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Coriolis effect vs common sense!

  1. Sep 28, 2012 #1
    Hello, I've long thought that the coriolis effect was something quite logical (i.e., things luck funny when in a rotating frame of reference), but interested in the mathematical reasoning behind it (because it, being a ficticious force is more about geomtry than physics), found that it was indeed one of the most counterintuitive things I know (at least in classical physics). My problem is the following:

    Imagine you have a spinning mass (holded by a massless string say). And you increase the radius of the string with a constant rate, at the same time than exerting a perpendicular force on the mass so as to keep the angular velocity constant. Then tangential acceleration (tangential velocity is w*r) of the mass should be w*r' (as w is constant), where w is angular velocity and r' is the time derivative of the radius.

    However, when I saw the same situation in Feynman's physics lectures book, I was dazzled. He, instead, considers the change in angular momentum, which is 2*m*r*r'*w, and that is equal to the torque, which is F*r. So te force exerted is 2*m*r'*w, and the acceleration, 2*r'*w.

    So..how come this is double the previous result?? Where have I flawed my reasoning?

    Thank you in advance
     
  2. jcsd
  3. Sep 28, 2012 #2

    haruspex

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    You are confusing tangential linear acceleration (d(rw)/dt) with angular acceleration, in the sense that applies to torque and angular momentum. If the radius were increased with the tangential velocity constant, there would still be a torque required.
     
  4. Sep 28, 2012 #3
    Thanks! Altho I am still a bit confused. Imagine the case you propose: you have to exert a torque equal to m*r'*v, meaning that the force you have to apply is m*(r'/r)*v. However, the mass doesn't gain any velocity? Is in this system, the effective mass equal to infinity?

    Also, in my case the 2*m*r*r'*w force, the Coriolis force, actually comes from the Coriolis acceleration 2*r'*w. So according to this acceleration, the tangential acceleration should be that, and not half of that!

    Reference: http://en.wikipedia.org/wiki/Fictitious_force#Rotating_coordinate_systems
     
    Last edited: Sep 28, 2012
  5. Sep 29, 2012 #4

    A.T.

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    I think in the case of constant tangnetial velocity and changing radius the mass still moves along a circular arc, but with a different center. The net force is still acting perpendicular to the velocity, so no work is done on the mass. But there still can be a tangential force component in respect to the original center, and hence a torque around it.
     
  6. Sep 30, 2012 #5
    To my regret I also find such complex motions confusing. However, here my 2cts: if you consider the acquired increase of kinetic energy when you stop increasing the radius, then I think that half of the force was used for the increase of kinetic energy of the mass and the other half for pulling the string outward. The mass is doing work on the string when it is moving outward, and much of that energy is provided by F. So, perhaps you did not sufficiently consider the radial motion and the work done on the string. With that you could lift a weight or activate a heater.
     
    Last edited: Sep 30, 2012
  7. Oct 3, 2012 #6
    Someone finally told me the missing piece in the puzzle: The tangential component of the force doesn't in general equal the change in the tangential component of velocity. (Because tangential and radial components are a non-inertial frame of reference, it doesn't work like using x and y coordinates) Think for example that in Uniform circular motion, the there is a radial acceleration, but the radial velocity isn't changing.

    So taking this into account, coriolis force finally reconciles with common sense: Imagine the case in which the tangential velocity isn't changing, in that case the tangential force is omega x v. This is the acceleration needed so that the velocity doesn't change (as A.T. says, it is the acceleration needed so that the acceleration is perpendicular to the force). Well then, in my original case, I need to add the extra omega x v to have the change in tangential velocity as well as the increasing radius.
     
  8. Oct 3, 2012 #7
    Yes, one can say the same thing in different ways. :smile:
     
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