SOLUTIONS TO CHAPTER 2 PROBLEMS

Transcription

1 SOLUTIONS TO CHAPTER PROBLEMS Problm.1 Th pully of Fig..33 is composd of fiv portios: thr cylidrs (of which two ar idtical) ad two idtical co frustum sgmts. Th mass momt of irtia of a cylidr dfid by a hight h, a radius R, ad mass dsity ρ is giv by: 1 1 J mr h R 4 cyl ρ π (P.1) whras th mass momt of irtia of a coical portio dfid by th radii R 1 ad R, i additio to th paramtrs itroducd i Eq. (P.1), is xprssd as: Jco ρhπ ( R1 + R1R + R1R + RR 1 + R1 ) (P.) 10 As a cosquc, th total mass momt of irtia of th pully of Fig..33 is calculatd addig up two idtical cylidrical portios of radius R 1, aothr cylidrical part of radius R, ad two idtical coical sgmts, amly: J h1 R1 + h3 R + h ( R1 + R1R + R1R + RR 1 + R ) ρ π ρ π 10 ρ π (P.3) ad th fial xprssio of J is: πρ J 10hR h( R1 + R1R + R1R + RR 1 + R ) + 5h3R 10 (P.4) For th umrical valus of this problm, J x 10-5 kg-m. Solutios: Chaptr 1

2 Problm. Wh chagig th plat width from w to w + Δw, th axial mass momt of irtia bcoms: ( + ) ' m' ρlw wh Jx ( w+ w) + h ( w+ w) + h 1 1 (P.1) whr l is th plat lgth (dimsio prpdicular o w i Fig..34). Usig Δw w/ (rquird by th maximum 50% width icras), chags Eq. (P.1) to: ( ) ' ρlw wh 3 m 9w Jx + ( w+ w) + h + h This mass momt of irtia ds to b 0 tims largr tha th origial o, amly: Equatio (P.3) rducs to: h 0 w + h m w m 133w 148h 0 ( ) (P.) (P.3) + (P.4) which is impossibl for w > 0 ad h > 0. Th scod modality is to calculat th mass momt of irtia with rspct to a axis paralll to th ctral axis x ad which is at a distac d from x. Th w mass momt of irtia is: J J + md J + mw (P.5) x x which usd d w (th coditio of maximum d). Th coctio btw th mass momt of irtia of Eq. (P.5) ad th dd momt of irtia is: Equatio (P.6) ca also b writt as: J x mw 0J x + (P.6) mw 19J x (P.7) Cosidrig that: m Jx ( w + h ) (P.8) 1 i cojuctio with Eq. (P.7), lads to: which is, agai, impossibl. 7w 19h 0 + (P.9) Solutios: Chaptr

3 Th last rsort is usig a combiatio of th two mthods attmptd thus far. O variat is to calculat th mass momt of irtia with rspct to a movd axis ad to icras th plat width at th sam tim. Assumig th width icrass by w/, lt us calculat th distac d btw th ctral axis x ad th w axis. Takig ito accout that th w mass is 3/m (with m big th origial mass), th followig quatio rlats th w mass momt of irtia ad th cssary o: + h + md 0 w + h m 9 w 3 m ( ) Aftr som algbra, Eq. (P.10) yilds th followig distac d: (P.10) 133w + 148h d (P.11) 1 This valu is accptabl providd d is lss or qual to w, which lads to: 11 h w (P.1) 148 Solutios: Chaptr 3

4 Problm.3 As show i th sprigs sctio of Chaptr, th stiffss of a torsioal hlical sprig is: k h 4 Ed G 1+ 64R E whras th stiffss of a spiral (plaar) sprig is: (P.1) 4 E d ks π (P.) 64l E is Youg s modulus, G is th shar modulus, d is th wir diamtr, is th umbr of activ turs, ad R is th hlical sprig xtral radius. Th lgth of ithr of th two sprigs is: ( ) l π R (P.3) Usig th rlatioship btw G ad E, which is giv i Appdix D ad which is G E/[(1 + υ)], it follows that: E G 1 ( ) 1 E + ν E (P.4) 1 + ν whr µ is Poisso s ratio. Th hlical sprig stiffss bcoms: 4 Ed + ν h + k (P.5) 3l 1 ν Th followig rlativ stiffss ratio is formulatd by mas of Eqs. (P.) ad (P.5) as: kh ks 3 + µ (P.6) k + µ h ( ) Figur P.1 shows th variatio of this rlativ stiffss ratio with Poisso s ratio, which idicats a rductio comprisd btw 70% ad a littl mor tha 7% by usig th spiral sprigs istad of th hlical os. Solutios: Chaptr 4

