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Angular Momentum - Precession

  1. Apr 6, 2005 #1
    Precession for me is one of the most confusing and interesting concepts in physics so far, I've thought about it alot and I still don't have it all figured out. Here's an interesting question:

    Consider a gyroscope that has the axle in a horizontal position and spinning so that the angular momentum vector is pointing horizontal. The angular momentum is large and the support is taken away so the gyroscope starts to pivot in a horizontal circle - this is a very familiar form of precession.

    Ok, the precessional angular velocity is constant, as well as the angular velocity of the rotor, correct? Therefore, neglecting friction, the total angular momentum of the system is constant.

    Now, torque changes angular momentum - If you have a spinning object and you apply torque so that the torque vector is in the same direction as the angular momentum vector, it will spin faster. If the torque is in the other direction, it will spin slower. Also, if you apply a torque continuously, the angular velocity will keep increasing.

    The torque on the gyroscope that causes the precession is the torque due to gravity. Gravity is always pulling down with a constant force, so the torque is constant and continuous. Therefore, with a continuous torque, the total angular momentum of the system should be changing. However, this can be experimentally disproved. How is this possible?
     
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  3. Apr 6, 2005 #2

    Meir Achuz

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    Angular momentum is a vector. The vector can change by changing its direction without a change in magnitude. As the gyro moves in a horizontal circle, the gravity torque (given by rXF) equals the rate of change of the vector. The algebra is somewhat similar to motion ar constant speed in a circle.
     
  4. Apr 6, 2005 #3
    Oh ok, now I see how it works out in terms of the vectors, the angular momentum vector is horizontal, lets say to the right at a certain point. At this point the torque vector is still in the horizontal plane, but perpendicular to the L vector, which causes the direction to change - the vector spins around in a circle.

    However, try to comprehend it just in "plain physics". If it wasnt spinning it would just fall down - the torque from gravity is pushing it down trying to get it to rotate towards the ground. If it is spinning, the torque is still trying to push it down, but since it has a high angular momentum it is much harder to change its motion. Therefore, it logically seems, to me anyway, that instead of precessing, the gyroscope should fall down in the same way it does when it isnt spinning, except it would take much, much longer. For example I did some calculations using Newtons second law in angular form to figure out how fast the torque from gravity should change the angular momentum. I plugged in some arbitrary values for mass and axle length etc. When not spinning, I figured it would take about 0.24 seconds to fall 90 degrees till it was pointing toward the ground. If it was spinning at 30000 rpm, I figured it would take 16.3 hours to fall the same amount, neglecting friction of course. This seems perfectly logical to me. I don't understand what makes it spin horizontally rather than vertically.

    Edit: I'm not trying to say that the physics of angular momentum/precession is wrong. I have seen many gyroscopes and tops and seen precession in many different ways. I'm not disputing that it occurs at all, I'm just being skeptical and thinking of what it seems like it should do logically. That's how you really get down to a good solid explanation.
     
    Last edited: Apr 6, 2005
  5. Apr 7, 2005 #4
    Actually, if you let a spinning gyroscope go from rest, it does start falling down initially. But this downward rotation immediately generates internal forces that act at right angles to this motion, so it starts moving sideways. The right angle forces keep acting until it the gyroscope starts rotating up again, reaching its initial height but displaced horizontally. And the process starts over again. This bumpy motion is called nutation. It sometimes has such a small amplitude that it's hard to notice. When you see a gyroscope precessing smoothly, that just means it has been going long enough for friction to dampen the bumpy motion. The vector algebra that Meir Achuz mentioned allows you to understand what's going on in this steady state case, in which the gyroscope has been precessing forever and there is no nutation.
     
  6. Apr 7, 2005 #5
    Ok, that makes sense, I've seen that before. If you push down on a spinning gyroscope, it does go down but then it starts going sideways. So this vector notation doesn't truely explain it because it doesnt account for nutation.

    So when you push on the gyroscope and it initially moves downward, what are these internal forces that create a perpendicular force?
     
  7. Apr 7, 2005 #6
    Right. The forces that cause this perpendicular motion are reaction forces on the "body" of the gyroscope resulting from the forces it must exert on the more massive "wheel" part of the gyroscope to make it go in a curved path. The reason I'm saying "forces" instead of just "force" is that there's really a different force vector applied to each small chunk of the wheel, and they add up to create a net torque that causes the whole gyroscope to rotate sideways.

