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- 323 Homework set 8, Spring
323 Homework set 8, Spring 2007 Problem 11.3 P=PR= qu sin? (wt)R (Eq. 11.15) > (P) = daqju*R. Equate this to Eq. 11.22: Liana poahPust _ Modu | ince wr = 2me pee Toe + Gre |r Ie = = GEASS Feme (5) = fete 10x10 ()' = eo (5) 0 =freoaarara] = Gee ae = er) = gran x 1077)(3 x 10°) x = 80 X 1 =) 789.6(d/A)? 2. Problem 11.9 At t = 0 the dipole moment of the ring is is ar Po [ova = [0 sin $)(bsin PF + bcos@R)bdd = Apb* (s[ ain? odp.+8 [ singcosdd) Ab? (mF + OR) = WHA SH. W As it rotates (counterclockwise, say) p(t) = po[cos(wt)¥ — sin(wt) X], so = —w*p, and hence (js)? = wpz. _ [pow bt carn = Hoa 2 Therefore (Eq. 11.60) P= 57 u' (nb? do) Problem 11.10 p=-eyy, y = bol? so p = —Aget?¥; §=-ge¥. Therefore (Eq. 11.60): P = EX (oe). Now, the time ac it takes to fall a distance h is given by h = tgt? + t = \/2h/g, 90 the energy radiated in falling a distance A 2 is Ura = Pt = totes) /2h/g. Meanwhile, the potential energy lost is Upoe = mgh. So the fraction is _ Vesa _ oat e* 2h 1 ue* [2g | _ (4m x 10-7)(1.6 x 10-)?_ /(2)(9.8) [276% 10] Upoe One Vo mgh ~|6xmeV h'|~ Gx(9.11 x 10-31)(3 x 10) \/ (0.02) [2:76 x 10°] Evidently almost all the energy goes into kinetic form (as indeed I assumed in saying y = }gé?). Problem 11.15 a 0 According to Eq. 11.74, the maximum occurs at — eB [el Bcos 8) Qsin@cos@ — Ssin? @(GsinB) _ _ = 5asin? 9 = ~ cos? 6): To Feasts 7 T= Boos aye 0 Pease Bcos@) = 58 sin? @ = 5(1 - cos* 6); 2cos@ ~ 28cos* @ = 58 —58-cos* 4, or 38cos*4 +4cosé— 58=0. So 0. Thus — 2 cos = —2e 4+ OORT vine = 3 (4Vi+ 152? — 1). We want the plus sign, since By, > 90°(c08 0, = 0) when 2 8 +0 (Fig. 11.12): | max = cos? (Ho For vs c, 6 = 1; write 8 = 1 -e (where € € 1), and expand to first order in ¢: (=) irs g VIF a= - 1] ] 2 3a +0 [vie isa =] 38 = hate (Vig 30-1] = hate [a Tea) - 1] =Ftl +e) i (1-3) -1| 1 1b 5 5 1 = -2e\l= ~ re) = e=1--«. gu t+9 (3 1 ‘) (l+e)(1- =e) Pl+e i ze Evidently Omax 0, 80 COS Omax 2 1~ 462g, = 1— fe => OF, = $6, OF Omax = /e/2 =| (1 — A) /2. (dP/dQe..Jue _ sin? Oran mt Let f= et f= TPjaNly,.) T= 8008 8max)® | . Now sin? @max @ €/2, and rest ur 5 (1-8 c086max) #1 (1 e)(1- be) S1-(1-e- de) = Se, So f= a = (3) a. But 1 Sex i. Therefore jet ld 1 "Vi-® Vien-— Vi-d-%) ve
