Lagrangian for simple pendulum

In summary, the Lagrangian for a simple pendulum of mass m whose point of support moves uniformly on a vertical circle with constant frequency \omega in a uniform gravitational field can be written as T = {1\over 2}m \left(\dot {r}^2 + r^2\dot{\theta}^2\right) and V = -mgr\cos \theta, with r and \theta as the chosen coordinates. The angular velocity \omega can be added to the equation of motion by incorporating it into the expression for T.
  • #1
Reshma
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Find the Lagrangian for a simple pendulum of mass m whose point of support moves uniformly on a vertical circle with constant frequency [itex]\omega[/itex] in a uniform gravitational field.

Let 'l' be the length of the pendulum string. Using plane polar coordinates:
Let T be the KE of the pendulum.
[tex]T = {1\over 2}m \left(\dot {r}^2 + r^2\dot{\theta}^2\right)[/tex]
Let V be the PE.
[tex]V = -mgr\cos \theta[/tex]
r = l = constant
I am wondering how to add the angular velocity [itex]\omega[/itex] to the equation of motion. Need help here.
 
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  • #2
how is omega related to theta?
 
  • #3
Reshma said:
Find the Lagrangian for a simple pendulum of mass m whose point of support moves uniformly on a vertical circle with constant frequency [itex]\omega[/itex] in a uniform gravitational field.

Let 'l' be the length of the pendulum string. Using plane polar coordinates:
Let T be the KE of the pendulum.
[tex]T = {1\over 2}m \left(\dot {r}^2 + r^2\dot{\theta}^2\right)[/tex]
Let V be the PE.
[tex]V = -mgr\cos \theta[/tex]
r = l = constant
I am wondering how to add the angular velocity [itex]\omega[/itex] to the equation of motion. Need help here.
As with your other problem, You need to pick a pair of coordinates. A pair of angles looks good here also.
 

Related to Lagrangian for simple pendulum

1. What is the Lagrangian for a simple pendulum?

The Lagrangian for a simple pendulum is the difference between the kinetic energy and potential energy of the pendulum. It is given by L = T - V, where T is the kinetic energy and V is the potential energy.

2. How is the Lagrangian derived for a simple pendulum?

The Lagrangian for a simple pendulum can be derived using the principle of least action, where the Lagrangian is minimized to find the path of least resistance. It can also be derived using Hamilton's principle, which states that the action of a system is stationary at the actual path taken by the system.

3. What are the advantages of using the Lagrangian for a simple pendulum?

The Lagrangian approach allows for a more efficient and elegant way to describe the motion of a simple pendulum compared to using Newton's laws of motion. It also takes into account all forces acting on the pendulum, including non-conservative forces, and can be easily extended to more complicated systems.

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

Yes, the Lagrangian can be used to solve for the motion of a simple pendulum by setting up the Lagrangian equations of motion and solving for the variables. This approach is often preferred over using Newton's laws of motion, especially for more complex systems.

5. How does the Lagrangian for a simple pendulum change with different initial conditions?

The Lagrangian for a simple pendulum remains the same regardless of the initial conditions, as long as the pendulum is in an ideal environment with no external forces. However, the resulting motion of the pendulum may vary depending on the initial conditions, such as the initial angle and velocity.

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