analytic mechanics 7E ch8

analytic mechanics 7E ch8 - CHAPTER 8 MECHANICS OF RIGID...

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Unformatted text preview: CHAPTER 8 MECHANICS OF RIGID BODIES: PLANAR MOTION . . . A m 8.1 (a) For each portlon of the VVII‘C havmg a mass? and centered at (421%), (9%) -am=i[-(%J(%J+O+(%J( yanfiKEJt—Stetéfl; (b) ds = xdy = (b2 —y2 )3 dy 1 , t i 5’ ~ :_ yb“_)"2dr ym m KPH ) ) ,1 p =” w 2 2 _ x ‘ _—ELO(b ~y)d(b —1) yon _ 1 —/'Ib2 4 ,0 _ 9 yCI” 4b From symmetry, xcm = — 37: (c) y The center of mass is on the y-axis. I ds = 2xafy = 2(by)3 a’y I l R } " ) 3 y : EM) (by)2 d) : Ky dy cm I I £2p(by)2 dy Eyzgfy _ 3_b yCI" 5 (d) The center of mass is on the z-axis. dv = 711'de = Jr(x2 + y2 )dz = rz'bzdz ix) _ I: pzzz'bza'z _ fzzdz at," — E pfrbzdz — fzc/z 2 26m =—b 3 (e) Z The center of mass is on the z-axis. a is the half-angle ofthe apex ofthe cone. 1; is the radius of the base at z = 0 and r is radius ofa circle at some arbitrary z in a plane parallel to the base. 9 tan a = i = r , a constant a I) b—z dv = zrrzdz = 7r(b — tan2 a dz l 7 1 a m = -/ni”b = —7r b3 tan' a p 3 a. 3 p £p:7r(b—z)2tan3adz 3 - 7 7 =—l———— =p (b'z—sz’ +z’)dz —7rpb3 tan2 a 3 - -2 “cm 4 8.3 The center of mass is on the z-axis. Consider the sphere with the cavity to be made 2 ofa (i) solid sphere of radius a and mass MS. with its center of mass at z = 0 , and (ii) a solid sphere the size of the cavity~ with u y mass —Mc and center of mass at z = —E . The actual sphere with X the cavity has a mass m = MS — Mr and center of mass 20". 0 = —l— M[ [—£)+mzm MI 2 3 4 3 4 a] M =—7ra , MC =—7r — s 3 P 3 [2 ,0 (d) [Isinz 0‘19 = Q — 2 ds = rdBdr , R = rsinH I: = JRZpa's 1: = p rdr fish}: ado =4 fisinz 6d6 4 4 Iz=pb 4 sin 26 , $15-1] Z 4 4 2 l 7 =- III)‘ m 4p ml)? 1 = 1r—2 , 47g ) Y2 5’ ds=hdx=[b—T]dr m“ Where the parabola intersects the line y = b. l I: (by)3 2 ii) 2 x2 ’ 2 x4 1y = £1 p[b—?]dl =p[b(bx —7]dr 4 1 =— b4 ’ 15p ’ I? 1 = l—— d =— b“ M) by 3,, l 7 1|,=—mb‘ ' 5 dv=27rthR 2 h=b—z @1») Vi y 3 R =(x2 +y2)% =(bz)g dR = dz 2 z I 7 l l 1 I=R‘d’= b22bzlb—z—— 2 31px 1mm 5(2) ‘1 7 l ' I: b' lz—z‘dz=— b’ = "P El) ) 6760 I J l 3 m=Ipdv=£p27z(bz)1(b—z)%[§) dz m = irpbf(b—z)d.; = £750}? I,=lmb2 ' 3 a is the half-angle of the apex of the come. i; is the radius ofthe base at z = 0 and r is radius ofa circle at Z A; some arbitrary z in a plane parallel to the base. tana : — = . a constant K b b—Z (b—z)R° dv = JIthR = 272' zdR b— R ( ° ‘ me = —%dz. and the limits ofintegration for R = 0 —> R“ correspond to z = b —> O — 2R2 19—- R Izszgpdi/sz 2 (#4 1?")le Since R = b‘ b I) R4 l .,=+2 —° b32—3b222+3b.3—z4 dZ=—7T be _ 72- b4 ) p _ l 1 m: —7rbe p3 . I,=imRn2 - 10 . 