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Force applied on a rotating disc

  1. Jun 26, 2012 #1
    I understand that if we apply a force on the axis of rotation of a spinning disc, the torque is in a different direction, and the axis is moved in a direction perpendicular to the direction in which the force upon the axis was applied, determined by the right-hand rule.

    But what if the force was applied directly on the disc? e.g. we have a functional gyroscope, and we apply a force (effectively a torque) on the disc, say, by blowing down on one half of the spinning wheel. What happens then? Would we still get gyroscopic precession or will the axis of rotation simply move in the direction in which we are blowing?

    Thanks!
     
  2. jcsd
  3. Jun 26, 2012 #2

    rcgldr

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    In the case of helicopters, when using the cyclic (pitch and roll control), the result is equal and opposing forces between rotor and air, resulting in the air exerting a torque on the rotor. The rotor's reaction to the torque is about 90 degrees later out of phase, so a pitch torque results in a roll reaction and vice versa. The pilot controls are advanced 90 degrees to compensate for this.
     
  4. Jun 26, 2012 #3

    Cleonis

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    In both cases the same happens.

    Regardless of whether the force is exerted on the axis or on the disk itself, either way a force is exerted upon the spinning disk. From there on it's the same case.




    A point of clarification:

    Given the way that gyroscopic precession is usually demonstrated one easily gets the impression that gyroscopic precession happens in response to a force. It's actually a sequence: a force exerted on a spinning disk causes motion of it, this motion then gives rise to gyroscopic precession. In the usual demonstrations the initial motion is imperceptably small, but without it the motion pattern of gyroscopic precesson would not arise. The initial motion follows the force; it's in the direction of the force. The direction of the motion of gyroscopic precession is at right angles to that initial motion.

    The reason I expand on this is to emphasize there is no lag in the response. Yes, the gyroscopic response is at a 90 degrees angle, but that 90 degrees angle is not due to any lag. The response is instantaneous. The 90 degrees angle is very counterintuitive of course, but the reason for it is known, and it's not some lag effect, there is no "later", no out-of-phase.
     
  5. Jun 27, 2012 #4

    rcgldr

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    The phase comment refers to the phase difference between the perpendicular component of acceleration versus the perpendicular component of velocity on a disk. The peak perpendicular component of velocity occurs when the perpendicular component of acceleration is zero, and vice versa. As an analogy, in simple harmonic motion (sine wave), like an oscillating mass on a spring, if you plot acceleration, velocity, and position versus time, acceleration is 90 degrees ahead of velocity, which in turn is 90 degrees ahead of position.

    Link to wiki article that mentions the 90 degree advance used on the swash plate for cyclic control on a helicopter:

    wiki_helicopter_cyclic_blade_control.htm
     
  6. Jun 27, 2012 #5

    Cleonis

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    I'm unfamiliar with the terminology 'perpendicular component of the acceleration', etc., so I cannot comment on that. (I mean, I don't know perpendicular to what is intended here.)



    In general, about the description of helicopter cyclic blade control:
    As described in the wikipedia article, the way the blade pitch is controlled has conceptually two components.
    One is adjustment in time: the blade needs some time to arrive at the desired pitch, so an advance in time is needed. This component is strictly about when the blade control acts, so that the desired pitch occurs at the desired point.

    Then there is adjustment of blade pitch to take gyroscopic effect into account. That is strictly about where the blades have their maximum pitch. This is at a spatial 90 degrees angle.


    Of course, since the rotor angular velocity is pretty much a fixed rate controlling when and where is indistinguishable from a control point of view.

    I wonder, what if there is a helicopter design where depending on circumstances the rotor angular velocity is varied a lot. Freight helicopters need to adjust for flying with or without cargo. Possibly there are freight helicopters that vary the rotor angular velocity according to the load.

    Let's say such a variable rotor speed helicopter exists.
    - At every rotor speed the adjustment for gyroscopic effect is the same: a spatial angle of 90 degrees.
    - Depending on the current rotor speed the time adjustment must be set differently, so that at every rotor speed the maximum pitch occurs at the desired point.


    The two adjustments are handled by a single mechanism, but it's still two adjustments, one is a spatial adjustment, the other a timing adjustment.
     
    Last edited: Jun 27, 2012
  7. Jun 27, 2012 #6

    rcgldr

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    I meant the component of acceleration or velocity of a point on a rotating disk that is be perpendicular to the plane of the rotating disk at an instant in time. The situation is dynamic since the plane is also rotating due to precession. Peak perpendicular component of velocity of a point occurs when perpendicular component of accelertion of a point is zero, during a transition from "positive" to "negative" acceleration or vice versa. (This situation is similar to harmonic motion of a point).

    As far as I know, all full scale helicopters have collective control, and the throttle is adjusted by pilot or by computer control to compensate for the collective input to ensure adequate rotor speed is maintained. Some (or most) gyrocopters don't have collective, but these rely on auto-rotation and forward speed to maintain rotor speed. For a freight helicopter, momentum of the rotor would be an issue if the pilot wasn't able to compensate for sudden changes in load during transitions with collective.

    There are model helicopters without collective, but the control is not precise, and transition into and out of forward flight involves normally unwanted changes in altitude due to the momentum of the rotor. The better model hellcopters have collective, and there's a throttle versus rotor pitch curve mix on the transmitter to maintain rotor speed (the "pilot" only has a single axis for combined throttle + collective). In aerobatic mode (enabled or disabled by a toggle switch on the transmitter), the rotor speed is kept at a near constant and fairly fast rate of rotation regardless of collective pitch (positive or negative).
     
    Last edited: Jun 27, 2012
  8. Jun 27, 2012 #7

    Cleonis

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    I retrieved this quote because this is the one that triggered the elaboration in my first comment. I addressed the suggestion that in some way there is a 'later', some sort of delay. There is no 'later'.



    I think that component-of-velocity/acceleation-perpendicular-to-the-plane of the-disk is not a causal factor.

    While in the motion pattern of gyroscopic precession:
    Velocity component perpendicular to the plane of the disk rises from zero to a peak and goes back to zero again. Acceleration component perpendicular to the disk oscillates between positive and negative.

    It seems to me these velocity/acceleration patterns are logical consequences of the motion pattern of gyroscopic precession, and not in themselves a causal factor.

    The reason I think that way is that in the convincing quantative derivation of gyroscopic precession rate (given spin rate and available torque) that I know, the velocity/acceleraton component perpendicular to the disk is a physical consequence, and not a cause.
     
  9. Jun 27, 2012 #8

    rcgldr

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    I used the term "later" to give a sense of the direction of precession in response to torque with respect to the direction of rotation of a disk (right hand rule using the cross product of vectors can also be used), and also to correspond with articles about the cyclic on helicopters that describe the cyclic control as being 90 degrees in advance to compensate for gyroscopic precession.

    If torque, angular momentum, and angular velocity are stated as angular vectors, then the equation for precession can be stated in the form of a vector cross product:

    {torque} = {angular velocity of precession} x {angular momentum}

    Some of the angular momentum is in the direction of precession, but if the angular velocity of a disk is much greater than the angular velocity of precession, it's a small effect.

    I didn't mean to imply a cause and effect relationship, just a co-existance of the relationship.
     
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