Accelerating Elastic Pendulum (Lagrangian)

[darkfall13]

Homework Statement

A pendulum of mass $$m$$ is suspended by a massless spring with equilibrium length $$L$$ and spring constant $$k$$. The point of support moves vertically with constant acceleration. Write the Lagrangian of the system and the equation of motion.

Homework Equations

$$L = T - U$$

$$\frac{\partial L}{\partial x_i} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x_i}} = 0$$

The Attempt at a Solution

We can draw a vertical and call the angle of the pendulum to this $$\theta$$. I know for an elastic pendulum we have (where $$l$$ is the variable length) $$T = \frac{m}{2} \left( \dot{l^2} + l^2 \dot{\theta^2}\right)$$ and $$U = \frac{1}{2}k\left(l-L\right)^2 + mgy$$ but I'm not sure how to involve the acceleration term. A hint in the right direction would be greatly appreciated :D

 

[Jhero]

Thank you for your question. To write the Lagrangian for this system, we need to consider the kinetic and potential energies of the pendulum. The kinetic energy is given by the sum of the translational and rotational kinetic energies, which can be expressed as T = \frac{1}{2}m\left(\dot{x}^2 + \dot{y}^2\right) + \frac{1}{2}I\dot{\theta}^2, where x and y are the horizontal and vertical positions of the pendulum, and I is the moment of inertia about the point of suspension. The potential energy can be written as U = \frac{1}{2}k\left(\sqrt{x^2+y^2}-L\right)^2 + mgy, where g is the acceleration due to gravity.

Now, using the Lagrangian equation L = T - U, we can write the equation of motion for the pendulum as \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0 and \frac{d}{dt}\frac{\partial L}{\partial \dot{y}} - \frac{\partial L}{\partial y} = 0. These equations can be simplified using the expressions for T and U, and the acceleration term can be included by taking into account the acceleration of the point of support. I hope this helps guide you in the right direction. Keep up the great work in your studies!