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\title{27.  Impact}
\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 27. \hfill Mechanics of Manipulation }

\begin{document}

\maketitle

%=======================================================================
%
% SLIDE 2
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\begin{slide}[R]{Lecture 27.  Impact}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC27}
\end{slide}

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

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

\begin{itemize}
\item Impulse-momentum equations
\item Impact of a particle
 \begin{itemize}
\item Plastic and elastic impact
\item Newtonian and Poisson restitution
\item Impulse space
 \end{itemize}
\item Impact of a planar rigid body
\begin{itemize}
\item Solution of the sliding rod problem
\end{itemize}
\end{itemize}

\end{slide}

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

\begin{slide}{Frictional impact}
\begin{trivlist}
\item
When rigid bodies collide:
\begin{itemize}
\item Discontinuity in velocity,
\item Infinite forces,
\item Zero time,
\item Even for simple cases there is no agreement on laws,
\item For multiple contact indeterminacies abound.
\end{itemize}
 We will adopt Routh's analysis of planar impact, which uses Poisson
 restitution.  
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Impulse momentum equations}
\begin{trivlist}
\item
Recall:
\begin{align*}
        \vv F & = \frac{d\vv P}{d\vv t} \\
        \vv N & = \frac{d\vv L}{d\vv t}
\end{align*}

\item
Integrating:
\begin{align*}
        \Delta\vv P & = \int_{t_0}^{t_1} \vv F \, dt \\
        \Delta\vv L & = \int_{t_0}^{t_1} \vv N \, dt
\end{align*}
where $\Delta\vv P = \vv P_1 - \vv P_0$,
$\Delta\vv L = \vv L_1 - \vv L_0$,
$\int \vv F \, dt$ is the \emph{impulse}, and $\int \vv N\, dt$ is the
\emph{impulsive moment}.
\index{impulsive moment}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 11
%
%

\begin{slide}[R]{Determining impulse.  Particle in plane.}
\begin{trivlist}
\item
  We need impulse and impulsive moment
  as a function of state and physical parameters.  
\item Start with a frictionless particle on plane.  Velocity changes
  discontinuously from $\vv v_0$ to $\vv v_1$:
\begin{equation*}
        \Delta\vv v = \frac{1}{m} \Delta \vv P = \frac{1}{m} \vv I
\end{equation*}
where
\begin{equation*}
        \Delta\vv v = \vv v_1 - \vv v_0
\end{equation*}

\item
The tangential impulse $I_t$ is zero, so
\begin{align*}
        v_{1t} & = v_{0t} \\
        v_{1n} & = v_{0n} + \frac{1}{m} I_n
\end{align*}

\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 12
%
%
\begin{slide}[R]{Plastic and elastic}
\begin{trivlist}
\item
How to get the normal impulse
$I_n$?  Two special cases: \emph{plastic impact}%
\index{impact!plastic}
\begin{align*}
        & I_n = - m v_{0n} \\
        & \rightarrow v_{1n} = 0  
\end{align*}
and \emph{elastic impact}%
\index{impact!elastic}
\begin{align*}
        & I_n = -2 m v_{0n} \\
        & \rightarrow v_{1n} = - v_{0n}
\end{align*}
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 13
%
%
\begin{slide}[R]{Newtonian restitution}
\begin{trivlist}
\item
Newton hypothesized a continuum from plastic to elastic, defined by a
\emph{coefficient of restitution}%
\index{coefficient of restitution}%
\index{impact!Newtonian restitution}
\begin{equation}
        e = - \frac{v_{1n}}{v_{0n}}
\end{equation}
so that plastic impact corresponds to $e = 0$, and elastic impact
corresponds to $e = 1$.
\end{trivlist}
\end{slide}

%=======================================================================
%
% SLIDE 14
%
%

\begin{slide}[R]{Poisson restitution}
\begin{trivlist}
\item
A competing definition of the coefficient of restitution was given by
Poisson.%
\index{impact!Poisson restitution|(}
Divide the collision into two stages.
\begin{align*}
  v_n &< 0\qquad\text{compression} \\
  v_n &= 0 \\
  v_n &> 0\qquad\text{restitution}
\end{align*}

\item Let $I_c$ be impulse accumulated during compression, and let
  $I_r$ be impulse accumulated during restitution.  Poisson's
  hypothesis is that the ratio of these two parts is governed by
  material properties:
%               Why did I used to have here "and initial conditions:" ?
\begin{equation}
        e = \frac{I_r}{I_c}
\end{equation}
\item
For simple cases no advantage over Newton, but for frictional rigid
bodies Poisson restitution is superior. 