5 rlativ chag i stiffss µ Figur P.1 Rlativ stiffss chag i trms of matrial Poisso s ratio Solutios: Chaptr 5

6 Problm.4 Th sprigs of stiffsss k 1 ad k ar coctd i sris; thrfor, thir quivalt stiffss is: which bcoms: k 1 kk 1 k + k 1 ' l kk 1 1 k1 3l 9 k1 k k ( + ) (P.1) (P.) wh trasfrrd to poit A. Similarly, th origial stiffss k 3 is trasfrrd at A ad bcoms: ' l k k k 3l 9 (P.3) Th two quivalt sprigs that ar ow locatd at poit A ar coctd i paralll (thy udrgo idtical motios), ad thrfor th total stiffss at poit A is: ' ' 1 kk 1 ka k1 + k3 + 4k3 9 k1+ k Its umrical valu is k A 3.17 N/m. (P.4) Solutios: Chaptr 6

7 Problm.5 Th traslatory dampig cofficit accouts for th latral frictio, ad thrfor, its xprssio ca b rtrivd from th first Eq. (.8) takig ito accout that th hydraulic rsistac is zro, amly: c t πµ Dl i fd D D v which abls xprssig th cofficit of dyamic viscosity as: µ o ( ) D D f i o i d π D lv Th rotary dampig cofficit is xprssd i th scod Eq. (.8) as: 3 Dl i md c πµ r ( Do Di) π 30 This last quatio abls formulatig th cofficit of dyamic viscosity as: µ ( D ) 60 o i d 3 π Di l i D m (P.1) (P.) (P.3) (P.4) By quatig th cofficits of dyamic viscosity of Eqs. (P.) ad (P.4), th pisto diamtr bcoms: D i 30mv π f d d (P.5) with a umrical valu of 13.3 mm, which is usd i Eq. (P.) to calculat th cofficit of dyamic viscosity µ N-s/m. Solutios: Chaptr 7

8 Problm.6 O ach shaft, th barigs act as rotary damprs i paralll; th dampig cofficits of th log ad short barigs ar accordig to th scod Eq. (.8): πµ Dl πµ Dl cl ; c 3 3 i 1 i s ( D D ) ( D D ) o i o i (P.1) Trasfrrig th dampig cofficit corrspodig to th short barigs, c s, from th origial shaft to th log shaft rsults i a quivalt (total) dampig cofficit: N c 3cl + cs (P.) N Takig ito accout Eq. (.59), th agular vlocity of th log shaft ω 1 is xprssd i trms of th rpm (rotatios-pr-miut) of th short shaft as: 1 N N π ω ω (P.3) 1 N1 N1 30 Th dampig torqu applid to th log shaft is: N 1 N π md cω 1 3cl + cs N (P.4) N 1 30 With th umrical data of this xampl, th followig rsults ar obtaid: c l x 10-8 N-m-s, c s 6.83 x 10-8 N-m-s, c x 10-7 N-m-s, ad m d x 10-6 N-m. Solutios: Chaptr 8

9 Problm.7 Figur P.1 shows th sid viw of th plat i a arbitrary trasvrs positio dfid by th coordiat z i th chal. h z g 0 l Figur P.1 Trasvrs positio of body slidig i a chal Frictio, which lads to viscous dampig, occurs i both itrstics, udrath ad abov th slidig body; accordig to Eq. (.7), th corrspodig viscous dampig cofficits ar: µ lw µ lw c1 ; c z g h z 0 (P.1). x c(n-s/m) z(m) x 10-3 Figur P. Equivalt viscous dampig cofficit as a fuctio of th body-chal wall gap Th total ffct o th body is that of two damprs that ar coctd i paralll, ad th quivalt dampig cofficit, as providd i Eq. (.31), is: Solutios: Chaptr 9

