# Non-Uniform Spinning Sphere Stabillity

1. Sep 10, 2009

### rkelley192

I am trying to find the best way to determine whether a spherical shell would stabilize in a certain orientation when falling.
Assumptions I need to take into account:
The sphere is spinning with some random angualr velocity for each axis
The density of the shell is higher at the +x extreme and decreases until the density becomes uniform for the -x half of the shell. This density gradient can vary.

It seems like it should be simple, but I must be missing something. That's what happens when a mathematician pretends to be an engineer.

2. Sep 10, 2009

### FredGarvin

I don't know if I would call this a simple problem. The again, dynamics was not a real strong suit for me.

As a first pass, have you calculated the principal rotational axes? But I am thinking if the angular velocities are truly random, then it can't be possible to predict.

3. Sep 10, 2009

### rkelley192

About the only thing that seems logical is that the sphere is going to be pre-disposed to orienting in a heavy side down configuration. It seems that looking at the x,y, an z components of the rotation is the best approach, but beyond that, I am at a loss.

4. Sep 10, 2009

### kote

If you ignore air resistance, it will stabilize in a spin around an axis through its center of inertia, such that the moment of inertia is maximized around the axis, and such that the initial angular momentum is conserved.

In your case, spin will stabilize around an axis orthogonal to your x direction (lets call it the y axis). The center of inertia, which is where the y axis and x axis will intersect, if we set this point to x=0, will be the point where the moments of inertia on either side of x=0 are equal. In other words

$$\sum \left(X\times\left| X\right|\times M\right)=0$$

Edit: Okay, I had to change the formula since X^2 obviously doesn't work. I realize it's terribly ugly now, so let me know if I need to reformulate it! I couldn't find any quick easy links for this one.

For the purposes of finding this location you can think of all of the mass condensing on the x axis and you can ignore the other coordinates. M here is each individual and equal element of mass. Please forgive the discrete notation. I'm half remembering and half making this up as I go .

This effect is seen in gyroscopes and bicycle wheels etc. A wheel spinning in a wobbly fashion will orient itself in a smooth spin around an axis at its center that is orthogonal to the plane on which r^2 x m is maximized, where r is the radius. And of course, the total initial angular momentum is always conserved.

Basically, due to centrifugal force, as much mass as possible wants to get as far from possible from your axis of rotation.

The actual torque that will cause this stabilization can be calculated as well. I'd start with a look at how precession torque works.

If you factor in air resistance then it will orient itself with the densest part down, as the less dense part will be pushed up more easily by the force of the air resistance. If you factor in friction, it will eventually stop spinning around any axis. If you factor in turbulence etc it will add some amount of somewhat chaotic motion.

Last edited: Sep 10, 2009
5. Sep 10, 2009

### Bob S

Every object has three principal moments of inertia. In general, these three moments are different. If your object is spinning about the intermediate axis, it is inherently unstable and liable to flip at any time. Read the section on Euler's equation in
http://en.wikipedia.org/wiki/Polhode
Watch the brick (with blue side up) spinning in the video
http://www.physics.rutgers.edu/~shapiro/classmech/polhode.mpg
"The Polhode Rolls without Slipping on the Herpolhode Lying in the Invariable Plane"

6. Sep 10, 2009

### kote

Thanks for the links. I hadn't heard of this before. I don't have enough experience to tell if this means what I said above is incorrect (although this information is certainly more detailed). It looks to me like this fits with what I said. I found this part particularly interesting:

If energy is dissipated while an object is rotating, this will cause the polhode motion about the axis of maximum inertia (also called the major principal axis) to damp out or stabilize, with the polhode path becoming a smaller and smaller ellipse or circle, closing in on the axis.

A body is never stable when spinning about the intermediate principal axis, and dissipated energy will cause the polhode to start migrating to the object’s axis of maximum inertia. The transition point between two stable axes of rotation is called the separatrix along which the angular velocity passes through the axis of intermediate inertia.

Rotation about the axis of minimum inertia (also called the minor principal axis) is also stable, but given enough time, any perturbations due to energy dissipation or torques would cause the polhode path to expand, in larger and larger ellipses or circles, and eventually migrate through the separatrix and its axis of intermediate inertia to its axis of maximum inertia.

http://en.wikipedia.org/wiki/Polhode

7. Sep 10, 2009

### Bob S

Here is a good DIY demonstration of the polhode rolling on the herpolhode.
Get a good textbook, preferably one with art on the front cover.
Put a rubber band around it to keep it from flying open.
Hold it using both hands, with the front cover upright and the binding on the left side.
Gently toss the book up, spinning it on axis across the front cover (rubber band axis).
Let it spin 3 or 4 times, and catch it. (This is easy)
Is the book binding always on the left side? Why not?

Once you've mastered the art of doing this, ask your ME professor why.