How Do You Derive the Lagrangian for a Pendulum on an Oscillating Cart?

[noranne]

Homework Statement

A simple pendulum (mass M and length L) is suspended from a cart (mass m) that can oscillate on the end of a spring of force constant k. Write the Lagrangian in terms of the two generalized coordinates x and $$\phi$$, where x is the extension of the spring from its equilibrium length. Find the two Lagrange equations.

Homework Equations

L = T - U

$$\frac{\partial L}{\partial x} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{x}}) = 0$$

$$\frac{\partial L}{\partial \phi} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}}) = 0$$

The Attempt at a Solution

So far I'm thinking $$T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}M(L\dot{\phi} + \dot{x})$$ and $$U = \frac{1}{2}kx^2 + -Lcos(\phi)Mg$$

I feel like that's wrong, like the potential energy of the pendulum might not just depend on its height below the track. And the kinetic energy might be wrong. Really, this problem is totally throwing me for a loop, after I was feeling pretty confident about Lagrangian mechanics for about 4 problems. Any help??

 

[Meir Achuz]

So far I'm thinking $$T =+ \frac{1}{2}M(L\dot{\phi})^21 + \dot{x})$$

This should be $$T =+ \frac{1}{2}M(L\dot{\phi})^2$$
The rest looks OK.

 

[Gokul43201]

That's not right either.

In the frame attached to the cart, the kinetic term is:
$$T'=M/2(\dot{x'}^2 + \dot{y'}^2)$$

where $\dot{x'}=L\dot{\phi}cos \phi$. In the lab frame, this becomes:

$$T=M/2((\dot{x'}+\dot{x})^2 + \dot{y'}^2)$$