Rotation Question: How to Calculate Force for Arbitrary Axis

[Chaser]

Hi,

I've been doing some research on rotating objects and moments of inertia, and I've run into a bit of a conundrum. It seems easy to do the math on how quickly an object is turning after a torque when that torque is around either the x, y, or z axis, but not when it is an arbitrary axis. In my experience, it seems that objects tend to reorient themselves around their principal axis when a torque is applied, but I don't know what this force is called, or how to calculate it.

The only way I could describe this would be if you had a plate hanging from the ceiling by a thread attached to its center of mass, where the plate could rotate freely on the end of the thread. If a torque were applied around the thread on the plate, in my experience, it would align itself horizontally. Why, and what are the equations for this?

 

[MasterX]

Hi,

I am just thinking out loud, but I believe that I am on the right track.

As the plate rotates around the thread, three different forces are exerted on it:
1) Force from the string
2) Gravity
3) Centrifugal force

The second force will force some parts of the plate to spin faster than some others. if the plate is not horizontal. This is because an object on a higher altitude has a large dynamic energy than an object on a lower altitude. Since, you are rotating it at a constant angular velocity, the plate has to align itself horizontally to eliminate the effect that the gravity force has on rotation.

 

[ZealScience]

This case is were direction of angular momentum follows the edirection of the torque. I think it is a similar case to gyroscope, maybe you can search for physics behind gyroscope.

 

[Quinzio]

Why, and what are the equations for this?

Let's say you twist the thread by some number of turns and then you leave it free to untwist. What will happen is that the axis of the disk will align to the thread.
As one poster said, the gyroscope theory is surely a way to deal with this object, but until you are familiar with it, you'll find it a bit confusing.

I think, but I'm not really 100% sure, that the formula you search is:

\tau = { I (\mathit{\Omega}_{\mathrm{P}})^2 sin (2\theta) \over 2}

where:
I is the disk inertia about its axis
\mathit{\Omega}_{\mathrm{P}} is the angular speed of the thread at the point where it's attached to the disk.
\theta is the angle between the disk axis and the thread or, if you prefer, the angle between the horizontal plane and the plane where the disk lies.
\tau finally is the torque that make the disk try to stay in an horizontal position, if it's tilted. The direction of the torque is the only diameter of the disk which is horizontal.

Now you may be curious where does this all come from.
One way is, as said before, to look at it as a gyroscope.
The basic gyroscope equation is
\tau = \mathit{\Omega}_{\mathrm{P}} I\omega \sin\theta
Just google for gyroscope to learn why this this formula works and how.

For our spinning plate, \mathit{\Omega}_{\mathrm{P}} is the angular speed of the thread at the disk conjuction, \theta is the angle between the disk axis and the thread.
\omega is a little tricky to find but it can be shown that \omega = \mathit{\Omega}_{\mathrm{P}} cos (\theta)
this can be in some way "seen" intuitively in this way: if the plate is already horizontal, the precession and the speed of the disk are of course equal. If the disk is near vertical, the spinning of the disk (about its axis) is almost zero, because the upper point of the disk always remain in the upper position. In between there are all the intermediate speeds which go with the cosine.

This leads to
\tau = { I (\mathit{\Omega}_{\mathrm{P}})^2 sin (\theta) cos (\theta)}
or more nicely
\tau = { I (\mathit{\Omega}_{\mathrm{P}})^2 sin (2\theta) \over 2}

the tilt angle where you will obtain the maximum torque is a 45°.
At the extremes, if the disk is vertical, it will not try to get horizontal, and the same of course if the disk is already horizontal.

 

[tiny-tim]

Welcome to PF!

Hi Chaser! Welcome to PF!

… if you had a plate hanging from the ceiling by a thread attached to its center of mass, where the plate could rotate freely on the end of the thread. If a torque were applied around the thread on the plate, in my experience, it would align itself horizontally.

What experience?

My guess is that it would choose an angle, and stick to it.

(I'm very reluctant to believe that you could fix a thread to the centre of mass of a plate accurately enough to carry out such an experiment )

 

[Quinzio]

Hi again Chaser and TinyTim,
it's actually not necessary to find the exact center of mass of the object.
This is because any rotating object "free" in the space will try to rotate about his center of mass. With a small string attached, the center of mass isn't virtually changed.

I've done a short video which shows this behavior.
I was lucky to find a metal plate which is the cover of a basket, which does nicely the job.
This is a scientific forum, and it's fine to try to demonstrate things, whenever possible.
I attached a short nylon string with some tape near the center of mass.
If I were able to find the center of mass, when the plate doesn't rotate, the plate would stay horizontal, or at any angle, but it's very hard to find the exact center of mass, the plate will almost always stay tilted.

Then I twisted the string with my fingers and the plate slowly starts to spin about a vertical axis.
As it's evident, as the speed increases, the disk try to stay horizontal, and finally is horizontal. The very little torque first makes it horizontal, then it's enough to maintain it horizontally.
If the plat is horizontal, the torque disappears (sin 180°), and the plate finds its equilibrium.
Any little angle that is created by chaotic movements, it's counterbalanced by a small gyroscopic torque.
As the torque goes with the square of speed, this is evident at relatively high speed.

