Thanks for the prompt reply. There's a bit there to think about. Hoyt
-----Original Message----- From: Stephen A. Lawrence [mailto:[email protected]] Sent: Sunday, December 14, 2008 9:09 AM To: Vortex Subject: Re: [Vo]:Angular momentum physics question Hoyt A. Stearns Jr. wrote: > Is it true that conservation of linear momentum implies conservation of > angular momentum? I don't think so. In mechanics, conservation of angular momentum is a consequence of a sort of "footnote" on the third law: ===================== If particle 'A' exerts force F on particle 'B', then particle 'B' exerts force -F on particle 'A' [footnote 1] [footnote 1]: Force F must lie on a line connecting particles A and B ===================== The footnote is not generally mentioned in casual (English) descriptions of Newtonian mechanics, but it certainly is mentioned when presenting the theory analytically. Newton, FWIW, presented all of his assumptions in what would today be considered very sloppy language -- but then, he was inventing the math as he went along so he can be forgiven for not using purely mathematical formulations from the get-go. Without the footnote, we could have two particles acting on each other as follows, where the force each feels is shown by the arrows: ^ | A B | V AFAIK that doesn't violate anything else in the three laws -- only the footnote is violated. Note, however, that it certainly *does* violate conservation of energy. Tie A and B together with a string and they'll merrily spin themselves up until the string breaks. If, instead, the force between them acts along the line connecting them, then returning the particles to their starting positions also returns the total energy to its starting value, and energy is conserved. But conservation of energy is also *not* an assumption of Newtonian mechanics; it's a theorem. > > On first thought, it seems that integrating the linear momenta of > infinitesimal elements of a rigid body implies that angular momentum is > just the sum of all the linear momenta. Angular momentum is the weighted average, weighted by the distance from the point about which you're evaluating the angular momentum. It's not a simple sum. You need to integrate the position vector dotted into the linear momentum of each infinitesimal region, not just the linear momentum itself. Remember, the angular momentum depends on the location of the origin. It's not conserved under an arbitrary translation. A particle of mass M moving in the positive Y direction at velocity V has angular momentum zero if it's located at the origin, or, in fact, if it's located anywhere on the Y axis. However, if it's not on the Y axis it has nonzero angular momentum. For example, if it's located anywhere on the line (X=1, Z=0), then it has angular momentum of magnitude MV pointing in the Z direction. > > In particular, if a device is shown to violate conservation of angular > momentum, does that imply that conservation of linear momentum is also > violated? I don't think so. A motor which causes a shaft to spin with no accompanying "back torque" on the motor does not, as far as I know, violate conservation of linear momentum. How you transfer the energy from some other source to the shaft without violating conservation of energy is another question, of course (but, again, conservation of energy is not an assumption, and if you change the rules to allow conservation of angular momentum to be violated you may find conservation of energy is lost as well). > > Hoyt Stearns > Scottsdale Arizona US
