Varying inclination on plane: Undetermined multipliers

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SUMMARY

The discussion focuses on determining the constraint force acting on a particle of mass ##m## resting on a smooth plane inclined at an angle ##\theta##, which is raised at a constant rate ##\alpha##. The Lagrangian is defined as ##L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-mgy##, and the constraint equation is given by ##f=\frac{y}{x}-\tan\alpha t = 0##. The method of undetermined multipliers is applied, leading to the equations ##m\ddot{x}+\lambda \frac{y}{x^2} = 0## and ##mg+m\ddot{y}=\frac{\lambda}{x}##. The discussion suggests exploring polar coordinates while maintaining both ##r## and ##\phi## as generalized coordinates without incorporating the constraint into energy terms.

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Homework Statement


A particle of mass ##m## rests on a smooth plane. The plane is raised to an inclination ##\theta## at constant rate ##\alpha##. Find the constraint force.

Homework Equations

The Attempt at a Solution


##L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2)-mgy## Lagrangian

##f=\frac{y}{x}-\tan\alpha t = 0 ## constraint equation

##\partial_y f = \frac{1}{x}##

##\partial_x f = \frac{-y}{x^2}##

##\partial_q L - d_t \partial_\dot{q} L + \lambda \partial_q f = 0 ## Method of undetermined multipliers formula.

##\to \boxed{m\ddot{x}+\lambda \frac{y}{x^2} = 0} \quad \boxed{mg+m\ddot{y}=\frac{\lambda}{x}}##

Using tedious manipulation I've gotten to the point where I can say

##\ddot{x}x+\ddot{y}y+gy=0##

And haven't found any other useful formula.

I know I could switch to a polar coordinate basis and find ##r(t)## there and solve ##x## and ##y## and indirectly find constraint forces, but I'm not interested in that. Unless I'm clearly using the Lagrange undetermined multipliers.
 
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Here's an idea: Write the Lagrangian in terms of polar coordinates, but keep both ## r ## and ## \phi ## as the generalized coordinates - do not incorporate the constraint into the kinetic or potential energy terms. Rather, incorporate it through the Lagrange multiplier technique.
 

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