details of equ 15_30
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Supporting derivation note in Phil's own words, dated 8.18.12, kept because it shows the brute-force method before he found a shortcut. It rotates the primed-frame position, velocity and acceleration components (eqs. 15.29-15.31) by angle φ, using the sin/cos angle-addition identities to get terms in φ+Ωt. It ends with an expansion of the x-acceleration that includes ω, Ω, V and b terms.
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This is true support material do not delete. It derives equation 15.30 by the brute force method. I later learned you can get the same results by a trick, but I want the brute force stuff kept. All the sinsin+coscos stuff is here. -- 8.18.12
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r'(t) = x' ' + y' '
where (15.29)
x' = (r'0 – Vt)cos(Ωt)
y' = (r'0 – Vt)sin(Ωt)
' = cosφ + sinφ
' = -sinφ + cosφ (15.6)
r'(t) = x' [cosφ + sinφ ]+ y' [-sinφ + cosφ ]
= (x' cosφ - y' sinφ) + (x' sinφ + y' cosφ)
= (r'0 – Vt)[ (cos(Ωt) cosφ - sin(Ωt) sinφ) + (cos(Ωt)sinφ + sin(Ωt)cosφ) ]
= (r'0 – Vt)[ (cos(Ωt) cosφ - sin(Ωt) sinφ) + (sinφ cos(Ωt) + cosφ sin(Ωt)) ]
= (r'0 – Vt)[ (cos(φ +Ωt) + (sin(φ+Ωt) ]
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v' = (v')'x ' + (v')'y '
where (15.30)
(v')'x = –Vcos(Ωt) – Ω (r'0 – Vt)sin(Ωt)
(v')'y = –Vsin(Ωt) + Ω (r'0 – Vt)cos(Ωt)
v' = (v'x cosφ - v'y sinφ) + (v'x sinφ + v'y cosφ)
(v'x cosφ - v'y sinφ) =
[–Vcos(Ωt) – Ω (r'0 – Vt)sin(Ωt)]cosφ
- [–Vsin(Ωt) + Ω (r'0 – Vt)cos(Ωt)] sinφ
= –V cosφ cos(Ωt) – Ω (r'0 – Vt) cosφ sin(Ωt)
+V sinφ sin(Ωt) –Ω (r'0 – Vt) sinφ cos(Ωt)]
= -V [cosφ cos(Ωt) - sinφ sin(Ωt)] – Ω (r'0 – Vt)[ sinφ cos(Ωt) + cosφ sin(Ωt)]
= -V cos(φ+Ωt) – Ω (r'0 – Vt) sin(φ+Ωt)
To get the other result, take cosφ→sinφ and sinφ → -cosφ. Do this in the second last line above
(v'x sinφ + v'y cosφ)
= -V [sinφ cos(Ωt) +cosφ sin(Ωt)] – Ω (r'0 – Vt)[ -cosφ cos(Ωt) +sinφ sin(Ωt)]
= -V [sin(φ+Ωt)] – Ω (r'0 – Vt)[ -cosφ cos(Ωt) +sinφ sin(Ωt)]
= -V [sin(φ+Ωt)] + Ω (r'0 – Vt)[ cosφ cos(Ωt) – sinφ sin(Ωt)]
= -V [sin(φ+Ωt)] + Ω (r'0 – Vt)[ cos(φ+Ωt)]
= -V sin(φ+Ωt) + Ω (r'0 – Vt)cos(φ+Ωt)
