By Derek F. Lawden
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Additional info for A Course in Applied Mathematics, Vol. 1 and 2
K = 0·7k. 11 0 kx4dx + i\ ydy 0 The power which is being developed by a force acting upon a particle at any instant is defined to be the rate at which the force is doing work at this instant. Consider again the particle which moves along the curve A B (Fig. 3). Let at be the time taken to move from P to P' The work done by F in this time at is F cos eas. Hence, the average rate of doing work over this time interval is as F cos 6 -· at Letting at -- 0, this expression approaches the value H = Fv cos e = F v, .
A COURSE IN APPL IE D MATHEMAT ICS assuming that the integration can be carried out. , at t = 0. 14) yields the value of v appropriate to this instant. 15) We shall normally find it convenient to choose 0 to be the position of the particle at time t = 0. 15} is satisfied by these values. As a particular case, F may be constant. The acceleration is then also constant and equal to f, say. Thus, = = dv = f. 16) v = u + jt, where v = u at t = 0. 17) x = ut + tft2 , having assumed that x = 0 at t 0.
R= a (i) (ii) 16 A COURSE I N APPLIED MATHEMATICS From (i), & From (ii), = [CH. (2 + y3)w. r = -law2 cos lwt. w2(4 + 2y3)0N = -w2(4 + 2y3)MN. = fr is therefore representable, both in magnitude and direction, by the --+ vector (4 + 2y3)w2• PN. fo is similarly representable by the vector --+ (4 + 2y3)w2• NM. The total acceleration of P is now given by the vector sum --+ --+ (4 + 2y3)w2 • PN + (4 + 2y3)w2• NM = --+ --+ (4 + 2v3)w2 • (PN + NM) --+ = (4 + 2y3)w2. PM. EXERCISE 1 P and Q are moving around two coplanar circles in the same sense and with the same angular velocities w.
A Course in Applied Mathematics, Vol. 1 and 2 by Derek F. Lawden