Assuming the ball is released from rest, what should its minimum initial height be in order to completely orbit the circular loop of radius $R$? This is a classic textbook problem requiring both energy conservation and the analysis of forces at the top of the loop. Neglecting the radius of the ball, $r$, compared to the loop’s radius, $R$, energy conservation from the initial position, $h$, to the loop’s top at $2R$ gives

$$\begin{align} mgh = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2 + 2mgR \end{align}$$

Now, we know that:

$$\begin{align}I = I_{sphere} = \frac{2}{5} mr^2 \tag{2} \end{align}$$

$$\begin{align}\omega = \frac{v}{r} \tag{3} \end{align}$$

Using equation (2) and (3), we have:

$$\begin{align}gh = \frac{7}{10} v^2 + 2gR \tag{4} \end{align}$$

At the loop’s top, the minimum force required for the ball to just remain in contact with the loop is supplied by gravity without any normal force, so

$$\begin{align} mg = \frac{m v^2}{R} \tag{5} \end{align}$$

$$\begin{align} g = \frac{v^2}{R} \tag{6} \end{align}$$

$(6) \rightarrow (4)$ provides: $h_{min} = 2.7R$ (note that this is $.7R$ above the loop’s top).