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Suppose a very long railway line goes from South Africa to Sweden, and then it's decided to move the entire railway line, sliding it 1 km to the north (leaving aside the difficulty of moving and the force required).

Would the railway bend because of Coriolis force?

I mean, every point of the railway would have different tangential velocity so when moving, we have different speeds at different time, then that's acceleration, then force, so if those are different forces applied to different points of the rail, then perhaps it would bend.

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Interesting question, but you'll need a little more definition. Most importantly, how fast is it moved north? Is it just a straight line, or does the railway turn at all? –  spencer nelson Feb 22 '11 at 23:47

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Of course that there would be forces that would try to bend the track but they would be tiny. Each segment of the track would be under the action of $-2m \Omega \times v$ Coriolis force. Note that the Coriolis force only depends on velocities, not accelerations as you stated!

In other words, there is the Coriolis acceleration, $-2\Omega\times v$, and you may see that it's tiny for realistic velocities of the moving track. Note that the direction of this acceleration is East-West at each point and its magnitude is zero at the equator where $\Omega$ and $v$ in your experiment have the same direction.

So the Coriolis force would mostly try to rotate the track around the axis going through its equator point. Obviously, the mechanical constraints would compensate this force and it would be largely invisible.

But if you consider a very similar setup to the tracks - namely rivers going to the North - they will eventually make their troughs asymmetric because of the Coriolis force. Albert Einstein was obsessed by this problem and wrote papers about it. ;-)

If your concern was that the shape of the track - along the length - doesn't change as you move it North, it doesn't matter. The laws of mechanics still recognize that the metal is moving and the Coriolis force depends on the actual velocity, and not just the velocity that is easily seen. ;-)

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