dumbbell satellite as rigid body problem v2

Dumbbell Satellite as Rigid Body Problem v2 PhL 3.26.17 do not read below, see summary doc Part I : The issue of Special Case #4 Status: I tried in v1 of this doc to solve the dumbbell satellite problem using these body frame equations for rigid body motion. (N)'1 = I'1 ()'1 - (ω)'2(ω)'3 (I'2 - I'3) (N)'2 = I'2 ()'2 - (ω)'3(ω)'1 (I'3 - I'1) (N)'3 = I'3 ()'3 - (ω)'1(ω)'2 (I'1 - I'2) . (I.4.3) Things were not working out very well, and now I think I realize why that is the case ?? Looking just above (I.1.1) I have stated that what follows applies in Special Case #4 where b=0 . But this is NOT the situation for the dumbbell satellite problem, where we have b ≠ 0 ! In particular, the equations of motion above do not apply!! So no wonder they led to problems. [ but see conclusion at the end ] Is it possible to "fix this up" in some simple way. I can track through Section I.1 and see how things would have to be generalized. The satellite problem in fact is Special case #1 which has ω axis passing through the Frame S origin (non swap notation, Frame S us the inertial frame). In that case we know that S ≡ (db/dt)S = ω x b . // Special Case #1 (4.4.3) Now staying in non-swap notation, we can try to update Section I.1 for Special Case #1 vα = v'α + ω x r'α + S (12.1.2b) (I.1.1) = 0 + ω x r'α + ω x b = 0 + ω x (r'α + b) = 0 + ω x rα (12.1.2a) = ω x rα Then the total Frame S angular momentum of the rigid body is L = Σα rα x pα = Σα rα x mαvα = Σα mα rα x (ω x rα) = Σα mα [ rα2ω - (rα ω) rα ] . //A x (B x C) = (AC)B - (AB)C (I.1.2) So this result survives! Hmmm. The following results also survive, Li = Σα mα [ rα2ωi - (rα ω) (rα)i ] = Σα mα [ rα2Σjωjδij - Σj(rα)jωj (rα)i ] = Σα mα Σj[ rα2δij - (rα)i(rα)j ] ωj = Σj { Σαmα [ rα2δij - (rα)i(rα)j ] } ωj = Σj Iijωj where Iij ≡ Σαmα [ rα2δij - (rα)i(rα)j ] = the inertia tensor (I.1.3) or Iij ≡ ∫dx1dx2dx3 ρ(x)[ r2δij - xixj] . // continuum notation We rewrite (I.1.3) as a vector equation, L = Iω . (I.1.4) Next, how about T = Σα(1/2)mαvα2 = Σα(1/2)mα vα (ω x rα) = Σα(1/2)mα ω (rα x vα) // cyclic rule = Σα (1/2) ω (rα x mαvα) = (1/2) ω Σα rα x pα = (1/2) ω L = (1/2) ω (Iω) . // (I.1.4) (I.1.7) This too seems OK. Li = Iij (ω)j Iij = Σαmα [ rα2δij - (rα)i(rα)j ] // rα2 = (rα)i(rα)i (I.1.6) (L)'i = (I)'ij (ω)'j (I)'ij = Σαmα [ r'α2δij - (rα)'i(rα)'j ] . // r'α2 = (rα)'i(rα)'i So maybe I should relax my condition to be Special Case #1 and not just Special Case #4 !!!! Let's continue into Section I.2 It looks totally fine! And I.3 looks fine, and I.4 looks fine. So I was wrong with my conjecture above! These equations SHOULD work for the satellite. But now have to pay attention that r ≠ r'. Part II : Are Euler Angles defined for Special Case #1 ?? Go back to their definition. In Fig (H.1.5) I show Frame S and Frame S' having a common origin, which means Special Case #4. But is this significant? I can always move the basis vectors to have a common origin even if there is some long b vector. The discussion of Section H seems OK through (H.3.5) where I am really only dealing with basis vectors. Now what about the transformation of Kinematic Vectors in (H.3.6) ? Consider this claim, (r)' = R r = Rz(-ψ)Rx(-θ)Rz(-φ) r (H.3.10) where I do NOT have anything called vector r'. My notation x' I think should be (r)'1 to be clear. Other than this sloppy notation, things seem OK through end of Section H.3. What about Section H.4 ? It just seems OK through (H.4.4). Everything is just using unit vectors, and origin location just does not matter! So I think this is after all OK (ω)'x = sinθsinψ + cosψ ≡ ωx' (ω)'y = sinθcosψ - sinψ ≡ ωy' (ω)'z = cosθ + . ≡ ωz' // Frame S' (H.4.7) That is, I think this is OK for Special Case #1 situations. Conclusion: I don't see any reason why the Euler angle stuff is restricted to Special Case #4. Part III : For the dumbbell satellite problem, what do I know about ω ? Looking at Fig (F.1.1) I would say that ω = ω' = ω at all times. I thought I wrote this fact down somewhere in Appendix F ? ***** Maybe I should add the above equation somewhere useful! I just don't see it written anywhere, but the picture makes it obvious. Implication: In the body frame we have ω = ω which says ω1 = ω ω2 = 0 ω3 = 0 This certainly simplifies the development in a way I did not take advantage of!! I will start a new v3 with this simplification installed from the start. Part IV : For the dumbbell satellite problem, body frame? In my dumbbell satellite treatment I have Frame S' at the Earth, and I have Frame S at the satellite. But the dumbbell moves freely within Frame S! So it is wrong to identify this Frame S with a "body frame" of rigid body motion theory!! So maybe this is the Big Error I have been making in this series of docs!