# Lagrangian Mechanics - Pulley System

1. Aug 27, 2008

### Newtons Balls

1. The problem statement, all variables and given/known data
Here's a little diagram I whipped up in paint:

Sorry about my sucky art skills.

The wiggly line is a spring with spring constant mk and natural length d. The actual length of the spring is y. The two masses to the left have mass m, the mass on the right has mass 2m. The springs are attached by a light string which doesn't slip over the light pulley, which has a radius R which is free to rotate. The angle theta is the angle by which the pulley has rotated since the initial conditions. The red line indicates the position of all the masses at the initial conditions. The initial conditions are y=0, t=0, $$\theta$$=0 and everything is at rest.

a) Write down the vertical displacement of each of the three masses from their initial positions in terms of theta and y.

b) Show that the Lagrangian describing the system is up to a constant given by:
L = 2mR$$^{2}$$$$\theta$$'$$^{2}$$ -mR$$\theta$$'y' + m/2 y'$$^{2}$$ + mgy - (mk/2)(y-d)$$^{2}$$

where ' indicates a first derivative with respect to time- can't do the normal dots it seems.

2. Relevant equations

L = T - V

3. The attempt at a solution
Well for a) I think for the mass on the right its just -R$$\theta$$
for the mass on the top left its R$$\theta$$
for the mass on the bottom left its -$$\theta$$ - y

I assume thats right- correct me if I'm wrong!

for b) I can't get what I'm supposed to.
I do not get the -mR$$\theta$$'y' term. Here's my thinking:

For the kinetic energy:
1/2mv$$^{2}$$ for each mass where v is the time derivative of their vertical displacement in a). This gives 2mR$$^{2}$$$$\theta$$'$$^{2}$$ + m/2 y'$$^{2}$$

For the potential energy: -mgy as the total loss in GPE is due to the mass on the bottom left going down by y, the other two masses cancel each other out. The last term is correct, just the potential energy of the spring.

So, my question: Where does the -mR$$\theta$$'y' term come from?
hopefully this is fairly clear :/
Thanks (again) :D

2. Aug 27, 2008

### Newtons Balls

lol. I'm an idiot. I finally see where its from. This will teach me to skip steps and do too much in my head.

The displacement of the third mass if $$R\theta + y$$
The derivative of this wrt time is $$R\theta ' + y'$$
This squared is $$R^{2}\theta^{2}' + y'^{2} + 2R\theta'y'$$

I really feel like an idiot :)
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