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{\bf Question}

\begin{description}
\item[(a)] Show that the force field ${\bf F} = -\kappa r^3 {\bf
r}$ is conservative.
\item[(b)] What is the potential energy of this force field?
\item[(c)] If a particle of mass $m$ moves with velocity $\bf {v =
\dot r}$ in this force field, show that if $E$ is the constant
total energy then $$ \frac{1}{2}m \dot{\bf r} \cdot \dot{\bf r} +
\frac{1}{5}r^5 = E.$$ What important physical principle does this
illustrate?
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
\begin{eqnarray*} {\bf F} & = & -\kappa r^3 {\bf r} \\  &  = & -\kappa r^3(x{\bf
i} + y{\bf j} + z{\bf k}) \hspace{.2in} {\rm where\ \ } r =
\sqrt{x^2 + y^2 + z^2} \end{eqnarray*}

\begin{eqnarray*} \nabla \times {\bf F} & = & -\kappa \left|
\begin{array}{ccc}  {\bf i} & {\bf j} & {\bf k} \\
\frac{\pl}{\pl x} & \frac{\pl}{\pl y} & \frac{\pl}{\pl z} \\ xr^3
& yr^3 & zr^3 \end{array} \right| \\ & = & -\kappa \left\{  {\bf
i} \left( \frac{\pl}{\pl y}(zr^3) - \frac{\pl}{\pl y}(yr^3)
\right) + ... \right\} \\ & = & -\kappa \left\{ {\bf i} 3r^2
\left(z \frac{\pl r}{\pl y} - y \frac{\pl r}{\pl z} \right) + ...
\right\}
\end{eqnarray*}

Now $\ds \frac{\pl r}{\pl z} = \frac{z}{r} \hspace{.2in} \frac{\pl
r}{\pl y} = \frac{y}{r}$ therefore $\ds z \frac{\pl r}{\pl z} - y
\frac{\pl r}{\pl y} = 0$

Therefore the ${\bf i}$ component of $\nabla \times {\bf F} = 0$,
as are the ${\bf j}$ and ${\bf k}$ components by symmetry.

Therefore $\nabla \times {\bf F} = 0.$

\item[(b)]
Potential energy.  $\ds {\bf U} = -\int {\bf F} \cdot d{\bf r}$
for all paths.

As ${\bf F}$ is conservative, choose  a radial path.

Therefore

$\begin{array}{rcl} {\bf U} & = & \ds -\int -\kappa r^3 {\bf r}
\frac{d{\bf r}}{dt} dt \\ & = & \ds \kappa \int r^4 dr \\ & = &
\ds \frac{\kappa}{5}r^5 (+ {\rm constant})\end{array}$


\item[(c)]
K.E. + P.E. = constant

Therefore $\frac{1}{2}m \dot {\bf r}^2  + \frac{\kappa}{5}r^5 = E$


\lq\lq Conservation of Energy"



\end{description}


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