# Equibrium point

1. Oct 5, 2014

### PeteSampras

I have a pendulum of large L , suspended at a disk of radius R, with angular velocity constant omega.

$x= L \cos \theta + R \cos(\omega t)$
$x= L \sin \theta + R \sin(\omega t)$

where $R \cos(\omega t), R \sin(\omega t)$ are de coordinates of mobile system.

The equation of motion is

$L \ddot{\theta}+\sin \theta (R \omega^2 \cos(\omega t)+g)+b \omega^2\sin(\omega t) \cos \theta$

The problem says : " the equilibrim at the mobile system is the horizontal position"

¿this means that $\theta=0$??

If the last is true, ¿why $\theta=0$ is equilibrium?

If i see the force at equation of motion $\sin \theta (R \omega^2 \cos(\omega t)+g)+b \omega^2\sin(\omega t) \cos \theta$, this is not zero when $\theta=0$...i dont understand

2. Oct 10, 2014