Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

CHAPTER 13 REVIEW 0 177
23. (a) r (t) = R coswt i + R sin wtj => v = r '(t) = -wR sinwt i + wRcoswt j, so r = R(coswt i + sinwtj) and
v = wR(- sinwti + coswtj). v · r = wR 2 ( -
coswt sinwt + sinwtcoswt) = 0, so v ..L r . Since r points along a
radius of the circle, and v ..L r , vis tangent to the circle. Because it is a velocity vector, v points in the direction of motion.
(b) In (a), we wrote v in the form wR u, where u is the unit vector - sin wt i + cos wtj . Clearly I vi = wR lui = wR. At
speed wR , the part1c . I e camp I etes one revo I ut10n, . a d. 1stance 2 R,. . T 2·nR 27r
1r m t1me = - R = - .
w . w
(c) a = dv = - w 2 R coswti - w 2 Rsinwtj = - w 2 R(coswt i + sinwtj ), so a = - w 2 r. This shows that a is proportional
dt
to rand points in the opposite direction (toward the origin). Also, ja j = w 2 jr j = w 2 R .
(d) By Newton's Second Law (see Section 13.4), F = ma, so jF j = m jaj = mRw 2 = m (~R) 2 = m ~~ 2
® 2012 Ccngage Lc;~mi n g. All Righls Rc:servcd. May not be sc~mncd. copied. or dupJicatcd. or posted to a publicly ncccssiblc websilc. in whole or In part.