1
$\begingroup$

I have some doubt about a really trivial and simple problem in which I have to use ELE.

Supposing I have a pendulum, in which the rope is a spring, so it's length may change in time. I have a mass M and the spring rest length is $\ell$. I have to find the Lagrangian of the system, and the equations of motions.

Then I start finding $\mathcal{L} = T - V$

$$T = \frac{1}{2}m v^2$$

and to find $v$, remembering that $v^2 = \dot{x}^2 + \dot{y}^2$ I simply perform a change of coordinates with

$$x = \ell\cos\theta$$ $$y = \ell\sin\theta$$

and I'll get

$$v^2 = \dot{\ell}^2 + \ell^2\dot{\theta}^2$$

which leads me to have the kinetic energy $$T = \frac{1}{2}m\left(\dot{\ell}^2 + \ell^2\dot{\theta}^2\right)$$

Hope it's fine until here

Then The first problem: the potential $V$. I know that for a pendulum I can use $V = mg\ell\cos\theta$ but considering the rope is a spring, should I add to that term the elastic recall force $kx = k\ell\cos\theta$?

This would mean

$$V = (mg - k)\ell\cos\theta$$

but now a problem arises: $(mg - k)$ cannot be calculated because dimensions of $mg$ isn't the same of $k$ so where am I wrong?

EDIT AFTER FIRST ANSWER - UNDERSTOOD THIS I was using a Force, whilst I had to use, of course, the potential $\frac{1}{2}kx^2$

The potential then becomes

$$V = mg\ell\cos\theta - \frac{1}{2}k\ell^2\cos^2\theta$$

Continuing

After finding out what $V$ is, I get $\mathcal{L} = T - V$ so I'm ready for ELE:

$$\frac{\partial \mathcal{L}}{\partial \dot{q}_i} - \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial q_i} = 0$$

I know we have Einstein convention about the $i$-index so here arises another problem: my generalized coordinates will be $q_1 = \ell$ and $q_2 = \theta$.

How will Euler-Lagrange equations be written? Will I have two coupled equations or one single chaotic equation?

ELE

So I cam up with those Euler-Lagrange equations. Taking into account the comment below, I don't really know if it's good to name $\ell = l - l_0$. Anyway, those should be the equations. Are they correct?

$$m\ell^2\dot{\theta} = \frac{\text{d}}{\text{d}t}\left(\left[mg - \frac{k}{2}\ell\right]\ell\sin\theta\right) = 0$$

$$m\dot{\ell} = \frac{\text{d}}{\text{d}t}\left(m\ell\dot{\theta}^2 - mg\cos\theta + k\ell\cos\theta\right)$$

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ @BruceLee is correct about the force/energy confusion, but here is another issue. You are using the rope length as both a constant and a variable. You define it as a rest lenght initially, but then allow it to vary later. You should make a distinction (ie $l_0$ for rest length and $l$ for the variable). Then the potential term to be added is $k/2(l-l_0)^2$. $\endgroup$ – Lewis Miller Jan 7 '16 at 14:03
  • $\begingroup$ Uhm.. can't I simply rename $\ell = l - l_0$? I guess nope.. Something will change in the polar coordinates then. $\endgroup$ – Les Adieux Jan 7 '16 at 14:31
2
$\begingroup$

Well you are adding force terms in an expression for energy. The potential energy of a spring is $1/2kx^2$ and that is what needs to be added.

And in this case since $l$ and $\theta$ are both variables you will have two coupled differential equations to solve for.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Dammin! How stupid I am, lol. I was using the elastic FORCE, gosh. Thank you for the explanation. I will proceed to the calculation of ELE, then I'll edit the answer. I may ask you to check if I wrote them good then ^^ $\endgroup$ – Les Adieux Jan 7 '16 at 13:58
  • $\begingroup$ it happens..nothing to worry much about.. $\endgroup$ – Bruce Lee Jan 7 '16 at 14:00
  • $\begingroup$ So, did I write them correctly? Assuming also my hypothesis by which writing $\ell = l - l_0$ holds.. $\endgroup$ – Les Adieux Jan 7 '16 at 16:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.