rigid boday app section

G.8 Comments on Rigid Body Dynamics In this section we assume the Special Case #4 version of Fig 1 where Frame S and Frame S' have the same origin. Assume Frame S is inertial and Frame S' is embedded into a rotating rigid body. At some instant in time Frame S' might have an orientation relative to Frame S according to the Euler angles of Fig (G.5.5). From (11.3.2) Newton's Second Law (angular motion) applied in inertial Frame S states, N = // = ∂SL where we set the torque and angular momentum reference point to the common Frames' origin. Using the G Rule (2.1) one finds, = ∂SL = ∂S'L + ω x L so Newton says (∂S'L) + ω x L = N . Note that (∂S'L) is not a "natural" object in the sense of Section 1.8 since L is a Frame S object, but the time derivative taken in Frame S'. We wish to take the Frame S' component of the above equation. For (∂S'L) we write (∂S'L)'i = ∂S'[ (L)'i] = ∂t(L)'i In the first equality we have used the commutation theorem (1.11.1) that (∂S'a)'j = ∂S'[(a)'j]. In the second equality we use the fact (1.10.1) that the time derivative of a tensor component requires no frame specification. The Frame S' components of ** are therefore ∂t(L)'i + εijk(ω)'j(L)'k = (N)'i (∂S'L)'i = ∂S'[ (L)'i] = ∂t(L)'i // using below 1.11.1 Then we really should write rigid body doc ∂t(L)'i + [ ω x L]'i = N'i or d(L)'i/dt + εijk(ω)'j(L)'k = (N)'i Meanwhile, the angular version of p = mv is L = Iω where ω is the angular momentum vector and I is the inertia tensor. Then = I ∂t(L)'i = Iij ∂t(ω)'j = Iij ()'j Question: Is I the Frame S inertia tensor? I think yes, since I have the Frame S L, and is the same object in both Frames from ***. But somehow you want it to be I' which is diagonal only in Frame S'. Have to mess with that a bit. This same question is fudged in my Goldstein section 5.5 notes! Then Iij ()'j+ εijk(ω)'j(L)'k = (N)'i If Frame S' is chosen so that the inertia tensor of the rigid body is diagonal, I = diag(I1,I2,I3), then Ii ()'i+ εijk(ω)'j(L)'k = (N)'i meanwhile again L = Iω (L)'i = Ii(ω)'i Then it is Ii ()'i+ εijk(ω)'j Ik (ω)'k = (N)'i I1 ()'1 + (ω)'2 I3 (ω)'3 - (ω)'3 I2 (ω)'2 = (N)'1 or I1 ()'1 + (ω)'2 (ω)'3[ I3 - I2] = (N)'1 I guess you then say (N)'1 = N1 (the true torque in both frames is the same) Then you have I1 ()'1 – (ω)'2 (ω)'3[ I2 - I3] = N1 This then explains the equations I see in Goldstein page 159. This is pretty much all I wanted to do in this Comment. Relate things to Goldstein equations, but show the Frame S' stuff carefully in terms of notation. This does bring in some frames doc facts! I guess I should also read through this Goldstein section first. Update 2.27.17 I have reviewed Goldstein on this topic, and he fudges the issue of what frame L and I are "in". He just says nothing about it. So I will need to look elsewhere. It seems to me right now that in terms of a spinning symmetric top, say, that I will only be diagonal in the body frame. So we have this contradiction that in writing L = Iω Goldstein would be using the space L and the body I, which to me does not make sense. I don't think that L and L' are the same. In my Special Case #4 summary I say L = L' + mr' x (ω x r') (f) for a point particle in the rigid body. So you cannot say L = L' and escape that way! I will need another source on this topic! I will start with the usuals: Marion: On page 365 (12.17) he writes L = mr' x (ω x r') summed over particles in the rigid body. How does this relate to my equation above?? I think he is implying that this is all in the body system. I think I need to resolve this question. For his L, he uses v = ω x r as the velocity of each particle relative to the fixed origin and just sums L = r x p relative to the fixed origin point. Resolution! In the body frame, the spinning top is at rest and so L' = 0, duh!! For each particle in the body frame, p' = 0 so r' x p' = 0. So I guess L really is the space L as I have assumed all along above. So now consider L = Iω again in the space frame. How can you just "select" principal axes? You cannot select Frame S to be aligned with the top axis because that axes is moving (rotating, not to mention nutating) but S has to be an inertial frame! and indeed it is the Frame S' components that you want!! This last equation appears as and you see how they are using a very compact notation. But I agree, their ωi really are the Frame S' components!!!