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If I roll a hoop forward, what keeps it up?

  1. Apr 30, 2010 #1
    I've seen the gyroscopic effect explained in terms of angular momentum, but I feel that there should be a much more tangible and easily understood solution - one that would make sense to a person without any understanding of algebra.

    Specifically, I'm looking for an explanation that looks at the hoop as just a bunch of moving particles, and explains why each particle doesn't fall down (as quickly?) when the hoop is moving. After all, angular momentum is just a math trick for combining multiple linear momenta. (Failing that, maybe you could explain to me exactly why you can't consider this on a particle-by-particle basis.)

    I've got a Physics BSc, and I've taught high school physics for two years now; I've posed this problem to everyone from fellow physics students to engineers to my university profs and never received a satisfactory answer. (I.e., one that explains it conceptually without invoking angular momentum, cross products, etc.)
     
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  3. Apr 30, 2010 #2

    rock.freak667

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    Well depending how the hoop is rolled, I would just assume its center of mass is not moving for it to be in unstable equilibrium. That's what I's say without knowing the gyroscopic effect.
     
  4. May 1, 2010 #3

    Cleonis

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    I agree that it should be possible to explain without invoking angular momentum. The explanation comes from the physics of gyroscopic precession.

    The thing is, the case of a spinning gyroscope is simpler, because the gyroscope wheel is gimbal mounted. The suspension point coincides with the center of mass, and that symmetry is quite helpful. I wil discuss only the case of a spinning gyroscope, followed by outline of how to translate to the case of the rolling hoop.

    attachment.php?attachmentid=24771&d=1270068885.png

    The gyroscope wheel is spinning along a horizontal axis.
    Defining the following axes:
    Roll - spinning of the gyroscope wheel
    Pitching - turn along a horizontal axis, perpendicular to the roll
    Swiveling - turn along a vertical axis.

    First start the gyroscope wheel spinning fast. Then add some swivel.
    Think of the gyroscope wheel as moving through four quadrants. In two of the quadrants the wheel's mass is moving towards the swivel axis, in the other two quadrants the wheel's mass is moving away from the swivel axis.
    In the quadrants where the wheel's mass moves towards the swivel axis a force is required to reduce the velocity of rotating around the swivel axis. If that force isn't present the mass in that quadrant will tend to move in the direction of the green arrows in the image.
    In the quadrants where the wheel's mass moves away from the swivel axis a force is required to increase the velocity of rotating around the swivel axis. If that force isn't present the mass in that quadrant will tend to move in the direction of the green arrows in the image.
    The inertial effects in the four quadrants combined give rise to a tendency to pitch.

    A more extensive account is in the article about http://www.cleonis.nl/physics/phys256/gyroscope_physics.php" [Broken] that is on my website.

    Now to translate it to the rolling hoop case.
    What I think is happening is the following:
    As the rolling of the hoop starts, a very slight precessing motion develops. This swiveling is so slow that it's not visible (or it drowns in other effects). Once the hoop has settled on a particular precession rate the tilt of the hoop won't increase further.

    Cleonis
    http://www.cleonis.nl
     
    Last edited by a moderator: May 4, 2017
  5. May 2, 2010 #4

    tiny-tim

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    Welcome to PF!

    Hi smallflight! Welcome to PF! :smile:

    If I roll a hoop forward, what keeps it up?

    I think it doesn't keep up. :redface:

    Like a coin, if a hoop is at an angle to the vertical when it starts rolling, that angle will gradually increase, and the hoop will move in a circle as it gradually falls over.

    (of course, if the cross-section of the hoop is stable, so that the stationary hoop is self-righting, then the moving hoop will be self-righting also)

    A bicycle is the same … if you only turn the steering wheel to the left, the bicycle will fall over … you have to "countersteer" slightly to the right first, to make it fall to the right, and then steer to the left to balance it again as you turn left (most people are never taught this, which is why learning to ride is so difficult :rolleyes:).

    If you roll a hoop, you need a stick, not only to keep it moving, but also to keep it upright. :smile:
     
  6. May 2, 2010 #5

    Cleonis

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    Re: rolling a hoop

    If I roll a hoop forward, what keeps it up?
    What you say is at odds with what I remember of seeing coins roll a distance. Then again, coins have a flat outer edge; on a sufficiently smooth surface a coin can remain upright and be staticly stable.

