newtonian mechanics - Pouring coffee in a rotating space station

Following upon of the question Throwing a ball on a rotating space station, I'm interested in understanding how to actually calculate the result of the Coriolis effect in a particular situation. I've seen the equation defined as:

$a_{cor} = -2\omega v$

But I'm really not sure how to solve it, or what the result would mean practically.

So, when I wake up in the morning on the inside of a spinning cylinder with a radius of 3.2km and an angular velocity of 0.52rpm (to simulate ~1g of gravity), and I pour my first cup of coffee, how much is the stream deflected?

Then, when I take the space taxi to the spinning Bernal sphere next door, with a radius of 0.25km and angular velocity of 1.9rpm (to simulate ~1g of gravity in the valley), when I pour my second cup of coffee, how much is the stream deflected?

Then, when I hop onto the Discovery for my journey to Jupiter, as I pour my third cup of coffee in the spinning habitat, with a radius of 30m and an angular velocity of 5.5rpm (again, simulating ~1g of gravity), how much is the stream deflected?

We can probably simplify the coffee bit to assume a spherical 1 gram droplet dropped 20cm, for a foreshortened, but dramatic, pour. :)

Answer

In another forum I've been told that the angular momentum and velocity doesn't actually matter for figuring the deflection, and it actually breaks down to a trigonometry problem involving ratios of radii. In a faster or slower rotation, everything just happens faster or slower. Because I'm a software engineer and not a physicist, please pardon the Python:

In [1]: import math
   ...:
   ...: def measure_coffee_deflection(dropped_from, landed):
   ...:     ratio_x = dropped_from / landed
   ...:     print("Ratio", ratio_x)
   ...:
   ...:     trajectory = math.acos(ratio_x)
   ...:     print('Trajectory', trajectory, 'radians')
   ...:
   ...:     habitat_rotation = math.sqrt(1./(ratio_x**2) - 1.) # radians

   ...:     print('Habitat rotation', habitat_rotation, 'radians')
   ...:     effective_rotation = habitat_rotation - trajectory # radians
   ...:     print('Effective rotation', effective_rotation, 'radians')
   ...:
   ...:     return effective_rotation * landed
   ...:

In [2]: # On the O'Neill Cylinder, radius 3.2km
   ...: radius = 320000 #cm
   ...: dropped_from = radius - 10 #cm

   ...: landed = radius
   ...:
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.99996875
Trajectory 0.007905714738315722 radians
Habitat rotation 0.007905879445677281 radians
Effective rotation 1.647073615586303e-07 radians
Deflection 0.0527063556987617 cm


In [3]: # On the Bernal Sphere, radius 0.25km
   ...: radius = 25000 #cm
   ...: dropped_from = radius - 10 #cm
   ...: landed = radius
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.9996
Trajectory 0.028285214141364843 radians
Habitat rotation 0.028292759782811733 radians
Effective rotation 7.5456414468898225e-06 radians

Deflection 0.18864103617224556 cm

In [4]: # On the Discovery, radius 30m
   ...: radius = 3000 #cm
   ...: dropped_from = radius - 10 #cm
   ...: landed = radius
   ...:
   ...: print('Deflection', measure_coffee_deflection(dropped_from, landed), 'cm')
   ...:
Ratio 0.9966666666666667

Trajectory 0.08167235558059345 radians
Habitat rotation 0.08185443645833014 radians
Effective rotation 0.00018208087773669002 radians
Deflection 0.54624263321007 cm

So:

• 
  

  on the O'Neill cylinder, the coffee deflects about half a millimeter.

  
  



• 
  

  on the Bernal sphere, just shy of 3mm

  
  


• 
  

  on the Discovery, about half a centimeter

  
  

Looks like I can continue to pour my coffee with confidence, no matter the rotating frame of reference.