1. The problem statement, all variables and given/known data A uniform thin circular rubber band of mass M and spring constant k has an original radius R. Now it is tossed into the air. Assume it remains circular when stabilized in air and rotates at angular speed ω about its center uniformly. Derive an expression for the new radius of the rubber band. Express your answer in terms of ω, k, M, R, and any fundamental constants. 2. Relevant equations Fspring = kΔx Fcp = mω2R ∑F = ma C/(2r) = π 3. The attempt at a solution I first considered a mass element Δm in the rubber band. This Δm has acceleration ω2R', where R' is the new radius. Of course, this means that the mass element has centripetal force of magnitude ΔF = (Δm)ω2R'. Furthermore, a rubber band stretched a distance Δx behaves just like a spring; so if a rubber band with initial length 2πR (original radius of the rubber band) is stretched to a final length of 2πR', the magnitude of the force on the rubber band is T = 2πk(R' - R). But I don't know how to relate the force T to ΔF. Since the center of mass of the rubber band only moves in the up-down direction, I don't think it's true that the sum of all the small ΔF's is T; I think the sum of all the small ΔF's is actually 0 since there's 0 net force on the rubber band in the horizontal plane. Because of this, I don't know how to even use these two equations together. What more information can I use to solve this problem? The correct answer is R' = (4π2kR)/(4π2k - Mω2).