Reader Exercise I_5_27
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Phil's worked exercise (I.5.27) from the Rigid Body appendix of his mechanics notes. It compares the angular velocity components in equation (I.5.26) with those from (H.4.12) in Euler-angle form, using Ω' = ω cosα (I'3-I'1)/I'1, L/I'1 = ω sinα/sinθ, and tanθ = (I'1/I'3) tanα. The algebra is reduced step by step until both equations hold as identities.
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This is a Reader Exercise in
STOP! How does this compare with
(ω)x = cosφ + sinθsinφ
(ω)y = sinφ - sinθcosφ
(ω)z = + cosθ . (H.4.12)
Well, = 0 and φ(t) = (L/I'1)t and = (- Ω') so the above says
(ω)x = sinθsinφ (- Ω') =?= ωsin(θ-α)sinφ
(ω)y = - sinθcosφ (- Ω') =?= - ωsin(θ-α)cosφ
(ω)z = (L/I'1) + cosθ (- Ω') . =?= ωcos(θ-α)
For this to be consistent with (I.5.26) I would have to show that
sinθ(- Ω') = ωsin(θ-α)
sinθ(- Ω') = ωsin(θ-α)
(L/I'1) + cosθ (- Ω') = ωcos(θ-α)
I know that Ω' = ωcosα (I'3-I'1)/I'1 so then have to show
- sinθωcosα (I'3-I'1)/I'1 = ωsin(θ-α)
(L/I'1) - cosθ ωcosα (I'3-I'1)/I'1 = ωcos(θ-α)
But ωsinα/sinθ = L/I'1 so restate these equations as
- sinθωcosα (I'3-I'1)/I'1 = ωsin(θ-α)
ωsinα/sinθ - cosθ ωcosα (I'3-I'1)/I'1 = ωcos(θ-α)
or
- sinθcosα (I'3-I'1)/I'1 = sin(θ-α)
sinα/sinθ - cosθ cosα (I'3-I'1)/I'1 = cos(θ-α)
or
- sinθcosα [ (I'3/I'1) - 1] = sin(θ-α) [ (I'3/I'1) - 1]
sinα/sinθ - cosθ cosα [ (I'3/I'1) - 1] = cos(θ-α)
Good grief! Now us
tanθ = (I'1/I'3) tan α (I'1/I'3) = tanθ/tanα (I'3/I'1) = tanα/tanθ
so then the two equations are
- sinθcosα [ tanα/tanθ - 1] = sin(θ-α) [ (I'3/I'1) - 1]
sinα/sinθ - cosθ cosα [tanα/tanθ - 1] = cos(θ-α)
or
- sinθcosα [ tanα/tanθ - 1] = sin(θ-α)
sinα - sinθcosθ cosα [tanα/tanθ - 1] = sinθcos(θ-α)
or
- sinθcosα [sinαcosθ/cosαsinθ - 1] = sin(θ-α)
sinα - sinθcosθ cosα [sinαcosθ/cosαsinθ - 1] = sinθcos(θ-α)
or
- sinθcosα [sinαcosθ - cosαsinθ] = sin(θ-α)cosαsinθ
cosαsinθsinα - sinθcosθ cosα [sinαcosθ- cosαsinθ] = cosαsinθsinθcos(θ-α)
or
- sinθcosα sinαcosθ + sinθcosαcosαsinθ = sin(θ-α) cosα sinθ
cosαsinθsinα - sinθcosθ cosαsinαcosθ + sinθcosθ cosα cosαsinθ = cosαsinθsinθcos(θ-α)
or
- sinαcosθ + cosαsinθ = sin(θ-α) // correct!!!
sinα - cosθ sinαcosθ + cosθ cosαsinθ = sinθcos(θ-α) // problem Houston
cosαsinθsinα - sinθcosθ cosαsinαcosθ + sinθcosθ cosα cosαsinθ = cosαsinθsinθcos(θ-α)
sinθsinα - sinθcosθ sinαcosθ + sinθcosθ cosαsinθ = sinθsinθcos(θ-α)
sinα - cosθ sinαcosθ + cosθ cosαsinθ = sinθ[cosθ cosα + sinθ sinα]
sinα - cosθ sinαcosθ + cosθ cosαsinθ = sinθcosθ cosα + sinθsinθ sinα
sinα - cos2θ sinα + cosθ cosαsinθ = sinθcosθ cosα + sin2θsinα
sinα + cosθ cosαsinθ = sinθcosθ cosα + sin2θsinα + cos2θ sinα
sinα + cosθ cosαsinθ = sinθcosθ cosα + sinα
cosθ cosαsinθ = sinθcosθ cosα yes !!!!