I was trying to solve the classical problem of finding when a mass sliding on a frictionless dome loses contact with said dome.

I got the lagrangian $$L=\frac{1}{2}M\left(r\,\dot{\theta}\right)^2-Mgr\cos\theta$$ and I got the 2 lagrangian equations (with the lagrangian multipliers):




I've now managed to solve the problem, but on my first attempt I thought it was a good idea to isolate $\dot{\theta}$ on the first equation to write $\dot{\theta}=\sqrt{\frac{g}{r}\cos\theta-\frac{\lambda}{Mr}}$. Then, I calculated the derivative:


The problem is that, from the second equation, I get:


How can this be?


1 Answer 1


The problem is in your first step, you forgot the implicit time dependence of $\lambda$. To get the correct result, you'll need the chain rule since you know $\lambda$ as a function of $\theta$. You'd get: $$ \ddot \theta = -\frac{\frac{g}{r}\sin\theta\dot \theta+\frac{1}{Mr}\dot\lambda}{2\sqrt{\frac{g}{r}\cos\theta-\frac{1}{Mr}\lambda} }\\ =-\frac{\frac{g}{r}\sin\theta+\frac{1}{Mr}\lambda'}{2\sqrt{\frac{g}{r}\cos\theta-\frac{1}{Mr}\lambda}} \dot \theta\\ = -\frac{1}{2}\left(\frac{g}{r}\sin\theta+\frac{1}{Mr}\lambda'\right) $$ using $\lambda = 3Mg\cos\theta-\frac{2E}{r}$ with $E$ the total (conserved) energy, you get $\lambda' = -3Mg\sin\theta$ hence the result: $$ \ddot \theta = \frac{g}{r}\sin\theta $$

Hope this helps and tell me if you need more details.

  • $\begingroup$ Thank you so much! I did find it weird that I'd lost lambda... $\endgroup$ Jun 14, 2022 at 8:40

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