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Bicycle Wheel Platform Without using Angular Momentum?

  1. Jan 21, 2012 #1
    So, the classic scenario: Person on frictionless rotating platform, holding a spinning bicycle wheel with vertical axis of rotation. Flips it over, and the platform spins in the direction the wheel was originally spinning. Angular momentum is conserved.

    My question is: Is there any way to understand what is going on here without using the concept of angular momentum? Are there forces or torques that we can follow from the wheel to the person to the platform that would explain the motion of the platform? I know this would be much more cumbersome and unwieldy an explanation, but I'm curious how you would do so.
     
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  3. Jan 21, 2012 #2
    Suppose you have a wheel in front of you, rotating around a vertical axis. The part of the wheel closest to you moves to the left, and the part thats away from you moves to the right. (the wheel rotates clockwise, seen from above).
    You hold the axis with your left hand below the wheel and with your right hand on top,
    and you try to move your right hand to the right, and your left hand to the left, to make the wheel rotate so that the part closest to you moves upward, and the part the furthest away moves downwards.

    If you rotate the axis through a small angle [itex]\delta \phi [/itex] the part of the wheel closest to you, wich was moving to the left with speed [itex]\omega r [/itex] will now get an additional upward movement [itex] \omega r \delta \phi [/itex]. This means there must have been an upwards force on the part of the wheel closest to you, and likewise a downwards force on the part the furthest away.

    The only way to produce such a force is to push the axle with your right hand, and pull it with your left hand. When the wheel is still horizontal this will produce a torque on your body that will make it rotate backwards (wich it can't, so nothing will happen), but as the axle rotates you will have to keep on pushing with your right hand and pulling with your left.

    As soon as the axle is no longer vertical, and your right hand is more to the right than your left hand, pushing with your right hand and pulling with your left will move the right part of your body backwards and the left part forward, thus giving your body a clockwise rotation as seen from above, wich is the same direction the wheel was rotating.
     
  4. Jan 23, 2012 #3
    Okay, that makes sense. One thing I'm having trouble with: The points on the wheel that are furthest to the left and right of you. As they move through dθ, the direction and magnitude of their velocities don't seem to change, which would mean they don't undergo acceleration and there is no torque in that direction. Yet the two points have ended up in different positions than they would have if they'd been left alone. Doesn't that require acceleration, and therefore a torque?
     
  5. Jan 24, 2012 #4
    A torque will be needed to move even a non-rotating wheel, and it's still present for a rotating wheel.
     
  6. Jan 25, 2012 #5
    Okay. So in order to flip over a bicycle wheel, you need to apply torque in two directions? The torque you would need even if it wasn't spinning, and then another torque perpendicular to that?

    If this is true, then I would guess that the ratio between those two torques depends on the angular velocity?

    But when I do the problem using torque vectors, I only need the second torque to get the motion I need. And I'm still confused about those extreme points. Their velocity vectors MOVE as the wheel tilts, but don't seem to change direction or magnitude (except due to the centripetal force) What kind of acceleration does that?
     
  7. Jan 25, 2012 #6

    rcgldr

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    No, just apply a torque about 90 degrees out of phase with the desired direction of flipping.

    Imagine a wheel rotating about a vertical axis. You apply a torque along an axis that goes directly outwards. The wheel will respond by rotating on a sideways axis perpendicular to the outwards axis.

    Precession can be explained at the micoscopic level, but it's complicated. My attempt at a simplified explanation. Imagine a frame of reference where your initial view of the wheel has it's axis "vertical" (rotating counter clockwise as viewed from above, using right hand rule), and that you apply a counter clockwise torque, which mean's it's axis points towards you (right hand rule). For any point on the wheel, the results in an upwards acceleration on the right side of the wheel (the part moving away from you) and a downwards acceleration on the left side (the part moving towards you), and no vertical acceleration on the front or back side. This would mean that maximum vertical velocity occurs at the front or back where vertical acceleration is zero and transitioning from upward to downwards or vice versa. The result is the front (closest to you) part of the tire moving downwards and the back (farthest from you) part of the tire moving upwards. In order to continue this movement, the axis of the counter-clockwise torque would have to rotate in the same direction as the wheel precesses. After the wheel precesses 90 degrees, the torque axis that was pointing at you is now pointing downwards, at 180 degrees it's pointing away from you, at 270 degrees, upwards, and at 360 degrees, back towards you again.
     
