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chipotleaway
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Background:
I've derived the equation of motion of the bead shown in the picture below using the Lagrangian
[tex]\omega^2Rsin(\theta)+r\ddot{\theta}=0[/tex]
The equilibrium points are at θ=0 and π. I'm now to show that only one of these points is stable.
Show that only one of the equilibrium points is stable against small deviations from equilibrium. Do so by considering small deviations [tex]\delta \theta=\theta-\theta_{equilibrium}[/tex] from equilibrium.
Rearranging the equation of motion for the acceleration [itex]\ddot{\theta}[/itex], we have [tex]\ddot{\theta}=-\frac{\omega^2R}{r}sin(\theta)[/tex]
For small angles [itex]\theta[/itex] deviating from 0, sin(θ)≈θ so the above equation becomes [itex]\ddot{theta}=-\frac{\omega^2R}{r}\theta[/itex]. Similarly for deviations from angle of π, sin(θ-π)≈θ-π and therefore [itex]\ddot{\theta}=-\frac{\omega^2R}{r}\theta--\frac{\omega^2R}{r}\pi[/itex].
The latter is larger in magnitude than the former so the acceleration with be bigger and hence it's unstable (but this doesn't really prove anything, does it?). Don't know what I'm doing here!
I've derived the equation of motion of the bead shown in the picture below using the Lagrangian
[tex]\omega^2Rsin(\theta)+r\ddot{\theta}=0[/tex]
The equilibrium points are at θ=0 and π. I'm now to show that only one of these points is stable.
Homework Statement
Show that only one of the equilibrium points is stable against small deviations from equilibrium. Do so by considering small deviations [tex]\delta \theta=\theta-\theta_{equilibrium}[/tex] from equilibrium.
The Attempt at a Solution
Rearranging the equation of motion for the acceleration [itex]\ddot{\theta}[/itex], we have [tex]\ddot{\theta}=-\frac{\omega^2R}{r}sin(\theta)[/tex]
For small angles [itex]\theta[/itex] deviating from 0, sin(θ)≈θ so the above equation becomes [itex]\ddot{theta}=-\frac{\omega^2R}{r}\theta[/itex]. Similarly for deviations from angle of π, sin(θ-π)≈θ-π and therefore [itex]\ddot{\theta}=-\frac{\omega^2R}{r}\theta--\frac{\omega^2R}{r}\pi[/itex].
The latter is larger in magnitude than the former so the acceleration with be bigger and hence it's unstable (but this doesn't really prove anything, does it?). Don't know what I'm doing here!
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