# Lagrangian and Euler-Lagrange of a Simple Pendulum

1. Sep 27, 2015

### Yeah Way

1. The problem statement, all variables and given/known data
A simple pendulum with mass m and length ℓ is suspended from a point which moves
horizontally with constant acceleration a
> Show that the lagrangian for the system can be written, in terms of the angle θ,
L(θ, θ, t˙ ) = m/2(ℓ^2θ˙^2 + a^2t^2 − 2aℓtθ˙ cosθ) + mgℓ cos θ

> Determine the Euler–Lagrange equation for the system.
2. Relevant equations

3. The attempt at a solution
I thought I could prove that l^2θdot^2 + a^2t^2 - 2altθdotCosθ was v^2 using relative velocities: v^2 = (at - lθdot)^2 = (l^2θ^2 + a^2t^2 - 2altθdot). But I've no idea where the Cosθ is coming from, so I can only assume I'm wrong somewhere.

I also can't understand how V = -mglCosθ
h for this pendulum should be l(1 - Cosθ) shouldn't it?

Any help's appreciated. Thanks.

#### Attached Files:

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2. Sep 27, 2015

### CAF123

It can indeed be solved via relative velocities, but it would be much simpler to just write the generic position vector for the mass, noting that it has an additional horizontal component of velocity of $at$.
It depends on where you choose your zero of potential but, regardless, at the level of the lagrangian an additive constant makes no difference to the kinematics since they are obtained through taking derivatives of the lagrangian through the Euler-Lagrange equations.

3. Sep 27, 2015

### Yeah Way

Are you saying the − 2aℓtθ˙ cosθ is part of the angular velocity? I'm sorry, I'm really at a loss with this.

4. Sep 28, 2015

### CAF123

The first two terms of the lagrangian can be attributed to a rotational kinetic energy of the mass about the pendulum pivot and the translational kinetic energy due to the additional horizontal component of velocity imposed on it. The interpretation of the term with cos θ in it is not so easy.

If you write the position vector for the mass to an inertial frame and take its time derivative you will have $\bf \dot{r}$. Then the kinetic energy to this frame is given by $\frac{1}{2}m \bf{\dot{r}^2}$. Does it help?