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Interpretation of Gyroscopic effect and rotation

  1. Feb 20, 2012 #1
    hello forum,

    we all know that a rotation wheel, a rotating football, etc.. maintain their direction of motion with more stability than if they were not spinning....
    A bicycle wheel is suspended from one of end of its axie by a rope, and spun up by hand. The wheel's axle is then placed horizontally and the free end of the axle processes about the supported end.

    here the MIT video:

    Why does that happen from a simple conceptual point of view? Why do the spinning parts of a spinning object give the object itself this stability?
    I understand the formulas and the angular momentum explanation but I am wondering how we can explain this interesting phenomenon, this dynamical equilibrium, from the point of view of the rotating parts....

    Last edited by a moderator: Sep 25, 2014
  2. jcsd
  3. Feb 21, 2012 #2


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    Why doesn't the Moon just fall down straight to Earth? Because it has a velocity perpendicular to the Earth's pull, which is merely modified by the pull.

    You can visualize the same for different parts of the wheel:

    Why doesn't the top part of the wheel just flip over to the side? Because it has a large velocity perpendicular to the flip direction, which is merely modified by the flipping.

    Draw the velocity vectors for different parts of the wheel, and add to them the small modifications caused by the tendency to flip over.
  4. Feb 21, 2012 #3
    The behavior of gyroscopes is fairly easily explained in plain english. I'll use a screenshot from the video you posted for a quick illustration.

    As you put force on the axel of the wheel, it will force the wheel in the directions indicated by the red arrows. Everything above the black dotted line will be forced in the direction of the top arrow, and everything below will be forced in the direction of the bottom arrow. As these red points rotate along the wheel to the green points, they're still being forced in the same direction, and by the time they cross the dotted line, they've got a small velocity in that direction. When these points keep rotating around beyond the green points, past the dotted line, they'll begin to be forced in the opposite direction, but this force is spent just counter-acting the small velocity in the opposite direction of the force, which came from being on the other side of the wheel. When the wheel is spinning very fast, it doesn't allow enough time on a single side of the wheel in order for a given point on the rim to build up significant velocity before it crosses over and gets cancelled out.
  5. Feb 21, 2012 #4
    Thanks, good explanation....

    So, when the wheel is spinning and vertical it has also to rotate around the axis of the wire (precession), correct?
    It cannot stay with its axle pointing in the same direction....or can it?

    When a football is thrown it follow a parabolic trajectory. But the nose of the ball, even if the ball is spinning, changes direction...why?
  6. Feb 21, 2012 #5
    me again:

    so you are saying that the top red point is being pulled to the left side but in the meantime it is rotating to the green point where it starts being pulled in the opposite direction....
    If the rotation speed is high enough, while the red point is still subject to the left force, it should move "a little bit to the left" and as it crosses the black line it should, in virtue of the force on the right to move to the right......but it does not...
  7. Feb 21, 2012 #6


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    When it isn't thrown perfectly, you are applying a torque that rotates the ball around its axis, but you are ALSO applying a force on the end of the ball in one direction. This force is analagous to the force that gravity has when a spinning top isn't spun with it's vertical axis exactly vertical. This article explains it. Read it closely and carefully and it should make sense. I had to re-read a few times to get it all. http://en.wikipedia.org/wiki/Precession
  8. Feb 21, 2012 #7

    D H

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    This is not right.

    Angular momentum is not conserved in those long football passes in which the football appears to always be pointed along the flight path (the technical term is zero angle of attack). A football thrown on the Moon will not behave in this manner. It would instead have a constant angular momentum, and the orientation of the ball will remain more or less constant.

    The orientation of this lunar football pass will remain exactly constant if the the football's angular velocity vector is exactly aligned with the football's axis of symmetry (a perfect spiral pass). If the angular velocity is a bit misaligned from the axis of symmetric the football will instead wobble a bit. It will not however turn end over end. This wobbling motion is called torque free precession.

    The perfect football pass thrown on the Earth instead exhibits gyroscopic precession. When the football's angle of attack is not zero, aerodynamic drag creates a torque on the football that under the right conditions act to bring the football back to (near) zero angle of attack. It turns out that that perfect pass is not quite a perfect spiral pass. A perfect spiral makes the center of pressure too close to the axis of symmetry. A tiny bit of wobble makes the restoring torque a bit bigger, thereby helping to keep the football on track.

    That wiki article is pretty lousy. A good sophomore/junior level physics text will inevitably have a chapter that covers both torque-free and gyroscopic precession. That wikipedia article only covers gyroscopic precession, and it does a lousy job at it (you had "to re-read it a few times to get it all.")
  9. Feb 23, 2012 #8
    Hello D H,

    so, in the Gyroscope, how do you explain that while a point is moving and pulled by a force on the left it is not moving in that direction? After 180 degrees the force on that same point changes and becomes to the right....

    Is the switching between the two forces so fast that there is no effect on the rotating point(s) at all?

  10. Feb 23, 2012 #9

    D H

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    It is torque, not force, that makes the gyro precess. Forces of course beget torques via [itex]\vec T = \vec r \times \vec F[/itex]. The torque is thus perpendicular to the force, so the precession is normal to the applied force.
  11. Feb 23, 2012 #10
    I see the formula and that the torque is normal to the force...
    but I wonder how each rotating portion of the wheel is not pulled, from a physical point of view more than mathematical, in the direction of the force....

    Some explain it with inertia and the tendency of the parts to continue to move in their rotating motion...but that does not explain the absence of displacement along the direction of the force....
  12. Feb 23, 2012 #11


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    Trying to undertstand what happens to "each rotating portion" of the wheel is hard, because you need to take account of the forces acting on that part from the rest of the wheel (i.e. the stress distribution inside the wheel) as well as the applied forces. In other words, you can't think about one part on its own, you have to think about everything together.

    Any explanation that talks about "the tendency of the parts to continue to move in their rotating motion" is probably nonsense. That idea went out of fashion in physics when people stopped believing that everything Aristotle wrote was necessarily true, simply because Aristotle had written it.
  13. Feb 23, 2012 #12


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    I thought that was what I explained. If the throw is not done correctly you force one end of the ball off to one side instead of keeping it on axis. This would cause the precession correct?
  14. Feb 24, 2012 #13

    I agree with you: each portion is constrained.....But do you personally have a conceptual explanation of why the wheel does not fall under the torque (only precesses)?

    Howstuffworks makes an attempt to explain in it but I am not clear on it:


    I have been asked to explain this concept, conceptually, and not just with the formulas for angular momentum....

    There must be some more intuitive explanation.... Another explanation, maybe a better one , is given at

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