 # "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?" we're looking for is a relationship tan(theta)=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. Jaxson Mack 2022-08-12 Answered
"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?"
we're looking for is a relationship $\mathrm{tan}\left(\theta \right)=\frac{{a}_{n}}{{a}_{t}}$, 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?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it Donovan Shields
That line will get Coriolis acceleration
$\stackrel{\to }{a}=-2\stackrel{\to }{\mathrm{\Omega }}×\stackrel{\to }{v}$
($\mathrm{\Omega }$ 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 $\mathrm{\Omega }$ and $v$ and the absolute value will be simply
$a=2\mathrm{\Omega }v$
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.

We have step-by-step solutions for your answer!