- #1
MisterX
- 764
- 71
Homework Statement
Homework Equations
[itex] \frac{\partial\mathcal{L} }{\partial q} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}}[/itex]
[itex] \cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)[/itex]
The Attempt at a Solution
The position of the center is
[itex]
\frac{D}{2}\left(cos(\omega t)\hat{x} + sin (\omega t)\hat{y}\right)
[/itex]
The vector from the center to the mass is
[itex]
\frac{D}{2}\left(cos(\omega t + \phi)\hat{x} + sin (\omega t +\phi)\hat{y}\right)
[/itex]
The position of the mass is the sume of these two vectors
[itex]
\frac{D}{2}\left[\left(cos(\omega t) + cos(\omega t + \phi)\right)\hat{x} +\left(sin(\omega t) + sin(\omega t + \phi)\right)\hat{y}\right]
[/itex]
The velocity is
[itex]
\frac{D}{2}\left[-\left(sin(\omega t)\omega + sin(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{x} +\left(cos(\omega t)\omega + cos(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{y}\right]
[/itex]
The velocity squared is
[itex]
\frac{D^2}{4}\left[\omega^2sin^2(\omega t) +2\omega sin(\omega t)sin(\omega t + \phi)\left(\omega + \dot{\phi}\right) +sin^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right][/itex]
[itex]
+\frac{D^2}{4}\left[\omega^2cos^2(\omega t) +2\omega cos(\omega t)cos(\omega t + \phi)\left(\omega + \dot{\phi}\right) +cos^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right][/itex]
[itex]
=\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)\left[ sin(\omega t)sin(\omega t + \phi) + cos(\omega t)cos(\omega t + \phi) \right] +\left(\omega + \dot{\phi}\right)^2\right][/itex]
[itex]
\cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)
[/itex]
[itex]
v^2 =\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right][/itex]
If I set [itex]\phi = 0 [/itex] and [itex]\dot{\phi} = 0 [/itex], I get [itex]v^2 = D^2\omega^2 [/itex], which is what I expect, since the diameter should just be rotating with angular frequency [itex] \omega[/itex] in this case.
[itex]
U = mgy = mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)
[/itex]
[itex]
\mathcal{L} =\frac{mD^2}{8}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right] -mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)
[/itex]
[itex] \frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}[/itex]
[itex] \frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{8}2\omega\left(\omega + \dot{\phi}\right)sin(\phi) -mg\frac{D}{2}\cos(\omega t + \phi) [/itex]
[itex] \frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{8}\left[ 2\omega cos(\phi) + 2\dot{\phi} + 2\omega \right] = \frac{mD^2}{4}\left[ \omega cos(\phi) + \dot{\phi} + \omega \right] [/itex]
[itex] \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right] [/itex]
[itex]
\frac{mD^2}{4}\ddot{\phi} = - \frac{mD^2}{4}\omega\dot{\phi}sin(\phi) - mg\frac{D}{2}\cos(\omega t + \phi)
[/itex]
[itex]
\ddot{\phi} = -\omega\dot{\phi}sin(\phi) - \frac{2g}{D}\cos(\omega t + \phi)
[/itex]
That doesn't look exactly like a simple pendulum to me. I'd appreciate some help.