A **solid sphere** rolling without slipping enters a vertical loop of radius $R$. The **minimum height** $h$ from which it must start (measuring $h$ from the bottom of the loop) so that it just completes the loop is:

At top of loop, minimum condition: $v^2/R = g$ ⇒ $v^2 = gR$. Energy conservation: $mgh = mg(2R) + \tfrac{1}{2}mv^2(1 + 2/5) = 2mgR + (7/10)mgR$ ⇒ $h = 2R + 0.7R = 2.7R$. (Compare: point-particle answer is $2.5R$; the extra $0.2R$ is rotational KE the sphere also needs at top.)