# Problem with a rotating frame of reference on the South pole

Consider this problem:

A high-speed train is traveling at a constant 150 m/s (about 300 mph) on a straight horizontal track across the south pole. Find the angle between a plumb line suspended from the ceiling inside the train and another inside a hut on the ground. In what direction is the plumb line on the train deflected?

I'm positively perplexed by it. I think that what we're looking for is a relationship $\tan(\theta)=\frac{a_n}{a_t}$, because for the plumb line in the hut $a_t=1$ but not for the plumb line in the train due to the Coriolis force.

However,

1. What is the frame of reference here? Is it rotating, fixed so that the train is standing still? Is it rotating with the earth?
2. Since the train is on the south pole, isn't that the same to say that the earth's rotation doesn't affect it?
3. Can you show me how to get the right answer? The right answer is supposed to be 0.13 degrees. In which direction?

That line will get Coriolis acceleration $$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$ ($\Omega$ is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between $\Omega$ and $v$ and the absolute value will be simply $$a = 2\Omega v$$ and the direction will be to the right (you can use some hand rule or vector cross to get that) if front is the direction in which the train moves.
• What frame of reference do you use? Is $\Omega$ the rotation of the earth? How can I calculate that the line deviates 0.13 degrees? – user714 May 18 '13 at 11:08
• Rotating frame in which train is going straight (as says in the exercise). Yes, $\Omega$ is. Try $\tan(\theta)=F_c/F_a = a/g$. Also $\tan(\theta) = \sin(\theta) = \theta$ (in radians) should for work such a small angle. – Džuris May 18 '13 at 11:14
• That works, thank you very much. Do you know why the centrifugal acceleration didn't matter here? It seems to me that the acceleration normal to the surface would be $g+R\Omega^2$? – user714 May 19 '13 at 11:19
• @Althalos, there is no centrifugal acceleration on the axis. In which direction could it be there? Your formula is correct, but the $R$ is distance from axis of rotation which is $0$ in your case. – Džuris May 19 '13 at 13:02