So we can then write
v' = (v'x cosφ - v'y sinφ) + (v'x sinφ + v'y cosφ)
= (v')x + (v')y
where
(v')x = -V cos(φ+Ωt) – Ω (r'0 – Vt) sin(φ+Ωt)
(v')y = -V sin(φ+Ωt) + Ω (r'0 – Vt)cos(φ+Ωt)
(v')x = (v'x cosφ - v'y sinφ) = -V cos(φ+Ωt) – Ω (r'0 – Vt) sin(φ+Ωt)
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a' = (a')'x ' + (a')'y '
where (15.31)
(a')'x = 2ΩVsin(Ωt) – Ω2x'
(a')'y = – 2ΩVcos(Ωt) – Ω2y'
a' = ((a')'x cosφ - (a')'y sinφ) + ((a')'x sinφ + (a')'y cosφ)
= (a')x + (a')y
where
(a')x = ((a')'x cosφ - (a')'y sinφ)
= [2ΩVsin(Ωt) – Ω2x']cosφ
- [– 2ΩVcos(Ωt) – Ω2y']sinφ
= [2ΩVsin(Ωt) – Ω2 (r'0 – Vt)cos(Ωt)]cosφ
- [– 2ΩVcos(Ωt) – Ω2 (r'0 – Vt)sin(Ωt)]sinφ
= [2ΩVsin(Ωt) – Ω2 (r'0 – Vt)cos(Ωt)]cosφ
+ [ 2ΩVcos(Ωt) + Ω2 (r'0 – Vt)sin(Ωt)]sinφ
= 2ΩV { sin(Ωt) cosφ + cos(Ωt) sinφ } + Ω2 (r'0 – Vt) { sin(Ωt)sinφ - cos(Ωt)cosφ }
= 2ΩV { sin(φ + Ωt) } + Ω2 (r'0 – Vt) { sin(Ωt)sinφ - cos(Ωt)cosφ }
= 2ΩV { sin(φ + Ωt) } – Ω2 (r'0 – Vt) { -sin(Ωt)sinφ + cos(Ωt)cosφ }
= 2ΩV { sin(φ + Ωt) } – Ω2 (r'0 – Vt) { cos(φ + Ωt) }
= 2ΩV sin(φ + Ωt) – Ω2 (r'0 – Vt)cos(φ + Ωt)
To get the other result, take cosφ→sinφ and sinφ → -cosφ. Do this in the 5th last line above
(a')y = 2ΩV { sin(Ωt) sinφ - cos(Ωt)cosφ } + Ω2 (r'0 – Vt) { -sin(Ωt)cos - cos(Ωt)sinφ }
= - 2ΩV { - sin(Ωt) sinφ + cos(Ωt)cosφ } - Ω2 (r'0 – Vt) { sin(Ωt)cosφ + cos(Ωt)sinφ }
= - 2ΩV { cos(φ + Ωt) } - Ω2 (r'0 – Vt) { sin(φ + Ωt) }
= - 2ΩV cos(φ + Ωt) } - Ω2 (r'0 – Vt) sin(φ + Ωt)
Therefore we find that
a' = (a')x + (a')y
where
(a')x = 2ΩV sin(φ + Ωt) – Ω2 (r'0 – Vt)cos(φ + Ωt)
(a')y = - 2ΩV cos(φ + Ωt) } – Ω2 (r'0 – Vt) sin(φ + Ωt)
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ax = 2ΩVsin(φ + Ωt) – Ω2(r'0 – Vt)cos(φ + Ωt)
– [– bcosφ + (r'0 – Vt)sin(φ + Ωt)] – ω2[bsinφ + ( r'0 – Vt)cos(φ + Ωt)]
- 2ω[–Vsin(φ + Ωt) + Ω (r'0 – Vt)cos(φ + Ωt)]
= [2ΩV– (r'0 – Vt) +2ωV ] sin(φ + Ωt)
+ [– Ω2(r'0 – Vt) – ω2 ( r'0 – Vt) -2ω Ω (r'0 – Vt)]cos(φ + Ωt)
+ bcosφ - ω2bsinφ
= [2(ω+Ω)V– (r'0 – Vt) ] sin(φ + Ωt)
+ [– (ω2+ Ω2)r'0 -2ω Ω (r'0 – Vt)]cos(φ + Ωt)
+ bcosφ - ω2bsinφ
= [2(ω+Ω)V– (r'0 – Vt) ] sin(φ + Ωt)
+ [– (ω- Ω)2r'0 +2ω Ω Vt)]cos(φ + Ωt)
+ bcosφ - ω2bsinφ