# Coriolis effect in gyroscopes

1. Jan 22, 2010

### vin300

How does the coriolis effect have role in the precession of gyroscopes?http://en.wikipedia.org/wiki/Precession" [Broken]
In the torque-induced precession section, it says masses move due to the coriolis effect, but it happens only in rotating frames.
I would want an explanation on why a torque applied to a rotating gyroscope should cause precession about an axis perpendicular to that of the applied torque and spin axis

Last edited by a moderator: May 4, 2017
2. Jan 22, 2010

### D H

Staff Emeritus
I assume you read the (IMO lousy) wikipedia article or precession, which mentions coriolis force in its explanation of torque-induced precession. That torque-induced precession has nothing to do with coriolis effect. The coriolis effect only appears in rotating frames. If you want to describe motion in terms of Newton's second law in such a frame, objects moving with respect to that frame will appear to be subject to a force normal to the velocity vector. There is a rotating frame in this problem -- a frame based on the rotating gyroscope. However, the gyroscope itself is stationary in this frame. No part of the gyroscope is moving; there is no coriolis effect in the gyroscopic frame. There is no coriolis effect in the external frame, either; it is presumably an inertial frame.

Forget about the rotational motion for a bit. Think about a moving point mass. A force that is parallel (anti-parallel) to the point mass' velocity vector will cause the point mass to speed up (slow down). A force that is normal to the velocity vector will not change the magnitude of the velocity vector. It will instead change the direction of the velocity vector.

The analogy in rotational motion is that a torque parallel (anti-parallel) to a rotating object's angular momentum vector will increase (decrease) the object's angular momentum. A torque perpendicular to the angular momentum vector will not change the magnitude of the angular momentum vector; it will instead change it's direction.

3. Jan 22, 2010

### A.T.

The easiest way is to consider two connected point masses spinning around their center of mass. Let's say spin axis is vertical Z, and the connection is horizontal X in certain moment. So one mass moves to +Y and the other to -Y.

Now you apply a torque around Y, by pushing the +Y mass up to +Z, and the other down to -Z. So they start moving to (+Y,+Z) and (-Y,-Z). Which means that the plane of rotation was rotated around X, which is perpendicular to the (old) spin axis Z and the applied torque Y.

Last edited: Jan 22, 2010
4. Jan 22, 2010

### Cleonis

Well, in the case of gyroscope physics there are inertial effects that come from moving towards an axis of rotation or moving away from it. The classic example fo that is one of an ice-skater who starts spinning with outstretched arms, and then pulls her arms close to her body. That gives a massive spin-up.

In engineering the way that inertia is involved in that spin-up is referred to as 'Coriolis effect'. In helicopter engineering one has to deal with the fact that when the blade bends up its center of mass moves closer to the central axis of rotation, which causes the blade to speed up. If you google the combination of 'coriolis effect' and 'helicopter' you will find that engineering usage.

(In physics the way that the expression 'Coriolis effect' is used has shifted. Originally the meaning in physics was the same as in engineering (it entered physics via engineering), but in the past decades the way the expression is used drifted to 'only in rotating frames of reference'.)

On my website I present a http://www.cleonis.nl/physics/phys256/gyroscope_physics.php" [Broken], illustrated with some screenshots from youtube videos, and some diagrams.

http://www.cleonis.nl

Last edited by a moderator: May 4, 2017
5. Jan 23, 2010

### D H

Staff Emeritus
One problem with this comparison: The inertia tensor of an ice skater changes. The inertia tensor of a gyroscope does not.

I did just that (googled 'coriolis effect' and helicopter'), and all I can say is "yech!"

This muddies two very distinct effects -- that of a changing inertia tensor ($(dI/dt) \omega$) and that due to velocity ($-2\omega \times v$) in a rotating frame.

