From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Sent: Monday, December 5, 2011 12:44 AM Subject: [math-fun] spherical pendulums?
Are they chaotic or just hairy? The Wolfram demonstration seems quasiperiodic.? Gene once disabused me of the folly of trying to resolve the motion into x and y. But suppose we hung the string from the inside of an upward cusp of a cycloid of revolution. Would the pendulum simply describe an ellipse? --rwg _______________________________________________
While the one-dimensional cycloidic pendulum is isochronous, the two-dimensional one is not. Assume a point mass pendulum. In the one-dimensional case, It is known that in this configuration, the motion lies on a cycloid, whose parametric equation can be taken to be x = a (u + sin u), z = a (1 - cos u). Doing the algebra, and letting s = sin(u/2), the kinetic and potential energies are T = (1/2) m ((dx/dt)^2 + (dz/dt)^2) = 8 m a^2 (ds/dt)^2, V = m g z = 2 m g a s^2. This has the form of a harmonic oscillator, T = (1/2) M (ds/dt)^2, V = (1/2) M ω^2 s^2, and solving for frequency, ω^2 = g/(4 a), verifying the isochronicity since ω is independent of amplitude. Now consider circular motion in the two-dimensional case. The centripetal force in general is m v^2/r = m r (dθ/dt)^2 = m a (u + sin u) (dθ/dt)^2, and for this case is dV/dr = m g (dz/du) / (dr/du) = mg tan(u/2). Then (dθ/dt)^2 = (g/a) tan(u/2) / (u + sin u). Now the frequency increases with amplitude. This makes sense. The mass in an ordinary pendulum moves on a circular arc, but the frequency decreases with amplitude. By bending the circular arc upward into a cycloid, we increase the restoring force, compensate for the decreased frequency, and make the pendulum isochronous. But for circular motion in the two-dimensional case, the pendulum is always in the region of increased force, and so the frequency is higher. -- Gene