\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Applying Poisson restitution for frictionless particle}
\begin{trivlist}
\item
Given $m$, $\vv v_0$, and $e$, first find $I_c$:
\begin{equation*}
        I_c = m v_{0n}
\end{equation*}
then restitution impulse $I_r$
\begin{equation*}
        I_r = e m v_{0n}
\end{equation*}
then total impulse $I$
\begin{equation*}
        I = I_c + I_r = (1+e) m v_{0n}
\end{equation*}
and resulting normal velocity:
\begin{equation*}
        v_{1n} = - e v_{0n}
\end{equation*}
The tangential velocity is unchanged.
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Frictional particle}
\begin{figure}[h!]
\includegraphics[height=2cm]{buggylaw}
\end{figure}
\begin{trivlist}
\item Consider frictional particle with initial leftward velocity.
  There should be some frictional impulse acting toward the right.
  Suppose we apply Coulomb's law, by looking at initial contact mode,
  and multiplying $I_n$ by $\mu$.  The result is absurd, and violates
  conservation of energy!
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{A better model}
\begin{figure}[h!]
\includegraphics[height=7cm]{ispace}
\end{figure}
\end{slide}

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

\begin{slide}[R]{The left-drifting particle}
\begin{figure}[h!]
\includegraphics[height=7cm]{piex}
\end{figure}

\end{slide}

%=======================================================================
%
% SLIDE 19
%
%
\begin{slide}[R]{Rigid body.  Kinematics.}
\begin{minipage}[t]{6cm}
\begin{trivlist}
\item
First, we
observe some kinematic relations
\begin{align*}
        \vv c & = \vv x + \vv r \\
        \dot{\vv c} & = \dot{\vv x} + \dot{\vv r} \\
                                & = \dot{\vv x} + \vv\omega \btimes \vv r \\
        \Delta\dot{\vv c} & = \Delta\dot{\vv x} + \Delta\vv\omega \btimes \vv r
\end{align*}
\end{trivlist}
\end{minipage}
\hfill
\begin{minipage}[t]{4cm}
\begin{figure}[h!]
\includegraphics[width=4cm]{rimp}
%\includegraphics[height=3cm]{*file*}
\end{figure}\end{minipage}
\end{slide}

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

\begin{slide}[R]{Impulse momentum laws}
\begin{trivlist}
\item
Now we can write the impulse-momentum laws:
\begin{align}
        m \Delta\dot{\vv x} & = \vv I \\
        \rho^2 m \Delta{\vv \omega} & = \vv r \btimes \vv I
\end{align}
\item
Substituting into the expression for $\Delta\dot{\vv c}$ yields:
\begin{align}
 \Delta\dot{\vv c} 
  & = \frac{1}{m} \vv I 
        + \frac{1}{\rho^2 m} (\vv r \btimes \vv I) \btimes \vv r \\
  & = \frac{1}{m} \vv I 
        - \frac{1}{\rho^2 m} \vv r \btimes (\vv r \btimes \vv I) \\
  & = \frac{1}{\rho^2 m} \left( \rho^2 I_3 - R^2 \right) \vv I
\end{align}
where $I_3$ is the three by three identity matrix, and $R$ is the
cross-product matrix.
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Compression line and restitution line}
\begin{trivlist}
\item
Substituting above and expanding, we obtain
\begin{equation}
        \left(\begin{array}{c}
                        \Delta\dot c_t \\
                        \Delta\dot c_n
                  \end{array}
        \right)
                = \frac{1}{\rho^2 m} 
                        \left(\begin{array}{cc}
                                        (\rho^2 + r_n^2) & - r_t r_n \\
                                        - r_t r_n & (\rho^2 + r_t^2)
                                  \end{array}\right)
                        \left(\begin{array}{c}
                                I_t \\
                                I_n
                                \end{array}\right)
\end{equation}
The sticking line is defined by the equation $\dot c_t
= 0$,
\begin{equation}
 \dot c_{0t} 
 + \frac{\rho^2 + r_n^2}{\rho^2 m} I_t 
 - \frac{r_t r_n}{\rho^2 m} I_n 
 = 0
\end{equation}
and the compression line is defined the equation $\dot c_n = 0$,
\begin{equation}
 \dot c_{0n} 
 - \frac{r_t r_n}{\rho^2 m} I_t 
 + \frac{\rho^2 + r_t^2}{\rho^2 m} I_n 
 = 0
\end{equation}
\end{trivlist}
\end{slide}

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

\begin{slide}[R]{Rigid body example}
\begin{figure}[h!]
\includegraphics[height=7cm]{impex1.eps}
\end{figure}
\end{slide}

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

\begin{slide}[R]{Analysis of rigid body example}
\begin{figure}[h!]
\includegraphics[height=7cm]{impex1is.eps}
%\caption{Routh's method applied to Example 1.}
%\label{fig:impex1is}
\end{figure}
\end{slide}

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

\begin{slide}[R]{Sliding rod problem}
\begin{figure}[h!]
\includegraphics[height=7cm]{improd.eps}
%\caption{A sliding rod problem with no finite force solution.}
%\label{fig:improd}
\end{figure}
\end{slide}

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

\begin{slide}[R]{Sliding rod solution}
\begin{figure}[h!]
\includegraphics[height=7cm]{improdis.eps}
%\caption{Routh's method gives an impulsive force solution to the
%sliding rod problem.}
%\label{fig:improdis}
\end{figure}

\end{slide}



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

\begin{slide}[R]{Next:  Impact.}
\vspace{-0.5in}
\includegraphics[height=3.0in]{TOC28}
\end{slide}
\end{document}



\end{document}