    Last edited: Jan 25, 2012
  8. Jan 26, 2012 #7
    And there's where you've lost me.

    So when you barely start to tip it, the left and front points have close to zero vertical velocity. But the front and back had high horizontal velocity that is starting to change direction. So they'll have much greater vertical velocity than the left and right points. Is that what you're trying to say?

    If so, then when you slightly tilt the right half of the wheel up, you're directing the velocity of the point closest to you slightly upwards as well, since it is moving to the right and must follow the curve of the wheel. And the point furthest from you will now be moving slightly downward. But in reality, the point closest to you moves downward, and the furthest point goes up.

    So I'm missing something, but I don't see what.
     
  9. Jan 26, 2012 #8

    rcgldr

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    The left and right locations on the wheel have near zero vertical velocity, while the front and back have maximum vertical velocity, so the wheel precesses about an axis perpendicular to the axis of the applied torque.

    Upwards or downwards depends on the direction the wheel is spinning and the direction of the torque. In the second post the wheel is spinning clockwise as viewed from above, so an angular velocity vector is downwards. My last post had it spinning counter clockwise. The torque in the second post was the same as my last post.

    I'll switch to the rotation and torque as stated in the second post to continue with this.

    So switching to the rotation and torque as stated in the second post, the right side of the wheel (the part spinning towards you) experiences upwards acceleration, and the left side (the part spinning away from you) downwards acceleration. Peak magnitudes of velocity occurs at the front (upwards in this case) and back (downwards in this case) during the transitions (zero acceleration) between upwards and downwards acceleration.

    The situation is similar to a sine wave created by the relationship acceleration = - position. Velocity will be 90° out of phase with position, and acceleration will be another 90° out of phase with velocity, with acceleration 180° out of phase with position.
     
  10. Jan 26, 2012 #9
    Sorry, this was a typo, I meant left and right of course.

    I agree that the peak magnitudes occur at the front and back, and that these points are the transition between upwards and downwards acceleration.

    But I still have the same problem. If the wheel is rotating clockwise as seen from above, so that the left side is spinning away from you and the right side is spinning toward you, then that means the front (closest to you) is moving to your left and that the back (furthest from you) is moving to your right.

    So when you tilt the right side upward, the point at the back, where acceleration is zero, now has upward velocity, since it is moving to the right and we have tilted the right side up. And the point in the front has downward velocity, since it is moving to the left and we have tilted the left side down.

    Which is backwards from what you said in this latest post. What am I not getting?
     
  11. Jan 26, 2012 #10

    rcgldr

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    The issue is that you're not tilting the right side upwards, you're applying a torque that produces an upwards acceleration on the right side, and this results in a reaction that is about 90° out of phase so that the front of the tire moves upwards when you apply a counter-clockwise torque to the wheel.

    If you want the right side to move upwards, you need to apply torque that is outwards on the bottom and inwards at the top, this will result in a counter-clockwise precession.

    Other than gyroscopes, the "cyclic" control for the main rotor on a full scale or model helicopter is set 90° off phase (90° ahead of the desired reaction). So a torque along the pitch axis is used to produce a roll response and vice versa. Yaw isn't an issue since that is rotation on the same axis a the main rotor.
     
  12. Jan 27, 2012 #11
    I understand all that, what I'm trying to figure out is what causes the out of phase reaction. How does applying upward acceleration to the right side result in the front accelerating upward instead? Why does the front, which is moving away from the region where an upwards acceleration is being applied and toward a region where downwards acceleration is being applied, have upward velocity, as you said? And if we're applying an upwards acceleration to the right side, why doesn't it move upwards? I'm trying to understand the mechanism that results in precession when a torque is applied by looking at how different points on the wheel accelerate.
     
  13. Jan 27, 2012 #12

    rcgldr

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    I'm not sure about what happens during transitions, where the applied torque transitions from zero to some fixed value. What I was describing was more of a steady state situation where the applied torque and rate of precession are constant.

