- #1
PeteSampras
- 43
- 2
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 don't understand
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 don't understand