# Tag Info

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Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference: $$\left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$ where $s$ denotes the space fixed frame and $r$ denotes the rotating frame. ...

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I think the answers are all correct, but it's worth pointing out that a lot of Earth's rotation came when it was hit by Theia. If Theia had hit the earth's other side, the Earth just might be spinning clockwise, against the spin of all the other planets.

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When our solar system formed it had a certain amount of intrinsic angular momentum. As it collapsed over time it began to spin faster like an ice skater that brings her arms in. Our planet, Earth, was formed in this cloud. It too is the product of that spinning gas cloud long gone. So the Earth retains the angular momentum of the matter that formed it. The ...

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If you're asking why the Earth began rotating, the question isn't particularly enlightening, but the answer is simple: Because some torque acting on the Earth (more likely its constituent particles before gravity pulled them into a single object) in the distant past caused those particles to rotate in the counter clockwise direction. Although as LDC3 ...

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The solutions these equations give for the angles are different because they are for different rotation sequences. The first set of equations has the subscript xyz and the second has the subscript yxz, so these angles depend on the rotation sequence.

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You are going in the right direction: Since on a different latitude the pendulum will be rotating with earth, it will change the rotation due to the coriolis force. As the pendulum being at a pole is an extreme case, so is the position at the equator: Here there's no reason for the rotation Foucault's pendulum is famous for. An intuitive guess would ...

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Your strategy is fine (though I haven't carefully checked if your rotations are correct). Since rotations don't commute, you need to be careful to apply them in the correct order. You say that your rotation about the x-axis, let's call it $R$, followed by your rotation around the z-axis (be careful whether it's around the old or the new z-axis!), let's call ...

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The same effect does occur, but the magnitude is very small compared to the natural curvature of the earth. The follow up question would be, could there be a planet on which the ocean formed a concave surface at the poles? Intuitive answer If the water wants the flow away from the poles to the outside so badly that it forms a concave surface it will want ...

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I think (and somebody correct my math if I get this wrong), there is a tiny effect, and, no, it's not concave ever. If the Earth was a flat rotating disc, then this would happen. The Earth's sphere shape makes the 2 dimensional math inaccurate, but lets look at the 2-d math anyway. Newton's bucket mathematics Source: ...

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No, it won't. In a bucket, the gravitational forces acting on each particle of the fluid are parallel (if the the bucket is small with respect to the Earth radius, which I think we can always assume). Without rotations, the fluid will form a plane surface inside the bucket (which is neither convex nor concave). This surface coincide with a surface of ...

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The water is not in a bucket it is free to move around. An it does move to the equator because of the earths rotation (Here also gravitational forces from the moon play a role). So no I don't think they would see a convex surface, not even at a microscopic Level. They would however see that the water surface is a tiny bit less concarve then at the equator.

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Let's make this concrete by using a $\text{spin-}\frac 1 2$ state in the $z$ direction, where we get to use the Pauli matrices, usually written as$$\sigma_x = \left[\begin{array}{cc}0&1\\1&0\end{array}\right]; ~~~\sigma_y = \left[\begin{array}{cc}0&-i\\i&0\end{array}\right]; ~~~\sigma_z = ... 2 I don't have that text, but I can find the table of contents on the internet. Somewhere in that text (most likely chapter 5 on non-inertial reference systems), there should be a derivation that for any vector quantity \boldsymbol q, the time derivative of that vector in an inertial frame and a rotating frame that share the same origin are related by$$ ...

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The statement is really about the transformation between inertial co-ordinates and co-ordinates fixed to the body. This is expressed by: $$D_t = d_t + \omega(t)\times\tag{1}$$ where $D_t$ is the "total" derivative, i.e. the time derivative in the inertial frame, $d_t$ is the time derivative in the frame fixed to the body. Since there there are no torques ...

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