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Compton Generator

  1. Dec 21, 2006 #1
    *I don't know if this belongs to homework section or not.
    I just found the description in my book interesting and want to know more. If this is inappropriate, please move it to homework section.

    Compton Generator is basically a device invented by Compton which is used to measure the Coriolis Force (a fictitious force in non-inertial frame). As described in my physics book, it works like this:

    First, you get a torus filled with water and flat on the ground (the water is initially non-rotating in the frame of the earth), then you flip the device 180 degrees through the east-west axis.

    My physics book claims that after the flip, the water in the torus begins to rotate. The book even gives a explicit formula for the speed after the flip, which is only dependent on the radius of the torus and the colatitude (assuming the cross section of the torus is very very small comparing to the torus as a whole).

    my understand of how the device works is that when one turns the torus, the water gains velocity in the turning direction. The Coriolis Force then changes the direction of that vector, resulting in a net tangential velocity.

    The device seems very interesting... and I've been trying to figure out the mathematics behind it... I tried to assume constant angular velocity when flipping the torus, but that integral turns out quite messy and is dependent on the angular velocity. I couldn't find any work-energy relationship either, since Coriolis Force is absolutely non-conservation.

    there might be some kind of parallelism between magnetic flux and this. But that seems to be true only if an initial "current" (flow of water) is presented (similar to how energy is magnetic moment dot magnetic field). I can express the torque in terms of the flow in the torus, radius and such but that doesn't seem to help much (since it is dependent on the flow/speed of the water)

    The speed of the water after 180 flip is as following:
    [tex]v=2\Omega R \cos\theta[/tex]
    where Omega is the angular velocity of the earth, theta is the colatitude.
    I wonder if the equation in my book really works. And what about the equation for speed for turning any amount of degrees (v in terms of the angular displacement). And what happen if the shape is not a torus, but a random closed loop (smooth) of water? And what if you turn the torus the other way around 180 degrees? (I suppose the water will stop rotating)
    Last edited: Dec 21, 2006
  2. jcsd
  3. Dec 23, 2006 #2
    Come on... anyway has any idea?

    I have been trying to figure this out for a couple of days... it is really frustrating how I cannot understand where the equation comes from..

    How can I derive the equation for speed of the water after the 180-degrees flip?
  4. Dec 23, 2006 #3
    I think the correct equation shouldn't contain the colatitude dependence. The coriolis force has equal magnitude everywhere on a circle (center to circumference), so a measurement at a particular (co)latitude (say the equator) should produce the same result as if it could be repeated with the same orientation closer to (or along) the polar axis. For this particular device a 180 degree flip seems to return it to its starting condition, that is, it doesn't appear to matter whether it is oriented initially flat on the ground (of the earth's sphere) or tilted arbitrarilly (eg. so as to be flat if the earth were a cylinder with axis aligned to rotation). If so then it should produce the same measurement at any latitude.
  5. Dec 25, 2006 #4
    Finally, I seem to have reasoned out the equation. It turns out that it doesn't really have anything to do with Coriolis Force.

    when viewed in the inertial that shares the same center as the earth, the equation can be easily derived.
    let the center of the torus be at latitude [tex]\theta[/tex]
    then the northern end of the torus would be at latitude
    and the southern end would be at latitude

    the torus is rotating as seen in the inertial frame. When the torus is quickly flipped, the speed of the water in either end is not affected. However, when the southern end is brought to the position of the northern end, the speed of the water is not rotating fast enough to remain stationary in the rotating frame. the difference in speed would be the difference in speed between the northern end and the southern end before the flip. That difference is:
    (R_e is the radius of the earth)

    simplify the expression, the difference in speed is:
    [tex]\Delta v=2R_e\Omega\cos\theta\sin\epsilon[/tex]

    a little bit of trig reveals that

    so, in the rotating frame, the speed of rotation is exactly the difference in speed, hence:
    Last edited: Dec 25, 2006
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