# Finding the Lagrangian for a wheel-pendulum system

1. Dec 9, 2015

### alivedude

1. The problem statement, all variables and given/known data

Ok so I need to find the Lagrangian $L$ for this system below, I have drawn some poor sketch in paint but I think its pretty easy to see what i mean

Its a wheel with mass $m$ and radius $r$ that rolls inside a big cylinder with radius $R$ and at the center of the wheel there is a pendulum attached. The pendulum has a lenght $l$ and a mass $m$ (same $m$) attached at the end of it and the string can be threated as massless. The angles from equilibirum is $\theta_1$ and $\theta_2$.

I think that is all that is needed

2. Relevant equations

$L = T- V$

$T = \frac{1}{2}mv^2$

$V = mgh$

3. The attempt at a solution

OK so this is what I have been thinking so far.

First of I did the substitution $l_1 = R-r$ and $l_2=l$

We have calculate the kinetic and potential energy for each mass separatly so for the wheel we have

$T_1 = \frac{1}{2}ml_1^2 \dot{\theta_1}^2 \\ V_1 = -mgl_1cos\theta$

and for the pendulum there is a little bit more tricky

$T_2 = \frac{1}{2}m[\frac{d}{dt}(l_1sin \theta_1+ l_2 sin \theta_2)]^2+\frac{1}{2}m[\frac{d}{dt}(-l_1cos \theta_1- l_2 cos \theta_2)]^2 \\ V_2 =-mg(l_1cos\theta_1+l_2cos \theta_2)$

After this I form $L = T_1 + T_2 - V_1 - V_2$ and just carry out the algebra. But when I use this later on for Lagrange equations of motion I get some factors wrong in the answe. The dimensions and everything else is right and I have checked so many times now so im starting to think that something might be wrong in the energys above, have I missed something?

2. Dec 9, 2015

### andrewkirk

I can see one thing that seems to be missing, which is the rotational kinetic energy of the wheel, which is $\frac{1}{2}I\omega^2$ where $I$ is the wheel's moment of inertia and $\omega$ is its rotational velocity.
The problem is incompletely specified because it doesn't tell us the shape of the wheel - eg is it a solid disc or more like a ring? I suggest you assume it is a uniform disk, and state that in your answer.

Last edited: Dec 9, 2015