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\title{24.  Rigid Body Dynamics}
\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 24. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
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% SLIDE 2
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\begin{slide}[R]{Lecture 24.  Rigid Body Dynamics.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC24}
\end{slide}

% ============
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%  Slide 2.5
%

\begin{slide}[R]{First a quiz}
\end{slide}


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

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

\begin{trivlist}
\item Newtonian mechanics of a single particle;
\item of a system of several particles;
\item of a rigid body.
\end{trivlist}

\end{slide}

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

\begin{slide}{Newton's laws}
\begin{enumerate}
\item
        Every body continues at rest, or in uniform motion in a straight line, 
        unless forces act upon it.
        
      \item The rate of change of momentum is proportional to the
        applied force.

\item
        The forces acting between two bodies are equal and opposite.

\end{enumerate}

Define \defterm{momentum} to be mass times velocity.
\end{slide}

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

\begin{slide}[R]{Consider a particle $\ldots$}
\begin{trivlist}
\item
 $\ldots$ of mass $m$,

\item with position represented by a vector $\vv x$,

\item total applied force $\vv F$,

\item momentum
   \[\vv p = m \vv v = m \frac{d \vv x}{dt}\]

\item so Newton's second law can be written
   \[m \frac{d^2 \vv x}{dt^2} = \vv F\]

\end{trivlist}
\end{slide}

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

\begin{slide}{Impulse, kinetic energy}
\begin{trivlist}
\item Integrating Newton's second law:
\[
        \vv p_2 - \vv p_1 = \int_{t_1}^{t_2} \vv F \, dt
\]
stating that the change in momentum is equal to the \emph{impulse}.


\item We can also define \emph{kinetic energy} $T$%
\index{kinetic energy}
\[
        T = \frac{m}{2} |\vv v|^2
\]


\end{trivlist}
\end{slide}


%=======================================================================
%
% SLIDE 7
%
%
\begin{slide}[R]{Power}
\begin{trivlist}
\item
Differentiating kinetic energy yields
\begin{align*}
 \frac{dT}{dt} 
   & =  \frac{m}{2} \frac{d}{dt} (\vv v \cdot \vv v) \\
   & =  \frac{m}{2} \left( \frac{d\vv v}{dt} \cdot \vv v 
     + \vv v \cdot \frac{d\vv v}{dt} \right) \\
   & =  m \frac{d\vv v}{dt} \cdot \vv v \\
   & =  \vv F \cdot \vv v
\end{align*}
stating that the time rate of change of kinetic energy is \emph{power}.


\end{trivlist}

\end{slide}

%=======================================================================
%
% SLIDE 8
%
%
\begin{slide}[R]{Work}
\begin{trivlist}
\item Integrating the power over a time interval,
\[
        T_2 - T_1 = \int_{t_1}^{t_2} \vv F \cdot \vv v \, dt
\]
or
\[
        T_2 - T_1 = \int_{\vv x_1}^{\vv x_2} \vv F \cdot \vv {dx}
\]
stating that the change in kinetic energy is \emph{work}.%
\index{work}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 9
%
%
\begin{slide}[R]{Moment of force; moment of momentum}
\begin{trivlist}
\item
Recall definition of moment of force about a point $\vv x$:
 \[  \vv n = \vv x \btimes \vv f \]
and about a line $l$ through origin with direction $\uu l$
 \[ n_l = \uu l \cdot \vv n \]

\item
Similarly, suppose a particle at $\vv x$ has momentum $\vv p$.
\begin{itemize}
\item Define \defterm{moment of momentum} about the origin
\[ \vv L = \vv x \btimes \vv p \]
\item and about the line $l$
\[ L_l = \uu l \cdot \vv L \]
\end{itemize}
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Rate of change of moment of momentum}
\begin{trivlist}
\item
Differentiating the moment of momentum:
\begin{align*}
 \frac{d \vv L}{dt} 
   & =  \frac{d}{dt} (\vv x \btimes \vv p) \\
   & =  \frac{d}{dt} ( \vv x \btimes m \vv v) \\
   & =  m\left( \frac{d\vv x}{dt} 
        \btimes \vv v + \vv x \btimes \frac{d \vv v}{dt}\right) \\
   & =  \vv x \btimes m \frac{d \vv v}{dt} \\
   & =  \vv x \btimes \vv F\\
   & =  \vv N 
\end{align*} 
which is essentially a restatement of Newton's second
law, but using moments of force and momentum.
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{So, for a particle $\ldots$}
\begin{trivlist}
\item Using either $\vv F = d\vv p / dt$ or $\vv N = d\vv L / dt$, we
  have three second order differential equations.

\item If $\vv F$ or $\vv N$ is uniquely determined by the state $(\vv
  x, \vv v)$, then there is a unique solution giving $\vv x(t)$ and
  $\vv v(t)$ for any given initial conditions $\vv x(0) = \vv x_0$,
  $\vv v(0) = \vv v_0$.
\end{trivlist}

\end{slide}

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

\begin{slide}[R]{For a bunch of particles}
For the $k$th particle
\begin{itemize}
\item Let $m_k$ be the mass,
\item let $\vv x_k$ be the position vector, 
\item and let $\vv p_k$ be the momentum.  
\item Let the force be composed of internal force (from interactions
  with other particles in the system) and external forces $\vv F_k =
  \vv F^i_k + \vv F^e_k$
\end{itemize}
\end{slide}

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

\begin{slide}[R]{Momentum and force}
We define the momentum%
\index{momentum}
of the system to be
\[
        \vv P = \sum \vv p_k
\]
and the total force%
\index{force!total}
on the system to be
\[
        \vv F = \sum \vv F^e_k
\]
(The sum of all internal forces is zero, by Newton's third 
law.)
\end{slide}