10 lw( g0 h) ( ) µ c c1+ c z g h z 0 (P.) Figur P. illustrats th variatio of c with th positio z of th body i th chal. For th plot of Fig. P., th miimum ad maximum valus of z hav b st as g0 h zmi ; zmax g0 h zmi (P.3) 0 whos valus ar: z mi 0.45 mm ad z max 8.6 mm. Th dampig cofficit rachs its maximum valu wh th gap btw th body ad th itral wall of th chal is miimum (which mas th othr gap is maximum) or maximum (which mas th othr gap is miimum). I ordr to fid th trasvrs positios for which c is miimum, w d to aalyz th first drivativ of c with rspct to z, which is: ( 0 )( 0 ) ( ) dc µ lw g h g h z dz z g h z 0 (P.4) g0 h Th drivativ of Eq. (P.4) is zro for z ' 4.5 mm ; it ca b chckd that c is miimum for that valu bcaus for z < z th drivativ of Eq. (P.4) is gativ (so th fuctio c dcrass) whras for z > z th drivativ is positiv (which idicats c icrass). Solutios: Chaptr 10

11 Problm.8 Lt us us Nwto s scod law of motio basd o Fig. P.1 blow. l/ m mg θ Mg l M Figur P.1 Rod-bob pdulum i arbitrary positio with gomtry ad forcs Th gravity forcs of th bob ad of th rod (which is placd at th ctr of gravity) produc momts opposig th rotatio θ, accordig to Nwto s scod law of motio: l J θ Mgl siθ mg siθ (P.1) with m ρlπd /4 big th total mass of th rod. For small rotatios, siθ is approximatly qual to θ ad, thrfor, Eq. (P.1) bcoms: l J θ Mglθ mg θ (P.) Th total momt of irtia of th mchaical systm about th rotatio axis passig through th pivot poit is: 1 J Ml + ml (P.3) 3 Substitutio of Eq. (P.3) ito Eq. (P.) rsults, aftr simplificatio, i: m m l M + θ + M + gθ 0 (P.4) 3 It is kow that Eq. (P.4) ca b writt i th gric form: θ + ωθ 0 (P.5) ad thrfor th atural frqucy corrspodig to Eq. (P.4) is: Solutios: Chaptr 11

12 m M + g 3 ( M + m) g 38 ( M + πρld ) g ω m 3 ( M+ ml M + l ) ( 1M + πρld ) l 3 (P.6) Wh th mass of th rod is ot cosidrd (which is quivalt to ρ 0), Eq. (P.6) rducs to ω * g l ; this is th kow quatio of th atural frqucy of a masslss rod bob pdulum. For th umrical valus of this problm, th atural frqucis of itrst ar ω rad / s ad * ω rad / s so th rlativ rror btw ths two valus is lss tha 0.08%. Solutios: Chaptr 1

13 Problm.9 Cosidr a coutrclockwis rotatio of th lowr lvr by a agl θ, as sktchd i Fig. P.1, which shows th a displacd positio of th two-lvr systm. θ A l/4 C B l β D l/ l Figur P.1 Two-lvr mchaical systm i displacd positio Th sprig is trasfrrd from its origial locatio (A i Fig..39) at th d poit B, cas whr its stiffss bcoms: l/ k k' k l 4 (P.1) Similarly, th mass is rlocatd from its origial positio at D to C (which is th sam as big movd at B o th adjact rod) ad th trasformd mass bcoms: l m' m 16m l /4 (P.) For small motios, th dyamic quatio dscribig th rotatio of th lft rod about its pivot poit is: ( ) θ θ θ (P.3) J fl or ml ' k' l l whr f is th sprig lastic forc. Combiig Eqs. (P.1), (P.) ad (P.3) yilds: 1 k θ + θ θ + θ (P.4) 4 64m 16ml kl 0 or 0 which idicats th atural frqucy is: 1 k ω (P.5) 8 m Solutios: Chaptr 13

14 Th mass m ds to b movd to th lft o th uppr lvr i ordr to icras th atural frqucy. Assumig th mass movd a distac x masurd from th right d of th uppr lvr, it ca b show that th modifid dyamic quatio is: ( ) 1 16m l x θ + kl θ 0 4 (P.6) As a cosquc, th w atural frqucy is: * l k ω (P.7) 8 l x m ( ) Th coditio of th problm actually rquirs that: By combiig Eqs. (P.5), (P.7) ad (P.8), rsults i x l/6. ω 1.ω (P.8) * Solutios: Chaptr 14

