A ball sliding on a rail: Conservation of energy and centripetal force

Question:
A ball is released at a certain height to slide on the rail as shown. It goes through the straight and circular track that is ended at the highest position C. The radius of the circular rail is $R$. The angle $\theta$ is taken from C to position of the ball. The mass and velocity of the ball are denoted as $m$ and $v$, respectively. We assume that there is no air resistance.
(1) What is the normal force so that the ball keeps sliding on the rail?
(2) Use conservation of energy. Find the angle when the ball gets off from the track.
(3) What is the minimum height to release to reach the maximum point C?

Answer:
(1) When the normal force, gravitational force, and centripetal force are balanced, the ball barely tracks the rail. From the Newton's equation of motion in a circular track is given as
\begin{equation}
\sum F = N + mg\cos\theta = \frac{mv^2}{r}
\end{equation}
Note that the direction toward center is defined as positive. Thus, we have
\begin{equation}
N = \frac{mv^2}{R} - mg\cos\theta
\end{equation}

(2) The height of the ball in the circular track can be expressed as $R+R\cos\theta$. From conservation of energy, the initial total energy must be equal to the final total energy.
\begin{equation}
mgh = mgR(1+\cos\theta) + \frac{1}{2}mv^2
\end{equation}
From (2) and (3), eliminate $v$.
\begin{equation}
N = mg \left( \frac{2h}{R} -2 -3\cos\theta \right)
\end{equation}
When $N=0$, the ball gets off from the track. Substitute zero into (4) and solve for $\theta$.
\begin{eqnarray}
& & \cos\theta = \frac{2}{3}\left(\frac{h}{R}-1\right)  \\
&\rightarrow& \theta = \cos^{-1} \left[\frac{2}{3}\left(\frac{h}{R}-1\right) \right]
\end{eqnarray}

(3) When $\theta$ is zero in (5), the ball falls off at C. Since $\cos 0 = 1$, the minimum height $h$ is
\begin{equation}
h = \frac{5}{2}R
\end{equation}

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