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\title{25.  Tumbling rigid bodies}
\subtitle{ Mechanics of Manipulation}
\author{Matt Mason}
\institution{Carnegie Mellon}
\email{matt.mason@cs.cmu.edu \\
http://www.cs.cmu.edu/\symbol{"7E}mason}
\slideCaption{Lecture 25. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
%
% SLIDE 2
%

\begin{slide}[R]{Lecture 25.  Tumbling rigid bodies.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC25}
\end{slide}

%=======================================================================
%
% SLIDE 3
%
%

\begin{slide}[R]{Outline.}

\begin{itemize}
\item The angular inertia tensor;
\item Euler's equations;
\item Poinsot's construction for a tumbling rigid body.
\end{itemize}

\end{slide}

%=======================================================================
%
% SLIDE 4
%
%

\begin{slide}{Last time $\ldots$}

 $\ldots$ we defined the 
\emph{angular inertia matrix}%
\index{angular inertia matrix}
$I$: 
\[
\label{eq:idef}
        I = \sum m_k \left( |\vv x_k|^2 I_3 - \vv x_k \vv x_k^t \right)
\]
so that moment of momentum is 
\[
\label{eq:liw}
        \vv L = I \vv\bomega
\]

% Forgot to go far enough last time
where Newton's second law gives
 \[
      \vv N  =  \frac{d \vv L}{dt}
 \]


\end{slide}

%================================================================
%
% SLIDE FU

\begin{slide}[R]{Differentiating $\vv L = I \vv\bomega$}
\begin{trivlist}
\item
$I$ is constant in the \emph{body} frame, not in an inertial frame.
\begin{align}
  \vv N & =  \frac{d (I \vv\bomega)}{dt} \\
  & =  I \frac{d\vv\bomega}{dt} + \frac{dI}{dt} \vv\bomega \\
  & =  I \frac{d\vv\bomega}{dt} + \vv\bomega \times (I \vv\bomega) 
\end{align}

\item
Zero torque implies constant angular momentum.

\item
Zero torque does \emph{not} imply constant angular velocity.

\item What \emph{can} you say about how angular velocity changes?
  First we need to look closer at the angular inertia tensor.

\end{trivlist}
\end{slide}


%=======================================================================
%
% SLIDE 5
%
%

\begin{slide}[R]{The inertia tensor}
\begin{trivlist}
\item
Let body be continuous with density $\rho$.
\begin{equation}
  I = \int \rho \left( |\vv x|^2 I_3 - \vv x \vv x^t \right) \, dV
\end{equation}

\item
In components:
\begin{equation}
\label{eq:contidef}
        I = \int \rho \left(\begin{array}{ccc}
                                x_2^2 + x_3^2 & - x_1 x_2 & - x_1 x_3 \\
                                - x_1 x_2 & x_1^2 + x_3^2 & - x_2 x_3 \\
                                - x_1 x_3 & - x_2 x_3 & x_1^2 + x_2^2 \\
                        \end{array}\right) \, dV
\end{equation}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 6
%
%

\begin{slide}{Moments of inertia; products of inertia}
\begin{trivlist}
\item
Diagonal elements: moments of inertia w.r.t. the coordinate axes:
\begin{align}
        & I_{11} =  \int \rho ( x_2^2 + x_3^2) \, dV \\
        & \hbox{etc.}
\end{align}

\item
Off-Diagonal elements: the \emph{products of inertia}:
\begin{align}
        & I_{12}  = I_{21} = - \int \rho x_1 x_2 \, dV \\
        & \hbox{etc.}
\end{align}

\item We could try to understand them, or we could get rid of them $\ldots$

\end{trivlist}
\end{slide}


%=======================================================================
%
% SLIDE 7
%
%
\begin{slide}[R]{Principal axes; principal moments of inertia}
\begin{trivlist}
\item
Inertia matrix is symmetric---diagonal in the right frame.  Define $A$:
\begin{equation}
        ^AI = \left(\begin{array}{ccc}
                        ^A I_{11} & 0 & 0 \\
                        0 & ^A I_{22} & 0 \\
                        0 & 0 & ^A I_{33} \\
                        \end{array}\right) 
\end{equation}

\item
$I$ in $A$-coordinates can be obtained by:
\begin{equation}
\label{eq:inxform}
        ^AI = A I A^T
\end{equation}
where matrix $A$ transforms to $A$-coordinates.  

\item
\defterm{principal axes:}  coordinate axes of $A$---eigenvectors of $I$.

\item
\defterm{principal moments:} diagonal elements of $^AI$---eigenvalues of $I$.

\item
Distinct eigenvalues implies uniquely determined principal axes.

\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 8
%
%
\begin{slide}[R]{Scalar angular inertia.  Radius of gyration.}
\begin{trivlist}
\item
Consider moment of momentum $\vv L = I \vv \bomega$.  When are $\vv L$ and $\vv \bomega$ parallel?

\item
Consider rotation about some fixed axis in direction $\uu n$.  Scalar angular inertia $I_n$ is
\begin{equation}
        I_n = \uu n^t I \uu n
\end{equation}

\item
\defterm{radius of gyration}
$k_n$ with respect to the axis $\uu n$:
\begin{equation}
\label{eq:radgyr}
        I_n = M k_n^2
\end{equation}

\item
The radius of gyration represents the distance of a point mass that would give 
the same angular inertia.

