For Frank Krygowski's helmet files

In article
,
Dave Lehnen wrote:

Peter Rathmann wrote:
On Jul 3, 3:08 pm, Dave Lehnen wrote:
Michael Press wrote:
In article ,
Michael Press wrote:
In article ,
wrote:
(To use the idealized case, the top of a solid rod toppling sideways
hits the ground as if it had fallen straight down from 3/2 its
height.)
How do you know that?
Does any one know what Carl is talking about here?
Yes, in the ideal case of a thin, rigid rod, toppling by pivoting around
its point of contact with the ground (no slipping, no friction of any
kind), ...

If there's no friction between the rod and the contact point then that
point will slip as the rod falls.
Conversely, if there is to be no slipping of the contact point then
there must be some level of friction.
The derivation of the 3/2 ratio assumes sufficient friction to prevent
slipping.

Of course there is friction. I was trying to state the conditions for
which the math is completely accurate, an ideal zero-friction pivot or
hinge joint, no friction in the hinge or air friction, etc. A real bike
falling over will pivot for quite a while almost as if about an ideal
hinge, but the tires will slip out at some point. The simple ideal case
is not a very good model of a bike falling over, for a lot of reasons,
but does show that under some conditions, a head could impact at
slightly higher speed than in a pure drop from the same height. It could
also hit much more slowly, depending on how the falling rider reacts,
and what he falls on.

In the case of the frictionless horizontal plane, the
vertical component of velocity of the free end when the
rod is horizontal is the same as in the case of the fixed
pivot case.

Bonus question: What is the locus of the instantaneous axis
of rotation in the freely sliding fall?

--
Michael Press