Analytical mechanics for students of physics and engineering . MOTION 81 PROBLEMS. 1. Find the path and the velocity of a particle which moves bo thatits position at any instant is ijiven by the following pairs of equal ii (a) x = at, V = ht. (b) x = at, y = at- \ gt°- (c) X = at, y = b cos ut. (d) X = a sin ut, V = bt. (e) x = a sin ut, y = a cos ut. (f) x = a sin ut, y = b sin u& (g) x = aekt, v/ = ae~kt. ve the relation v = Vx2 + y 2 + i2. 81. Radial and Transverse Components of Velocity. — Themagnitude of the velocity along the radius vector is, accord-ing to the results of the preceding s
Analytical mechanics for students of physics and engineering . MOTION 81 PROBLEMS. 1. Find the path and the velocity of a particle which moves bo thatits position at any instant is ijiven by the following pairs of equal ii (a) x = at, V = ht. (b) x = at, y = at- \ gt°- (c) X = at, y = b cos ut. (d) X = a sin ut, V = bt. (e) x = a sin ut, y = a cos ut. (f) x = a sin ut, y = b sin u& (g) x = aekt, v/ = ae~kt. ve the relation v = Vx2 + y 2 + i2. 81. Radial and Transverse Components of Velocity. — Themagnitude of the velocity along the radius vector is, accord-ing to the results of the preceding section, a) Vr ill The expression for the velocity at right angles to r is ob-tained by considering the motion of the projection of tin-particle along a perpendicular ytor. When the particle movesthrough ds, its projectionmoves through r dd, Fig. 50,therefore the required velocityis rdd dt do -nr— = (2). The components vr and vp may be expressed in terms of xand y by differentiating the equations of transformation r2=a;2+ya (3) A (4) and 6 = tan 82 ANALYTICAL MECHANICS with respect to the time. Differentiating (3) we obtaindrV=dt_x dx y dyr dt r dt= i cos 0 + ?/sin 0. (5) Differentiating (4) we get dd v = r — p dt _rxy-yx x2 + y2 = ycosd-x sin 0. (6) These components satisfy the relation v = Vr2+ r-b\ (7) ILLUSTRATIVE particle describes the motion defined by the equations x = a cos kt, (a) and y = a sin kt. (b) Find the equation of the path, the velocity at any instant, and the com-ponents of the latter. Squaring and adding (a) and (b) we eliminate t and obtainX2 -f if = o* Y for the equation of the (a), we have . dxX = dJ = —ha si i kt = -ky. Differentiating (b), we obtain . dy = ka ens kt= fee.
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