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What would be the period of a pendulum with an infinite length, assuming a non-rotating Earth? The answer is not infinity.

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    $\begingroup$ Well, the limiting case of a mathematical pendulum of finite length would suggest that the period would tend to infinity. If you say it's not infinity, what kinds of assumptions are we allowed to make here? I mean, an infinitely long pendulum would rupture under the weight of its string. Is that supposed to be the answer? $\endgroup$ – Jonas Greitemann Sep 16 '13 at 20:42
  • $\begingroup$ I presume you are expected to calculate the result not in a uniform field as we do with reasonably sized pendulums (in which case the period does grow without bound) but with the full spherical field and neglecting the mass of the rod. $\endgroup$ – dmckee --- ex-moderator kitten Sep 16 '13 at 20:47
  • $\begingroup$ @dmckee how does the field falling off with height help? Wouldn't this "softening" just make the problem worse? $\endgroup$ – Brian Moths Sep 16 '13 at 20:51
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs It's not that it falls off with height, it's that there is a residual change in direction of the field. You can't use the analysis that we use on the finite length pendulum because the rod always points in the same direction. $\endgroup$ – dmckee --- ex-moderator kitten Sep 16 '13 at 21:17
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    $\begingroup$ Heck, while we're at it, what if the Earth is rotating infinitely fast instead? $\endgroup$ – Alfred Centauri Sep 16 '13 at 22:17
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Well the pendulum bob, moves in a straight line, and if we presume that the bob is effectively in a vacuum, or is heavy enough for air resistance to not matter, and the path is basically at the earth surface, it would seem that the period is effectively independent of the pendulum length for lengths many times the earth radius.

I can't be bothered to work out the answer, which should be relatively simple; but I am prepared to give you a WAG answer.

I believe the period is 84 minutes or thereabouts, provided of course the amplitude is a small value compared to the earth radius.

Now, 84 minutes just happens to be the period of a simple pendulum, whose length is equal to the earth radius. It is also the orbital period of an earth satellite orbiting essentially at the earth surface (no air resistance).

Moreover, if you drill a hole between any two points on the earth surface, close enough together that the center of the hole is not too much below the surface as a fraction of the earth radius, then the transit time for a round trip between those two points in say a straight evacuated frictionless tube, is also 84 minutes.

So I don't distinguish between the tube case, and the infinite pendulum case, so I say the answer is 84 minutes.

People who make inertial stable platforms (gyro stabilized are familiar with a "hum" instability in their systems, that has an 84 minute period.

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I choose to put the "centered" position of the bob over the north pole, and have the z-axis point upward through the pole. The bob swings in the x-z plane, with it's point of closest approach at radius $r_c = r_e (1 + \epsilon)$.

The gravitational force acting on the bob, in the limit of small oscillations is $ -mg(\hat{z} + \frac{x}{r_c}\hat{x}) $, and the rod provides $+mg\hat{z}$ leaving a residual of $F = -x\frac{mg}{r_c}\hat{x}$ so this arrangement is a harmonic oscillator with $k = \frac{mg}{r_c}$.

The usual solution applies giving $\omega = \sqrt{\frac{k}{m}} = \sqrt{\frac{g}{r_c}}$. That gives me a bit over 5000 seconds for small oscillations near the surface. This agrees with George's analysis.

Cute problem. As usual the process is more important here than the result.

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