Re: Sunday Morning Puzzle.

[Bernard Cleyet <anngeorg@PACBELL.NET>]

I thought so too -- then as an exercise I tried doing the integration, first
using the formula for a disk for E with the proper substitution for gravity -- I
failed -- not a math person.  So I "really" cheated and found the answer (and how
to get it) in Marian;  substituted b.w.'s values and didn't get his result.  So
either the center of mass is not the center of "gravitational force" or I
goofed.  However, I also tried a counter example to the assumption one may put
all the mass in the center of mass and, for points outside the mass, get the same
g force.

Take two point masses 1 and 10 units separated by 11 length units.  The c of m is
on the line connecting them, one unit from the 10 unit mass.  Right?  The force
field one unit from the 10 unit mass collinear, not between the masses is G*10m/1
+ G*m/144.  If one places all the mass at the center of mass the g field is
G11*m/4,  obviously considerable less. The sphere is a unique? example where the
center of g force and mass are the same.  Is the disk also?

bc  willing to pronounce self dead after the pair of docs. is explained.

P.s. Other than one obtains an inverse r force from an effectively infinitely
long rod, I don't understand the "... idea of inverse r squared forces versus
inverse r ...."

brian whatcott wrote:

> At 22:38 10/5/01 -0700, Bernard wrote:
> > Last week brian w. suggested that a disk of the same mass and radius as
> > a sphere would have a greater attractive force at the center of its face
> > (toward the sphere on the pendulum).  Rather than stick my neck out, I
> > pose the Sat. morn. question.  (Since most of you won't read this 'till
> > then.)
> >
> > What are the respective g fields at the surface of a sphere of radius R
> > and the center of a face of a disk also radius R and length 4/3 R.
> > (These better have the same volume, or I'm dead!)
> >
> > Express in terms of rho (volume density), R, and G
> >
> > bc
> >
> >
> >
> > P.s. If you don't want to do the triple integration for the cylinder,
> > the formula is in the solutions book for the latest ed. of Marian.
>
> I'm not the kind of spoil-sport that wants to squash peoples' natural pleasure
> at doing triple integrations over the volumes, but if you'd rather
> find results on sight there is the attractive concept sometimes derogated
> on the list, called 'center of gravity'. Add to this the idea of inverse r
> squared forces versus inverse r forces, and you have a pretty contrast.
> (These had better be relevant, or I'm dead too??)
> :-)
>
> brian whatcott <inet@intellisys.net>  Altus OK
>                 Eureka!