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Draft scrap notes dated 3.26.05 from a folder on reference frames in mechanics, apparently by Phil. They summarize the inverse-problem results (10.1)-(10.5) for r', v', a', L' and (dL'/dt)S', then substitute v' and a' into the torque-type equation using vector triple-product identities. The expansion gets messy and the text remarks there must be a simpler route via the rule (dL'/dt)S = ω x L' + (dL'/dt)S'.
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This is the Title PhL 3.26.05
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For the (dL/dt)S equation, we start with its alternate form (9.10)
(dL/dt)S = ω x L + ω x (r x v') + r x ( x r) + b x a' + (dL'/dt)S'
to find that
(dL'/dt)S' = (dL/dt)S – ω x L – ω x (r x v') – r x ( x r) – b x a' (10.5)
which we leave "as is". One can insert (10.2) and (10.3) for v' and a', but the result is unpleasant.
We may now summarize the solution of the Inverse Problem:
r' = r - b (10.1)
v' = v - ω x r (10.2)
a' = a + ω x (ω x r) – x r – 2 ω x v (10.3)
L' = L – r x b – r x (ω x r) (10.4)
(dL'/dt)S' = (dL/dt)S – ω x L – ω x (r x v') – r x ( x r) – b x a' (10.5)
******************
We now have to replace both v' and a'. First v' :
– ω x (r x v') = – ω x (r x[v - ω x r]) = –ω x (r x v) + ω x (r x (ω x r)
From (9.13) with r ↔ ω we know that
ω x (r x (ω x r) = - (ω r) (ω x r) = (rω) (r x ω)
so that
– ω x (r x v') = –ω x (r x v) + (rω) (r x ω)
As for the a' term, we have
– b x a' = - b x [a + ω x (ω x r) – x r – 2 ω x v ]
= - b x a - b x (ω x (ω x r)) + b x ( x r) + 2 b x (ω x v)
Inserting these two results gives
(dL'/dt)S' = (dL/dt)S – ω x L – ω x (r x v') – r x ( x r) – b x a'
= (dL/dt)S – ω x L –ω x (r x v) + (rω) (r x ω) – r x ( x r)
- b x a - b x (ω x (ω x r)) + b x ( x r) + 2 b x (ω x v)
or
(L'/dt)S' = (dL/dt)S - 2 ω x L + (rω) (r x ω) – r x ( x r)
- b x a - b x (ω x (ω x r)) + b x ( x r) + 2 b x (ω x v)
There must be something simpler! The G rule for L' says
(dL'/dt)S = ω x L' + (dL'/dt)S'
which we c