 \end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 9
%
%

\begin{slide}[R]{Inertia ellipsoid}
\begin{trivlist}
\item
Consider the surface described by the equation
\begin{equation}
        \vv r^t I \vv r = a
\end{equation}


\item In principal 
coordinates, since moments are positive, we get an ellipsoid:
\begin{equation}
        I_{xx} r_x^2 + I_{yy} r_y^2 + I_{zz} r_z^2 = a
\end{equation} 

\item  Let $\vv r = r \uu n$.  Then 
\begin{equation}
        I_n = \uu n^t I \uu n = \frac{1}{r^2} \vv r^t I \vv r = \frac{a}{r^2}
\end{equation}

So distance to ellipsoid surface is inverse of radius of gyration.
\begin{equation}
        M k_n^2 = \frac{a}{r^2}
\end{equation}
\end{trivlist}
\end{slide}

%=========
%
% Slide 10

\begin{slide}[R]{Cylinder and its inertia ellipsoid}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=1\linewidth]{iellips}
\end{center}
\end{figure}
\end{slide}

%==========
%
% Slide 11

\begin{slide}[R]{Principal axes by inspection}
\begin{trivlist}
\item
There are theorems:  

\begin{itemize}
\item
Any plane of symmetry is perpendicular to a principal axis.

\item  
Any line of symmetry is a principal axis.

\end{itemize}

\item
If you start in the principal frame, you know the products of
inertia are zero, so you can get the inertia tensor by just doing
three integrals.
\end{trivlist}
\end{slide}

%==========
%
% Slide 12

\begin{slide}[R]{Rigid body tumbling}
\begin{trivlist}
\item
\begin{align}
  N &= \left(\begin{array}{c}
      I_1 \dot \omega_1 \\
        I_2 \dot \omega_2 \\
        I_3 \dot \omega_3 \end{array}\right)
  + \left(\begin{array}{c}\omega_1 \\ \omega_2 \\ \omega_3
      \end{array}\right)
    \times \left(\begin{array}{c}
        I_1 \omega_1 \\
          I_2 \omega_2 \\
          I_3 \omega_3 \end{array}\right) \\
    &= \left(\begin{array}{c} 
        I_1 \dot\omega_1
        + (I_3 \omega_2 \omega_3 - I_2 \omega_2 \omega_3 ) \\
        I_2 \dot\omega_2
        + (I_1 \omega_3 \omega_1 - I_3 \omega_3 \omega_1 ) \\
        I_3 \dot\omega_3 + (I_2 \omega_1 \omega_2 - I_1 \omega_1
        \omega_2 ) \end{array}
      \right) 
\end{align}

\item
If $N=0$ we get what looks like Euler's equations:
% ``what looks like'' is my cue to discuss body frame versus inertial frame
% Also ask what happens when omega is an eigenvector.
\begin{align}
        \dot \omega_1 & = \frac{ I_2 - I_3}{ I_1 } \omega_2 \omega_3 \\
        \dot \omega_2 & = \frac{ I_3 - I_1}{ I_2 } \omega_3 \omega_1 \\
        \dot \omega_3 & = \frac{ I_1 - I_2}{ I_3 } \omega_1 \omega_2
\end{align}

\end{trivlist}
\end{slide}

%==========
%
% Slide 13

\begin{slide}[R]{Poinsot's construction}
\begin{minipage}[b]{.6\linewidth}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=\linewidth]{poincon.eps}
\end{center}
\end{figure}
\end{minipage}
\hfill
\begin{minipage}[b]{.38\linewidth}
\begin{trivlist}
\item
Rigid body tumbling:  inertia ellipsoid rolls without slipping on a plane.
\end{trivlist}
\end{minipage}
\end{slide}

%==========
%
% Slide 13

\begin{slide}[R]{Proof of Poinsot's construction}
\begin{trivlist}
\item
If $N$ is zero, then kinetic energy $T$ is constant:
\begin{equation}
   T = \frac{1}{2} \vv \bomega^t I \vv \bomega \hfill\;\;\; \hbox{is constant}
\end{equation}
That is, $\omega$ is on the surface of the inertia ellipsoid.

\item
What is the \emph{tangent plane} normal at $\omega$?
\begin{align}
 \nabla \frac{1}{2} \vv\bomega^t I \vv \bomega 
 & =  \nabla \frac{1}{2} (\omega_1^2 I_1 + \omega_2^2 I_2 + \omega_3^2 I_3) \\
 & =  ( I_1 \omega_1, I_2 \omega_2, I_3 \omega_3)  =  \vv L
\end{align}
The attitude of the tangent plane is constant!

\item
How far from center of mass to tangent plane?
\begin{equation}
        \frac{\vv \bomega \cdot \vv L}{ | \vv L |} = \frac{ 2 T}{ | \vv L |}
\end{equation}
which is also constant!

\item So the tangent plane is fixed: the \emph{invariable} plane.  The
  ellipsoid rolls without slipping on the invariable plane.
\end{trivlist}

\end{slide}

%=============
%
% Slide 14
%

\begin{slide}[R]{polhodes}
\begin{minipage}[t]{.38\linewidth}
\begin{trivlist}
\item Take an ellipsoid, hold the center a fixed distance from an
  inkpad, and roll it around.

\item
Near the pointy end you get little loops.

\item
Near the center of mass you get little loops.

\item
Near the third principal axis, you get sent away.
\end{trivlist}
\end{minipage}
\begin{minipage}[t]{.6\linewidth}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=\linewidth]{polhodes}
\end{center}
\end{figure}
\end{minipage}
\end{slide}

%=======================================================================
%
% SLIDE last
%

\begin{slide}[R]{Next:  ?}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC26}
\end{slide}
\end{document}



\end{document}