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Let $T := 24\times 60\times 60 [s]$.
Let $r [m]$ be the radius of the Earth.
Let $g [m/s^2]$ be the gravitational acceleration at the equator.
Let $m [kg]$ be the mass of Mr. A.

Mr. A is standing on the equator.

Centripetal force is $mr\omega^2 = mr(2\pi/T)^2 [N]$.
Gravitational force acting on Mr. A is $mg [N]$.
Let $N$ be the normal force acting on Mr. A.

It must hold that $mg - N = mr(2\pi/T)^2$.

Therefore, $N$ must be $mg - mr(2\pi/T)^2$.

If Mr. B's mass is half of Mr. A's mass, then the normal force must be adjusted to half, and Mr. B will also perform the same circular motion as Mr. A.

Isn't it strange?

The faster the Earth rotates, the more the normal force decreases. And when the necessary centripetal force becomes $mg$, the normal force becomes $0$. If the necessary centripetal force exceeds $mg$, the normal force can no longer be adjusted, and the circular motion cannot continue.

In reality, this is an impossible scenario, but let's suppose the necessary centripetal force did exceed $mg$, and Mr. A continued to perform circular motion. In that case, would physicists consider that a negative normal force is acting?

What exactly is the normal force?

Is it something that cannot be known without experimentation whether Mr. A will continue circular motion when the necessary centripetal force exceeds $mg$?

And as a result of the experiment, if Mr. A continues the circular motion, will physicists conclude that a negative normal force is acting?

Furthermore, if Mr. A stops performing circular motion as a result of the experiment, will physicists conclude that the normal force is always positive?

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If Mr. B's mass is half of Mr. A's mass, then the normal force must be adjusted to half, and Mr. B will also perform the same circular motion as Mr. A.

The gravitational force and the centripetal force is also halved, so the mass is a red herring. It makes no difference.

What exactly is the normal force?

The normal force is an upward reaction force of the ground, in response to the force of an object acting downwards. The normal force is a contact force and cannot become negative when an object moves away from the surface.

Is it something that cannot be known without experimentation whether Mr. A will continue circular motion when the necessary centripetal force exceeds mg ?

It is best to analyse this situation in the accelerating rotating reference frame of a person standing on (or holding onto) the surface of the Earth, so that the centrifugal force is considered real. Let the test object be a loose ring on a vertical pole that is free to slide upwards if required. As the Earth rotation rate is increased, the normal force decreases and as a result, the centripetal force increases by just the right amount to keep the object moving in a circle with the same radius as the Earth. Once the centrifugal force equals the force of gravity, the normal force goes to zero and cannot reduce further to compensate. At this point the object is effectively weightless. Any further increase in the rotation speed of the Earth will cause the ring object to start sliding up the pole, because the centrifugal force exceeds the the gravitational force. The centripetal force stays constant, but is not sufficient to keep the object in an orbit with the same radius as the Earth at this increased rotation speed and the object moves out to a larger circle. The object will move away from the surface of the Earth. The normal force does not become negative.

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  • $\begingroup$ KDP, thank you very much for your answer! $\endgroup$ Commented May 25 at 3:43

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