rotation debug 4

Rotation debug 4 PhL 7.7.12 Review Comments: This is just more flailing due to my bad G Rule understanding. I compute v and a directly by differentiating r in Frame S and my results disagree with my interpretation of the G Rule. Probably my differentiation math is all OK. It turns out that there is a LOT more to this stuff, as now shown in frames doc, than meets the original eye. 1. Derivation of the Goldstein Rule: Suppose we have some vector VF being viewed in some frame F of interest. In some arbitrary frame f this vector will have coordinates [VF]f, and we could select f = F if we wanted. If we apply to this vector in frame F a small rotation R(dφ) within frame F, then we get dVF = [ -i dφJ ] VF = dφ x VF . Now suppose the frame in which V lies happens to be "the body frame" S' which is instantaneously rotating by dφ in time dt relative to the space frame. What those words mean is that the basis vectors of the body frame, as viewed in the space frame, move a little according to e'n(t+dt) = R(dφ) e'n(t) If we identify frame F as S', then our equation above would read dVS' = dφ x VS' Then if the vector V is static in the body frame, it has a change in the space frame given by dV = dφ x V But how to we indicate that the V appearing above is in the body frame? [ endless confusion] ________________________________________________________ Problem 1: An ant is glued to a phono disk at some radius r with θ = 0 at t=0. The disk starts rotating at time t=0 with some ω(t) that varies in time. Describe everything that happens and do it in polar coordinates. d/dt = -ω and d/dt = ω ω = dθ/dt θ(t) = !Syntax Error, Iω(t)dt = known. = rcosθ(t) + r sinθ(t) r(t) = r = -rsinθ(t) + r cosθ(t) v(t) = dr/dt = r d/dt = rω a(t) = dv/dt = r + rω (-ω) = r - rω2 da/dt = r + r( -ω) - 2rω - rω2(ω) = -3rω + (r - rω3) Now, how does this relate the Goldstein's various "rules". Here they are: dr/dt = ω x r = ω x r = ωr // agrees with above dv/dt = ω x v = ω x [rω] = -rω2 // disagrees with above da/dt = ω x a = ω x [r - rω2] = -rω - rω3 // disagrees with above [ I guess my thinking here was that since the ant is glued down, then vbody = 0 and abody = 0 and then one has dvbody/dt = 0 etc so that is why there is only ω x q sitting there in the last two G Rule lines. The first line would have drbody/dt = 0 because rbody does not change. If this is done correctly, the second line would be ∂Sv = ∂S'v + ω x v and ∂S'v is a cross acceleration that has to be computed and is not 0. I just don't yet have the tools needed to do ant problems! ] Now to understand why, consider these facts: r = r dr/dt = rω r dr/dt = 0 v = rω dv/dt = r - rω2 v dv/dt ≠ 0 a = r - rω2 da/dt =-3rω + (r - rω3) a da/dt ≠ 0 When we say r is "instantaneously going in a circle", we mean that r dr/dt = 0, the change in r is exactly perpendicular to r. So you can see that when ω is a function of time, neither v nor a goes in a circle in this instantaneous sense. This is why the Goldstein rules give wrong results in those two cases. This is the first time I have realized this simple fact. ________________________________________________________ Problem 2: This time the ant moves in radius with some r(t). Resolve the problem. d/dt = -ω and d/dt = ω ω = dθ/dt θ(t) = !Syntax Error, Iω(t)dt = known. = rcosθ(t) + r sinθ(t) r(t) = r = -rsinθ(t) + r cosθ(t) r(t) = r v(t) = dr/dt = r d/dt + = rω + a(t) = dv/dt = r + rω (-ω) + + ω = (r + ω) + (- rω2) Now we have r = r dr/dt = rω + r dr/dt ≠ 0 v = rω + dv/dt =(r + ω) + (- rω2) v dv/dt ≠ 0 Therefore, when ≠ 0, even the simplest Goldstein equation will fail. As a check dr/dt = ω x r = ω x r = ωr // not valid since missing the term. ________________________________________________________ Problem 3: Solve Problem 2 using the rotation plus extra piece idea. dr/dt = ω x r + = vr "in the body frame" dr = dθ x r + dr // same thing, note extra piece. Now what about dv/dt = ω x v + extra = ω x [rω + ] + extra = rω2(-) + ω + extra = -rω2 + ω + { + ω } extra = { + ω } Here we are able to "prop up" the dr/dt = ω x r equation with a simple extra term. Can we somehow interpret this extra term in this way Goldstein would like: (dv/dt)body = { + ω } Well, maybe we look back at v = dr/dt = ω x r + = vrot + vbody vbody = It seems to work! ________________________________________________________ Problem 4: Solve the problem with ant doing some arbitrary body motion. rbody = r ______________________________________________________ dr = dθ x r + drbody v = dr/dt = ω x r + (drbody/dt) = ω x r + vbody = ωr + vbody vbody = (drbody/dt) = d(r)/dt = r ω + v = dr/dt = ωr + [r ω + ] = 2rω + a = dv/dt = d/dt(2rω + ) = dt(2rω) + dt( ) da/dt = d2t(2rω) + d2t( ) I have failed to provide the ant with some general trajectory on the disk. Needs more work. I think problem 4 is one worth studying a lot. We used the first basic Goldstein extra thing to get v and then we just diff'ed to get a and da/dt. ______________________________________________________ Now try the next level Goldstein extra rule: dv = dθ x v + dvbody dv/dt = ω x v + (dvbody/dt) vbody = dt(r) = r ω + (dvbody/dt) = dt(r ω ) + dt( ) Now let's see if it works: a = dv/dt = ω x v + (dvbody/dt) = ω x [2rω + ] + dt(r ω ) + dt( ) = 2rω2(-) + ω + dt(r ω ) + dt( ) In order for this to "work" (ie, match a above) we would need to have this be true 2rω2(-) + ω = dt(r ω ) Well RHS = ω + r + rω (-ω) and we see it does NOT work because RHS has a . So why does this not work?