For vertical circular motion, why does the minimum speed for a full cycle depend on what the ball is attached to? Suppose we have the following situation and I'm asked for the minimal speed to be given to the ball so it can continue a full cycle.

If the ball is connected to the middle point with some hard material (like Iron) then the answer is: $v_{min}=\sqrt{4gh}$ (using the law of conservation of energy we calculate the minimum speed so that the ball reaches the top with a speed of 0 and from there it will continue a full cycle)
But in case we replace the iron with a normal rope then why the answer changes to $v_{min}=\sqrt{5gh}$
 A: If the ball is attached to a hard support then it is constrained to move along the circular path, so no matter what the speed, as long as it is non-zero, circular motion will be achieved.
In contrast, if the ball is attached to a non-rigid rope, then this constraint is no longer valid. If the speed is not large enough at the top of the circle, then gravity will pull the ball down off of the circular path; the ball will fall. Contrast this with the hard support from earlier where even if the speed is small the rod will support the ball and a circular path is still possible.
Taking a more quantitative approach, at just the top of the circle$^*$, we need the following equation to be true for circular motion to occur:
$$mg+F=\frac{mv^2}{R}$$
where $m$ is the mass of the ball, $v$ is the speed of the ball, $R$ is the radius of the circle, and $F$ is the force exerted by the support/rope.
Now, if the speed is small enough so that $mv^2/R<mg$, then in order for this equation to be valid it must be that $F<0$, i.e. the force from the support points outwards. The hard support can exert an outwards force, but a rope cannot (try pushing something with a rope).
This is why there is a discrepancy between the minimum velocity. For the hard support $F$ can point outwards, and so $v>0$ at the top is sufficient. However, the rope cannot push the ball outwards, so we need $mv^2/R\geq mg$

$^*$These concepts are valid at other points on the circular path as well, but the equations are a little more complicated. Additionally, if the ball can make it past the top of the circle then by the symmetry of the situation the ball will make it back down along the circle. Therefore, just thinking about the top of the circle is sufficient here.
