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I have managed to derive the equation of motion of a simple pendulum under the influence of gravity using the Lagrangian, but since that only tells me what the angular acceleration is, I now want to derive the $x$ and $y$ components of acceleration. The formula I have derived is the following: $$\ddot θ = -(g/l)\sinθ$$

And the $x$ and $y$ are these: $$x = l\sinθ$$ $$y = l(1-\cosθ)$$

So now I'm not sure what to do next... Do I use the chain rule and differentiate $x$ and $y$ with respect to time twice? I don't really know what to do here...

If you can, please give me a hint instead of a full solution, and, if possible, stick to these formulas and Euler-Lagrange equations, instead of a solution that would incorporate something not directly related.

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  • $\begingroup$ Instead of writing the Lagrangian parametrizing as a function of $\theta$, try writing it in terms of Cartesian coordinates $x$ and $y$. The E-L equations will then give you the acceleration along each axis. Another way to do it is to rewrite $l$ and $\theta$ in terms of $x$ and $y$. $\endgroup$ May 25 '16 at 17:04
  • $\begingroup$ Hmm... How would I rewrite l in terms of x and y, since l is a constant independent of x and y? I'm gonna try determining the Lagrangian like you said though to see if it works. $\endgroup$
    – Andreas C
    May 25 '16 at 17:05
  • $\begingroup$ Look again at the formulas you wrote. Remember that in Cartesian coordinates $L = \frac{1}{2} m(\dot{x}^2 + \dot{y}^2) - V(x,y)$. $\endgroup$ May 25 '16 at 17:07
  • $\begingroup$ Yeah, and V=mgy. When I try to find the partial derivative of L with respect to x, it's just 0, and it follows from the E-L equations that the x component of acceleration is 0 as well, but that doesn't make much sense, does it? $\endgroup$
    – Andreas C
    May 25 '16 at 17:15
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As you noticed, if we use Euler-Lagrange equation on $L= \frac 1 2 (\dot x^2 + \dot y^2) -mgy$ we get

$$\ddot x=0$$ $$\ddot y = -g$$

Something is clearly missing: gravity is not the only force acting on our mass: we have to take into account the tension of the rod/string. But why doesn't it come out from the equations?

The point is that system only has one degree of freedom: $\theta$. If we use $x$ and $y$ we are using the Lagrangian formalism as if it had two. But it doesn't: one degree of freedom is taken away from the constraint $x^2+y^2=l^2$. So the usual formulation of the EL equation is not going to work properly.

This was the hint. If you want the solution, you'll find it here: http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node90.html

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  • $\begingroup$ Huh... Ok, that makes a lot of sense, thank you. I now know that using the Lagrangian like that is not the answer. However, I don't want a full solution, and I still don't know where to begin... Hmmm... I thought about differentiating x and y twice with respect to time, but I'm not sure how to do that... $\endgroup$
    – Andreas C
    May 25 '16 at 19:15
  • $\begingroup$ If you want the $x$ and $y$ components of the acceleration without using EL equations and not as a function of $\theta$, you can try differentiating twice the expression you wrote with respect to time and then substitute in the final expression $\theta = \arctan(x/y)$. $\endgroup$
    – valerio
    May 25 '16 at 19:25
  • $\begingroup$ Oh but that's what I want to do, it's just that I don't know how to differentiate l*sinθ with respect to time... Also, if I get the x and y components of acceleration as functions of θ, well, that's just fine by me, it's just that I don't know how to do that either! $\endgroup$
    – Andreas C
    May 25 '16 at 20:42
  • $\begingroup$ Ooh, ok! I thought you knew how to do that and you were trying something a bit more complicated... ;-) $\endgroup$
    – valerio
    May 25 '16 at 20:57
  • $\begingroup$ Ah, no... I just realized how to do it, it's fine now. Now all that's left is to find θ as a function of time, but I think I know how to do that. $\endgroup$
    – Andreas C
    May 26 '16 at 8:35
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Everything is a function of the angle $\theta$ and its derivatives $\dot{\theta}$ and $\ddot{\theta}$. From there use the chain rule of differentiation.

$$\begin{align} x & = \ell \sin \theta & y & = \ell (1-\cos \theta) \\ \dot{x} & = \ell \dot{\theta} \cos \theta & y & = \ell \dot{\theta} \sin \theta \\ \ddot{x} & = \ell \ddot{\theta} \cos\theta -\ell \dot{\theta}^2 \sin\theta & \ddot{y} & = \ell \ddot{\theta} \sin \theta + \ell \dot{\theta}^2 \cos\theta \end{align} $$

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  • $\begingroup$ Thank you, it wasn't that complicated after all! It makes perfect sense now! $\endgroup$
    – Andreas C
    May 25 '16 at 20:45
  • $\begingroup$ Also, to find the angular velocity I just integrate the angular acceleration once with respect to time. One of the borders (or limits, I don't know how you call them in English) of that integration will be the amplitude. What will the other one be? The phase? $\endgroup$
    – Andreas C
    May 25 '16 at 20:52
  • $\begingroup$ Oh, uh... Forget about it, I think I figured it out... $\endgroup$
    – Andreas C
    May 25 '16 at 21:00

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