\begin{problem*}{7.62}
A roller coaster car is released from rest at the top of the first
rise and then moves freely with negligible friction.  The roller
coaster shown in Fig.~P7.62 has a circular loop of radius $R$ in the
vertical plane.
\Part{a} First, suppose the car barely makes it around the loop; at
the top of the loop the riders are upside down and feel weightless.
Find the required height of the release point above the bottom of the
loop, in terms of $R$.
\Part{b} Now assume that the release point is at or above the minimum
required height.  Show that the normal force on the car at the bottom
of the loop exceeds the normal force at the top by six times the
weight of the car.  The normal force on each rider follows the same
rule.  Such a large normal force is dangerous and very uncomfortable
for the riders.  Roller coasters are therefore not build with circular
loops in vertical planes.  Figure P5.24 and the photograph on page 134
show two actual designs.
\end{problem*} % problem 7.62

\begin{solution}
\Part{a}
Because the riders ``feel weightless'' at the top of the loop (point
$T$), we will assume that they are in free fall with a centerwards
acceleration of $g = v_T^2/R$.  Conserving energy between $T$ and the
release point $A$
\begin{align}
 E_A = mgh &= E_T = \frac{1}{2} m v_T^2 + mg(2R) \\
 h &= \frac{1}{2g}v_T^2 + 2R = \frac{Rg}{2g} + 2R = \ans{2.5 R}
\end{align}

\Part{b}
If the release comes from a higher point, there will be some normal
force at the top $N_T$ and at the bottom $N_B$.  Summing forces at
both points
\begin{align}
 \sum F_{cT} &= mg + N_T = m \frac{v_T^2}{R} \\
  v_T^2 &= gR + R\frac{N_T}{m} \\
 \sum F_{cB} &= -mg + N_B = m \frac{v_B^2}{R} \\
  v_B^2 &= -gR + R\frac{N_B}{m}
\end{align}
And conserving energy between the top and bottom
\begin{align}
 E_B = \frac{1}{2}mv_B^2 &= E_T = \frac{1}{2}mv_T^2 + mg(2R) \\
 v_B^2 &= v_T^2 + 4gR = -gR + R\frac{N_B}{m} = gR + R\frac{N_T}{m} + 4gR \\
 N_B &= N_T + 6mg
\end{align}
\end{solution}
