PDA

View Full Version : Question about solid object rotation

rotate
01-03-2009, 12:38 PM
When a solid body rotate in space and is free of any external forces (including gravity), is the following possible.

1. can it rotate about an axis which does not go through the center of gravity?

2. can there be two different axes of rotation, i.e. can it wobble?

I think the answers to both questions is no, but my friend thinks that it's possible since the earth rotates in a wobbles.

websterz
01-03-2009, 01:33 PM
How are you going to free an object from ALL external forces? However faint, even in the deepest reaches of space gravity WILL affect an object. I believe the question(s) to be physically impossible to answer.

rotate
01-03-2009, 01:53 PM
This is a physics question and not an engineering one. Along the same line, solid objects don't exists in reality either. All real objects bend, twist, and compress under it own gravitational force, but "solid object" used in dynamic analysis does not.

Astronowanabe
01-03-2009, 02:09 PM
hmmm...
can an object have a center of gravity if it is free of the force of gravity?

a wobble is a change in the axis of rotation over time
so maybe it is easier to describe as an axis of rotation
having its own different axis of rotation
or more likely, orbit or path of movement.

but I think any two rotations an object *may* have can be combined into one
(not longer) rotation just as any two connected line segments can have a
not longer line segment connecting them. (triangle inequality)

Rich Carlstedt
01-03-2009, 02:09 PM
1.
yes, all "continuous" rotation must occur through its "apparent" center of gravity (center of mass)
This is confusing because of structure appearance. when its a ball, all mass lies within the structure and so no rotation occurs outside the structure
If the structure look like a ring, then indeed the rotation is about it's apparent center of mass,
Consider a Boom-a-rang for example
2. Yes, but the key words are "free of external forces"
what about internal forces. consider a airplane in a Lomchevik which is erratic tumbling. You may say that "air" causes the tumble and that is a outside influence. But the engine provides a "center of thrust" and is internal and also does not coincide with the center of gravity, and depending on direction, changes the internal forces.
The difficulty of a correct answer is compounded by the object. For instance, is it a waterballon ?. such rotation is dramatically affected by the struture, which can be solid, or liquid, or.....a gas.
Defining the exact nature of the mass, and the conditions as needed, like in a vacuum, or in air...or water ?.
It's a trick question of sorts, as a "solid" does not imply a "rigid" structure ie. one that cannot change

By the way, the earth may wobble, but it is never free of influences

ckelloug
01-03-2009, 02:14 PM
If the system is an object like a rock rotating in space, I believe it can rotate about an axis that is not through the center of gravity but that motion won't be stable. IIRC, the motion will eventually stabilize to a motion around one of the principle axes but my memory is a touch fuzzy on this right now. (added after I originally posted) It is however true that you can resolve any complex rotational motion into rotations about the principles axes which do pass through the center of mass. Because moment of inertia, torqe and angular momentum are actually tensors, you can set up the coordinates so as to consider the rotation as being about any set of orthogonal axes located anywhere in space. Any tensor formulation is equivalent to any other tensor formulation however so any rotation can also be regarded as rotations about the principle axes.(end addition)

See this abstract of a paper about the mir space station.
http://library.wolfram.com/conferences/conference98/abstracts/rotational_dynamics_of_mir.html

In the case of 2, there is instantaneously only one axis of rotation although this axis may vary continuously leading to a wobble.

See this fascinating movie from the International Space Station dug up with a bit of accidental google-fu. It demonstrates a rotating physics book in space and the stability about the maximum and minimum principle axes and the instability around the third axis.

http://mix.msfc.nasa.gov/abstracts.php?p=3873

Regards,

Cameron

rotate
01-03-2009, 02:25 PM
I should clarify that the question is posed as a thought experiment, done in space (vacuum) devoid of any other matter or external forces.

Rustybolt
01-03-2009, 03:18 PM
I would think the answer would depend on the mass of the object. Wouldn't the rotation eventually wind up rotating around the area of the greatest mass? Cross sectional density?

barts
01-03-2009, 03:43 PM
An rigid object free of external forces will rotate about its principle axes; these are the 3 orthogonal axes about which the object is has no products of inertia - ie is in dynamic balance. These axes intersect the center of mass.

