Lagrange Multipliers: Deriving EOM & Conditions for Contact Loss

[ct1993]

Homework Statement

An object of mass m, and constrained to the x-y plane, travels frictionlessly along a curve f(x), while experiencing a gravitational force, m*g. Starting with the Lagrangian for the system and using the method of Lagrange multipliers, derive the equations of motion for the bead, and then derive the condition under which the bead would lose contact with the surface

Homework Equations

L=T-V, as well as several other

The Attempt at a Solution

T=.5*m*v^2
=>.5*m*((dx/dt)^2)+((dy/dt)^2)
V=m*g*y
=>m*g*f(x) (When the object is on the curve)
L=.5*m*((dx/dt)^2)+((dy/dt)^2)+m*g*f(x)

I think I'm correct up to this point, but I'm not sure how to apply Lagrange multipliers to the Lagrangian and reduce to the equations of motion. Also, I know intuitively that the object would lose contact with the surface when dy/dt is greater (less negative) than df(x)/dt, but I have no idea how I would find this condition using Lagrangian mechanics. If someone could help, I would be really greatful.

 

[Orodruin]

Lagrange multipliers are used to solve variational problems with constraints. In classical mechanics, there is a one to one correspondence between the Lagrange multiplier and the constraining force. What part exactly are you having trouble with? Do you know how to use Lagrange multipliers in general?

 

[ct1993]

Lagrange multipliers are used to solve variational problems with constraints. In classical mechanics, there is a one to one correspondence between the Lagrange multiplier and the constraining force. What part exactly are you having trouble with? Do you know how to use Lagrange multipliers in general?

I believe the constraining equation is 0 = f(x) - y and that the condition when the bead leaves the surface is a condition which cause it not to be true. Using that constraining equation, I think the new Lagrangian would be L=.5*m*((dx/dt)^2)+((dy/dt)^2)+m*g*f(x)-k(f(x)-y) where k is the multiplier. I would then set the condition such that the gradient of the new Lagrangian is 0, which means dL/dx=0=m*g*df/dx-k*df/dx, dL/dy=0=m*g-k, and dL/dk=0=f(x)-k. I'm not sure if I did that right and I have no idea where I would go from here. Any help would be appreciated.

 

[stevendaryl]

I believe the constraining equation is 0 = f(x) - y and that the condition when the bead leaves the surface is a condition which cause it not to be true. Using that constraining equation, I think the new Lagrangian would be L=.5*m*((dx/dt)^2)+((dy/dt)^2)+m*g*f(x)-k(f(x)-y) where k is the multiplier. I would then set the condition such that the gradient of the new Lagrangian is 0, which means dL/dx=0=m*g*df/dx-k*df/dx, dL/dy=0=m*g-k, and dL/dk=0=f(x)-k. I'm not sure if I did that right and I have no idea where I would go from here. Any help would be appreciated.

You have an original Lagrangian: L = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} m \dot{y}^2 -V(x,y) (where \dot{X } = \frac{dX}{dt})

You have a constraint: f(x) - y = 0. What's better for Lagrangian approach is the time derivative of this:

\frac{d}{dt} (f(x) - y) = \frac{df}{dx} \frac{dx}{dt} - \frac{dy}{dt} = f' \dot{x} - \dot{y} (where f' = \frac{df}{dx}).

At this point, we can insert the lagrange multiplier k(t): (Note: it's a function of t, not a constant)

\tilde{L} = L - k(t) (f' \dot{x} - \dot{y})

Now, use the usual Lagrangian equations of motion with the new Lagrangian \tilde{L}.