    The upwards velocity continues to increase as long as there is any upwards acceleration, even though that upwards acceleration is decreasing. Peak velocity occurs when the upwards acceleration is zero, just before it transitions into downwards acceleration.

    Still this doesn't directly expain why the torque and the reaction to the torque on a wheel is 90° (or nearly so), out of phase. That because the wheel is spinning, the upward momentum of a point on the wheel on the right gets "carried" to the front of the wheel as that point moves to the front of the wheel. The velocity of a point at the right side of a wheel is mostly towards you and a bit upwards.

    As I originally stated, the actual situation is more complicated than my explanation, but I was trying to provide some idea of what is happening without using angular momentum.

    If you applied a very large amount of torque, then you could get the right side of the wheel to move upwards as well as precess. In the case of a gyroscope with a horizontal axis supported at one end by some pivot point, the torque is relatively large and the initial reaction is downward rotation, but then it starts to precess, with the rate of precession increasing and downwards rotation decreasing, transitioning into upwards rotation and the rate of precession decreasing as upwards rotation increases. If this process isn't self dampening, then the movng end of the axis of the gyroscope would continue to cycle up and down. To prevent this, you would need to initiate a rate of precession that corresponded to the downwards torque to get a steady state rotation.

    Considering the gyroscope situation, my guess is that there is some initial upwards movment of the tire when torque is first applied, but it quickly transitions into precession and counters the upwards movement, and the amount of oscillation in the rate of precession and reaction along the torque axis is too small to be noticable in the case of a bicycle wheel.
     
  14. Jan 27, 2012 #13
    EDIT: Sorry, I was typing while you posted. You actually moved on to my next question before I even asked it! I think what I was getting at before might have been the oscillation in the rate of precession you just mentioned. Now that I can see where the bulk of the precession actually comes from, that seems to fit with what I was looking at earlier, since the "backwards" prescession I was looking at would actually act to dampen the initial upward acceleration. And these oscillations would be very small compared to the actual precession. Anyway, here's my actual post, although you can probably ignore a lot of it now /EDIT

    *****

    Okay, never mind, I think it's starting to click. When you provide the acceleration at the rightmost point, you're going to change the velocity to a slightly upward velocity. If it weren't moving, it would just develop a upward velocity, but since it has a high initial velocity, it's inertia carries it toward you as well. So as it moves, it tends to drag the wheel with it, pulling the front of the wheel upward. When it reaches the front of the wheel, it no longer has any upward acceleration, and at this moment continues along the path of the wheel. But the left side of the wheel has downward acceleration, so as it moves into the left side this all repeats again, but in opposite directions.

    Does that sound right?

    I think what I was doing before, without realizing it, was turning the wheel the way I thought it should turn, and then looking at the acceleration that would have to be applied in order to achieve that motion, instead of applying acceleration and looking at what it would do. Seems kind of dumb now.

    I guess my next question would be this: When the wheel is rotating slowly, it is still easy to turn it in the direction in which the torque is applied. So is it just a question of which effect dominates, or is there a speed at which the wheel moves entirely perpendicular to the torque? Because if I do math with torque vectors, it doesn't seem to matter what angular momentum I choose, an applied torque always moves the rotational axis perpendicularly.

    Maybe the rotational axis does precess at slow angular speeds, but is not very noticeable, and the person flipping it can easily compensate for it by applying the necessary forces without even realizing that they are doing so?
     
    Last edited: Jan 27, 2012
  15. Jan 27, 2012 #14

    rcgldr

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    Yes, based on my simplified explanation.

    There just needs to be sufficient speed so that precession can counter the effects of torque. The slower the rate of rotation, the greater the rate of precession needed to counter torque and vice versa. Wiki article states that in steady state precession, rotation is at right angle to torque (which would mean entirely perpendicular), with the end of the primary axis sweeping through a cone. The combination of rotation, precession, and torque could result in this cone being a flat plane (zero offset in direction of torque). It seems that it should be possible to achieve steady state by adjusting the torque input during start up with an already rotating wheel.

    http://en.wikipedia.org/wiki/Precession
     
    Last edited: Jan 27, 2012
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