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Precession question

  1. Oct 2, 2009 #1
    Probably an easy one...

    So, I understand the idea of precession on a gyroscope: we've got the angular velocity vector, the angular momentum vector (same direction but times the inertia tensor), then a torque is applied => change in L in the direction of torque.

    I'm thinking of the example where you have the spinning bicycle wheel, with one end of the axle attached to a rope hanging from the ceiling (the axle is horizontal, the wheel vertical). The wheel, instead of falling down, stays upright and rotates around the rope. Let's say the L vector points away from the rope, and the torque vector (caused by the rope's offset from the wheel's center-of-gravity) points perpendicular to the rope and the L vector.

    So, as time passes, L moves along in the direction of torque. What I don't understand is, why does the center-of-gravity of the wheel move? It seems to me that since L is basically the axis of rotation, changing L should result in the axis becoming offset from the axle, with the effect that the wheel should start to have an increasingly wobbly spin - but this shouldn't necessarily cause the "body" (the wheel/axis) to move linearly.

    I'm guessing this has something to do with the end of the rope being somewhat fixed, but can somebody give me a more robust explanation?

    Thanks
     
  2. jcsd
  3. Oct 2, 2009 #2
    That's actually a good question. I think it has something to do with friction in the system. From what I understand the wheel/gyroscope wouldn't rotate in a frictionless environment.
     
  4. Oct 2, 2009 #3

    Cleonis

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    Gold Member

    Yeah, the evergreen of physics demonstrations. The lecturer has the bicycle wheel initially supported on both ends of the axle, allowing him to spin up the bicycle wheel, and then he removes support from one axle end.

    I will borrow some terms from aviation: pitch, roll and yaw. Here, 'yaw' is unambiguous; it's rotation around a vertical axis. (The direction of vertical is set up by gravity.) I will call the spinning of the wheel 'roll', leaving 'pitch' for turn perpendicular to the roll.

    What happens at the very instant that the lecturer removes support from one axle end?

    The very first thing that happens is that the spinning wheel pitches ever so slightly. The pitching gives rise to a torque at right angles to the pitching: the torque causes yawing motion. The energy to build up the yawing comes from the pitching.

    Now, the yawing itself gives rise to a torque just as much. So now there are two counteracting torques: the torque from gravity that tends to pitch the wheel down, and the torque from the yawing motion that tends to pitch the wheel up. Initally the yawing motion is not fast enough to completely counteract the torque from gravity. But very quickly the wheel has pitched enough, and then the torque from the yawing motion opposes torque from gravity precisely.

    In the demonstration of the spinning bicycle wheel the yawing motion is the factor that keeps the axle horizontal. (More precisely, the combination of fast spinning and slow yawing gives rise to the required up-pitching torque.)

    So what would happen if you would prevent any yaw? What if the wheel is free to roll and pitch, but barred from yawing? Then there is nothing to prevent the wheel from pitching down all the way.

    Cleonis
     
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