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Slow spinning gyroscope

  1. Jul 1, 2014 #1
    how does a slow spinning gyroscope behave?
    does it behave as a normal fast spinning gyroscope would?
    if the mass of the gyro is large so that the angular momentum equals a fast spinning gyro does its precession equal a fast moving one?
    can anybody explain?please.
  2. jcsd
  3. Jul 1, 2014 #2


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    Depends on what you are doing.

    If the force is the same: yes. This can be seen from the equation for the precession speed.
  4. Jul 1, 2014 #3


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    This holds so long as the precession rate is small compared to the rotation rate. If the moment of inertia of the rotating object is made larger and larger while angular momentum and applied torque are held fixed, the precession rate remains constant, but the rotation rate decreases.

    The equation for precession speed is an approximation. If rotation rate is low compared to the computed precession rate, it fails to hold, even approximately.
    Last edited: Jul 1, 2014
  5. Jul 2, 2014 #4
    so does it fall down if suspended by one end of the axis under the action of gravity or does it precess to one side?
  6. Jul 2, 2014 #5


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    It will simply fall down.

    On reflection overnight, I realized that what I wrote in post 3 is not correct. MFB is right The equation for precession -- as a rate of change in angular momentum under a torque at right angles to that angular momentum is exact. It is not an approximation.

    A problem is the description of the movement of the gyroscope as a rotation about a fixed axis of symmetry where that axis is itself moving according to a precession. If the gyroscope is precessing rapidly enough to matter then its angular momentum vector will not align perfectly with the axis of symmetry. It will have a component corresponding to the "precession". In other words, when it precesses, the angular momentum of your gyroscope only approximately lines up with the axis that goes through its bearings.

    A torque that you apply perpendicular to the axle of a precessing gyroscope will not be at right angles to its angular momentum.
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