\begin{problem*}{7.2}
A $F_g = 400\U{N}$ child is in a swing attached to $r = 2.00\U{m}$
ropes.  Find the gravitional potential energy $U_g$ of the child-Earth
system relative to the child's lowest position when
 \Part{a} the ropes are horizontal,
 \Part{b} the ropes make a $\theta = 30.0\dg$ angle with the vertical, and
 \Part{c} the child is at the bottom of the cirvular arc. 
\end{problem*} % problem 7.2

\begin{solution}
$U_g$ is given by
\begin{equation}
 U_g = mgh = F_g h
\end{equation}
Therefor, we need to determine the vertical distance between the
child's location for a given part of the question and the child's
lowest position.

\Part{a}
The child is one radius above the lowest position, so 
\begin{equation}
 U_{gA} = F_g r
        = 400\U{N} \cdot 2.00\U{m}
	= \ans{800\U{J}}
\end{equation}

\Part{b}
The child has height $h_b = (1-\cos\theta)r$, so
\begin{equation}
 U_{gB} = F_g (1-\cos\theta) r
        = U_{gA} \cdot (1-\cos\theta)
	= 800\U{J} \cdot (1 - \cos 30.0\dg)
        = \ans{107\U{J}}
\end{equation}

\Part{c}
The child has a height of 0, so $U_{gC} = \ans{0\U{J}}$.
\end{solution}
