Rotating Pendulums

My reasoning was that if a pendulum would be close to the Northpole then the rotation of the earth would have no influence on it, it would just swing in the same plane - so, after 6 hours it would be swinging in the North-South direction. I think the sun comes up in the East (doesn't really matter), so looking on top of the northpole the Earth rotates towards the East or counter-clockwise and for an observer the plane in which the pendulum would swing would seem to rotate clockwise. On the South pole things are the same but the observer would be upside down and think that the pendulum rotates counter-clockwise (if you look on top of the southpole then the Earth rotates towards the East and thus clockwise). Therefore, my reasoning was - if you move the pendulum from northpole to southpole it has to change rotation from counter clockwise to clockwise, which is only possible when half way it doesn't rotate at all. So it seemed logical to assume that close to the equator it would rotate slower :*). Yay! No formulas!

This problem is about the so called "Foucault pendulum". The angular velocity of the rotation of the pendulum's plane of motion is

$\omega= 360^{\circ} \sin \phi /day$

where $\phi$ is the latitude. See more here: https://en.wikipedia.org/wiki/Foucault_pendulum

G force at the pole is generally lesser at /near poles than compared to other parts of earth. Also, time period of a standard pendulum varies inversely with the square root of g. Using these simple facts one can come to the answer.

the Earth rotates around its axis which means that if you're standing directly on the north or south pole, you're only rotating with the Earth's spin while if you're on the equator, you're not rotating at all and instead you're moving with the Earth's spin, what I just described were the two extremes on how you move while standing on the Earth. If you're on the equator and you're conducting this experiment then the pendulum will never rotate because you're not rotating at all but as you travel more north or south, it will start to increasingly rotate with the Earth thus making it possible to mark a point in time where you can see the pendulum swinging along the North-South direction.

Time to turn is inversely proportional to the sin of the latitude. At the poles the direction of a Foucault pendulum swings a full circle in one day. At the equator, it never changes direction.

In simplified terms, the period of rotation , T ( in hours) of the plane of oscillation for a Foucault pendulum is T = 24/ sin ϕ where ϕ represents the latitude of the place Here for Pierre , T = 24 /sin48.9° = 24/0.7535 = 31.85 hours

For Sydney , T= 24/ sin33.9° = 24/ 0.5577hours = 43.03 hours Longitude does not have much bearing on the oscillation time