Correction below: said "rotating", meant "vertical". 2011/12/12 Bill Gosper <billgosper@gmail.com>
On Mon, Dec 5, 2011 at 1:44 AM, Bill Gosper <billgosper@gmail.com> wrote:
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
Gene>
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 _______________________________________________
A spherical pendulum that wishes it weren't: http://en.wikipedia.org/wiki/Foucault_pendulum : "A Foucault pendulum requires care to set up because imprecise construction can cause additional veering which masks the terrestrial effect. The initial launch of the pendulum is critical; the traditional way to do this is to use a flame to burn through a thread which temporarily holds the bob in its starting position, thus avoiding unwanted sideways motion. Air resistance <http://en.wikipedia.org/wiki/Air_resistance>damps the oscillation, so some Foucault pendula in museums incorporate an electromagnetic or other drive to keep the bob swinging; ..."
Questions: How do you electromagnetically pump it without introducing spurious torques? How do you synchronize with the swinging?
How feasible would it be to simply weld the hinge of an ordinary pendulum to the bottom of a freely spinnable [vertical] shaft? Emphasis on the "freely". --rwg