15 Problm.10 Th followig coordiat coctios ca b formulatd cosidrig small motios occur i Fig. P.1: y lϕ x lϕ Rθ (P.1) As a cosquc, th coordiats x, y ad θ ca b xprssd i trms of φ (th rotatio agl of th horizotal rod) as: l θ ϕ R x lϕ y lϕ (P.) Th origial mchaical systm is partitiod i two subsystms as show blow i Fig. P.1 with f big th forc coctig th two sparatd portios. f k k 1 φ m x y θ R m, J x x l l l f Figur P.1 Fr-body diagrams ad gomtric paramtrs Th o-slippag rollig of th whl about th istat ctr of rotatio (th cotact poit with th vrtical wall) is govrd by Nwto s scod law of rotatio motio ad rsults i th quatio: J + mr θ fr (P.3) ( ) Takig ito accout that th whl momt of irtia about its ctr is J mr /, th forc f is xprssd from Eq. (P.3) as 3mR f θ (P.4) Rotatio of th horizotal masslss rod about its pivot poit is similarly xprssd as Solutios: Chaptr 15

16 ( ) ϕ ϕ (P.5) ml k1 ky l fl whr th mass momt of irtia i th lft-had sid of Eq. (P.5) rsults from th poit mass m which lis o th horizotal rod. Usig th coctio rlatioships of Eq. (P.) ad f of Eq. (P.4) i Eq. (P.5), trasforms th lattr quatio i: ( 1 ) 5ml ϕ+ k + 4l k ϕ 0 which is th mathmatical modl of this mchaical systm. (P.6) Solutios: Chaptr 16

17 Problm.11 Th quatio of motio of th systm was drivd i Problm.10 as: ( 1 ) (P.1) 5ml ϕ+ k + 4l k ϕ 0 which yilds th atural frqucy of th systm ( k1+ kl ) 1 4 ω (P.) l 5m with a umrical valu of ω 16 rad/s. A rductio of 10% i th atural frqucy ca b achivd through a icras of m, which chags Eq. (P.) to ( k1+ kl ) 1 4 ω 0.1ω (P.3) l 5 m cm ( + ) whr c m is th, ukow as yt, fractioal icras i m. Solvig Eq. (P.3) for c m lads to c m ( k1+ kl ) lm ( ) ω m (P.4) which is c m This mas that a dcras of 10% i th atural frqucy is obtaid through a 3.46% icras i th mass m. Solutios: Chaptr 17

18 Problm.1 Th quivalt stiffss trasfrrd to th middl shaft is N N k k + k (P.1) 3 1 N1 N4 ad its umrical valu is k N-m. Th total, quivalt mass momt of irtia corrspodig to th middl shaft is N N J J + J + J + J (P.) N1 N4 with a umrical valu of J 0.01 kg-m. Th rsultig atural frqucy of th rotary systm is ω k / J (P.3) which is ω rad/s. Accordig to th problm, th altrd atural frqucy is ω ω + 0.0ω 1.0ω (P.4) * Usig Eqs. (P.1), (P.) ad (P.3), th altrd frqucy of Eq. (P.4) is xprssd as N N3 ( k1+ ck k 1) + k N1 N4 1.0ω (P.5) J whr c k is th fractio icras i k 1. Solvig Eq. (P.5) for c k rsults i N 1 N 3 ck 1.0 ω J k 1 (P.6) kn 1 N4 whos umrical valu is c k I othr words, a icras of 38.31% i k 1 is dd to produc a % icras i th systm s atural frqucy. Solutios: Chaptr 18

19 Problm.13 Th forcs actig o th body m o th icli ar show i Fig. P.1(a), whil th origial positio ad a arbitrary positio spacd at x from th origial positio (which coicids with th rfrc fram) ar idicatd i Fig. P.1(b). f m x origial positio α mg f t x gric (displacd) positio (a) (b) Figur P.1 Mass-sprig sigl-dof mchaical systm o a icli: (a) schmatic rprstatio with forcs; (b) two positios o th icli Applyig Nwto s scod law of motio to th body m rsults i: mx f f (P.1) whr f is th lastic (sprig) forc ad f t is th tagtial compot of th gravity forc; thy ar xprssd as: f f t t kx mg siα (P.) Substitutig th forcs of Eqs. (P.) ito Eq. (P.1) yilds: mx + kx mg siα 0 (P.3) which rprsts th mathmatical modl of th mchaical systm of Fig. P.1(a). Solutios: Chaptr 19

