analytic mechanics 7E ch2

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

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

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: T@v = 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 ...
View Full Document

Page1 / 10

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

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online