Section 11b rewrite

This material has been installed and updated within frames doc, do not view again. (b) Expression of L(c) and (c) in terms of Frame S' objects We replicate Fig 1.4 which now includes a single torque reference point which is c is Frame S and c' in Frame S'. Fig 11.3 where the three equations shown are obvious from the drawing. Again mass m = 1 so momentum p = mv = v. For arbitrary reference point c one has from (1.36), L(c) = (r-c) x v (c) = (r-c) x a – S x v L'(c') = (r'-c') x v' '(c') = (r'-c') x a' – 'S' x v' . (1.36) (11.1) The reference point c might be moving. We know from Sections 6 and 7 how r,v,a and r',v',a' are related, r = r' + b (6.1) v = v' + ω x r' + S (6.6a) a = a' + x r' + 2 ω x v' + ω x (ω x r') + S . (7.6a) (11.2) Part of the goal of Section 5 is to express L(c) and (c) in terms of Frame S' quantities. In the following all algebra is shown to provide an easily traceable path since there won't be any result verifications: L(c) = (r-c) x v = (r'-c') x v = (r'-c') x [ v' + ω x r' + S ] = (r'-c') x v' + (r'-c') x [ (ω x r') + S] = L'(c') + (r'-c') x [ (ω x r') + S] . Here was my previous result: L(c) = L'(c') + (r' - c + b ) (ω x r' + S) + (c'- c + b) x v' In this previous result replace c'- c + b = 0 and b-c = -c' to get L(c) = L'(c') + (r' - c' ) (ω x r' + S) and this agrees with the result above obtained much more quickly. Next: (c) = ∂SL(c) = ∂S'L(c) + ω x L(c) G-rule = ∂S'{ L'(c') + (r'-c') x [ (ω x r') + S] } + ω x L(c) = '(c') + ('-') x [ (ω x r') + S] + (r'-c') x [ ( x r') + (ω x ') + ∂S'S] + ω x L(c) Now I use ∂S'S = ∂SS - ω x S = S - ω x S so that (c) = '(c') + ('-') x [ (ω x r') + S] + (r'-c') x [ ( x r') + (ω x ') + S - ω x S] + ω x { (r'-c') x v' + (r'-c') x [ (ω x r') + S] } = '(c') + (v'-') x [ (ω x r') + S] + (r'-c') x [ x r' + ω x v' + S - ω x S] + ω x { (r'-c') x v' + (r'-c') x [ (ω x r') + S] } = LA (checked) Notice that two of the terms are (r'-c') x [ω x v'] + ω x [(r'-c') x v'] = not cyclic so cannot combine A B C B A C A x (B x C) + B x (A x C) = (AC)B - (AB)C + (BC)A - (AB)C = [(r'-c') v'] ω - [ωv'] (r'-c') - 2 [(r'-c') ω] v' = does not help much In a special case where c' = 0 and ' = 0 This says (b) = '(0) + v' x [ (ω x r') + S] + r' x [ x r' + ω x v' + S - ω x S] + ω x { r' x v' + r' x [ (ω x r') + S] } = LA (checked) I just do NOT see any good way to simplify this. But maybe .... Note that v = v' + ω x r' + S = ' + ω x c' + S . I like the idea of c' and ' showing up in the results because we wanted to have things expressed in Frame S' variables. My previous result was (c) = '(c') + (c'- c + b) x a' – S x [v' + ω x r' + S] + 'S' x v' + (r' - c + b) x [ x r' + 2 ω x v' + ω x (ω x r') + S] . In this previous result replace c'- c + b = 0 and b-c = -c' to get (c) = '(c') – S x [v' + ω x r' + S] + 'S' x v' + (r' - c') x [ x r' + 2 ω x v' + ω x (ω x r') + S] . Now replace S = 'S' + ω x c' + S to get (c) = '(c') – [ 'S' + ω x c' + S ] x [v' + ω x r' + S] + 'S' x v' + (r' - c') x [ x r' + 2 ω x v' + ω x (ω x r') + S] . ≡ LB Let's try Maple to show the two expressions are the same. I need to have names v', ', ω, r', S, c', , S = vp,cpd,w,rp,bd,cp,wd,bdd After I fixed many things, Maple finally says LA and LB are the same! _________________________________________ These results may now be summarized: L(c) = L'(c') + (r' - c + b ) (ω x r' + S) + (c'- c + b) x v' (11.3) (c) = '(c') + (c'- c + b) x a' – S x [v' + ω x r' + S] + 'S' x v' + (r' - c + b) x [ x r' + 2 ω x v' + ω x (ω x r') + S] . (11.4) Equations (11.3) and (11.4) express L(c) and (c') entirely in terms of Frame S' objects, for the general case where angular momentum reference points c and c' are arbitrarily selected. Equation (11.4) can be rewritten as '(c') – (c) = – (c'- c + b) x a' + S x [v' + ω x r' + S] – 'S' x v' – (r' - c + b) x [ x r' + 2 ω x v' + ω x (ω x r') + S] . (11.5) (c) Fictitious Torques and Newton's Rotational Law in a non-inertial frame In analogy with Newton's Law F = (dp/dt)S, Newton's Rotational Law in inertial Frame S is given by, N(c) = (dL(c)/dt)S = (c)S = (c) // true Newton's Rot Law in Frame S (11.6) where L(c) is the angular momentum of our Particle in Frame S relative to reference point c, and N(c) is some externally applied torque about that same reference point acting on the Particle. In Frame S' we want to find some effective ("bogus") Newton's Rotational Law, N'(c')eff = '(c') // bogus Newton's Rot Law in Frame S' (11.7) N'(c')eff = N(c) + N'(c')fict // defines N'(c')fict (11.8) where N'(c')fict is the fictitious torque that mysteriously appears in non-inertial Frame S', so then N(c) + N'(c')fict = '(c') . (11.9) We can then apply this bogus Newton's Rotational Law in non-inertial Frame S' as long as we include the fictitious torques N'(c')fict along with the real torques N(c) which are just those present in inertial Frame S. This is in complete analogy with the use of fictitious forces as reviewed in Section 8. Subtracting (11.6) from (11.8) gives, N'(c')fict = '(c') - (c) , (11.10) and using (11.5) , N'(c')fict = – (c'- c + b) x ma' + S x [mv' + mω x r' + mS] – 'S' x mv' – (r' - c + b) x [ m x r' + 2mω x v' + mω x (ω x r') + mS] (11.11) where we have quietly reinstalled the mass m. Recalling now the fictitious force from (8.6), F'fict = – mS – mω x (ω x r') – 2m ω x v' – m x r' (8.6) (11.12) frame centrifugal Coriolis Euler the fictitious torque can be written as N'(c')fict = – (c'- c + b) x ma' + S x [mv' + mω x r' + mS] – 'S' x mv' + (r' - c + b) x F'fict . (11.13) As noted earlier, a common situation is to assume c ≡ 0 and c' ≡ 0 so that N'(0)fict = – b x ma' + (r' + b) x F'fict . // c = c' = 0 (11.14) If in addition the origins of frames S and S' coincide, then b = 0 and the result is simply N'(0)fict = r' x F'fict . // c = c' = b = 0 (11.15)