Inverted pendulums
August 17, 2018. I discuss the physics of an inverted pendulum, and prove (using a judicious combination of hand-waving and the Hill determinant) that if you wobble the pivot fast enough, the pendulum will settle into equilbrium upside-down.
Introduction
Prerequisites: basic classical mechanics, Fourier series, differential equations.
Tie a rock to a piece of string, and suspend it from a nail. If you give the rock a little nudge, it will oscillate, with its period going as the square root of the length of the string. Gradually, it will lose energy to air friction and return to its freely hanging, equilibrium position. So the equilibrium is stable. All this is familiar and not particularly interesting.
But here is a much weirder fact about pendulums. If you turn the rock and string arrangement upside down, and wiggle fast enough, the rock will stand on end! Not only that, it will be stable. Someone can nudge it, and it will slowly return to the upright position. Similarly, if I take an ordinary rope, and wobble it fast enough, then not only will it stand on end, but a suitably light fakir could climb up. So classical mechanics can almost explain the famous levitating Indian rope trick. (I say “almost” since “rope” here really means “finite chain of rigid pendulums”. I will discuss this restriction in the next post.)
Single pendulum
Let’s start by turning a normal pendulum upside down.
We have a mass
We’ll use Lagrangian mechanics, which is a bit neater than the
Newtonian approach here.
Recall that the Lagrangian
We have a gravitational potential:
and kinetic energy
The Euler-Lagrange equations yield
Cleaning up, we find the equation of motion:
Note that this is exactly the equation we get for a forced pendulum
hanging down, but with
Since we are considering perturbations around
We can make things a bit neater by rescaling time
This is Mathieu’s equation. It appears all over the place in applied mathematics, from vibrating drumheads and forced pendulums to rocking boats and radio antennae. (See Lawrence Ruby’s article for further discussion of these examples.) This is a tricky equation, with no closed-form solutions built out of elementary functions. In the next section, we will use some clever approximation methods (and a liberal amount of hindsight) to come up with a sharp stability criterion. But before we do that, we can make a guess based on dimensional analysis.
The parameters for the forcing frequency of the pivot are the
amplitude
We will derive a condition of exactly this sort below.
Parameters and stability
There is a bit of theory needed to rigorously understand the stability of the
Mathieu equation.
Instead, we will make a sequence of well-motivated guesses which turn
out to give the right answer!
Here is the basic idea.
When we wobble the pivot with a periodic driving function
These solutions will be stable, since they decay to the equilibrium
configuration we want,
These resonant solutions are unbounded and hence unstable.
Whether we get unstable or stable solutions depends on the parameters
of the
problem, in our case, the constants
Our strategy will be to figure out what values of
Periodic solutions and the Hill determinant
So, let’s consider an arbitrary solution of period
for some unknown, infinite set of coefficients
where on the last line, we shift the dummy variable so that
For each
This linear system only has a nontrivial solution if
We will use a shortcut.
Let’s restrict to the regime of fast, small-amplitude oscillations of
the pivot, where
Setting
using the fact that
After all this work, we learn that wobbling the pivot of an inverted
pendulum at sufficiently high frequency and amplitude will indeed stabilise
the pendulum around the inverted position
This only finesses our dimensional analysis by a factor of
Exercise 1. To evaluate the Hill determinant
Let
The Hill determinant is obtained, in principle, by solving the
recurrence and taking
Two other points are worth mentioning.
First, there are additional regions of stability and instability in
Second, we can use the Hill determinant to solve a generalisation of
(
where
Next time
In the next post, I’ll look at a beautiful theorem of David Acheson which uses different methods to show that a chain of rigid pendulums, oscillated at suitable amplitude and frequency, can be stabilised in the upside-down position. I’ll also talk about the experimental realisations of this phenomenon. They reveal something that our perturbative analysis cannot tell us: how far the pendulum can fall and still return to the inverted equilibrium position. It turns out to be surprisingly large!
References
- “Inverted Pendulums”, §VI.4 Princeton Companion to Applied Mathematics (2015), David Acheson.
- Non-equilibrium statistical mechanics (2016), Andrew Melatos and Andy Martin.
- “Mathieu’s equation” (2016), Richard Rand.
- “Applications of Mathieu equaton” (1995), Lawrence Ruby.