analytic mechanics 7E ch9

analytic mechanics 7E ch9 - CHAPTER 9 MOTION OF RIGID...

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Unformatted text preview: CHAPTER 9 MOTION OF RIGID BODIES IN THREE DIMENSIONS 9.] (:1) [Au = “y: +22 )dm dm 2 pdrdy and m = 202 p = E x=0 a = [zapyzay E 2 4 ma2 I 1 : _ a = _— g “ 3 p 3 ' - ---- -- 1 i ’ 2a .— 2 2 _ ya: (:21: ‘2 _ ' I J I)“T — +z )dm— LO LO .\ pain]; .4, a V 3 1 8a p = dv L" 3 . I _ 804p _ 4mg: "t" — 3 _ 3 From the perpendicular axis theorem: 5 2 I :1“ +1"). + ma 2 . , 3 y=1l 1:20 I”, = I” = — Ixydm = — L0 L0 xypdrajz 2 If!” -pa‘=-% In =1J ——J-xzdm=0=1_w =1, 2 l (b)cosa=— cosfl=— cosy=0 From equation 9.1.10 ma? 4mu2 ( I ma2 2 1 I = — + — + 2 — —- — 3 5 3 5 2 J3 5 =£ma2 15 (c) (I)=(u(fcosa+jcos,5)=%(2i+}) From equation 9.1.29 [Ni 7 2(1) ma7+ a) ma2 + ~. 20) may]+ (a 427102 :1 _. — _ —_ _ _. J3 3 J3 2 J J5 2 J5 3 (mm:2 A ~ [+2 ' 6J3 ( J) ((1) From equation 9.1.32: 7 I" — to wma‘ l 7 7 (u-L=—- (2+2)=—ma'a)‘ 5 2J3 6J5 Z: T: 9.2 (a) (a (i+]‘+li) m(2a)2 2 . c : 7 : : — F = : 1 -1 , 2 12 3 ma I U I“ In. =— Jirydm = 0 since. for each rod. either x or y or both are 0. The same is true for the other products of inertia. _a a From equation 9.1.29: _ 2 a) r A. . c L=—maz-—( +_/+k) 3 f3 From equation 9.1.32, _ l (a 2ma2a)(lq muza)2 ——— “+12+12 = 2J3 3J3 ) 3 (b) From equation 9.1.10, with the moments of T 7 2ma' 3 inertia equal to and the products of inertia equal to O: 2ma2 2 I = 3 (cos2 a + cos2 ,6 + cos2 7) = Ema2 (c) 9 For the x-axis being any axis through the center of the x lamina and in the plane ofthe lamina, and the y-axis also in the plane ofthe lamina In = 1», due to the Similar geometry of the mass distributions with respect to the x- and y-axes. From the perpendicular axis theorem: Ix : In + In Ix = 2]” ma2 From Table 8.3.1, I: = 6 I” : ma2 " 12 21 9'3 (a) From equation 9.2.13, ta1126 = I If _ ’1}: From Prob. 9.1. I” = "111’ ’ I“. = 4W?“ , 1n = _ mu“ 3 ~ 3 . 2 tan 261 = 1 219 = 45" 19 = 22.5: The l—axis makes an angle of 22.5” with the x—axis. (b) From symmetry. the principal axes are parallel to the sides ofthe lamina and perpendicular to the lamina, respectively. 9.4 (a) From symmetry. the coordinate axes are principal axes. From Table 8.3.1: (0 Il =—l”§1[(2u)2+(3a)2]=§ma2 I? = ’—”[:a2 +(3a)2] =Bma2 ' 12 12 _____ __ I m 3 z 5 3 ) I3 =El:a +(2a) J=Ema (z) e,+._e,+ e . 7: 3.3 m From equation 9.25, 1[13 ,afi 10 74a)? 