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If the type of object matters, consider the human body. If the situation matters, consider standing on the inside wall of an O'Neill cylinder compared to standing on the surface of Earth.

"Differ in their effects on objects" means: Would the object be able to tell the difference? That is, is there an instrument that could tell whether it is placed in an O'Neil cylinder or on the surface of a planet from the effects (acceleartion, I suppose) of centrifugal force and gravity alone?

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    $\begingroup$ Wiki: People would, however, be able to detect spinward and antispinward directions by turning their heads, and any dropped items would appear to be deflected by a few centimetres. $\endgroup$ – Jasper Nov 17 '18 at 22:29
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General Relativity is compliant with the Strong Equivalence Principle. According to this principle:

The outcome of any local experiment (gravitational or not) in a freely falling laboratory is independent of the velocity of the laboratory and its location in spacetime.

This implies that locally gravity is indistinguishable from acceleration. Thus the answer is that locally the effects of gravity and the centrifugal force are the same. Here "locally" means a region small enough where the force is uniform. For example, if the rotating cylinder is large and you are confined inside an elevator, you would have a very hard time telling gravity from acceleration. However, in a larger region, many different experiments and observations would easily reveal differences between the centrifugal force and gravity, as justly stated in the comments and the other answer.

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What we normally think of as “gravity” on earth is actually a mix of gravitational and centrifugal force: plumb bobs don’t hang toward the center of the earth, but rather slightly toward the opposite pole. They are both static body forces, so it’s not possible to directly tell them apart locally.

But any rotating frame also has Coriolis force, which is detectable locally.

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Yes. An instrument that can sensitively measure the force gradient (for example, the difference between the force at one spot and the force at a nearby spot, say a foot away) could tell the difference. This “tidal” force will be greater for the O’Neill cylinder.

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  • $\begingroup$ I don't understand. Why should there be a difference between centrifugal forces at different spots inside the O'Neill cylinder at all? $\endgroup$ – user213272 Nov 17 '18 at 23:24
  • $\begingroup$ @user57423 The centrifugal force at your head height is slightly lower than what it is at your feet. $\endgroup$ – PM 2Ring Nov 18 '18 at 2:28
  • $\begingroup$ The centrifugal force points away from the axis of rotation. At a particular distance from the axis, it has the same magnitude but different directions. And as PM 2Ring pointed out, at different distances from the axis it has different magnitudes. $\endgroup$ – G. Smith Nov 18 '18 at 3:57
  • $\begingroup$ The centrifugal force on a mass $m$ is $m\omega^2 \rho \hat{\mathbf{\rho}}$ where $\omega$ is the angular velocity of the cylinder, $\rho$ is the distance from the axis, and $\hat{\rho}$ is a unit vector pointing away from the axis. $\endgroup$ – G. Smith Nov 18 '18 at 4:02
  • $\begingroup$ @G.Smith I see. But doesn't that hold for gravity as well? On Earth, gravity at my feet is different from gravity at my head, isn't it? And if the O'Neil cylinder had the same diameter as the Earth, the difference would we equally minimal, wouldn't it? $\endgroup$ – user213272 Nov 18 '18 at 8:20

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