    Suppose you spin up the gyroscope very fast and hold it fixed in place. All you have to do is support its weight. The only curved motion in this case is the circular, spinning motion of the wheel on its axis. The wheel itself and the "spokes" (See pic of Tedco model below) provide the centripetal force required to keep each small chunk of the wheel moving in a circle. There is no net reaction force on the "body" because of the symmetry of the gyroscope--all the little centripetal force vectors add up to zero. Let's say you were holding the gyroscope in place horizontally. Now if you push one end of the gyro down, holding it firmly so that it doesn't move sideways, you feel the gyroscopic force pushing your hand sideways. By rotating the gyro, you are causing the individual chunks of the wheel to take a curved path, but not just curved in the plane of the wheel. There has to be a force to cause this curved motion of the wheel chunks, but this time it is not directed toward the center of the wheel. You are providing this force. It is transimitted through the body and the spokes to the wheel. The force you feel is the reaction to this.

    If you support one end of the spinning gyroscope and let it go, the gravity supplies the force necessary to rotate the whole body of the gyroscope downward, but it cannot provide the horizontal force necessary to keep it moving this direction, so the gyroscope begins to move sideways.
     

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  8. Apr 7, 2005 #7
    If the plane of rotation of the wheel is turning, it seems like the centripital force would still be provided by the wheel itself since the axis of rotation is still at the center of the wheel (if you hold both ends and push up on one and down on the other).
     
  9. Apr 7, 2005 #8
    I'll try to find a picture to illustrate what I'm talking about. It is tricky to visualize. But if you follow a particular chunk of the wheel as the plane of rotation of the wheel turns, the entire path of the chunk does not lie in the instantaneous plane of rotation. It is a "centripetal" force that causes this motion, but it is not directed toward the center of the wheel (it's perp. to that direction). Of course, since the chunk is still going in a circle around the axis of rotation, there is also the real centripetal force that is directed toward the wheel center.

    BTW, this stuff is explained in detail with drawings in Feynman Lectures Vol I, if you have a copy of it handy.
     
  10. Apr 7, 2005 #9
    Hmm. I looked around on the internet a little bit and didn't come up with much. I attached two pics below...sorry they are so small. These are practically the same pictures given on pages 20-6 and 20-7 of Feynman Lectures on Physics, Vol I. It basically shows a spinning bicycle wheel being torqued.

    edit: the one on the right shows the curved path that I was talking about.
     

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    Last edited: Apr 7, 2005
  11. Apr 7, 2005 #10
    In the second picture, wouldn't the centripital force needed simply be suppied by the axle?
     
  12. Apr 8, 2005 #11
    Great, you CAN see the picture. I'll translate the tiny words if you need me to.

    To answer your question, yes, ALL forces (ignoring gravity) get transmitted through the axle to the rest of the wheel. If you are holding on to the axles and applying a torque, there's no other way for that force to be communicated to the rest of the wheel except through the axle. However (and this is the strange part), it is not necessary for the direction of all those forces to be toward the axle. In fact, you can see in the second picture (a bird's eye view) that the path of a chunk of that wheel curves to the left, in addition to curving up and down (since the wheel is spinning). In order to make that chunk follow a path that curves to the left, there must be a leftward force applied to that chunk. But from our bird's eye view, the direction of the axle is downward from the chunk. The solid wheel, a rigid body, is perfectly able to transmit such perpendicular forces through the axle and the spokes.

    I went back and read part of the discussion in the Feynman Lectures to make sure what I'm telling you is okay. Actually, it is more correct to say that the gyroscope starts moving sideways because of the ABSENSE of an applied force, rather than mysterious internal forces. Because as I described a couple of posts ago, an applied sideways force is needed to rotate the gyroscope downward. So really, the gyroscope is "falling" in the direction opposite of the absent applied force. I'll say more about this tomorrow.
     
  13. Apr 8, 2005 #12
    (to eliminate confusion, im talking about the case of the drawing you just posted where you are trying to rotate it horizontally and it rotates down instead) ok, so your saying that the force vector needed for the chuck points perpendicular to some empty point where there is nothing, so the vector must be pointed downward so the axle can provide the force? like if you have a ball spinning on a circle attached to a string, the ball is trying to "run away" from the circular path but the force tension of the string is keeping it from doing so. with the gyroscope the chunk of the wheel at the top is trying to run away from the circular path that you are trying to turn it in. since the axle pivots in all directions, it can only provide a force along the length of the axle. this force is the x compontent of the force vector needed. since there is no y compontent, the gyroscope starts to spin downward. similarly, if you think about it in torques. if you think of the centripital force needed as an imaginary force in the other direction - centrifugal force - and you have the force of the axle in the opposite direction......but their tails arent at the same point, so they create a torque to move the gyroscope downward, perpendicular to the direction you are trying to turn it. i think im starting to get it now, except going by what i said, it seems like although it would rotate downward, it looks like it should rotate horizontally also. but in actuallity it, for the most part, only rotates downward.....?
     
