analytic mechanics 7E ch2

# analytic mechanics 7E ch2 - 2.1 2.2 CHAPTER 2 NEWTONIAN...

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Unformatted text preview: 2.1 2.2 CHAPTER 2 NEWTONIAN MECHANICS: RECTILINEAR MOTION OF A PARTICLE l (a) .’\3=—(F:l+ct) m 1 F 7 .i’=_[——(E+Ct)c1’t=—”t+it’ m m 2m I: 1 F " x: [[41+i1‘Jd12—‘14it‘ ' m 2m m 6m (Vb) f=£sinct m F . F I F A“:[48111C1dl=—-;COSCIIO =—°(1—coscl) ’m cm cm F F 1 . x:[4(l—cosct)dt=;(1——smci] '02; cm c N I; (‘I (c) ,\‘=—e m F ,I * . Ag: (ell cm 0 cm F l I F \=—[-LI———I]= ,° (ea—l—ct) cm (' c 0'”? _, cli' di' dl’ .(lx't (a) x:_:_._:rx_ d! dx dt (it jg:i E+cx) ctr m l .i'di‘=—(l~f +cx)d\' m if = i Ex+£ 2 m 2 x =[i(21u;+cx)]_; "I (b) = = (’1 m 2.4 m 1 7 F _ F _ —x‘=——° eu—l =—1—eu 2 cm( ) cm< ) I 2F 3 r: —” l—e" [Cm( ' l (c) x:.t£=—(E,cosczt) dY m F .i’di‘=—°cos cx (if m 1.2 E, . —x 2—smcx 2 cm I 2F . 5 x: —°Slncx cm *~ . ~_ ‘2: (@ V(x)=—£(}g+c1ﬁb-— E1 2 +C (b) V Z - J" lit" "(11‘ = e'” + C I (r. (C) V (x) = “I E cos cx u’x = —_5m 6x + C K (; @)FQQ=—d:S9=—h V(x)= Ekxdr=ékr2 (b) To =T(x)+V(x) -1 -0.5 0 0.5 1 -1 x 1 mm-1/(x)=gk(A_xz) (Q E=z=ébf (d) turning/Joints @ T(.vcI ) —> O x, = iA ‘4 .. I ,.s .4 SO V(_r):L[kT_k\2 Przlkx2_ik\2 A 1 , 1M4 b T. =T—V, =T—~k'+— , () (Y) ° (0 "2 Y 4A‘ (Q E=Z 2.6 2.7 2.8 2.9 (d) I/(x) has maximum at |F(xm —> 0 3 ’j: =0 1 1 M“ 1 V x =—kA2—— =— ( m) 2 4 A2 4 If E < I/(xm) turning points exist. iA lcrm— kA'l Turning points @ T(x,) —> 0 let 21:17,2 E—lku +1 k”, 2 4 A? solving for u , we obtain =0 I 0.25 7 3 0'2 11:14” lip—[VF] 0.1 or i J 2 0 o '2 '2 a _ a2 1—37"? FZMgsina F = mi; = mic (iv A? = bx"3 ﬂ = —3bx"1 (it F = m (bx-3 )(—3bx—4) F : —3mb2x_7 (a) V = mgx = (.145kg)(9.812’](1250ﬁ)[ S .3048’—” ft =541J 7 I 2 (b) T = lmvz = lmv; = _m "18’ = i m g7 2 2 2 c: 2 .22D‘ )1 (.l45kg)2(9.8—2 N S ——_J2kg = 87.1 (2)(.22)[(2)(.0366)] IN I F dx = I—cvzdx = —c Jv3dt = —c {(—v, tanh [ijsdt z. = cvfr[—l tanh2 + Itanh 2 2' 2' r = cv,‘r[—l tanh2 + 1n cosh 2 1' 2' ~. I ‘ Now tanh‘(—]El tor I<< 2' r T: / Meanwhile X = Ivdl = [(-v, tanhtLDdt = v,r In cosh 2' z' I .l‘ In cosh — = r v,r "I x = (1250]?)[3048Z] = 381m l i (.145kg)[9.81,’) ' ’ = k = 34.72— 02 (.22)(.0732)2 —g 5 I71 _ ‘F— = 3.543s (.22)(.073.2)2 —g(9.8\$) S- "I r_[ m )5 _ (.145kg) 3.81 dex = (.22)(.0732)2 (34.72)3 (3.543)[-.5 J: 454.1 V-T=54lJ—87J=454J F 1F. 2.10 For 03131,: v=—”t, x=——”12 m 2 m F F For 431521,: vc=—°t,, x,=—°t,2, 10:1I F 2 E 12E 7 r-—t, +—t,(t—t,)+——(t—t) 2m m 2 m F9 2 Fu 2 E 2 5F 2 At t=2tlz x=—z‘I +—t| +—tl =—‘tI 2m m m 2171 v dv dx dv z. 3 2.11 =d—=—-—=v—:——v2 dt air dt d\ m | -— c v 3dv=——ci\ m l —, Jam C [v 11'sz ——-cb( ‘ m l c _2v«2 =__\‘ma\ m ‘ 1 211112: xmxz c 2.12 Going up: F =—mgsin30°—/1mgcos30” I 3h=—g(gn30°+01cos30j==—57493§ s‘ v=m+m at the highest point v = 0 so [up = —i = 0.] 74vas a l a '7 7 , 36",, = vatup +§at,jp = 0.1741); —.087vﬂ' = 0.0871277? Gomgdownzxj=0xm7ui uf=0.cx=—93(05—008mg xww==0=0087n?-%405Lyiw Idem, = 0.207VQS floral = tup + tdawn = Ives 5 2.13 At the top v = 0 so 6' mm“ = k g + Us? k Coming down 3:“ = xmax and at the bottom at = 0 2 . has] _1_(1)= A k [g+mj §+u2 U! "1‘1 _ ’8 _ [mg v:..