# Lagrangian of Pendulum with Oscillating Hinge

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1. Nov 22, 2015

### Xyius

1.) The Problem Statement:

a.) Find the Lagrangian of a pendulum where the height of the hinge is oscillating in the y direction and is is defined as a function $y_0=f(t)$

b.) Add a function (a gauge transformation) of the form $\frac{d F(\theta,t)}{dt}$ to the original lagrangian $\mathcal{L}$ to obtain a new lagrangian $\mathcal{L}'=T'-V'$ where $T'$ is the "usual kinetic energy of a pendulum" and $V'$ is the potential of a time varying gravitational field. Compare their Hamiltonians

2.) Relevant Equations

Lagranges equations

3.) Attempt at a Solution
Part a I got. The Lagrangian I got is the following. (I can go into more detail about how I got this if required)

$$\mathcal{L}=\frac{1}{2}m\left( l^2 \dot{\theta}^2 + \dot{y}^2_0(t) + 2 \dot{y}_0(t)l \dot{\theta} \sin\theta \right) - mg \dot{y}_0(t)+mgl\cos\theta$$

I know this answer is correct because for when $y_0(t)=a\cos\omega t$ it yields the equation of motion that is the same as the result in the text with a function of this form.

Part b is where I am stuck. When I see "potential of a time varying gravitational field," I think something of the form $m g(t) y$. The only idea I can come up with is to write the function $\frac{dF}{dt}$ in a form that cancels out most of the terms of the original Lagrangian, but only keeps the kinetic term, $\frac{1}{2}l^2\dot{\theta}^2$ and other terms so that I can write the potential in the form I desire. But I cannot seem to make any headway with this, and I am very iffy on if this approach is correct or not. I think I understand the premise of the problem, instead of viewing the pendulum to have an oscillating hinge, instead cast the problem as a normal pendulum in an oscillating gravitational field. Does anyone have any thoughts or suggestions??

2. Nov 22, 2015

### TSny

I think you have the right idea. You just need to play around with constructing the appropriate function F(θ,t). Take it one term at a time. For example, it is easy to construct a term in F such that dF/dt yields a term of the form $mg\dot{y}_0$.

3. Nov 22, 2015

### Xyius

I made a typo in my original post, the one term at the end of the lagrangian should be $mg y_0$ not $mg \dot{y_0}$.

I figured it out! I was correct in my reasoning as you pointed out, but the trick was to realize that part of the integral of the third term can be written in the form $-ml \dot{y_0(t)}\cos\theta.$ So you can re-write it in terms of it's integral and cancel out the extra terms by the gauge transformation.