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

\begin{slide}[R]{Newton's 2nd law for system of particles}
Newton's 2nd law for $k$th particle:
\[
        \frac{d\vv p_k}{dt} = \vv F^e_k + \vv F^i_k
\]
Summing:
\[
        \sum \frac{d\vv p_k}{dt} = \sum \left(\vv F^e_k + \vv F^i_k\right)
\]
Hence
\[
        \frac{d\vv P}{dt} = \vv F
\]
\emph{Newton's second law extends to the system of particles.}
\end{slide}

%==========
%
% Slide 15
%

\begin{slide}[R]{Center of mass}
Define total mass:
\[ 
    M = \sum m_k 
\] 
and the center of mass,%
\index{center of mass}
\[ 
    \vv X = \frac{1}{M} \sum m_k \vv x_k 
\] 
Then
\[ 
    \vv P = M \frac{d\vv X}{dt} 
\] 
and 
\[ 
    \vv F = M \frac{d^2\vv X}{dt^2} 
\] 
which 
means that the center of mass behaves just like a single particle. 

\end{slide}

%===============
%
% Slide 16
%

\begin{slide}[R]{Moments for systems of particles}
\begin{trivlist}
\item
Define $\vv L_k$ to be the angular momentum of the $k$th point, 
\item


Define the total angular momentum to be the sum, 
\[
        \vv L = \sum \vv L_k
\]


\item Define the total torque,
\[
        \vv N = \sum \vv x_k \btimes \vv F^e_k
\]

\end{trivlist}
\end{slide}

%==============
%
% Slide 17
%

\begin{slide}[R]{Rate of change of moment of momentum}
\begin{trivlist}
\item
Now for the $k$th particle
\[
 \frac{d\vv L_k}{dt} = \vv x_k \btimes \vv F^e_k + \vv x_k \btimes \vv F^i_k
\]
Summing over all the particles,
\[
        \frac{d\vv L}{dt} = \vv N + \sum \vv x_k \btimes \vv F^i_k
\]
By Newton's third law the sum of the internal 
moments is zero, so that the second term vanishes: 
\[
        \frac{d\vv L}{dt} = \vv N
\]
which is grand, 
but six equations is not enough to determine the motion of several particles.
\end{trivlist}
\end{slide}

%=================
% Slide 18
%

\begin{slide}[R]{Rigid body dynamics}
\begin{trivlist}
\item
  A rigid body is a bunch of particles, but with all distances fixed.
  Six degrees of freedom.  Wouldn't it be keen if the six equations
\begin{align*}
\vv F &= d\vv P/ dt \\
\vv N &= d\vv L/dt
\end{align*}

were enough?
\end{trivlist}
\end{slide}

%==================
%
% Slide 19
%

\begin{slide}[R]{Angular inertia, part one}
\begin{trivlist}
\item
For a rigid body, velocity of $k$th particle is
\[ 
\label{eq:omegacross} 
    \vv v = \vv v_0 + \vv\bomega \btimes \vv x 
\] 


\item Substituting into moment of momentum
\[
        \vv L_k = m_k \vv x_k \btimes (\vv v_0 + \vv\bomega \btimes \vv x_k)
\]


\item Summing to obtain the total angular momentum,
\begin{align*}
  \vv L &= \sum m_k \vv x_k \btimes \vv v_0 
  + \sum m_k \vv x_k \btimes (\vv\bomega \btimes \vv x_k) \\
  &= M \vv X \btimes \vv v_0 
  + \sum m_k \vv x_k \btimes (\vv\bomega \btimes \vv x_k) 
\end{align*}

\item Place origin at center of mass to eliminate first term on right
\[
        \vv L = \sum m_k \vv x_k \btimes (\vv\bomega \btimes \vv x_k) 
\]


\end{trivlist}
\end{slide}

%==========
% 
% Slide 20
%

\begin{slide}[R]{Angular inertia, part two}
\begin{trivlist}
\item How can we get that pesky $\vv\bomega$ out of the sum?
\[
        \vv L = \sum m_k \vv x_k \btimes (\vv\bomega \btimes \vv x_k) 
\]


\item

Applying the identity $\vv a \btimes (\vv b \btimes \vv c) = (\vv a \cdot \vv 
c) \vv b - (\vv a \cdot \vv b) \vv c$, 
\[
        \vv L = \sum m_k \left[ (\vv x_k \cdot \vv x_k) \vv\bomega 
                                - \vv x_k(\vv x_k \cdot \vv\bomega)\right]
\]


\item
Represent each vector as a 
column matrix, and substitute $\vv x_k^t \vv\bomega$ for $\vv x_k \cdot 
\vv\bomega$: 
\[
 \vv L = \left( \sum m_k \left( |\vv x_k|^2 I_3 
                                - \vv x_k \vv x_k^t\right)\right) 
                         \vv\bomega
\]
where $I_3$ is the three-by-three identity matrix.
\end{trivlist}
\end{slide}

%-========
% 
% Slide 21
% 

\begin{slide}{Angular inertia part three}
\begin{trivlist}
\item
Define the 
\emph{angular inertia matrix}%
\index{angular inertia matrix}
$I$: 
\[
\label{eq:idef}
        I = \sum m_k \left( |\vv x_k|^2 I_3 - \vv x_k \vv x_k^t \right)
\]
Substituting above,
\[
\label{eq:liw}
        \vv L = I \vv\bomega
\]
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Next:  Poinsot's construction.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC25}
\end{slide}
\end{document}



\end{document}