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Why does a horizontal (wheel vertical) gyroscope eventually fall?

  1. May 21, 2012 #1
    I was hoping someone could clear this up for me. I've been thinking about it a bit, and I am thoroughly confused. Here is what I have come up with so far.

    The situation: A horizontal (wheel vertical) gyroscope spinning counter clockwise held up at one end by a string. Now change views so we're looking down on it from above. Looks like •--|-- with • being the point where it's held up and --|-- being the gyroscope (from this angle the wheel looks like a line). Positive y and x are |_ respectively (positive z is out of the page).

    My thoughts:

    Angular momentum is in the positive x direction via right hand rule. The force of gravity is in the negative z direction and R is in the positive x direction, so using the cross product of R x F gets the positive y direction for the torque. So the gyroscope moves counterclockwise around the point where it's held up by the string.

    For friction: The friction force will act on the wheel and point in the opposite direction of the tangential velocity of the wheel. For simplicity just think about this as if it where in the orientation of a skidding wheel. In this situation that direction is the negative y direction. For the friction force the R will be the radius of whatever it is acting on. Lets just say that the radius points in the negative z direction because this works for our purposes. R x F for friction gives us with a torque that points in the negative x direction. As we would expect it opposes the angular velocity.

    But wait!

    The torque from friction is in the x-y plane! The torque from gravity also acts in the x-y plane. How can the gyroscope drop into the negative z direction if its net torque has no z component? I can think of only two answers, air resistance supplies this downward torque or there is some underlying physics that I am unable to comprehend.

    Air resistance:
    The force from air resistance points in the negative y direction because the gyroscope moves counterclockwise, and it operates over an R that is in the positive x direction. The cross product of these vectors gives us a torque in the negative z direction.

    If you kept a gyroscope in the orientation described in a vacuum would it forever remain "defying" gravity? That just seems wrong. There must be something else to it.

    Sorry for the really long and confusingly phrased post...

    -Alex
     
  2. jcsd
  3. May 21, 2012 #2

    K^2

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    The torque that brings the gyro down is the friction torque acting on the point where the gyro is suspended. (• in your diagram) It works against the direction of precession, so the torque points straight down, which means the angular momentum vector rotates to point towards the ground.
     
  4. May 21, 2012 #3
    I see what your saying, but what about in a situation like 2:32 in a video on youtube titled Gyroscope by ScienceOnline (sorry but I can't post links yet)

    The frame around the gyroscope isn't moving (probably because friction from the string is keeping it from moving), and because of this the frame of the gyroscope is putting friction on the spinning gyroscope. This friction opposes the rotation of the gyroscope, but I don't see how it opposes the precession of the gyro as the procession comes from the torque by gravity. Eventually, I see the the angular momentum being annulled by friction and the torque being unable to change the angular momentum because there no longer is an angular momentum to be changed. In which case gravity would exert a torque on the frame+gyro that would cause it to fall. Also, I don't understand how this transition of torque would take place. The gyro obviously falls slowly, which supports the fact that friction is the culprit.

    If you think about it purely as friction opposes movement, and the precession is the movement then it makes sense how the gyro would fall. I just have trouble getting from where I am right now to that answer. There should be another way to describe the situation that essentially means the same thing.
     
  5. May 21, 2012 #4
    Perhaps a good way to think about this would be in terms of the Second Law of Thermodynamics?
     
  6. May 21, 2012 #5

    Cleonis

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    Let me restate that question with a more general scope: if you can achieve a perfectly frictionless setup (vacuum, frictionless bearings) would the same state of precessing motion remain?

    Sure it would.

    In the case of gyroscopic precession the only source of motion decay is friction. In the theoretical case of a frictionless setup precessing motion continues without decay. This is a general principle of motion. In the absence of friction there is no decay of motion. For example, in their motion around the Sun the planets are subject to negligable friction; planets remain in pretty much the same orbit. Conversely, when you do notice a decay of motion then there is some form of friction at work.



    Trying to think about this case in terms of elements that involve the right hand rule is quite error prone, I think.
    It is in fact possible to understand gyroscopic precession in more direct terms.
    Some time ago (november 2010) I posted a discussion, illustrated with two diagrams, of the underlying mechanics of gyroscopic precession.

    In that post there is also a link to a more extensive discussion on my own website.
     
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