__—, v _ _. _ _ I 1 k 2 2 3 C2 (1’, +1)“ )' 2.14 Going up: I”; = —mg—c2v2 d) a z a=v—‘=-g—kv', k=C— dr m , ,d I [#:de -—g—/w' 1 2 v . —§ln(—g—kv )r =1 I+A 3 . is "’ :eizu g + Ina2 v2 =[§+v02)emr —£ k k . 2 (30mg down. Ii = —mg +c2v (Iv v—=—g+kv2 c/x ’ 71 1' #17: I (it ) I 2 v iln(—g+kv )0 =x—xc l__/£v7 _ ellae-ZLI g v2 =§_[§e—2Lx Jean k k I} ) 2.15 mg=mg—clv—c,v' d! ' ﬂ_ dv m mg — clv — 62v2 Using I air _ I In 20x+b—\/b2 —4ac V a+bx+6x2 Vb: —4ac 20x+b+xlb2 —4ac “ V 1 ] —2czv — c, — Jclz + 411ng2 =———ln ’7 ‘lclz +4mgc2 —2c2v— c, + ‘lclz +4mgc2 0 (2c2v+cI + ‘fc,2 +4mgc2 )(c, —"clz +4mgc2) (2c2v+ (i, — dc]: +4"ng2 )(q + 40,2 + 4mgc2) as t——) 00 , 202v, +6. —"clz + 4mgc2 = 0 l 2 3 7 v, = _ C! + _C1_ + I 1g 202 2c2 c2 Alternatively. when v = v, . I I 3 ' —(c, + 4mgc2 )2 = ln m dv O m cv 2 m—= = — —c,v, dt g It _ l \'__i+ 6; +ﬁ' 2c2 262 c? 216 azvﬂ——£\7 dr m I kdt W” - L‘ . . , Smcexsb,say =sm't9 mb3 I = .1. [ 2k ] cosﬂ mb3 1 : 8k B‘ld ul— sinﬂ 23' 6‘ 0(16 2 3 5 ————( m L08 )=[ 12b J Elsinzﬁdf) ‘7 l 2 71’ 2.17 mg: = mvﬂ = f(x)-g(v) (/1 (IX mvdv : f dx 5'0”) By integration, get v = v(x) =§ If F(x,l) = f(,r)og(t): d’x d (dr)=f(x).g(t) "1—— = m— — dt d! 2 d! This cannot. in general. be solved by integration. If F(v.!)=f(v)-g(t): dv nzZ=f(v)og(t) mdv =g t dt .f-(v) ( ) Integration gives v = v(t) dx —: ) I (It i( ) dx = v(t)dt A second integration gives x = x(t) 2.18 CI =(l.55x10“)(10'1)=1.55x10*‘§ S c2 = (0.22)(10'2)2 = 2.2x10‘5k—g S 1.55x10'° 1.55x10'6 2 ( V, =_———_5+ '—_‘ + 2x2.2x10 2x2.2x10' v, =0.179’—" S . . (10”)(98) Usmg equation 2.29, v, = —‘__ 2.2x10 3 2.19 F(x) = —Ae""i = mi‘ or Let u = e’" du = acuvdv dv = av ae 10‘7)(9.8) 2.2x10'5 =O.211— F(v) = —Aem' = mi} dz! _ du au T (IV A m = — — dt e m du a A '. , = —— d! u‘ m lnte grating l l A — —— = —a! u 116 m and substituting e‘" :11 A , (a) v=vz—iln[l+—emat] a m (b) r: [email protected] = 0 av= = ln[l +/—1e‘" QT] m ea" =l+ﬁemaT T=”—1|:l—e'”"] m 01.4 z ' . d (C) vi=v=—iem var=—ﬁclr (11‘ m e m . m. d1! I again. let u = e (IN = audv or dv = — v = — In 11 an a [ 1 ] du —lnu — 1 a—ﬂ = —i dx Integrating and solving 1/ m m x = 1— Have 6"" d F = (mv) = mv+ vm = mg B0 (It | | 4 1 I but m 2 pa 3717'3 m = p,ﬂ')"v if s" ‘ :1: ' ‘ - i - lvdl 4 a 7 4 7 4 I I so (1) —7rpvr3v+7rplr“v“ =—7r‘ =—7zp°r3g : P1 ii 3 3 3 I : .0: -3 - - - l4 —’ Now — z 10 so, second term IS negligible-small ‘ ‘ ,0; hence v z g and speed act but 7 . l l 17': = p: 47rr2i- = pI 7n"v or r 5 —ﬂ v Hence r z —ﬂgt and rate of 4 p. 4 p0 growth at! The exact differential equation from (1) above is: 4 4 4 = 2 4 —Irp,r if +413. W = were 3 pl pl 3 which reduces to: i‘ +— = _ g Using Mathcad. solve the above non-linear d.c. letting ﬂaw 10'3 and R z 0.01mm (small raindrop). Graphs A Show that voci‘oct and reucr2 Radius vs Time Rate of Growth vs Time N D A Hadius[mm] —:| c: dB Idt [mmis] m 0 5 10 0 5 1g TimeIs] Time [s] 10 ...
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analytic mechanics 7E ch2 - 2.1 2.2 CHAPTER 2 NEWTONIAN...

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