Classical mechanics - Lagrange multipliers

In summary, the disk moves on an inclined plane without friction and perpendicular to the plane. The constraints are that the velocity is always at the same direction as the plane and that the Lagrangian is always zero. The initial conditions are that at t=0, x=0, y=0, \dot{x}=0, \dot{y}=0, and \varphi=0.
  • #1
LiorE
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0

Homework Statement


A disk moves on an inclined plane, with the constraint that it's velocity is always at the same direction as it's plane (similar to an ice skate, maybe). In other words: If [tex]\hat{n}[/tex] is a vector normal to the disk's plane, we have at all times: [tex]\hat{n} \cdot \vec{v} = 0[/tex]. Also, it's free to move without friction, and always perpendicular to the plane. (as seen in the figure.)

I need to get and solve the equations of motion for certain initial conditions that I'll write promptly. We set an x-y coordinate system at the top-right corner of the plane with the y-axis going downwards, and denote that angle between [tex]\hat{n}[/tex] as [tex]\varphi[/tex].

Homework Equations



The constraint is:

[tex]c_1 = \dot{x}\cos\varphi + \dot{y}\sin\varphi[/tex]

and accordingly the Lagrangian is:

[tex]L = \frac{1}{2}m(\dot{x}^2 + \dot{y}^2) + \frac{1}{2}I\dot{\varphi}^2 + mgy + \lambda(\dot{x}\cos\varphi + \dot{y}\sin\varphi)[/tex]

The initial conditions that were given are that at t=0:

[tex]x=0, y=0, \dot{x}=0, \dot{y}=0, \varphi = 0, \dot{\varphi} = \omega_0 [/tex]

The Attempt at a Solution


The obvious way of solving is to use Euler-lagrange and get the equations of motion. The problem is that I can't solve them! They're too damn complicated. There is a hint that I should try to find constants of motion by setting t=0 in the equations, but I can't seem to find them.

I would appreciate any help...

Thanks in advance!
 

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  • #2
ok well if [itex] y=0 [/atex] and [itex] \dot y=0 [/itex] then [itex] y=0 \forall t [/itex] which will simplfy your Lagrangian
 
  • #3
The same would happen for [itex] x [/itex] so I get [itex] L = \frac{1}{2} I {\omega_0}^{2} [/itex] Then solving the Euler-Lagrange I get [itex] I \ddot \psi = 0 [/itex]. This could be massively wrong but I though I'd give it a shot anyway.
 
  • #4
Thanks, I solved it by variating the action integral - the TA said that Lagrange multipliers give here the wrong answer, and he doesn't know why. I guess it's a question for mathematicians to answer.
 

Related to Classical mechanics - Lagrange multipliers

1. What is classical mechanics?

Classical mechanics is a branch of physics that deals with the motion of objects under the influence of forces. It is based on the laws of motion and gravitation as described by Sir Isaac Newton in the late 17th century.

2. What are Lagrange multipliers?

Lagrange multipliers are mathematical tools used in classical mechanics to find the extreme values of a function subject to certain constraints. They were developed by Joseph-Louis Lagrange in the late 18th century.

3. How are Lagrange multipliers used in classical mechanics?

In classical mechanics, Lagrange multipliers are used to optimize a system by finding the minimum or maximum values of a function while satisfying a set of constraints. This allows for the determination of the most stable or efficient state of a physical system.

4. What are some practical applications of Lagrange multipliers in classical mechanics?

The use of Lagrange multipliers in classical mechanics has many practical applications, including determining the trajectory of a projectile, finding the minimum energy state of a system, and optimizing the motion of a pendulum.

5. Are there any limitations to using Lagrange multipliers in classical mechanics?

While Lagrange multipliers are a powerful tool in classical mechanics, they are not always applicable. They can only be used for systems that can be described by a set of equations and constraints, and they may become increasingly complex for more complicated systems.

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