Last decades? Last century. I have several thirty to fifty year old physics texts (I'm an old fart; yesterday was my 55th) and they all refer to the Coriolis effect as the ω×v effect rather than the Idot effect. In fact, that the Coriolis effect is the ω×v effect rather than the Idot effect is far older than my aged flatulence. Here is a 1912 book: http://books.google.com/books?id=zXkRAAAAYAAJ&pg=PA317. The discussion starts at the bottom of the page and continues for several pages.

Nice job! One critique: You might want to rethink this statement:

Last edited by a moderator: May 4, 2017
6. Jan 23, 2010

### Cleonis

No need to rethink; the statement is correct:
The following two cases are distinct:
- a force that sustains a speed by compensating for friction
- a force that sustain a dynamics configuration

- If you're milling flour, then you are exerting a tangential force to sustain rotation by compensating for friction.

- If you have two objects in the vacuum of space, connected by a cable, circling their common center of mass, then the radial force that is exerted by the cable does not sustain the angular velocity. Of course the centripetal force does sustain the dynamic configuration in which there is an anguler velocity in the first place, but that is quite distinct from sustaining velocity itself.

Let me add how this translates to the case of gyroscopic precession.
The precessional motion does not experience friction, therefore no sustaining force is required for the precessional motion itself. That said, sustaining the dynamic confuguration in which there can be precession in the first place does require a torque. But that is quite distinct from sustaining the velocity of the precession itself.

Last edited by a moderator: May 4, 2017
7. Jan 23, 2010

### A.T.

I second that "yech!" :yuck:

You can actually go back to Gaspard Gustave de Coriolis himself:

Page 377, last sentence of 3 paragraph: Coriolis describes

$-2m \omega \times v$
with words as the second type of inertial forces in a rotating frame (first type being centrifugal forces). That second type was later named "Coriolis force".

8. Jan 23, 2010

### D H

Staff Emeritus
That statement is incorrect.

That is just wrong.

Think of it this way: If what you are saying is correct, continued application of a torque normal to the angular velocity vector would cause the precession rate to build up linearly over time. That is not what happens. What happens instead given a torque normal to the angular velocity and (ideally) zero friction is that the gyroscope precesses at a constant rate that is proportional to the external torque.

Another way to look at it is that sans any external torque the angular momentum of the gyroscope must remain constant. A precessing gyroscope does not have a constant angular momentum vector. The angular momentum vector, like the angular velocity vector, is precessing.

That said, there is a torque-free precession due to (Iω)×ω. The angular velocity and angular momentum of a precessing top are not parallel to one another. Take away the external torque and the top will undergo this torque-free precession. This torque-free precession is similar to the nutation sometimes observed in a spinning top. It is a much smaller effect than the torque-induced precession.

9. Jan 23, 2010

### Cleonis

Yes, I agree those are fundamentally distinct things, but there is something interesting.

If a mass is circumnavigating a central axis, and a force pulls the mass somewhat closer to the central axis, then how much will the tangential velocity change? To calculate that we can apply conservation of angular momentum. We have that angular momentum is conserved.

$$\omega r^2 = constant \qquad \Rightarrow \qquad \frac{d(\omega r^2)}{dt} = 0$$

Differentiating:

$$r^2 \frac{d\omega}{dt} + \omega \frac{d(r^2)}{dt} = 0$$

using the chain rule to obtain an expression in terms of dr/dt:

$$r^2 \frac{d\omega}{dt} + 2 r \omega \frac{d(r)}{dt} = 0$$

Rearranging, dividing left and right by r, and substituting dr/dt with vr, (vr is velocity in radial direction) we obtain the sought answer:

$$\cfrac{d\omega}{dt} r = - 2 \omega v_r$$

You can evaluate this expression using an inertial coordinate system or a rotating coordinate system. Either way the result is the same, because:
- the radial velocity is the same in both coordinate systems
- the angular acceleration is the same in both coordinate systems

Interestingly, the derived expression is identical in form to the expression for the Coriolis term.

