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Gravitational torque on a ring mass

  1. Dec 8, 2014 #1
    What is a gravity tidal torque on a simple circular ring, inclined at some angle i?

    I can't find a solution for this simple problem, despite the ring's idea is frequently used
    in the precession problems, for example in the Earth's axis precession case.

    How this can be computed effectively?
  2. jcsd
  3. Dec 8, 2014 #2


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    Where's the center of mass of the ring? What is the gravitational force on an increment of mass dm on the ring? What is the orientation of the normal to the plane of the ring to the direction of gravitational force on the ring's center of mass?
  4. Dec 9, 2014 #3
    The ring is flat and circular, with a mass m, and a radius is r;
    the ring is inclined at angle i to the z axis (like a moon's orbit, or the equator of the Earth);
    and it can be rotated at angle along z axis: f.

    The second mass M is placed at a distance d to the ring, and we can assume: d >> r, say: d/r > 100
    And question is: what is a tidal torque due to the mass M action on the ring?
  5. Dec 9, 2014 #4
    I wrtie some equation for this ring geometry.

    a simple unit circle |r| = 1, in a plane x-y:
    [tex]r = (\cos t, \sin t, 0)[/tex]

    then I must incline it at i angle to the z:

    r' = r A; where A is a simple rotation matrix around y axis:
    [tex]x' = x\cos i - z\sin i; z' = x\sin i + z\cos i[/tex]
    [tex]r' = (\cos t\cos i, \sin t\cos i, \sin i)[/tex]

    now we must rotate this around z jet:
    r'' = r' B, where B is now rotation in x-y plane with angle f:
    [tex]x' = xcosf - ysinf; y' = xsinf + ycosf[/tex]
    thus the final ring is:

    [tex]r = [\cos i\cos t\cos f - \cos i\sin t\sin f, \cos i\cos t\sin f + \cos i\sin t\cos f, \sin i][/tex]
    Last edited: Dec 9, 2014
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