# Lagrangian of a 2 mass rotating rod

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1. Mar 11, 2015

### Physkid12

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

A long light inflexible rod is free to rotate in a vertical plane about a fixed point O. A
particle of mass m is fixed to the rod at a point P a distance ℓ from O. A second particle
of mass m is free to move along the rod, and is attracted to the point O by an elastic force
of strength k. The system is subject to gravity g. The constant k is given by k = mg/2ℓ.

Find the Lagrangian and equilibrium points

2. Relevant equations

The potential due to the elastic force is $V = 1/2 kx^2$

3. The attempt at a solution
Let $X$ be the distance of the second particle from the origin O. Then the Lagrangian I got is
$L = \frac{1}{2}m[\dot{X}^2 + (X^2 +l^2)\dot{\theta}^2] - mg[X(1-cos\theta) +l(1-cos\theta)] - \frac{1}{2} kX^2$
This gives the equations of motion for $X(t)$ and $\theta (t)$ as
$\ddot{X} = X(\dot{\theta}^2 - k) -g(1-cos\theta)$
$\ddot{\theta}(X^2+l^2) + 2\dot{X}X\dot{\theta} = -gsin\theta(X + l)$.

Now to find the equilibrium points I believe I differentiate the potential and set it equal to zero giving me:

$2 k r + mg (1-cos\theta) = 0$
$-mg(r+l)sin\theta = 0$

Is my Lagrangian correct and how would I solve these equations if so?

2. Mar 11, 2015

### BiGyElLoWhAt

=[
I miss mechanics class...
There's a problem in you X double dot equation. You cancelled your masses improperly, and there should be a 1/m in there somewhere.
For the theta dot equation, 1: watch your negative signs attached to the cosine terms, 2: watch the half that is attached to the
The lagrangian looks good on setup, but there are some errors in the equations of motion.

As for the equilibrium point, is that the point where the system is at rest? If so, why would you differentiate the potential? Why not just set your equations of motion equal to 0? (The velocities as well as the accelerations).

3. Mar 11, 2015

### BiGyElLoWhAt

I forgot to mention to watch the product rule in your theta dubbs equation.

4. Mar 11, 2015

### BiGyElLoWhAt

Upon further analysis (hehe) your negative sign was right.

5. Mar 11, 2015

### Physkid12

Hi there thanks for your response!

I've updated my eq. of motion based on your feedback, please let me know what you think

$\ddot{X} = X(\dot{\theta}^2 -k/m) -g(1-cos\theta)$

I'm struggling to change this one though as I thought the 1/2 cancels with the squared term when you differentiate with respect to theta dot. And also is my product rule wrong?

$\ddot{\theta}(X^2+l^2) + 2\dot{X}X\dot{\theta} = -gsin\theta(X + l)$.

Once I've got my eq of motion right I will try your equilibrium method!

Last edited: Mar 11, 2015
6. Mar 11, 2015

### BiGyElLoWhAt

Ahhh! haha. So you fixed the 1/m part in the X dubbs, but now it's not attached to the X anymore.

[-----
With the theta dubbs you have (distributing the important stuff)
$\frac{d}{dt}\frac{1}{2}(X^2 \dot{\theta}^2 +\ell^2\dot{\theta}^2)$
So when you do the first term you have:
$\frac{d}{dt}\frac{1}{2}(X^2 \dot{\theta}^2)= \frac{1}{2}(\frac{d}{dt}X^2 \dot{\theta}^2)$
Sometimes going through and distributing things can make it easier to see. I think it's clear that the half gets distributed through to both terms in the product rule.
Also, check the product rule, mine looks differently than yours does. (There's a squared that disappeared)
----]Retracted.

Last edited: Mar 11, 2015
7. Mar 11, 2015

### BiGyElLoWhAt

Oh wow... -.-
Maybe I need to go back to mechanics class. I think you're right about the factor of 2. My apologies.
Also, for the same reasons, that squared doesn't need to be there like I have it...

8. Mar 11, 2015

### Physkid12

So are my equations of motion correct now? :)

9. Mar 11, 2015

### BiGyElLoWhAt

Yes they are =D

10. Mar 11, 2015

### Physkid12

Brilliant! I'll try your method of getting the equilibrium points now!

I really appreciate your help, you're a maths hero!

11. Mar 11, 2015

### BiGyElLoWhAt

XD,
Let's just pretend I didn't make any math errors in this thread.