# Constructing lagrangian and phase portrait

1. May 4, 2015

### CAF123

1. The problem statement, all variables and given/known data
A non–uniform disk of mass M and radius R rolls without slipping along a horizontal plane in a straight line as shown in the attachment. The centre of gravity G is displaced a distance a from the centre of the disk. Let θ be the angle between the downward vertical and the line from the centre of the disk to the centre of gravity, as shown in the diagram. Show that the Lagrangian of the system is $$L = \frac{1}{2}[I + M(R^2 + a^2 - 2aR\cos\theta)]\dot{\theta}^2 + Mga\cos\theta$$ You may assume that the rotational energy of the disk about its centre of gravity is $(1/2) I \dot{\theta}^2$ , where I is the moment of inertia of the disk about an axis through the centre of gravity (and perpendicular to the disk).

Sketch the phase portrait for $-\pi \leq \theta \leq \pi$

2. Relevant equations
L=T-V
T = kinetic energy about co.m + kinetic energy of c.o.m

3. The attempt at a solution
I nearly have the right expression for the lagrangian except one sign. The position vector of the point G, is relative to the coordinate axes $$\mathbf r = (x + a\sin \theta)\hat x + (R - a\cos \theta)\hat y,$$ where $x$ is the horizontal distance from origin to the centre of the disk. Taking time derivative and squaring I get $(R^2 + a^2 + 2aR\cos \theta)\dot{\theta}^2$ where I took $\dot x = R\dot{\theta}$ since no slip. So I have a minus error in the third term there.

To sketch the phase portrait, I should find the stability matrix, compute eigenvalues and determine their nature. I found elliptic points at $(\theta^*, p^*) = (\pm \pi, 0)$ and hyperbolic at $(0,0)$. I thought the centre of gravity would maybe behave similar to a pendulum however this is not reflected in the nature of the critical points. Really I am just looking for some help on how to decide the orientation of the arrows in the phase portrait. When $\theta = \pi$, the COM is near the top of the disk, so maybe expect a lower momentum when it is there, so as approach $\pi$ want to be decreasing in momentum so have the ellipse around $(\pi,0)$ going clockwise and similarly for the one at $(-\pi,0)$ because of symmetry. Then draw in hyperbolic lines for (0,0) so that the chart arrows are consistent. Is that ok? Thanks!

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2. May 4, 2015

### TSny

Check the signs in this relation. From the way that $\theta$ is defined, does $x$ increase as $\theta$ increases?

I'm a little hesitant to comment on your phase portrait as I have not worked much with this topic. But it seems to me that $(\theta , p) = (0,0)$ should be a stable elliptic point and $(\theta , p) = (\pm \pi, 0)$ should be hyperbolic points (saddle points?).