20 Problm.14 This particular applicatio is quivalt to a forcd udampd cas wh cosidrig th Coulomb dampig forc is quivalt to th (gativ) xcitatio agt. Figur P.1 shows th fr-body diagram for th body movig to th right. x f f f m mg Figur P.1 Fr-body diagram of th sigl-dof sprig-mass systm udr th actio of a dryfrictio forc Th solutio to this problm follows a mixd approach, as show xt. Th fr-body diagram of Fig. P.1 abls formulatio of Nwto s scod law of motio for th body udr th actio of th lastic (sprig) forc f ad th frictio forc f f, amly: mx f f or mx + kx µ mg (P.1) f Ituitivly, th motio cosists of vibratios whos amplituds dcay gradually dow to a trmial poit whr th lastic pottial rgy stord i th sprig i a xtrm positio is o logr sufficit to coutract th frictio forc. Figur P. shows th qualitativ profil of th rspos curv, capturig th iitial motio portio (th amplitud gos succssivly from x 0, through x 01 dow to x 1 ), as wll as a gric portio (with th dfiig amplituds big x -1, x -1, ad x ). x A 0 A 1 A -1 x 0 x 01 A 01 x 1 x -1 x -1, A -1, A x t Figur P. T T Dcayig vibratios of th mass-sprig systm with Coulomb dampig Th solutio to th homogous part of Eq. (P.1) is harmoic (siusoidal for istac), as s i th fr udampd rspos sctio. Morovr, thr is a priod for th Solutios: Chaptr 0

21 vibratory motio which is T π/ω π (k/m) 1/, ad this xplais th tim itrvals that ar qual to T btw ay two coscutiv amplituds havig th sam dirctio i Fig. P.. Lt us aalyz what happs btw poits A 0 ad A 01 i Fig. P. i trms of rgy. At both poits th motio dirctio chags ad thrfor, th vlocity bcoms zro. As a cosquc, it ca b statd that a dcras i th lastic pottial rgy (sic, ituitivly x 01 < x 0 ) is qual to th work do by th frictio forc, amly: 1 kx0 1 kx01 µ mg ( x0 + x01) (P.) Th right-had sid of Eq. (P.) idicats th distac travld by th body btw poits A 0 ad A 01 o th plot of Fig. P. is th sum of th two coscutiv amplituds. Equatio (P.) simplifis to: 1 ( ) µ (P.3) 0 01 k x x mg A similar quatio ca b writt for th motio btw poits A 01 ad A 1, amly: 1 ( ) µ (P.4) 01 1 k x x mg By addig up Eqs. (P.3) ad (P.4), th followig quatio is obtaid: 4 mg 4 g g x 0 x µ 1 T k µ µ ω π (P.5) It ca simply b show that th diffrc x x -1 has th sam valu as th o of Eq. (P.5), ad, actually, ay diffrc btw two succssiv amplituds that ar associatd with motio i th sam dirctio is th sam, which ca also b xprssd as: x0 x x 1 1 x x 1 x... g µ T (P.6) T T T π Th amout o th right-had sid of Eq. (P.6) is a costat ad thrfor, poits d to b locatd o a li sgmt, as illustratd i Fig. P.3. Solutios: Chaptr 1

22 x A 0 A 1 A A -1 x 0 x 1 x A x -1 x T T T t Figur P.3 Amplitud poits o dcayig vibratios ar locatd o a li sgmt Th small right-agl triagls of Fig. P.3 ar similar ad thrfor: x0 x1 x1 x x 1 x... costat (P.7) T T T Equatio (P.6) also idicats th sam coclusio as Eq. (P.7), ad thrfor, th dcayig vlop of th vibratory rspos is a li sgmt it ca asily b show that a sgmt li that is mirrord with rspct to th tim axis is th vlop for th plot portio corrspodig to gativ valus of x. Solutios: Chaptr

