2
$\begingroup$

Lets say we have a gyroscope that has a flywheel spinning with constant angular velocity. The gyroscope is held in place in a position where the massless metal rod connecting to the flywheel is parallel to the ground. We now let go and are asked to predict the gyroscoped motion.

When researching gyroscopes the predicted motion of precession at constant vertical height was explained by saying, in essence, that the precession causes the angular momentum to change in the correct way, the derivative of angular momentum being equal to torque, based on the gravitational torque acting on the system. This question is not so much about gyroscopes specifically but getting a better grasp of the way one reasons in physics.

To me the above explanation seems incomplete. My reasoning would be> if the normal force is not equal to the weight, the body will move vertically causing a change in angular momentum in a direction in which there is no torque => contradiction. Now the only possible motion that remains to ensure that torque is equal to the derivative of angular momentum is precession. Does my reasoning make sense? I am right in my feeling that the first argument is not satisfactory or am I misunderstanding something?

Improve this question
$\endgroup$
1
  • $\begingroup$ The pure precession model is an approximation in the limit of unbounded angular momentum. A more complete model is given in, for instance, arxiv.org/abs/1007.5288 which references Feynman's explanation in the The Feynman Lectures where he addresses exactly how you should be reasoning about such a system. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 18, 2017 at 20:28

1 Answer 1

Reset to default
2
$\begingroup$

If you hold a gyroscope supported at both ends of the axis, and let go of one end, then it will initially "fall" a little bit, and this falling motion is what makes it start to precess.

That motion is called nutation and it is (unfortunately) often ignored when gyroscopes are first introduced - but your intuition is exactly right! See for example this paper which quotes Feynman as saying

It has to go down a little in order to go around

This paper describes in some detail (and demonstrates with experimental results) that there is initial vertical motion which is what generates the necessary horizontal precession.

Improve this answer
$\endgroup$
3
  • $\begingroup$ I don't understand why the gyroscope going down isnt a contradiction. In an idealised model the gyroscope initially has zero vertical angular momentum and zero vertical torque. If the gyroscope changes its vertical position it seems to me that it will definetly have vertical angular momentum. Could it be that the observed initial drop only takes place in a system that is not"ideal"? $\endgroup$
    – fibo11235
    Commented Jun 18, 2017 at 22:27
  • $\begingroup$ No. If the gyroscope had no angular momentum you would not question that it falls. If it has just a little bit of angular momentum why wouldn't it still fall? The degree to which it drops will depend on its angular momentum - the higher the angular momentum the smaller the drop. Did you read the paper I linked? $\endgroup$
    – Floris
    Commented Jun 19, 2017 at 3:05
  • 1
    $\begingroup$ I read it again, this time thouroughly, and I have a much better grasp of the whole process now. Thanks for bearing with me. $\endgroup$
    – fibo11235
    Commented Jun 19, 2017 at 10:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.