\documentclass[11pt]{article}
\usepackage{graphicx}
\usepackage{color}
\usepackage{html,htmllist}
%\usepackage{/usr/local/share/texmf/tex/latex/graphics/graphicx}
\textwidth 456pt 
\textheight 650pt
\hoffset -44pt 
\voffset -52pt 
\pagestyle{headings}
%\makeglossary
%\makeindex
\begin{document}
The main aspects of this exercise is to combine  motion  in a 
conservative force problem with a secondary boundary condition problem.

 \medskip

QUESTION: An object of mass $m$  is released at a height $h$ above the center of
a circular loop and slides down without friction.  How high must the
drop be so that it just barely completes the loop-the-loop?\medskip



\begin{enumerate}
\item Since this is a conservative force problem the potential energy
(gravitational) plus the kinetic energy of the mass is a constant of 
the motion. Thus at the start $mgh + KE_i = \mbox{constant}$.

\item The KE+PE energy at the end point (the top of the loop) hasn't changed 
although any changes in kinetic energy are due to changes in the potential energy (work
done by gravity).

\item The constraint is that the object must remain in circular motion 
throughout.  If the ball is released at $h=2R$ it will rise again to
a height of $2R$ but will depart from the loop when the component of the
acceleration due to gravity (normal to the loop) exactly equals the centripetal
acceleration. 

\noindent
(This should happen when ball is at a height of 2/3 $R$ above the loop center.
You should try problem this yourself!  To further test yourself you might
wish to locate where the ball impacts the loop.) 


\noindent 
This constraint implies that at the top $F_c=mv^2/R \ge mg = F_g$ or $v^2 \ge gR$.

\item The above constraint implies the KE at the TOP must be greater that
$1/2~mv^2$ or $ mgR/2$.  Thus $mg \Delta R = mgR/2$ or $\Delta R = R/2$ to 
give $h= 2R + R/2 = 5R/2$.


\end{enumerate}
\begin{center}
\includegraphics[height=3.1in]{loop_fig1.eps}
\end{center}





\end{document}
\end
