Re: transfer of momentum

[Brian Whatcott <betwys1@SBCGLOBAL.NET>]

At 01:29 PM 11/19/2003, Joel R you wrote:
> > -----Original Message-----
> > From: Bob LaMontagne
//
> > I find this to be the case with my non-science majors. They never fully
> > understand the concept of 'acceleration'. But they can easily grasp the
> > idea that when an object falls, it picks up 10 m/s in speed every
> > second.
//
> > Bob at PC
> >
>
> They need to explicitly consider a small but finite interval before and
> after the maximum height (straddling the moment in time that the object is
> at its apogee) and do all the stuff you say they can do above, and presto
> they have derived that the acceleration at the top is g downwards.
>
> Of course, they have to do it, not watch an instructor do it.


Which reminds me..... it used to be a boyhood experience to whirl
a firecan  in a vertical plane, where it glowed brightly at dusk.

 One could feel that the greatest force on the can's wire tether, was
at the 6 oclock position, and that at the 12 oclock position, the tether
could be almost forceless, for an instant. If the rotation rate were
decreased further, the hot embers departed the can.

A person with this particular experience might well suppose that the
relevant forces add for maximal force at the nadir, and either oppose
 each other, or disappear at the top.
In favor of the disappearance choice, at least of "weight" at the top,
one could imagine there is no opposition to gravitational acceleration
there.

Brian Whatcott    Altus OK    Eureka!