Finding the Lagrangian for a wheel-pendulum system

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SUMMARY

The discussion focuses on deriving the Lagrangian for a wheel-pendulum system consisting of a wheel with mass m and radius r rolling inside a cylinder of radius R, with a pendulum of length l attached at its center. The Lagrangian L is expressed as L = T - V, where T is the total kinetic energy and V is the potential energy. The user identifies the need to include the rotational kinetic energy of the wheel, represented as T_{wheel} = \frac{1}{2}I\omega^2, and suggests assuming the wheel is a uniform disk for clarity in calculations.

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alivedude
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Homework Statement



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

2qd3c08.jpg


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 length ##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

Homework Equations



##L = T- V##

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

## V = mgh ##

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 I am starting to think that something might be wrong in the energys above, have I missed something?
 
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alivedude said:
have I missed something?
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:

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