Re: [Phys-l] NYT article: Centrifugal force

[Brian Whatcott <betwys1@sbcglobal.net>]

I'll try:
On a perfect sphere of radius r fix a point at 45 degrees North.
Considering a non-ratating sphere, assume that the normal force is 1 g
Rotate this sphere at one revolution per day.
In the rotating frame, there is a centrifugal force of  ?cos 45
The resultant of a normal 1g and a centrifugal ?cos45 at 45 degrees
is  ??g at 0.3 degrees to the normal. 
(where ??g is less than 1 g as defined above)

Does this help?

Brian W
p.s. this ignores the operational definition of g as the ACTUAL force
at any point on Earth. Ah, well....

alex brown wrote:
> How would one go about calculating this one third of a degree?... I would just like to know the starting point so I can have a go.  Thanks
>
> ALex
>
> --- On Fri, 3/7/09, John Denker <jsd@av8n.com> wrote:
>
>   
> > From: John Denker <jsd@av8n.com>
> > Subject: Re: [Phys-l] NYT article:  Centrifugal force
> > To: "Forum for Physics Educators" <phys-l@carnot.physics.buffalo.edu>
> > Date: Friday, 3 July, 2009, 2:16 PM
> > On 07/02/2009 08:11 PM, Anthony
> > Lapinski wrote:
> > In temperate latitudes, the direction "up/down" differs
> > from the
> > direction "straight toward the center of the earth" by
> > about 1/3rd 
> > of a degree, due to the centrifugal contribution to the
> > universally-
> > accepted practical notion of "up/down".
> >
> >     
>
>