Figure: The Foucault pendulum and the rotating frame of the Earth explicated
Image 1 Caption: A Foucault pendulum at Monash University in Australia. You see a the pendulum bob and its arm, a compass plate (i.e., a compass without a needle) across which the penulum bob oscillates, and a background image of the Earth centered on Australia.
But if the Foucault pendulum had been at the North Pole and you looked down on the pendulum and Earth from north celestial pole (NCP), the pendulum would oscillate in a fixed plane relative to the observable universe and the Earth would rotate counterclockwise relative to the observable universe. In an absolute physical sense the Earth is rotating.
However, if you are taking the Earth as a rest frame, the plane of pendulum oscillation rotates clockwise or, in other words, pendulum is in precession). The precession period at the North Pole is a sidereal day = 86164.0905 s = 1 day - 4 m + 4.0905s (on average).
Further explication:
But note that very strong gravitational fields (like those very near black holes) may cause inertial frames to be intrinsically in rotation relative to the observable universe, but this is a tricky point for which yours truly CANNOT find a clear explication. The best so far (and it does NOT say much) is Wikipedia: Inertial frame of reference: General relativity.
The effect is the precession of the plane of the Foucault pendulum's oscillation relative to the the frame of the Earth.
The precession is caused by the torque of the Coriolis force.
Torque is the twisting manifestation of a force.
The pivot the Foucault pendulum is frictionless, and so the pivot CANNOT exert any torque on the Foucault pendulum. Even a relatively small torque by friction would overcome that of the Coriolis force which is rather weak in this case. Of course, any rigid-direction pivot would completely stop the precession.
Another reason for NOT seeing precession is that even for a Foucault pendulum, the precession period is rather long. The formula for precession period is
P = ( 1 sidereal day )/sin(L) where L is latitude. = 1 sidereal day at L = 90° . = [sqrt(3)/2] sdays = (0.8660 ...) sidereal days at L = 60° . = sqrt(2) sdays = (1.4142 ...) sidereal days at L = 45° . = 2 sidereal days at L = 30° . = ∞ sidereal days at L = 0° .(Wikipedia: Foucault pendulum: Examples of precession periods). Note, a Foucault pendulum needs some kind of driver to keep it oscillating, but a driver that exerts NO torque.
In the Northern Hemisphere (Southern Hemisphere), the precession is clockwise (counterclockwise) relative to the rotating frame of the Earth when looking down on the Earth from north celestial pole (NCP) (south celestial pole (SCP)) (Wikipedia: Foucault pendulum: Mechanism).
The simplest locations for a Foucault pendulum are at the poles. There a Foucault pendulum oscillates in a plane fixed relative to the observable universe.
The second simplest location is the equator where the plane of oscillation does NOT precess at all relative to the Earth. The above period formula shows why this is so mathematically: when L = 0, sin(L) = 0, and the period goes to infinity.
Large size scale non-inertial frame effects are evidenced by weather (particularly anticyclones and cyclones: see Mechanics file: coriolis_force.html) and long-range artillery ballistics.
It had length 67
meters
and was released at a distance of 50.25 meters (3/4 times its length).