Gyroscopic motion

  • #1
Päällikkö
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The classical(?) problem:
Suppose you have a ring, where practically all the mass is concentrated on the edges (like the tire of a bicycle). There is, however, a massless rod placed at the central axis.

The ring's given some [itex]\omega _i[/itex] amount of angular speed (about the "logical" axis).

The system is then placed by the end of the rod to lean horizontally on another rod that is placed vertically. The system will now rotate about the second rod too at an angluar speed [itex]\Omega[/itex].
In case you found my description confusing (I sure did), here's a diagram I drew:
http://img473.imageshack.us/my.php?image=img0157us.jpg

Anyways, my question is, how much initial angular speed [itex]\omega _i[/itex] must be given in order for the system to execute the described motion?

If you still think my desciption is confusing, please ask for more precise info :smile:.

This is what I tried:
1) I assumed that as the rod's placed it is given some initial [itex]\Omega _i[/itex]. For [itex]\Omega _i[/itex] to stay constant, [itex]\Omega _i = \frac{rmg}{I_\omega \omega}[/itex] (this I derived and it should be right). [itex]I_\omega[/itex] is the moment of inertia of the ring about the axis that was first set to motion.
----
2) So, the [itex]\Omega[/itex] solved above will be the final [itex]\Omega _f[/itex] for the case where [itex]\Omega _i[/itex] = 0.

I assumed that energy is conserved (is it?), so:
[tex]\frac{1}{2} I_\omega \omega _i ^2 = \frac{1}{2} I_\omega \omega _f ^2 + \frac{1}{2} I_\Omega \Omega _f ^2 [/tex]
[itex]\omega _f[/itex] and [itex]\Omega _f[/itex] are connected by the equation solved in 1). So plugging that in gives me:
[tex]I_\omega ^3 \omega _i ^2 \Omega ^2 = I_\omega (rmg)^2 + I_\Omega I_\omega ^2 \Omega ^4[/tex]

If I solve for [itex]\Omega[/itex], I can have real answers only if discriminant [itex]\geq[/itex] 0:
[tex]\left( -I_\omega ^3 \omega _i ^2 \right) ^2 - 4I_\Omega I_\omega ^3 (rmg)^2 \geq 0[/tex]
[tex]\Rightarrow \omega _i \geq \left( 4 \frac{I_\Omega}{I_\omega}(rmg)^2 \right) ^{\frac{1}{4}}[/tex]

Is it all wrong? :smile:
(If) so, could someone give me a hint to the right direction?

EDIT: Corrected a typo.
 
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Answers and Replies

  • #2
Astronuc
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It's been a few decades since I have done a problem like this, but I will give it a try. Perhaps this belongs in the Advanced Physics section, because gyroscopic precession is more of an upper class problem.

I'll be back later after I cut some vegetables and clean up a spill. :biggrin:
 
  • #3
Astronuc
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It seems this problem is asking one to find the precession angular frequency as a function of the angular frequency of the rotating ring (and other appropriate variables). The torque generated by the rotating ring causes the gyroscope to rotate, or precess. The acceleration is simply going into the change of direction of the rotational vector of the ring.

If the centerbar mass was considered, it would be manifest in ring system's moment of inertia.

Here is a simple derivation of recession frequency, but it may not be what you want. http://scienceworld.wolfram.com/physics/GyroscopicPrecession.html
 
  • #4
Päällikkö
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That's pretty much what I did in 1). It being pretty trivial (meaning that one can find the derivation all over the internet and books) I decided not to include the derivation. I did, however, include the result (which itex-tags make a bit small).

It's part 2) that I'm unsure of.

The problem is just one that I made up, so if you're unsure what I'm after, please let me know.

Thanks for replying, though, I was sure nobody would :smile:.
 
  • #5
mezarashi
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I have rarely done gyroscopic motion in the past. But from what I understand, there can be multiple precession frequencies, so I'm not quite sure what you're trying to find. There won't be one frequency. Depending on the spin frequency, there will some amount of 'droop' associated while it precesses.

Applying some quick fundamentals:

[tex]\frac{d\vec{L}}{dt} = \vec{\tau}[/tex]

You can do some successive approximations (which I'm not sure will always be applicable). The torque here is due to gravity.

[tex] \tau \approx \frac{\Delta L}{\Delta t} , \frac{\Delta L}{L} = \Delta \Phi[/tex]

[tex] \tau \approx L \frac{\Delta \Phi}{\Delta t}[/tex]

[tex] \tau \approx L \omega_p[/tex] while [tex] L = I\omega_s[/tex]

You'll sorta have a relationship between torque and the precession, spin frequencies.
 
  • #6
Päällikkö
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mezarashi said:
I have rarely done gyroscopic motion in the past. But from what I understand, there can be multiple precession frequencies, so I'm not quite sure what you're trying to find. There won't be one frequency. Depending on the spin frequency, there will some amount of 'droop' associated while it precesses.
I assumed that there'd be only one precession. My physics book does mention something like what you said (and that's actually what I wanted to examine closer after I get this problem sorted), but leaves it there. I thought I made a valid assumption, or do you disagree?

In the problem I meant to ask what must the initial angular speed be for the system to execute gyroscopic motion.
If the system is not given any initial angular speed in the precession angle's direction, some energy must be used to gain the precession angular speed, right?

I hope you understood what I meant :smile:.
 
  • #7
mezarashi
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I guess your asking for something like, what is the minimum [tex]\omega_s[/tex] so that this precession will still occur?

I would go back to the 'droop' idea. As the angular velocity of the wheel decreases, the wheel will continually droop such that [tex]\vec{F_g} \times r[/tex] or in scalar form [tex]Mgr sin\theta[/tex] will be the torque [tex]\tau[/tex] to maintian the equation of gyration. The smallest possible value of theta would be such that the wheel is about to touch the center pivot. This will probably correspond to your minimum speed you are looking for.
 
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  • #8
Päällikkö
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mezarashi said:
I guess your asking for something like, what is the minimum [tex]\omega_s[/tex] so that this precession will still occur?
Yes.

mezarashi said:
I would go back to the 'droop' idea. As the angular velocity of the wheel decreases, the wheel will continually droop such that [tex]\vec{F_g} \times r[/tex] or in scalar form [tex]Mgr sin\theta[/tex] will be the torque [tex]\tau[/tex] to maintian the equation of gyration. The smallest possible value of theta would be such that the wheel is about to touch the center pivot. This will probably correspond to your minimum speed you are looking for.
As in videos on the internet (eg. MIT's OpenCourseWare video lectures), I'd like to place the ring with the axis in a 90 degree angle to the other rod (which is vertically on ground), like in the diagram in the first post. So, [tex]\theta[/tex] = 90o.
 
  • #9
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hi everyone...

I've got a question to you. It may seem a bit stupid but I have not been able to figure out WHY the gyroscope precession happens.

In fact, why doesn't the torque of the applied force just make the object fall? Assumpting that this torque does not make the object falling but contributes (almost) exclusively to its rotation around a vertical axle (precession), it is very easy to do the necessary calculus and get the function which makes a relationship between angular precession speed and the angular speed of the ring (check out this link: http://physics.nad.ru/Physics/English/gyro_txt.htm)..

But WHY? WHY does the percession motion happen? I found this webpage that makes it clear for a simplier example: http://www.howstuffworks.com/gyroscope2.htm . However, I can't still explain clearly why this precession motion happens in this kind of examples.


I hope I have been clear enough. I am not an English native speaker and I have some difficulties expressing myself in this language :confused:


Please answer me as soon as possible.



Farewell!
 

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