Varying inclination on plane: Undetermined multipliers

  • Thread starter Thread starter davidbenari
  • Start date Start date
  • Tags Tags
    Plane
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 1K views
davidbenari
Messages
466
Reaction score
18

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.
 
Last edited:
Physics news on Phys.org
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.