As depicted in the figure, a rough track features a portion shaped...

For the sphere to completely traverse the circular track, it must not lose contact at the highest point. At the highest point, the minimum required speed $v_{top}$ is obtained when the normal force is zero.

The equation of motion for the center of mass at the highest point is:
$mg = \dfrac{mv_{top}^2}{R - r} \Rightarrow v_{top}^2 = g(R - r)$

The kinetic energy of a rolling solid sphere is:
$K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2 = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\left(\dfrac{2}{5}mr^2\right)\left(\dfrac{v}{r}\right)^2 = \dfrac{7}{10}mv^2$

The height of the center of the sphere at the highest point relative to its initial height is $h = 2(R - r)$.

Applying the principle of conservation of mechanical energy between the initial horizontal segment and the highest point of the circular track:

$K_{initial} = K_{top} + \Delta U$

$\dfrac{7}{10}mv_0^2 = \dfrac{7}{10}mv_{top}^2 + mgh$

Substituting $v_{top}^2 = g(R - r)$ and $h = 2(R - r)$:

$\dfrac{7}{10}mv_0^2 = \dfrac{7}{10}mg(R - r) + mg(2(R - r))$

$\dfrac{7}{10}v_0^2 = \left(\dfrac{7}{10} + 2\right)g(R - r)$

$\dfrac{7}{10}v_0^2 = \dfrac{27}{10}g(R - r)$

$v_0 = \sqrt{\dfrac{27}{7}g(R - r)}$