Last edited: Jan 23, 2010
10. Jan 23, 2010

### Cleonis

You are attributing a point of view to me that isn't mine. Your disagreement is not with me.

Obviously, continued application of a torque normal to the angular velocity will not cause the precession to build up linearly over time.

Think of it this way: if you have piece of string, and you swirl an object around then there is a continued application of a force (perpendicular to the instantaneous velocity). The velocity of the object will not build up over time. That is the appropriate analogy.

The analogy between linear dynamics and rotational dynamics can be very helpful. But it is unclear how you frame that analogy. You offered an: "If what you say is correct then, [...]" and what followed was in contradiction of what I had stated about the subject.

What is happening here is that you are criticizing a point of view that is not held by me.

11. Jan 23, 2010

### D H

Staff Emeritus
If you cut the string the object will fly off with a constant velocity (momentum). It will cease following a circle path. Continuing the analogy, if you remove the external torque the gyroscope will now have a constant angular momentum. It will cease precessing.

12. Jan 23, 2010

### Cleonis

This is interesting.

The http://www.youtube.com/watch?v=zLy0IQT8ssk" video is the most vivid one that I know of. To reproduce what is seen in that video we must add a moderate degree of dampening in the process of introducing a torque (and when removing that torque again). At 38:20 into the video a demonstration starts in whichProfessor Lewin adds and removes torque gingerly, so as not to cause violent rocking (a nutation). What he does has the same effect as applying critical dampening; just enough dampening for quickest settling to a steady state.

The full story is in the article on my website. (You can read the discussion on my own website, or I can copy and paste here, but the article is illustrated with diagrams.)

Establishing conventions:
Spinning of the gyroscope wheel => Roll
Rotation around an axis parallel to gravity => swivel
Perpendicular to both above => pitch.

To establish basics, check out the following http://www.youtube.com/watch?v=NQSYERWASE4" that (as far as I know) was uploaded by Glenn Turner.
The gyroscope wheel is spinning, and using just his fingertips Glenn is gently swiveling the gyroscope. The gyroscope swivels readily. What the gyroscope wheel actually does is that it adds a motion to the swiveling motion; a pitching motion is added

What that illustrates is a key property. The response to torque can be distinguished in two responses. The direct response to a torque around the swivel axis is swivel motion. As a consequence of the swivel motion there is pitching. The pitching is only indirectly a consequence of the torque. There will be pitching if and only if the torque around the swivel axis actually induces swivel motion.

Now back to the Professor Lewin demonstration:
At 38:20 into the video the bicycle wheel is precessing at a rate of about one revolution per 10 seconds, then an extra weight is added, the extra weight is added gingerly.

Placing the extra weight applies a down-pitching torque. The direct response to that down-pitching torque is down-pitching motion (The down-pitching motion is very short-lived.) As a consequence of the downpitching motion the existing swivel motion is increased. In other circumstances that extra swivel motion would result in pitching up. But here there is the down-pitching torgue from gravity that started it all in the first place. The steady state of precessing is one where gravity is prevented from pitching the wheel down further by precession-caused tendency to pitch the other way.

The process is self-adjusting. Let's look at what happens when Professor Lewin removes the extra weight. At that instant there is less down-pitching torque, but the same (faster) precession rate is still there, so the bicycle wheel will pitch up. The motion of pitching-up decreases the existing precession rate. As the precession rate slows down, it reaches the point where the precession rate is once again the amount that is required to prevent gravity from pitching the wheel further.

This explains why precessing motion of a gyroscope wheel comes to a complete stop when the external torque is completely gone.

Working out the analogy for a point mass circling a central axis is straightforward.

Cleonis
http://www.cleonis.nl

Last edited by a moderator: Apr 24, 2017
13. Jan 23, 2010

### A.T.

Not for the entire Coriolis term, just its tangential component, which depends on the radial velocity component in the rotating frame. The Coriolis force has also a radial component which depends on the tangential velocity component.