Enjoy.

https://www.youtube.com/watch?v=41kHq7v777Q

 

[Chaser]

Er... Wow, that has to be the most well-explained and proven answer I've ever received to a question. Massive thanks for this, it had been giving me quite the headache for about a month now. I'm actually rather surprised you knew exactly what I meant, thanks very much!

 

[Cleonis]

In my experience, it seems that objects tend to reorient themselves around their principal axis when a torque is applied, but I don't know what this force is called, or how to calculate it.

The only way I could describe this would be if you had a plate hanging from the ceiling by a thread attached to its center of mass, where the plate could rotate freely on the end of the thread. If a torque were applied around the thread on the plate, in my experience, it would align itself horizontally. Why, and what are the equations for this?

In effect what you describe is a setup that involves friction, with the friction having a significant effect.

If the connection between the plate and the tread would be perfectly frictionless (and the hinge point coinciding exactly with the plate's center of mass) then there would be no tendency of the plate to shift to rotation around its axis of symmetry.

Case 1
- Thread attached to plate center of mass. Thread attachment hinging with friction.
Then what I expect is that the friction will selectively dampen any motion component that is subject to that friction.

Case 2
- Point of thread attachment slightly outside the plane of the plate, but concentric
Now the point of attachment does not coincide with the point of gravitational attraction.
Under frictionless circumstances I expect a sustained precessing motion (gyroscopic precession)
With some friction there is some decay of the motion, and over time the axis of rotation will become aligned with the direction of gravity.

Case 3
- Thread attached slightly eccentric.
I'm not sure here, but my guess is that if the spinning is sufficiently fast the effect from eccentric attachment will even out. It then reduces to either case 1 or case 2.


You wrote:
"[...] the plate, in my experience, it would align itself horizontally [...]"
Such self-alignment can't be due to some gyroscopic effect, or other solid body mechanics effect. I mean, in the absence of friction gyroscopic precession is perfectly cyclic. You get some decay or 'aligning itself with' only if there's opportunity for friction to do some selective damping.

I think in textbooks only the idealized frictionless cases are discussed.

 

[Cleonis]

[...] in the absence of friction gyroscopic precession is perfectly cyclic. You get some decay or 'aligning itself with' only if there's opportunity for friction to do some selective damping.

It occurred to me that there's probably also an influence from air friction. That too will tend to shift the motion to rotating around the axis of symmetry, as rotating around the axis of symmetry has the least air friction.

 

[Chaser]

Quinzio's video summed up what I meant quite well. I find it hard to believe that any frictional forces from a piece of string or the surrounding air could have had a significant effect on the plate stabilizing around the axis of rotation. The pan he used is quite heavy, so the forces applied would be negligible in comparison. I mean, you can do this yourself as well. Hold an arm out at a 45 degree angle and do a spin. You can feel your arm go horizontal, and it's not from any wind or frictional pressure, since you would be able to feel either of those forces. Quinzio's equation sums that up quite nicely, and he proved it very well.

 

[Cleonis]

I've been doing some research on rotating objects and moments of inertia, and I've run into a bit of a conundrum. It seems easy to do the math on how quickly an object is turning after a torque when that torque is around either the x, y, or z axis, but not when it is an arbitrary axis. In my experience, it seems that objects tend to reorient themselves around their principal axis when a torque is applied, but I don't know what this force is called, or how to calculate it.

The only way I could describe this would be if you had a plate hanging from the ceiling by a thread attached to its center of mass, where the plate could rotate freely on the end of the thread. If a torque were applied around the thread on the plate, in my experience, it would align itself horizontally. Why, and what are the equations for this?

Quinzio's video summed up what I meant quite well.

Well, in your original post you specified the idealized case: a disk is suspended, and the point of suspension coincides with the point of gravitational attraction (thread attached to center of mass). That is, the case specified in your original post was equivalent to the no-gravity case, which is the case of a disk (or any solid object) free floating in space.

Solid body rotation
As it happens the physics of spinning free floating in space has been quite relevant recently. Understanding the mechanics of spinning solid objects was crucial for the Gravity Probe B mission. The relativistic effects were swamped by several classical effect due to imperfections of the gyroscopes. To extract the relativistic effects from the raw data the imperfection effects had to be accounted for with a level of accuracy better than had ever been done before.

What the Gravity probe B team had to deal with was that when there is dissipation of energy, then over time the spin axis will settle down to alignment with the principal axis that has the highest moment of inertia. (And to the team's surprise there was an unanticipated amount of dissipation.)

In the idealized case of no dissipation of energy this settling down will not happen. In the idealized case of no dissipation of energy there will be a socalled polhode motion that is perfectly cyclic. On the Gravity Probe B site Bob Kahn explains about the polhode story

So, if you do see a decay in the motion pattern, if you do see a gradual shift towards a particular alignment, then it follows that there must be some dissipation at work.