  14. Apr 8, 2005 #13
    I found the essay that you got those pictures from, its a good essay, but the part it talks about the curved path of the 'chunks' of the wheel confused me. It was said that the path of the top part caused a centripital force needed which then causes a reaction force in the opposite direction. Then it says that a piece on the bottom has a reaction force in the opposite direction, causing a torque perpendicular to the original torque. This would make sense, but the path of the bottom piece curves in the same direction, so wouldn't the reaction force be in the same direction? This would balance the forces in that plane so it wouldn't rotate perpendicular, it would only rotate the direction you applied the torque. Obviously this doesn't happen, I am missing something...eventually it will all just click....
     
  15. Apr 8, 2005 #14
    Yes, I am saying that the force needed to make the path of the chunk curve to the left points out in space somewhere, oustide the wheel. The force vector (with its tail located at the chunk) does not need to point toward the axle. If instead of a rigid wheel you had a spinning wheel of Jello, if you tried to torque the axle the wheel would break apart because the Jello can't transmit these sideways forces to the outer parts of the wheel.
    That is exactly right. You provide a centripetal force to make the ball go in a circle. The reaction you feel is the centrifugal force pulling on your arm. The same thing happens with the gyroscope.
    Actually, yeah that sounds about right. I'll need to think about it some more though. For example, if you were trying to swing a bowling ball on a rope around you in a circle, you are going to get jerked around unless you weigh a ton. Even though you are the one providing the centripetal force, there's no reason the ball can't cause you to move with its reaction force. Of course, you can still explain your movement in terms of the centripetal force that you apply (you are pulling yourself toward the heavy ball), but the result is the same.
    It rotates both horizontally and downward. Initially, you provide the necessary torque to start it rotating horizontally, and it does just that at first. But in order to keep it rotating horizontally, you have to simultaneously torque it "upward". If you don't provide enough of this "upward" torque, the wheel starts to rotate downward (in addition to horizontally). Weird, huh?
     
  16. Apr 8, 2005 #15
    Actually, if the piece on the top curves to the left, the piece on the bottom curves to the right (sorry, don't have a pic for this one). If you are looking down on the wheel from above, the pieces on top are moving away from you, while the pieces coming around the bottom are moving toward you. This is what makes the difference in the direction of curve.
     
  17. Apr 8, 2005 #16
    I see what your getting at, but its hard to visualize it without a picture. I'll have to think about it some more.
     
  18. Apr 9, 2005 #17
    Perhaps a better way to see it is this. Suppose you are looking down on the wheel and you twist it to the left. To a person underneath the wheel looking up, you are twisting it to their right.

    edit: for clarification, I am assuming that you are both facing the same direction, so that his right is also your right.
     
    Last edited: Apr 9, 2005
  19. Apr 9, 2005 #18
    I thought about it for a while and I think I've got it now. Imagine the gyroscope wheel not spinning, and you try to spin the body in the horizontal plane. As you spin it faster and faster, there is a force on the top and bottom of the wheel (as well as every other part but just to keep it simple) that keeps getting bigger. Each force creates a torque trying to rotate the gyroscope in the vertical plane. However, the forces are of equal magnitude and in opposite directions, so they cancel each other out and the gyroscope does nothing in the vertical plane. If you kept spinning it faster and faster, you would see the wheel start to bend (concave to the right) under the forces that exist. Now you can visually see that there are forces at the top and bottom of the wheel. Actually it is the absense of force causing it to bend, just like you said, but it is easier to imagine it as a force. What would happen if the bottom force was in the opposite direction? The wheel would start to rotate in the vertical plane - precess.

    When the wheel is spinning rapidly and you try to rotate it in the horizontal plane, forces will exist on the top and bottom of the gyroscope just as in the other situation. Consider a piece of the wheel exactly at the top. It has a velocity forward (looking down from above) from the spinning of the wheel. When you try to rotate the wheel counter clock wise in the horizontal plane, you are trying to change the direction of the velocity of the piece of the wheel. You are trying to change it from directly forward to pointing slightly to the left. In order to do this, you need a force to the left at the top of the wheel. Since the gyroscope can pivot in all directions and you are pushing forward with your right hand and pulling back with your left hand in attempt to rotate it horizontally, there is no force to the left. As you try to rotate it, the piece at the top is resisting this change - hence the imaginary force to the right.

