Steady precession of a gyroscope

[dyn]

Hi
I am trying to understand the concept of why a gyroscope that is not spinning drops down but when it is spinning then it precesses. I have looked in "University Physics" by Young + Freedman and Kleppner , Morin and Gregory but i am not getting anywhere. Does anyone know of a relatively simple explanation in some online notes or videos or maybe a textbook with a clear explanation
Thanks

 

[dyn]

Hi
This question arises due to my confusion from trying to use Kleppner to understand gyroscopes. If i have a gyroscope or even simpler a light rod with a mass at one end which is pivoted at the other end. If the rod is held horizontally and then released then obviously it swings downwards and applying torques about the pivot shows this. But if i use vertical forces then i end up with the overall vertical force being N-W where N is the normal force upwards applied by the pivot and W is the weight of the mass. If N=W then there is no overall vertical force and the centre of mass situated at the mass at the end of the rod should not move. Why does using torques show that the mass will swing downwards but using forces show that the mass can remain horizontal ?
Thanks

 

[nrqed]

Hi
This question arises due to my confusion from trying to use Kleppner to understand gyroscopes. If i have a gyroscope or even simpler a light rod with a mass at one end which is pivoted at the other end. If the rod is held horizontally and then released then obviously it swings downwards and applying torques about the pivot shows this. But if i use vertical forces then i end up with the overall vertical force being N-W where N is the normal force upwards applied by the pivot and W is the weight of the mass. If N=W then there is no overall vertical force and the centre of mass situated at the mass at the end of the rod should not move. Why does using torques show that the mass will swing downwards but using forces show that the mass can remain horizontal ?
Thanks

The fact that the net force on an object (which is not point like) is zero is a necessary condition to have equilibrium but it is not sufficient. Put a light rod on a table and apply 10 N perpendicular to the rod at one extremity and 10N perpendicular to the rod at the other extremity but in the opposite direction to the first force. The rod will move, despite the sum of the forces being zero.

(By the way, how do you know that N=W ? Do you assume it or do you prove it?)

 

[phyzguy]

I always come back to my old favorite, "The Feynman lectures on Physics". This link, to Volume 1, Chapter 20, is what you are looking for. Section 20-3 discusses the gyroscope in some detail. I suspect your question is something like (quoting from section 20-3), "It is very strange that when one suddenly let's go of a gyroscope, it does not fall under the action of gravity, but moves sidewise instead! Why is it that the downward force of the gravity, which we know and feel, makes it go sidewise?" I'll let you read Feynman's answer.

 

[dyn]

In the example above would the rod not just rotate about its centre of mass which would remain stationary ?

 

[dyn]

I looked in Feynman and that explanation did not work for me

 

[hutchphd]

It is not trivial. Maybe Walter Lewin?

 

[Dr_Nate]

Have you seen the demonstration of precession of a bicycle wheel from a string? I think this is the easiest way to understand:

• The wheel is spinning and hung on the string. Let's just consider two infinitesimal points on the rim of the wheel at the top and the bottom at some instant. I guess this model is actually an idealized dumbbell.
• Gravity will cause a torque on the wheel, which normally rotates it downward. It does that by giving the top and bottom points linear momentum in opposite directions perpendicular to the plane of the wheel.
• But in this case the wheel is spinning.
• Let's say you blink and the wheel is rotating fast enough that it has rotated 90 degrees (you'll have to do some thinking to generalize this to all situations).
• Which way will the momentum cause the wheel to turn?

 

[nrqed]

In the example above would the rod not just rotate about its centre of mass which would remain stationary ?

Sorry for not being clear. I had in mind the rod in your question with a mass at one end.
My point still holds: consider a rod of uniform mass and now apply a force of 10 N at the center of mass and a force of 10 N in the opposite direction at some other point. The rod will move even through the net force is zero. There is equilibrium only if in addition to the net force being zero, the net torque with respect to any point is zero.

 

[A.T.]

Why does using torques show that the mass will swing downwards but using forces show that the mass can remain horizontal ?

In general you have to use both, forces and torques. In special cases you can choose your pivot such that torques alone will do.

 

[anorlunda]

[Moderator: Duplicate Threads Merged]

 

[phyzguy]

I looked in Feynman and that explanation did not work for me

So after reading Feynman, what exactly is it that you do not understand?

 

[phyzguy]

If N=W then there is no overall vertical force and the centre of mass situated at the mass at the end of the rod should not move.

