stick tension T calculation

How do you get the tension in the stick? PhL 1.5.17 This is a very important question for two reasons: (1) will the tether stay taut? (2) I want the tension T in the rest vertical position in order to verify my claims in the main text where the factor 2 or 3 is in question. The only way I know to proceed is in the way I start off Section D.10 which is sitting right now in App F v 6. Keep in mind that I am in swap coordinates where Frame S is rotating. A. COMPUTE TRUE FORCES The only forces on a dumbbell mass are gravity and stick tension T. From ** we then write F1 = - (GMEm1/r'13) r'1 - T 1 = - (GMEm1/r'13)( b + r1) - T 1 F2 = - (GMEm2/r'23) r'2 - T 2 = - (GMEm2/r'23)( b + r2) - T 2 Given that b is the distance to CMS, the above equations are exact no matter where COG lies. B. COMPUTE FICTITIOUS FORCES We must now compute the fictitious force on each mass seen in Frame S: Ffict,1 ≈ – m1S' – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 // swap Ffict,2 ≈ – m2S' – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 // swap What is the meaning of ω here? Frame S is rotating about the Frame S' origin, but in non-swap notation we would say that Frame S' is rotating around Frame S origin, so this would be Special Case #1, but I have to translate the results! Here are some translated results from frames doc: S ≡ (db/dt)S = 0 seems OK (swap) (4.4.2)s S' ≡ (db/dt)S' = ω x b . // Special Case #1 (swap) (4.4.3)s I then have this non-swap claims in (7.13), S = x b + ω x (ω x b) . // Special Case #1 (non-swap) (7.13) which I can translate to S' = x b + ω x (ω x b) . // Special Case #1 (swap) (7.13)s If this is correct, then I can restate things above as Ffict,1 ≈ – m1[ x b + ω x (ω x b)] – m1ω x (ω x r1) – 2m1 ω x v1 – m1 x r1 Ffict,2 ≈ – m2[ x b + ω x (ω x b)] – m2ω x (ω x r2) – 2m2 ω x v2 – m2 x r2 or Ffict,1/m1 ≈ – [ x b + ω x (ω x b)] – ω x (ω x r1) – 2 ω x v1 – x r1 Ffict,2/m2 ≈ – [ x b + ω x (ω x b)] – ω x (ω x r2) – 2 ω x v2 – x r2 or Ffict,1/m1 ≈ – [ ω x (ω x [r1+b])] – 2 ω x v1 – x (r1+b) Ffict,2/m2 ≈ – [ω x (ω x [r2+b])] – 2 ω x v2 – x (r2+b) or Ffict,1/m1 ≈ – [ ω x (ω x r'1)] – 2 ω x v1 – x r'1 Ffict,2/m2 ≈ – [ω x (ω x r'2)] – 2 ω x v2 – x r'2 Now remove the Euler term to get Ffict,1/m1 ≈ – ω x (ω x r'1) – 2 ω x v1 Ffict,2/m2 ≈ – ω x (ω x r'2) – 2 ω x v2 Here then are the F = ma equations of motion in Frame S" m1a1 = { - (GMEm1/r'13) r'1 - T 1} - m1 ω x (ω x r'1) – 2m1 ω x v1 m2a2 = { - (GMEm2/r'23) r'2 - T 2} - m1 ω x (ω x r'2) – 2m1 ω x v2 or -m1a1 = (GMEm1/r'13) r'1 + T 1 + m1 ω x (ω x r'1) + 2m1 ω x v1 -m2a2 = (GMEm2/r'23) r'2 + T 2 + m2 ω x (ω x r'2) + 2m2 ω x v2 or -a1 = (GME/r'13) r'1 + (T/m1) 1 + ω x (ω x r'1) + 2 ω x v1 -a2 = (GME/r'23) r'2 + (T/m2) 2 + ω x (ω x r'2) + 2ω x v2 or -a1 = (GME/r'13) r'1 + (T/m1) 1 + ω x (ω x r'1) + 2 ω x v1 -a2 = (GME/r'23) r'2 – (T/m2) 1 + ω x (ω x r'2) + 2ω x v2 Here a1 is the acceleration of mass m1 in non-inertial Frame S! I am hoping that I can use just the a1 equation and somehow obtain T from it! But I have to compute these cross products, and I am not sure I have done them elsewhere. So put this off to tomorrow! I don't offhand know a simpler