Re: [Phys-L] quickest route

[David Bowman <David_Bowman@georgetowncollege.edu>]

Regarding John D's comment concerning Anthony L's multiple choice
problem and answer (related to the brachistochrone problem):

> On 07/26/2013 07:31 PM, Anthony Lapinski wrote:
> > ... wins since ball has highest acceleration at start.
>
> Consider the right triangle shown here, with vertices
> A,B,C facing sides a,b,c respectively.
>
>
>        B
>        |\
>        |  \
>      a |    \ c
>        |      \
>        |________\ A
>       C     b
>
>
> The base b is horizontal and the altitude a is vertical, 
> aligned with the local gravitational field. 
>
> The ball starts at rest at point B.  The ball then moves
> under the influence of gravity along some path to A.  The
> only forces are gravity and the forces of constraint.  The 
> question is this:  Is it faster for the ball to approximately
> follow path BCA, with a reasonably sharp 90 degree turn near 
> point C, or is it faster to follow the direct path BA?
>
> For simplicity, assume the "sharp" turn is rounded enough 
> to keep the forces finite, but sharp enough to have no 
> significant effect on the timing.
>
> This question is well within the scope of the introductory
> algebra-based physics course.
>
> And yes, I am quite aware that there is something ill-posed
> in the question as I have stated it.  Students should learn
> to deal with this sort of thing.  Some hints can be found at:
>   http://www.av8n.com/physics/ill-posed.htm#sec-how-to

I think another issue arises besides the one John discusses above,
which I believe, is alluding to a tradeoff in the timing of the path
traversal between effects of increasing the average speed along the
path by descending vertically as far as possible as quickly as
possible versus the concomitant effect of the resulting increase in
overall length of the path to be traversed when doing so.  This
other issue relates to the effects of the details of the contact
force between the surface and the ball.  Although I doubt any
problem poser would be perverse enough to draw a sufficiently
extreme set of three possible paths for the ball to follow that
would actually change the correct answer from the concave path to
another path, a slightly rounded variation of John's counterexample
above for the concave path notwithstanding, it still gets to the
issue of properly thinking through all the relevant issues necessary
to make a correct choice in the matter.

The other issue I bring up of the nature of the contact force is
relevant to the issue of how the local slope of the path affects
the local acceleration at that point.  We can't just assume that
the local scalar acceleration along the path is a monotone
increasing function of the magnitude of the slope of the path at
that point independent of the past history of the ball's motion.
This is because the value of that scalar acceleration depends on
whether or not the ball is slipping on the track as it rolls down
the ramp/path.  The acceleration when it skids is different than
when it is rolling without skidding.  But the issue of whether it
is slipping or not is a complicated one that may involve the ball's
past history of its motion up to that point (at least in the concave
path case, if not the other two cases).

If we model the behavior of the contact force as a simple friction
coefficient model suitable for lower level classes we find that the
scalar acceleration in units of the gravitational field strength,
a/g, is given by a/g = sin(θ)/(1 + 1/p) when the ball is rolling
*without* slipping, and a/g = sin(θ) - μ_k*cos(θ) when the ball *is*
slipping.  Here θ is the tilt angle of the local slope of the path
w.r.t. horizontal; μ_k is the kinetic friction coefficient for the
ball/ramp interface; and p is the square of the ratio of the ball's
radius of gyration to its radius (p = (k/r)^2).  The value of p
depends on the internal mass distribution inside the ball.  The
criterion determining whether the ball is slipping or not is a
complicated one for the concave path.  But for the straight
constant slope path and the convex path the criterion is relatively
simple.

For those cases (straight and convex) the ball is not slipping if
the path is locally tilted such that tan(θ) < μ_s*(1 + p), and the
ball *is* slipping if the path is locally tilted such that
tan(θ) > μ_s*(1 + p), where μ_s is the static friction coefficient
for the ball/ramp interface. The reason for this is in those two
cases the tilt angle never decreases along the path so the ball
never has the opportunity to transition from a slipping situation
to a rolling/without slipping situation.  Since the ball is released
from rest in those situations at most the only transition possible
is from non-slip to slip, and the transition occurs (if it occurs)
at the place where the value of tan(θ) crosses the μ_s*(1 + p)
threshold.

Things can be a lot more complicated for the concave path
situation.  If the initial slope is so tilted that the slipping
threshold is exceeded at the top of the ramp the ball is already
slipping from the get-go.  In this case as the ball descends its
scalar acceleration is decreasing along with the slope.  But if
the slope descends below the previous threshold that will not
suddenly make the slipping stop.  It will take some time for the
mismatch in rotational speed and translational speed to come into
balance as the ball's over-spin helps accelerate the translational
motion as its looses spin angular momentum until the no slip
condition (angular velocity = velocity/radius) again obtains.
*After* this point the ball continues to move without slipping
along the rest of the concave path.  The problem is that the
location of where the slipping may stop, if it stops, depends on
the ball's past history, and the shape of the already transversed
part of the ramp, and the particular value of p, and the particular
values of μ_s and μ_k.

Of course a lot of trouble could have been avoided if the problem
statement required that the moving object be a frictionless particle
that always slides without friction along the whole path, rather
than be a ball capable of rolling down the path as well as slipping.

One thing about this issue, however, is that if the ball is slipping
at some place along the path its scalar acceleration is necessarily
greater at the place than if it is not slipping there.  So, somewhat
ironically, if the ball is behaving dissipatively, it is hurrying 
to its destination faster than if it is behaving in a conservative
non-dissipative manner at each point of its path.

Dave Bowman