[...]Hold an arm out at a 45 degree angle and do a spin. You can feel your arm go horizontal, [...]

A rod and a disk don't compare well. Observations of what a rod does don't transfer one-on-one to what a disk will do.

 

[Cleonis]

In my experience, it seems that objects tend to reorient themselves around their principal axis when a torque is applied [...]

Let me put this experience in a wider context.


You get decay effects only when the physics involves some process where the entropy increases, such as grinding or radio-active decay.

With the mechanics of a spinning solid body there is no opportunity for increase of entropy.
(That is, in the idealized case of perfect rigidity. In the real world there is very often some non-rigidity, offering a pathway for dissipation.)


A short push from some arbitrary torque sends a solid object spinning. The spinning solid object will exhibit a polhode motion. The polhode motion is cyclic. That is, the polhode motion does entail continuous reorientation, but the polhode motion doesn't settle down; it's cyclic.

In the real world we don't ever get to experience that idealized case. We always see friction effects mixed in. I think it's easy to underestimate the friction effects. The effects from the friction are cumulative. We can get a feel of the instantaneous amount of friction force, but how that adds up over time is another matter.


Summerizing:
A tendency to settle down towards a particular orientation is a red flag. Decay can happen if and only if there is opportunity for increase of entropy. If you see some settling-down-towards, you're seeing friction at work.

 

[Morcam]

A short push from some arbitrary torque sends a solid object spinning.

Alright, I can understand this in a system where there is no outside torque/energy remains constant, but what about when there is a constant outside torque being applied? Will there be a mixture of polhode motion and motion induced by the torque?

 

[Quinzio]

The torque applied by the thread at the beginning increases the angular speed, then it's just enough to maintain the speed.
Because the disk is slowed down by friction. Of course there is friction in action.
But the friction is not the cause of the horizontal position of the disk.
The cause of the horizontal position is the vertical direction of the thread and the vertical action of the gravity. The gravity plays an important role here because the thread is not perfectly attached at the CoM but near it.
This particular configuration make the gyroscope/disk having a precise relation between precession and spinning speed, that is w_{thread}\ sin \theta = \Omega_P and w_{thread}\ cos \theta = \omega, the speed about the axis of the disk.

The gyroscopic torque is applied very very very gently and smoothly. If we call y the axis of the disk parallel to ground, w_y is aòlmost zero. That is the gyroscopic torque doesn't produce any significant change in the precession speed and direction, but slowly and smoothly challenges the gravity torque. The effect is that the disk slowly goes horizontal, and the only reason here is a very little gyroscopic torque.

In addition, the precession of the disk/gyroscope doesn't produce any complicated polhode behavior. The pohode path is just a circular motion of the angular velocity vector.
As is clearly explained here:
If the rigid rotor is symmetric (has two equal moments of inertia), the vector \omega describes a cone (and its endpoint a circle). This is the torque-free precession of the rotation axis of the rotor.
http://en.wikipedia.org/wiki/Inertia_ellipsoid

And this is precisely what the plate axis does in my video when it's rotating. It's axis describe a cone. Of course when the disk rotates fast, the cone is narrow and it's impossible to spot it. As the plate spins faster, the cone narrows.
The real example of the plate I gave in the video is slightly more complex, because the cone is larger due to the gravity torque.
The angle of the cone is relatively simply com calculate for standard objects. For example, a a spinning sphere. Let a spinning sphere have a material massless rod in its axis. If two objects hits the rod such as to create an angular precession, the rod will describe a cone, whose angle is: \theta= arctan({\Omega \over \omega}).
For a disk, since the inertia about its axis is the double respect a perpendicular axis, the cone described will be narrower, like \theta= arctan({\Omega \over 2\omega}).

The formula I gave for the torque that keeps the disk horizontal against the gravity toruqe is exact and it finds references in books about the gyroscopic motion.

The same torque can be obtained by computing the centrifugal forces that acts on the spinning plate of the video, and the result is exactly the same.
It's an esay exercise to compute the centrifugal forces on the spinning tilted disk.
In real honesty, I cannot understand why the motion of a simple disk can create so much perplexity.
The polhole mode is not present here because the torque is increase very slowly as I said before, and the almost inevitable friction present immediately stops the very small movement that make the disk horizontal. This is evident because the reorientation of the disk is not perceived as a motion, but as a very slow change of position.

And finally, frankly, comparing my plate to the probe B gyroscopes is really funny.
The probe B gyroscopes are the most precise gyroscopes probably ever built with has an eccentricity of no more than 1/1.000.000. That means that they are perfect spheres. The ration y the biggest and the smallest ratio is, as said, 1.000001.
The plate I used has nothing to do with those objects, and every comparison will just lead to useless discussions.
The centrifugal forces on the probe B gyro is so small that it take 14 months for the polhode path to complete.
The plate has a ratio 2 between its classic inertia axis, and the polhode path is so quick and little that it's impossible to see it.
I think some people should revise the basic of the gyroscopes before lauching, useless and pointless discussions.