Is there any way to harvest large amounts of energy from the Earth's rotation?


This is what tidal power generators do - if earth did not rotate relative to the moon, there would be no lunar tidal motion (there would be a distortion of the earth's equipotential surface but it would not move). Since earth does rotate, you get motion of the oceans. At Fundy Bay, for example, significant power is extracted from this motion. See http://fundyforce.ca/

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    $\begingroup$ See also: physics.stackexchange.com/q/6400 $\endgroup$ – dmckee Feb 14 '15 at 18:50
  • $\begingroup$ another excellent places are the Gibraltar straight, and at Panama/Nicaragua to explore the altitude difference between the Pacific and the Atlantic Ocean (it is due to Earth's rotation). $\endgroup$ – Helder Velez Sep 19 '15 at 9:15
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    $\begingroup$ @HelderVelez the mean height difference across Panama is disappointingly small (about 23 cm: most of the locks are there because you have to "climb a mountain" to get to the other side; there is a tidal component but that would have to travel a long way from coat to coast); the tidal current at Gibraltar is also much less easy to capture than at Fundy (and I believe much ofthr current there t is due to differential evaporation / salt levels, in other words a one way component due to solar power, not rotation. See tidetech.org/news/NEW_Straits_of_Gibraltar_tidal_model.html) $\endgroup$ – Floris Sep 19 '15 at 11:55

The Foucault pendulum motion is induced via the Earth's rotation.

Because it moves it can perform work.

I answer Yes, but irt others it may depend on the more or less flexible definition of vast amount and the size and number of pendulums .

EDIT ADD to 'debunk' myself:

The above statements are correct,imo, but I will decompose the situation.
Some energy is given to the pendulum to put it in motion and it is only this amount of energy that can be obtained from it. There is no transfer of energy in virtue of the motion of the Earth.

The pendulum will keep in motion in the same plane in relation to the distant stars, obeying to the 1st Newtons law and keeping the same momentum (under the constriction of maintaining the same distance to the pivot point), summed along the full cycle of forward-backward periodic motion (it is not the instantaneous moment because there is a constant change of potential versus kinetic energy).

The Earth only transports the pendulum without exerting force on it and without any transmission of energy, in what is called 'parallel transport'. Nevertheless the pendulum is subjected to the normal balanced forces at Earth's surface: inward gravitational force and outward centrifugal one.

So. Lets rewrite the first sentence in a more correct way. The Foucault pendulum motion is induced via the Earth's rotation.
The motion of the Foucault pendulum plane irt the ground is induced via the Earth's rotation. But there is no energy transfer between the object Earth and the pendulum and so it will stop as soon we get back all the energy that was given to it in the initial moment to force its motion.

Conclusion: It is hopeless the use of the Foucault pendulum to extract energy from the Earth's rotation.

IMO, it is pedagogic to leave this wrong answer and the correction.

  • $\begingroup$ Your answer contradicts the preservation of the impulse moment. $\endgroup$ – peterh Sep 18 '15 at 20:07
  • $\begingroup$ Sorry, but I smiled. I do not have to explain because IT MOVES. Anyhow there is an explanation in that link and I'm sure that no principle is broken. I bet that it is the preservation of the impulse/inertia that makes it to move, imo. $\endgroup$ – Helder Velez Sep 18 '15 at 21:25
  • $\begingroup$ Sorry, maybe I am bad but I don't really see, where. The impulse moment of the Earth + pendulum system preserves, it doesn't matter, where is the pendulum, how it works, etc. And if it preserves, you can't take out energy from it. This is what I thought. $\endgroup$ – peterh Sep 18 '15 at 21:30
  • $\begingroup$ Second remark: the Foucault pendulum changes only the direction of the plane in which it moves. Its amplitude (in ideal case) doesn't change. So you won't have more energy in it, as before. $\endgroup$ – peterh Sep 18 '15 at 21:34
  • $\begingroup$ I will EDIT and debunk myself. $\endgroup$ – Helder Velez Sep 19 '15 at 8:28

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