What is fictitious torque for tether

What is fictitious torque for tether situation? PhL ~12.6.16 New material, keep track of it. Aside: What is the fictional torque for the tether situation? Ignoring diff between cog and cms we have b = RE S = ω x b S = x b + ω x (ω x b) (7.13) Special Case 1 We also have c' = torque center in Frame S' = 0 so c = b and = S . Then Ffict = – m[ x b + ω x (ω x b)] – mω x (ω x r') – 2m ω x v' – m x r' (8.6) N(b)fict = (r'-0) x Ffict + (0 + ω x 0 + S) x ( v' + ω x r' + S) – 0 x v' = r' x Ffict + S x ( v' + ω x r' + S) = r' x Ffict + S x ( v' + ω x r') = r' x Ffict + (ω x b) x ( v' + ω x r') At least it can be written down! Also ω x (ω x b) = (ωb)ω- (ωω)b = 0 - ω2b = -ω2b ω x (ω x r') = (ωr')ω- (ωω)r' = - ω2r' so then Ffict = – m x b + m ω2b + mω2r' – 2m ω x v' – m x r' = – m x (b+r') - mω2(b+r') – 2m ω x v' = – m x r - mω2r – 2m ω x v' Euler centr Coriolis Is v'1 in the plane of paper? It is constrained to be along the circle of particle 1 if it lies in the plane of paper, so that is what I will assume for the moment to get maximum simplicity in searching for an EOM. Now. (ω x b) x (ω x r') = [ω,b,r'] ω - [ω,b,ω] r' = [ω,b,r'] ω = r' (ω x b) so then N(b)fict = r' x Ffict + (ω x b) x ( v' + ω x r') = r' x Ffict + [ω,b,r'] ω + (ω x b) x v' = - r' x [ m x r + mω2r + 2m ω x v'] + [ω,b,r'] ω + (ω x b) x v' But now write r' x r = r' x (b+r') = r' x b N(b)fict = = - r' x [ m ( x r) + mω2b + 2m ω x v'] + [ω,b,r'] ω + (ω x b) x v' Now I will of course set = 0 to get N(b)fict = = - r' x [ mω2b + 2m ω x v'] + [ω,b,r'] ω + (ω x b) x v' Consider now r' x (ω x v') But ω x v' is in the r' direction so this term does nothing (thinking r1 for example in my tether picture). So N(b)fict = = - r' x [ mω2b] +[r' (ω x b)] ω + (ω x b) x v' How does this compare to the non-fictitious torque? N(b) = r'1 x F1 + r'2 x F2 We know first that m1r'1 + m2r'2 = 0 since cms is center of circle, we are staying r'cms = 0 m1r'1'1 + m2r'2'2 = 0 m1r'1'1 - m2r'2'1 = 0 m1r'1 - m2r'2 = 0 m1r'1 = m2r'2 '1 = - '2 Second, we know that, F1 = -(Gm1ME/r13) r1 F2 = -(Gm2ME/r23) r2 So then N(b) = r'1 x (-Gm1ME/r13) r1 + r'2 x (-Gm2ME/r23) r2 = r'1 '1x (-Gm1ME/r13) r1 + r'2 '2 x (-Gm2ME/r23) r2 = -(GME) [m1 r'1 '1 x (1/r13) r1 + m2 r'2 '2 x (1/r23) r2 ] = -(GME) m1 r'1 [ '1 x (1/r13) r1 + '2 x (1/r23) r2 ] = -(GME) m1 r'1 [ '1 x (1/r13) r1 - '1 x (1/r23) r2 ] What about this cross products? What have I done before with them? '1 x r1 = α1 α2 I think that β1 + (θ' + π/2) + α1 = π so that β1 = π/2 - (θ'+α1) '1 x 1 = sin(β1) ' = sin[π/2 - (θ'+α1)] ' = cos(θ'+α1) ' On the other hand (π/2 - θ') + (π-δ) + α2 = π or (π/2 - θ') -δ + α2 = 0 so δ = (π/2 - θ') + α2 = π/2 - (θ'-α2) '1 x 2 = sin(δ) ' = sin[π/2 - (θ'-α2)] ' = cos(θ'-α2) ' Now we can go back: N(b) = -(GME) m1 r'1 [ '1 x (1/r13) r1 - '1 x (1/r23) r2 ] = -(GME) m1 r'1 [ '1 x (1/r12) 1 - '1 x (1/r22) 2 ] = -(GME) m1 r'1 [ (1/r12) '1 x 1 + (1/r22)'1 x 2 ] = -(GME) m1 r'1 [ (1/r12) cos(θ'+α1) ' - (1/r22)cos(θ'-α2) ' ] = -(GME) m1 r'1 [ (1/r12) cos(θ'+α1) - (1/r22)cos(θ'-α2) ]' As expected, the two forces oppose each other with a small