8.5 Consider the sphere with the cavity to be made ofa (i) solid sphere of radius a and and mass Ms. with its center of mass at z = (l , and (ii) a solid sphere the size of the . . a . . cav1ty. With mass —ML, and center of mass at z = —E. The actual sphere With the C8Vlty has a mass m = My — MC and center of mass 20". Z 3 4 4 a 7 4 m=M —1M_=—7ra3 ——7I —- =——7ra3 x i 3 P 3 (2) P 8 3 P y My =§m and Mr :1»: ' 7 7 " From eqn. 8.3.2. Ix = If +1 7 , 2 2 nga- =7ML.[£) +1 3 3 2 2(8 2[1 Jazz 31 3 [=— —m a‘—— —m —=—mu 5 7 5 7 4 70 8.6 The moment 01‘ inertia about one of the straight edges is I: = IRlpdv where R3 = x2 +y2 . From Appendix F (IV = 1'2 sin 0 dr d0 61¢ R2 = x2 +y2 = r2 sin2 6 Leta = radius of sphere 1: = r2 sin2 9,0,3 sin6a'rd0 d¢ fl' =1) 1: = p— =0 f; r“ sin3 0mm =0 2 l g '7 . 3 L:— zr‘ 331110516 _ 10p a J; 3 [ISM 0cm = C0: 9 —cos€] I —3—20pn'a m 147a3 limp :——/ :— 83 p 6 1 =Ema 5 8.7 For a rectangular parallelepiped: (iv = hdxdy h is the length of the box in the z-direction l R = (x2 +y:)2 12 = J‘Rzpdl’: LEAK + yz) phdtdy 3 / I: = ;7/1[u[2b1'2 +%]ci\' = glowibe + b2) m = p(20)(2b)h = 4pab/1 I: = %(a2 +b2) l<2a >l For an elliptic cylinder: Ey Again dv: lzcivdv. and R = (x3 +y1)3 l x2 V: On the surface. T +'—7 : I a" b“ .rzia 1—): - b- r=b ‘=“['i: ’ 1_, = JR2dv= I):pr 1’ {(13 + y2 ) phdx‘dy I .l'z—u [JV—e b. 3 | 2 3 .3 3 Izzphfb 2 057 From a table ofintegrals: “()2 —y2 aiy =£(b2 —y2 +2b:y(b2 —y2)% +251 sin’I 4 8 8 b | 3 2 | 4 1 -2 [fl “2 21,4 I); —i 3 2 __— pzb 3b? 8 II + 8 fl' —4plzirub(a +b) m = ph(7rab) I: 2141((12 +b2) For an ellipsoid: dv = hdrdy. R2 = x2 +y2 and on the surface. :7 +3; = I a‘ ' c‘ l h = 2|z| = 2c-[1 — 42%]: In the xy plane + y; :1 a“ b“ l x=ia 1—; - z b- l 3 2 7—4] am b. From a table ofintegrals: 3 “kg _x2 xzdx : _%(k2 "372)2 +%(k2 “352 'l'égisin‘l “k2 —x2 dx='—;(k2 —x2)12 +E2:Sin‘l 2 2pc [Iv a4 ’2 1 a2 y2 1_=_ _ 71-4... "_ " a b[ 8[ b‘) y 2[ 12' ‘az 2y2 y‘ 2 y“ Izzpiracrij 1— b2 +b4 +y ‘b7 05’ 1: = épiz‘abda2 +b2) 4 m 1 7 For 2m 6111' soid, m = —Imbc , so I, =— a“ +b’ p p 3 , 5 ( ) 8.8 (See Figure 84.1) Note that l'+l is the distance from 0' to 0 , defined as d From eqn. 8.4.13, k3," = [l' k§m+12 =11'+12 =1(/'+/) It;x +12 2 Id From eqn. 8.4.9b, It2 = k3," +12 k2 = Id From eqn. 8.4.6. T0 = 27; gl 1 z MF 8 Period ofa simple pendulum: T = 27A Period of real pendulum: I = 2/: —l— (eqn. 8.4.5) Mgl Where I =moment ofinertia / = distance to CM of physical pendulum a = distance to CM of bob 1) = radius of bob location of CM of physical pendulum: 111(a_b)+(]l/I-m)a I: 2 m+(M—m) .M Moment of inertia: 7 [bob =(M—m)az +%(M—m)b2 =(M—m)[a2 +§b2] 2 =Ma?