A Course of Mathematics for Engineers and Scientists. Volume by Brian H. Chirgwin, Charles Plumpton

By Brian H. Chirgwin, Charles Plumpton

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Corresponding to eqn. 47) where xAdoes not vary with time. Coordinates in the frame OX, X2 X3 are related to those in Ox1 x2 x3by the matrix R, through the equations X = R,x , XA = Ri xA . 48) In eqn. 47), XA varies with the time because of the variations in R1. The coordinates of P in 0X1 X2 X3 are X = Ri xA R i Rg • Differentiating this relation gives the resolutes of the velocity of P in the frame 0X1 X2 X3 . = iti XA (h i R2 = R1 xA (R1 R2 =Rl Ri`Y A = it,R;XA RA)g Rik)R2 (x — x A) + R1 112) — XA) (kRj R1ll2R2R1) (X — XA).

First we consider the equilibrium of the whole system by taking moments about the line AB. R3 . CN — (W + W3) . GN — (2 W + + W2) . GN = . Since CN = 3GN, CG = 2GN, this equation gives 6R3= 4 (W + R3 = W + + (2 W + + + W2). + W3. § 2 : 2] 53 SETS OF FORCES : EQUILIBRIUM Similarly, R, = W + g w, + w2 + 1472, R2 = w + + w2 + W3 Effectively, we have used three conditions of equilibrium by taking moments about three different axes. No fresh information can be obtained by resolution since R1 ± R2 + R3 = 3 W + + W2+ W3.

30) we have quoted eqn. ] But R = R, R2 . R = 11,R, + R1/12 . RR' = (111112+ R1R2)1:4111 = 141R; + R1 R211A. 31) Eqn. 31) shows that when a rigid body has two angular velocities simultaneously we add the angular velocity vectors to obtain the resultant angular velocity. Alternatively, the angular velocity of one moving frame relative to another moving frame is obtained by a formula similar to that for linear velocities. Example. , find BcoA . The relation between the frames is = x. The observer B will calculate the angular velocity of A's frame using 11, whereas A would use R1.

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