Hi all, "When the top end of the Slinky is dropped, the information of the tension change must propagate to the bottom end before both sides begin to fall; the top of an extended Slinky will drop while the bottom initially remains in its original position, compressing the spring. This creates a suspension time of ~0.3 s for an original Slinky, but has potential to create a much larger suspension time." https://en.m.wikipedia.org/wiki/Slinky#Levitation https://en.m.wikipedia.org/wiki/Hooke's_law#Rotation_in_gravity-free_space Suppose you are spinning a Slinky around some point A in absence of gravity. The links above seem to imply that perhaps the far end of the Slinky will momentarily maintain a constant distance from point A after its release and perhaps even maintain its angular momentum. This seems wrong to me. The law of conservation of Momentum implies that the Center-of-Mass of the slinky would immediately follow a straight line after being released. Furthermore the COM would maintain (temporarily) a constant position relative to the far end before the slinky contracts. All this should result in the far end not only not maintaining an angular momentum around point A, but also immediately increasing its distance from point A. The difference here being that unlike dropping a slinky which would entail the COM staying aligned to the gravity vector throughout the free-fall, in the spinning scenario the COM will immediately move away from the radial path to a tangent path. A video and discussion of the free fall version can be found at http://blogs.discovermagazine.com/badastronomy/2011/09/26/slinky-drop-physics/ Link credit to: 01101001 Am I completely off? Thank you in advance for any comments and insights.