Re: L2-"Negotiating" a curve.

[John Denker <jsd@MONMOUTH.COM>]

At 12:02 AM 11/4/99 -0500, Ludwik Kowalski wrote:
> Yes, the Ffx component must be negative. And I do not know
> why. The fact that the wheel is rotating must have something
> to do with it.

The following step-by-step approach may help elucidate what's going on.

Lemma #1:  A wheel can follow a curved path with arbitrary little friction.

Imagine the arm of a centrifuge, pivoted in the middle and driven from the
middle.  The outer end of the arm is supported by a wheel.  The axis of the
wheel is constrained to be always perpendicular to the instantaneous
direction of motion.

The wheel need not support any sideways load;  it need only support its
share of the weight of the arm.

In the limit that the wheel is thin (in the dR direction) there's no reason
for any irreducible dissipation in this situation.

======================

Lemma #2:  Work is force dot dX, not force times dX.

For starters, consider a cart supported by wheels, moving in a horizontal
_straight_ line in the direction this thread has been calling the Y
direction.  Consider the force due to gravity, in the Z direction.  This
force does no work (dissipative or otherwise) since it is perpendicular to
the velocity.

Next consider a modified cart rolling in the corner between the floor and
the wall, as shown:

      |  ____
      | |    |
      |+|____|<-- force
      |__+__+___


where the cart is moving perpendicular to the paper.  The two lower + signs
are the main wheels and the third + sign is a wheel rolling against the
wall.  This allows the cart to resist a sideways force.  Again this force
does no work, and the wheel that resists this force does no work.

=====================

Lemma #3:  Wheels can resist a sideways force.

Let's get rid of the extra wheel:

      |  ____
      | |    |
      | |____|<-- force
      |__+__+___


This force is still perpendicular to the velocity.  It still does no work.
 (Nitpickers note:  as always, when I say "no" work I mean "arbitrarily
little" work;  there is no nontrivial lower bound on the work that must be
done, neglecting irrelevant imperfections blah blah blah.)  The main wheels
can resist this force.  They do this by refusing to slide in the X
direction.  They continue to roll in the Y direction.

The following subtlety may explain why this point is hard to see:  We've
got *static* friction in the X direction even though the wheel is rolling
in the Y direction.

======================

Putting it all together:  An ideal wheel *never* dissipates energy.  If you
put a force on it in the direction where it can roll, it just rolls (no
force).  If you put a force on it in the other two directions, it resists
the force (no motion).  No matter how you slice it, there's no force dot
displacement.

In ideal conditions, in a steady circular path, each wheel of the tricycle
is always rolling with its axis perpendicular to its instantaneous
direction of motion.  Each wheel creates a sideways force of the type
discussed in lemma 3.  These three forces combined are just enough to
  *) In the lab frame, to accelerate the tricycle around its curved path.
--or--
  *) In the rider's comoving frame, to overcome the effects of the
centrifugal field.

In nonideal conditions, such as on a rain-slicked road, the wheels may be
unable to statically resist forces in the direction perpendicular to their
natural rolling, and you will skid out of the turn.  This is grossly
dissipative because there is now a force dot displacement.

OK?

______________________________________________________________
copyright (C) 1999 John S. Denker             jsd@monmouth.com