (in terms of its acceleration and amplitude) then the inverted pendulum can recover from perturbations in a strikingly counterintuitive manner. If the driving point moves in simple harmonic motion, the pendulum's motion is described by the Mathieu equation.

Equations of motion

Stationary pivot point

The equation of motion is similar to that for an uninverted pendulum except that the sign of the angular position as measured from the vertical unstable equilibrium position:

$\ddot \theta - {g \over \ell} \sin \theta = 0$

When added to both sides, it will have the same sign as the angular acceleration term:

$\ddot \theta = {g \over \ell} \sin \theta$

Thus, the inverted pendulum will accelerate away from the vertical unstable equilibrium in the direction initially displaced, and the acceleration is inversely proportional to the length. Tall pendulums fall more slowly than short ones.

Pendulum on a cart

The equations of motion can be derived using Lagrange's equations. We refer to the drawing above where θ(t) is the angle of the pendulum of length l with respect to the vertical direction and the acting forces are gravity and an external force F in the x-direction. Define x(t) to be the position of the cart. The Lagrangian L = T − V of the system is:

$L = \frac{1}{2} M v_1^2 + \frac{1}{2} m v_2^2 - m g \ell\cos\theta$

where v₁ is the velocity of the cart and v₂ is the velocity of the point mass m. v₁ and v₂ can be expressed in terms of x and θ by writing the velocity as the first derivative of the position;

$v_1^2=\dot x^2$

$v_2^2=\left({\frac{d}{dt}}{\left(x- \ell\sin\theta\right)}\right)^2 + \left({\frac{d}{dt}}{\left( \ell\cos\theta \right)}\right)^2$

Simplifying the expression for v₂ leads to:

$v_2^2= \dot x^2 -2 \ell \dot x \dot \theta\cos \theta + \ell^2\dot \theta^2$

The Lagrangian is now given by:

$L = \frac{1}{2} \left(M+m \right ) \dot x^2 -m \ell \dot x \dot\theta\cos\theta + \frac{1}{2} m \ell^2 \dot \theta^2-m g \ell\cos \theta$

and the equations of motion are:

$\frac{\mathrm{d}}{\mathrm{d}t}{\partial{L}\over \partial{\dot x}} - {\partial{L}\over \partial x} = F$

$\frac{\mathrm{d}}{\mathrm{d}t}{\partial{L}\over \partial{\dot \theta}} - {\partial{L}\over \partial \theta} = 0$

substituting L in these equations and simplifying leads to the equations that describe the motion of the inverted pendulum:

$\left ( M + m \right ) \ddot x - m \ell \ddot \theta \cos \theta + m \ell \dot \theta^2 \sin \theta = F$

$\ell \ddot \theta - g \sin \theta = \ddot x \cos \theta$

These equations are nonlinear, but since the goal of a control system would be to keep the pendulum upright the equations can be linearized around $\theta \approx 0$.

Pendulum with oscillatory base

File:Pendulum-osc.png

A schematic drawing of the inverted pendulum on an oscillatory base. The rod is considered massless. The pointmass at the end of the rod is denoted by m. The rod has a length l.

The equation of motion for a pendulum connected to a massless, oscillating base is derived the same way as with the pendulum on the cart. The position of the point mass is now given by:

$\left( -\ell \sin \theta , y + \ell \cos \theta \right)$

and the velocity is found by taking the first derivative of the position:

$v^2=\dot y^2-2 \ell \dot y \dot \theta \sin \theta + \ell^2\dot \theta ^2.$

The Lagrangian for this system can be written as:

$L = \frac{1 }{2} m \left ( \dot y^2-2 \ell \dot y \dot \theta \sin \theta + \ell^2\dot \theta ^2 \right) - m g \left( y + \ell \cos \theta \right )$

and the equation of motion follows from:

${\mathrm{d} \over \mathrm{d}t}{\partial{L}\over \partial{\dot \theta}} - {\partial{L}\over \partial \theta} = 0$

resulting in:

$\ell \ddot \theta - \ddot y \sin \theta = g \sin \theta.$

If y represents a simple harmonic motion, y = Asin ωt, the following differential equation is:

$\ddot \theta - {g \over \ell} \sin \theta = -{A \over \ell} \omega^2 \sin \omega t \sin \theta.$

Plots for the inverted pendulum on an oscillatory base. The first plot shows the response of the pendulum on a slow oscillation, the second the response on a fast oscillation

This equation does not have elementary closed-form solutions, but can be explored in a variety of ways. It is closely approximated by the Mathieu equation, for instance, when the amplitude of oscillations are small. Analyses show that the pendulum stays upright for fast oscillations. The first plot shows that when y is a slow oscillation, the pendulum quickly falls over when disturbed from the upright position. The angle θ exceeds 90° after a short time, which means the pendulum has fallen on the ground.
If y is a fast oscillation the pendulum can be kept stable around the vertical position. The second plot shows that when disturbed from the vertical position, the pendulum now starts an oscillation around the vertical position (θ = 0). The deviation from the vertical position stays small, and the pendulum doesn't fall over.

Applications

The inverted pendulum was a central component in the design of several early Seismometers.

See also

• Self-balancing unicycle
• Segway PT
• Double inverted pendulum
• Inertia wheel pendulum
• Furuta pendulum

References

• D. Liberzon Switching in Systems and Control (2003 Springer) pp. 89ff

Further reading

• Franklin; et al. (2005). Feedback control of dynamic systems, 5, Prentice Hall. ISBN 0-13-149930-0

External links

• YouTube - inverted pendulum
• A dynamical simulation of an inverse pendulum on an oscillatory base
• Inverted pendulum modeling with several control solutions in Matlab