Perhaps this is your problem? The center of mass of the gyroscope is not at the end of the rod. If it were, there would be no torque and the gyroscope would not precess.

 

[dyn]

With a simple gyroscope the entire mass of the gyroscope is situated at the end of the rod so its centre of mass is there assuming the rod is light.. The torque is due to the weight acting downwards from this point

 

[A.T.]

If N=W then there is no overall vertical force and the centre of mass situated at the mass at the end of the rod should not move.

No net vertical force means that the centre of mass has no vertical acceleration. It can still move and accelerate horizontally, as is the case when it moves along a horizontal circle during the precession.

 

[dyn]

If a horizontal uniform rod is pivoted at one end the overall vertical force could be zero as N-W = 0 if N=W but the centre of mass will still drop due to the torque caused by gravity. I think this is a case of a textbook totally confusing me

 

[Dr_Nate]

If a horizontal uniform rod is pivoted at one end the overall vertical force could be zero as N-W = 0 if N=W but the centre of mass will still drop due to the torque caused by gravity. I think this is a case of a textbook totally confusing me

Why do you think the normal force at the pivot is equal to the weight? You have a contradiction here, don't you? It looks like you've started with Newton's second law:
$$\Sigma F_y = ma_y,$$
and jumped right to setting the acceleration to zero.

 

[dyn]

I'm using the example from the Kleppner book which started my confusion.

 

[Dr_Nate]

Am I correct in reading that in this merged thread that post #1 is about a spinning gyroscope and post #2, while mentioning your want to understand gyroscopes, is actually about a rod pivoting downward?

I read it differently the first time I saw your post before today because of your mention of gyroscopes. My response today is based on my new understanding as a pivoting rod.

In the case of a pivoting rod, you can probably intuitively understand that the normal force won't equal the weight with the following example. You have a friend with a 9-kg sledgehammer. You and your friend are on either end of the sledgehammer holding it horizontally. Your friend is holding the head while you are at the end of the handle with it resting on your palm. Your friend let's go and the hammer pivots swings down while pivoting in your hand. Do you feel 9 kg worth of weight pushing down on your hand?

 

[A.T.]

...if N=W but the centre of mass will still drop ...

If the net vertical force is zero the CoM will not accelerate vertically.

 

[dyn]

Hi. I will break my confusion down into smaller questions and use for my example a gyroscope consisting of a light rod with the "spinning bit" situated at the end so the centre of mass is at the free end. The normal reaction at the pivot is N and W is the weight of the gyroscope.

1 - if the gyroscope is not spinning then the overall vertical force is N-W. If N=W then there is no overall vertical force and the gyroscope would remain horizontal( if that was the initial position) but in reality W > N so the centre of mass accelerates downwards and the gyroscope drops. Is that correct ?

2 - When the gyroscope is spinning it precesses in a horizontal plane so N=W and there is no vertical acceleration. Does the spinning of the gyroscope cause the normal reaction to increase and become equal in magnitude to W ?

3 - When the spinning gyroscope precesses , it does so in a horizontal circle so there must be a centripetal force ? What is the force causing this ? Is it tension in the rod ?

Thanks

 

[hutchphd]

I agree with (1).
For (2) one needs to be careful with causality. The combination of the spinning gyroscope and its mandated precession makes N=W on average (see Feynman or ?).
For (3) the centripetal force is not really interesting (the axle of the gyro is a rigid bar) but the fact that the rigid bar constrains the axis of the gyro is exactly relevant.
Hope that helps.

 

[Swamp Thing]

I have sometimes felt that a thought experiment based on your point (2) would clear up much confusion and help to introduce the gyroscope in a better way. Rather than thinking of precession as an effect of gravity's torque on the spinning gyroscope, let's think of it as a given angular momentum (which of course doesn't need any torque to sustain it).

To start with, imagine that the horizontal axis rod is not pivoted, but rather is locked to a vertical shaft, which in turn is free to rotate around a vertical axis. But we have a torque sensor at the locked pivot, so we know the net torque that is trying to tip the gyro axis down. And let's start with the gyro not rotating either. Obviously, in this case the torque sensor will show only the gravity-induced torque, and will continue to do so whether we (manually) set up some "precession" type motion or not.

