Re: BEFORE "Negotiating" a curve.

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Yes, I am more comfortable about the curve negoatiation problem
now than when it presented. Both of your presumptions, Bob, (see
below) are justified. Your second comment is very important,
the static frictional force has to do with the direction of attempted
motion rather than with v.

But that phrase a "wheel which offers unlimited resistance to any
motion other than pure rolling" is not really very different from the
phrase Brian used, "it rolls because it must".  For the time being
I am satisfied but I may not be so satisfied when I face this problem
again next year. I will keep paying attention to what others have to
contribute. Thanks to all who helped so far. Rolling friction should
be explained better in textbooks, or avoided.
Ludwik Kowalski

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To understand is to find a satisfactory causal  relation.
To explain is to express that understanding.
To teach is to promote understanding.
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Bob Sciamanda wrote:

> John Denker wrote to Ludwik:
> > . . .There is a notion of "ideal wheel" which suffices to
> > answer your questions about negotiating a turn.  I say
> > motion wheel again an ideal offers unlimited resistance
> > to any other than pure rolling.  If this model isn't good
> > enough, please explain why not.  . . .
>
> Ludwik,
> Forgive me for presuming, but I think you may be missing the point
> that the force of "unlimited resistance to any motion other than pure
> rolling" is transmitted (through the bearing) to the frame.  This gives
> rise to a large transverse force of static friction.
>
> In sum, the frictional road force on the turned wheel is resolved into
> tangential and transverse components by the bearing.  The tangential
> component affects rolling motion; the transverse component
> (perpendicular to the plane of the turned wheel) is transmitted, through
> the bearing, to the frame and provides a "sideways", turning force.
>
> I think you are also misled in the assertion that the frictional force
> is in the -v direction.  That may be true of kinetic friction, but
> static friction is in the direction opposite to any ATTEMPTED motion.
> Here we have a combination of both, along with rolling friction.  It is
> the attempted, directly forward motion that gives rise to a large static
> frictional force transverse to the plane of the turned wheel.  In a FBD,
> all of these external forces are shown acting at the CM, to illustrate
> the LHS of F_ex=mA_cm .
> Hope this helps.