Re: [Phys-l] Turning effects of a moving bike

[John Denker <jsd@av8n.com>]

On 08/19/2008 12:00 AM, Lim Hwee Ke wrote:

> Usually we say that the rider leans inwards so that there is rotational
> equilibrium about the CG of the rider right? The frictional force F produces
> a moment about the centre of gravity G such that the normal contact force
> produces a counterbalance moment. However, while taking the moments about G
> solves the issue of rotational equilibrium, if we take the point of rotation
> to be anywhere else, (say the point of contact of the rider and the road),
> the turning effects no longer appear to be balancing. There'll only be one
> moment due to the weight of the bike and biker. Where's the other force
> ensuring that the biker does not fall to the ground? 

This is an interesting and tricky situation.  There are half
a dozen misconceptions that could come into play here.  Let
me pick one and discuss it; see if this helps:

By way of background, recall that the rotational physics depends
on /where/ forces are applied relative to the center of mass.
Nevermind the equations;  think about the fundamental qualitative
physics, as in the following example:

Use a yardstick as the deck of a bridge supported by the backs of
two chairs:

         ===============================  <-- yardstick
           /     ^      ^      ^     \
          /      L      M      R      \
         /                             \


Now whack the yardstick from below.  If you whack it at point L,
it will spin to the right.  If you whack it at point R, it will
spin to the left.  If you whack it in the middle (point M), it
will go more-or-less straight up, without spinning.

This is the unalterable physics.  It is not affected by what datum
(if any) you use for quantifying the torques.

There is a famous theorem that says the torque around one datum
is the same as the torque around any other datum ... *provided*
the system is not undergoing any overall center-of-mass motion. [1]

The proviso in this theorem is important, as you can easily see
in the yardstick example, by whacking it at point L and choosing
that same point as your datum.  Then you get zero torque, but
nevertheless the yardstick spins.  Relative to any other datum,
there is a nonzero torque.

  This proviso is a notorious source of confusion.  In introductory
  courses, students are exposed to lots of systems that are not
  undergoing any overall center-of-mass motion, so this term drops
  out of the problem ... and soon drops out of mind altogether.

  On top of that, students are exposed to lots of highly symmetric
  systems, such as ideal rockets, where even though the system is
  accelerating, the rotational term vanishes by symmetry.  (This
  corresponds to whacking the yardstick at point M.)

  On top of all that, if students habitually choose the center of
  mass as their datum, it conceals the general relationship between
  rotational dynamics and straight-line dynamics.

There are many ways of dealing with this.  One conventional and
reasonable approach is to introduce the idea of a /couple/, i.e.
a _pair_ of equal-and-opposite forces separated by a lever arm.
A couple has the great virtue that it is independent of the choice
of datum.  It is a conceptually "clean" quantity.

There is another famous theorem that says any rigid dynamical
system can be described in terms of:
 a) A simple force /applied to the center of mass/, plus
 b) a couple.                                             [2]

Obviously in special cases term (a) and/or term (b) may vanish.
Term (a) describes the straight-line dynamics and term (b) describes
the rotational dynamics.

Obviously theorem [1] is a corollary of theorem [2].

I suspect that if you restate the problem in terms of couples, i.e.
using the "datum-free" approach, the worst of the difficulties vanish.
(If not, please re-ask the question.)