23 Problm.15 Aftr displacig th body a distac x 0 from th quilibrium positio, it will travl aothr maximum distac x 1 o th othr sid of th quilibrium positio. Du to frictio, kitic rgy is lost through frictio, such that th rgy variatio is qual to th work do by th frictio forc, amly: 1 1 kx0 kx1 µ kmg ( x0 + x1) (P.1) Similar quatios ca b writt coctig succssiv poits of maximum displacmt with rspct to th quilibrium positio: 1 1 kx kx µ mg x + x... ( ) 1 k 1 (P.) 1 1 kx 1 kx µ kmg ( x 1+ x) Lt us assum th body stops aftr vtually travllig th distac x from th quilibrium positio. At that poit, th static frictio forc is at last qual to th lastic forc, which mas: rsultig i: µ mg kx (P.3) s smg x µ (P.4) k Additio of Eqs. (P.1) ad (P.) rsults i: 1 1 kx0 kx kmg x0 + ( x1+ x x 1) + x µ (P.5) By usig Eq. (P.4) i cojuctio with (P.5) yilds th total distac travlld by th body: which is d m. kx µ mg d x0 + ( x1+ x x 1) + x µ mgk 0 s k (P.6) Solutios: Chaptr 3

24 Problm.16 Th quatio of motio for th cylidr is dtrmid by applyig Nwto s scod law of motio which ca b writt as whr J θ c θ kθ (P.1) t 0 t θ + ξω θ + ω θ (P.) ct ξω J (P.3) kt ω J Th quatio systm (P.3) is solvd for th atural frqucy ω ad th dampig ratio ξ, which yilds ω rad/s ad ξ 0.3. Th dampd frqucy is calculatd as ωd 1 ξ ω (P.4) ad its valu is ω d rad/s. Th amplituds of th fr dampd vibratios of th cylidr corrspodig to a tim t k ad to a subsqut tim t k+m, ar xprssd as Θ k Θ Θ k+ m Θ ξω k t ξω ( t mt ) k+ d (P.5) whr T d π/(ω d ) is th dampd priod. Th ratio of th two amplituds of Eq. (P.5) is Θ k ξωmtd Θ k+ m Usig th atural logarithm i Eq. (P.6) givs m as πξ m 1 ξ (P.6) 1 ξ m l( ) (P.7) πξ which has a valu of m.79 for ξ 0.3 ad 100. Thrfor, th amplitud dcrass 100 tims aftr approximatly 3 vibratio cycls. Solutios: Chaptr 4

25 Problm.17 For th fr vibratios i vacuum, th atural frqucy of th systm ca b dtrmid from th kow atural priod as: ω π (P.1) ad its valu is ω rad/s. Th traslatory-motio stiffss of th hlical sprig is calculatd as: 4 Gd k (P.) 3 64R with a umrical valu of k 0 N/m. Th lumpd-paramtr mass ca ow b foud as T which yilds m 0.5 kg. k kt m ω 4π (P.3) Basd o th rlatioship btw atural ad dampd frqucis, ω 1 ξ ω, th dampig ratio is calculatd as d ω T ξ 1 1 ω d Td (P.4) ad its umrical valu is ξ Th viscous dampig cofficit of th liquid is ow dtrmid as: k c ξmω ξ ω Th umrical valu of th dampig cofficit is c 1.6 N-s/m. (P.5) Solutios: Chaptr 5

26 Problm.18 Th quivalt stiffss corrspodig to th thr actual sprigs is k kk 3 k1 + k + k 3 (P.1) bcaus th sprig of stiffss k 1 is coupld i paralll with th srial combiatio of th sprigs of stiffsss k ad k 3. Similarly, th quivalt dampig cofficit corrspodig to th paralll combiatio of c 1 ad c is c c + c (P.) 1 Thir umrical valus ar k N/m ad c 17 N-s/m. (a) Th mathmatical modl for o xtral forcig is simply: which ca b writt as cx + kx 0 (P.3) k x c Th Simulik diagram is show i th figur blow. (P.4) x Figur P.1 Simulik diagram of th sprig-dampr mchaical systm with o forcig ad ozro iitial coditios At th itgratio stp, th iitial coditio of 0.0 (corrspodig to th iitial displacmt x(0) 0.0 m) ds to b spcifid. Figur P. illustrats th tim rspos of this systm as plottd by th Scop. Solutios: Chaptr 6

27 Figur P. Simulik plot of th tim rspos for o forcig ad ozro iitial coditios (b) Wh a forc f acts at th poit whr x is masurd, th mathmatical modl bcoms: which is writt as cx + kx f (P.5) k 1 x x+ f (P.6) c c Th Simulik diagram is show i Fig. P.3 ad th rsult of th tim-domai simulatio i Fig. P.4. Figur P.3 Simulik diagram of th sprig-dampr mchaical systm with forcig ad zro iitial coditios Solutios: Chaptr 7

