Spring 2003 16.614-2 The approach is to find the motion of P with respect to frame 1 P1 1 7+("r+ P1 7+3 27+(372+2(2×3)+21×37+2×(2×) While the notation is a bit laborious, there is nothing new here this is just the same case we have looked at before with one frame moving with respect to another Steps above were done ignoring the motion of frame 1 altogether Now consider what happens when we include the fact that frame 1 is moving with respect to the inertial frame I. Again P1 +-T+ +1y Now compute the desired velocity P1 PI P+(2×PF) I 2 7+2×7)+1×(27+7) 1 2 F+27+2+1×27+(1+2)×7Spring 2003 16.61 4–2 • The approach is to find the motion of P with respect to frame 1. P1 ˙
r 1 = 2 ˙
r 1 + 3 ˙
r 1 ≡ P1
v = 2 ˙
r 1 + (3 ˙
r 2 + 2
ω × 3
r ) and P1¨
r 1 = 2¨
r 1 + 3¨
r 1 ≡ P1
a = 2¨
r 1 + (3¨
r 2 + 2(2
ω × 3 ˙
r 2 ) + 2 ˙
ω 1 × 3
r + 2
ω × ( 2
ω × 3
r )) • While the notation is a bit laborious, there is nothing new here – this is just the same case we have looked at before with one frame moving with respect to another. – Steps above were done ignoring the motion of frame 1 altogether. • Now consider what happens when we include the fact that frame 1 is moving with respect to the inertial frame I. Again: P I
r = 1
r + 2
r + 3
r ≡ 1
r + P1
r Now compute the desired velocity: P I ˙
r I = 1 ˙
r I + P1 ˙
r I ≡ P I
v = 1 ˙
r I + P1 ˙
r 1 + (1 ω
× P1
r ) = 1 ˙
r I +
2 ˙
r 1 + (3 ˙
r 2 + 2
ω × 3
r ) + 1
ω × ( 2
r + 3
r ) = 1 ˙
r I + 2 ˙
r 1 + 3 ˙
r 2 + 1
ω × 2
r + (1 ω
+ 2
ω) × 3
r