me351_lec04

In summary all we need to nd are the two angular

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Unformatted text preview: f link 4 is determined. In summary, all we need to find are the two angular velocities ω3 , ω4 . The actual computation is quite straight forward. We start again from the vector loop equation: R2 + R3 + R4 − R1 = 0, Rj = rj eiθj , j = 1, . . . , 4. Recalling that the derivative of a complex number ˙ ˙ Z (t) with fixed modulus, |Z | = const, i.e., Z (t) = |Z |eiθ , is Z = iθZ , we find upon differentiating the loop equation ˙ ˙ ˙ i θ 2 R2 + i θ 3 R3 + i θ 4 R4 = 0 . The real and imaginary parts of this complex equation yield two real equations which can be written in matrix form as ˙ r2 cos θ2 ˙ θ3 r3 cos θ3 r4 cos θ4 θ2 =− ˙ r2 sin θ2 θ4 r3 sin θ3 r4 sin θ4 Assuming the position analysis has already been done, angles θ3 , θ4 are known (and of course, θ2 ˙ ˙˙ and θ2 are given). Thus we have two linear equations in two unknowns: θ3 , θ4 . We can make a small ˙ simplification in these equations – essentially by dividing out the given quantity θ2 – the advantage being that the resulting equations will no longer depend on time. Thus, we define k3 = ˙ θ3 dθ3 =, ˙ dθ2 θ2 k4 = 1 ˙ θ4 dθ4 = ˙ dθ2 θ2 and we have the equations r3...
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