A mass at the top of a vertical hoop. (Lagrangian Mechanics)

1. The problem statement, all variables and given/known data
A heavy particle is placed at the top of a vertical hoop. Calculate the reaction of the hoop on the particle by means of the Lagrange's undetermined multipliers and Lagrange's Equations. Find the height at which the particle falls of.

2. Relevant equations

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

3. The attempt at a solution

$$L = \frac{1}{2} m (\dot{r}^2 + r^2 \dot{\theta}^2) - mgr cos(\theta)$$

Euler-Lagrange Equation with respect to $r$ becomes:

$$m \ddot{r} - m r \dot{\theta}^2 + m g cos(\theta) = \lambda$$

And, with respect to $\theta$:

$$mr^2 \ddot{\theta} + 2 m r \dot{r} \dot{\theta} - mgr sin(\theta) = 0$$

As the radius is constant, it's derivative is zero. So, we have:

$$- m r \dot{\theta}^2 + m g cos(\theta) = \lambda$$
$$m r^2 \ddot{\theta} - m g sin(\theta) = 0$$

Now, I don't really know what's the next step.
What should I do?

Any help appreciated.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 $\lambda$ is actually the reaction force of the surface of the hoop on the particle. We can include the reaction force as a potential in the formulation of the Lagrangian: $$L = \frac{1}{2} m (r^2 + r^2 \dot{\theta}^2) − m g r cos(\theta) - V (r)$$ Comparison with your E-L equations will then give $- \frac{dV(r)}{dr} = \lambda$ Hence, the reaction force, taking the boundary condition r = R, is $$F_{R} = - \frac{dV(r)}{dr}|_{r = R} = \lambda |_{r = R} = mg cos(\theta) - mR\dot{\theta}^2$$ As for when the particle falls off, that means that the reaction force must be zero since it is no longer in contact with the hoop.

 Quote by Fightfish $\lambda$ is actually the reaction force of the surface of the hoop on the particle. We can include the reaction force as a potential in the formulation of the Lagrangian: $$L = \frac{1}{2} m (r^2 + r^2 \dot{\theta}^2) − m g r cos(\theta) - V (r)$$ Comparison with your E-L equations will then give $- \frac{dV(r)}{dr} = \lambda$ Hence, the reaction force, taking the boundary condition r = R, is $$F_{R} = - \frac{dV(r)}{dr}|_{r = R} = \lambda |_{r = R} = mg cos(\theta) - mR\dot{\theta}^2$$ As for when the particle falls off, that means that the reaction force must be zero since it is no longer in contact with the hoop.
I guess I could use Newtonian Mechanics to find out that $v^2 = 2 g R (1 - cos(\theta))$, and then substituting in the equation, which gives me:
$$\lambda = mg (3 cos(\theta) - 2)$$
But, is there any other way to find $v^2$ or $\ddot{\theta}^2$? I.e., by using only the Lagrangian Mechanics.