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I can't quite fathom the source of your confusion (I think it might have something to do with a focus on the notion of rotation here---angular momentum does not require rotational motion), so I'm having trouble writing a really clear response. For the moment I would rather offer a program for practicing the right skills rather than reinforcing the mistaken ...


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You can describe Coriolis effect If one walks along meridian, strange force pulls him aside. If one walks along parallel, gravity appears stronger or weaker. If one jumps, he lands into different place. All these effects the grater the faster planet rotation. If it rotates to make Earth-like weights, effects will be noticeable.


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One issue is since it's a sphere, the spinning is only like gravity along the plane of spin. At the poles, there will be no such effect, and part-way to the poles, the force will be less, and at an angle to the surface. That's why a hollow "ringworld" is slightly more plausible than a hollow sphereworld. This would be a welcome question with ...


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Suppose you're standing at the edge of an enormous merry-go-round. You'll feel a "force" that seems to be pushing you outwards. This is the centrifugal force. The centrifugal force is the force you feel because you're at a certain distance away from the axis of rotation. It is written as $\vec{F}_{\rm centrifugal} = m~\vec{\omega} \times (\vec{r} \times ...


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Look in Wikipedia http://en.wikipedia.org/wiki/Coriolis_effect. For understanding intuitively the Coriolis force effect, assume an object moving according to a static (inertial) frame of reference, in the plane perpendicular to the rotation axis, and along the radius, In the rotating frame, see the animation in Wikipedia, the Coriolis force imposes an ...


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The Coriolis force $\vec F_{\text{coriolis}} = -2m \, \vec \omega \times \vec v$ only depends on velocity. The centrifugal force $\vec F_{\text{centrifugal}} = -m \, \vec \omega \times (\vec \omega \times \vec r)$ only depends on position. Finally, if the object is not rotating uniformly ($\dot {\vec \omega} \ne 0$), then yet another fictitious force comes ...


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what does the commentator mean by concavity of the floor? He or she means that the surface you walk on is in fact the inside of a cylindrical surface. Like a very large version of the inside of a wedding-ring. The curvature of this surface can be measured.


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That the floor is not flat but circular, an the centrifugal force acts radially, and is the same at the same radius. If you cover the concavity of the floor with a planar surface or either make the stations of planar segments (a polygon instead of a circle), then you will measure different accelerations at different points of the floor, because they are not ...


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The "rotating particle" picture does not work for $L=0$, unfortunately. Both the electron and the nucleus are waves who create a standing spherically symmetric wave of their relative motion.



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