Helium Balloon and Compton Generator

[Sculptured]

From Classical Mechanics by Taylor:

"The Compton generator is a beautiful demonstration of the Coriolis force due to the Earth's rotation,... A narrow glass tube in the shape of a torus or ring (radius R of the ring >> radius of the tube) is filled with water, plus some dust particles to let one see any motion of the water. The ring and water are initially stationary and horizontal, but the ring is then spun through 180 degrees about its east--west diameter. Explain why this should cause the water to move around the tube. Show that the speed of the water just after the 180 degree turn should be 2 \Omega R cos[\theta] . \Omega is the Earth's angular velocity and \theta is the colatitude..."

Alright, so I've figured that in the rotating frame if you make the 180 spin in the direction of the Earth then the southern portion has a force to the west and the northern portion has one to the east. This causes the torque upon the water and its subsequent motion. Now, the only way I can solve the problem is to get out into the inertial frame but I'd like to know how to find the speed using the effect of the colioris force. Any advice on where to begin in this direction?

 

[member 553083]

The Coriolis force is a force that acts on an object moving in a rotating frame of reference, and it can be used to calculate the speed of an object in this frame. To calculate the speed of the water after the 180 degree turn, you need to consider the change in velocity due to the Coriolis force. The Coriolis force is given by F = 2m \Omega \times v, where m is the mass of the water and \Omega is the angular velocity of the Earth. The change in velocity due to the Coriolis force is then given by \Delta v = 2\Omega \times v. Thus, the speed of the water just after the 180 degree turn can be calculated as v= \Delta v/2\Omega = 2Rcos[\theta].