way to get T. Back on Fri 1.6.17. Go back to an earlier statement of forces F1 = - (GMEm1/r'13)( b + r1) - T 1 // true Ffict,1 ≈ – m1[ ω x (ω x b)] – m1ω x (ω x r1) – 2 m1ω x v1 // fictitious Feff,1 = - (GMEm1/r'13)( b + r1) - T 1 – m1ω x (ω x b) – m1ω x (ω x r1) – 2 m1ω x v1 Feff,1 = m1a1 // Newton's law in rotating frame I really need to know all of these terms in spherical unit vectors, then I can look at the 1 equation and determine T. So dot Newton's effective law into 1 . Here is Newton's Law: a1 = - (GME/r'13)( b + r1) - (T/m1) 1 – ω x (ω x b) – ω x (ω x r1) – 2 ω x v1 Dot Newton into 1: a1 1 = - (GME/r'13)( b1 + r1) - T/m1 - 1 [ ω x (ω x b)] - 1 [ ω x (ω x r1)] - 21 [ω x v1] We know that ω = ω and b = b Work from left to right. First a1 1 = [a1r 1 + a1θ + a1φ ] 1 = a1r = - r12 – r1 2 sin2θ = r1(2 - 2 sin2θ) Next: b1 = b 1 = bcosθ Next: A x (A x C) = (AC)A - A2C ω x (ω x b) = (ωb)ω - ω2b = ω2b - ω2b = - ω2b - 1 [ ω x (ω x b)] = - 1 [- ω2b] = ω2b 1 = ω2b cosθ Next: A x (A x C) = (AC)A - A2C ω x (ω x r1) = (ωr1)ω - ω2r1 = ω2r1( 1) - ω2r11 = ω2r1[ sinθcosφ - 1 ] - 1 [ ω x (ω x r1)] = -ω2r1[ sinθcosφ 1 - 1 ] = -ω2r1[ sinθcosφsinθcosφ - 1 ] = -ω2r1[sin2θcos2φ - 1] Next: ω x v1 = [ω ] x [v1θ + v1φ ] = ωv1θ x + ωv1φ x = ωv1θ[ sinθcosφ + sinφ 1] + ωv1φ[ -sinθ cosφ + cosθcosφ 1 ] = [ωv1θ sinφ + ωv1φcosθcosφ] 1 + [ -ωv1φsinθ cosφ] + [ωv1θsinθcosφ] Then, - 21 [ω x v1] = - 2[ωv1θ sinφ + ωv1φcosθcosφ] = - 2ω[v1θ sinφ + v1φcosθcosφ] = -2ω[ r1sinφ + r1 sinθ cosθcosφ] = -2ωr1[ sinφ + sinθcosθcosφ ] Now finally put all the pieces together: a1 1 = - (GME/r'13)( b1 + r1) - T/m1 - 1 [ ω x (ω x b)] - 1 [ ω x (ω x r1)] - 21 [ω x v1] says : r1(2 - 2 sin2θ) = - (GME/r'13)( bcosθ +r1) - T/m1 // GME = ω2b3 + ω2b cosθ - ω2r1[sin2θcos2φ - 1] -2ωr1[ sinφ + sinθcosθcosφ ] Again: r1(2 - 2 sin2θ) = - ( ω2b3/r'13)( bcosθ +r1) - T/m1 + ω2b cosθ - ω2r1[sin2θcos2φ - 1] -2ωr1[ sinφ + sinθcosθcosφ ] Divide by b : (r1/b)(2 - 2 sin2θ) = - ( ω2b3/r'13)( cosθ +(r1/b)) - T/(bm1) + ω2 cosθ - ω2(r1/b)[sin2θcos2φ - 1] -2ω(r1/b)[ sinφ + sinθcosθcosφ ] Maybe I can just drop all terms which are proportional to (r1/b) . The result would then be 0 = - ( ω2b3/r'13)( cosθ) - T/(bm1) + ω2 cosθ or 0 = - T/(bm1) + ω2 cosθ [ 1 - b3/r'13] or T/(bm1) = ω2 cosθ [ 1 - b3/r'13] or T = bm1ω2 cosθ [ 1 - b3/r'13] So this is my first rough estimate for stick tension T. I know that r'12 = b2 + r12 + 2br1cosθ (r'1/b)2 = 1 + (r1/b)2 + 2(r1/b)cosθ (r'1/b)3 = [ 1 + (r1/b)2 + 2(r1/b)cosθ ]3/2 b3/r'13 = [ 1 + (r1/b)2 + 2(r1/b)cosθ ]-3/2 Now in our smallness limit this is roughly b3/r'13 = [ 1 + 2(r1/b)cosθ ]-3/2 ≈ 1 + (-3/2)2(r1/b)cosθ //(1+a)n = 1 + na + ... [ 1 - b3/r'13] = 1 - 1 - (-3/2)2(r1/b)cosθ = (3/2)2(r1/b)cosθ = 3(r1/b) cosθ So then I get T ≈ bm1ω2 cosθ 3(r1/b)cosθ = m1ω2 cosθ 3r1cosθ = 3m1r1ω2 cos2θ For an aligned dumbbell have θ = 0 so get T ≈ 3m1r1ω2 // T = ML/T2 , rhs = M L/T2 check What did I get inside frames doc where the Swedish controversy lives? If I replace GME/b3 = ω2 the above becomes T ≈ 3m1r1GME/b3 = 3(m1MEG/b3)r1 which compare to frames doc = 3 (mMG/r03)(Δr) . (8.6.9) I think I have finally recovered what I wanted!! More pondering is needed, but this is very promising.