Constraint Force and Lagrangian multiplier

Question

I can't seem to find my mistake in this problem and I think it stems from not understanding how to correctly form constraints and the meaning behind the Lagrangian multiplier.

So first of all I learned in class a nice problem and how to solve it using Holonomic Constraints.

Q. Consider a mass starting at the top of a quarter of a circle and sliding down, at what point $(x,y)$ will it detach? note: as a 2D problem

we solved this by using the constraint as follows:

a. Formulating the constraint - $$x^2 +y^2 - R^2 = 0 $$ b. Writing the equations of motion using the constrained Lagrangian - $$L' = \frac{1}{2}m(\dot x^2+\dot y^2) - mgy - \lambda(x^2 +y^2 - R^2)$$ c. Moving over to polar coordinates - $$x=Rcos(\theta)$$ $$y=Rsin(\theta)$$ d. Substituting this into the Lagrangian

d.1.Using energy conservation we have

$$(\dot\theta)^2= \frac{2g}{R}(1-sin(\theta))$$ d.2. When the mass detaches the constraint force vanishes so $$\lambda = 0$$ e. We plug all those things into our equation of motion and get $$sin(\theta) = 2/3$$

My Questions are:

1.In my homework assignment I have the exact same problem but the curve is now defined :

$$y(x) = 2 - cosh(x)$$

I worked in a similar manner:

a. $$L' = \frac{1}{2}m(\dot x^2+\dot y^2) - mgy - \lambda(y+cosh(x)-2)$$ b1. $$\dot y = -sinh(x) \dot x$$ b2. $$\ddot y = -(\ddot x sinh(x) + \dot x^2 cosh(x))$$ c. Wrote E-L equations for x and y substituted b2 into y E-L equation, some algebra, and got to this equation :

d.$$\lambda = \frac{m \dot x^2 cosh(x) - mg}{cosh^2(x)}$$

e. Used conservation of energy to get:

$$\dot x^2 = \frac{2g(cosh(x)-1)}{cosh^2(x)}$$

Solving this I get that cosh(x) = 2 at the detachment point, basically meaning the mass never detaches and follows the whole curve to the bottom which is impossible. I think it's because I force lambda = 0 at the detachment point. I don't understand if it's wrong or not, judging by my lecture notes and the way I understand it, the mass will detach when the constraint force is zero meaning lambda = 0.

If lambda is indeed not equal to zero (at the detachment point) in this case, why is it and how should I find it's value ?

2.What is the point of calculating the partial derivative :

$$\frac{\partial L'}{\partial \lambda}$$ if it always just returns the constraint that we wrote by looking at the system and it's properties

Answer 1

If we have an action given by: $$I=\int_1^2\bigg(L+\lambda f\bigg)\;dt,$$ the Euler-Lagrange equations are given by: $${d\over dt}\bigg({\partial L\over\partial \dot x}\bigg)-{\partial L\over \partial x}=-\lambda{\partial f\over\partial x}.$$ The Lagrangian is $L={m\over 2}(\dot x^2+\dot y^2)-mgy$, and the constraint $f=y+\cosh x-2=0$. We may write the Lagrangian in terms of the constraint as: $$L={m\over 2}(\dot x^2+\dot x^2\sinh^2 x)-mg(2-\cosh x).$$ So we have an Euler-Lagrange equation in the variable $x$. Calculating the relevant derivatives gives: $${d\over dt}\bigg({\partial L\over\partial \dot x}\bigg)=m\ddot x+m\ddot x\sinh^2 x+2m\dot x^2\sinh x\cosh x;\; {\partial L\over\partial x}=m\dot x^2\sinh x\cosh x+mg\sinh x.$$ And: $$\lambda{\partial f\over\partial x}=\lambda\sinh x.$$ So the equation of motion is: $$m\ddot x+m\ddot x^2\sinh^2 x-m\dot x^2\sinh x\cosh x-mg\sinh x=-\lambda\sinh x,$$ from which we can obtain the multiplier as: $$-\lambda={m(\ddot x+\ddot x\sinh^2 x+\dot x^2\sinh x\cosh x-g\sinh x)\over \sinh x}.$$ From Jacobi's integral (conservation of energy) we have: $${m\over 2}\dot x^2+{m\over 2}\dot x^2\sinh^2 x+mg(2-\cosh x)-h=0,$$ where $h$, is a constant. If we differentiate the first integral of the motion we get: $$\ddot x={\dot x^2\sinh x\cosh x+g\sinh x\over \dot x+ \dot x\sinh^2 x},$$ which we can substitute into the earlier obtained expression for the Lagrange multiplier to get: $$-\lambda=m{\dot x^2\sinh x\cosh x+g\sinh x+\dot x^4\sinh x\cosh x-g \dot x^2\sinh x\over\dot x^2\sinh x}.$$ Setting $\lambda=0$ we get: $$\cosh x=g{\dot x^2-1\over \dot x(\dot x^2+1)}.$$ The next logical step, to find where the particle will detach, is to find a way to integrate this equation and solve for $x$, however, just how to do this eludes me for the moment.