There was no need to focus on the fact that the center of the ellipse
does not lay on the center-line of the (circular) cone.
Instead I focused on the much more practical and easy problem
of determining empirically whether or not the near edge of a circle
(or ellipse for it is the same result) does indeed pull "off-center".
For that, we have the following perfectly accurate diagrams,
which will work equally for an ellipse as for a circle.
Try to follow the simple argument below.
I would say it is self-evident "by inspection", but clearly that seems
not to be the case for some:
5. How the Center of Mass Theorem Fails The balance of forces from all parts of the disk is only strictly maintained along the axis of rotation. Only from a spot along this axis does the distance to moving parts stay constant.  For test-masses facing the spin, the movement can be roughly represented simply as a pair of point-masses rotating around the GC in the same plane.  Although changes in distance seem balanced, the forces acting on the off-center test-mass are not balanced at all, since they follow the inverse-square law. To show this without mathematics, you can simulate the gravitational field using a compass and a ball of unmagnetized iron in the same horizontal plane. If the force between ball and south pole, and between ball and north pole were equal regardless of compass swing (corresponding to tilt above), the needle would sit wherever you placed it, instead of turning toward the ball. Magnetism also decreases as the square of the distance, and gives us a feel for gravity. Gravity is simply too weak between small objects to measure at home, without extreme cleverness. 6. Watching the Direction of Force change We can easily show the change by using the parallelogram rule. The final vector will always point toward the true location for an equivalent point-mass ( EPM ), and this is not usually at the GC. During a tilt, the closer part (a) will always be a greater angle from the test-mass, since both (a) and (b) are always an equal vertical distance from the GC.  For the final vector to point along the GC line, the furthest part would have to give the greatest force, to counteract the smaller angle. This is clearly impossible.  This is independent of the inverse-square law, and is true for any force which decreases with distance. 7. Tilted Disk Pairs on Sphere Don’t Balance Out Could disks balance their forces in spite of tilt? Yes, but only between two (infinite) uniform parallel planes, where the tilts cancel. Then masses between the planes can indeed experience zero net force. Newton's claim for hollow spheres actually turns out to be true (in theory) for parallel planes.  Opposing disks on a sphere have the wrong orientation when they tilt:  |
What I would bring your special attention to is the following diagram:
Please note that here we can substitute an ellipse for the 'barbell' idealization.
Here the Ellipse would then be rotated on its axis, preserving its distance from its own center to the test-particle.
This does NOT mean that this line remains in the center of the 'cone' in other diagrams.
In fact, 'by inspection' you can see that the angles above and below the horizontal
indicate that this line between the center of the ellipse and the particle CANNOT
fall on the centerline of the cone.
The original purpose of the diagram is strictly to prove that the "center of mass"
idea (theorem/hypothesis/approximation) does not hold, since force from the near
half (yes half!) mass increases disproportionately to the decrease in the far half.
And it does prove that, in conjunction with the second diagram below:
Here the Vector Addition Law is used (albeit abstractly) to prove graphically
that the DIRECTION of the pull cannot be along the line to the geometrical center.
But in doing so, it ALSO proves graphically that the DIRECTION of PULL
cannot be along the cone centerline (since then the two angles between
the near and far vectors would have to be equal if the central summed
vector fell on the centerline of the circular cone).
That is, by inspection you should be able to see that the ratio of the two angles
between the vectors added and the final vector can be anything besides 1.
Since this ratio is VARIABLE (based on the relative pull of each half),
the final vector can swing all over the place inside the cone,
depending upon the tilt of the ellipse.
Note there are several lines under discussion:
(1) The centerline of the cone.
(2) The line from particle to the center of the ellipse.
(3) The final vector line indicating the summed direction of the pull.
These three only coincide when the disk is a circle, and is perpendicular
to the line of pull (which is the line from the center to the particle).
At other times, the DIRECTION OF FORCE may vary, always toward
the nearest end of the circle/ring/disk/elliptical disk.
Because the direction of force veers off the centerline of the cone,
AND ALSO the line to the center of the ellipse,
and always in a predictable direction (toward the nearest end of object),
It means that for forces to balance, the line of pull from the opposing
circle/ring/disk/elliptical disk must tilt the same number of degrees
in the opposite direction (still passing through the particle point),
exactly the same amount.
The forces can be balanced if the planes are parallel, whether or not
the cones cut circles or disks. It is in part a question of "equal effective
area"/distance = force, but also quite visibly a question of geometrical
symmetry and balance.
The theorem concerning equal force from disks cut by cones
on parallel planes ALREADY accounts for the fact that the disks
will be ellipses when the cones are on an angle to the planes.
This point was never disputed. It was always accepted both by the
originators of the parallel plane theorem and myself, that the cones
would actually cut ellipses, and that the GCs of the ellipses would not
fall on the axis of symmetry of the cones themselves.
It is worthwhile to speak a bit more on this important theorem:
(1) It is a statement about the ANGLE of the planes to the cones,
not particularly the DISTANCE of the ellipses from the particle in the center.
That is, it is a theorem that is rather elegantly, INDEPENDENT of DISTANCE.
The ellipses are so defined by the cut of the cones on the surface, and nothing more.
The particle can be anywhere between the two parallel planes,
and naturally unequal distances from each plane.
___________________________
Your main objection to the observations we have made here seems still to be this:
You are having a problem with the idea that the ellipse in these illustrations
is rotated on a line through its own Geometric center, instead of say,
on a line across the ellipse that falls upon the line of axis of the cone.
Lets think this through together carefully.
It is most natural to rotate the disk/ring/barbell/ellipse through
its OWN geometric center, in testing the Center of Mass Idea.
The Center of Mass is always (for any shaped object) defined
as the weighted average of the masses and their position in the object,
because the total force is the vector sum of all the individual forces.
This coincides always with the Geometrical Center (GC) for all
regular and symmetrical objects of constant density.
Thus if the Center of Mass idea holds at all, it should hold for
a rotation around the object's GC.
If this fails, the Center of Mass idea fails.
Remember the whole point of the Center of Mass idea,
is to simplify the problem of force by replacing the object
with an Equivalent Point-Mass (EPM) at an appropriate position,
so that the simple Gravitational formula can be reliably used.
It so happens, that in testing the Center of Mass idea,
it is shown by inspection via Vector diagrams,
that the direction of pull will not stay on the axis of symmetry of
a circular cone from the ellipse.
The rotation of the ellipse on an axis through its own GC
is not relevant or remarkable to the independent fact that
that near end will always change the direction of pull away from the GC.
One can by a successive series of approximations (replacing portions of
the ellipse with point-masses) or by calculus show that
indeed the VECTOR changes direction when the ORIENTATION
of the Ellipse departs the perpendicular position relative to a line
drawn from a test-point to its own GC.
This is entirely independent of any cone one might care to draw or consider.
Any ellipse in any absolute position in Euclidean space can be wrapped in a
cone of absolute radial symmetry such that the point of the cone is at the test particle.
This 'fixes' the position of the cone, and if the test-particle is not on the
axis perpendicular to the plane of the ellipse, then neither will the
axis of symmetry of the cone pass through the GC of the ellipse.
The Axis of symmetry for the cone will ALWAYS pass through the near side
of the ellipse off-line from the GC. But the axis perpendicular to the
Axis of symmetry, piercing it, and cutting across the near end of the disk
is not an axis of interest in testing whether or not the near and far halves
of the ellipse balance.
The axis crossing the GC of the ellipse is chosen instead, because it
neatly divides the mass into equal halves (on a uniform density ellipse),
and so the near/far hypothesis can be tested, independently of other issues.
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