    At the bottom of the wheel the velocity is in the opposite direction (towards you when looking from above). As you try to rotate the wheel to the left, you are trying to change the direction of the velocity of the bottom piece to pointing slightly more to the right. Just as with the top piece, this causes a 'force', except since you are trying to move the direction of the velocity towards the right instead of the left, the force is in the opposite direction. Now you have two forces in opposite directions at the top and bottom. This creates a net torque in the same direction - in the vertical plane. Now the gyroscope beings precessing in the vertical plane.

    What do you think?
     
    Last edited: Apr 9, 2005
  20. Apr 10, 2005 #19

    Andrew Mason

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    Because one end of the gyroscope is fixed (eg. by a string or a surface), the force of gravity on the centre of mass of the gyroscope creates a torque about the fixed end and is constantly adding angular momentum to the system. The torque is:[itex]\vec\tau = m\vec{g}\times\vec{r}[/itex] where [itex]\vec{r}[/itex] is the vector representing the displacement of the centre of mass of the gyroscope from the fixed end.

    This causes the gyroscope to move in a horizontal circle about the string that can only be understood by vector analysis of the changing angular momentum.

    Explanation of the horizontal circle prescribed by the gyroscope:
    In small interval of time [itex]\Delta t[/itex], the torque [itex]\vec\tau = m\vec{g}\times\vec{r}[/itex] creates a change in the angular momentum of the gyroscope system, [itex]\Delta \vec L = \vec\tau\Delta t = m\vec{g}\times\vec{r}\Delta t[/itex].

    The direction of this change is perpendicular to the gyroscope axis, which is the direction of the angular momentum of the gyroscope. Assuming the gyroscope angular momentum is pointing to the right (at 3 o'clock) and the free end of the axis is to the right from your position, the change in momentum is directly away from you (ie. at 12 o'clock).

    This means that the resulting angular momentum of the gyroscope system (ie. the axis) is now pointing right and a tiny bit forward of you (ie. a little before 3 o'clock, say 2:59) so the gyroscope has rotated a tiny amount counterclockwise around the axis made by the string. This changes the direction of the torque [itex]\vec\tau[/itex] so the change in angular momentum in the next small interval [itex]\Delta t[/itex] is perpendicular to the new angular momentum vector and the gyroscope rotates a little more. This keeps repeating. The result is that the gyroscope prescribes horizontal circular motion about the fixed end.

    AM
     
  21. Apr 10, 2005 #20
    It looks like you've pretty much got it. However...
    This part is not quite right. If the wheel is not spinning, and you try to turn it to the left about a vertical axis (that is, you are turning it horizontally), the only forces involved are directed toward that vertical axis of rotation. That is because each piece of the wheel is simply moving in a circle about the vertical axis. The necessary centripetal force is provided by the solid wheel's tendency to hold itself together rather than fly apart. And of course, by symmetry of the wheel, these forces balance each other so there is no net reaction. But my point is, if the wheel isn't spinning (about its axle), there are no perpendicular forces on the top and bottom of the wheel--not even ones that cancel out. Also, I should point out that what I'm saying here applies to a rigid wheel. If you turned a non-spinning Jello wheel, it would bend, but not at the top and bottom. It would bend at the sides.
    This part looks good. You now understand what causes the strange perpendicular reaction when you push on a gyroscope.

    Now I should say something about the terminology. Originally I said that when gravity pulls down on the gyroscope, it generates internal forces that cause the perp. motion. This is misleading and isn't really true. In your inertial (non-rotating) frame of reference, you don't actually need any new forces to explain why the gyroscope moves the way it does. It is actually falling toward the earth the best way it can. What we did establish in the above analysis is this: To make a spinning gyroscope rotate downward and not horizontally, one must apply both a downward force (gravity) and a horizontal force simultaneously. The horizontal force is necessary to make the pieces of the wheel take curved paths. But since gravity only acts downward, it cannot provide the necessary horizontal force for this motion. Thus you cannot expect a spinning gyroscope to just fall over. The only option is that it begins to "fall" in the direction of the absent horizontal force. I think you understood this already, but I just wanted to emphasize it because I had originally said it wrong. When we see such a strange motion, we feel the need to make up a force that's responsible for it. What is actually happening is that the gyroscope is falling in such a way that the paths of the pieces of the wheel do not curve away from the plane of the wheel. I hope that made sense.

    AM: thanks for providing the vector analysis for precession.
     
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