Homework 4 Problem 2

Problem

A mass $m$ hangs on a massless rod with length $l$ from the center of a wheel with mass $M$, rotational moment of inertia $J$, and radius $R$ as shown:

Use Routh reduction to obtain the equations of motion for this system in terms of $\theta$, $\dot \theta$, and the conserved quantity $\mu = \left((M + m) R^2 + J\right) \dot \theta - m R l \dot \phi \cos \phi$.

Solution

First you need to find the Lagrangian.

(1)
\begin{align} KE_{wheel} = \frac{1}{2}M\left (R\dot\theta\right)^{2}+\frac{1}{2}J\left(\dot\theta\right)^{2} \end{align}
(2)
\begin{align} KE_{pendulum} = \frac{1}{2}m\left[\left (l\dot\phi\right)^{2} +\left(R\dot\theta\right)^{2}-2Rl\dot\theta\dot\phi\cos\phi\right] \end{align}
(3)
\begin{align} PE_{pendulum} = -mgl\cos\phi \end{align}
(4)
\begin{align} L = \frac{1}{2}\left(M+m\right)\left (R\dot\theta\right)^{2}+\frac{1}{2}J\left(\dot\theta\right)^{2}+\frac{1}{2}m\left(l\dot\phi\right)^{2} -mRl\dot\theta\dot\phi\cos\phi+mgl\cos\phi \end{align}

Since $L$ is independent of $\theta$, $\frac{\partial L}{\partial \dot \theta}$ is conserved according to the Euler-Lagrange equations. Take this conserved quantity to be $\mu$:

(5)
\begin{align} \mu = \frac{\partial L}{\partial \dot \theta} = \left[\left(M+m\right)R^{2}+J \right]\dot \theta - mRl \dot \phi \cos \phi \end{align}

Next, define the Routhian to be:

(6)
\begin{align} R^\mu(\phi, \dot \phi) = \left[ L(\phi, \dot \phi, \dot \theta) - \mu \dot \theta \right]_{\frac{\partial L}{\partial \dot \theta} = \mu} \end{align}
(7)
\begin{align} R^\mu(\phi, \dot \phi) = \frac{1}{2}ml^{2}\dot\phi^{2}+mgl\cos\phi-\frac{1}{2}\frac{\left(\mu+mRl\dot\phi\cos\phi\right)^{2}}{\left(M+m\right)R^2 +J} \end{align}

Finally, compute the Euler-Lagrange equation for the Roputhian wrt $\mu$. The equations of motion are:

(8)
\begin{align} \frac{d}{dt}\frac{\partial R^{\mu}}{\partial \dot \phi} - \frac {\partial R^{\mu}}{\partial \phi} = 0, \mu =\frac{\partial L}{\partial \dot \theta} \end{align}
(9)
\begin{align} \frac{d}{dt}\frac{\partial R^{\mu}}{\partial \dot \phi}= ml^2 \ddot \phi - \frac{\left(mRl\right)^{2}\ddot \phi \cos^{2}\phi-2\left(mRl\right)^{2}\dot \phi \cos\phi \sin\phi - \mu mRl\dot \phi \sin\phi} {\left(M+m\right)R^2 +J} \end{align}
(10)
\begin{align} \frac {\partial R^{\mu}}{\partial \phi} = -mgl\sin\phi + \frac{\left(mRl\right)^{2}\dot \phi \cos\phi \sin\phi + \mu mRl\dot \phi \sin\phi} {\left(M+m\right)R^2 +J} \end{align}
(11)
\begin{align} ml^{2}\ddot \phi +\frac{\left(mRl\right)^{2}\left(\dot\phi^{2}\cos\phi \sin\phi -\ddot\phi \cos^2\phi \right)}{\left(M+m\right)R^2 +J} +mgl\sin \phi \end{align}
(12)
\begin{align} \mu = \frac{\partial L}{\partial \dot \theta} = \left[\left(M+m\right)R^{2}+J \right]\dot \theta - mRl \dot \phi \cos \phi \end{align}

I'm pretty sure that I screwed something up somewhere, as there is no $\mu$ in equation 11. I'll try and figure out where I made my mistake, but if it's obvious to anyone, I would really appreciate you helping me out.

Discussion

page revision: 7, last edited: 18 May 2007 06:24