28 Figur P.4 Simulik plot of th tim rspos for forcig ad zro iitial coditios This tim, th iitial coditio ds to b zro i th itgratio oprator to coform to th problm s rquirmt. Solutios: Chaptr 8

29 Problm.19 Nwto s scod law of motio is applid to th rod rotatig about th pivot poit udr th actio of dampig ad lastic forcs, which rsults i 1 m ( l) + m( l) θ c( l) θ kl 1 θ k( l) θ 3 r (P.1) ad ca b writt as 1 4 m + m θ + 4c θ + ( k1+ 4k) θ 0 3 r (P.) Equatio (P.) rprsts th mathmatical modl of th rotary mchaical systm. To b usd i Simulik, Eq. (P.) is rformulatd as with a θ a θ aθ (P.3) c ; a 1 1 mr + m 4 mr + m 3 3 k + 4k (P.4) ad thir umrical valus ar a ad a Figur P.1 shows th Simulik procss ablig itgratio of Eq. (P.3) to dtrmi θ. Th iitial coditio for th scod itgratio is 3 o, which is th valu of θ(0). Th tim variatio of θ is plottd i Fig. P.. Th two gais ar itroducd as gativ valus. -K- -K- Gai a_1 Scop 1 1 s s -Kd^th/dt^ dth/dt th (rad) th (dg) Itgrator Itgrator1 Gai Gai1 a_ Figur P.1 Simulik diagram itgratig Eq. (P.3) Solutios: Chaptr 9

30 Figur P. Simulik plot of th rotatio agl θ i trms of tim Solutios: Chaptr 30

31 Problm.0 (a) Figur P.1 shows th fr-body diagrams of th two pullys with th wir tsio f t, th sprig lastic forc f ad th xtral forc f A. θ θ 1 f R C f t R A R D f t R B f A Figur P.1 Fr-body diagrams of th two pullys Nwto s scod law of motio i th rotatio vrsio is applid to th two pullys, which rsults i: whr: J θ f R fr J θ fr fr AB 1 A A t B CD t C D RB θ θ1 RC RR B D f krdθ k θ1 R C (P.1) (P.) Substitutio of Eqs. (P.) ito th scod Eq. (P.1) ad combiatio of th rsultig quatio with th first Eq. (P.1) yilds th quatio of motio: R B RR B D JAB + J CD θ1+ k θ 1 fara (P.3) RC RC Th complt mathmatical modl of th rotary mchaical systm cosists of Eq. (P.3) ad th first Eq. (P.). For th umrical valus of this problm, Eq. (P.3) ca b writt as: θ1 1,061.9θ f A (P.4) Figur P. shows th Simulik block diagram basd o th two aformtiod quatios, ad Fig. P.3 cotais th plots of th iput forc f A, ad of th output agls Solutios: Chaptr 31

32 θ 1 (t) ad θ (t). I ordr to obtai th rquird forcig fuctio a Ramp iput (from th Sourcs library) with a slop of 80 is combid with a Wrap to Zro fuctio from th Discotiuitis library with a Thrshold valu of 80 as wll. This particular typ of fuctio is a uit ramp fuctio which drops to zro past th thrshold valu ad kps that valu aftr that. I th Cofiguratio Paramtrs if you choos a Stop tim of s, you will obtai th plots of Fig. P.3. Figur P. Simulik diagram of th two-pully mchaical systm (a) Solutios: Chaptr 3

33 (b) Figur P.3 (c) Simulik plots: (a) iput forc (Scop 1); (b) plot of θ 1 as a fuctio of tim (Scop); (c) plot of θ as a fuctio of tim (Scop3); (b) Th systm is a sigl-dof o, as th variabl θ (t) ca b dtrmid i trms of θ 1 (t), s th first Eq. (P.). Cosidrig that f A 0, chags Eq. (P.3) to: ( ) which shows that th atural frqucy is: R J R J θ kr R θ (P.5) C AB + B CD 1+ B D 1 0 Th atural frqucy valu is ω 3.59 rad/s. k ω RR B D (P.6) RJ + RJ C AB B CD Solutios: Chaptr 33