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Here we consider a simple pendulum that is being analyzed by Lagrange Multipliers. Shown in Fig. 1 is the pendulum of length $l$ and mass $m$. Let $U=0$ on the $x$-axis. Let the constraint equation be $f(x,y)=\ell=\sqrt{x^2+y^2}$.

                          

The Lagrangian becomes, $$L=\frac{1}{2}m[\dot{x}^2+\dot{y}^2]-mgy\mathrm{.}$$ Applying Lagrange multipliers, we get $$F_x=m\ddot{x}=\lambda x/l\mathrm{,}$$ and $$F_y=m\ddot{y}=\lambda y/l -mg\mathrm{.}$$ By just comparing these results to Newton's second law, we can conclude that $$\lambda=-T\mathrm{,}$$ and $$\lambda=T\mathrm{.}$$

My confusion comes from the fact that one result negates the other one. I am certain I have made a mistake, but I cannot seem to find it.

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So you should re-write your Lagrangian $\mathcal{L}$ as the following

$$ \mathcal{L} = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) -mgy + \lambda (x^2 + y^2 - l^2), $$

where the final term is your constraint equation which normally takes the form $f(x,y)=0$ or in generalised coordinates $f(q_i,t)=0$ (this is a holonomic constraint equation). So the next step is to do what you have done above. Hope you can take it from here. P.S a nice reference on how to approach the question is given here and here.

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  • $\begingroup$ Dear Rumplestillskin, thank you for your response. However, I noticed the problem is more subtle. It all comes down to the wrong coordinate system I chose. Working with the conventional xy-coordinate system (as shown in Fig. 1), the constraint equation is NO longer $\ell=\sqrt{x^2+y^2}$, but some other function. To fix this, I simply let the top of the pendulum be the the y-axis, and had the axis pointing down. Then, the given constraint equation is correct and everything works out. $\endgroup$ – Ptheguy Mar 14 '17 at 16:58

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