[1—fl) 1+Zb_2 11/1 50 /,,,,=lm(a—b):=lMa2 fl[1_2] ‘ 3 3 M a 2 3 .-.1=1,m,,+/m,=Ma2 [1—3] 1+3b—2 Jib—3] ‘ M 5a 3M (1 . b letting a = 21- and fl = — [W a 1W(12[(l—a)[l+—§fl2)+%a(l—fl)2:| 2 1Wgu[ -g —%:| 1:2” 1 (a) z l—l—ja to 1” order ina (b) m = l0g M = Hg (1 =1.27m b = Scm a = 0.01 fl = 0.0394 T l . . ?° z l—l—ja = 0.9992 (actually 0.9994 usmg complete express'ion) 8.10 The period ofthc "seconds" pendulum is T, :23 L =25 ‘ Mg] The period of the modified pendulum is T'=27r 1 =2[ ” J Mgl' 11—20 where 1' , M ' , 1' refer to parameters of pendulum with m attached and n (= 24x 60x 60) is the number of seconds in a day. _ M + mlm 114 where [m is the distance of the attached mass m from the pivot point. 20 ‘2 7I21' 7rzl +7r2ml2 So 1—— = — = —m n M'gl' (M/+mlm)g _ Mg] +7r2mlnzI (MI+mlm)g 2 Thus [LE] =___(M'+’”i~)§ (Mg/+7r'ml,;) n a] 1+—"' 40~ ( I ) m Or l—— ~ —,;— a = — n zr'al“ M 1+ ”' g1 Solving for 0: gives the approximate result 1'=I+m[:I l' Letting I”, = 1.3m: I: 1.0m; we obtain a 21.15-10—3 v ‘) u u 8.11 (a) 11m = ma” (all mass In run) 9 _ 2 2 _ 2 [Mm — ma + ma — 2ma 1 ‘ 2 --x 1:274 i’"”=27r’—a mga g z. (b) I: =11+ly =21Hm =ma2 (= 1m") ma2 .. 1W = 2 ma2 7 3 7 hence I'm," = 2 +ma” =—ma” 2 .'.T=27r [Wm =27: 3—a mga 2g 8.12 mi‘cm = mg —T f {ma} = aT Sim = an) [m = 2 ma2 5 __ [mm 1 [2 2 2cm) 2 ,, nut”, = mg — = mg —— —ma — = mg ——mxm a a 5 a 5 7. .. 5 3 mg xm = 7 g x mg 8.13 When two men hold the plank. each supports When one man lets go: mg —R = mme and R1 = land) 2 2 From table 8.3.1. In” 2 1% . RI 12 6R a) = — o a = —— 2 ml” ml 10 ll .. 1. 3R xv =—a)=—— 'm 2 m R mg — R = m[3——) = 3R m R : £4: 4 . 6R fend = [(0 = I— = ml m 4 xend = 8.14 For a solid sphere: M‘ =flira3p and 13 = Eli/Iva? . 3 5 i ' '1 2 7 k,‘ = —a' ( LI" )5 5 For subscript c representing a solid sphere the size of the cavity. from eqn. 8.3.2: 15 = I +16 ,-3 3m: 3(2)’ [gjizigms ‘53‘ 532p25332’0 4 3 4 a’ 47 3 mz—rm p——ir p:—-—7ra p 38 8 3 3 5 7 l 2 31 31 2 Ix' =—=—-—-—a —a m m 5 32 7 70 From eqn. 8.6.I 1, for a sphere rolling down a rough inclined plane: gsinfl 33cm =7 a- 1+ , r: a- " 1+kgz' (1' 2 7 1+— — : 5=_5_ 31 101 1+— — 7O 7 ; 2*. f: 101 11 12 8.15 Energy is conserved: l ,3 l ,, l .r ‘ —ml.\’ +—mzx'+—I — +ng.\‘—mlgx:E 2 2 2 2 m,x.