But you are right, these are just two ways to describe the same inertial effect:

- In the inertial frame the cause of the angular acceleration is the change of the moment of inertia.

- In the initially co-rotating frame the cause of the angular acceleration is the tangential component of the Coriolis force.

Yet it is important to keep them apart, and not to talk about Coriolis force outside of the rotating frame context. The same applies to centrifugal force.

14. Jan 23, 2010

### Cleonis

Let me derive the general case. I found it helpful to notate the equations in parametric form.
Let a moving mass, free to move in 2 dimensions of space, be subject to a centripetal force $$-m\Omega^2r$$, where $$\Omega$$ is the angular velocity of the rotating system. Note that the centripetal force is proportional to the distance to the central axis and directed in the negative r-direction, directed back to the center. As is well known, a force with that property is called a harmonic force.
Then the following parametric equation is a solution to the equation of motion:

$$\begin{matrix} x = a \cos(\Omega t) \ \ \ (1) \\ y = b \sin(\Omega t) \ \ \ \ (2) \end{matrix}$$

In the general case the motion will follow an ellipse (and the case where a=b is of course motion along a circle.)
The above parametric equation can be decomposed as follows.

$$\begin{matrix} x = \left(\begin{matrix}\frac{a+b}{2}\end{matrix}\right)\cos(\Omega t) + \left(\begin{matrix}\frac{a-b}{2}\end{matrix}\right)\cos(\Omega t) \\ y =\left(\begin{matrix}\frac{a+b}{2}\end{matrix}\right)\sin(\Omega t) - \left(\begin{matrix}\frac{a-b}{2}\end{matrix}\right)\sin(\Omega t) \end{matrix}$$

This shows that motion due to a harmonic force can always be decomposed in two circular motions. One with radius (a+b)/2 and one with radius (a-b)/2

I attach an animated gif that illustrates the decomposition. I will refer to the small circle as 'the epi-circle'.

So, for the motion relative to the co-rotating coordinate system, what is the magnitude of the acceleration towards the center of the epi-circle?

re => radius of the epi-circle
ve => velocity of moving along the epi-circle
ae => acceleration of moving along the epi-circle
(ve and ae are relative to the co-rotating coordinate system)
For the motion relative to the co-rotating coordinate system there are two cycles of motion along the epi-circle for every overall cycle of rotation.

$$v_e = 2\Omega r_e \quad => \quad r_e = v_e / 2\Omega$$

and:

$$a_e = -(2\Omega)^2 r_e$$

substituting re :

$$a_e = -2\Omega v_e$$

This proves that if there is a centripetal force with strength given by $$-m\Omega^2r$$, then for velocity in any direction the acceleration (relative to the co-rotating system) is given by $$2\Omega v$$, (where v is velocity relative to the co-rotating coordinate system).

Conclusion:
The radial component and the tangential component of the acceleration factor are identical:
- both are given by to $$-2 \Omega v$$
- both are perpendicular to the instantaneous velocity

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15. Jan 23, 2010

### A.T.

I'm sure you mean the right thing. But it sounds a bit weird when you talk about the "components of a factor" which are both perpendicular to velocity. Aren't they perpendicular to eachother, while their sum is perpendicular to velocity?

The way I visualize it:
- The tangential Coriolis acceleration is proportional to the radial velocity.
- The radial Coriolis acceleration is proportional to the tangential velocity.
This implies that the Coriolis acceleration is perpendicular to the combined velocity.

16. Jan 23, 2010

### Cleonis

Yeah.

What happened is that I was juggling all sorts of decomposition. The parametric equation in x and y is a decomposition, and tangential and radial is another decomposition.

The interesting observation is that what we know as the 'Coriolis term' arises identically in two contexts:
- In the case of working out coordinate transformation to a rotating coordinate system
- In the case of working out the dynamics of motion under the influence of a $$-m\Omega^2r$$ centripetal force.