\documentclass[12pt]{article}
\usepackage{graphicx,wrapfig}
\textwidth=17.5cm
\textheight=23.5cm
\voffset=-2.cm
\hoffset=-2.cm

\newcommand{\hatbf}[1]{\mathbf{\hat{#1}}}
\newcommand{\vecbf}[1]{\mathbf{\vec{#1}}}
\newcommand{\pder}[2]{\frac{\partial #1}{\partial #2}}


\begin{document}
\pagestyle{empty}
\begin{center}
\vspace{0.5truecm}
{\large{Solution for \# 29}}

\end{center}
\vspace{0.5truecm}

Defining $\rho$ as the mass density, we use polar coordinates
$$ x\ =\ r\cos\theta\,,~~~~~~y\ =\ r\sin\theta\ . $$
Given a coordinate system $(x',y')$ which rotates with the disk (see 
the figure), the location of the CM is $\bar{x}_{CM}=0$ and
\begin{eqnarray*}
  \bar{y}_{CM}&=& \frac{\rho}{M}\left\{\int_0^R drr\int_0^{\pi}d\theta 
     r\sin\theta+2\int_0^R dr r\int_{\pi}^{2\pi}d\theta r 
     \sin\theta\right\} \\
  &=&\frac{\rho}{M}\left\{\frac{R^3}{3}2-\frac{R^3}{3}4\right\}
   = -\frac{2}{3}\frac{\rho R^3}{M}=-\frac{4R}{9\pi}.
\end{eqnarray*}
Where in the last line we have used
\[M=\rho\frac{\pi R^2}{2}+2\rho\frac{\pi R^2}{2}=\frac{3}{2}\rho\pi R^2.\]
Again referencing the figure, we see that the relationships between the 
lab coordinates and the coordinates of the center of mass are
\[\Bigl\lbrace\begin{array}{ccccc}
  x_{CM}&=&R\theta-|\bar{y}_{CM}|\sin\theta&=&R\theta-(4R/9\pi)\,\sin\theta\\
  y_{CM}&=&R-|\bar{y}_{CM}|\cos\theta&=&R-(4R/9\pi)\,\cos\theta.\end{array}\]
\[\Downarrow\]
\[\Bigl\lbrace\begin{array}{ccc}
  \dot {x}_{CM}&=&R\dot\theta-(4R/9\pi)\,\cos\theta\dot\theta\\
  \dot{y}_{CM}&=&(4R/9\pi)\,\sin\theta\dot\theta.
\end{array}\]
The kinetic energy of the disk is made of 2 terms: 
\[T=T_{trans}+T_{rot},\]
where $T_{trans}$ is the translational kinetic energy of the mass 
$M$ with coordinates $(x_{CM},y_{CM})$ and velocity $(\dot{x}_{CM},
\dot{y}_{CM})$, and $T_{rot}$ is the rotational kinetic energy about 
the center of mass. Thus
\[T=\frac{1}{2}M(\dot{x}_{CM}^2+\dot{y}_{CM}^2)+\frac{1}{2}I_3 \dot\theta^2.\]
Here $I_3$ is the moment of inertia about the CM, with respect to the 
z-axis, perpendicular to the plane of the figure. To find $I_3$ we will 
calculate it with respect to the center of disk (since it is easier!) 
and use the parallel axis theorem, (32.12) of Landau and Lifshitz, to 
obtain it with respect to the CM.
\begin{eqnarray*}
  I_3\Bigl\vert_0&=&\rho\int_0^R dr r\int_0^{\pi}d\theta (x^2+y^2)+2
  \rho\int _0^R dr r\int _{\pi}^{2\pi}d\theta (x^2+y^2)\\
  &=&\rho\frac{R^4}{4}\pi+2\rho\frac{R^4}{4}\pi=\rho R^4\frac{3}{4}\pi
  =\frac{1}{2}MR^2\\
  I_3\Bigl\vert_{CM}&=&I_3\bigl\vert_{0}-M\bar{y}_{CM}^2
  =\frac{1}{2}MR^2-M\frac{16}{81}\frac{R^2}{\pi^2}\
  =\ \frac{1}{2}MR^2\left[ 1-\frac{32}{81\pi^2}\right].
\end{eqnarray*}
Then we can calculate the kinetic energy
\begin{eqnarray*}
T&=&T_{trans}+T_{rot}=\frac{1}{2}M(\dot{x}_{CM}^2+\dot{y}_{CM}^2)+\frac{1}{2}I_3\Bigl\vert_{CM}\dot\theta^2\\
&=&\frac{1}{2}MR^2\dot\theta^2\left[\left(1-\frac{4}{9\pi}\cos\theta\right)^2+\left(\frac{4}{9\pi}\sin\theta\right)^2\right]+\frac{1}{4}MR^2\dot\theta^2\left[1-\frac{32}{81\pi^2}\right]\\
&=&\frac{1}{2}MR^2\dot\theta^2\left[1+\frac{16}{81\pi^2}-\frac{8}{9\pi}\cos\theta+\frac{1}{2}-\frac{16}{81\pi^2}\right]\\
&=&\frac{1}{2}MR^2\dot\theta^2\left[\frac{3}{2}-\frac{8}{9\pi}\cos\theta\right].
\end{eqnarray*}
The potential energy follows from the center of mass position,
\[U=Mgy_{CM}=MgR\left[1-\frac{4}{9\pi}\cos\theta\right],\]
and the Lagrangian is the sum of these two terms,
\[L=T-V=\frac{1}{2}MR^2\dot\theta^2\left[\frac{3}{2}-\frac{8}{9\pi}\cos\theta\right]-
MgR\left[1-\frac{4}{9\pi}\cos\theta\right].\]

\begin{figure}[h] \begin{center}
\includegraphics[scale=0.5]{HW8-2b}
% \caption{} \label{fig:2b}
\end{center} \end{figure}

\end{document}