\' + mzxx + —3 xx + ml gx — mI gx = 0 a- ,, (mI — m2 )g x = —— I mI +m2 + —, a- 8.16 While the cylinders are in contact: I f; = mvc’" = mg c056 — R Q * ‘ r = a + b, so mv‘“ = mg c056 — R ‘ van a + I) From conservation of energy: 1 7 1 mg(a+b) 252721;" +%I(u‘ + mg(a+b)cos() "A? From table 8.3.1, I 2%»er2 Vt'm a) = — a 1 , 1 1 w v? —mvf +— —ma' U," = m I a +b 1— cosf) 2 in, 2[2 a >( > mv2 4 "’" = —m l— c089 (1+1) 3 g ( ) When the rolling cylinder leaves, R =0: 4 mgcosfi=§mg(l —cos(9) 10036 = 3 3 i9 = cos" 7 8.17 m? = N2 my = Nl — mg 12 8.18 l 2 1 2 E mvm + 5 160 + mgy = mgvn x=10056, .{'=—16)Sin€, c059+ésin6) 2 2 2 v-isinfi '-léc036 V—L(—ézsin0+écosfi) y 2 ‘ y 2 ’ ' 2 2 2 2 '2 V:m=.f2+y2=[-éésin0] +(éécosfl) =1: 1: mr . and (1) =9 12 2 '2 2 ~1—ml 9 +ifl07+mgisinfi=mg—Csinf)c 2 4 2 12 2 2 £92 =g(sin(}D—sin0) l 0.:[Elg—(sin60—sinfi):|2 J (9:1 3—g(sin(}D-sin6’)] ' (3—g](—0050)9=—3—gc059 2 l l 2/ N2 = mi: = —fll:c056(3—g)(sin0° —sin0)+sin 0[—3—g]c059] 2 1 Separation occurs when N2 = 0 : sich—sinQ—lsinflzo, 2 Ric:an R’, — mg 2 m}? I . . I . .. x=—sm6’. x=—¢9c050, x: 2 2 yzicosfi. Vim—£95110, 2 2 —l—mv2 +1192+m 10036—271 — 2 m 2 g2 g vcm=19. I=£ 2 12 3 l 2 I y=—§ l 2 13 f) = sin"I sin 0°] _(_92 si116+5c059) (92 cos€+5sin€) l3 [21/202 1171/2 -2 l — +——(9 =m — l—COSH 2 4 212 g2( ) E=g(l-c056) 3 l 19'=[3[g(I—cos())]2 ,1 0—1 3—g(l—c050) '[3—gjsm66—3—gsm6 2 l l 21 RK =fl (—sin6)[3—g)(l—cos:9)+c05c9 3—gsin9) ' 2 l 2] R1 = 33g sinH(3c056—2) Ry =mg—fl c030(3—g)(1—cos€)+sin9[£sin§] ' 2 l 21 RV = mg — 3’:g [cos 6 — cos2 6 + 5m; 9] Ry = "f (3 cos 0 — l)2 The reaction force constrains the tail of the rocket from sliding backward for R“ > 0 : 30050—2>0 8<cos*'— 3 The rocket is constrained from sliding forward for R1r < 0 : 0>cos '3 3 8.19 mi = —mg sin 0 — ymg c059 = -g(sin6+,ucos€) .1 . . . . I .. . Slnce acceleration lS constant, x = x0! +—xt2 : 2 gt2 . x = mgr—7(31n6+,u0030) . . 2 7 . Meanwhile (,umg 0056)a = [(1) = gmu‘a) 5 0-, PM 2 (1 ,ug cosflt 5 a):— 2 (1 14 14 The ball begins pure rolling when v: ua) V: ‘1 +551: V,,—g(sin0+ycos())z = angcosg, a 2v° g(231n9+7,ucos€) At that time: 2"? g 4%2 (sin0+ycos€) .T:—————__ g(2Sin()+7/JCOS«9) 2g2(28iI10+7}1COS(9)2 _2vo2 (sin6+6ye056) .r— ’————., g (23infl+7ycos€)' 8.20 mi: = ymg f=#g 1 .1" = ygt, and x =Epgt2 . 2 7 . [w = g mam) = —,umgu 5 w:__fi§ 2 a w=i_éa§, Za Slipping ceases to occur when v = a0) 5 #gt=awg-§#gt 8.21 Let the moments of inertia of A and B be I“ [=%Maazjand lb (=gMbb2]. The angular velocity of A is (2 while that of B is —d +¢ (remember that in two dimensions. angular velocity is the rate of change of an angle between an line or direction fixed to the body and one fixed in space). For rolling contact. lengths traveled along the perimeters of the disks A and 8 must be equal to the arc length traveled along the track C. .