Hula-hoop physics What are the important parameters to consider when trying to keep a hula-hoop in the air by spinning it around the waist?
For instance, when shopping for hula-hoops, one can commonly find its diameter and weight, but how do they affect the actual ease of keeping it in the air? What's the physics behind it?
 A: The ratio of the circumference of the body and the hulahoop (HH) is an important parameter. The greater the ratio, the more the HH will rotate - and the more it rotates, the greater its angular momentum. When you have an object spinning with its axis vertically, and apply a torque, it will precess - instead of falling in the direction of the torque, its angle of rotation changes. As it does so, the angle of the HH relative to the body changes - and in fact, it will "climb" a little bit on the body.
That is the key of HH physics, I think: the axis of rotation is precessing. As long as the ring is "thin", the angular momentum scales with $r^3$ - assuming that mass is linear with size. The torque, on the other hand, scales with $r^2$. That means that the rate of precession will be slower for larger HHs - and that a larger hoop is easier. But also, the ratio of diameters of body and hoop will change the rate of rotation. The faster you rotate, the less far the angle of the hoop needs to tip in order to maintain precession.
