Lagrangian of Simple Pendulum with Fixed Masses and Horizontal Bar

In summary, a system consisting of two fixed masses, m1 and m2, connected by a rigid rod of length l is considered in a uniform gravitational field g. The mass m1 is attached to a horizontal bar that allows it to move freely in the x direction but not in the y direction. The Lagrangian of the system is determined using Euler-Lagrange equations, with the only issue arising in the kinetic energy terms. The correct solution includes an additional term, \frac{1}{2}m_{2}(2l\dot{x}\dot{θ}cosθ), which accounts for the x velocity dependence. Expressions for the x and y coordinates of m2 are found in terms of the x coordinate of m
  • #1
cpsinkule
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Homework Statement


2 masses, m[itex]_{1}[/itex] and m[itex]_{2}[/itex] are fixed at the endpoints of a rigid rod of length l. mass m[itex]_{1}[/itex] is attached to a horizontal bar so that it may move in the x direction freely, but not in the y direction. let θ be the angle the rod makes with the vertical, what is the corresponding Lagrangian of the system if it is assumed to be in the uniform gravitational field g?


Homework Equations


Euler Lagrange eqns


The Attempt at a Solution


The only issue I am running into with this problem is in the kinetic energy terms. My kinetic terms are : [itex]\frac{1}{2}[/itex](m[itex]_{1}[/itex]+m[itex]_{2}[/itex])[itex]\dot{x}[/itex][itex]^{2}[/itex]+[itex]\frac{1}{2}[/itex]m[itex]_{2}[/itex]l[itex]^{2}[/itex][itex]\dot{θ}[/itex][itex]^{2}[/itex] but the book proposes a solution the same as mine except with the added term [itex]\frac{1}{2}[/itex]m[itex]_{2}[/itex](2l[itex]\dot{x}[/itex][itex]\dot{θ}[/itex]cosθ). I am not understanding where this term comes from, I thought I took care of the x velocity dependence in the first term.
 
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  • #2
Find expressions for the x and y coordinates of m2 in terms of the x coordinate of m1 and θ.
 

What is the Lagrangian of a simple pendulum?

The Lagrangian of a simple pendulum is a mathematical function that describes the motion of the pendulum based on its position, velocity, and acceleration. It is derived from the principle of least action, where the motion of the pendulum is the result of minimizing the action (a measure of the energy) of the system.

How is the Lagrangian used to describe the motion of a simple pendulum?

The Lagrangian is used to describe the motion of a simple pendulum by using the principle of least action. By minimizing the action, the equation of motion for the pendulum can be derived, which describes the position, velocity, and acceleration of the pendulum at any given time.

What are the advantages of using the Lagrangian to study the simple pendulum?

One advantage of using the Lagrangian to study the simple pendulum is that it provides a more general and elegant approach to analyzing the system. It takes into account both the kinetic and potential energy of the pendulum, making it easier to understand the dynamics of the system. Additionally, it can be applied to more complex systems with multiple pendulums or non-uniform pendulums.

How does the Lagrangian of a simple pendulum change with different parameters?

The Lagrangian of a simple pendulum can change with different parameters such as the length of the pendulum, the mass of the bob, and the initial conditions. These parameters affect the potential and kinetic energy of the system, which in turn, affects the equation of motion for the pendulum. As a result, the Lagrangian may change depending on these parameters.

Can the Lagrangian be used to solve for the motion of the simple pendulum?

Yes, the Lagrangian can be used to solve for the motion of the simple pendulum. By using the equation of motion derived from the Lagrangian, the position, velocity, and acceleration of the pendulum can be calculated at any given time. This allows for a more comprehensive understanding of the system's behavior and can be used to make predictions about its motion.

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