Setting Up Lagrangian, David Morin 6.25

In summary, a rigid “T” consists of a long rod glued perpendicular to another rod of length l that is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ω, and a mass m connected to the intersection of the rods by a spring with spring constant k and equilibrium length zero. The position of the mass along the long rod, r(t), can be found by setting up the Lagrangian with the correct kinetic and potential energy terms. The special value of ω is when it allows for a simplified expression of the Lagrangian and easier calculation of the mass's position.
  • #1
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Homework Statement



A rigid “T” consists of a long rod glued perpendicular to another rod of length l that is pivoted at the origin. The T rotates around in a horizontal plane with constant frequency ω. A mass m is free to slide along the long rod and is connected to the intersection of the rods by a spring with spring constant k and equilibrium length zero. Find r(t), where r is the position of the mass along the long rod. There is a special value of ω; what is it, and why is it special? Figure 6.27 in David Morin Classical Mechanics

Homework Equations


L = T - V

The Attempt at a Solution



I am having trouble setting up the Lagrangian for this system so that it is in appropriate coordinate system.

So far T = ½mv^2 + ½ω^2 + ½kx^2

V = -kx - mghsin(ωt)
 
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  • #2
Your expression for the Lagrangian is wrong. I think that the kinetic energy term should be:
(1/2)Iω2 + (1/2)mv2 and the potential energy should be
-(1/2)kr2. I is the moment of inertia of the rod and v is the velocity of the mass in the horizontal plane. I suggest writing the velocity in terms of the rotational angle and the x and y components in the plane, i.e.
x = rcos(ϑ), y = rsin(ϑ),
dx/dt = -rdϑ/dt sin(ϑ) +dr/dtcos(ϑ),
dy/dt =rdϑ/dt + dr/dtsin(ϑ),
and v2=((dx/dt)2 + (dy/dt)2)
 
  • #3
okay thanks! Except l is changing with time as well based on where the mass is on the rod, so l = sqrt(l2 + r2)
 

1. What is Lagrangian?

Lagrangian, also known as the Lagrangian function, is a mathematical function that describes the dynamics of a physical system. It is named after the Italian mathematician and astronomer Joseph-Louis Lagrange.

2. Why is Lagrangian important in physics?

Lagrangian is important in physics because it provides a more elegant and powerful approach to solving problems in mechanics. It allows for the use of generalized coordinates, which simplifies complex systems and makes it easier to apply the laws of physics.

3. How do you set up Lagrangian?

To set up Lagrangian, you need to identify the generalized coordinates of the system and the kinetic and potential energies of the system. Then, you can use the Lagrangian function L = T - V, where T is the kinetic energy and V is the potential energy, to describe the dynamics of the system.

4. What is the difference between Lagrangian and Newtonian mechanics?

The main difference between Lagrangian and Newtonian mechanics is the use of generalized coordinates in Lagrangian mechanics. In Newtonian mechanics, the equations of motion are described in terms of Cartesian coordinates, while in Lagrangian mechanics, generalized coordinates are used, which can simplify the equations and make them easier to solve.

5. How can Lagrangian be applied to real-world problems?

Lagrangian can be applied to a wide range of real-world problems, including classical mechanics, electromagnetism, and quantum mechanics. It has been used to study the motion of celestial bodies, the behavior of fluids, and the properties of particles at the quantum level. It is a powerful tool for analyzing complex systems and predicting their behavior.

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