# Non-dimensionalization using Buckingham-$\pi$: example of the “bead on a rotating hoop, with viscous damping” problem

I am trying to find a way to non-dimensionalize known equations, using the Buckingham-$\pi$ theorem.

Consider the "bead on a rotating hoop, with viscous damping" problem---if you are interested in the details, you can check out Strogatz's Nonlinear dynamics and chaos. For now, I just present the equation:

$$mL^2 \frac{\textrm{d}^2\phi}{\textrm{d}t^2} = -mgL\sin(\phi) + \Gamma -b\frac{\textrm{d}\phi}{\textrm{d}t},\quad (1)$$

where $\Gamma$ is an applied torque, and $b$ is the viscous damping factor, and $\phi \in \mathbb{S}^1$ (radians and unitless).

I want to non-dimensionalize it into this form:

$$\epsilon \frac{\textrm{d}^2\phi}{\textrm{d}\tau^2} = -sin(\phi) + \gamma -\frac{\textrm{d}\phi}{\textrm{d}\tau}. \quad (2)$$

I am going to try and use the Buckingham-$\pi$ to determine how to produce $(2)$ from $(1)$. First, I note that $mL^2\frac{\textrm{d}^2\phi}{\textrm{d}t^2}$ is the dependent variable. Second, I identify a complete set of independent variables, which in this case will be:

1. $\phi \thicksim 1$
2. $mgL \thicksim [M][L]^2[T]^{-2}$
3. $\Gamma \thicksim [M][L]^2[T]^{-2}$
4. $b \thicksim [M][L]^2[T]^{-1}$

Question: is $\frac{\textrm{d}^2\phi}{\textrm{d}t^2}$ dependent on $\phi$? I am thinking yes, because for $t > 0$ (assuming that $t=0$ is the initial condition) the velocity of the bead will depend partially on the location of the bead on the hoop.

I note that $[M][L]^2$ appears in every independent variable, including $\phi$ where it occurs to the power of $0$. So, I let $[M][L]^2 = [\Lambda]$, and identify two basic dimensions in the problem: $[\Lambda]$ and $[T]$.

Noticing that $b$ and $mgL$ do not appear in $(2)$, I take them to be the dimensionally independent variables out of the complete set of variables.

Now I calculate the $\Pi$ groups:

$$(mL^2) \quad \quad \Pi_0 = [\Lambda][\Lambda]^{\alpha}[T]^{-2\alpha}[\Lambda]^{\beta}[T]^{-2\beta} = 1$$

$$\implies 1 + \alpha + \beta = 0,\quad -2 -2\alpha -2\beta = 0$$

So, this is a condition that cannot be satisfied. Maybe I have to consider:

$$(mL^2\frac{\textrm{d}^2\phi}{\textrm{d}t^2}) \quad \quad \Pi_0 = [\Lambda][T]^{-2}[\Lambda]^{\alpha}[T]^{-2\alpha}[\Lambda]^{\beta}[T]^{-2\beta} = 1$$

$$\implies 1 + \alpha + \beta = 0,\quad -2 - 2\alpha -2\beta = 1 + \alpha + \beta = 0$$

So, this time we only have one equation for two unknowns, so we cannot find a unique solution. Let's choose $\alpha=-1, \beta=0$.

Question: After getting this far, I am a little confused as to how to form the final dimensionless equation. I know that $\Pi_{0} = f(\Pi_{1}, \Pi_{3})$ (the numbers corresponding to the numbers in the list). My thought is that I should be doing something like this:

$$\frac{mL^2}{mgL}\frac{\textrm{d}^2\phi}{\textrm{d}t^2} = \epsilon \frac{\textrm{d}^2\phi}{\textrm{d}\tau^2}$$

...but actually working that out causes issues, obviously. for instance, what should be $\epsilon$ (the coefficient in front of $\frac{\textrm{d}^2\phi}{\textrm{d}\tau^2}$ isn't dimensionless. What should actually be done next?