Appendix D questions v1
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Phil's working notes dated 12.6.16 on restarting the equations of motion for a tether in motion, as a check on his frames document. They restate the force and torque relations between inertial frame S and rotating non-inertial frame S', including centrifugal, Coriolis and Euler terms. He questions whether torques about different reference points c and c' can be added, and decides to rewrite Section 11(b) so both vectors point to the same point.
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Appendix D questions PhL 12.6.16
Previous doc has good results, but here I am going to just start over on the equations of motion problem for the tether in motion. Lagrangian approach would be best probably, but I want to do it high school style. The first aid is the new picture
Frame S is inertial, Frame S' is non-inertial for two reasons. (1) satellite is moving around the earth (2) the cms is moving relative to the cog. I am hoping to ignore item (2) completely. This problem serves as a check on my fancy frames doc equations involving torque which I claim are exact.
From my summary of the forward problem, I know that for forces,
F = ma // true Newton's Law in inertial Frame S (8.1)
F'eff = ma' // fake Newton's Law in rotating Frame S' (8.3)
F'eff = F + F'fict . (8.5)
F'fict = – mS – mω x (ω x r') – 2m ω x v' – m x r' . (8.6)
frame centrifugal Coriolis Euler
and for torques
N(c) = (c) // true Newton's Rot Law in Frame S (11.6)
N'(c')eff = N(c) + N'(c')fict // defines N'(c')fict (11.8)
N(c) + N'(c')fict = '(c') . (11.9)
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)
My presentation needs work, it is not parallel and the summary is incomplete. I don't say F = for the linear case. So maybe
L(c) = (r-c) x mv
(c) = m(r-c) x a – S x mv = (r-c) x F – S x mv
The symbol ω is already used up. I want to say L(c) = I(c)ω in parallel with p = mv but that implies I think that ω here is about point c. If I were to pick c = the ω axis point in Fig 11.3, then I could use ω the same variable, but that is not very general-case. So I have to stick with the above. So I could restate like so
F = ma = N(c) = (c) Newton
F'eff = ma' = ' N'(c')eff = '(c') effective Newton
F'eff = F + F'fict N'(c')eff = N(c) + N'(c')fict
F'fict = F'eff - F = ' - N'(c')fict = N'(c')eff - N(c) = '(c') - (c)
F'fict = – mS – mω x (ω x r') – 2m ω x v' – m x r' . (8.6)
frame centrifugal Coriolis Euler
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)
Is this the best I can do? Has several confusions I think.
Item 1: Look at equation N'(c')eff = N(c) + N'(c')fict . Normally you add two torques with respect to the same reference point! Does this make sense with arbitrary c and c' vectors as shown in Fig 11.3 ?? Forces don't have reverence points, so this issue does not arise.
Suppose we start with Frames S and S' with different origins and S' is rotating and we have this
N'(c')eff = N(c) + N'(c')fict
If we were to first turn off the rotation (set ω = 0) and then align the axes, it seems we should have
N'(c')eff → N(c) r - c = r' - c' c - c' = b
But in this case you need to have c' → c or it does not make sense. So I don't think I can set these two points arbitrarily as I have done, so lets now rewrite Section 11 (b) (for starters). There I will have c' and c be vectors to the same point in space!