In the case where the inertias about the principle axes are such that

Ixxx > Iyyy > Izzz, the object will not rotates in a stable fashion about Iyyy; this can be shown using a tennis racket - it is impossible to get a tennis racket to continue spinning about the axis that is parallel to the face of the racket - it will always start tumbling. This instability about the second axis shows that wobble is an inherent part of free-space rotation; rotation can and will change from axis to axis; energy is still of course conserved, but stability is not a requirement; torque-free precession can and will occur in non-symmetric objects.

- Bart

Jpfalt
01-03-2009, 04:09 PM
Engineering is the application of physics.

Answer to question 1: No. If an object is not acted on by external forces, It sill rotate around the center of mass.

An object can rotate around several axes at the same time, However, each axis must pass through the senter of mass.

The earth wobbles because it is part of a larger system (the solar system) with forces acting between ek]lements of the system. Also, because of the external acting forces, the senter of the earth's mass is also shifting and oscillating due to tial forces from the moon, sun and etc.

jacampb2
01-03-2009, 04:27 PM
Since the hypothetical situation can never exist, at least not with any method of observation, as that would impart some form of external force, then what is the point in the question, even as a thought exercise? If it can never happen, then who the heck cares what would happen? I am sure someone does. It is kind of like the old "if a tree falls in the forest and..." bs, it doesn't matter, and even if it did, there is no way to prove your hypothesis.

BTW, the ultimate answer is 42

Later,
Jason

Fasttrack
01-03-2009, 05:09 PM
Now wait a minute guys... you mean to tell me that you really believe that the minute traces from gravity are going have any effect at all? Take a rock of, say 50 lbs, traveling through space. The strength of the force of gravity falls off very very rapidly, and, within a relatively short distance, space will again become flat. Observing the object is going to have nearly no effect either, the momentum imparted by a low intensity beam of light is miniscule compared to its motion. You have to be realistic when talking about this stuff.

What you have to realize is that the effect from these little tidbits is not significant enough to hurt the accuracy of the answer. When I say that the object will eventually decay into a rotation about a center-of-mass axes, there is a certain amount of uncertainty in the posistion of the center of mass, since we cannot measure it to an infinite degree of accuracy. The error involved with that measurement is far greater than the effects of those little tidbits will have on the motion of said rock.

So it is a perfectly valid question and assumption. Physicists are often accused of over-simplifying problems by saying "assume it is in vacuum" or whatever. The truth of the matter is, we just view it realistically. Engineers over-complicate the situation. :p :D

But no, it will not expierence a real wobble or a rotation about any axis besides its CoM axis. Since there is not frame of refrence, we can choose the set of orthogonal axes such that the rotation is only about one and the origin of the set is at the CoM. Think about throwing a ball, no matter how hard you try, you can't make a baseball go in a full circle. It might curve, but its motion decays to a straight line path and, if it spins, it spins about an CoM axis. To keep it in a circular motion, there must exist a centripedal force to bend its path (or at least space must be non-flat).

edit: just for clarification, in keeping with what others have said, the object may appear to be tumbling if it is non-uniform, but we can always choose an axis of rotation that is stable and passes through the CoM

Evan
01-03-2009, 05:17 PM
All rotation occurs around the center of mass. It cannot be otherwise since if it were the center of mass would be subject to acceleration with no source.

An isolated object may have spin in all three axes at the same time. As far as stability goes as long as there are no external forces present and the center of mass is fixed it will remain stable regardless of shape. Precession and nutation as displayed by the Earth are the result of force couples between the Earth and the moon with contributions by the Sun, Jupiter and the rest of the mass in the solar system, galaxy and universe in that order.

To correctly describe the motion of the Earth around the sun in it's orbit requires the consideration of about 130 periodic terms that are present because of the influence of other local celestial bodies. These terms can fortunately be summed and averaged to reduce the numbe to a more manageable level. Years ago I wrote a stand alone program to calculate the position of the Earth for the purpose of celestial navigation and was able to produce numbers that agreed with the US Naval Observatory over a 5 years span within a error band that results in navigation accuracy of about 1000 yards.

rotate
01-03-2009, 05:32 PM
Since the hypothetical situation can never exist, at least not with any method of observation, as that would impart some form of external force, then what is the point in the question, even as a thought exercise? If it can never happen, then who the heck cares what would happen? I am sure someone does. It is kind of like the old "if a tree falls in the forest and..." bs, it doesn't matter, and even if it did, there is no way to prove your hypothesis.

