Well, I just couldn't hold back any more so here goes: my full explanation
of the "friction" problem. I contend that mu(k) < mu(s) between a tire and
the road but the phenomena that were given as evidence that the reverse is
true can be explained retaining the relationship above. But first a
reminder or two: i)the static friction force, f(s), can take on ANY
magnitude up to f(s)max = mu(s)N and can have any direction; ii)while the
kinetic friction force has a "fixed" magnitude given by f(k) = mu(k)N and
its direction is always opposite the direction of slippage (e.g., if the
object is sliding over the surface in the +x direction, the f(k) on it will
be in the -x direction).
To keep things simple I'll assume a flat (i.e., non-banked) turn of fixed
radius but the condition of the road (i.e., dry, wet, icy) will only change
the values of mu(s) and mu(k) still retaining the relationship above. I'll
slowly build up to the "race car slipping" in three stages.
I)Uniform Circular Motion & No Slipping
If you go around the circle of fixed radius at a constant speed the static
friction force is what keeps you going around the turn since I'm assuming
no slippage of the tires on the road. This force will point towards the
center of the turn with no component tangential to the turn (i.e., parallel
to the velocity vector) since the speed (magnitude of the velocity vector)
remains constant.
For a given radius turn as you go around the turn at a higher speed the
f(s) necessary to "get you around" will increase. The maximum speed you
can go around this turn will occur when f(s) becomes equal to f(s)max. If
you increase the radius of the turn you can go around the turn faster;
hence the maximizing of the radius of the circle by starting and ending at
the outside edge of the road while touching the inside edge (the "apex") in
between.
II)Non-Uniform Circular Motion & No Slipping
If you wish to increase the speed of the car as you negotiate the turn the
static friction force (still assuming no slippage) must now have a
component parallel to the velocity vector (tangential to the circle) as
well as a component towards the center of the turn. Since the MAGNITUDE of
f(s) is constrained to a maximum value the maximum speed you can reach
during the turn will be smaller than going around the turn at constant
speed and the larger the rate at which the SPEED changes the smaller the
maximum speed before slippage occurs under these conditions.
To partially compensate for this race car drivers will not really take the
corner in a circle of constant radius but will open up the radius of the
turn as the car speed up so that the necessary radial component of f(s)
will go down.
III)Slipping
Once the driven tire start to slip (i.e., the necessary f(s) to continue in
a circle would have to be > f(s)max, an impossible situation) those tires
will no longer be going around in a circle, i.e., they will be slipping
over the road towards the outside of the turn. This is why front wheel
drive cars tend to "plow ahead" or understeer while rear wheel drive cars
tend to oversteer when the driven wheels are at the limits of adhesion. In
addition to the sideways (relative to the road) slippage the tires are
slipping towards the rear of the car because they are still spinning. The
actual direction of slippage of the tires over the road will be the vector
sum of these two slipping motions and can be controlled, to a certain
extent, by the driver of the car since he/she can adjust the speed of the
spinning of the wheel. The f(k) force has a fixed magnitude but, since its
direction is opposite to the direction of slippage, which is partially
under the control of the driver, the direction of that force can be varied
with the direction generally being towards the center of turn and forward,
increasing the speed.
Under the right surface conditions (i.e., mu(k) isn't too much less than
mu(s)) it may be possible that the component of f(k) that is parallel to
the velocity vector and therefore causes the speed to change could be
greater than the possible component of f(s) in that direction (the radial
component of f(k) wouldn't have to be as large as that of f(s) since the
tire is no longer going in a circle without slipping). Note that all of
this can be true even though mu(s) > mu(k) since we are dealing with one
component of the force rather than the force's magnitude. What a race car
driver calls acceleration only considers that part of the acceleration
parallel to the velocity vector; the radial part of the acceleration vector
is a completely separate (although admittedly important) item and the two
aren't related.