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Physics guy

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- Thread starter Physics guy
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- #1

Physics guy

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- #2

You've got the right intuition, but let's iron a few things out.

For a minute, forget about the Earth. Just imagine an idealised flat, rigid surface of infinite extent. Impose a uniform gravitational field ##-g \hat{\mathbf{y}}##. The book sitting on the table has its weight exactly canceled by the normal force exerted by the table, to satisfy the equilibrium condition.

If the Earth wasn't rotating, that would be a pretty good local model. In fact, even though the Earth is rotating, the rate of rotation is sufficiently slow that this is an*approximately* correct model.

But we might want to think about a different, maybe more accurate, model in which the Earth has non-zero rotational speed. To simplify the description, imagine the table is at the equator. As you correctly deduced, the net force on the book must now be non-zero and pointing toward's the Earth's centre, in order to provide the necessary acceleration ##\rho \omega^2## toward the centre. In other words, the magnitude of the weight will slightly exceed the magnitude of the normal force.

These two descriptions aren't in contradiction, simply because they're not describing the same scenario. The assumptions that underly the two models are different.

For a minute, forget about the Earth. Just imagine an idealised flat, rigid surface of infinite extent. Impose a uniform gravitational field ##-g \hat{\mathbf{y}}##. The book sitting on the table has its weight exactly canceled by the normal force exerted by the table, to satisfy the equilibrium condition.

If the Earth wasn't rotating, that would be a pretty good local model. In fact, even though the Earth is rotating, the rate of rotation is sufficiently slow that this is an

But we might want to think about a different, maybe more accurate, model in which the Earth has non-zero rotational speed. To simplify the description, imagine the table is at the equator. As you correctly deduced, the net force on the book must now be non-zero and pointing toward's the Earth's centre, in order to provide the necessary acceleration ##\rho \omega^2## toward the centre. In other words, the magnitude of the weight will slightly exceed the magnitude of the normal force.

These two descriptions aren't in contradiction, simply because they're not describing the same scenario. The assumptions that underly the two models are different.

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- #3

A.T.

Science Advisor

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Newton's Laws apply to inertial frames. The rotating rest frame of the Earth is not exactly inertial. To make Newton's 2nd Law applicable there, you have to introduce inertial forces (here Centrifugal force):

Rotating rest frame of the Earth:

Inertial frame where the Earth rotates:

See also:

https://en.wikipedia.org/wiki/Rotating_reference_frame

- #4

Dale

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