In the "collision" (better, interaction) between the linear and rotational
modes, the changes in linear and angular momentum are correlated because the
same force f effects both changes:
fR = -(d/dt) (Iw) and f = (d/dt) (mv)
As v increases from zero, w decreases from w_o and
mv = -(Iw)/R + (Iw_o)/R
This correlation plays the same mathematical role as the conservation of
linear momentum in the two particle collision.
If you require v = Rw =>
w = w_o / (1 + [R/k]^2 ), where k=rad of gyration.
For the simplest case, take R=k (the analog equivalent of equal particle
masses) and get w = w_o/2, the analog of the halving of the particle
velocity in a totally inelastic particle collision of equal masses.
Just as in the particle collision, this forces a dissipation of kinetic
energy. As in the equal mass particle case, in the k=R case exactly 1/2 of
the original energy is dissipated.
Bob Sciamanda (W3NLV)
Physics, Edinboro Univ of PA (em)
trebor@velocity.net http://www.velocity.net/~trebor
----- Original Message -----
From: "Brian Whatcott" <betwys@DIRECTVINTERNET.COM>
To: <PHYS-L@lists.nau.edu>
Sent: Tuesday, October 29, 2002 11:14 PM
Subject: Re: When Physical Intuition Fails
| I find it easy to visualize a spur wheel spinning at 1 rpm suddenly
engaging
| a fixed rack with no friction, in a fully elastic way.
| Is this a limitation of my imagination, or of yours, Bob? :-)
| . . .
| Brian W
---
Outgoing mail is certified Virus Free.
Checked by AVG anti-virus system (http://www.grisoft.com).
Version: 6.0.404 / Virus Database: 228 - Release Date: 10/15/2002
This posting is the position of the writer, not that of SUNY-BSC, NAU or the AAPT.