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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 |
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