By Thomas R. Kane

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SEC. 3 Diff. of Vectors] 25 where 0 is the angular displacement of a line d fixed in D, relative to a line c fixed in C, 0 being regarded as positive when generated -C FIG. 3a by a clockwise (as seen by the reader) rotation of d (or D) relative to c (or C). A and B are points fixed in D. Draw a sketch showing the time-derivative in C at t* of the position vector p of B relative to A. Solution: Let n be a unit vector parallel to line AB. 5) dt t* dt\ Then p Let k be a unit vector perpendicular to the plane of the disc D, choosing the sense of k such that D rotates clockwise during a k rotation of D relative to C (see Fig.

2) to be determined. P Λr ^ * 1 B 1 Q ► H FIG. 5b Express φ in terms of 0: From Fig. 5b, ~PQ = r sin 0 and TQ cotan φ + r cos 0 = rV2 Hence sin 0 cotan φ + cos 0 = V 2 so that φ = arc cotan and V2 — cos fl\ 0> /'V - cos \ / sin 0 άφ _ V 2 cos 0 - 1 άθ dt 3 - 2 Λ / 2 COS 0 * When P crosses line AB, (D) 0= 0 Thus, using Eq. 5 rad min - 1 2V2 (The minus sign means that the follower is rotating clockwise at the instant in question, as may be seen either by substitution into Eq. 6 The theorem stated in Sec.

FIG. 5) of v at z* + h and at 2*, where z* is a particular value of z and h a scalar having the same dimensions as 2, can be expressed in terms of values of derivatives of v with respect to z in R at z*, as follows : R h Rdv\ h? RcN\ *+A "~ V U* 1! dz\ 2* 2 ! dz2 + 1, 2, 3, be unit vectors (not parallel to the Proof: Let nt, i same plane) fixed in R. 3) and, by Taylor's theorem for scalar functions, WdhJ , « _ h^dvj\ v i\z*+h ~~ ΌΑζ* ~ Wfa 2! dz2 + +. The values of v at z* and at z* + h, expressed in terms of their nt-, i = 1, 2, 3, components, are 3 3 1=1 Diff.