The very nature of scientific investigation (at least in physics) is that the observer does not participate or influence the the outcome of the experiment. Assuming that the observer's mass is not part of the picture is both a standard practice and necessary.

As to who would cares, this is something that would have been covered in undergraduate engineering or physics courses (I slept through them), and as the video shows, it's very relevant in understanding kinematics in places like space station and satellites.

Evan
01-03-2009, 05:48 PM
Since the hypothetical situation can never exist, at least not with any method of observation, as that would impart some form of external force, then what is the point in the question, even as a thought exercise? If it can never happen, then who the heck cares what would happen?

When investigating physical phenomena it is very useful to pose questions in a way that eliminates factors that have nothing to do with the matter under consideration. This is true even if those factors can never actually be eliminated. Most often this will be manifested in scenarios that propose at the outset that the normally unavoidable losses do not exist. The simplifies calculations by removing what would otherwise cause very small but important sources of error in the calculations. In a wide variety of problems disposing of the inconvenient but unescapable physical realities introduces no error of fact to an idealized situation.

Peter.
01-03-2009, 05:49 PM
Since the hypothetical situation can never exist, at least not with any method of observation, as that would impart some form of external force, then what is the point in the question, even as a thought exercise?

The only requisite for asking a question is curiosity. That a condition can never exist has no bearing on it if a person is curious about a hypothetical condition.

barts
01-03-2009, 09:05 PM
An isolated object may have spin in all three axes at the same time. As far as stability goes as long as there are no external forces present and the center of mass is fixed it will remain stable regardless of shape.

Actually, depending on the object's symmetry the rotation may not be stable. All rigid objects have 3 orthogonal principle axes of inertia, about each all products of inertia are zero, e.g. the object is in dynamic balance about each axis. If the magnitudes of the inertia about each axis are such that Ixx < Iyy < Izz for arbitrary selection of x, y, and z, rotations purely about Ixx and Izz will be stable, but any rotation about Iyy will be unstable, and results in precession of the angular velocity vector; e.g it will appear to tumble.

- Bart

Evan
01-03-2009, 11:15 PM
It may appear to tumble but the motion will not be random. It shouldn't be chaotic either as there is no attractor although it may appear chaotic.

J Tiers
01-04-2009, 12:46 AM
All rotation occurs around the center of mass. It cannot be otherwise since if it were the center of mass would be subject to acceleration with no source.

Bingo.

That is the ONLY possible answer.

ANY rotation aside from around center of mass MUST have some external perturbation affecting it.

As Evan points out, where IS that external perturbation? Nowhere. It is disallowed by the conditions of the problem.

I don't think any possible tumbling affects that.... it is subject to the same rules.

barts
01-04-2009, 01:51 AM
It may appear to tumble but the motion will not be random. It shouldn't be chaotic either as there is no attractor although it may appear chaotic.

This is correct; the precession is simple for symmetric objects; the angular velocity vector is seen to describe a simple cone. Of course, since this is the velocity rather than position, the position of the object vs time becomes a rather interesting function, and will be very difficult to predict, I think since any error in initial conditions or inertia measurement will completely change the result. For less regular objects, the path of the angular momentum vector will be more complex.

- Bart

Evan
01-04-2009, 05:05 AM
One of the areas of mathematics that interests me is that of intractable problems. These are problems that have definite solutions but are too difficult to solve for some reason. One such problem is calculating the orbits of three non infinitesimal bodies orbiting each other. We can solve for some special cases but there is no approach that has been found to predict such orbits in the unrestricted general case. A solution was found and verified to be correct in the early 1900s but it is only a mathematical curiosity since it would take nearly forever to converge on a solution.

The interesting part is that there are quite a few intractable problems that can be easily solved without using mathematics. One such problem is finding the point on a plane that represents the location that has the shortest possible distance to n points on the same plane. As the number of points grows the number of calulations that must be done increases as the factorial of n.

However, if we stick pins in a board to represent those points and tie them all together using equal lengths of rubber bands the are hooked over a pin at one end and tied together with the other bands at the other end something unexpected happens. The junction where all the bands are tied together automatically settles at the point which represent the optimum solution.