Next set the gyroscope spinning, with its axis rod stationary. Again, the torque sensor will still read just the gravity induced torque. Now, give the axis a little nudge to set it revolving around the vertical pillar. Once we give it that nudge, it will continue to revolve through sheer angular inertia. The word precession would be somewhat irrelevant at this moment, being a descriptive label rather than an explanatory term or a result of gyroscopic behavior.

At this point, the torque sensor would read less than before, because you now have your W-N effect. We can nudge the axis a couple of times more, and the torque sensor will read less and less as the gyrosopic upward torque increases with each nudge. At some point we can get the torque sensor to read exactly zero, and we now have the gyroscope "precessing" with some corresponding angular momentum. Again the "precessing" is just a descriptive term, not an explanation of why it is rotating. It is rotating because we have been nudging it, and now it has to continue to rotate because of inertia.

Finally, we remotely unlock the pivot while the thing is still "precessing", and we know it is going to remain horizontal because the torque sensor is reading zero torque. So in this picture the precession is the cause of the upward torque that compensates exactly for the gravitational torque. And the precession is the effect of the past history of the little nudges that we have been applying.

 

[A.T.]

2 - When the gyroscope is spinning it precesses in a horizontal plane so N=W and there is no vertical acceleration. Does the spinning of the gyroscope cause the normal reaction to increase and become equal in magnitude to W ?

3 - When the spinning gyroscope precesses , it does so in a horizontal circle so there must be a centripetal force ? What is the force causing this ? Is it tension in the rod ?

As @hutchphd notes, the causation questions are not necessarily physically meaningful. If you are asking how the forces are transmitted, then yes, it is differential tension and compression in the material.

Here some good explanations of gyroscopic precession:

 

[mgkii]

Hey A.T. - I've watched plenty of Gyroscope vids in my time and I have to say that first one is superb - I've seen the Veritasium one before but not the first one. Excellent find!

 

[mgkii]

PS... the spinning/flipping nuts in Zero G you can see linked bottom right in that first video to is also fabulous; makes a very unintuitive topic quite reachable

 

[mpresic3]

My credentials for the gyroscope problem are very good. I have simulated the motion of the gyroscope, called the symmetric top in a uniform field with reaction force at the base. I used elliptic functions and theta functions a la Landau and Lifshitz or Whittaker.
I have looked at many elementary treatments at the freshman/ sophomore level and found them unsatisfying as well. I confess I do not remember Feynman's argument and I may read them tonight if I am not too tired. After reading several elementary treatments, I came to the conclusion, that if I were teaching an freshman/sophomores, I would never test them on the gyroscope.
The bad news is I do not know of any good treatment of the gyroscope that doesn't involve the Lagrangian, the Hamiltonian treatment, and roots of the cubic equation. (I'll get back to you if I like Feynman's treatment) That is why I wouldn't ask a question of anyone but very good physics undergraduates (Junior/Senior) at a strong college/university.

One professor, perhaps Ed Purcell was asked what the hardest things physics undergraduates learn. The questioner expected Purcell to say Quantum mechanics, or Relativity. The professor agreed these are the most novel areas, but in his experience the hardest things undergraduates are expected to learn is rigid body mechanics. After working with many physics, math and engineering graduates, I agree. (I might get back to you to the link to this article)

 

[vanhees71]

The message is that learning Hamilton's principle of least action and analytical mechanics is well worth the effort for two reasons: (a) mechanics becomes much easier than when sticking to pure Newtonian methods and forces (for me Hamilton's principle was a revelation when I first learned about); (b) Hamilton's principle underlies all fundamental physics, and Newtonian mechanics is the most simple example to learn about it. The most important feature is that it makes Noether's theorems on symmetries and conservation laws possible, which is the most important and most beautiful concept of physics.

 

[A.T.]

Hey A.T. - I've watched plenty of Gyroscope vids in my time and I have to say that first one is superb

Yes, I have seen this type of explanation in some books, but for a long type there was no animation of it. I was very happy when this one appeared, so I didn't have to make one myself.

The book "Thinking Physics" uses a rectangular pipe loop, in which a heavy fluid circulates. If you try to rotate the pipe loop around one axis Z, it is obvious that fluid on opposite sides along Z will react against the pipe in opposite directions perpendicular to Z, because it is moving in opposite directions in these segments.

 

[Swamp Thing]

PS... the spinning/flipping nuts in Zero G you can see linked bottom right in that first video to is also fabulous; makes a very unintuitive topic quite reachable

It's here :

www.youtube.com/watch?v=1VPfZ_XzisU