    So I looked around in my house for some objects that are circular and not staticly stable: a thin ring (in english that kind of part is called a 'washer') and a key ring, both several centimeters across. Both of them flop over instantly when they are not rolling; no static stability.

    I push them off, and sure enough, they easily make it to the other side of the room. Something is stabilizing them; they remain upright far longer then when not rolling.

    When I deliberately release them with a bit of a tilt angle, then they don't fall over either, they keep going with a steady tilt and along a path with a steady curve. If the tilt is to the right the path curves to the right.

    But now I wonder: the rolling ring spins very slowly, the gyroscopic effect will be quite small. Quite possibly the gyroscopic effect is negligable. Moving along a curved path a centripetal force is required, and with the inward tilt gravity provides that inward force. So now I think the case of the visibly curved path is more complicated than I thought. So thank you tiny-tim, for making me reconsider.

    Let me return now to the case where the rolling ring makes it to the other side of the room, not visibly deviating from moving in a straight line. I think in that case gyroscopic effects keep the ring upright.
    I think that the very instant that any tilt develops the ring will turn in the direction of tilt, moving the point-of-contact-with-the-floor right under the center of mass again. I think there's a true self-righting effect going on there.
     
  7. May 2, 2010 #6

    tiny-tim

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    Hi Cleonis! :smile:
    But then why doesn't a bicycle self-right? :confused:
     
  8. May 2, 2010 #7

    Cleonis

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    To my knowledge: in the physics of bicycle steering gyroscopic effects are negligable.

    A few researchers have actually put that to the test. The way to test that is to build a bicycle with two counterrotating front wheels. One wheel for the actual cycling, it rests on the ground, the other one just there to cancel the angular momentum of the main wheel.

    The men who build such a bicycle (I think it was done twice, independent of each other) report that the neutral-angular-momentum bicycle had the same ride characteristics as a standard bicycle, it felt the same to them. That is good evidence that gyroscopic effects are negligable. The theoretical models indicate the same thing.

    Bicycles are in fact self-righting, but that arises from the steering geometry.
     
  9. May 2, 2010 #8
    If we imagine a cart on wheels (like a well oiled grocery cart) and we give it a good shove, it will keep moving along in the direction we shoved it. That's the conservation of linear momentum. It follows directly from Newton's laws, meaning that it's a primitive concept. You can't do much more to explain it than say that Newton's laws are just the way things behave.

    The case is the same for angular momentum. I think you are incorrect to claim that angular momentum "is just a math trick for combining multiple linear momenta". That's not true because angular momentum is fundamentally different from linear momentum. Strictly speaking, angular momentum is called the "moment of momentum". It may appear naively similar because it looks like you just take a cross product, but what you get is a fundamentally different quantity. Newton's laws show that angular momentum is conserved, just like linear momentum, but these two laws -- the conservation of linear momentum, and the conservation of angular momentum -- are distinct.

    This is why I believe that the traditional (conservation of angular momentum) explanation is the best you can do in this situation. Angular momentum is something that occurs even on the atomic scales -- it is a fundamental, primitive quantity, which is derived from basic experimental knowledge.
     
  10. May 2, 2010 #9
    Have you never seen conservation of angular momentum also derived from Newtons laws? (Please inform yourself.)
     
  11. May 2, 2010 #10
    Conservation of angular momentum is derived by taking the moment of the force equations about some coordinate axis. However, conservation of angular momentum and conservation of linear momentum are distinct.
     
  12. May 3, 2010 #11

    rcgldr

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    The primary reason the hoop stays up is friction. Roll a hoop on ice and it will fall over much sooner. There's a yaw to lean response due to gyroscopic action, and the change in yaw results in a lateral friction related force at the contact patch which helps keep it upright while rolling. It only works for range of speed, too slow and it falls over, too fast and it may speed wobble.

    On the side topic of a bicycle, you could replace the wheels and tires with curved skating blades, and glide it on ice. It would remain stable without gyroscopic effects. The key is in the steering geometry, specifically the trail, where the contact patch is aft of where a line from the steering pivot axis would intersect with the ground. Again there's a speed range sensitivity, depending on the amount of trail, the greater the trail, the slower the minimum self stability speed.
     
  13. May 3, 2010 #12

    Cleonis

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    Then again, we have the distinction between 'being different' and 'being independent'.

    Angular and linear momentum are different, but not independent.
    If some physics taking place can be accounted for in terms of angular momentum it can also be accounted for in terms of linear momentum. This is always valid; it's always possible to use linear momentum only.


