Homework Help: Mass moving on hoop (with hoop itself rotating)

1. Apr 8, 2013

MisterX

1. The problem statement, all variables and given/known data

2. Relevant equations

$\frac{\partial\mathcal{L} }{\partial q} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{q}}$
$\cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$

3. The attempt at a solution

The position of the center is
$\frac{D}{2}\left(cos(\omega t)\hat{x} + sin (\omega t)\hat{y}\right)$
The vector from the center to the mass is
$\frac{D}{2}\left(cos(\omega t + \phi)\hat{x} + sin (\omega t +\phi)\hat{y}\right)$
The position of the mass is the sume of these two vectors
$\frac{D}{2}\left[\left(cos(\omega t) + cos(\omega t + \phi)\right)\hat{x} +\left(sin(\omega t) + sin(\omega t + \phi)\right)\hat{y}\right]$
The velocity is
$\frac{D}{2}\left[-\left(sin(\omega t)\omega + sin(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{x} +\left(cos(\omega t)\omega + cos(\omega t + \phi)\left(\omega + \dot{\phi}\right)\right)\hat{y}\right]$
The velocity squared is
$\frac{D^2}{4}\left[\omega^2sin^2(\omega t) +2\omega sin(\omega t)sin(\omega t + \phi)\left(\omega + \dot{\phi}\right) +sin^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right]$
$+\frac{D^2}{4}\left[\omega^2cos^2(\omega t) +2\omega cos(\omega t)cos(\omega t + \phi)\left(\omega + \dot{\phi}\right) +cos^2(\omega t + \phi)\left(\omega + \dot{\phi}\right)^2\right]$
$=\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)\left[ sin(\omega t)sin(\omega t + \phi) + cos(\omega t)cos(\omega t + \phi) \right] +\left(\omega + \dot{\phi}\right)^2\right]$
$\cos (\alpha - \beta) = cos(\alpha)cos(\beta) + sin(\alpha)sin(\beta)$
$v^2 =\frac{D^2}{4}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right]$
If I set $\phi = 0$ and $\dot{\phi} = 0$, I get $v^2 = D^2\omega^2$, which is what I expect, since the diameter should just be rotating with angular frequency $\omega$ in this case.

$U = mgy = mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)$

$\mathcal{L} =\frac{mD^2}{8}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right] -mg\frac{D}{2}\left(sin(\omega t) + sin(\omega t + \phi)\right)$

$\frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}$
$\frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{8}2\omega\left(\omega + \dot{\phi}\right)sin(\phi) -mg\frac{D}{2}\cos(\omega t + \phi)$
$\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{8}\left[ 2\omega cos(\phi) + 2\dot{\phi} + 2\omega \right] = \frac{mD^2}{4}\left[ \omega cos(\phi) + \dot{\phi} + \omega \right]$
$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right]$
$\frac{mD^2}{4}\ddot{\phi} = - \frac{mD^2}{4}\omega\dot{\phi}sin(\phi) - mg\frac{D}{2}\cos(\omega t + \phi)$
$\ddot{\phi} = -\omega\dot{\phi}sin(\phi) - \frac{2g}{D}\cos(\omega t + \phi)$
That doesn't look exactly like a simple pendulum to me. I'd appreciate some help.

2. Apr 8, 2013

TSny

Nicely written!

The problem says that the system rotates about a vertical axis. So, does that imply that the system rotates in a horizontal plane?

Check to see if you might have made a careless error when combining terms to get
$\frac{mD^2}{4}\ddot{\phi} = - \frac{mD^2}{4}\omega\dot{\phi}sin(\phi) - mg\frac{D}{2}\cos(\omega t + \phi)$

I'm not sure the first term on the right should be there and I think you overlooked a term that should be there.

3. Apr 8, 2013

MisterX

Thanks for noticing this. I had interpreted the rotation as being in a vertical plane. So with a horizontal plane I may have U = 0. I also left out a term when connecting both sides of the Euler-Lagrange Equation. Part of the term I omitted cancels the "first term on the right" that you weren't sure should be there.

$\mathcal{L} =\frac{mD^2}{8}\left[\omega^2 +2\omega\left(\omega + \dot{\phi}\right)cos(\phi) +\left(\omega + \dot{\phi}\right)^2\right]$

$\frac{\partial\mathcal{L} }{\partial \phi} = \frac{d}{dt} \frac{\partial \mathcal{L}}{\partial \dot{\phi}}$
$\frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{8}2\omega\left(\omega + \dot{\phi}\right)sin(\phi) = - \frac{mD^2}{4}\omega\left(\omega + \dot{\phi}\right)sin(\phi)$

$\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{8}\left[ 2\omega cos(\phi) + 2\dot{\phi} + 2\omega \right] = \frac{mD^2}{4}\left[ \omega cos(\phi) + \dot{\phi} + \omega \right]$
$\frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}} = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right]$

$\frac{\partial\mathcal{L} }{\partial \phi} = - \frac{mD^2}{4}\omega\left(\omega + \dot{\phi}\right)sin(\phi) = \frac{mD^2}{4}\left[- \omega sin(\phi)\dot{\phi} + \ddot{\phi}\right] = \frac{d}{dt}\frac{\partial \mathcal{L}}{\partial \dot{\phi}}$

$- \omega\left(\omega + \dot{\phi}\right)sin(\phi) = - \omega sin(\phi)\dot{\phi} + \ddot{\phi}$
$- \omega^2sin(\phi) -\omega\dot{\phi} sin(\phi) = - \omega sin(\phi)\dot{\phi} + \ddot{\phi}$
The $- \omega sin(\phi)\dot{\phi}$ may be removed from both sides.
$\ddot{\phi} + \omega^2sin(\phi) = 0$

Last edited: Apr 9, 2013
4. Apr 9, 2013

TSny

Looks good to me.