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Pseudo Force Direction Confusion.

  1. Dec 19, 2013 #1
    Q1) What angular velocity would be required for your effective weight at the equator to be third of that the north pole? You may neglect all inertial forces except the centrifugal.

    I understnad the majority of my books solution which is that:
    At the equator, the centrifugal force is -mRw^2k, gravitational is mgk, where k is the unit vector pointing perpendicalry out of the Earth's surface.

    At the north pole there is no centrifugal force, so we want to solve:

    I am struggling to see how we have reached this equation, sign-wise.

    The main equation of this topiic is ma[rotating frame]=ma[inertial frame]-inertial forces *
    where here inertial forces is only the centrifugal .
    So here, as other inertial forces have been neglected, I think we take the north pole as the interial frame and the equator as the rotating frame?

    So from the question we want to ar , where ar is the accelertatin in the rotating frame, to be -g/3k on the LHS of * . And on the RHS we have aI= -gk, and Fcentrifugal
    = -Rw^2k, which gives **.

    But, (assuming the above procedure is correct), I don't understand the sign choice of the centrifugal force. I thought it always points radially outward, so it would be +Rw^2k?

    Q2) The direction of the centrifugal force points directly away from the axis of rotation... I can clearly see that it must be perpendicular to both w and wxr, (by F=-mwx(wxr)but I am struggling to see how this implies it must point away from the axis.

    Many Thanks to anyone who can shed some light on any of this. Greatly appreciated :) !!
    Last edited: Dec 19, 2013
  2. jcsd
  3. Dec 19, 2013 #2
    If you take the positive direction along the radial direction to be outward, then the centrifugal term is positive and the g terms are negative.
    But you can choose the other way. The result will be the same.
  4. Dec 19, 2013 #3
    But then from * doesn't this yield: -g/3=-g-w^2R.
    st all terms have the same sign?
  5. Dec 19, 2013 #4
    Which way do you take as positive?
    And why 1/3? The problem does not say this.
  6. Dec 19, 2013 #5
    Sorry the question should read a third instead of a half,have edited it now.
    'k is the unit vector pointing perpendicularly outward...' so radially outward is positive.
  7. Dec 19, 2013 #6
    OK, then what is the sign of the centrifugal term with this choice of positive direction?
  8. Dec 19, 2013 #7
    positive. as its always directed outward?
  9. Dec 19, 2013 #8
    Yes. It's positive.
    And the "g terms" will be negative.
  10. Dec 19, 2013 #9
    which yields -g/3=-g-w^2R. when I use * with the justification of the frames as explained in my first post.
  11. Dec 19, 2013 #10
    You just said that the centrifugal term is positive. Why do you write it negative??????

    Oh, I see, you are referring to that equation in bold.
    It that a vector equation in original? And what is the context?
    Is the transformation of accelerations when you change the frame of reference?
  12. Dec 19, 2013 #11
    Because equation * has any inertial forces subtracted.
  13. Dec 19, 2013 #12
    I don't see how that equation applies here. Do you understand the significance of the terms in the equation?
    What is a[rotating frame]?
    What is a[inertial frame]?

    In your case, neither g nor g/3 represent accelerations. The acceleration of body on the equator is
    v2/R in an inertial frame, fixed, attached to the stars
    and zero in the rotating frame (the body is not moving relative to the rotating Earth.

    I would recommend to write Newton's law in either system (fixed or rotating) and use it to solve the problem. Do you understand what do they mean by "effective weight"? This is the key to this problem.
  14. Dec 20, 2013 #13
    The equation discussed in this book's chapter and it's derivation are pretty similar to the stuff on wikipedia here: http://en.wikipedia.org/wiki/Centrifugal_force_(rotating_reference_frame [Broken]) under the headign 'Force'. Sorry the equation is the exact same.

    Sorry I have divided the whole equation by m, the mass, as we are comparing the same body at the equator and north pole.

    The problems in this chapter are based on solving around this equation.

    The book defines a 'local' coordinate system for in the region of a point on the surface of
    the spherical earth as taking the positive x direction as east, the positive y direction as north and the positive z
    direction is out from the centre of the earth.

    We tend to make no reference to fixed stars, but use a 'local' coordinate system. We tend to take the Earth as the rotating frame.

    In this case, as justified in my first post, I deduced that the rotational and inertial frames could be the equator and north pole respectively . I mentioned that this stage could be flawed, perhaps causing the sign confusion...but if not, the next obvious error would be the sign of the centrifugal force.
    Last edited by a moderator: May 6, 2017
  15. Dec 20, 2013 #14
    I did not question the validity of the equation.
    Just the relevance or usefulness for this problem.

    You did not answer the key question.
    Do you understand what is the "effective weight" means here?
  16. Dec 20, 2013 #15
    My course requires me to solve this problem via this equation. which is why I included it and based my method on it.

    I believe effective weight is used to describe the phenonoma when the weight as measured in a rotating frame is less than that measured in an inertial frame. And this can be explained by pseudo forces, which are not real but occur due to the non-inertial nature of the frame of reference.
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