    Angular momentum is a more abstract concept, and in particular circumstances the concept of angular momentum is a more efficient tool. But angular momentum is a less intuitive concept than linear momentum. In formula's angular momentum is put to use in expressions with a vector cross product. It is very hard (if not impossible) to develop a visceral sense of the physical meaning of that vector cross product.

    In the case of gyroscopic precession thinking in terms of the instantaneous linear momentum of individual parts is more readily visualized.

    Anyway, talking about how angular momentum and linear momentum relate to each other is a move away from the original subject of this thread, which is hoops.
    I have started a new thread.
    Mordechai9, I hope you'll be willing to discuss this subject with me.
     
  14. May 3, 2010 #13
    It is much simpler to stay upright on a bike with rotating wheels. Lets say the bike starts to fall to the right perpendicular to the motion of the particles in the wheel. As a particle is moving along the tangent (+z direction) at the top of the wheel it gains some momentum in the +x direction, when the same particle is at the bottom of the wheel it gains some momentum in the -x direction though the original momentum fights this movement.

    The faster the wheel is turning the sooner the particles are transferring this energy (momentum) and the wheel stays upright.
     
  15. May 3, 2010 #14
    No, angular momentum and linear momentum are independent. There are simple mathematical arguments to show this. Anyways, without needing to get into that, even the simplest rotational problems involve the Newton's law of the form [tex] I \alpha = M [/tex], which is obviously an equation of angular momentum. You absolutely cannot use "only linear momentum" to solve these problems. Or in the sense that you can, I am convinced it would be much more of a painful calculation, without gaining anything extra from the solution.

    Personally I feel that angular momentum is perfectly intuitive. You can think of the static cases for example, like with a wrench. When the handle of the wrench is long, you can apply more force with it, and this makes sense from all practical experience. The same principle is behind the placement of a door knob on the far side rather than the near side to the hinge.

    This is the same principal with seesaws and other such objects that we all know and understand. Angular momentum is simply the extension of this same idea to dynamic motion. There are many simple cases which are very intuitive and easy to understand, such as with the figure skater who pulls their arms in to spin faster. Mathematically, the theory is classical, and extremely elegant. The vector cross product just pulls out the component perpendicular to the moment arm -- it's expression is extremely simple, as just rFsin(theta).

    I agree that linear momentum is a simpler concept, but angular momentum is not exactly rocket science.

    Gyroscopic precession arises directly due to conservation of angular momentum. So I just can't agree that it is easier to understand in terms of instantaneous linear momentum, which is an indirect interpretation.
     
  16. May 3, 2010 #15

    rcgldr

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    The gyroscopic reactions mostly dampen any reaction along the steering axis or along the bikes roll axis. As I mentioned before, the wheels of a bicycle could be replaced with rounded skate blades while gliding on ice, no rotation, but still stable. There are uni track ski vehicles made from modified dirt bikes, but due to the lack of grip in the snow, they aren't quite as stable.

     
    Last edited by a moderator: Sep 25, 2014
  17. May 4, 2010 #16
    Unlike the demonstration of a spinning bicycle wheel suspended (from one side of the axle) by a string?
     
  18. May 4, 2010 #17

    rcgldr

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    Similar, a bicycle wheel will be stable while rolling on the ground at a much lower rate of rotation than the rate of rotation required for a spinning wheel suspended by one side of its axis to precess at a relatively slow speed while it's axis remains horizontal.
     
  19. May 4, 2010 #18

    Cleonis

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    It's not clear what you mean by 'independent' here.

    In classical mechanics conservation of angular momentum follows from the laws of motion. That is, if you take the laws of motion as axioms, then conservation of angular momentum is a theorem. Putting it differently, angular momentum is reducible to elements.

    It seems to me that a phenomenon can be counted as independent if and only if its description has the status of axiom.

    My best guess is that you actually mean to say that angular momentum and linear momentum are different, which of course they are.


    Perhaps you have quantum physical spin in mind. Quantum physical spin has properties that have no counterpart in classical mechanics, and I suppose quantum physical spin must be regarded as an independent phenomenon, not otherwise reducible.

    Anyway, this thread is about hoops.
    My request is that you reply in the https://www.physicsforums.com/showthread.php?t=400562"
     
    Last edited by a moderator: Apr 25, 2017
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