Using a chain of rubber bands, I swung a ball around in a vertical loop. This demonstration shows how the tension in an elastic band changes according to the position of the ball, by referring to the length of the elastic band.

When the ball of mass $m$ is at the bottom of the loop, the centripetal force is given by the difference between tension $T_{bottom}$ and weight $mg$, where $T_{bottom}$ varies depending on the speed of the ball $v_{bottom}$ and the radius of the curvature $r_{bottom}$.

$T_{bottom} – mg = \dfrac{mv_{bottom}^2}{r_{bottom}}$

When the ball is at the top of the path, it is given by

$T_{top} + mg = \dfrac{mv_{top}^2}{r_{top}}$

As the weight is acting in the same direction to tension when the ball is at the top, a smaller tension is exerted by the elastic band to maintain a centripetal force. Therefore , $T_{bottom} > T_{top}$.

The GeoGebra app below shows a simpler version of a vertical loop – a circular path with a fixed radius $r$. Consider a ball sliding around a smooth circular loop. The normal contact force varies such that

$N_{bottom} = \dfrac{mv_{bottom}^2}{r} + mg$

$N_{top} = \dfrac{mv_{top}^2}{r} – mg$

It can be shown that the minimum height at which the ball must be released in order for it to complete the loop without losing contact with the track is 2.5 times the radius of the frictionless circular track.

If we were to consider the rotational kinetic energy required for the ball to roll, the required initial height will have to be 2.7 times the radius, as shown in the video below:

Many thanks to Dr Darren Tan for his input. Do check out his EJSS simulation of a mass-spring motion in a vertical plane, which comes with many more features such as the ability to vary the initial velocity of the mass, graphs showing the variation of energy and velocity, as well as an option for a mass-string motion as well.