Using Lagrangian and Euler to Analyze the Falling Stick Problem

In summary, the problem is to find the time it takes for a meter stick to fall from an angle of 1 degree from the vertical, on a frictionless surface and against a frictionless wall. Lagrange and Euler methods are to be used to solve this problem. The potential of the stick is mgL and the kinetic energy has both rotational and translational components. The rotational kinetic energy is \frac{1}{2}I\omega^2 and the translational kinetic energy is \frac{1}{2}mv^2, with v being x-dot squared and y-dot squared. The stick has two degrees of freedom, with the center of mass describing a circle in its motion. The potential is mglcos(theta).
  • #1
Raziel2701
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Homework Statement


A meter stick stands on a frictionless surface and leans against a frictionless wall as shown. It is released to fall when it makes an angle of 1 degree from the vertical. Use Lagrange and Euler to find how long it takes the stick to fall to the ground.



The Attempt at a Solution


The potential of the stick will essentially just be mgL correct? Since the stick starts from an angle of 1 degree.

The Kinetic energy would have two components, a rotational and a translational. For rotation, I guess I'll just use [tex]\frac{1}{2}I\omega^2[/tex]. I guess I could look up the moment of inertia later on but I want to get the setup right.

For translational kinetic energy, when the stick falls, the center of mass of the stick falls down but it also moves horizontally, so there's an x and y component to the velocity so the KE would be [tex]\frac{1}{2}mv^2[/tex]

What concerns me is that v should be x-dot squared and y-dot squared but then it seems like I have too many variables for an object that has two(?) degrees of freedom.

How do I set this up?
 
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  • #2
Ok I realized a few things that are very hard to grasp for me. Mainly, understanding how we can describe the problem requires such a feel for the problem that I request some links, or books that deal with this method(Lagrangian method).

So the center of mass describes a circle in its motion so the only degree of freedom is governed by the angle correct? The potential would also be different now, it should be mglcos(theta).

I'm going back to work on this, I'll come back if I need more help but I'd still like some more references to read up and study if you'd be so kind.
 

1. What is the Lagrangian of a falling stick?

The Lagrangian of a falling stick is a mathematical expression that represents the energy of the system consisting of a falling stick. It takes into account the kinetic energy and potential energy of the stick as it falls.

2. How is the Lagrangian of a falling stick calculated?

The Lagrangian of a falling stick can be calculated using the equation L = T - V, where T represents the kinetic energy of the stick and V represents its potential energy. The kinetic energy can be calculated using the stick's mass and velocity, while the potential energy can be calculated using its height and the force of gravity.

3. What does the Lagrangian of a falling stick tell us?

The Lagrangian of a falling stick tells us about the overall energy of the system and how it changes as the stick falls. It can also be used to determine the motion and behavior of the stick as it falls.

4. Can the Lagrangian of a falling stick be used to solve real-world problems?

Yes, the Lagrangian of a falling stick can be used to solve real-world problems related to the motion and energy of a falling stick. It can also be applied to other systems in physics, such as pendulums and springs.

5. Are there any limitations to using the Lagrangian of a falling stick?

While the Lagrangian of a falling stick is a useful tool for analyzing the energy and motion of a falling stick, it does have some limitations. It is based on certain assumptions and may not accurately represent all aspects of a real-world falling stick. It is important to consider other factors and sources of error when using the Lagrangian in practical applications.

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