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Gyroscope doubts. Please answer

  1. Aug 17, 2011 #1
    Gyroscope doubts. Please answer :)

    Consider the case of a common demonstration which you would find in videos. A bicycle wheel mounted on a shaft is set into rotation, axis is made horizontal and the shaft stays horizontal for sometime and eventually goes down as the wheel spin slows down due to friction.

    Like this one : http://www.youtube.com/watch?NR=1&v=8H98BgRzpOM

    My questions are

    1) What torque balances the gravitational torque vertically and the precession torque horizontally (applying Newton's second law) ?

    2) Let the torque resisting the vertical fall of shaft be R and torque due to gravity be T. Since the shaft finally tilts down is R more than T and why does this happen?

    3) How is the resisting torque (against gravity) R dependent on
    a)Rate of spin
    b)Gravitational torque T ?

    Please answer these questions :smile:
     
  2. jcsd
  3. Aug 18, 2011 #2

    Bill_K

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    Science Advisor

    Re: Gyroscope doubts. Please answer :)

    Shivanand, This question shows that you have a bit of misunderstanding with the concepts. Let's see if we can clear it up. Here's some questions for you:
    a) What is the general relationship between torque and angular momentum?
    b) There is a torque on the gyroscope caused by the downward force of gravity and the upward reaction force of the support. In what direction does this torque point?
    c) What is the angular momentum of the spinning gyroscope? In which direction does it point?
    d) Is the angular momentum constant? Remember it is a vector.

    Also you mentioned Newton's Second Law. Does Newton's Second Law apply to torques or to forces?
     
  4. Aug 18, 2011 #3
    Re: Gyroscope doubts. Please answer :)

    No, even if the shaft continued rotating at the same speed (for example, if it were motorized) it would still slowly be pulled down.
     
  5. Aug 18, 2011 #4

    DaveC426913

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    Gold Member

    Re: Gyroscope doubts. Please answer :)

    It is against the rules for PF members to provide you with answers. You must show your attempts at the solutions before we can help you.
     
  6. Aug 23, 2011 #5
    Re: Gyroscope doubts. Please answer :)

    Hi,

    Sorry for the late reply. I tried hard to understand and I thought I need help from people like you. I attempted to get a solution but I am too confused. Please excuse me for asking straight away and not posting my attempts in this thread :redface:

    I will make my questions simple. If I could get an explanation, I would be able to understand. Please bear with me if my questions were/are annoying and if I seem like a noob. I am still learning.

    1) Does a spinning bicycle wheel supported at one end of its shaft fall down if the rate of spin of the wheel (angular velocity) is constant?

    2) If it does not fall down, is there a critical angular velocity above which the shaft does not fall? Or do the wheel and shaft fall irrespective of angular velocity (high or low) of the wheel? :confused:

    It was my mistake to have told Newton instead of Euler. I meant Euler's equations. Sorry for that :redface:
     
    Last edited: Aug 23, 2011
  7. Aug 24, 2011 #6
    Re: Gyroscope doubts. Please answer :)

    It depends. But not on the spinning speed. Someone before gave you the answer. It is important you understand why to master properly the gyro behavior.

    It can always fall down, without anything you can do against it by adjusting the angular speed.
    Try to understand why (it's not so easy), then eventually come back.
     
  8. Aug 24, 2011 #7

    rcgldr

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    Re: Gyroscope doubts. Please answer :)

    It depends on the initial condition, but assuming no friction or losses, it won't truly "fall", but instead cycle up and down. Also it depends on what you mean by supported at one end. If the support is a string that is suspended from above, the end of the string will follow a circular path that varies in radius (see video #9 in the link below). If the support is an unmovable rod (or at least unmovable relative to the earth), like a gyroscope placed on a "tower", then the initial condition can be one where the wheel axis remains horizontal as it rotates about the rod support.

    Video #9 on this web page shows the complicated motion of a gyro suspended at one end by a long string mounted to the roof of the room. The key point of this video is that even though gyro is losing it's rate of spin, once it establishes the cyclical pattern shown in the video, it's center of mass will not significantly move vertically (the professor mentions it remains on a horizontal plane). As the gyro slows, the cyclical pattern will just speed up, keeping the center of mass from moving vertically until eventually almost all of the energy is lost and then it will "fall". (The friction interference with the rate of precession will determine where the center of mass will tend towards).

    http://www.gyroscopes.org/1974lecture.asp
     
    Last edited: Aug 24, 2011
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