difference. b + r'1 = r1 b2 + r'12 + 2br'1 cos(π/2-θ') = b2 + r'12 + 2br'1 sin(θ') = r12 b + r'2 = r2 b2 + r'22 + 2br'2 cos[π - (π/2-θ')] = b2 + r'12 + 2br'1 cos[π/2+θ')] = b2 + r'12 - 2br'1 sinθ' So we have r12 = b2 + r'12 + 2br'1 sin(θ') r22 = b2 + r'22 - 2br'2 sin(θ') But I need to get rid of α1 and α2 somehow. sinα1 = r'1cosθ' / r1 sinα2 = r'2cosθ' /r2 Here is our result so far: N(b) = -(GME) m1 r'1 [ (1/r12) cos(θ'+α1) - (1/r22)cos(θ'-α2) ]' r12 = b2 + r'12 + 2br'1 sin(θ') sinα1 = r'1cosθ' / r1 r22 = b2 + r'22 - 2br'2 sin(θ') sinα2 = r'2cosθ' /r2 Now we can approximate noting the αi are very small angles for tether, so cos(θ'+α1) ≈ cosθ' + α1 [-sinθ'] = cosθ' - α1 sinθ' cos(θ'-α2) ≈ cosθ' - α2 [-sinθ'] = cosθ' + α2 sinθ' sinα1 ≈ α1 = cosθ' (r'1/ r1) = very small sinα2 ≈ α2 = cosθ' (r'2/ r2) = very small Then N(b) = -(GME) m1 r'1 [ (1/r12) [cosθ' - α1 sinθ'] - (1/r22) [ cosθ' + α2 sinθ' ]' Now what is the size of (1/r12) - (1/r22) = [ r22 - r12] / (r12r22) ≈ [ r22 - r12] / b4 = [(b2 + r'22 - 2br'2 sin(θ')) - ( b2 + r'12 + 2br'1 sin(θ') )] / b4 = [b2 + r'22 - 2br'2 sin(θ') - b2 - r'12 - 2br'1 sin(θ')] / b4 = [ r'22 - 2br'2 sin(θ') - r'12- 2br'2 sin(θ')] / b4 = [ (r'22 - r'12) - 2b(r'2+r'1)sinθ' ] /b4 Then (1/r12) cosθ' - (1/r22) cosθ' ≈ cosθ' [ (r'22 - r'12) - 2b(r'2+r'1)sinθ' ] /b4 But the second term here is much larger than the first term since b >> r'i so we then get (1/r12) cosθ' - (1/r22) cosθ' ≈ - 2b(r'2+r'1)sinθ' cosθ' / b4 = -(r'2+r'1)sin(2θ') / b3 Meanwhile, the other two terms are (1/r12)[ - α1 sinθ'] - (1/r22) [ α2 sinθ'] = - sinθ' [ α1/r12 + α2/r22 ] ≈ sinθ' [ cosθ' (r'1/ r1) / r12 + cosθ' (r'2/ r2) / r22 ] = sinθ'cosθ' [ r'1/r13 + r'2/r23 ] ≈ sinθ'cosθ' [2(r'1+ r'2) /b3] = + (r'2+r'1) sin(2θ') /b3 Yikes! The two pairs of terms exactly cancel leaving no torque at all. dimension check: N(b) ~ -(GME) m1 r'1 [ (1/r12) = F L = r x F = correct N(b) = -(GME) m1 r'1 [ -(r'2+r'1)sin(2θ') / b3 + (r'2+r'1) sin(2θ') /b3 ]' Here are the two terms cancelling. Conclusion if I did this right: To order r1'/b3 there is no torque on the dumbbell. So if the dumbbell satellite were just hung from a string attached to some distant star (satellite does not orbit earth) this would be the situation. You would have to examine higher terms to find a non-zero torque. Question: If the satellite does orbit, so there is a fictitious torque, how does that torque compare to what is computed above? N(b)fict = - r' x [ mω2b] +[r' (ω x b)] ω + (ω x b) x v' Well add these for the two particles N(b)fict = - r1' x [ m1ω2b] +[r'1 (ω x b)] ω + (ω x b) x v'1 - r2' x [ m1ω2b] +[r'2 (ω x b)] ω + (ω x b) x v'2 Now we have to throw in the fact that GME/r02 = ω2r0 or roughly GME/b2 = ω2b or ω2 = GME/b3 dim =(F/m)/L = (ma/m)/L = a/L = T-2 OK Then we have N(b)fict = - r1' x [ m1(GME/b3) b] + (GME/b3)[r'1 ( x b)] + ( x b) x v'1 - r2' x [ m2(GME/b3)b] + (GME/b3)[r'2 ( x b)] + ( x b) x v'2 or N(b)fict = - r1' x [ m1(GME/b2) ] + (GME/b2)[r'1 ( x )] + ( x ) x v'1 - r2' x [ m2(GME/b2) ] + (GME/b2)[r'2 ( x )] + ( x ) x v'2 = - ( m1r1'+ m2r'2) x [(GME/b2) ] + (GME/b2)[( ( r1'+r'2) ( x )] + (v'1+v'2)