Find the Lagrangian of simple pendulum?

Answer 1

The generalized coordinate for the pendulum is the angle of the arm off vertical, theta. Theta is 0 when the pendulum arm is down and pi when the arm is up.

M = mass of pendulum

L = length of pendulum arm

g = acceleration of gravity

\theta = angle of pendulum arm off vertical

\dot{\theta} = time derivative of \theta

What are the kinetic and potential energies?

Kinetic energy: T = (1/2)*M*(L*\dot{\theta})^2

Potential energy: V' = MLg(1-cos(\theta))

V = -MLg*cos(\theta)

--note: we can shift the potential by any constant, so lets choose to drop the MLg

The Lagrangian is L=T-V: L = (1/2)ML^2\dot{\theta}^2 + MLg*cos(\theta)

Answer 2

The Lagrangian of a simple pendulum can be given as: [ L = T - V ] where the kinetic energy (T) is given by (\frac{1}{2} m l^2 \dot{\theta}^2) and the potential energy (V) is given by (mgl(1 - \cos(\theta))), where (m) is the mass of the pendulum bob, (l) is the length of the pendulum, (g) is the acceleration due to gravity, (\theta) is the angle of the pendulum with the vertical, and (\dot{\theta}) is the angular velocity.