- a¢=bfl=(a+2b)(a—¢) + 2b +2b 2(u+b) 2b(a+b) After some algebra the angular velocity of B is found to be - . - ad a) = —a+ :— 3 l3 (15 2}) For A . we take moments about 0 and for B we take moments about its center. Call TA and '1'}, the components of the reaction forces tangent to A and B (the “upward-going“ TA acts on disk B. The “downward-going” TA acts on disk A) Thus K — TAU = [Ari —TAb—T,,b=—IB(,8—d+g}5)=—1,,a% 1 I) —T,, = M,, (a+l))(d—¢') =§Mflaii Eliminate TA and TB _ d — 4/)2 Integrating this equation gives : K (4b21A + Msazbz + (1218) Kt = 57(41):]‘4 + Mnuzbz + an”) Putting 60,4 2d at 1:15 gives 4133100 (U = A (4191’I + M3021)2 + (1213) Putting in values for If, and 1,, gives 4K4 (12(2MA +%MB) (0.4 = Since the angular velocity of B is (08 = —d +415: a 2100 (93 = —(0A = 2” ab[2M,, gm] 16 (Torque on Disk A) (Torque on Disk B) (Force on Disk B) ad — , we have 2b 16 17 8.22 From section 8.7 (see Figure 8.7.1), the instantaneous center of rotation is the . . . l . . pomt 0. If x IS the distance from the center of mass to O and 3 IS the distance from the center ofmass to the center of percussion 0'. then from eqn. 8.7.10 t l _/m,_MI2 i _E ' 2 M 12 M 12 I x:— 6 8.23 In order that no reaction occurs between the table surface and the ball, the ball must approach and recedc from its collision with the cushion by rolling without slipping. Using a prime to denote velocity and rotational velocity after the collision: , P : 1’C’" _ _ m 13 I (um - (um — 7(11 — a) The conditions for no reaction are vm = mum and v' = awfim. 3 . awm —L = a[a)m —§(h—a)] "I 1 l ——=—a h—a m I ( ) 1:3»ch 5 Zma2 7 h: _ +a=-a Dma 5 d a:— 2 7 h=—d 10 8.24 During the collision. angular momentum about the point 0 is conserved: m'vJ' = 10 0 : m'vcl' 1 After the collision, energy is conserved: 1 '3 1 1 —16“—m ——m' 1'=—m —cos€ —m' l'cosél 2 g 2 g g 2 t. g e l m'zvzl'2 ml — ° = l— C056 —+ m'l' 2 1 g( a)( 2 j 17 18 1 = mlk +m'l'2 3 l 1 ml ml2 , 2 v=—2 l—c056 —+ n'l' —+ ’I" n m'l'[ g( I )[3 m 8.25 $- A 81 c [:H:I The effect of rod BC acting on rod AB is impulse + i". The effect of AB on BC is —P’. mv| = P+P va, = —P’ 1w,=1(13_iw) 1a), 3113' 2 ' 2 hi 12 6 A - 6 * a) =— P—P' a) =——P' 1 ml( ) 2 ml 1 I vB=vl—§wl VB =V2+E(02 MP 1 6 ~ ~ v = —— — P—P B m 2(ml]( ) VB =_:+i[_£f)'] m 2 In] fi+13'—3(13—P')=—13'—3P 813'=2P 13' 5 4 5/3 13 v =— v, ——— 4m 4m 913 313 a), = (02 ——— 2m] 2m! P v” ——— m 18 ...
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analytic mechanics 7E ch8 - CHAPTER 8 MECHANICS OF RIGID...

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