5 29(1):] T—~— ——ma'-—+—ma'- +—ma -—— 212 14 12 14 12 14 7 2 2 =—mua) 24 (b) From equation 9.2.4, i-é g"102- a) J+é [Erricrz-EQJHé irizaz-flj '12 J11 212 m 312 fl - mazw . . A L: l“e +20e,+15e 12J1_4( 3‘ ‘ 3) “Em—M _W _ _ l 1 ML] (12+23+32)2.(132+201+153)2 (11.116)? 6:21.6" 9.5 (a) Select coordinate axes such that the axis of the rod is the 3-axis. its center is at the origin. and (2) lies in the l, 3 plane. i 2 : From Table 8.3.1, 1, = 12 =%, 13 = 0 ‘7’ a3=w(élsina+é3cosa) From equation 9.2.4. 2 2 L zfl—(Mé, sina + 0+0) : él "11 a) sina ’ 12 12 ’1 mild) . I is perpendicular to the rod, and : sma 12 (b) Since (I) is constant. from equations. 9.3.5 NI 2 0+0 . ml2 N2 =0+(wcosa)((usma) F N3 = 0+0 N = £1. "II—w- sinacosa ‘ 12 N is perpendicular to the rod (é3 direction) and to I: (él direction), and INI = “dim. sinacosa l2 2 a) . . . 9.6 From Problem 9.4 r = e +26, +3e M( l V 3) w 1 7 - II zgnza'. I, =—qma‘ , and I3 =ima2 12 ' 12 12 From eqns. 9.3.5: N! =0+g(2)(3)l710" (5—10):_”1(1-(0~ l4 I2 28 a)2 ma2 maza)2 N,=0 — 3 1 13—5 = 4 _ +,4()()l2( ) 28H (02 ma2 maze)2 N =O+—l 2 10—13 =— . 14m ) '2( > 28 lNl: masz (52+42+12)é : musz m 28 28 9.7 Multiplying equations 9.3.5 by a), , a)2 , and a)J ,respectively 0 = Labia), + 0402(1)3 (I3 — 12) 0 = Izabza)2 +wlcuza)3(1, — I3) 0 = kayo} +a)la)2a)3 (I2 —1,) Adding, 0 = I,a'),a)‘ + [2503(02 + 13(1)3w3 a’ 1 , 2 0 =E[§(I.w,2 + [3602' +I3a)3 From equation 9.2.5, 0 = $20,] C Tm, = constant Multiplying equations 9.3.5 by Ila)I . 12602, and 13a)3 , respectively: 0 = 15am, + Ila)la)za)3(l3 — [3) 0 = 12202503 + [3Q)l[£)2(!)3 (II — l3) 0 = [32(D3w3 + 130,0)!(93 (I2 — 1.) Adding, 0 = [farmI + Izzaiba)2 + 132(113023 0 = 3250.101 + 12(1)?2 + 13(1):) From equation 9.2.4. 0 = (It 7 L“ = constant 9.8 From equations 9.3.5 for zero torque O = Ila), +a22a)3(13 —12) O = 12602 +a)l(z)3(1l ——I3) From the perpendicular axis theorem. 13 = Il + 12 0 = 1.6;), + 002(1)}!l O = 12032 —a),a)312 (0‘ and w? , respectively: [I I2 0 = (Ola), + wlcuzwj Multiplying by 0 = (02(02 — (ulmza)3 1 d 7 7‘ Adding, 0 = cola), +w2a')2 = 35((15' + (02') a),2 +6022 = constant If II = 12. from the third of Euler‘s equations (13(1)3 = 0) (03 = constant 9.9 (a) From symmetry 5 = 13 and II = II = I From the perpendicular axis theorem. I3 = II +12 or Ix = 2] From eqn 9.5.8, Q = (2 —1)a)cosa UI For 05:45“. Q=£ fi For 2—”=13. w T. is the period of precession of (I) about é}. 27r_ T,= fis=r414s From equation 9.6.12, J04 s = 0.6323 T2 is the period of wobble of £3 about 1:. (b) I.=1.=1=fl a2+£ =lZ-"m' ‘ 12 l6 16 12 2 m 3 _ ma 13:15:1—2(a +a )—2 12 . 2 1 15 a) Frome uation 9.5.8, Q: ——1 (u—=—— q 2 J5 [ME 16 T,=2—”=1—7\/25=1.603s Q 13 1 . . 2. 2 l 33 From equation 9.6.12, ¢=w 1+ 2 126 A] —— 17 I J5 ¢=1.5072w 7 7g=£= I 5:0.6635 ' (13 1.5072 9.10 From equation 9.6.10. tanf) = — tana . I 5 Since 15 > 1 . 0 <51 and the angle between the axis of rotation ((23) and the invariable line (L) is a —9. From your favorite table of trigonometric identities mm (a 0) _ tuna — tang ‘ 1+tanatan0 I l—— tana 15 tan(a—6) =——— 1 '7 1+—tan' (1 IS 17 —1 tana a _g = tan—1 MT—, + 1 tan“ a . . , Iy . 9.1] 61—6 1.9211111le1111111] for —‘ a max1mum. I I, tana For —“=2. tan(a—0)=—, 1 2+tan' a dtun(a—0) _ 1 2tanza 2—tzm2a " ——2_ ——fi = ——_2 (“and 2+1“ 0‘ (2+tun'a) (2+t11113a) _ dtan (or—6) , At the maxnnum, —— = 0 = 2 — tan” a dtana a: tan” JE=54.7° tan(a—6)g_[_:_]£= ‘5 l+l(\/§)2 T 9.12 (a) From Problem 9.10. for 15 > I , the angle between 67) and I: is I —I tana a —6 = tan'l M,— IJ + 1 tan“ a From Problem 9.9(a), II = 21 and tana = tzm45° =1 a —9 = tan" 1 = tan"I l = 18.49 (2+1) 3 ma2 l7 ma2 (b) From Prob. 9.9(b), 15 = 2- and 1 =—- 12 16 12 [2~%) 15 01—49=t2m‘l ———)— = tan"—=l7.0” 2+u 49 I6 9.13 From Example 9.6.2. a = 0.2" and = 1.00327 1 It follows from equation 9.6.10 that, for I: > I , the angle between (2') and i is: I Y . 1 05—6 =a—tan l[—tanar] For a so small. , 1 I la tanazamnd tan '[—tana z—tanaz— 13 IS IS a—Oza—I—az 1“ _]a 15 l“. 7 a —6? z '003"7 0.2 = 0.00065 arcsec l.00327 9J4 merwmssJ.g= f mm1=”§u 3 (I)? = we) and l3 = "1:60 e3 Selecting the 2-axis through the point of impact. during the collision mam . maza) . 33 2 el 4 4 2 The only non-zero component of IV is N,. Ith=FxE=aé2x During the collision (02 = 0, so integrating the first ofEulcr‘s equations 9.3.5 [Mm=na Immediately after the collision, 7 ma'a) 4 a), = = a) 2 ma 4 ((03 still is equal to a). and a)2 = 0) After the collision, 1‘7 = 0. From Prob. 9.8, with II = I2 , for N = 0, 7 ’7 (03 = constant and a); +(uz‘ = constant . 2 2 g (a), +w2)' tan a = = constant (03 Using the values of (1),. immediately after the collision I ((02 +0)? 60 tana= =1 a=45 From equation 9.5.8, with (I) = (50 +wlél : a)(é, + 673): Q = (2—1)(\/§(0)(cos45°)= a) From equation 9.6.12 é: \/§m[1+(22 — i)(cos2 49)] ul— =coJ§ 9.15 N = —ca") = —c(a),é, +w2é2 + (03%) Say the 3 axis is the symmetry axis, 13 = 15. II = I2 = 1 For the third component of equations 9.3.5 _ 3 —ca)3 = Isa)3 + 0 a) (b3 _ c l E ‘7, (I) C 3 ——t In ((03 1,: Cl (93 = ((03) 67‘ b For the first two components of equations 9.3.5 N —ca)l = 1a)| +m2(03(13 — I). and —ca)2 = [mg + (0,603 (I — Ix) Rearranging terms and multiplying by a), and (02 respectively c (0,0)l +760: +a),a)2a)3 (I: —I) = O . c (026:)2 + 7a): — (ulw2(03(13 — I) = 0 . . . C 2 . Adding, (olw, +w2w2 +—(a)l +co§) = 0 l 1 d 1 2 +(U§)E((uf +a22” ) = -Tc mmbg, (ml2 +5032 1 2C! 2 2 2 2 I (a), +w2)=(a)l +012) 8 l l 7 - 7 1' ' — ((0,‘ + (1);)3 = ((0,‘ + a); )2 e’ | ’ 2 — l I (a),‘ +a)2 )2 “[77] tana= =(tana°)e ‘ (03 For I". > I. > 0 and a decreases with time. 9.16 (a) From table 8.3. l I = mu» about the symmetry (spin) axis. 5 2 7 a‘ . lanai“. = m— about an axrs through ' 4 the center of mass of the disk and J. to the spin axis. From the parallel axis theorem. equation 8.3.21. 2 2 INC] (1 . 1m, = ——+ m[—] about the pivot : 4 2 ............................................................ .. point. y m a2 I“, = — — moment ofinertia of r 2 3 the rod about the pivot point. "Unis I -1 +1 - ma: +m (—7 2 +(2 i-Ema2 - diu' rod — 2 2 3 3 I 155 + (1552 — 4Mgl] cos a)? From equation 9.7.10. = 21cosO | 3 , . . 3 {I}: 1 [—58i 52—4M51L056 2cos6l I 1 I ma2 1 7 3 ,. 1 _S= ~ =—,c056=cos45 =— I 2ma2 4 \/§ 3 +11 2 m 2 2 9 9 4 _= 1 :—=—Cn7 1 Zma' 8a 80 3 . 1 3 I 3 2 2— i? V, E L 60rpm 3 3 (15:3 2(9001pm)i[(3] (900) 4(80c n )(980 32 2% 2/754 ) J ¢ = 15.1Ipm or 939rpm 10 (b) From equation 9.7.12, 1352 2 4Mg11 m2 L4MI_2 2_i_3 ,. :_ —_ a - ——cm I: 3 I, ma‘ 20 20 2 7 4M] I 3 _ J . 2 S"Z g ._.:4[__Cm IJ(980cm_x fl 60lpm 1‘ I" 20 3 272'3’I 52267110271 9°17 (Nfiglcctingthepencilhead)FromTable 8.3.1, b b 3 m 5 mb? .............. .. [5: : 2 8 The moment ol‘inenia ofthe pencil about an axis thru its center of mass and .L to its symmetry axis from equation 7 (I 191 7 2 I 2 8.3.26 Ic=m “ +5— :1): 1+1 4 12 16 12 e . a 2 b2 u2 from the parallel axns theorem. I=lc+m — =m —+— 2 16 3 . . 4 I] From equation 9.7.12, S2 2 a b2 a2 4mg — m —+— , 12] 116 3] 5‘ Z——24—— m b 64 l I I 2 5 980 20 2 2 3 .g_a 2.4.1 : 167 iii L+2i rudugkl b” 2 16 3 2 16 3 S 218.2941‘ad-s'l = 2910rps 9'18 15:]rim+lspokm+1hub Irim = "Irma? : Ina- 2 3 2 1W,“ = in:ka % = "11: , assuming the spokes to be thin rods 1th = 0. assuming its rad1us 15 small 11 From the perpendicular axis theorem, equation 8.3.14, 1 = 1—‘ 2 [Inga From equation 9.l0.l4, S2 > ‘9 15(15 + ma‘) S> i mga _ 2 7/1102 3 — 19(1 +ma 12 For rolling without slipping, v = (15 | ' 6x32x[3—0] 2 6ga 3 2 e, . _ v> — = “— t-s =3.55 to ' i 19 J 19x12 f j 5 ma2 Ifthc spokes and hub are neglected, I: = 2 l i 1 S> 1 mgu 2(27102) 2 3a +ma 2 l 1 32%?) ‘ ga 2 - _| —l v> — = —— (-3 =3.65 t's [ 3 ) 3x12 / f 9.19 From equations 9.3.5 with IV = O 1,0), +(l'3 —12)a)2(o3 = 0 [2(1)2 +(1l — 13)a),a)3 = 0 13033 +(12 — Il )wla)2 = O Differentiating the first equation with respect to t: I,(2i,+(13 —12)((0:,a)3 +c£)2(z)3 ) = 0 From the second and third equations: 1 —l I —l' (b2=( 3 ')(u,(03,and (b3=( ' 2‘)a)|(z), 12 I, - 12 I —I, 3 1 _ Ilél+([?_]2)|:(l -)wtw§+( 3 1')(0.0)32:|=0 . [3 _ 12 é),+Klw,=0,Kl=—w§ H032 - [113 1,12 (a) For (03 large and (a2 = 0. K] > 0 so (2), +Kla)I =0 is the harmonic oscillator equation. a), oscillates. but remains small. Motion is stable for initial rotation about the 3 axis ifthe 3 axis is the principal axis having the largest or smallest moment ofinertia. (b) For a)3 :0 and a)2 large. K] < 0 so a), +K,a)l =0 is the differential equation for exponential growth of a)| with time: a), = Ae‘m' +86%" . Motion is unstable for the initial rotation mostly about the principal axis having the median moment of inertia. 9.20 I”, = Zmixiyi = 0 since either xi or yI is zero for all six particles. Similarly. all the (0 0 C) other products of inertia are zero. Therefore the ‘ coordinate axes are principle axes. [u = Zmi<yi2 +43): m[0+0+bI +(—b)2 +c2 +(-c)::| I = 2m(b2 +62) 3 (0,b,0) I”, : 2m(az +02) 2 2 (a,0,0) I: =2m(a +b ) b2 +c2 0 0 i=2»: 0 a2 +(;2 0 0 0 a2 +1?2 a F=—Ll(aél+bé2+cé3)=—————w I b (away): 6 From equation 9.1.28. Z = in) b2“l"C2 0 0 a i :Jw—I 0 a2 +c2 0 b ((12+b2+c2)2 0 0 az+b2 c 13 z =_—2’”a’ 14am!) 1 (a2 +b2 +C2)2 C(az +b2) From equation 9.1.32. T = lib-Z 2 a(b3+cz) T=1L[a b c] b(a2+c‘_’) 2 (a2+b3+c2) ' C((l2+b2) T = ————1 2m?)- , (02b? + azcl + 1936) (1' +b‘+c~ 9.21 For Problem 9.1: l -1 0 3 42 2 "=maz —1 — 0 , (az—(g— I 3 fl 5 O 0 0 — 3 1 _ -1 0 3 2 2 Z=fcb=maw —1 3 0 1 J3 2 3 0 0 0 3 3 3-1 1 3 2 6 l _maza) 4+5 _maza) l _maza) 2 J3 3 J? 3 NE 0 O O 1 1 ’ l 4 a) ma'a) T=—(T)-L=——— 2 l () 2 2 2J3 6\/§( ) 0 . 2 2 2 2 :maw (2+2+0)=maa) 60 15 I4 For Problem 9.4: 13 0 0 1 ~ ma CL) 1: 0 10 0 . (D=— 2 12 o 0 5 fl 3 7 13 0 0 1 , 13 A ~ _ [INT (U "MI “(0 L: a): 0 1o 0 2 = 20 12M 0 O 5 3 12M 15 1 1 ° 13 fl (2) "10.10 T=—a"-L=—— 1 2 3 20 2 2M12Jfi( ) 15 = """a" (13+40+45) =lma2w2 24(14) 24 9.22 Since the coordinate axes are axes of symmetry, they are principal axes and all products of inertia are zero. b- 1-: : I710) I 3((12 + b2 + c2 )S a z : ma) I b 3((12 +b2 +02); 6 From equation 9.1.32. T: From Table 8.3.1. 1n = {lg—[(212)2 + (20)? = 3(b2 +02) 1, =fl((12+c:). 1,.=fl(uz+b2) m ” 3 " 3 If“! 0 0 “ I7! 7 3 [=3 0 (I-+C— 0 0 0 a3+b2 f: w l(aé,+bé2+cé3)= 2 (a + (02+c2) ((12 +122) CDT-Z l 2 15 (a [22 +62) (1 b c a(b2 +62) 2 1 ma) ______— 2 .2 T_2 3(a2+b2+c2)[a b C] b(a +c) C(az+b2) (2 a 2 2 2 2 2 T=fi+7>[a‘(bz+o )+b2(a2+c )+C (a +1) T 2 We? +bzc2) With the origin at one corner, from the parallel axis theorem: 2 ‘2 1_‘,,=%(az+cz). gigging) I I}, = —>J’xydm = — Jlxypd V 1 = ‘P I: if: I: = —8pa2b2c m = p(2a)(2b)(2c') = 8pabc , SO I”, = —mub 2 —mac. 1): = —mbc $5 + c3) —ab —uc i = m —ab gm? + c2) —bc ~ac —bc %((12 + b2) 9.23 (See Figure 9.7.1) L: = (Lx.): +(Ly. +(L,)z L: = L),. sin 6 + Lg. c050 = (Igfisin 6)sin6+ (15$)c036 9.24 (See Figure 9.7.1) lfthe top processes without mutation, it must do so at 0 = (A where l/(Q) is a minimum of V(0) dV(0) d6? 0=9 (L: — 14:.cosl9)2 2] sin2 6 V (0) = + mg] 0056 (See equation 9.8.7) 16 dV «30500 L, —L,. c0361 2 +L,sian L, —L-.cos€° _ _ = )_3' (' ‘ )—mglsrn(2,=0 d0 6:0 15111 09 let 7 = Lz —L_,, c0500 Then 72 cos 65 —}/L:. sin2 6: + mg/I sin4 Q = 0 and solving for y y = LZ. sin2 90 Ii 1_ 4mglljcost9° 3 2 cos t9c L; now L:. is large since g'u is large and the precession rate is small. so we can expand the term in square root above and use the (-) solution since 7 must be positive L,. Sin2 0“ 2mg]! cos 6“ y z ~—_ 1 — 1 + —2— 2 cos 9 L . mgll sin2 6” L, From equations 9.7.2, 9.7.5 and 9.7.7 L: =lqizsinzfln+IS<¢c0360+y2)cos(90 and 7 = L: —L_,. 00500 x L__. cos 60 = 15(gicos 60 + 1/?)cos 00 so y = (1 sin2 67“ + 15 cos2 6’: + [X y) c059: — 1x (fi'cos2 6° — I: 1/) c0360 mg]! sin2 00 _ mglIsin2 0“ = [qlginz 0c 2 , and since ([1 >> 0 , we can ignore Lg. 1x (y}+¢cos a) the term in the denominator and we have m 'l . . . = —g the top Wlll precess Without nutation 1.1/7 9.25 b,+h, =71? b—'=sin30°=l b2 2 a (I b =— b =— l 2 J3 az=HZ+bz2 Hi-aj—bZ-az—i—Z—a2 17 2 H=\/:a 3 Thus, the coordinates of the 4 atoms are: Oxygen: (0.0,H) a a Hydrogen: (—1), .§.0);(—b. 75:0) Carbon: (b2,0,0) (a) The axes 1, 2. 3 are principal axes ifthe products of inertia are zero. -1”, = Z nnxiyl. = 0 + [—b. + [—bl + 0 a 0 —I}: =0+0+0+OEO —lxz = 0 + 0 + 0 + O E 0 The 1,2,3 axes are principal axes. (b) Find principal moments 7 7 7 7 89 9 7 v 02 a2 7 14 Hm 6—7 0 0 6 1: 0 Q 0 a2 6 0 0 ii 3 18 ...
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analytic mechanics 7E ch9 - CHAPTER 9 MOTION OF RIGID...

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