# Homework Help: 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.

the follow position equation

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

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

### Staff: Admin

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted