a a/ .I a\' '1 -,.1 { fL L L L L; L fL L L L L tfL L L L L L L L L L L u COURSE NIO. : AMLLL0 ENGilNEERilNG MEGHAffiIG$ By tr- c." eEr 4,E-ga tj" i;iE'' " ''::\ '' . -.' - ',, 6X P' C"D U t/riR for finite bodiesoisize much greater MECHANICS ' .. ' wlid Classical 1t{cchanicsis not valid for point rnasses (particles) but shall cover in this course than the atomic size moving with speed less than 1/10th of the speed of light' We 2. Axiorns and force systerns - 3' Rigid body dvnamics i- fio"*"tics of i point and a rigid body work aird stability 4- Equilibrium 5. tagrange's equations 6. Principle of virtual 1. KINEMATICS' frame' A' rcfcretce fi'o'me Kinematics is the study of geometry of mo0ion with respect to a reference rlith--Euclidean geometry them between is a set of locations in three dimensions having invariant distances disLance betrveen any the which in being valid. Reference frames can be attached to any igid bodg (a body axes ,' yt z can be imbedded in a tso rrraterial poinis remains the sanre fcr all tinie). ,Rectangular Carcesian be the unit vectors in the directions given frame .F rvith its origin and orientation being arbitrary- Let !,j,k to frame F is defined b}' nith respect oy(r)i+ o,(t)k :,;:r- The derivarive of a vector A(t) : Ar l"!r)L+ = dl.,ldtii-;-,y lizcd = The rules of derivatives of vectors are the i"i* dci+ (I'l) a'k' same as those for qcalars excePt that the order of the cr F terrrr should be Preserved' L1 ICTNEMATICS OT'A POINT point at lirnes t ],er P and' P' (F.ig.1.1) be the locations of frame r. occupied by a moving are defined by f' frame to and t * Ar. The ietocity qr and acceleration q5..rvith respect gr = jil. *- =Ji:. * = er , 4r - 4r= 4r ' -- - -BRr $flir:.ie Frw of the a-\es x' y' i gr and qp depend on F but are independen! of the choice of origin olr the orientation for simplicicl'' yields embedded in the frame F. lniegrating (1.2) rv-r.t. t a1d dropping ( )tr '[tf' r(t):r(0) s(t):s(o) + !o' {t')dt, For r.niforrn accelerrrtiott e= 9t,eq(1-3). + u(t) = g(0) + srt, + Jrv1)dt' (1"3) dt) = f(0) +g(O)t + est?lz' (1'4) path coordinates' 8 \Are derive expressions for u,g in Carrrsiau, cylindrical polar and kt%rg :o 1.1-1 Cartesian soordinates "))1--: Then-eft Let a(t),y(t),2(t) (fig-1.2) be the Cartesian coordiuatesof the moriug point at tirne t' F;s Ii-. - ; --: + ,Ur:ii+vi+rk, + r(t)=r(t)i+v(l)i+z(tl\, t!"=i, ar=i=i1r, oc=i, lz=i, *3{ s(t)=;i+ii+ or=y='.ity, ,:=i=t:.,. (tls1 (1'6) gstt) . - L ,rg,r4* the llV f zor movingpoiniaitimet.Lete..(t),g1(t),e,(t)betheunitvectors/ffi;N'*-.,^ 1.1.2 Cylindrical polar coordinates r,et r(r),{(r),2(r) (Fig.l.3) be the cyliudrical polar coordinates of g i (z) */. ffi'o-.! il".*"#;;;;;;;ti;i;:.:'#Xtiar ;t9$t F-+-../\., t' ,$j$*:1:g:19 ,.r', \ l '' (d) and axiar i,Yr triad rvith positionThese form a righi-handed c.rordinate lines at this e,(t) = $tr = (-sindi+ cosdrd = In.tlresequel, (') = ( )1r is %(t) = 4e+, ir-nfilied. The position g(t) = k' fig. t.3 c*lr = g' (1'7) g, vector r is differentiated rv.r.t. F to obtaiu a using (l'7): t, = i = igo * r9r + i9r,* 29., ecl,, = g = ci-Y b"Sg-, + ( (-aosCi- sindj-)i = aip *'t 6)ef+Zezt 1 I ,L -6%, (r.8) + =) -sin{!*tosfi, + ffiq,,F1- :, : cosdi+sinCi 'z t1.e) t 1 ! i l :i _;: _- l: ;t a.-- :r = t6..':,':,r,: r,., . .o: = i, @d, ar=2, aadi , , -2i6+t6, ;. i and a, * i-, oo i*' nosince g., 94 arcnot constanc tectors in-i;rime + Note that "* = 1tr1"'6) 1-1-3 Path cooidinates Let s(r) (Fig.l.a) be the coordinate along the trajectory of the moving point at time t- A right-handed triad of rinit veciors g, gn, g! at the location s is defined s= dr de. &r Ar Jf45=* ;-':>;Sg, a: Eo yn as: = frri= o :+ ,*';:0" 'T I t , '*r."+ I QrX9.=9C, g is the fangent l)ectortothe curve in the direction of increasing s since in the limit as As of the chord Ar becomes tlre tangent and the magnitude of-@!" becon:es 1' Notc tlrat -e,) (r.1i ) 95=Qxq' = w=;%' tq.Xel=9.3r gCXq=9.' ;; = Jito a,' ftr, (1110),.,,- ud -' (1.12) * 0, the directiorr dfr d*re.. .= '= ueclor oi t'he cun't'Tlrerefore the unit vecto.r e.n.in thJ-iirectio,r of der/Jis defined as the princip al normal q. and-dirccled ioirards cen!rc of [here being infinite set of unit vecrors normal to-the curvg ai P. en t l/p, :urd ccrtalarc C (f+ = g, #g^Lof the curvl. The mignitude of dglds is delined as the c-uarctare the radius of crtta:,nnr- of the curve at P. The cross-product oll and lc" ftq* the first ts'o p iscalled equations of (l-f2) yields ( e5 ic is rrorrnal to both q and 9,"- The pliure of g, ard g, at P is called the (1'12): the plane through P which is closesi to the trajector-r'. \lte obtairr g, a using is called thb.6itornral osctlatitg plo*e. ltis da '- dt- aectora-s dzd.s ,-u=ie':i.a -__!_< -. r_. I* fU rU frJ U u U IJ dsdt da ds o=i=ia+6|]fi, + .+a- k (1.1.1) 'a4 (l.lri 4=3St+fre'r => Ua=u5=$, ilr:i, = 3.. =u: izl1r' ac = 0. directed torvards the centre of curvature. If the trajec0ory is gisen paranretrically as != d2r ,dr rz dr d2r *, dr d.r dr ds and (1.13) yields For 01 77=77t7;) - dt ds' :p :t#*P*rr*r or aT.' dild, - i+ r'ii (l-17) + -. | dg f dx = f' ,d9^, rdt i : l;!l =- l;fl, , - 4z4d* = f"!,. .rp.= , q = cos 1i! f sin s,.7 (dslds) I Ve,ft')= For a circrlar trojcctorX of radius fi i.tci tJrcn -EE' =t#*fitrrffr (I , .l7i and rvith ( - (- dSld" sin lii { cos *trt* (-qi!{i+ ce/i}t*) / l;,g $l = *t-rtn i{., Oand (Fig.1.6) i:-j;:: r - g =-sq . . i:l i rp!+ cos r;!) *}:... -ve sign to be used for point A '-'r fi'! 0; &e--:ie r -n "/\ 'G9 . p=R, et=?' 9.=-9-: "=E' :=0,..=* X=0, i:'i=0.s- i l. ts) riyld!lds, re'=t si#iaity, {t). E +:Stzltl? 111uft {'j, nilrere *r,e sign to be used for point P rvhere ( r= e. r,rerobtain pusing(l-17): l(drldt) x (d?ildx1ll= lI"l. l&,1&l- (1+ I''l't', aplcte lrajeclory(Fig-1.5), r=ei *f(a)j, taag = 1.13) r-S .a * -9n = _nire,+ Itd%. 2 :L i.e .9t e p--t L ,;. t,'.':./ ({ I t .*.j\- fr:- _ ..-- Define"rclotiacaclocitygeelp-andor.iu:acxlerutioitgg,.elrlfpointP.relativetopoint/ tfrtr = - irrlr wzrlF = irelr; 6(re -: :.:a. ;r, : i. -d t',ttr) s.PlF-4.^lF' aptlF = ip1r-- fr(urlr - 3e1r)lr= ;;" t2ele' pararrcters of independeat sysrem are defined as rhe nunrber ot a rigid bodlr are of .o.* of freedom orrreedom a"g,* - ,'" (.".:f"l:f:::iT:'"',X$iJ::r#;.'J;;;;"-. t'hree rotational)' "t* "_*; ,- t." and 1t1.." translational 1.1.4 Index-Notation . ^- -r^-^ro rlrp rnn. .Indevelopmen[oftheoryit,iscont'enierittouseintlexnotoliorr,i.e..todenotetlrecoordinatGZ,V':.tlre drra?'ag (ai); 9t'91'9s'' (g); resPectit'ely-. ttre unit t-"ttoo f'!,L by e,'a2'23 ('i); aslayto.-and components indoi such as j i' a t'erm of ' following the "tgeu'"it use we For brevity "o-*ot;ori-'lono"or;on " ""p""tud anexpressionimpIiestlratthesumof,t,.*;;;;';tobetakeuwitlr1.takingvalues1,2,3.Forexample :+ 6; * dircr * dizcz * daca = a; 6.; * d;c; = a; a2' b3+ dsrcl + dszcl'!"!!*ca = tl3 ' O:'* = 0:'*-d1r : or'' d13ca dP,-h * "' : i-e-, 0r *.'drrcr {riinde''c:Di,+d''rc& ,i:::' The repeated i.dex The is ind& ii". fr". *igr**y argebraic repraced by other.co*r-enient irdcr and.-ci* be values 1'2'3 to obtain rli-rrrutiiau oo'u.'"*Jta,t'"oi;td i- :-q : = ecu4lo"* 1n=e q :-- .5{' " *::jilJ*.:.": :: ; :H:;r:jl:-T ":fflx" q1t(t)' .@.\' &p., t - Le is governed b1'' of rn relative to F at time The rate of ctrange of orientatioa -r ,\t:},{ I:: c191(t)*c:u,(t)+caq(r){a} _ ,.\ 0rsr(r)*0"9(t)+0sss(t). r - ,r\ r r-- /rl_r-fi-a-(rl= esls(r) e.,-(r) = Clgr(ar-rc3=?\&/TLrEr-r\-, salr(r) = oler(l)*c2sr(r)*a3s"(r), erlr(r.) = orthogonal uuit vectors' independent since g are nrutually not aIe oi,0f -compo[ents nine The (6) 'ci cs=oSinilarly :+ 2o1=$' i'e'' c1 =$' er(r)-gr(t)=1. + 2grir(4'e1t-t):o,' :+ or*6r =0' 'tJi"ty q1(t).g{t)=0, + 'ertr(r)'g(t}+e1(t)-Q1r(t}=0' a3 ' (c) -.b3' = -cr' 1-- Thu-" a-b-c-a<and tr-'2+J+cros-product bL : -a?- i.e-. -02=0' c2= The sinrilarrelationshaveSeenobtainedbycyciicchanges (a) witht'e r'trs'expressed as a ioatpu"a"ot' ll',e iewtite only three compoDents o3, 63, c13te using (b),(c): x gt = I X 9t" 9r;r atgz'-ct9g = [619r * crgr *4?9s] = g?lF =--439r fogr ='cr9.1' - * bs93 6391 = := [0a9r [63gi * crgu * t a3gj] x 92 ='t'' x g'' x 93 = gi x k' < + crgs + "'!b] (r-'20) Ltr i-e r's'r- =g*x€i g x (l'20): far q u by bv forming formi'g q' 'o for factot g- we soh'e The underlined ternrs are added to ger a common (grx 9i)'= (ei'et)gl-Gi1er.)e= $'t-tt=2s' e, xitr:g;x =uiei=a,rsr +u"s2+."3qj =-?ei.s =9.er + %-e2+" * =_::.:;r," since Bence Z of franre rn relative to frame is defined as the angula'r velocit'y 9L4p ! e .a- i ' (121;' < 1- rr :' F' i1- and i1r f,etAr(tlando;(t)bethecomPonentsofvec[orA(t)retativeto!'andg:ta;(t)q' = 4(t) = Ar(t)& + AI.Q)D_+ Ae(t)&-= or(tlsr +iz(r)er-+as(t)h:i:::'* 4^=4q="""'=@ig.ir ' : 4r={[-=.o.rr--'{iE' : ' 1.2.1 Relatiou Betweeo I , ! i q d - -r L I t _.--- -9 =-, rt -:..-t__*-;j,: : .i i, 4r=G-q.)ii.::"*cier:":_-.fi]?,;;,,,6. r-€'r * - hin; Io particular, if L-. A.'ts A' !:+e-x vector in m' then "tinffi ":;:; 4*=Q^tand (1'22) br--s.x ,.1 (1.22) : yieldg A- 1t:] r'-*'Consideraniiherodhogonalri8ht-handedunittriad{' of ql' qmsrqef arel'e' the definition we ptove ihe co*srslcncy of - 6r"d-; we obtain ,rr- U=i"g (1'20) and (123)' .g4d.l=f,tu-d=* r x (e x el)l = I 4lr : 5bl .- :4x |tt*'*1"- te - ,4.' - - , ::x"ffi ':i:l:1::::":':ffi Lra le directiorrs of 9i in m. It clrcrge ot is a mea-.ute of t.he rale.of ;::j,i}:.'iit!::::lJ*;:::;'i:ilJ;* ::::lIflll5f}fi ##*;.1*fffi -t-- I L: |;#r*;l;H|;:j:.:j:;#.'ff-'l,1ll,*T.il,HrffH;,;;;Gsr+is+sv-rh€ierore;::, \='il?*ee1r) x L x i!= 3P +{llsp 4= 5{3T x L-. = = a1, = - L- We r. respectivel, t.- u \\te take ,n" &l?' x f''*'l.'ni"i}'o i"i-"u ,r*" (1.?4) (I jzs) ? F -f:E< v tLt and to P and A be fu:red to nr' Thus &0) = &i(4+!p.{(r)' telocitY relaticn: F and use (l '22) tr) ;-ield derivative of eq(a) rv'r't' frame +; x lr'rl' + = 8rt" + Lpslr =jl,.,rt Er^1(t-264) !e1r = 9e1r *t'r x I're * 9P1'''' to f ield acceleration relarion: frame F and use (l'22) t' w'r' eq(l'26a) of derivative \te take the tirnb t-... tj lJ t&r,. lrel, -rtr u tj L 1-. erlr:9e1r +elrf I t*t*ex d Writing !p,,= {L, eqs(126} can be--written .e.p g1i f i' = ipelr + ft(arr*)rr as V- ietgxtSflrt. i yu71; .grlF-l 9 a;r +sgxtf + IJ n &: , -f;*"t + l*gtr-trn +sr xsPlml' x = g.rlF 4gx rPa *gl [i,p,tllplm * Qr1'a' gzlr = 4r1r *t't" x Lp,t{t'r x (grx LP;,l+?4x l*. L L. dt"lt =ezlr *{ii1z +:*lL * *12 *&arg"t" *d gP|- & erlr and 4'fl13-2 ltela.tions betweea w't't' franres f' and m by I denote the position vectors L_ f.-- = *l'anriag: ritt, L* u I: *1r 4= Gf"*ql1: 4)+*lI li.*g{qr x A= Az*g&lt x.'9lsrr . =9s1:*9lz1, _--r_+ x -,-.--< * 6\r. erir+ *.iqr ---:(eblz)1r' Gil , = 6,, (es1r)1s (1.260) (r.27ol (1.276) 9Q'*(se'xff)+&g!ix1e1"+sef" .J ,. \, I -i :'t.. . \..a:r _ - 1.3 RIGID BODY KINTIMATICS _ Eg-r.ro . FormaterialpointsPand.l{ofrigidbodym(Fig.f.l0),g,,l,a=Oeel--gandeqp(1-26),(l-27)vi:td (r28a) 3 plr= 3 afp +co x lPa +S9X(Slx len) ! rfr'- ! e1.+ g') X lpR , epir = e71lr +a x .AP * s-x (sx AE)' 9-plr = 9,t1n * gx AP. .i 1.4 KINEMATICS RELATTVE TO TRANSLATING FRAME jr Let frame m translating w-r.t,. frame F be named as ?. Frame f is siid Lo tzvr.slatc rv.r-t. P i[displacements of all its poiuts A,B,C,-.. are the same in f' (Fig-f .11)- Eence F 9s- B(1-2s6) ll. - :lt )Fi? I L&a lsnlt = ilat1r = L-BIF :+ &sr b a constant vector in f- -rr1r:& Hence the displacements, velocities, accelerations of all of T are constant in .F. Therefore gr= (9.. x g'lr - ioints of T rv.r.t. F are same and all the line elernents 0 and eqs(1.21) aud (1-22) yield {p)/,2=Q, :+ gL=0,,', ()tr=()tr. Using (129), the kinematic retatioas (L-26) [cr'a rnoviag point !,te1r = u-aJp * u.4a, P and a point gplF = g.a1p.* g,pp. .d trted to (t-2e) f becorne (1.30) 1.5 ANGULAR VELOCITY OF m F cqr =Q. A body 2 is said to have plate rnotion relative to f if all its n'raierial points have plane trajectories in parallel planes. Choose g attached to rn in a direction norrnal to planes of trajectorl', then qj moves parallel to itself and Slr = 0. A rigid body rotating about a 6xed a:iis and a rigid cylinder rolling rvithout slip dos'n an inclined plane have plane motion relative to ground- ' The motion of a rigid nut on a fured screrv is not a plane motion. An a-tis q attached to the nut in the direction of the screrv a-tis does not chaage its orientation wJ.t. ground F, i.e., islr = 0For such cases, we choose & : e. (Fig-1.12). Then et,g, have no cornponent along $" Let d(t): be.the angle betrveen e1 iuid Er at time t: pr, e?(t) - - sin d(t)E, + .*O1r1qr, ft(t) = 83, *1r =Q!r1r=(-sin0E1+cos0E2jd=0e, 9:1r=(-cos0E, -sindQ)0- -r9.,, eq(121) + sr=(e.- xett')/2: [9r xirlr*e:xiztr*ss x4'F1/2=[e1 x iez*92*(-der)+esxq]/2, er(t) = =) cos d(t) E, + sin d(t) - - 0€t__ol- .=0- a e t 1-31) tr equals the rate of rotation of a (any) line g, (f fu) fixed in m relative to a (any) line E1 (a &) fuced in f. In this special case t.l is the rate of change of a f.ritc iaglc. For general spatial motion of a rigid body t: of a fi.nite angle. F;nite onqutcr dGplaternestt rt'.o. fr Pra"rirg t<<g:{.!*& does not equa! r cnl directr'trr, is not a tectov c3 thc! do sg't <.ae -g(r<.o<dilg *ro g..^ra\\oft1rca L"o qf olAiH,rn one o 4Aili.'^ \s nob <aqrutolri-e . on trg'l'ta 5r STANTANEOUS AXIS OF ROTATION OF A RIGID BODY A rigid body is said to have an r-nslcalcteous cris o! rctatiott at time t if all iis points (its material pgints or points on its three-dimeusional rigid extension) on such,an a:iis have zero velocity at that in<tant. .exists if It does rol always crisf. For given g-. of iG point A and u at t, the instantaneous a-ris of rotation lve can find a point B subh thai 9a =9e *u)xAB =Q,'i.e., if utxAB = -9^, i-e.,if u1 =0org^ICf,sincetrx.4B isnogllaltoq i.e., only if q. .!l = 0. Hence insiantaneous a-xis of rotation Liists onlyl/ u,t .r^, = 0. 1 .JJ J J JJ J,J JJ 'j :J ,.J .J j - ,-I -I -.-I ":I :iJ 1i ,l ,:. the instaniai-&gs.€ras- of !&"tioo ahvays exists for plane mocion 'but'ii'genei.l :f " l*d..b:d1 silce ue f IA t (Pig-f -13): !^: k +o, x IA= kellJAlsing0og = ld(I6)g . \ for. plane er'#9- The velocity of auy point ,4 can be expresbed in tcrnb of its radial distancc f.A from ls.,{l l-gr f _ls.al =lgql =ulIB IC I=I Fi3't'r3 (1.32) The magpit,ude of uo o[ a point A equals the product of lol and its radial distance Jrl from the-instaetaag'ousa-xig,of rotation through the instantaneous cent,re f, and i-s directed uornral to /A in lhe sense correspondingto r.r about f. For a rigid body in plane nrotion, if the directions of velocities of its trvo points /. I haviug ' tle same plane of motion are knorrn at a giveu instant, then the iusLantaneous centre .[ can be located in that plane at the intersec[ion of the nornrals to velocity veciors at ,{ and B (Fig.l.l3)- The instantaneous centre of a rigid body in plane translation is at infinity. The instantaneous centres ft of bodies A are shorvn in Fi9.1.14. ,;\t, -.1 \s ug .\ t o vn f,1 f, Fg-t.t+ \, >.d! 1.? ROLLTNG IvITHOUT SLIP Let the material points P1 and ?2 of bodies t.id* rt.*a,t,tS),Ue i1 The kinematic contact conditioii^iie defined as follorvs conraci ar time ,.. I l- stip at, t ir ci:i'" f .si,tr;, IJ L- s"jt."l t inpty that slji]- = 3- in'rpending slip at I if sp:,iii iil] in contact, L L L L r l- does uot hold at !f, for the corresponding points i L L L L. L L L. L. f.. = uili; at t,!,bu[ this equaliiy r-e- !pr1g (tr*) rb"fil*;l€(t.') no St"1' I I }.: etoi Pins:{ and B move in a 6xed elliptical tt-tt:'t rvhrich translates as in a straight slot : lvith Exarnple 1.1. and pin" o.r-t. (I) ground re*rence acceleration ay=_rozr6cosotasshg*rnr.il.Ji.r".Thepositio"ril"r""tr*rdresrotatt=0are-6and tt or alcebraiion a.d (a) velociry rate of change " ".rrtrlr from each other and't'he zero, respectirelv. Finiihe rshich- the ceirtres of pins ."0""* at from the ,"te reactions u.ar, tui bodv' (c) the (2) the translating trl.r.*"*,ng (1; .r-..t. ;;;;;y of this rerarive ,0""o oiooararion on the pin.,{ ornras.s..'.t i11;"::tl'!-;:'*:t'""X,:::1*"':1lt{fff ::'"'i'": smoorhsrors of ihe to the right of the centre Q -- {ri \=__ ulyy' L -1. t/g ((\) Solution jr#" r--t- ! €ig . tl D'of the slot z1 be the locatior r of rlre Point ? "1 [[encc:' Let (c,V) be the locatiou'of'ttre u1(0) = 0. ar = -u)-:rcrGosrd' *e' e1(0) lta*e *'e = D' poi*t for anrd ;;;i*a-u.rl)' ul -- it The coordinates Jt d' = =ur(0) + l" "'(') r,g :r1= c1(0) * -r^r*o siurat' -t .: G) l.i cent-re Nl 1t ordl ! =uocosut' ar' (t) : slot: of the ellipse and tlre straight are relatcd by the equiations L_q 7+ F= 1, y= (" - cr)tau0 for z'Y' Relations bctrteen Equations (2) arePotved i'i' Relat'ious'betsreu i'i (2) = DE taln0l t eqs(2): areobtaiued b''- dilferentiatiug v= ?*H:u' Equatitxrs (3) ate soh'ed for i,g IAE (i - i1)lau (3) 0' eqs(3): are obtait rred bY diltereutiating 1 (4) I + Equations(a)aresolr,edro,;:'ri.FortJregivendataiattlrepositi:n"',,..o..].. a;t cm;' :+ ""cosurl = 0'6' + sirl =t0,'l' l,r, = ?o cosurf, = 0;6ro = 6 : Eq.tr}+..i1:-80(10x0.8)=-6{0cnr/s.}1=-(80)l(roxos)=-38e0G,cm/s2. v = ('- 6)/5' 400' Eqs-(2) :l ,t become zo t44s into trre lsr equation yierds 1Br3 Substit,ting y from the 2nd equati,oo of .A and B, respectively' ,iu 0.2269- These a."o .ooo-"." ('?) +4!2= * 32 = 0 rvith roots 10'85 I a'd "*oo.dinates Y-(10'85-6)/5=8'400' + > i =-53g.d cm/s. ! = L74-2cm!s, 640h/5' : 0' i G+ -8'4v/100 * eqs(3)- + 10-85t/400 i=(i+38400)r/5' (539.43+10.8$)/400+(1?4.2?*8.4f)/100=0. eqs(a)+ cnr/s?' 7= -38338 crn/s?; ! = 10?'99 + t0?'99! cm/ir' a1 = ii*li! = -3833Si + cm/s' 1?4'2j ' '+ g^ = t! rj = -539r{i* (u) For P.oitt A: 9c(2) E u L u u IJ $qaily, Fy,.t*tii:*=o-zzo{canob.!\ yelocrtv yd acceleration'1d,3;.' ,",',,.. ' ls =-onz-f i- i-6'tj cui/s, s; = -32654i+ 10125i cm/sz' . ' : 201'1 cm/s, 3r.= fiacrxs*A = 10?39coeec60o = l.2tL7 <arrls2' tt':::--= 11691 cm/s?' cwcl = l0l25cosecOb" sB = em/s. 3s ia co:ecu: =gB = -3-Mcosec60" : -4'2 cmls. ,; LAt? = 3'{ 9= 124J9cmfs2' lelr =ir €:201-1ecm/s , gtlr =3n9= 11691 gcm/s?lalr = sp e = -4-29cmls, ;1 = yac&c i = u u, u L u L 39 = ggccec0 <' +-. 174-2covec60o flence (b)Thepositionvector-of r{relatinetoIissgwiththeseparatio65='48=(y:-ys)coaec8- i = (rir-tia)cosecg r -- r= 201.1-(-{2) = 205'3 cm/s' t:, --ii^lr i= - (!a-!3)cosec0 = 124'7-11691 == -1156? cnr/s3' as (Jkt*a eguirt s \ w.r.i- ground reference are egual -- ; c ..:n!lrtinr frar The ra.tes of separation rv-r.t. translating frame f and z-atis' rvith '' (") ftt the tangent io ihe eltipse at (2,t, ye) make-augle e ,.;'pao+y?/t00 = L. I :+ 2e1480+zwl fiaa.=0. li : (-)i'' + { = avfl 7 -'{:f!" Fig'El-lc' The cquation of nrosion of pin The FBD (free body diagrarn) of the pin is shorvn in 3 is 031(-383-3Si+ I!: aal1 + Nr(-"c 30. i+sir 30'i) + IE(cos 7!-!e lasiu ?zrli) - 0'0rsj = 0'1089' j : 0'55r * 03516N: = + i: -0.8660rVr +0-3o?4rY2: -3-8338, 1'0?99i) direction of reactr2u nrz N- The negative sigu ia If2 inrplies that the actual o{:r is opposite to that shou'n in Fig-El'1c' screrv of pitch I cm rvith e*"orple 1.2 A rod $,ith a threaded hub rotates on'a righfhand the time { is in seconds. one side of rhe rod has' a telescopic o - o-uz rad (Fig;El.2), rvhere crrr/s3' s'irh its tip A at rest at arm mounted on it which has an outr*ard acceleration of 0-4t side of the roc] has a riglrt-ltand thread a d-rstance of 20 cm frorn thc a-xis at t = 0. The o0fter o!rto the.thread aL Lltc rate of r = 0'2t rad/s o pitclr of 0.5 cru- A disc rotat€s relalive t\ llu a n-ith the o[ rotation of fronr the axis and a point B on its a-xi! a0 | = o has a distarrce of 30 cnr F'1"it'z \9e I = l0 s'' -a< at I and points'4 rod- Find the velocity aud acceletal-ion of point A: cau be easilv co'rputed' co*Pon€nts since r,i' i,i,6';'i ;" ;; ;;*;t;; + llr : 3.?65 1q, ir1 - -rso4 "rtina.i"a i=o-4tcm/s?, =) i(t) = i(0)+ !o'o*dt=o'zt|cm/s, d=0.{r2rad, + +. i=0-8lra'd/s' at'=20+0-2r3l3 cm' r({) = r(0) + lo'olzf + o=0'8rad/s3' (2r), the rod adences arially by one pitch (2 cm)- lleuce i=16.l2r -f7tr cnr, :+ '' i -- ]fu cn'tls' :+ ! = $l* cmls?' .1 l.aft fi ; air=t0sr'"=.i0'i0.2(ig)3/3=86.6?crn, i=0-2(10)3=20cm/s' i=0'4(10) =4cmls2' For)o-ne rotatiron + - :+ ' *:t6'i"d/",,ja : OS rad/s?, i = tlr= 2.$,tG cnr/s. i = 0-8/r = 0.2546 crn1s?' i '')'-': --; I+t'+- il:is,+rCe++ i:g=20q.*G93.aeo+2.546so cm/s, A= ' O.t(rgt ,..:*.-;9t:,=-,ffr':aO]rl*f1z;O+r;)e++ieo =-55{3q'+389'3ea+0'2546scm/s2 .=.;{.='E.}o.4qrjeHp:} .=' -.i..rp 1o.s€iii**Fg : where g., Cr, i ."!9 -,T-,:-!,9.1y" at pornt 4 in Fig'El2' rr a{ I f,, .' ya cosec0' i'. il,t : gis theunit vector alongdB' rvith sr = Rdafiretoframe?., the pcitionrectorof r{ is crtg' where L L u ru u ru u rU L u u U U u Li -'.-lr) :;-: u aL - (0.5'cm)' Eeace for a rdta0ion For one rotation 1irf1"1*a'O-r*, point B advznccs re.dially by one pitctr by given A0 of the disc, the incredse Ar ia the'radial coordinate of I is 0's/t0/2r cm. +- i -- o'si[z* = - 0'5'.i2r- + i = 0'5dul2r cmfs2' = "*1"' u=02rrad/s, + 6=L2rad/s?,r(t)="(0)+ I"'*n-30+ lr'o''Ly'or=30*o.a2ilt"ftcm. * aLt- 10 s : r = 30.80 cmr o, : 2 tad/s, i = O-5(21/2n = 0.1592 cm/s, f = 0' (O'2)l2r =O'01592 cm/s3 A,r lflre rralues of i.d,i.! for poinc B are the ".*" ". for point''t- Hence !.s = igoa .Oe++ ;9* = 0-f5929 +246'4ea+ 2-5469. cnr,/s' +(2;O+ "O)+ * !e, - -19?1 1*27'194+ 0'25a69, cm/s?' ae = (i"d2)q' ,:.,-i"i; ffi:1:::l;::r;: iu a II:ll 8,,'r-"\-l i and vettical plane' A pin is Iy=r.ar' " consrrained io follorv the path:,r:(o) ='0-2+,0icosd rvith r€sPect I \44' ro the plate. Firrd the velocity atrd accelera[iou of point D and l, *ffi;l the centre P of tlre piu relative to the,ground'at'the'instant 5t1en'n w--:T'r loning parabolic paths.(Fig.El-3) rv en. e : l't : =-.i::. :l' : 1,1 11 :'^ 1,'.::', :T::. ::,:'; Js' pression is 2 cm and the speed of -.{ is 0.2 nr/s rvhich is decreasingga\ rr i,':t'-"* rt'^ 'f Li -} -^.;-^f attIretateo[10nr/s?.Firrdtherrrilrinrunrsti{Irlessoftlrespring< ; ;;;;i; *" ", m*, o-r kg does follo*' the sive' pat'n zt tlis ,/ 'tt?are stnooth. 'tt?the curved sut)port of the piu arestnooth. The slot.and thecurvedsut)portof insranf,. Theslot.and ,:1T,.^---- JJ dfr,,J 4, r-l J f\ , r-Z | 'Z-.+W '{ {.%_ff' : : .f Lr //i 2o -d \u.5" rtrr+ t"t=f Fi3.€l-5 Fi3. in nriud the type of data gir.en. rve use path coordinates ro find the acceleration of "1 of tlre plate' Since lrr and cytindrical components to find tlre acceleration of P s'-r-t- trartslating franre ? say i. jlv.r-t- ground is the sunr of these accelerations, u'e need to express each of rhem in a common bas'rs. m/3, .i = -10 m/s?, .g"= L2t2, + g! = 2'42, :+ y'' = 2'4 m-l ' = y{=tamc-1.2, y''=2-4tr-I. P=(l +itzftzlly*l=l'588m' o:50-lgo' :+-arr=0.5rn; (l) ga = sg = O2er nr/s, se = Se + (33/p)e. - -r0e. + 032'5t9g." m/s2. (2) e, - cosa!*sinai- 0-M02i+0-76S2j, 9. = -sinoi* cosc'j = -0'7682i+0'6402j: o-. - -6.il2li-7.666j m/s? Substituting (2) in (1) yields u.e = 0.1280i+ 0-1536j m/s' 0.2 t satne :+ llo = !!e, ,., = o ^' Forcomputing9.p17..6=0,ii'=i=3.rad1s'6=ij=2rad|s2,i=0.J=0and For all poiuts of [ranslatiug lranr1 u a.ud g are -J ,..:J Solution Bearing 6 fi,J 1-l m. :+ i= -0-lsind0 rn/s, .+ f : r=0.2866m, i=-0-l5nr/s, f =-0-879'tm/s2' + atd=30o, ! llplr = i9.- + r/'d +:g, = -0.15e.- + 0'&59Ego.trr/s, gptr= (i- rC')9" +(?r:f +"d)%*?9" =-3.4599.. -0.32689o m/s3, =+ rvhere g. = cosdl*siuqij :0-866i+0.5j, ea - -sinpi+ cosoj = -0-5!+0'866j. J .J JJ Jil Di <J Ir ii-t-J d r=0.2*0.1cos0 Er .tgt (4) (5) "-J J f,r v,j .J :J Substituring (5) in (3) and (4) yietds ltptr= -0-5598it0-6696j nr/s. aptr -2.832i-2'013j mls?' Hence w-r.t. ground,!.p = ge * !?lr = -0.4318i+ 0.S232j mf s,9.p = ge * g.e;r = -9'253i- 9'679j m/s3' Tlre FBD of the pin is shown iu Fig.El.3, where F1 , .lV1 , N2 are spring force. normal reaction froin the slot and normal reaciion fron'r the curved support,. For the given conditiou A'2 =0. Equationof motion of pin: F - ntap, :+ -Fre.aNr9+ q -O.1si=0-l(-9.253i-9.6?9i) (6) ; _d J ': -!i J .J J be ottaincd '^-:--' "q-'= -'-;'(u)' ' - :oa llr can irom L iegngonen3s ilr ,1er, r'l: ' t_., _ o.i1ssr11O5) :o-r(_9.253x0 s66)+(-9,?9xg.5[, can be obtained ,r,,.u*t"*.,"o I Exanrpre 1.4 by of t1e springis given & = F1l6 directly by fcrrnt *, . & =9-z9ls (a) Apoinr-",T-1.:1:::Y:1+;,:Xf*::i,-#;: It is A*.'ii"i o ez u ]- -,, c6 x: fZJl. ':ff-e" Ir Gtg-Er-.+ r = i' =(-Rsinder + Eccd "qe)d' * $=-al{R2 +"}'/" +"'l=7iii;+ct}r/2', 6,' (E?+c"}r/? 1'={gl= . - Itsindee)d?' a"o'de'+.':)i;i!;:/c:sesr * (-Esincqr ir tr= g: + ' qqs(r)_(3) + :=11}|?J,ff5J (b) We express 1in * tetms r= of e'" e': R. t - o' "::{:i;;_r,r;, r = Rcosdel * eqs(a)-(6) (1) (2) (3) + c?), _ + (_Rc6de, nsin431u:11.4i Rsin 4gz+ c|ps = R9'- * cae-' : = cQ' "r' "f.' ,'^= '.= ^t,= + c2)Lt2' +-";.t't'' :^:"''o' + Roea+i cde'' * ie' + -R62 r6)e' = (z;d + + ot ;: ;j(;7.u,f';l'*r;."!"i'n' -= ;- i"* : =O:g4&;p'02=:11'+l/:' - -' + Esin ;:ffif J#;;;;;;"rr""qtexpres'onsdirectrv i, o, = -;;""u e=' R.cs 6s, + soturion .. +t')t/'' =+ g= (Ecd+ce')u/(82 ;7 (4) [] al ii 1 ! + ce',lol(R" + "')t/t' e= -ll.o2l(R2+ "')lg. - lft+ . + + Rcosegr +c93)dE/ds (-Rsir{g' drldsq o, > (c) Given d$ld.s 1, + d$lds --.l!ot * "t)'/'' le,l = (E' a "'1't21a61<lsl = (7) g=(-Esindg.r*Ecos@g+ce.)/(R;*t'f'.;- D ac;o,-'siroq,) - rlsin og.\l(R'+'=)' i)'t' - (-Rcosee' I?sia{q' Xiii''itt'U+ (-.R'cos{e1 <Iqlds= p=(R?+'"\/n' rlp=ldgldsl=R/('R?+c2) + ' g,= P(tgrlds\= 1-cosdgr -sinS92) = -9" +tt)] 9"' g - -oq +{u'lp)e^='aga+[r'?t?7141 u = ua' . '*n';,: :1,iil',X$":1tr|,:l lJtrl?;' one roratiou berween tr," t"ng"ni ono ocis: p =2c.tr.Tlre angle cosa =q' 9d - Rl(Rz+ ,. t')'/'' 1= 1 (8) (9) pitch p or trre the axis is carre-d' trre angle o: l:.:1.'0"' i *'oer) is called the helix -.inrJ, = tauo = cl R= pl2rB- Helix is pri'cipal radius of currtaturea.d augre rre*x pitcrr, of constant varues The given circurar herix has animportantcurvewhiclrisusedirrdrills.scrervdrives'scrervfeeds.etc.i ru ,l -.- u U. 4o i :a i l I :, REMARK egit€ oft€u in connected systemilbodies in 3D motion, the information is provided for the relative rates of rotation s*li, ,;lili of body i relative to body j. about a:ies fixed in body y'. It is preferable to name the body w.r-t. which the inform-aiion is soughC as F. The body rvhce rotation is giren w.r.t- F is called body l, the body wlrose roation is given relative to body I is named bod-v 2, and so on. Then 4aV -- Clrlr * 9f-e1e * uzlr \ er?lr t!4r=9lz1r*911r 9s1r '-r = Artr * (cr{r + !4rlF x gqr} tihr = s;qr * ()42* szlr x zstz) grlr = *-qr * (ri:<s + 94s1r-x_144s) _,etr=Lh1p*€z1r ltqr: ;- -F99slz \ *9{els : t4.rs g--J i llttr uay= 6rlf +(Oq, * lrrtr x glztr)+(c&f? + grztF x ssr:)+(el.rtg+-?t".x !4ts) I \:J ,' u2 an-. -.....,...--i ,4= itr+ (c.?tr +91x us = glr * gl:tr *?*42 *tr)*(ri3:+ g* u41J + (it1e+ eb x €r1r) ( rli = drr + (estr 4 glr x laetr)+ (dtr, + !{r x !4qr) = 94r * (Or1r + s.r x !&gr)+t=rqz *(4r + qlr) x <t:l rrhere F has beeu.dropped for the entities rv.r.t- F for convenience- Stud-r; carcfull!'the uudcrlited {irst ierms in the cross-products. The firs! teuns irr the cross-products in r-- arc the arrgulirr rclocilies relatiae lo thc basic fiv;mc F and uot relative to the prcvious body- If tlre bodies are natued as described and the aogular velocities are added up starting $'ith body l. body 2 rv.r-c. l- body J rv.r.t. 2. etc-. then the first Lerrn, in the cross-product appearing in the derivatir.e of nth ternr, equals Lhe sutn of all the terms to the lett of tlre ath term in the or expression- Ttris procedure has been follot'ed iu all. exanrples. The velocity and acceleration of any given point P is found by successively finding the velocity and 4 rv-r.t. 2. including the retative acceleration of points on tlre a-xes of rotat,ion o[ 2 s'-r.t. l, 3 rv.r.t. .2, velocity, relative acceleration and the coriolis acceteration ternrs if [here is relative moti-on o[ such poirtts rv.r.t- [he previous. body. If P is either a point of body i or its motion retative to bod-r' i.isokno.vn. then up ar.d ap are determined in ternrs of a poiut (say Q) of body i. rvhose Sa.qO hai,e already becn deternrined: l+ lp: le + cl' x QP * !-rti. 9p = 9e+ it x QP + eix (st x V) + 2*- x gpli -f gpl;, rvith gp1; = jplig, Qpl; = Splie. * (i"rlr/ prle^- The last ternr in the cxpression of gi ind the last, tio terms in the expression of gp being inchlded only if the point P has rnotion relative to body i. governor (a) (Fig.EI.Sa) is used to couirol the speed of a rotating body. It is A fgba[ Exaruple 1.5 rotating aboui a vertical axis at aogular velocity qr and angular aceleralion rir. The displacemeat of the sleeve controls the angular speed by a connection to a feedbacli device- Find the angular acceleration of bar .l{B and the veloci[y and acceleration of poiut B of the ball at t = al(p, if the angle 0 made by the arms with the vertical varies as 0 - 0o*01sinpt. (b) A heticopter turirs in a lrorizontat circle (Fig.Eti-5b). Thc point, O in its body moves in a circle of radius ,l? at speed u, which is increasing ac the rate of i at 1 = r/4p. The helicopter blades oscillaCe rvith A = 0o * 01 sinl. Firrd.the velocity aud acceleration of lip B of the blade relative 0o O at, t = rf 4p. (c) A spacecraft, is rerolving about its _aris OD. having a fixed orientatiou. It a I .r -t ! t tj tj u tj l-. L. l_-, l-- t; t-- :.1 :-:r:ir '- :-' aL a rate (Fig.El'Sc)' Its solai r.r l-- u L. rl-- ru IJ u u r_ E E u U L, 'b-aoets are opeoing at a programmabte5ale ci'".Tt i1lhg,so.P!: ,t I . '<:' -Find'the : ."""i*.""j"."rlo..oftl,".ol".panel2justbeforeandJus0aftcr0eQuals0o.-,..'..., b oi .i' ?- -6 )o ti\) u Lord - -_-A_s n H -----r^.--;j * - ' -'t 5 \-t rLr \- (c) F,'g.€I-5 <- *.be 6xed to body 1- Body I problerns are the same kinematically' t*ti 'l' t r physicat rrrs 3 l/rr-Y''v'q Solution sotutron (a) \o', The or-ic ii fixed 6va, to tn body h.dv l. r -r-.-r about ^L^..t atis rotaies rv'r't' to bodyl rotates rv.r.t. grouud about the fixed axis !' Bodl'2 L- lj ]l-l-f-- --iP-ri.\::i..i,.i.. :,., ,-;.i${*-i:i}I+ =trh, * =-rt go =0. -b= grr x0il gr o, A arcon body =0, OA=q: t and points -{. I !-t=la*.:txQ.A. lta=!a-f -:zx-E. (2) +;!'',' t3) .::l i $'e successively cornpute are otl body-2' Lrsing eqs('t) rc (3)' gB qh * r/, x OA *=r x (*r x QA-l' x (91: x A-D)' = a,1 * Q, x .48 + *ir (4) = qt Q, tt,x -78- -ar6sind!+0bcos0j+06sin0L * (1) ;{B =6(sin0j-cc0b) -tik+6'i+-'h.* rii-0i+"'ai rr+{0i+ The points go x(szxAB) = -u,ibcos1!-(0?+&'2)t'sind (5) I j+O?f cos0 k (6) t = !a=-(o+6siud).i+00(cos0j+sin0!) as:-12!106cos0*t,(a+6sino)lr+ t-r'a*0silld) -b62sina+&dcosdJj+(602cos0*66sin0]!(7) (8) rvhere 0 = 0o* gs sinpt :+ i - fllcospt :+ 6 - -p2l1sinpt (9) 6=-p'o'l{2 - at / -zrl4p, 0=0s+0r/'/2' i)=patlrt' 4*;jc.o1lin q6 u-ufR,b=itlRand,a,eBreptacedbvt'so,aao-,, i;;";;;;;;;;;;.';;,',"asinpart(a)rvirtr ... , -r---i-r)itto\ttntrtrl G'-:$:,1 a u0ld0)(d0ldt)- G i =.,. *o.:i obtaiue,r e:. rvirh (a) 0 fi':::ffi;;;,,." as i' = ";i"g = j Stud1. the various tenns in eqs(6) and (?). The ternr l:l at 0 =0i, aOlaO =,,to/Aa,' +: i) =ti1Os' !: is tto! an.obvious oue' at0=0J. aila=0, + d.:o' t rc.l/S F"i/s'' ! i Exaruple 1.6 A box rotates about a fixed vertical axis A and a yoked artn rotates relative to the box (Fig'81-6)' Trvo disk rotates relative to lhe arnr at the tt1o th611'n' sliders move relaLive to the disk in straight aud circular have siors with cosd = 0.8. The Points P and Q on these relative speed of 2 cmf s deceasing at the rate of I cm/sz poiuts of acceleratiou and velocity the to the disk. Eind P, Q and a Poini r{ of the disk' a solution In order of successive rotatiot'ts, starting rvith body roiating Let t' be fixedrb-the box l as 5[ewn' 3' i' i-' and 2 l, rroiio arm and the disc as uurr ,i""*oed we narrre L.e 1;,., l!'"tff""rn" . 6r - ---:- ^a;.-+i^6 "'''.:::: :,:".,.--;:; rvhich is fi:."d ti thd ground.,The a:<is of totation k is grou[d to the relat'iYe 1 box of of rotation axis Th€ OO,l a-tis of t:tLl1on of arm 2 rela[ive to box t is along ! which lt i*ud to bJv 1' The ];9i:::.*. :'* IL i .l E - the arm 2 is 6xed to tnelri'e. et tU" glven insl'ant 3.- l: \fle successively compose the angular vetocitie" rad/s, Or = -k tad,lsz @e= tlt-4L= -4i*5k rad/s (-4$ =-2L-20j- E radls2 4= tix-zt+e{r x({!)--k-2i+5Lx 9s= 9z-3i= -4i-3i+ir..a7" , a.\ ,6: 13 I 20 j - $ + i + (-4i+ 5 k) x (-3j) = Qt i,itt + !* ttzx (-3!) : (-2i go = 0, QE= 40! cnr : Poirrfs o' B are on body 2 rvith co = 0, :(-4i+ 5$ x aOi= 2fi)j cm/s Ba = !2o * 9le x OB as - as *,'t, x OB + gz x (9i x @) qr = 5k ,n: lei + ll L radls?' .]I cru-ls? 20'2L cuVsi u*pro"t ,4,gein ter,is of poirrt B of bodf' 3: =-15i - '-.LJ lL U-J \ i' *:-L BQ=-10!+5kcm. sets=2cn1/.s, ;eF:-1cm/s3' "= (;ery/5)9* = - i- 0-s! crn/sz ; = SqpQ = 2i cm/s, oQF = islr& + cm/sz 30j 30 k cm/s, *t x BQ-= -95i- l?si - 190! x BQ sj r.:J J I r -J sb x rve '::J !.:J e=-cos0i*sin0!--0'8i+0'6k' gPl3=2e=-l'6i*1'2kcnr/s' lle1z=-e=0'8i-0'6kcm/s? q.3x BF= 50j*30Lcm/s, Qrx BP = tlbj+ 190k cm/s" {!$ x En= -340i* 120i -200L c1/s? (--l-6i+ 1-2t) - -7'2L-6'4i- 9'6E cnr/s: 2,.,.s, x,els = 2(-4 i - 3j + 5$ x ' llp : sa + ss * EZ! !r1s = -r-6i+ 250j + 3l2k crn/s x se : 98+ db x B,P+'e"x (cb x 8.Pl +2er3 {r'1a *8r1s = -1346'{!+r$'6i ,::J ,tJ lrodl'3: E-Z:1o!crn pointphastelalivemotionrv.t-t- body3=+$,eqipress1rl'1epirrternrsof PointIof e has relative .rotio* *,.r-t- body 3 + il,J -J =(_21_20i_k}x40i*(-4i+5k}x200i_-1000!_40jcm/s?. m: Points B, A areon body 3 ivith bA=l0k i\'.-,40j- cur/s cax BA = ('-si- 3i+ 5LI x t0!= -30i* , fu xBA=(l3i-rgj+ll!)x lok:-Tt-130!crn/s3 150i-250h !k x (ss x BA't:(-{i-3i+5k} x (-30i+,{0i} = -2001+ lle = !!a+ sis x BA :-30i+240j on/s 320j -'*':"1:t' s,r.= eB+ db x BA+!* r(ek x EA\= -1390i- point il r1 .d , tl$ i*j - l ,.k 1u' ,<l so = sB + ss x 9Q.+ s4r. - -r3i+ 1-?0i - 30k cnr/s x (!4 x W * 2,esx !4F * Lets = -856i e(2 = s s + L3 x BQ-+g r-- - 390i - 103'8 k cmfs?' ions of a space vehicle to study tlre tolerance sir t'he ftight condit to.sitnulate used to 'ont rifrrac i< is rrsa<l A centrifuge at angular speerd r.r1 (Fig-81-?)- The astronauts. Its main arm I rotates about, a fixed lertical axis Example ;rJ;; i.: 1.? to t'rz relat'ive to 1' The pilot is strapped cockpit 2 rota.tes about an axis fixed to tlre arm 1 al, an$llaq sPeed pilot is moving a btolk' of mass o-8 kg lreld a chair 3 which ro0ates.ai angular speed'ar3 relative to 2. The iai,r" R relative to the ctrair in the plane of in hls hand, such thac iis centre oi mass Q tio"ribes a circle of 'rind the acceleration o[ point P at the instant shorvn' symrnetry of the chair. p is a pcfrnt in the pilot's eye. i. _ .1A : .,\J : - ,,,1- +- CI .l< Lt,- J*:.I - t-- *fiff iil{;:r":".$f . *r".,.ry. *f,,'fi$,-*Iilt'"tffi \,r/ i\-\2 :------:- 7/(Y;) t ; ; t<e't z ;s <.\fBi-cLj=Ji...---- c ','';m:=l;*:5;T"':if:.1i".#":'f ;i*i sorurion*.0*l,i,Jl.*1*1ffi t "il';;;i* =,"?.#; * 5-:l:*:,:l:: j{'i:l1':rytl$tlll**:t*:' o:'* Ji:;:r:r;;-, : - relati*e Tlr" .o.kpit, 2 rotates z L .toug ff Ifl :,,:ffi -- or rr" trt" augulat velocities ;,;r:* \ :: I^- = _cc,itsioo!=;i-0-6i+0-s! " g=sio0!*cosoL=0'8i+0'68' co"1=0.6, sino:0'8' - i;=2..:,1;';--'i,'5'-^'.*,::'3 - The points o, -lE- sa zreou =::;':il;'i;i:['ja:'ii. 3' c. p body t aad the poi*ts rn' CP =^'oL+ =cj = 10j ,'*6c+-' * t*t' - OG ateou body -02i:' 6s= 0 s L) tj t_ t- + k'1'r"'. 7 e6 .'-:^, ib and gi'conrr].* . 02j + g'6k nr 1(0Si+ 0'6H = 0-8irofl = -a0!+sL*/'3 roiizi.i"- Ti=;;;x bodv 3' Hence knorvir relative to The morion of Q'is =':':' \\'e -ruccessi*err (gb xe$ g, = gc+ rit x CP + ca x a'zi-+isza$'+ (-'?'ssi+5-32i 5 $ + (o'032i+ (-40i+ = 03eE */"'' = -r.rnr, -=*r*:-* L- 've {i-16$ - expres5 l ---^*ie*r poinr co'l'err c ..t :i Tl:::..:a 6(03i+0-6u=0,1i+0'1!-0.]E*,,,,, 3err = n 0'8i+0:6b' *ll"t u* =' *1"' '=,'= : :i:' : r'- =.;A,:;#::i:iXX.;:::.i!+,:2!m/s3 i:=;:: * i *|-b*erj t-- : (s * 9l +Zwx tQlr-t coF 130i t'u L} se = 2'152 $ + (-1's{ I i * j + s t-'s*' ro,oi-u"tj-12ji3EY-a0 -* H =\-= "ii rj l.-. x r + (14'4i IJ Ia,the FBD of the t- ' 1-. L L L L =9, c s,=soa L-* 1- * ii: i=?"*; L j=0'5!-i*61:.0^: L=msa :+ 0""0, E1 - 1'6 j - 19'2E )+(r.6i-26.6?i+1ril= shorvq is the !n Fig'Ell?' Fr * -0'8s!-0'8(10'20i-'6?'8j--re'zg$ tt{ J J a :r'ted force e-te'o=- ei '-t -=-.. equaiion of nroti'xr' Tlrc gilotthe -- ' -- - s3.g2j - 6ai{b f' = 16'01!-oo'or1 N. *fr'=xy"- x ' y't v\a/ - y."rv ,LX -. n- incline'(Fig'EL'P *-""f"d Exarnple 1.8 A tank'iihvTrilio-* *r.z r"tfrt'* the given of 1g km/h *rri"r, r-J"""i"rating-ai t'c-..r. --_^r-_ i"l*: : *" :i:: ffiJ:fi,."#"i*,"*.r, odls rvhich is increasing at the rate o[ 1 rad/s2' The ! "0".a "f barrel is rotating, .l. tO' - \.g-rt" - tt+ :"_1ri,:i,',;:::::'Wffi\L"*,? r relative to the Lurret' rvith angular speed of 0'8 a-- -*;:.":tlfr *T:i:.'::"#{*,i!;ll#:{',,i'*:l',$ffi;':;t'* the 3m' Find The poinu.4 is at the'end-of-the-gun-barcl-of lengttr H6ZE-*"-",." to the ground' velocity and acceleratiou of points r{ and P relative about an a-tis fixed in the ground' Solutioa [n order of successive rotatio.s. starting rvith body rotating 6xed to the turret as bodies l ' 2 ancl 3' I*t i' i' E be we name the tank body. the turret and the gun l"trel ortrre o"*", *,.r.r. ro rrre rurrer a.d k arong trre a-tis orrotation the plale' is a vector in the inclined plane and tu6et rv-r-t- the tank tlody. L is lornral to t[e ipcliped .j accclerati'ons' conrpose the tngula' velocities' a*gular I vertical plane t'rouglr the gun barrel- lte successi'ely velocitiesarrdaccelerationsasexplained.irltlre.retnark.above: ;:fif:"i:;[":;i'J;;;;;;;;i. e='sin15'i+cosl5oj--O-2588i+O-9659j. so =(l8x 103/3600)9=5s= cro - -0.2x 103/3600)e =;2e = -0'51?6!' t'932!ru/s? =0. :1. = 0The tank body I translates relative ro the ground + I / I -JJ I ,, .. . \r_i 'i t.-t l'294i+4'830j m/s .gt tt2.= ltr*2h:2k 'rad/s izt= gr+f k+i4r x2k:8+ k+0x'2k= krad/s? 0-8i* 2 k rad/s ut = I4+0-8i= x o'8!= 1.2!+ db = i.z+ 12i+ r^r: x 0-$!' = h+ l'2i+2k - -. l points O and B aie lixed ou ttre turr€L 2 r*'ith OB = 0.6j nr : ;, --<I e2xQB=2*xO-6j: -1'2im/s. ' ib'1 OB-- !x 0'6j=-0'6[ur/s? szx (* x @l = 2\ x (-l'20 = -24j nr/s! la = b* eh x OB =-0-09{i+4'8!0i m/d l rn/s2- ,-"a=go* rizz1@-+.:r: x (,*a-: x gg)=-r-llti-,{3:l2j r._ -r.!:: . : -1r: _- PointsBandr{arefixedotrbatrel3rvith e'=sin{ltcosdh=tl'Sj+0'6L' &=3,g-=Zfitd+l'8knr: *tx BA.= (0.8i+ 2$ x (2-qj + l'8U = =4.'8!- 1-44j + l'92! m/s 1'152 k ur/s? ; - e . ;4 = (0-8i+ 2E) x (-a-8i - l'44i+ l'e2u = ?'tsi - rl'r36j ; Q"x BA= (1.2i+ r-Oi+ \) x (2-'{!+ l'8!} =0'48i-2'l6i+2'88k rn/s3 m/s 1,92L ' gr = ea + c! x BA=.4i00it3.-!9i+ aa.: eB+ tib x ItA+ !$ x (srt x B/ill =2242i- l?'63i+ t;t8l' m/s?' , poiqt p of camnon ball lras known motion r.r.t. ba,rrel 3+ .EE-= !ZA=2ij+1.8k, :;:;:; Notice that : ff : gp = we cxpress 1gp,gp eep=600C =480j+300k m/s, in ternrs of point gp = -e+ZCax S,"l.f +91n' ,.1 ,...,;.1 . ' {5 - - I of uodr = 120E: - 96i+72! epls "= -,,,,, -,,s? ;::;il','f-.l;:Uf ?T !A+s.pl3-, .- .1 1.6j+ Lrad/s? !t 6 m'/s 3.. J ,.q i .< .i .< ,i t2 :,< L .< .-, :: -{ .-l i + e4 r ? r -/"= . < r* klj- L. I .., L^ l*.I;: L-.-, I Exarnple 1.9 (") ifr"\gio,b"Isupport o[a gyroscopc rotates r l?,a1s l.J a.5rls (loiqs- \ 1, o' '7c tl s tspi. r*c) o 'i1=2,1s '6;^frq tr:€ ilt L3 (bdaf,z) zu tlt i I t' - (a) Aff I L L L AB at thc rat ss;hown ,ls' 'l'i=rrtrl gi-bal l_--' IJ ]J abouc a fixed axis (Fig.Ef .9a). l'he frame m supporting the shaft of the gyroscope rotor rocates rglative to the grnbal at the rates shown. The gyroscope rotor rchr€s telative to frame m at, the given races rvith 0 = cos-l : O-& Find at this instant, tlre arrgular vclo<ity and angular acceleration of tlre grrocope ro0or and th€ velocit5r and acceleration of its point P relative to ttre ground. The kinematically equivalent problems of a spinnint top and part of a robotic device ate shorn in Figs.El.9b,c. (b) The telescopic arm of a robotic device, shorzr in Fig.El.9d, is increasing its length at the given rates- Fiud the velocity and acceleration of a 1rcint Q on it 'Jszis' relative to the cround. *"ffitI''ffit?fq tG e=atY/s, ' o?= o.5 m O?= o'5 .--- y ! e=at,/s'' -t-P8:o'i'p..u.e,.9 jf -Lr (bJ tl.. Solution (a) In order of successive rotations, starting with ginrbal :'"5u -a" o! (dr tcI srtpport rotating ';+tls 'r.l-'- in the ground, rve name the gimbal support, thc franre ar anrd. thc g-vroscope rotor as bodies 1, 2 aad 3. Choose *-". i; !, E. attached to body t, s'ith k along its fixed a-ris of rotation. and ! along the ads of 1_ lL L u L. L; L, L_i l J J 5 rl 5 er.7 rotation of body 2 relative to body 1- \lte successively compee the anguiar telocities. angular accelerations, velocities and accelerations as epl&ed in the renrark above, 1'hc spiu of the rotor is abouG the a,\is e - sin0j * cosd k = 0.6i + 0-8k- llr:4k taA./s, (2r: -0-5k rad/s3 93= ttr-0.2i= -0.2i+4k rad/s ,Lz=gt_0:Ii+al1x(-0.2[}__05L.-0;ri*4.kx{:0.2,=-0-li-0.sj-0.5hrad/s" sb = !h * 20e, = (-0.2i+4u +an{0.6i+ 0-8 k} = -02i+.12i +20& radls .t ,.-- 3 = 4 - 2 e. + u, x 2O e = (-0-1 i - 0.Sj - 0-58) - 2(0-Oj + O.St) + t -0.2i + !) x 20(05j + 0-SU - -48.1 i+ t.Z5 - 4-5 k rad/s"Points P and O are on bod-r'3 rvith g -0, 9o = 0, OP =0.5e = 0-5(0.6j +0.S!) gsxOP: (-0-2i+ rzj1"2051x (0,3i+03!1,=.,:l2i*0.0Sj-0.06k nr/s ,, =0-3j+0-4k . rr: I ,:.i{i' !43x(sb xQZ)=(-0.2i+l2j+20t)x(-1.2i+0.08i-0.06k)=-2,32!-24.012i+14-3&1km/s? c,t,rxOP = (-48.1i+ 1.2j- 4-5!) x (0.3j+0.4k) = 1.83i+ 19.24j - t.l..t3k rrr/s3 W -- b + !q3 x OP = -L-2!+ 0-08j 0.06\' mA sp=!b*drsx OP+q3x(r.ls xOP) =-0.49!- 4..772i-Q0.l6k m/s3. Points P and O are on body 2 also- Eence reworking, rvith g4", gi5 replaced by e42, 1i1s in the above relatlons, would yield the same values fict g4,,gp- Chect this fact, as an exercise. (b) Point P, on che rigid e"xtension of arnr 3, has given motiou rv.r-t. bod-v 2 + s'e espress -.;,..gp in tcrnrs of point O of body 2 with g, = q, !Lo, =0, OP =5$9= 0.5(0.0j +Q.SL) = 0.3i+ O.aE T: x OP = (-0.2i+ 4tr) x (O-3j +0.4!) = -1.2i+ 0OSj - 0.06! m/s t*x (ttz x OPI - (-0-2i+ak) x (-1.2i+ 0.08j -0.06L) = -0-32i - 4.Sl2j - 0.016k mr/* !5 x OP= (-0.1!-O-Sj-O-SL) x (0.3j+0.4tr) = -O.tzi+0.0{j -0.03k m/s! t tr2 r6 i i ! :? : t i l rq0-45,= 0'4(0'S j'**OSS 024i+ = = ;;;=o.oe=0.6(0.6i+0.8k)=.o3,6i+0j8Em/c.... ii**p= (-0'2!+a!) x (024j+o'azg = -1'-n'i:o'128i'-o.os6L m/s3 032j +026k m/s e.p = lo + 92 x OP * Wu- =1'2!+ gPl2 * ePP = -2ALi- 4'2s4i + 0'338k */"'' ee = et +,iz x Q2.+ sz x (t!z x QP) * i'rx : Points P and Q are on body 3 with PQ =O'li m eep % x fQ= (-0-2!+ 12i 0'32!m'/s- +20$ x 0-'l i = 2!- 12k m/s; l|-3x(!b*PQ\=(-0.2i+12j-+208-x-(2j--12E)--5{':ti-0'24i-0'4L k ru/s? fu x PQ = (-48.1i+ l-2i- 4.5!) x 0'1!: -0'45j - 0'12 03{L m'/s 94 = 9r +.g5 x PQ = -L.Zi+ 2'32i 4'974: - 0'182k nr/sz' s4 = sp+. tizs x PQ + s.sx (* * PQ', - -56'81i- :a 't-.t'f .( _{ __ : .- h m/s? t t,rr, S -^i L D fixed.c.l.lirrde'"i:'::::::.(:;':-".1.T* radius r rolls lr.itlrout 1'o^":. l- Fiud the.radius.of curr:ature ott -. --^t^-^-.r^... ^r'r.-.^-i poirrts -{,O' C,D r cvlinder Find tie..,elocity a,,d acceleratiotr of'material of the oath of points C, D in this positiorl' - l'of r'rv A Exarrrple r'xz.'rlPrs 1.1o " cylinder -- -- r -'-rlb\t (gl h. (=oo U",V6?&c D t),. i---5-' l't.to ,a * A. .,--_ T, -11t'"q I' ia :-. -_ : -q -- '8 '-\ (eJ . F i3-€l.ro Solution Let point B of body 2 be in cotrtac[ rvith'po'int r{ of bo y l'-C-:iroose i, i il plane of motion- Let u6audig=rlgbet,lrespeedarrdrateofcharrgeofspeedofpoirrto.Ilerrce gB=0, rrg=9, ur=utk, ar1 ='r.l1k, O':1 =-r! OD=ri' OG=-t!' Po=R+r- lrodr- I taki[g tlrc geomcirk csefrc as velocity and acceleratiog relations of trvo poitrts on the tigid ,all points of intsestis the satng 91 x (r^r xg) = -r.,?c- fire the reference pohrt. since the planeof.urotion of kinematic condition of no slip,;g^ = 9.8, * _ W.e use Da.=b* th x OA=tioi*a'rhx (-f0= (r,6l --'tllj-=!:a =g' (r) - _<<I .q ? :+ W = (,lrJ' :&e - gO.= ga1* Q1,4'OA oO!, + (ub! p.l - ofilftlR+ l/'lL "tiOl1,= gcl s,- ;["ir'3 I lR + r)l [ + rir rj' (2) *6rb x (-r':il - -";(-'il (3J ttrat though the velocities of trvo The acceleration of point z{ is not zero but is independent of .i,,s. Notc since diflerent material points in contact are tlre same at er.ery iustant, theL accelerations are no! equal is nornal to the tangent o.f points_ make contact at different instanrts: se * ga. Horvever. the acceleration I B it conlacl haoc lhc same planeof contac[. I{ence, for thc cose o/rolling w-itbout slip'O1 poirfi '4 4,rl! retocity'of the con[act point A- is zero componettt of accclemtiot ia tlrc tal.lgenlial plr." oS corlacL The poitrt'on thc cllitdcrshodd be but its accclcrcliott is tlol zcro orL"J rrr = 0. Hence ltc accclcnilion of a conlact- ' ' c?trcssed in tctms of lhc aconctic ccttrc, rcthct lhan lhc.poirIgl .tt p= s* gI x O D = utpi *t.,r'( *'r'1 =-'l-i'i. 17 _ .. _ <{ _ -: ._ < .I _ tj It t LJ tj x (-r!) 1c'11i+.--rri' . =s, * r=.x OL'>riiiJr'rrk j'?; i = - iwir(R'+ 2r)/il +')l i 1 ? L D I atlQD :s, + a, t1 ; i " ,e.D=**?*? =p,r-;;"r/(E+rli+pri+-f";1 (-Il ="r'.|,;:t;#;";11' * *rirti ' 'i(:'l? g. et+ eirr x AS'-vio!: sa ltc l... tj oilco=.=o'oi=eD'(-il='^'fr'(E*2r)/(E*r)' (4) + p, =4r(R+r)l(R+2r1, ; -r- iltrt' ;\/Jf. gn6-= ?i+ ' l- (il itrt azs =2wlr", sc = vll r)''+ t" = 2Jir(n+ r)/(E+ 2105) !\1fr :''l'f*'*"tli(li + ,Z! p" = lb' *rZ*'i-,* lj slip: For tlre case of impeirding ;""iot of ortact or" oot .,quoi ;' A" t* IJ ,. ffiffi::T,:il:,1il:'ii:Trx'nxed ' ":':t':'' .':;, = oi=4ulf, *;-;' u ll-- ntl !'hc-comPate cornDottcttls of 9;o ge - "t'B' afld 'hc ap an'' ie" to = t"tr amd * nar surface {rig-er.rib1, rrre' a. gr = (l,irlr gO = t':1t j, uC = @lrjt IJ ii lhe tatgell'tia'l plate oad og R oo and e<is(r)-(5) = 2. t:']'jl:: rf cllinder I '; e.p=ugi-, so- lu'!r2/(R'. rIi+nrri' r[ radius L l,-.- 3- If L at J{ 1- If cylinder turns rvitlrout slip .-:r,3 +acorDer the fixed axis at A, R= 0 and ep(1){5) o^ E1' (Fig-i'l0c]"i${pg- R' =' L. I L IL L L L L L L L g4t=ttgj, , : -R1 1'";E*(rl<.1 * ,,,, =ti1lt1r- l/Rrli' ' a fixed cyliudtr 2 (Fig'El'l0d)' then r = oo and eq(3) on slip rvithout rolls surface body 1 wiih a ftat {3}$ For poi,t':D tFig'El'l0d) q. = ,i*i.,,[p* oe tltisrcsult c6i*it *- "nir"ro"irg so =eA+ 6-tr ur1 ! x d!= c'r1di' v p = 9-e* rar X ;![ = yields + =4r,. Pq=Lfrr' lrs iss::e a'6xed.':"t!.:; rous rvirtrour ptarte as-that of body. -sllF 2 is on ttr. s.me=iae of tlie tanger* as centre of curvature of body lJ ii ;'l,l,i, .il1;:l;i;:l.t ] -a'D -uig='i1R='d)!+;rdi (Fig'El'l0e)' then ttre cyli$der is in-rotation about G--1 .<t' ao=Lr1rj, gi =g, . Po--2'' o,=u";.rrrno=fra--.'.ir1i.irrt' I;;'i; 'i'd l""t""'io" X';,* *:;: Hi.';;ift::!:i::;':': t'he *alues o[ or body 2 *ir*';1 ,u1 beirrg rsoutd expressions given.ih this exi'.rple !q *,-r.1. refere'ce f...r'," theangularvelocityandangrrlaraccelerationofbodl-lrelativet.ofratrreofbodl-2. of a lathc Tlrerollingrvitlroutslipistheprefertedrrrotiorrofvehiclesqnr,rhegls.Forflatroad.tlre.uS"gl5[1avt: A piniou' i'''ou"tttl ou rhe carriage trroiion' tralslat'ory has <hassis t,he arrd general plane n'rotion maclrine,rollsrvitlroutsliporrafixedracktolrovidetrarrslatiorrcotlrecarriagt.. velocity and angr|lar accelerat-iort shorvn in Fig'El'lla' !.nd:t]re angular [rechanisrn the For 1.11 Example istlleinstantane:us ittst'aul' of rotation of body'3 at the given of rvheel2 and the insta[taneous ceatre )]:::t franre of body 1? c *"a.. of.tl6tqtion of wheel 2 rv'r'G- refereace Vc ti : c or_r-$'-.':r, \tdt; :il-'- _'/ .o,-f: ,, + . \\drr r 7':u ;FLe {, i<- f r r- ' 5:r.ffr*;ilffi1*, 5 /\ l!(a ^. rair (bl i" a foxr,ar crrain since ir co'sists members l' 2 Lhe dth member is the fixed of { ruenrbers- ""u"aolJ;T"I'#;,"T:::*:::::"i:1,ffi -.^ri^n ::il:"x',tr',X,"1fi orbodies *;:5:;.:.1 poi',.c (Fis.E1.11b) ;:::?13:"ffi ffi$.,fil"l.llT;l"i,lT:f[:ffi;;::;:,:#;;:;";;.,,"* U. i- o,'.1 ----' 2 and 3 in two ways in oi'JtU* i;;L .hc same plaae cf, convenient points of these bodies havi'g r;i !4=9-s=9r 4=9e=9t ?:lb ds:-rt' =-121' qs="l3k' rirr=tirl' 9{s=o'sB' (I) ec=eD+sbx pc-=t uiirx2Dl+!&x_*=5*ezx99' _,,3 De = so * (sg * ta1 x E D ui E D| + * x D8 . tLc __ s.o * & x D9-,3D9= cr-n, ED = -3L' OC-3j cm' OD =4 cm. re= -4i+3j &a x o C - -3ae' p) =1Sz-321r/z to eq(l] and cir2,63 are so*'ed from trre [rto scarar equations co*espondiug fro* tfr" trvo scatar equa[ions corresponding to eq(2]' - The unknowrs !r3,re,3 are sorved :+ akx(-30+abk.x(-+i+3j)=o2kx3j (4)' --3-si-.\:'?)j=-*'2! -12-*'ts=0 (3) i" -&.,3=-3o1 u'2 : of eqs(3),(4) is solution The ''3: -3 rad/s' Eq(2}: -12Lx(-3il-4?(_3!)*.i,skx(-4i+3il-(_3)?(-4i+3i)=t^r:kx3i-(_3)2(3i} (48-3dra+36)i+(36-4t3-Zili=-lut2i-27j :+ 'j : 36- 4r,rs -ziti,=-;-2,' (6) ., ,.. ,.: \... (s,: :_3i,2 : !: L;;r6 t'(6I i'' - sr'''-'.-19'rad/sii' ti'g = 9 *'"; uc (Fig'E1-l lb)' 4 to Lo gp vD and ehc ar' dre inl'crsection of.eni 'orrrrals The instantaneous centre / of body 3 is 1:l::: - Eq(l): :+ [: Fort.tregivenconfiguration..Iisat.o.oszanrd,*;3.catr]alsobe.olrt,aitredusingl: k"sr:3' #=#=Frr W:Tffi:p:l .u2;u3... d..idid ori t'e tasis J and of link 2 a'bout :-t,=W':Y'labout:,r-rr=3; of rrre direction ofiotariouif rirrk 3 =* j to !. i.e.. about _ k. This rotates alrout r iu the directiotr [rotn link 3 t,hat inplies ofgp direction The 0. k- Thercfore lb. lleuce link 2 rotates fronr j r'o i. i'e', about yierds the direction of gs as srro*,n in Fig-Er.r Trre sig.s of and t.r3 are negative' of bodl' c ie.tre Ia1 body I is a[ D, and r]re itrstantaneous The instanta*eous centre /:r of body 3 rv.r.t. in-<tamtaneous gclr alld galr ttu u5 5llontl in Fig'81'l1c' a'he 4 rr-r-t. body l is at E. Hence directions of D of the tlorutals lo1lclr and-51t' centre I21 of body 2 w-r.t- body L is at the intirsection lr'a,rs applies for acccleratio, of a Poiu[ irt tito dilTereirt Trre above ptocedure of writiog trre 'elocity a'd ilr (1): the mechanismsshorvn in Fig-El-1ld- For mechatristrrs {7) r..r2 uB = ge+{,rzk x AB = (ye *t'trk x o"l)*o:k x AB --vse" x)E-:;&:'::*t"}lr1l1'fll ss=et*ti,zL xAB-luiLf =(sa+tirrL *ql-'?QA)t'i:k andeq(8) Equatiorr (f) is1lv;d foru2'tts rvhereg =0,qo =Qandp, =ooif pathof Bisastraiglittine' vu*"vY "t'""j,." -" f : t as lugl/8, F all tt &no Llrerr sr8lrrs 'rs the.u obtained J" *usnitudes of o3.r.6 are trt1" for it,ita.The t. .rh" rnugni;;;"or-..;. ir'toin"d contact contact' \":lt!.'Y"^lt,:.ili::::.':Hffi::#f o[ point'of point "'.' tlte the is 3 body bod-v of ot cenLre , 8,4 itlstanl,aneous centre the insranrarreous tha[ rhe knorving that ug s{, il*;; Hff'"'lfi ections of up g, ;J"?ffi;;.il;,;;;, .A^a o,s'!L;T, n/t". t'oo, 't, i ! I I I I lt I ,l; te7 r {eI AG r . \" ,2\--**3{+^ffist d ! oa1 - X$l-;ed '!#R\a ,,i*.---€, trj"- For mechanisms in (2) g,ffi^ k17,,, e- g,.*r=15;, urrt,l, ,fri : *#Y# tt.a,* aB = 9o =faa{ *trrk x AB t'\1-: --ttsg' -:*:-*''t1 t 1 { I : A o..,i€, .,,ffi:@=.?I] F'.e.rld gv ffi.r-'e1 (2r + (ul/p,)g,',J+ 61\x AB : -'ie} - uaQ +{u'alp,l*' a -(9) (10) i \ i \ l. '"1 ,^-,1 + ' .:. i-" Equation (9) line. and P^ = @ if gatlt ot -,4 is a straight, = @ ii.p+Aof-{ is a straight line solved,for<.,1,us*d"oit;;;-igi,*:;n; irl=iatR"t;';l=psllRartd"!:eirsisnsaseobtainedfrom point of coniact' where pe - is , i";iliins Jt i;*ot*** centre of bodv 2 is the ariis radius R, rvhich rotates about a fuied A planetary gear train consists of a sun gear S of r?r' radius of each P' *"t"."tion a, (fig'BilZ'1' Three planetarv gears r, ;;;a;i* r, *,i, e., u";Lg thau the Example 1.12 ar angular velocity velocity' g.u Rorritt ir."iau' T[e ri.g Sear rotates 'vith anguiar mesh wiih the sun gear as rvell as the ring called the spider' (a) planets are nrounted on a fraure F ut2 atdangular acceleration dr2. The beari,gs of the the acceleration the planets and the spider' (b) Find Find the angular velocity anil angular acceleration of systetlls transurissiou irl used gear' Planetary gears are of points A, Q ona planet and point, B on the ring one or n}\::eats' jssion has to be altered. 6uev[rv'E"YE by D accornplished -' stopping rlrrs is an[ereq' This rchere power ttlgglti:rgq-has o+. ,zj fr :-ffi,'Gr\)w l'.--'>rR 5.-R 611 4Rr t \ ijI l!ri \,3$.*_-1 ,.\ .1\% tz8, t V' -_=q{g sorrrrio' i;H?;i"raing o' r.r r ot:" .",t (b) coudition implies tixerl axes i,. ris."-Er-rzb.c. t1e ,o slip Rt : --oZ R:. t^r3.R3 (1) = urlli.{ : ttedonlltesatnerotatittg6ody,theneq(l)appliesrritlrtbeangularr.elocit.icsbcirrg relative to that bodYof sun l' plauets be rd3 aud ara' The axes of rotatio0s Lct the angular velocities of the spider a$d the to relarive 1'2'4 3' The angular velocities of bodies ring gear 2 and the planets 4 are nroutrted olr spider at "l and D f ield: body 3 are &rt - o3,'.)7- t"3. g.l{ - t'r3' The uo slip cooditions (2) ' (<.rs -tr3)R1 = -(*rl -ul)Rl (.2 --rX& + 2.R3) = (ara -r'r3).fiL3'. (:l ) r^ra = [r2(It1 +21?a1'tqRil2Rz u3=[orBr +ura(I?r +2I?3[ lZ(fu* 8zl' + { + ci3=[ti1R1 *ri'z(8r +24?)] l2(fu * *a = li,|la1+ 282) Rzl' nrotion, u, Since all pointsof interes. have the same,plane of Q.A onbody 4 are obtained in ternrs of poinc 6 of {: 9a k)=scjr^rrLx 9Q-=,iQ, l oiiU CQ = -Rzi, @=:'83i;;$ince x (gx r,) - ntBi | 2Rz of points = -,.'3r' Tlre accelerations = 9ciroLx C.:.{ -'iQA, I e^= -ri2(r?1 + zRlrL-tW l--.) I iu lr -- ff J,J i) r-rl- € i - ori body 2 rvith @-={fu+ so=9' 2n?)i-- Hence buc t5e components of q, es'aJ9ng the tangent direction be, to sacisfy the conditi'on of r'olting rvithout' slip' L""t" ""-.,.;fU;i"hould i: (6) * lqdtu +2k) - "r:.RJ? I i' r'r -:,..., sB=qr*orgE*U--"id=-tfi(frr+2fu1i-rilfaizR:)i', th"t;ir,*-g,,. tL 'are (') g. and G'are poi,*t'sof:bod'v'3"9c is otitairl$J+;,' wit'h 'W-=lL:aErI!' k=go+u3k xoC-"'ioc, + Eg)li- t'{82 +..'3(EI + n?Ui' Eq{3) t tO) '+ so=l"fiRz-tr(& Tbc pcrintso and B (.1 ) (8) i at the point of contacr are . .':'(- particvlar ca-.e..:'ThF&illo*iqg results follorv from dre general relations (3). 1. Ifarl =e, i.e.,thes.unisetationar5r,then ur3 -ttz(Rt*2P'z\/2(fu*Ri, c'ta'=u2(P'1*2Rzll2Rz' 2.. lf or2 = A, i.e.; the ring is stal,ionary, then &rs = ur& | 2(h * Rz), ul = -o)1R1 / 2Rz' 3. r43 = Q, i.e., the spider is slationaty if u:(Er * 2R2) = -utBr' 4. .,{ : 0, i.e-, the planets do not rotate but have circular lntslation it uz(fu *2Rz) = r'r181- jr pxa'nple 1.13 Qrick rc,hrrr. mcclrcnisn is conr- monly used in meial cutting'to speed up the idling part of the return stroke s'hen no cutting is done' Sudr a mechanism is shorvn in Fig'81'13a- The pin P is fixed to the wheel 1. At the instant rvlren f = cc-l(0:8), find (a) tlre angular relocity arld atrgular acceleration of link -48, (b) the velocity arld acceleration of ram D. Solution The cenire P cf the pin is comnron to the slider and the-rvheel. The extretue posit'ions o[ i'ire link 2 are shorvn in Fig-81-13b. Tlre forrtard stloke of the tool occurs during rotation of s'heel I througlr a larger angle d1 cornpared to angle 61 trarerscd b-r' it during the return siroke of the tool- Ileuce n 3[qrrer cutting stroke and a rapid returu slroke is oblained' cosg:0-8, sin0 = 0.6, --F -l I '-----t L ' :)- Z ,"-'1." fi' .'0Az AO (dr', d- : tb) cuttingtool '// fi: (OJ f;g..zt A ,j .I ! AO = ti inr 3 ?-: 1CC.* ED-- tct+ 1L"q11fnosrls "r" q*xAP+t,p,r, :+ 2L.x(8i+6i) =,r:lix(8i+2aj)+o(0.3162i+O-9{S7i) tp:b*u1xOP=ee* j: . l6=8ta?+0'948?u Ql :+ !: tl) -12=-24u2*0-3162u u = crn/s- r.t = 0.650 rad/s=9..t * gz* AE---iAE-+2gz x grt:* 11.38 Lep + 9.p=go+ v;xOP --iOp j) 24i) + 2(0.65 !) x 11.38(0.3163 i+0.9487i) + n(0-3162 i+ 0'e{87i) (s -2?(8 r+ 6i) = r.4 ! x l+24 - 0.65'(s i+ ({} -:+ i: -24=&it-10.14+4-6?9+0-9{87t' -32=-2*)z-3.38-r4.O4+0.3162' (3) j: i: -\2-20 ctn/s3, t[ = 0-3150 rad/s?' The solution of eqs(3), (4) is: (l) of rype Oace a4, &3 are known,'gr, qD are delernriued by the procedure explairred for the mechauisms in Exl.Ll, since t,he rnechauism ABD is of this type. It is left as att exercise for the'student- Example 1.14 A vane pump, shown in Fig.Dl.14, is rotaiing at augular velocity ar and angular acceleration 6 about a fixed vertical axis. A material watei flows ou0 along a vane with speed o[ u arrd rate of increase of relatire to the vane. Find the velocity and acceleration of P relative Neglect the thickness of the \ranes- The problerns of fluid (lorv over blades of turbines and centrifugal pumps are sirnilar. . 7t rT rJ ttre u)2,o ateobtained by rvritingg.p ia ts'o rval's and equatiag correspondinS !,iconrponenG' Siroilarl-r-' j All conrpone{rts,nfro;* &2,i arcobtaired.by .rriting gp iu t5,o ways ind equating correpouding !, pirrr" nrotion. 1{e consider poiuts in the sanre plane of tuotion + g x (- x f} = -n2!' bodies t The solutiou oteqs(1), (2) is: * AR-- 3(.c- -L oP =}0(cos0!4'sindi) :8i*6j cnr, .4P = Ao*oP= l8j+oP=8i+24j cm e= AP llApl=(8i+24i)/(8? + 24')'t" = 0.3162i+0'948?j' ql=2L, lilr=g, 91=-u4k, .rz=t12k, go:9,r=9, 9o=9t=9.' Ttrepa.thof Prelativetobody2isthestraight lirre.4B. I{ence 9p11 =ue, Lrp=it9' rr) point P of speed of u ,I :-,.J J ,;-J --J ,,1 .::J -:J -:J --J .-...l .:J to ground. the curved F.'r. e i- lt J ;-J ':j J F ! Solution With th'e*&rice oferes 'g-riarli' as shorvn iu Fig'E1'14' we have -l Lt=6t, 9, = sia{!* "gud!' so se =0, =9-, . go t cosdi-sindj, QL= r(1 -cos{)i4(6+rsinf)j' radiusr. Iletrce !p11 =o9' Thepathof p.relativetothevane'1is acitcleof 9e1t = i;'+(a2 lt)g'' gr=9o+qhxOP*Y-e,:., ap:9o*r.rr x oP + qr x (grr x QA f 2t'r1 x 9r1r *9r1r :+ vp: f*u(b+ rsiud) * usindli+ ["1r - cos{) * u cos{Ji cosd) - 2t'ta cos $+ usin'd + Q? l')cos 9l! ap = [<ir(t, + rsine) :c]ttisin { + ir cos d - @' /') sin eJ] + [tir(l - :.= d) - '"(bt r siu d) + 2r"o : - -" - t__ t_ L- L L L-. ti LJ {i E.-, Lj U L t-. L L e i G -) .J. , .l I J -zL ilf ,,ffi ',.ri tr^.-^h^F.-...=;AxIoMSANDFoRcESYSTEMS.,.,,,.1W 2r or vecror ..*", q.I :",,.1r, 1: no'o,, **iln,3iw1 J;;ffi)ffi l?"TI; a,'d *,,**" .o.&ce or line-' The distributed over voluc is called fota' f I ):.tit rtc,forces f::'il,:,T**:Y5""ffifii**,?S'::,*".Tt[i1t;i.tr l*iigffi* -g;', tr force over vorumer ;"L L L L. L. "i,i^i*'i^;;;.#;;;;;;; Mt=1'ptx L L L raw (axiom) of fonts- ( ?" -:*y:\poro,rctosram io"tpoi*^J:'"rl' A,forces arc sorctttei! tr F :JI*':o k j- = AP-xF:(.ttP)Fsin0n ;i = Fdn= lr" rv ) r: l' lr" iu F,l , ! wtereF=F'i*&i+r.'L,LPt:i'i+rvi+r'k*dqistheunitvectornormaltotheplaneofEand''1' G tf,e relati*e ottained from = AQ x Lwhere:{Q givea by the cross-ptoduct. Note that !{1 ";; f'e eq x f' = oq 4,'sinF/Q x L= {aF"+ position vector of aay c<invenient point Q "" ,i. ft"" "t Y"" r fr:i":;,,{;..n:y:;,*,o..,,;,et!lite"**unitvectore..lo1git(Fig.2.2)f is defiired us the "o*porrJ-t L L L orit= nrornent /r4e about cnypoint Aon L alongg ,! : I 1". O: f A{"=tr'lo's=APxF-e=lF' et! e"ll' l"r L. L. r- = er!+ erj B momenl Q of the couple' about every point is the same and is called the * L= itr*" r7^'=' Af x F + AQx t-E) 1ap /' of (Q.p)F"in o = Q' anaTQP)Fsia0 n is independent " rvhose directiol is sat \4rrench is a system of force F and a couple C : L U I I 9 F F;e e.2) i i * erk' M. is s'ell-defined since for any other point BAx F'e* rlf" - M., (8alld' el+ lzl " r'e.l Its moment Conple is a se[ of fiorces f aad -f (fig'2'3)' rvlrere e il ii onL,MB:9= BPxf'9= -9 1 -6 -F {4* E: 9t | E3 a ac A F'g.2'3 2.2 MASS -49 Massm(B)ofabodyBistaliena.xiomaticallyasapositiverealscalar,whichremairrsirrtariantrvitlr the contin$X5tSry'$t*t{ the-mass By its parts &' rinre and equals thesumof the masses nr(P;) of all 'ts I kg/rn agp"-"|h.g oiffiLttiliution of mass kg/rn2 or d ersityat a point exisis and talren ,. p kg/*t, o nrass c of. a body (Fig'2'4) is defined by over a volume, an area or a line- The cenlre of L L L L f r- l* Lrd*=frb, + f.=ffii, l=Y#'* dtn W' or ffil of tz'er mass drn equals pdt' odA' lds for distribution rvhere P is a typical point in the nrass element drn and and curve arc deEhed by . over volume, surface and line, rcspectively. The ccatroids C' of tiolt"t'"' sudace " s respectively. c Jira" [*dA - _TT, !C: [*d" lc. = -TT. c- coincides with (2-4) {d' orsevet.at patts with for uniform bodies-'For composite bodies consisting the ith part, having mass mi aud centre of mass G; (2'3) Vields r,c = [t- tiY+o o5 being-negative for.a cuiout rur]l [t*1, i.e-, tu!4 = I*'tr' (2.5) llar[- Aro=fai I !iIE {c't" * g rig-zl.tr rt62\J i?. -**t$i t, t+{i'i?:. ' :{,?:,*:1r"* ,,a *O,T* ipt-ol lwncrrtzrm ir= l^*1Fdm, ' \p=mlbltr': since 4, =GT*1r,= Errler's Axioms: ([1" There exists "'1.' flap ,:,'L^ir= !*u".1", about point ri w-ri. frame f (Fig.25) are defined by I^*^-gpaly'*, = mrcrr. a frarrrc f such thaf for dvdrf, velr (2.7) m .A trg,r'5 any sgskm (2-8t.l) (2.8c) 4r=L' (2.6) whete O is a point fixed in .I and F is the sum of all Lb.e cxterzol forces from the surroundings on the s1'slenr arrd M-o is the sum of the moments about O of at cxlcr:r.al lcads frorn lhe suroundings on the system- The frame f in which Euler's a:cioms are valid is called an dnertiol ftume. It can be proved that a franre ? which translites with unifonn velocity relative to an inerLial frame / is itself an inertial frame (Galilean principle of relativi[y). The 6 scalar equations correspondirrg to eqs(2.8) are not suficiert lor .determining tlle orotion of a gcaetzJ syslen. Bolvever, these 6 scatar equations are ,tcccssary ard, suficical for tomplete determiuation of the motion of a rigid Dod3tsince it has precisely 6 degrees of freedomequatiou of nrotiou of centre of urass C of any systenr is obtained using (2.7), in (2'8a): , ]he F F, : tnac, = rni6, p. = mo6.- m(ic i.e-, - E*rq.rt, , : mic, tnag,Fu'- (2-g) mgs11: - ""OL), F4 -'ma6o = m(2icic + rcdc), fi,=niag.=m5c.,$-lIl46o:#Llp",}E*=,noc0=0.(2.l0) sYsTEMs F, - ma6, - '?2!cl F. = ma6, = rrri6:, Fis.2.6 2-4 EQITTVALENT FoRcE lfwo force systems are said to be cquiualent if they have sanre total force s-uin E and sanre total nroment. sum M^ about onc point..A. A sin'rpler force system equivalent to a given forcesyst'enr is callcd its resullatl 1- f,{omeni sum .itfa of trvo equivalent force systems about cay noint B is the same. Pt@f. Coasider a force systern consisting of n discretc forces sittr the ith fotce I actiug at point i. anrd m . couples with momenLsfli,i = 1,..-,m (fig.2-6). : W:f i ' si* grlD e, *4 B-: -fu.N.qj +I Qi = U*Lf-+(f x&+t e, ) = B Ax L * t{-a i Heace the moment sum i4, for eguivalent, force systems is same siuce F and M lor them are cg-ual^ slstenr can be replaced I{ence. for the.purpose of finding the mornent sum and fiorce sunr. * given fotce by asimpler equivalent forceiystem, i-e., its resultant. 2- Two-equivalenttforce systenrs cause the same motion of a sizgle n'gid 6odysince it is completely decernrincd by the total force sum { aud the total momen[ sum Mo, rvhich are'ttre sarne lor the lrvo s.YsterrsTwo equivalent,force systems; in geueral, cause diferent motiorr of a defotinable's1'steur, 4- Rt*Itaat (cguiralcn!) o! o gioen forc,e syslcm ol a giact poinl A consists of a force {4 and an associatgd couple with moment ep^ @ig.2-?), rvhich are obtained from the tirc conditions of equivalence: i .i = t(EA+,q;) i . &=De i i (2.rtc)':-l-i, . Q.a^= i .1r Ma=!e;x4+Eet (2-u0) I 5- frc - Ptwf;, simplcs{ rcsultort {simptest, caaipgkn!).of a given force system is a sirenctr. Let the.resultant at'pbint ,,t be Fp, en., The *iuttant at, point B i" fa, en, with (c) MB=C-nB;en" +BAxER=(qR^h*(Cn^)r *BAxfu., , Flcte ( )U, ( ).r- are the"o*p,io**tf 1 1rtiifei and normat to fs. tt b possibte to choose an appropriate point A suCb.t-ha,t the last trvo tcrrns r.hs. of eq(a) cancel each other since !! x fp is .L FB- For '-,.: .-E.=J .\ ol = ,, .--^ '- *- l.rh '^ --t .::?: C.Bire rt F.,,kllt -i ,al1-"r\.^q --t-^ .q' :R-> C; 'CotsiJ .i'. B !" '-d ,Fig:1.7' 'a* .J {ij J ,rJ ,-J .-l ;J ,"'J ..J *.J cJ G' I GJ cI GJ GJ cJ *J L'-I O'.-| J -J '-J -l-I i-t ,--,J -J J J - L':- rri. t- l L. IJ f LJ u TJ tJ L L l_ IJ r- ' | v' Llrt (G.)ll, isisting of \sA rrr ls B vrreusu rLurtD'er'rb resultiurE k b B 9,t9 iesultaot . , . En, "-T:&'::l-..1,--t 2-., ingte fotrc/ g sirglc couplc/ a",ll syslem. j.o -- -'{QR^'-ER:u-r8c'r"'srrrr'r'sc' .Fa= rr(Cn^)U +oij.i, ;;A^il 0,C8^=9ca^ = ernifir .Fa =.01i:1., = e, a nul syst€m ;'."noti"yrt ::::,:'::ii:T:':'6: g, a single coupte en^ if & = g' c. g^ ir !'r b' 1:l_1I {.*^'* 4-R 4 *t "' - "*tl---t-t ;, :A;r: A ,-].--r- f^.aa throu(h the point of concurtence' ,1L;;"#il;;';,6;I_:.::::':::'f:"-*1T*T"1,81ii, il::-ff :fL:ff ;::$;:;:"::.:: ffi"Y,:fl',-,i:*':::?:::;:ffi 3T:$i',"#.il:H"* k iompoaent aod & hi 3'' YRo auu' :51;;;laR. aLe=.0 "J* e*" Plan'e tr'e'r normal tD thls ],ff system' d""E momeots along -"",3ITIJ:L',Ji:1; its simplest only and components' Hence L l,, L L L. null ln cel a.sinrle couple/ nutl Arnl.single '"rv z-axis) and couples. rn e line (say s*' = 0 since epo has onrv Q n " i ::.tt3:-:j::::1, :T:ffili,;"il*11,:;:::1,I"IT"il;il;;;,rl',r:':"9'::il,lll:*:l'; *' i' singre l """Jlt"nt Ii":ti: -:;ffi ;ffi;il':il*i ^t-]., torcet a single force/ -"*::['fj.::,' ffl"J 'oir"ni i:: iTil']']ilil;xH;;;;:;;;a' ffi";:"rr'*[ couple/ null ! has single sinsre is a sYstem' "''J;rT"T.:"f:::i:;n The resurtant :+ {p cons'stins o'rv or rorces ofsuch a force s],strm g' at ri' i = " "' n such that & # "::** Tu-a poini 1;; calted lhe ccttrc of parcllcl forces' 1 irr*ogt ,*"i**"* E^=r^l{I&}e' a i t" = (Isr') ir /(trt)' * t(trr}rn-(E. 'v g- lxe=g' .:,:. ve' (2.121 ' Fordistributedforcethesummati.onin(2.i2)isrePlacedwitlrinrcgrat,i,on.ThecentreofparallelzaiJonn rtith thc centre of mass g graeitatiotto,f:* dtn o'la body, lc = LR=(T**) of ..n"a "r*io'os g*n;tg,c, curve parallel force norural to a plane The simples ! rcstllta*lor " aoa.iuo,.a **'ii*r-ou-iir= ""nt*;arf thc toodinss*focelif area t' 0) aud !:' e-quals .F For example, d,he a triang't* simplesc.resul[ants of a uniform and --4ft$# tltt-&"brtic atca J 1l ff;}t:{1" Ol5|{t{?ffi laen=an' _= "ffj* taT : il;; 6;''{ t"=:-T-:-F .* "oir..iii l(te) = f;rar t looo = l *ta*t l dF=fds-dA, E*= 1 IJ IJ l- '''-''-"tlg point uch a p"int such 1 [-ft ;a* '-""^L:-t::::t:k:t; --! thc arsebraic'ootume rorce normar co a prane areaequars ?;: *;;X,':*::ff:'ffi,i"*i'o**er toffirF:f--{t and ack thrc.,gh thc cctttrvid'ltu'" .,*ti"X' of tie pttssate spoce(if volume * 0) d.F=pd*,=dv; El: lv re {- fG.o + !:,r= lrav'=t't'. tL;+Wf' o'spsD a'F' = I lzdn + ( E =,& =.rn *( )!- glone. creo NEWTON'S THIRD LA\jl/ OF MOTION t t '''XJ:,::;;#;;;;;t""i''qarts)travee::1"lTi'1"^:::1#:*':*:::'; ;'al momel.t rn !{6'" bout o' ;::""H;;; *,,.,r{r, ::'ffil$,ffi :,ff il-'t then *. fr*;;i:fiTfi",fJ aharrt o O (Fiq2-11)' Fis,'..}''ihen ffi;n. --.-- t., - +ou! f: ::;:':L"j f ?, ;::'J.:i:J;; a su and ( 2 l+ Mtp=-Miz1 o'1. Mir|= -Mozt, rnomeat sum tfb, about of ;;1 and 82 be aforce ium [!'-ind * surrounorngs urr. to u due pnof,. t*L exterual load (24) to Ar axionu ('z4'\ r .-^ ,t_ annlv Euler's axio'rs ",rr.o,rnutnr. ^ ^_ ,,- G" ;Yff:"Tr:Il3.:::":"';XIH::.T:H'l4;;&;;il I .9 to BtU Bz:' alone, to Be alone and 3,,!- f,s i t furers Y c*i t I ! I a --/- \e-f;, $, L -.-> !-zl H^ =l <l; = MoLz+ &* i{or='Mcr1* W4 Lrtr:&2+Ei' Lsr:&r*&' P^r,:.Qt-r.{-12l -ilt,u",rr, = Lrv*l-tt= Ei+4' fup,ua,1lt = iI (c)' t; Ftg. z-t[ f\ The paror futdraora rorce = Y{'or+aitGl ._< "+I{-o2 l^!ze1sdm. point s'ith relocitv 4t) i" 1;' C- =F1u1 =F13. doue bv a rorce 'lrprr " T':'lT:::' _ J,, A Frg,2-lZ (2.13) ii; :.;: Jr, ,__ ,, -rr (2-14e) (2-lac) (2.r4d) - ;-1\r .._ :1q v . < {2-r5} \< {_ Ife(t)actsondiffererrtmaterialpointsatdillerentirrstants,thenr,(,)d,t'dlrvheredlistlrcdisplace.nrett g$)dt = d1 rvhere d1 is the ma/'c;,al poh;- fot ail iaoe t' lhcrof a oaterial poiut. I/ E(r) acts on the some trajectory Cand (2'1a) and (2'f5) yield displacement of itris material point rvith rr(rz) trt = | L(t)'dr Jc'4tr1 Fotd6 * *( F'dz)=/"1-.,"* Aq (2'166) C (Frg2'12) as: expressed in ternrs of centre of nrass conveniently W Entities 2, La, T,t !i -"*rbe -r ' If^=fu+rc^xm! t'ca (2'17) p-nlb iY = F 'tb = E* = JM[^r,^ -"/ ' ,: * l^nl +tr'4 (2'19) 1*41 \ [,,tr" d* Tlreseareproved ''" {^u'cdrn =ll -"--; J- :'pcdm= t(cp - gsld*.=bnt- g6rn =Q' =) x *^, *lt**'+ !rc^) a', = !;'e x u'6 dnt * tce f^o"" - ;r:;: =' i,:*.- + d* = f, 1,,* * =iL'u = t&' i-fi'' < (2-160) l.;"t"t (F'dx+ Fv d! * F' d;) = Jct's'z(t'l [:" :u-:.t(r"a' + - Jctt.v;'1r.; i f, )- J*rr.dmxgao*lcn ' ,, (*c+ o',. _ r-l (2-Laal feY + F'; = F,o, * &r, * Fztr = F.i + . = F.ts, * F4o5* F'u'= F;i* Fori+ F'i ,re i:.\ defined by fjffii;1;;**tu ! _{ (6) The result is obtained by forming (a) + (6) IN TER,MS OF CENTRE OE MASS 2.6 EXPRESSIONS OF SOME ENTITIES ? to frame .F (Fig2'12) is defined by T-he l'inctic cncrgyof a body relative rw = L ,1.:-- .-r .\%l r;_ t'< . d'n J^ 1 T' "" u\ =tlel' ea *Ie'n" = --I rq (2'rs) (2'20) *%^ { d^=E-c*uaxfn!car;. -.-'t +^+ii :, I"{s-i,"+pc),' (grc ., F.'!b+T&'uc AILBITRARY POINT E 2.7 r}uLr;R,s sEcOND AXIOM (iL^ _MA RELATION) IIOIL *bo x nlboy ='EcV +Ic x n!bl['* Using P-f8), L4r = EcV +bt x mtlct1t, IIoV = Eay f-44t: lbV'l,c x mtb1 * be x tnlzt1' x tttlb f-at : ibp - ?blt x rnlclr - tc x nlbv * lb 4r x m!4 Alt * b t ^lr x m9elr = llp' ya x bb lx'qr.lr M o - rs x E-*b t x m(!b,l r - o ali = ![e - y; x L * u $ L-ul -{ I I rr << < _v -q -r _ -q l .-_ Litr! a6 'J< _it -t ll I' Ii ) :i i; Ii il: i :i L-- ,dl,ro"i1no, =, bi- frcae, W,, mtby = + b- ""Eilt.= Mt Er x >1 rii rnQag. [ = Mi Gig2:13): -i r i provided Point Asatisfiesatl.eastoneofthefollowingthreeconditions: : i-e., .A has zero acceletation in [, : 1- 4,t1r Q, 2- c-.tgftrcr' ie-, acceleration of .'[ is along AC. -r.e 3- rc^:\ , -. \ e\rt j Lr\sS\s\-'-rs - Fig.z.\3 A->=*i;.^r+(2-23) .r Ip,-feA' - il"t, = W' 1 ! , IfaPu{sclofforceEanditscagzlcrimpulsc4.a^aboutpointAfiortimeinterval\tat2aredefinedby f" lrz - {-.*^(rrltrr= lr,,'*i*'-ro' r, _L, ,:L .l{, aclilat if 11 ,:''n"' poirit ' tcou angtila'r;*p'it"e toe'(") about iaslcala-r tnd irnTulsc l(tr) ,;.,. Iasta.rrtaacoqs .' t'z (2-2,t0) g' \--- - ' fz dt * L*s^lti = ,l'i,', J,.M,(tl 'r ; ,,' r(rr) = ,lI}, /: a(t) dt # q, L{trttr.).= Jr,E(r)dr, first axiom implies Alt entities are w.r't' inertial frame- Euler's r(tr. t 7= f"'Eat= l,:,,'r-*= $i,"'o,' {2.25) (-a niiagci ' {t yt2} = 1(t2} - g{rn) = 4Y= mL96 = L ' i-e- r-mpnlic of exter:a al fo;rccf cpals cha*gc it mbmc'rrtun' For instanlcneoas m;As'- ' I l(tr) : ag: e(ti) - d{} = m[Yaj I impulse (2.26) in Position' P A?, 6rlb,Ag6. are. iastantaneous changes lvithout momer.htm: o! If f(lr.t2) = q]i"o (2'?5) impties io*niotiou g'hang nihere 4(lz) =g(lr), ga:(r:) **;;; ef i*;;; If E(t) =g, then' i=L If a paiticular gi =P(0)' + :+ =k(tr)' zero, then ,[r" oc=g' * zt t1, (2-25) ' f,'uuc'(r') = t':*:(r')' "o,topondinS comPonet'.::*""""tum (2'27) is conserved' vt f,rnreci = Q + !runc;(t) = f,:nia'(0)' =s'(0)' '.t't i) = g, L**r0) i,r'' (2'28o) (2'286) tz:sol :g,tt)=r(o} I*'r-'(tl:tj:'10i'' ',-l'-.llu' *uzeroforall]t;then(2.28)'{2.29}holdonlyfortft*i.empoaerrt. j RsLATroNs TMPULSE-MOMENT O: MO.YEN:',* or e{(t} b point such t'hat ee(t} = $' or r{ = C' All cotities.are s't'i':furertial frame' If {-r1 a ' . 2.LANGULAR. atong (2.30) .i ail-' !4 I 3z j -..-l --: of l'nr^(t1,t2) alon( a f,xed :-. a point / nxed in f, if a component Since the results , is zero, then the oiirponent 9I.4^ in thit direction is conserved. Similartn if a -omponent . Los. along a fixed direction of f is zero,itheu the componenL of Hc in that-direction is coascrrcd- Note direction in of ':i.an{3 1".-"Ua for' thatforamass-point rn,'Ho, =(re.+rc,)xm(ie" +r&i+ig,).9.=qll.2dlf an instantan@us.ngui"ri-pirl"" Jo * 11, then wiihout change in position there is an change in moment of momentum about centre of mass C and about a point O fixed in f: (2'31) +-".:i'*1i;.1?;*...-. C slavning .. aoo36(., a-^. \MORK-ENERGY RELATTON rOR CENTRE OF MASS *;i{"';.o?'*c 2.1O Iiq.(2.9) =+ F . lctt -- ,te,,tr i.e., ic .t!c11 Q32) rslrere W' = F.yc= raLe of s,ork done by the forces as i/acting at C [1 fiz in general] (2-33o) at C fl T in generalJ (2-336) \= i*u211= kinetic energy as if all mass is concentrated ?6 in configurations I and 2 of the s-vstem and Wi-z b the value of srork done from configuration I to 2 by the e-xternal forces as if tliey were actitrg at C. 2.11 AXIOM OF COUL.OMB. ERICTiON (dA 2,r At a point P1 on.{he,surface"of bod;r'l:rvhich is.in'contact rvitlr point' P1 of surface force intensity can be resoh'ed into a norrnal force conrponent pN/m2 (prcsszre) and a tangenL\al (frictioaal) cornponeut rN/m2 (Fig-z-la). According to Coulomb's a:iiom of friction: Tc,,Tc, are the values.of dqt-e<rc&'d fa bif..ekr'm o^a *-b.at*lo { E o< <{,8?1r.'rs) <$ ' <'tr'r' (2'34) T = {tP,onj i-5 d.ireattd, qp,oosits to aqpoo€ok * 9.'.9. (t3) r-q 1i\Q- +o'^shtr-\ p?c-re. r Ar = PtP, ood iS dtrectc.d .pi.Eih. to Vi.p, for no slip betrveen Pr znd Pz: r 1ltrp, for impending slip betrveen'P1 and P2 : for slip betrveen P1 and P2: the s[atic and kinematic coefticients of friction. These are independent of ?, gl,, p. and the area of contact. The resultanrt of the contact force s1'stem is, in genera.l. a u'reuch- [f the contact surface of area ,{ is plane and the direction of r is the same (e-g-, when I translates s'.r-t. 2). thea (234) implies that the total uormal force iI = f^pdA,and total frictional fiorce F - l^rdA,are related by n'lrere *" I .CJ : GJ l t-J ,l- I J i:t.-I yctitr = !li*'Lt,it,, {e**, lUi-2, Tc, =+ -Tc, = = W' = - ' iastaataaeous e-l Ft. lrt are called '- I ( -thl J =-J *l ,-J ::J ,_.:J il LJ fil "J &l t'--,1 (2-35) F = ptNand for slip F = y,'N, :ql Eqs(2-35) are also valid rvhen there is a discrete point contact betrveen the bodies (fig.2.15) with .lV = Nr9,+N3er,/V=(/Vi +N?)tt2 forasmallb".donawireoraslideriuaslot,andF-Frg, *Fzg,,'f =op-,i!P' I (Fr= * Fi|rtz for a srnall slider on a surface. \€\ for no slip F < p,N, slip for impending .?t fa- oF ELECTR.MAGN;r," - l.rrAxroM "oo.h\+ g iu cledric Force F on a poinr charge s moving n"ld A?JJ7 a nragnetic fietd B,i,s : ac an . F -- qE +qq x a. 2.13 AxroM oF GRAvTIATToNAL FoRcE The total gtavitational force {p, on body 2 of mass m2 due to body Ezr = 'WJ''ffin, ,.*3 -T ._X t^' f6 / \S. asr.,s 9b \ 1 of mass m1 (2-i6) 3'l ^ 'N9,. i (Fig.2.l6) is giveu by i:' I I -J I I .,1 r '-_J (237) - l^,,[l^,c{ar,;}a,,r, -rs universal gravitaiional constant. The mass. appearing in (2.37) is calted the gruuitaligacl mcss wheteas the mass appearing in the Euler's a)iioms (2.8) is called itcrtial rnass. Experimentally, no {ifference has been observed in the measure of these trvo masses. The total force S1 is called the ucight of lody 2. Tlre simplcst ttsdtont of the general spatially distributed gravitational forcc.dErz is, in generat, a rrit:clr'cltFollowing results can'be proved from (2.37). wlrere G ,i.9 I ' dE' -- Fi9.2.t6 a8 -_J l=_J ,-] \'+-J E--1 ,\. : =l l; ,--:-t -= > ] -I L? _ 1. For H,' c "odiarsrr*iir;agi;r,.,** 2- mass-pointo:,"u""'- IU' ;"ffi-qt ?Y:';,;ror L- :'u-I-'"" :-' $'h( L.- -Garrn$ rZ =-ctt i -,,f its L-- L l^1*o*' from the centre of *T In general, the location of G differs L-- d-rstance from the sphere, theu r sp:r*r+"I-I*.;t;1) approxiinated at C' 4. For troo bodies of arbitrory shapcs thosc sizcs :^X:l-^6i' P, (2'3s) U*e si* 9 - ]l + betweeo tbeir ln can be coocentrated so that as the uuique point where ihe T:he a:rrGi of-gntityG is defned mass tn: if* *.a"f [o... on it equals that on the actual distributcd L. Ll+' s'a'ita"T"l.r":" o":*^":::-T';:'t'A1J' o: '..1 r''*"'ib) is a singre rorce F throush 2 -''--'-a? the concu""nt L:,:;:::;;;;;';;;';;;;';- LE a of radius r' is the mass of tbe'ii* of spheti inside a sphere of rnasses M and m rfith distance Ttre force F between tuo radidllyt-*"O;" "rlcrcs where h-\. of mass M and r4-^. r $tg:'l7zu'q;]i of thc bodgis small compared to ;" rcittl,.llff c'ca ue ^*::K;;;";;; t::':': =tc'i'e''tL" , :i;;;:.:;;;i'o1"'*F21=-G*a""-[1li1"*" i;,.l"i#::::',!#:Wf&iff ;l a] \l ; .ffiY = :T;;" :-" or b.dv.is (,.,1( :,neat : -:the^::!:;, :-5,i::: :,T,,I;il;:,':T.J surface of the eartb' Ri n'-t < distancs 1" through small For bodies of small Iir. *oriog I is the local b".o*siorritrr t"r"oti" ""a direction: f --'ngg'1hetesf Rrad 0-1" tlre gravitational force is modelled to 30 km' a = = 6400 km, even for s = l0 km' 1 = vertically dorvn direction. .Noie that as-8 = 039r,e' c M ml R2 = (1 * h| R1.. "'o and F - GM nl (R+ ;t' : (l + n/a)-3 \ri 2- -h- 2.14 E'[{EE BODY -,1 I L-.i -) ir.- -,1 ) L_. t L t, I L-- LJ LI --, t- of a seL of n bodies or of Euler's axioms to a system consisting For the purpose of propir applicatioa part of a body' i't is necessary that particles, a body, a finite or an infinitsimat jsolotiorr t:-t::ttundings' and its sketch U" i*1n in T"t" rhe systern should be rocll-identiJtcd and 1- should be drarvn on ttexerted by the surroundinSs on -the sysiem bv one part of the BoD1. ;;Gra;M (FBD). The forces exerted Such a diagram is caped a FREE in the FBD.ltJ;o kntal forces and these should not be shorvn ".. 2.L4-L Comluon Supports and Eteactions to rvp€ or consrraint provided bv it by a supporr on.a body depends on rrre |.],i]lol;;.I;a constrained by ::j.ITtrj:|i:il:#i;fi *i;:;i:l':::i::::::H:::::,;1,ff :#*":: ff::::ffi:T.::X.:;;;;;;"r* ^ -r i-^l- rampnt .omDolrents ^ rotatiou comporenrs consrrai.ed':b{dt''R'eactions r .,,"J1i,'f f::f] for several types of supports are giveo io Fig-2'19' "' i:ii'"1;;:H:::r":'.:ff;;ffi;;. U. L thesection can-T:::':::.i I ,t I i I f bv its resurtarit rvith opm€nt 'le rf-ll i:::::*;1f.:"i:::ffi;: ",.rlT"fr.lll?iIi'j"i*";""1,J;';;.*q"L"r'*'t*" a:iis: ilT.E:?::i;:X:fiffi;,X;.:qil;;;r:::'::::H:ilf orirs cen*oidal H:red'io'!he *j,'i;;d,":,,X;";.J;";;;;#.-r-*,troruanded .[n= u.it triad Qa=CtU.+C^gn (2-a0o) a: ; 1 LJ,: LJ.: Li,, l .l- r- io'* '"'o'" Lj" lj.; 'z'G DIAGRAM sysiem on anoLher part of thesystem I n'1 2. the cztcraauorres E -. !-,- fo, ,6*ing them ,/O ate'smattconrnil,{r.1,,:l:T'5;#::fr':t*n./rn*1" ! o1 rj U I (2-406) A systein is said to be in efti[brium if its every maCerial poinL P continues to remain at test, in art inertial frarne f. If a systcm is ra cgur'li,0rirm, lhcn lor its cocry'poti f=O Prcof,. lf O is a point of f, then s.qr oqr= and Euler's axioms - M,=0. (2.41) 9, spolr = 0, V t- Hence rl J^*11dm=0, yield F = q, :0- v t, J^teolrx!!potr1-=g' Mo=iIoV =9, :+ Mt= l[8+AO x F =Q- Ho11= These are necessary conditions of equilibriunr, but not suflicien0 conditions even for a rigid bodyT,he cquations of motiott of aa incdialcs.s syslem are also.(2.{l), since * = + \, = 0, EaI:oO V l. 2.1s.1 Two Force ' Mernber L T;G..-'-"JjI"r*-:=I; fl$Ilffl t-r--=-:A ts'o.forie ,rr"r,Si1fq8-2-20) is irvo force nember is a ruernber subjected to only t*ro fo..o]lf " ia cguilifrium or a trvo force iacraialess mernber is in motion then the trso forces have cgucl lml.gaifudes, opposilc dittrtions and act along. the line joiring thcir points of applicatiottProof, M = AEx& = 0 :+ Q acts through ,4 along.AB. Similarly F, acts along ,tB. F = & +& = Q A ^ = fr--&' a- Srnooth balt and socketjoiat L2- TJ r\'fu,*, d-s end (3D load) '?ry?ffi:f-fr{ 9e f.z j. a-a "#'F tnternal force,resultants in a bar k- Coplanar load on coplanar beam l- Coplanar badbn strafult beam 9'\r,z9t Ai FBD's of front rvheel assembly (m1), rear wheel assembly (m2), chassis (m3), and the completc rcad-roller with dtiving tarque M on the reat wheels, assuming no slipro.e giito in 11.r.l9: .:k; -{:'Rq -sB 8 tr x. z'1<jr 30 :: :J _J J J J J U J J J u L-I f- l. Internal section of a body ^T,Ei. 8i 'J U ,J Pin in smooth slo[ b. U U Fr il LI U U :J LJ J J Jil J J il'r I J ed- *ffifrur*H ;Le? i .'..i...,,,,...,.'.".x21"..'.9,,i,,*",".;2.'*,,.,,. 'v.1]5, tt' * FnrcrroN epif, pt aod ''";:[,Tr:::::'-iortg,/*J*.lr1',:*1:i5::X***:llffi;'::f .o"ffici"nts frictior ;: ;;;"rt: z.ro are of and sratic i"*i"siv an angre tens'ron in the bdt and ,rrP N/m and t'e ;;;;rm iif.i"tiona gravity' the FBD of p". Ar locarion {,, ler *;r".*., belt at {' bc7' Neglecting ;;t centroidJ; parh coordinares rs: be T. I*t the radius ***"."*.rtbe ,.,"*r, * anis elcmenr io ir",t'""'lt Fig.22t. in * o" an erement of rength "n. "i-*" conracL over an tf *c AB*a. get and sirlrilarly lle catr pullcv' tlren (1'42} hold 's for slip the bclt is on a ro{aiiag for impending sliP : lf the inertia of rlre belt forslip : TrlTz=eP'e Lf"' :./.0 . i1-:f,;t = - ' -rs negl'ected' then for no sl\r I : rr-}_: 2 " ",.d fr17 -' : (2-{3) - t' reduce to = 0 and (2'a2)'(2':13) I fornoslip: TtlT\<e"0' forinrpendingslip: T1!12=e"e' , l.l! (2-4{) ll as in (2"14]slip and.no slio are the 'sanre irnpe[ding of to the condigions the asd o l"t ot Ou/' be the force nornrat Eor a stctioncry bblt, o = itt t' ***-t ' papfot semi-vertex with p and of ropc """' belt or a N/m instead f's' tl i l. llii !! i7* -ta iut't * and q directio"lt"."";#;: 6.'vee' bett (F.q.2,21). Then "rJ. .noiu"rt.tt"n"".q.(z-i:-);;'i;'Y::'.1;l'l*:k*it:jIg;:;"":.ilFT'1 trso pullel= is is bel! better fot porver'transmrs of lap on t* s-ollct of tt't-*Eio ,rr1/sino. 'Vse' Ler Tq be ttu *-*iJu'* J,o.."rt" P lransmitted' in the expression of the pos'er i ;" ;;t*'n'i*io" rve get i1 Lhe t-h"'. sarne fior both tbis value' transmission, if p's are the are used to itlcrease pulleys Idler "*" belt' Suljstituting to be used in the auo"e c'iott=tntrension :! l1 7i frorn I li it (2{2)' ;i :J ri I . t, i: li1i p-T1.v-T2o=.$1.,:lU2)U(1.-e-,.sc).Fortna.timuttrP.dPldv-_0-.?|.1=3lrr:. '! trassmitted when u' = ?i/3iIlence the maximuqo'power'is " ': THIIUST BEATuNO A AT TORQUE 2-1? FRrcrroNAL surfare (co1i1L Aar' -- axisynrnretric --,. *itrr auotlrcr o1,er ao contact in t body tb.,,.t'P consider an axisymmetric of iodv I under an axia! coatact r{ lhe torque 1"*1'"" of ring spherical, etc-)- The ixial "*1"1 '*;;;;"*ia ii'Z'zz' Consil^er an elementary '"t i o*r,'it ,t i. r oocmal body the of Let FBD is to be determined- The *!?:" rvith the axis' norr.',.r'rr*un!-"o r frorn its *ith r distance radi.s "t'gtt and at a radial sutface of width ds t" * itt**tJ'"ntial-dircction is force t rir.l/*' frictionat Thc force be p N/*r. yicld for axial fot"" *d axial monrcnt the a:tis. Equations "t to"ir:*lrur , --: ; i 'l'1 . L L l*.. l-f*- t-, l- "*.' u L. r = /tr.i"."Xz"")a"' .-i[ffi\l u= MI P f!r,nl(2zrlds' ,9 .- I :!! : i - I I r,,'t *lL [ /'n"i"ga{ r']" * ,t" R t-.^ i;-Y,t, f .Lflt h V" r.€ffi iiS% I I , itic ; -'-* I I I i u&=fr/( F II ,, :F .31 , ,.,: . ::. .:., {l (a) Conical bearing: ->,i1+ 1- for uniformpressure - drlsinc, eq(l) * Ml P = i !," t,,'odr'l"iool t I f'*a'lel '(3) (2) + - Ml | = (2y,l3sinoX"i - fi) l(rl - r!,)- p=const' r2)12' 2. for uniform rvear p = Clr (2) :+ M lP = (p,/sin o)(r'r + to (b) Flat bearing , u -tllzregd2x{) for various casesr respectively reduce Mlp.= {2p,fni(rl- "?t/bi - "?t, MIP Mlp oa4 | I = I l,.it ,f {,'w*1, r=Rsind, MIP ds.=RdL' s: rl2-d' and eq(l) - P,(t +.)12' (5) '+ - Rl Jr,o"inz 0 d01t [,fr'u' 0 cosa dll, (6) r*lZ r*12 sin'ceE?odfl :F'R' cosedlltlJo *t"Ilr.sintA MIP= +' l. if pqc 'd "l 2. fcr unifotnr pres5ure p:const:':'t?o," an (6)' *':MlP'.- rP,Rf2' thenp=.{^Rcosg' and"(6}" 2.ISE.TiICTIoNALToILQUEFoR.ASQUARE.THREADEDScILEw porver-iransmission-(as in lead-screr*' of square-threaded scre$'s ate comlxlnly used iu screw-jacks and the axiat totque Ir' needed to impend lathe machine) and in tesling machioes and presses- we compute tt]t:10 of the screiv in the length nrotion of a sctew in a fixed nut agaiast an axial tlrrust P' Let 't: hetlx angle c is giverr The { (Fi5'2'23a}' contact with the nut be s and it-" rneao udius and lead be r and distaoce of Zat (Fig'2-23a)- The by taa c - ll2trsince the hellr a.tiall.v adraaces by I for a circumferential thread be r l\l/1 al.an angle c s'ith FBD of rhe screw is shorvn in Fi5223a- Let the normal fotce on the to the dirction of irnpeadiug the a-ris- The frictional force is r,p N/m aloog the helix in direction opposite slip. tfence the- torcc dF ori'an elerueat of length ds ai a dis[auce r fronr the a:cis 1 dL: pd-'(-dncq +ccag€ ) * prpdsl-Goso'gd - sino6.)' Equatioos of equilibrium for a-xial fotce aod atial moment yi'eld e= lt n, It = Jor(ps$o*p,pccc)ds' Jo-@"*c-;r,psina)ds,. M / P = r(sin a + p, cos o) / (cos o - p, sin c) = r(tatr o + P') I $ - g, tan c)r; (r) The rorque iy'r needed foroimpending motion in the direction'of the a-xia].rnTti."'.!:i62'23b) lil P,= t(p, is scf-locting if it fhe "...* Mt ) O, i.e., if tanar( P,- tan c) / does not advance under (l * P in the absen ce of. M1- Hence it is self-to<'king if P} :1 usPdS e.+ | I ...-l (ol (2) P, tan o)- rqiset- :rE F; tt: :' since the present directioni of'the'torque and tlre obtained from (1) by replacing 11/ b-v -Mr and p, by -p, -frictiou force are opposite to those of t'he previous case: 6iv--'1a ;J J J J J.,J r-l J ;-J J J J J J J l U ':J '*':J Ir \\|.-J -l j-J hds R'3. 3p_ -t (4) motncnt evaluated for t'he corresponding Equarions (2)-(5) imply that the mornent fot conlal- be*,19: :lt" : p, I sin a' lrence conica'l dutche flat bearing with r, replaced by an elfect ive coefficient of fric0ion Pefr pe6 ] p'' arb more efiective for power traasmissiou, since fc) Spherical ball bearing: ,. 16 r.z --J il :J :l I a- t- t- L-- t- :,. :".s\ - ExAMPr,Es z point the moment of the force system applid on the hinged bar (Fig-E2.la) about aad the resultant (equirralenl) of this force system at point ,,1- L-., Erampte-2,r 'Fini tt--- tj Fr=z-s roklr tsi sB, L.- FO= 5 k* € :1 tt r.l.{ j1 I t<-t L*_ LJ tj The force system is coplanar@mponent.s as shown in Fig.E2-lb: -So6tioa h It is convenient to resolve the forces and the distaaces t into suitable F3 = lOcos 30o = 8.660 kN, - 5 cos 60o = 2:5 kN, .F2 = 5 sin 60o = 4-330 kN, fa= 10sin30o =5kN, ri = [4/(3? +42]rt2l4=3-2kN, F';=[3/(32*4111214=2'4kN B$- L5cos20o =1.410m,' CE =l-5sin20o =0'5130m' CP=0'5cos20"=0'4698m' FD=0-Ssin20":0.12i0 m, DG-'EF --CE -CF = 0-0432 m, :{G; 1+ l-410'+0i!?r0 = -z;st1 t1" The rrapezoidal distributed force s-vsiem is decrmposed into rectanglhl Td trianqllql {!:t1i:::-"": Sr.pf*a resultant ,r7 of the rectangular distribution equals its ar^ea- L":, t- = !'0 1 Z :-I ? kN' a'n-dacts Fs of the tnrough its centroid, at a distance of AH = 0.2 * o-612 = 0-5 rn froar {- The simplest resultant at a centroid' its through triangutar distribution equals its area, i-e-, Fe = |(0.0 x 3) = 0-9 k\, and acts distauceofAl=0.2+(3x0.6)=0-6mfromA. : .. product of its rnagaitude rvith .F1 t-. L_ t: lt l' \-_ L L* L; l lt LJ LL- The moment of each forcc component a.bout point r{ is conrpuled as tiie tends trcr rotate its perpeudiCular distance from ,4 and is assigned positive or oegatile sign according as it Ttrus couples' the tle bar in the setrse , (D t V (i) or y (j-) to r-(!)- The sarne applis to the sign of tti =[2-5 x 0.5+ 4-330 x 2+ 8-660 x 0-5 x I +3'2 x 0-0432 2'4 x 2'581 * L'2 x05+O'9 x 0'6+3 - 2Jk= 0'9938k kN-m a forie Fn The lesultant (equivalent) of the given force system at .A is a brce-couple system, consisting of aad a couple Cn given bY Ca=I&=(-2.si-4.330j)+(8.660i-5j_)+(3.2r-2.4t)- l.2j-o.sj=e-36i -.:3-83ikN t ls Mi - 0.9938L kN.m. .-_\ p;ample 2.2 Find the moment of the force shorvn in Fig.E2-2 about 7/\ poiat,{ aod about the axis of the bolc. Find its resultant at r{-JiA.4Li Solution Lei qand g' be the unit vectors along r?C alo.d AB: a. i -=-.1k : e= (-5j + 12k)/(52 +L22)tt2 = (-5j + 12k)/13 i', -g' - (-4!+ 3B/{42+32}1fz=,-0.8i+0.6t , ---J /: .rx--: ,,.' ' er : (3g* 4ll I @2 q4zlrtz - 0.6e+ 0.8i :f o: .t, s't? _ _:?_6q -_ .!:ZGq=2.6(0.09+0.E!):I-569{2.U6! ./ //1 , =2.6(0.6e+0.80 1-56e*2.08i Q_a= lt L_. tt- tlr ' It Llt : t-.- kN Ur,l; lTii' :soi-ssj * 13ok mm -"^:.t4C=AE-+pQ1QR+RC=-30i+50i+?0k+65s I glt^ Ll- L56(-5i+ 12k)/13 +2.08i = ?.08i- 0-6j + 1-44k L..HI ^lj :;.1 at':''txu;,lf& ill I L-l- L = L-. L-. I r ': t+ L: Li, lr L. .4-ft*t . -, <i- le L,.- 31'* "t -12i+ 198.4j + 84.4k kN-mm = -1.2!+ 33 198.4j + 84-4 L N-tt llllf i,/ f \-Y \e, ' hS-E2.2- * { r- :':-';i'': '; '' ' . ,', ''- - ', 1 ' Moment about the 8xis oiih!:boit .*i i-ae'. N'm' 51'6 x 0'6 (-f2X-tiS1 +Aa'n = Mta = Me' - fu ' € = sVsta, /B cons-istinS The resultairt, of the given force. at 'A is a forcecouple coupl'e Qa + of a force & : E = 2.08 i - 0-6 j + l''4{ k kN and a &' : \e 198.4i +84'4k N'm' -1-2i+ its A rectangular plate of rverght 2 kN is hinged along ie Gig.E2-3)- Find the total moment of the force system sho*'n "agu (eguivalent} of tlre about po,int.A and aboulline-,{8. tin<tTh-eresultant given force system at point Ai o 7 Lm",P - 60o'?: ""'',9^:,1."]; list Lhg'fi Solution Let n be the normal to thc platc as shorvn' We t*-z-'l coordinates and poition vec0ors of various points: Example 2.3 | -r, t, { m, rs =3! m, cp - 4!* 2! m. s6 -^ i rf(o' 1'3) G(4sin30"0"1cos30o) = (2'0'3'464)' /(4,0,0), E(0,0,3), D(4.2.0)' 'g(0'2,3), s^-= 4! F;5.-c l--3 r \e, = 2!+3&-T:-.* = 21+3i648 mr rtr - i +3k ; l '*2"-j::ln g=&-lla;1'=,(-qr+rt) l(42 +121r/? = -0-8i+0.6k, alx ADit-<i+3,&)1'6i +2i+ I-8! m 0'6k) 3(-0-s1+ 2i + = 3s =,41* DG = tc - LD - -2L-2i +3-464 b m, !F - Iq + n: (AB-x AD) tlAg x ADI- (-Gi- s$/$2 + 82)t/? = -0'6!- 0sk' : t* ,*-=:.""rr* # t,h" firrc forces are given by Cr - 2s=2(-0:6i'- O.8k) = -lJi- 1,6! kN'm e:{es=4(cos30o!*sin3oos}:a[0'866i+05{-0'8i+0'6$]=-1'-6::3'46{l+r2h'tN'o = -4.472i - + 7'746 k kN - lagq-ii+2$/(6" [3=14e2 : ' f, 4-472i eu" = s[€o6affE: -sin3-oo bJ +32 : +z?f/z: -6i* 12j+{!&N ' 8[0i866(cos60o i+ sino0oil - 0'5hl = . T^- . -- Ee:5g+=S{cca!*cospj*coszl}=r5(cosI20?i+cos60oi+cos45"}=-251+25j+3536tkN Thcpcitionvectorsofthepointsofapplicat,ionoftheforcesrv.r.t.l.arc: m, AE= g.e- La=-ai+2j+3krn''4F =L"-!t= AD: lo -Lt=2j AH:t* -tA=-li+tjf L* AC-=-2i+j+15t' : +(-28t: AC.x'.&''+ Qt'+ +(-2i-4j)+(-l'2i- 1'6E)+(-r'6i+3'464j+1'2k) ,.. T.he moient w'r'L''{ yi't'h't}e force: Q ^ ^.'2j'' 3oh)'+i:18:8i: &365'i-'2L"t, +(-1'96ain-f6;6aii---t:iS Lrir,e:et.4E:l +'fr'x.\+ M-.q= AD x (rs.49.i+ES44k) -2'4i+2j+l$km m of its Pcition^r'ector Moment of each force abouh r,t:is cornputei: by' cross-product of theforce system about line 4Il :. = -40'0?i+0'743i-'56'2eEkN'm ' is given by M.z :-l'718 kN'mr,' e = (-40-0at-b'i* to'*'ltol * (-'u'ntt0'6) = * at .A is a ficrcecouple'systern' coosisting of a force & The resqltant (equivalent) of the given force M"= and a couple en M. g,ren bY "yrt r* =E& =.(4A721_ 4.472i+?.?468+(-6i+12j+4$+(3,..[|1i+oi-1$ i d'+(-2'5i+2'5j-+3'530$+(-?H=-e50bi,L1633i+e'282Ekr{ 0'?'t3j e* = W= -40'0?i* j 5629 k kN'm' ,j:-- . 34 I :- &=-2LkN 3'4sd1*'6i-ot I : -. -!-\ Earaple :i"fi:"$i'ffiT;?.-?Lrr'1."cu:rili"::l::.'-"-,:L'oo'*i:"'' ffi that l a' i""* in t-. F- direction to "i i",O t'"i"' outai"J i" Ca Q^a.:Cnk- t- I L. slampte 2.8 t'L: Eind the centre or ! #"t*-i zl d I Y. 4 \ ,! * *:,o'01::::1j:::t::r1;1ll :|] ;:t" aa sl t ffi1**",1"inl***"*fi"']'"';;iy; *;::ff YJ-'--- 4' a deretion or a 7 aseml-crrcur4r-Dvr's J' shell cylindrical 'vri"d"' taken L f#:::"rJ:riffiffiTil;,"-T;x""";;-;ii semi-circulat d cuboid 5 is taken-. Tr-^ *-.o af deleted u-,11'"-'11llo'"t' sphere -. q"*Y of size Rlz x Rlzxzi""i " cubo'ld ',*s = -(Rl2)(Rl2\(2R\p = L-.. m5 t_- IJ tc4= -ffp12, =(4rnl fi)lqlp=tffpll, i 3 ur='(ZRlr)i+ BL, ..oDr- m3=(r.R)(2&)o=2*ffc, *a= (rr2l2l{znle: uf p, T| )'- 5 antre or mass of each is given bv: :X";[-;;:;;""d LJ talll)-i*:ot' t,, \y'ry1y in*' Ilr-r i" = ulo'1.':. j:ji'::r'jr: 'l'=,5 t:u = (3818)i+(48+38/8)k' systems given in-Pigs-E2'6a'b'c'd:following bodies for t'e the.free body diagrams of the All coatact surfaces 3 + 4 + s' ia; J' 2' 3' AB' o' '" /'E: ('} light' ir{ake LJ *r]rrrr" iu*'r.", i' \12: i' {z) ABID, AB/:;E.&);, '' 'l memb.s' rot 'utti"tt no'*t"t cables'are are smooth. The belt "lil it " 'irr'i'-ti" which are trso-force mernbers' if any, due to the presence of supports t- tJ IJ tJ are t:*:' Ttre loading on s}'stems simplincatioas, LJ Lj L L tj relative rotation and shomn, becatrse thdte is cgmpletc "onrtr-uint-on tj t-. ) I ) -i a*r-;1":s;;::il-n X;-;Jr.g.r.!" it is drawn for b-ecomes convenient if iu ris.fz.6s. The FBD of rhe puuev n"ri", ; pal of the belt which overlaps it' . 35 :i ., ,i I 1 :l il '!! l. e1'e'bolt' but in rotations are allowed at an are sho*'n in Fig'82'6e' AII relative exerts tlvo FBD's (a) Solrrtioa .,The of the bott are preve4led- Hence, 'l: "',-oo"tt the. to normal the relative displacements "*i" to the axis of its eye- The supportinc,,Iembers PQ and dLections e-xerts 3 force componeats of force in,the.two ";;ao;r'"e Rs o"lrt But the memb..eqfl: ro.* .:t ""u as, u"iog two-force members, B in theufi.r-s'F'1fBD a4d at the 1 compoo ntsl'iie shotr:n' because';' a,two-{<irce components, since it is::not ""J"t]'O''the'6xe!''Su1P"t' 3 i..""-**0.."i* .r,a,s-.ouple FBD, secpnd tbe in E at force sy'stem ts ht€raal section d'splacenrent and the apptied ;il"ttt* ,ot"t" relative on thete is complete coustraint thtedimensional' siucr it is not a trvo' (b)TheFBD,saredepictedinFig.E2.6iThesupportingmembersPQand*'o_::1,.:o-forcemembers, exerts 2 force components' ?J *"*ler *" U* only.89 and cxerts forccs along PQ i in the first FBD and at the internal suppor fixed the At coplanat. is ' "' one out-of-plane couple component are force s*mbe" qrd the loa.ding and ;;**poot'ot" i"ph" 2 the applied sectiou at'b ia the second FBD, rerative d'splacement aud FigF-E2-6b'c is coPlanar' L. 1- TT" e:z-e z'u. - -t:':*HlJT:Tff *."'"J:ri'*.tlTr:il;:ffi . rL:- ---i-rircular ring 2, a t "":ffi il;;*"'rlu""'""densities\'oa'.dp' LJ I t' t-LJ u l-. tt* is' a couple the-simplesu resuliaat and thickness dr: element' of radius r rR .^ Cn=Mo= JOlrr(Za rldr= Il'',(T)(2rr)dr++ l*,,mlzT)'".= L- 1*. force' o[the tangentiat distributed ,.o F;i1t1*;*n''* "-tte"t t'he l 't .i ,i ,i i I tiI __:-&, NOTE: Once a sei.82,of siiplorf, rqictions on body *t slodd shor the ai""-t" *" FBD of Br due 0o body 82 have O:" :fu"^tl a reactions but rvith opfoe-i1 sense' FBD dBr' ""*" "rrppo.t joint' as at joint (i' tbco than 2 members meet at a hinge moie If in Fig-E2-6h. (d) with t&e preferably, the FBD's of the- members be drawu' either the FBD of the pin be drawn separately, or + le_ T ;".;"" part of one of the rpembers Both these procedure are illustrated' d! The FBD'8 are ahown (o) ,; ;;; J J I :l -l J 6 (c) ,l I '-+/ (dt I J n'*,ffgo " *'i-? r:ffi!:*,, ffi F i ol'-o. o.i,:u F1 tq3 ')i! *' F7. -J \ - sn e.1f*.' . r&g - -{ ,i 'f .If i - r'r .J). zc s.,Rn Jncs 8s.".s(i_ffirffi* m.3 - Iu'ls *.esl (qg c \-vl F3 atl {; _ Exarnpte 2.7 D.t*-f...-boaI-ailtt- F. 6-ls =tL:lolt' l-:F>. 4n/srz4s' of bodv | I of mass -t> > 4 6/3 Snls- -fiB lt . L *i"rtno;]" Fia V) 7t Tt" ate 8.d.4' ^--^ ^L^.,.- it" hE.E:j}-ziJ::*:n ,(.],.,-.ts;5{5'tsr ,.1f1 - ffir l-rs'L'" Tt ,sl ':-'' 'r ' . , . : -q Y '- :h9 If ::i !*ti ::. b determined t"ndSotution Lct.A and I be the points in contact. The rJative velocity 9es pilrt and is dirccted ytst' 0, then there is aciual slip :+ the friciional force F has magnitude of Fl = 1- ir L ::H-#ffiff{; ' I^!r 6Aea;tl -t r- 36 .,_ -r - l* u I --i..1 ', a*--l - there'is ]il its . 9t - (Iwhere q is any vector in the tangent pl*l,oJ contact''tlcu ;:';;., are ""aforce directirq'which arbiirary E has atbitrary a"gnit,rdu (bounded by f',/V) and ,; "UpTUi"tIoA cascs' ao slip lhc zcrc is trol fot detcrmined by solving the equaiions of motion. Most oftitt, ltictional fotce then there is contact' of plane the-tangent tl"to.-in 3: If q^s - Q, anc 9ts - 9f 0 where q is some has a component which is directed L I'- p,N and impending slip :+ the irictional force F has magnitude of .F = ofien, the direction of frictioa forces are oppositc to the component of gre in the tanglnt plane. Quiie assumption' For example' for the case of the same a.s those obtained io .-!.iot'sotuiioul based on ao slip about a fixed vertical axis with angular impending slip of a small block on a horizontal platform rotating radially inward as well as the circumferential acceleration, there would be frictional fcrce components in thd directions with the total magnitude of f being p,/Vvalue of gs'A'q as well' are shonrn in Fig'82.7b For each case, the .,r"to"lf 31^s and if gas"- 0, ,h"o the of the case' Thus there is slip in cases and FBD\ completed as described above, depending on the nature ofk for cases^1,5 and opposite to it in cases 2' 6' 1, 2,5,6 with the friction force acting in the direction tn"L r impending srip ror cases 4, r, k u L L IJ IJ L ; ;#;;'il;. Bx;mple 2.8 i "r,a (a) Chick rvhether the forces: (1) .f' = 2z(-cosz+ y2)k, (2) '?)vi+ ;:?;;;;;n"ra.-'tul L r_- rJ IJ 8- 8. .; lz +v? + :2 sin r) ?|: tr' 11'('I A^ i r"l,+ (v2z -' * :^)l*^':::; frj Find the.uork done bv these forcesin a close^d ','] 4: (2x'1, (c) Find the work done by these forces along the curve U: ,/ (') o-u=l- ^ fr+ zzsinz) , - \ 2(z+&-,,-,',,-.ftt*rl 2z(-cosr*9 l(-2, f ^,- + lJ f,-- =.\4zs - 4"v) i - (22 sin c - z2)s j* 2zsin z) (2v - 2v) ! = Q' Eence this force is.conservative(b) The work done over the clced path is zero, since F is conservative. first lind the potential enersi l'- of I ' (c) tn order to find the work done along any path, it is convenieut to the datum at Io - Q- Using eq(3'28) The x'ork done equals the negative of the cLange of 7' We choose IJ lJ IJ IJ IJ tr_ Y(d = I p-drJ; - li" dr=:/ ct',o,ol a'- lo"r"ti'v'0)dv - lo''''''v'z)dz '- - f,"tt ,l ar- fr'zrvav- lo'zrt-cose* v2\d'=" -'v' -t'(-n*'+v2)' C: {r)=2ri-rsj+r2k. :+ rr =dl) -2i- j+rkr.'rz= r{2) =4i- 8i+4I9F I(cr) = V (2, -t, t) = 22 - 2(-1)' - l2[- cos 2 + (- 1)?l = I * cos2 cos4 V(cz) = v(4,-8,4, - 42 -4(-8)2 - 42[-cos'I + (-8)'J - -1264 * 16 1275 units' Uf(& *s2) =.:[y(rz) - y(rr)] = V(2,-1.1) -V(4,-8,4) = 1265 +cc2- 16cos4 = (2) F =(2x-y+yz2)i* !f":2!l;zzli*2tvz\ Fr=2xyz'' ", F.=22-y*!12, F" :fz-t+xi2,' u IJ u ]*; (") VxE- o v i ! ,a .fr (2r,-g+vzz) j kl & * l=v2u+ (f"-t,*tz2l 2xvzl I 37 o ..j ', lt 1 t 'C. ='=' .. Eence this force i" non-iolootir, l a;;d;ij;d,r,4;?:iii(i-0) = (y;o)/(o-oit1 = 1;:g1p-0) =+ z = b,y: 0,ir, =, Jc;o ahe integral is evaluatecl by replacing y,z in f, in tcrms of - fot poiqts on Cr; z,e in f, ia terrns of y aod t,y il I| in terms of z. Similar procedure-is adopted for erraluatioa of intcgrals ficr curves Cz ad &. 1Aflfi wo-t = | (F,dz *-Fydy + F,dz) = I 12"- 0+(0)(2a12laz+'Jo| pqrpllolrldz Jcr:o Jo - r- C2: A(1,0,2)* B(0,3,0), (z- 1)/(0- l): (v-0)/(3-0) = (z-2)t@-2) + z=L-y/3, i ' fB zl2, y=-3r*3, y:-32/2+3, .7O 18 -=2t, z=-2y/3+2 - \. Fih = I Q, -y+:sz\dz = | 1zr-(-3'+3)+(-32+3)(2e)2ldr Jc*t Jt I Jcr.A . z*- = 'Jtfor-rrr"*L2,:t5r-3)dz:-e.5 .18tA-13 l_ Frdy-- JCz:.q I @'"-z+azzrd:g= Jo-lt'|:[-vl3) -(l - cl3l+e-yl3)(-zylT+z)2lds -' : JCz:A = Jo[l'-ny"pz +7y2rs- uy/3+3]dy = 3-?5 {' 323l2ldz = -2 + ttzldz: Jt' Jct.rr-'= {henleszn fis,, {i3r,rr, trqtl F,dz 1B W-a = JICztA@;ar* -2= 125' z=0, :+ d:c:O. d'z:Q Frdv + F.dz)= (a: 9(0,3"0)-O(0,0'0),- l:0 wa - o : fo -J * -05+3'75 Frds * F"dz) ",f,o* W^-a * Wa-o : Wo-a-o-o = Wo-t :/b.trl - 0 + 0(o)"tdy = 0. 1 * 1.25J0 = 2-25 uoits. " The work done for this closed path is non-zero. The work done by - a non-conservative is, in generatr;non-zero. r . z-;72, I (c) C: a(r)=.2ti-r3i:+r?k.-.+', t::':2r,,1=:f, * ,l-:Q) = J;"ir{f"dz * F,dy * F,dz) 12 = JII l8r * 8r3 - 1617 12 = lQF, - 3r2 F, fo-rce in a closed path ,.,, dz=2dr, .dy.='$i2dr, dz:2rdr' - 6fldr = -10818 units the clced path shorvn in Fig.E2-9- &) find the work done by this force for the circular path from F Lo C. (c) Is this force consenrative? C and 6 are constants- a----:->tF FiaZz.9 . 38 .-I l ':l J J .J *ZrF.)dr {J "t J J J I ':-:-- \,1 tt r\,r Solution' -) (a) (1) -) l3clslo'*'uoo7 - For path -r{B: L b(dz - 6i Rzb(62f+' *'uo1= P^b rt6 - R 1t1/e-c = l) Yields wB-c= r=82, dr=0 aadeq(l)yt"ld" J!" o' r0t wD-::J;,^","::='Br6(dr-dr) simirarry, wrl,-o: li'*r*\ds=-clll*i-Ll4ll2' Forparh Bc: - 7\f - a-e-:s-D-t = Wt-a * Wa-C {l (b) For path : fG: wr -c - r= = [* !t-r dcg{ Kc t l and eq(l) yields + rD ")dr dd] = li : no to t'31dr + ! u *" 4 de -c i zB + bd I J'' -- o- 'r- :- --''t f'\ is t" non-zero' It can also be q tlffir ' ua!" ts* - \-' l_ l^-^ "' '_ " the'\r'c since non-con1rvati1":"",:1T:::: ::1'"::;ilY,:::iJ::If] rhe force is non-conservative (c) The F it tlri. .*"y. f -.-->1=- L Ii -{ " "" at l:i'ff :,j:".i :"::;ffi';:':;:;':1""';:;; i:*,'sy s' P6-X' t= I";-Jti,';l'..o,ll'o (Fis'E2-'10) ":'.n"*i { io the horizont' *"-., .-','-*o Lrre angurar q<I ffi;;r e'rrF-;ist""':,i:':"f;:T::lj; " 0'; ;51J:3'?:::'"1': ::J:;coe-ot " I + rt,li -acceleration of the u,heel is u., = "r' : j"-:T:ao :,::;.::l:;:t sriP during the time = o' ;";;" Lr- the work is obtained oo t. + ! ,.a= us *,;[j"""-"' -,,''l &= = ,JoJ, ! a a, L = /oi*t*+ '+ t([ L *he1 cr) dt = co(l -: 2-ot)'la "=;i; "=+, : ^.-: (1) .. , .: +,crr3]lJ.i' :_: "*i*u*c]mr+m -'rz) -aolzla-orIs}cos0*{(Ii'-E)cosO+osrnc}tt'cors{-'+o}"-r' ,'rz= 1',a^o=*, o,= !o'n"'o'=Lj#:+-" ";il;;; : i ...; .., _-, !^- -rf L .. : to the distance equal the nrodult of I and case of constant Iltt "= ":l Norice that the work done for tbis of application of forcli.* poiuts initial and final ihe .. =.. . ,", betseen ) -, -) :'- . .'+i-:"- -' ='3,t::.+ . \ ag +41!r - -ot1ls - a$ "-otllat AlthoughVx[=Q,yetthisforceisnoa.con".,*j:i,esinceitbtimedependent. ;e *a o' 6 = o,o = 0'-'cs(1) rieldsi' ' (u) For the case : !-. ts ra - !) cosf'+6sin - - 1,12=Ps[{(uo',1-as/a)fr L j "rith W=L'9r-Fl{os+-(L-A)}cos6+r"6sinfl G ) . w = /'"ott(**?f-"-*e:?l'-cltedlcosc - ba 7" FiX-a 2'io * cf o' (Ir- 8)jJ: [uo+r'r(h - R]i-1461 x ttp - !o+ c-*tb-:ooi+t'rk [-6i bL- rrr { ='0' and t'o = _l o,lo 5i::::i;';"r:.T;;x;;:*."i" instauts' materiar point P at different *o::Y::"ut m*: ffirlH::];1i"#::i-"4'i-sind1 rclpect to trnle' by integrating lar - -, [t' rt' acts at 0 to 1 tima interval time intcrYs'l a fixed inclination of l-. : ""*';-*i , '' flj '/., L L L L |-.- rj' L l_ [- r{ lJ L L f-. I.lj f- rj u IJ, tu rJ l.--' f I I i I '-=;= -{\.':..'.. . .'. 3. ., DYNAMICS OF RIGID BODY I)dne Kronecker's delta 6;1 G;:{; I :11.}, by 9tP-- ,.e" 611'=522 =6s =1,612=66 =631 tg.:J =5s-6t=6a=0,1q'q'=6rr'(3'r) of arigid body (Fig.3.l), Wt = ae - 9e = t t x Let. I*t u = @i*t tpt = z.'q, \ta-- t,Pa' : t2 = 4. + ,tr* zl and let '9= tr"rr * u'2u'2 + ,,3t/i3 -- 'jd,i' l-. r f For pcints P and A then En= J *nxap^dm= J^rrnx (s4x zptldtu= J\Ar^'tpt).-(tp^'s)tpold'n, :+ rici) dnc,t; (z;ut;pildrn = f ,{r'6r, l:r"r, t;r;)drn, Defining t6 = l*{r'A;; go, Ilwi = Iirr+ Ilq+ IAq, = i-e-,,Ifa. = I{rur * If2u2]- f*.r, 7{tz = IAq +1*uz+I$aa. + = Ifil] [i* I"43u3,' I{zu2+ Ittq * .H*, "A [-; = g-9; = {u;c;} -g - H,t. = t.,;6t-;} (3.2) (3.3) it] tIl '+ i!{at=u'r&q}' rfrl [-'l ,= rhete Lo,r= lflll , rr ^,= Ln,;J Lrrii I:A r*.| L;.| . l-rr' 4' (34) (3"5) I* = li. The real syrnmefric matrix [IaJ is catled incrria'.molriz at A relatite to the ares rr(q) at A- In glneral, fi.f b not in the direction of 1g' If ';l is not a point of the rigid body, then fu: flc*bt x mgs^ rvith |fla.]= tI"lklNotc thar t AT ,{ ", ^=to*-r"*too of the mass element drn from a1-axis at .i4- The elemetts of [/Al are given by (3-2): -rua,dbtance - F;=i;;-;;= in;l,z*,sr- x!td,,,= i:'i+z!)dn= l{rr;a^ 4'^= f d2rd,rr1r2)dra : t',, = l rrr, d*., tlr'013 - r;: loi+x!)dm, Ifr: I!r=- trrrza*, $, = t"1"2+ ri)dm, €r= l,!,i+21)dm, tS: tlr=- rlr-r{"--1,",,au,, f ,rr"d*, t, = !(o'+ 22)dnt= *1rj12. ,*: I:,'+,\in--,,,(tf r!:ii=- f ,va*, (3.6) )2. ': = J-!" + v2ldnt= rn(tf )?' I$,IA,I{3, 1A,1"n" are called Lhe produck of idirtia of the body w.r.t. axes ri ar A. k!,k{,&f are called IIrc ndii of gymtion about axes t,y,z aL,{- Fo'r the case of scss dis{fiDa liott ott a pldnc arca ilt e,y planc with point A in Lhis plane, (3.6) yields I} = t!" +.lf;u ' I{L,I*,1!3 x; are called the momeats ol incdia and I(!, Lcr-us'estabtish the trrrslortnatioa rclalionsbetween the elenreats of the inertia matrix [-[^] rv-r-t. aies at, A.aad the elements o[ the inertia matrix [I'r] w.r.t. axes a.l (d') at ,'1 (Fig-3-2), where rJ d :"t.it {oiz9z*aisgs =Qp$. =+ 4 = oirsr, + 6ri :4'di = orr9r'aiq\ = a;oo;$s,(3'7) (3-8) - a+ - component of 9! atong % = 9.1 . g = cos[Z(e!,$)J = d.c- of g!' w-r.t- E, *1r^-=ee=a.lg!,, (") :+,?1 =zet -15.t = apgp..9lr=o;prp, :+ z! =airxr. llr9, t r-:sing (3J), (3.?) and (a), Ill ' = f^(r25i; - {xl)dm = aipaiq {dr"6o, - trtr)dm, i."., S\ ,-9u d-.o\ ,'Y5)-\ E l; Xrr9, .t- L. I l''- = ar( .lv* Fi-. -2 .1 - r.L r Dj=' *' "'io;'I*' summiti""'*tt";iiaiilni; *"t ti" t'n"' "i*iq 9) i" th" sumr / of\, 9 terms: ,- - ^r:-.^1., ( respectively' il;"; the elements of an-cotity w-r.t. bases 9i snd 4 by ( ) and )" zcw o.rdcr kasot' l- A scalar, like temperature ?, sltU So(f) rilement', su& that T =T' is called t otdet ico"o'' such that rli = a;e,,p,is called a first 2. A vector, [r. gp^, ;Jii itiil "L"r"";: otdcr tctsor' t:"::d calS{ is 3- An entity, tite fii1, with 3? (9) elements' such that lll = a;ra;rff " tcasofotdcr callel is 4. An entity D. with 33 (2?) eletoenrs, such that Di;r :i.rfir":.!of i'l'l : tt": atP n' l'e'' using (3'9) with 9't = otpge = The -- Ifl ' 9: ne* at ,tl, is obtained If" = r{f - avau!* = nunll{r, 121{2np2 + 2l{p2n3+ 2Iil E3!,i. = ,;r* : r1;; + I{ati+ '{!n3 about an a:iis i.e., " If, about othogonal axes n and s I* = \t = olpozqllo - t o"r{' sre{ at :4, is obtaiued using (3'9) rvith gl -9zs\ = g' i-e-' c3q : (3'10) s?: tj"--tlorrro=/frn1s1 *I$r.2s2*I$n3s3+Ifr(41s2*nzsr)+I$(42s3*n3s2)+I;{r(n3s.+41s3}-(3'11) 3.2.1 Principal Axes of Inertia at z{ n' The corresponding I*' i" called g An a-ds a is:called a pitcipal atis of ittertia at' A lt I:i :0 V lpincipal mor.cr,,Ii',.of iner-tia- i-e-, lrct u" : $rno = I{rne in (3'11): 4f;"=/flarq::tss{-t,'+=0 VgaE, * 1=^a,. + u6=}ni; It*r- lnr, * {fi., + I*nr+ Il1nr = }rro rfra1*rfrn,+rfin3=Ia1', r4t rr, ["'r =^[:il , 14 I I t* {i"r i.e., I{n1 * I*nr+ I*"zrs: lnz, *- | "' | ["'J 'li L";J liil = r*,,r +r$o, 't -;;:=;": + tr^]tar:r[3], (3.12) (3.r3) '?') of [IA]' since z a8d 'l' are defined rvhere [eJ = [r, nu nalr. Eence n and I ate cigca- lucilot ond eige,.':oal.'e (3J0) and (3-12] yield as the eigen-r.ector and.eigen-ratue pair of malrix B if 8u = lt- Eguations I*!* : If;rnrr.e = Iosr.q = l' oJ ,;ntettia at A = l4' (3.14) g arc' *spcclite'ly''ti'c cigce--wctors llence, tlrc prirrcipcl axes of iaaaia aad pincipal momcttls least 3 ort'hogooal cigenottd eigea-,alucs o{ *c;ocrt;o trrertri:-l'i'Al at.4. tr' reatsymnretric matrix has at principal azcs of iacfliogi vectors g1 rvith 3 eigen-values l;- Eence, there aluays crisi st least 3 orthogonal ot A uith rzspect to ubich lhc i?;c lia matix lal'es a diagono.l fonn| [), o ol [l;f o ol rr'^k:13 Let s1 : r.ri{. Using (3.4)' g.^ in terms t f,l =13 "f',,'fl is of principallx&of inertia at' 'A (3.15) ,', ."i:, If ut isirthedircctiotof agir<cipalatisof LrterlcioatA,fhenLoi"ir.thcdirectionoflv9,e'g''if<^r:<'''9!, then cri =u, ui=o! = 0, and Ha =Iifrgi = Ii.lg. In Fig.3.3, ry-plane is a plane olmass"yr,t*etry, "+ P(x,y,zl = p(x.y,-z\ :+ I*- f*zz'dm'= 0, I:! = j^"yd*=0, =+ z-a-xisisaprincipalaxisof inertiaat.A. Hence alir.cpcrltcttdiczlartoapla,'e of *ar" igi^ct ! is a pincipal ar:r of il.cr'ria at a poinl lohcl il':':*::? yz and zt ale pl'aocs of rn11 (since!::_n_'::": A, \e"'vv at 'ri principal o+a a1ces qe ate PtrrsrPq t and y .,{c (Fig3-a), rg-,J-.rr' z anq revolution tt of revoluElon Dooy ol For tOr a body aprincipalaxisof inertia z-axis isalso symmetry) + I*= I!, -0, rf-: i; =0, + I!r=Ilr:0,i.e., at,a. H#e ori'poiotonthca:crsof symmerrgof abodgof reooldiott, lhcaxisof rctolaliotctilottyluo I _r -, , by symmctiY f* = Ifyr.or1/rtalJo it oaslittrle a sct of orthogonal ptitcipol ates. axes orlhogonal - ._17 : .J ./\ t-r'1tZ) 1 crrgr-?) Fi 9.3.j ' 4r J T J J l J r (3.16) -:r ,t ,! \>j I I tt '"* 'g& 't_!i J .-l -J .-.J 7ai -r- i Ti'b\tt-"'iro Teram of C ; Consider rwo set of parallel oo]r, at i and i; 3-2.2 Prrrallel Axes :'i:rr1r!)drn+ t Lr*,'il f t$ =-t^rrrra* =-l @.+ ,crXiz :-j'rrrro*-,:or,* a*.+z'c, * L "1T*j I f *a'o+z'c", f ---+ c(-c-J (Fi&3'5) ad'',-=: 7,", !ar 7.r9rFie-35 f;'+*(""'*'Lzl' xsr)&n l:*-'", f 'ra'n-'c, I ' Ifi = lfi + ,o(r'"r+ "3"), $, = \9r+o.("Z"arLr\. /S=€+-('3, +rZr), - L. at centre lye+r!)dm = fito"+,,c2}2*(te+ t,= - ' tcig' then z; - i: * tci and J*;;dm = 0' xs.l\dm I*-L 71,*: aigir Lpc = i;gi,4t,: L !e. *, *tE, z'a*= I?'-mzctzc?' I?2-,rltc.zc2, tr= tfi= t$- (3-17) ntz.cze,ca, tt = I$-Dtz,Q3:,Qlt inertia matrix at A due ts + discrete nass rn at C' the bodies and.adding;the individual Inertia matris ro. "o-o*ir" uoai* ate obtained by decomposing tul"tio"" for rotation' if d&aeacontributions using parallel axes theorerrs and the transformaiioo /I + [I^J = tlc] + 3.2.3 Inertia Elements of a Cuboid "d a Cylinder -L-. L-- : !-t.L-. i t, I L-. i- ti = L *tP, l_ii,t{ R2 L? 'nR2 t' ' I:, = -7. . * ;), - t?v = rn(| = A RTGID BODY OF 3.s rvror{ENT OF MOMENTUM EQUATIONS OF MOTION ' 3.3.1 Motion of a Rigid Body with q = corts0ant' i'e', or = a'e" (3'{) =+ If A ,s a poittt of fhe rigid b_odyor.its 3-dimensional massless extension, then tc ti" ru^t=l',X IX ffi] Lrf' 's""jfil tl] = . [iA;] i-e-, E^= Ifrr'e1 +rtzuzz+rs'e"' to t*e *l fr ei ro nvenience, *'e ttzgi Ia.feneral, I/.e is not parallel-to tr, even for plane motion. For convenie"Tl 'l',j "i';: isid th;i.1:0.iiiaad,i!itionAcoitlcidcsu;thc/g^|r=9./eN&alorr9/C,then = iLttgi="'o,,t.+a, x H,+=Ii56e, + I*6e2:+ Ifual%+u'3 x (I$t'rg, * = (I*t - I$rf)e, + (ISa + rfiar?)e2 + /$ri'e., Ma': i-e-, ilJlt 1u{or=I{rA-It.laz, Mti= tt* + Ifr-', =+ Mt. - I:,& - If,.', tr{tt = Ii,,;t + I!,r'., Ma. = tA '. lvl A3 = l33u I L-. Lt' !-4t *0" = 19,:r.C#- I L: (t2 d;i* T* "1"^T'":"'v z?dm,"t5"oj':-=,"('y:11'.::,*ii*":":l':i I!. = R2drn/2. Hence for the cvlinder: f, : R2dm/4+ rf. n-,."";;;i";rt.'1J". For the disc *,'tio= =,n(+ * Lrr), ,f. = I t.- G - :n," .r'nna", (r'ii.s.ou) is divided L. L l_r:rY tf"'=rref!, I \- "l'^,(blz)' +, z"bl2l+d,:1fr1", -, *t"pt +otl?t' '{rr* I +g' I!&- t!3u9:-+ (3:]s) bods * ri!4es)' .i (3'19) r -.T J In thissection,a witi tidt.poiat-of,body and in addition A=C / selr =91e,t4r is along 'AGIf'aris 3 is a pritcipal aris of i,;c/.ia at,4, then I$= I$ = 0 and (3'19) + M1" = I$b, i-e. M,q, = I{2,o. I If z-a:<is is principal axis a[ i., then M,*. = I!,q-l- Equation (3'20) is centre of mass C and any point;{ on 93 through C for the follorving bodies (Fig-3.2): f 1- Body having two plones of mass sym11.elry rvith s3 along their line of intersection' 2. 3- M-.* = * -1f",,t291 iJ-J L l9z Its-292- Fii.3.7 (") Mo b needed for lrotion ai constant t r, since 91 , 92 rotate rvith the body- flos'ever, ,/q * a principol axis o! inertio a! A, then (a) inrplies M.a= Q". /A - Itr= 0, i.e., no rno,tr?.ent M^ is g;1, needed to mointain motiott at constotttar. Thus if es is principal asis at ? :"0 ?"f81.9*!::Eq.: A body ro.raring abour a ftxed a-xis (Fig.3.S) is said to be bala,nced if thc 9#;"7E-* beoritg reacliotts al O at& B arc zcro,:urhaa it rololes at co:nslanl aiigular ulocitlll :+ + rc -0o tngc = -mr,frsg.= f :0 -I?"n'Cr+ I?"r'*= Mc =9 i-e-;'C is on the axis of rctation and the a:cis of rotaiion is a principal a-tis at C. + If:. = If"+ azrzcr2c3 = O, I*: I=9"+mtco1,c, = 0, :+ axis of rotation is a principal atis at, everl- point on it- I{ence, thc necessary Hence, a variable moment : and suficield caulilioas oJ balancing are: I-.Centrc of moss C lics oa atis of ';1tatior-- 2- Aris af mtatioa is For angular impulse about axis qj thrcugh -'{: o pri*cipal axis at A of iticf- or.e point 1 (3.2r ) fonr."(tr.ru) = I$[a.'(t2) -,r(tr)]3-3.2 Euler's Equations for Three-dirnensional Motiou of a Rigid Body .I*r A bc o poitr.t of lhe igid body and eitbcr .4 Z C or ga1 = 9. or gs1t is alottg,{C, then tr'-r-iprincipal a:res of inertia g zt A: . r,^t= ['f + [] [I] tl3[] Chocse g. fixed to the Ma = E-^v . = [fi6r = tigid body so that i$ = E^g=*o, * * * il-e : O. = Ii'&*1 + I!r6-er+ I$n3e. * ez ltr l s3 r[l, ri:-r r#". -$Nr-l*)-+""]er *-.. Ifr6, - Q$MA" = I*zbz - (IS It{a, = i-e-, + rrr,=rnr"1s1 *Ig1u2e'+r*-s% = 1.43),t2-3, If1)t,3r.r1, a, 16^r=I*,i, - (If1- l{2lop2. Th€se are Euler's equations for ptincipdl axes of inertia at .l{. 3+ 3-3.3 Direct Equations of Momeut of Momentum if I{1= I{2 q b6ing the principal axes at 4- Every axis n(- cos0g,*sin0g3) 31 I g is a principal a:iis Id = t{r, siuce for !.(= - sin 0er * ccdg) I n & /f, = rfin1s t * I*r.zlc* I$n3s3 = (-/fi * /$) cos0sin 0 = 0' J- 9r, eqs(3.10), (3-11) yield: Ifr = If, rr,.0 + ISa2.0 + I$n3.1 = 0, I d.c.'s of g If;^ = Ifynl + I{szi+ I$nl = rf, cos? d + I$sinz 0 = If1. 43. JJ ffiffi J L:t-l I JJ JJ Bodg of rcuoldion with e, along a:ris of rerolutionSlab like cylizdncal body rviuh g" parallel to generator. If angular velocity is constant, then a' = 0 and (3-19) yields . (3.20) are 0,0, l] J J J J J J J J J J j J J J JJ J J J J J J J - kr Lil L., l-., LjLJ L_' LJ L. L: L. LJ LJ LJ L-. L_ f LJ L- LJ LJ L_- L. LJ LJ L t_- L-. LI L-- -' Li,-, L:- LL. Hence the a:<esg are.q[rcetsg t-hat9" is aligned with the principal a:cis 3 at,{ (Fig3.9) rb6eas 91,'9 could rotate relative.to the rigi(LMy about q3, stilt i7, = itr = i$ = 0. The al.grlar oeldtg {l o! thc fmme 9i dtfcrs fmm thc a;egzler oetociej io:of lhcTgid bodg in its 93 componcnt: Q.=Orer+Oz9z*Qag, + ", = Ht = Itrurgr* I-42,,t29r+/*atgj = ffl(trrer *utzgz)* I$ar39", M,c'= Lev = f.,11g.=.-,.-*o x tt^= I{rQ)r9r*t:29)+ I$ti39" !.r* r glo.cedu-.1g u2g2+ a4g3, urrgl revolurion. can definitcly_be-used for a body of Or = *ax ulr {lz = @2, Q: # -s, H^- (3.23) &c- .. 5'.i:#T".#'*of;"ai't;"?il;;"tff":r:*."-;;=';;e,*;.e-*i.s,+&^"r. '6;> -1 ' .e ) A translating system behaves as a rigid body- Equation (3.22) + M-c = Q, s!4_c,e.g1_;';Q. The equivalent force system at C is a force F and a coupte of nioment M c. Herce Mt = M-c + rct x F : aa, x ,ngcv. Hence for a translalirg sysletn, Mc ,: Q o:ad M a = 0 iJ AC is along 9cg. The moment about cocrg point is ao{ zero. For the translating system shorvp, Ma = Ma : Mo = Q, but I{s t' Q- 3.4 CONSERVATryE FORCES A force F is called .conscrualire if the *'ork done by i[ fron'r time f1 to [2, during l,he motion of its nraterial point of application from location !r b L2, is i*dcpend.ent of the poth Ct colllectfiig, thcsc. locations-, wr-z: lj,'r-ool= tJ,r:-rr, vc1 andvr,rr. (3.24) I{ence conservative force acls oa thc same malcrial point of the body and, is iadependeint of ik oelocity and time, i.e., F = &r). Equation (3-2{) implies that the integrand is a perGcL differential of a position dqpendenL scalar function l/(d. called the potettial cncryl!, such thar : dW (325c) -,F - dr = -dV, rI'(t") W\-z = -. I (3.256) dV -lV(a2) - t,(rr)I, = Jue,) i-e-, urorl' dora from L1 lo r, egaals ncgatiac of the change ir potential cncrgy from y, !o 1r- Eeoce roortdone by a cotserttaliue force in aay closcd path (rz = rr) is zerc. Let 7e be the dalum for V, i-e- y(ro) : O. Equation (3.25a) V(r)=- "+ i-e-, potential energy at a given !\'riting F ' d1= -dV , ia terrns- r!- I r.ar, Jb (3-25c) positionI equals negativeof the rvork done fronr the datum to this positionof the componenk of { and dr \r..r.t . various coordinates yiel&: f,dz* Frdy * F,dz = -#* - ff* - Ynr, =+ ," = -#, 1 -X, ,, = -#, F.d,r q F6r dg * F,dz = -{* ort Er av F- - -ol" ^ - #ot - #or, 34 ,."-Ar, fo=_fi, L=_*, AV ' :+ P, = -Y, F,ds= -7;*' Os + av. av' F=-#L-ffi-ffx--v% where F, = ( ) =i#*i#*t# =Gradient ( )- I ) (3.26) r*-*t* ".u=l+ lr" e Fe +F el ++ =r-#.ffii,* =Q ii tI ? 1b.zzy 1': Jt in direction g is obtained as . -: ;; I'. = F 'g= -gf'-e- -(directional derinaiive ot.V in thedirectionof 9). . .2i, Thus'if.F iscoasentotiocthca F--YU- Itscoaoer:sc knolhzc,e.g.,forT=fgrccrrrl, F:-gY-is, .I'-,,t notconservalive. IIF rs conscractio.th"oVx r. =g. The force corhponent t,j I A c E: Q itt o stmpty'cottt'cc'cs "s'e"' a clced path C (Fig'3'10)' w 10), W from Stoke's theorem: since for foll,ows Proof iiv x gJl zf,-dl!- (xrrl,z) tcr 'oL= [tYx - i"E.dt= I r(x Conservative Forces 3.4.1 Potential Energies of Sorne the piecewise linear curve C1 (Fig3.1r) Yiel& Using (3-25c) and integrating along or -edmk (Fis'3'12) is "'o^::!t,:f*:":!:=*:"' ;:,;;";;jffi"H, ";3;;;',:";;; -E = F(r)s'- (Fis'3'13): h€. l,u r,ou L, a,'l o2, totith Er="': r. -(Ft'drr * L,'dxz) = -& 'd(r, - r) = -Lr'dt -P(r)dr' : -F(r)!-dr= -F(r)qP 2r = -4..4,*= "r v=For datum at r = r'o : l,'r1rya,. t fc, 1. Fig.3.t). L dV = F.g.3rl gr (3.2s) F.3.3.i3 - : -GMmlr' datum -GMmlr3 *'ith For rnutual sravitatiou force f(r) 1-"1-=rTI'r^'"tf:SJ an extension'e = r - Lo' for be f(e) tlre'spring pirll on (lr). For a pair of sprirlg fo..e" (Fig-3-ralJ.t:trn" (a). rr.lrereIoisitsunstre[.t,".at".,gtLH",,."f(r)=-J{e},d,e=drand(3.29}yields Z..WP;#% eiq' 5'i.+ .-r=!;l-€i v=!o"rkta", ": (330) (3.30) . r:-.--- -6F sp;.,,g of sti{in'{ss f' /(") = &e, and a- li*ccr witlr datum at unstreic5ed conliguration 16 = f,,. For extension ke2 12- k62.'2 rvith 6 (=e) being the the -,-ields V j"l Mr = !@\ for telatite axif lvist,of 0 bet$'een couple (c). Siaitatl y,tor lorsional spnag."itl vith ( k,,02 and 12 &r0 = torsionalstiffness kt' l*= ends, V : I:t(0)d0. For a ;iicar,o"siooollsPringof untrvisted configuration as the daturn' 3.4.2'Workless Forces pait of forces rshich together are wotkless though lVe lls*, sonoe forces s'hich .are workless and some individually each may perfQrnr trcrk' . i- ft"go"ti" force F = 9gx B since W =E-'9--- qvx B'o= 0 slip' since li = A, 'fle-,.; &1;0 = 0' 2- Rr&tioa ft, at conf,acr ,P1 ivith a 6xed U"aV *itt-n" slip' siace W = N\' g' = IVu* : 0 as 3. R.eacti,on at smooth contact rvith a fixed body rvith or rvit'hout' Idsa no separation' eirher rrr :'0 for impending separation or ?o':0 for 4. Reactioo R 4L Aat smooth ball and socket or smooth joinr with afixed !ody, sincelil'= hinge gr^- -, __=_.2 .{ H:i'tr j:";-,fi:'i5#:i.:::f i*i:'i^:Y;"'*:' qej !i:!t*=,*t,0: l_, -^^,.^. i. fY ,-vv g_ N *_ ?, ' Y?r Nr P2 with noslipsince *itft n -\9-----&r w = &r-gpr +(-&r)' w, = &t'(qe. -!rp,) = 8t'0= 9-"" !!.r, =.!e'' u' (gr-.zj<{ T.Pairofreactionsatasmoothcontactwithorrvithoutslip,since14/= 0 8' /VrE,'9rr+(-Nra) '!p,7/rrr(,,'!rP. -4'cpr) = Nr('a)" -(up')'l = no'separatton' for (up.)o = o,'('o1^ 1i::)" 0 for impending separation or - (opr)^ as either nrr = o.lt*ll"oi.t*JL." . r* l'no'seRarati1".',, a$' ,tt g- pair of tcnsions in light inextensible cable since since W =T9' cr * (-7 g) 'ge Y -g-;rffi."llU, inr for impending slacluess or 9 ' 9t = 9' W for tight = T(e - 9r - e - gz) = 0 as either T = 0 e. pair of teacrions {Be,t:,lt$u':L4gl* ".rt #rt_ball and *Y i"]':: . x:":x':: s x ( A a ) s = 0' N. = F e -:::1ff::x1 ]; rfi L"i"*"Xi;"1#ffi;-i--"6 . = r r. @ -*.J=', * 6. Pairof reactions Rr-and aicontaco"f ffi;#;,: @ ] t I I I -.,,b- -) v'E .iJ --Eg. Z.rS ,< iru I I I l - S I lr I { !:' siis&€ix EeTPssIoN"or A RrGrD :oDY rccanexPressthekineticcnergyf ofarigidbodyas O"iO6 1r.19)and lpc:gifsa.'' 3.5 t rcrNETrc x rt7'6dtt rPc)d 6 = \ma/+ *gr' l:'" i:?.?: **o71i l^*t(crx gs ' s= I{ciui = I;c;"';"';l (3-31o) y = lmul + +; s: **rZ + lt$.r.; fusing (3'3) (3-316) 2lf'46r"1!' i.e., ; =i;; *itt?r.?+ $"$+ €-3 *2lf1't1,t2*21$wt'tt+ (33rc) r- lmtfq* i . L- where ri,";'1, imi,;' i;i';'i T -- lnrf6* r reGrs to principal axes of inertia at C' tf pohrt r{ of the'rigid body has 9't.= Q' then gp : 9P, = ll$-i-;' (3'31d) ia^'s-= r = i !,'?,ai,= i ! *^'1,ax=,,^|d*= fo-' l:r^* :^o:= r -= ;iijrri + I*lrtr+ I$"'f * 2l{+'tr'tz + 2l!*uzut * 2llvl-tl' r = lfli{ ui2 + t;! w;2 + r#-i'J' - i.e-. - can be expressed in ternrs wbere + refers to principal'axes of inertia aL A'' T rigid'body through A: of th6 instantaneous axis of rotation.(Fig-3-[6) otthe L , Q (3-319) L L - ra ar,\ (3.31e) For the case of 9' may or- nray aot be a pdncipal 7 = !rno26 + ll$"tz ' g = -g", ;; axis ac C' (3'31b) becomes (3.31n) RELATION F.OR A RIGID BODY 3.6 - - - WORK-ENEILGY ener&v T The rate of ruork I4l of the e-xternal forces and the kinetic 'ot a rigid body can be expressd as 'F. -l t L -L-E- +'i = tnec,b i7 rr"' e,"=dtn: = F'ec +r' ''a 1 1 *' l rr" E-' o, r'.a'ol * ! *"'u ='L' vc *'''it- lfence rve ob[ain the following uork caergy rclalio* 'i : iv, Lr x it apgdn = F - w +z' l^r" -- F' vq + Mc'q' [rncc mtc Jon:n a:l.d' in x gesdta - E- Ec : Mc] integraled fornt Tz-Tr - I'{/rr:' (3-33c) tnd 2' DenotinS. the rvork done by the where fi, T2 *ethe values of kinetic energy in configuration"-rl + I'iz-",, 1-3''33a1 reduces to consei'ative and nonconservative forces by i, *dwn", iI/.,=l'v"+li,r- -'it (3.336) (72+V)= (T1+ t'i; = Wn,r-|, I and 2' If 3ll forces are con1lati5'tfren where y,, Vz are the values of pot-ential energy in configurations ener''r): .oo"".-..io.i"rmecu;'cal ener'y (sum or kinetic and potential #". =;'",'i A;;;) ;;;; i+v =i{n., Tz.*Vz=71*Vt' t+ir-o,, F. t- L-.. System of Rigid Bodies 3.6.1 - - Work Energy Relation for Interconnected aad added up to yield relations for the oholc sgstcm Borr"rl"". tt.iri are applied to iudividual bodies the system since some of the' forces-t"'.tltn which involve work done by the intcrnal and the external forces ou the s'trole sysiem' De[o[ing the cootrilut'ion arc external for an individu-al body are ir fact internal fiorces for relations corresponding to (333) of internal and external forces by the s.bscript int*ext, the work-elergy teduce to the following form fot various cases: * L_-;, L--; L_ L. (3'33c) 44 F:3 "/lc siig no stip '"\ - i = i\lint+cst, T + Vnt+cEt = (77 Wac!n1+l"=i, * Vat+e"rzl 'i-*Var+e,,=0, 4 -Tt = Wiar+c:rt-2r - (Tt * Via**t1) = W?t,,.+crtr-r' T2* Vat+a2=Tt*Vntgutl' (334o) (3'340) (33ac) q'ork of internal internal forces of interaction are togeiher r,r'orkless (Fig-3-18), then in eqs(3'34) the conveoient to are form integtated forces is eliminated and these reduce to eqs(3.33). The rare form and provided the obtain acceleration (or angular acceleration) and velocity (or angular velocity) respective\', a priorill'- frr general, s],=tem of connected rigid bodies has one degree of freedom and. W can be obtained and rsork energr relations are conveuient to obtain velocity as a function of position. Impulse-momentum velocity obtaining and for problems, angular-impulse-momeut of riromentum relations are useful for in'rpact as a function of time provided the force is a given function of tinte. If all 3-z NECESSARY CONDITIONS OF EQ1}ILIB;11UM 01. A RIGID BODY TIre rrecesscry artd safficicttt cordiliors of cquilibrirm of a rigid bodg are.: l. E(t) = 0, 2' i{-r(') = and 3- the body is itritiallg in equilibriutn al t = 0- Q Proo-f Let q. be the principal:.a-\es of inertia at C- . Ma=A, + 'i:w=F'r'c+Mc'u-=O V t T -- lnu[ + +I?t-?+ Lrrg*i + iI&-3 = coustant -.(0) = 0: F=0, Me=O, + - - g, :+ u;;(t) = Q, i-e-. the rigid body remains in equitibrium for alt time. e{t) =,9a-6 !ac(t) = q' T-Agd 3.8 CENTRE OE PEBCUSSION e -P 'WY=\a rrr[gc(O+) - Frg,=.tSCt ,J jcy'G' - -(0- ]!co = t= I.s. + (Ie + Iar"4, I' = 0, t6 + Iq = nrlr[<.r(0+) - -(O-D- si(0-)] = rn&!.,(0+] (o) (6) (o*) -.(0-)J = Iaaeo3- (&+d)rqThe solution of (a), (b) yield: A'*r = (Ir + dllel i?s and, (") 16 = {h(h+ d)/(&&)3 - rlra- . Thus, in general there is a1. inrpulsive reactio-n at,O. The locotiott of poitrt of applicelioa'Q of e 'roTsucrse ;ropulsipi force for ultich there is no impulsiutEaction al the beoring al O * callcd,.lherccn1.;v o{ percassion' €t It -rs obtained [rorn eq(c) as [(h +d) = (rg)'= 1t$12 + liz, :+ (3.35) ad = (&$)2- For a thin uniform rod of leng[h I, hinged at the end, eq(3.35] yields d - L/6;and for a thin uniform circulir disc of radius R with hinge a:iis normal td its plane at the periphery', eq(3.35) yields d = Ri2' 2- Consider afree igid body at restsubjected to instantaneous impulse Ieed L CQ al Q (Fig3'f9b) with principal axis g" at C. The impulse-monrentum and angular impulse-nronrent of momenium rclations yield c,(0+) = dlol&, Iansc = d Ig = Iaca[u'(O+) -'(0- )] = /$ r'r(0+), + a L: 4 ix t n h 4 e 1 t : i Ieea - *[gc(O+) - !c(0-)] - mpc(O+), :+ i = Igeatn' Eence the distance r of the instantaneous centre of rotation 01 from C, at t = 0+ is grven by comparison of eqs(3.35) and (d) imply t$/md + rd=(t$)2' YctoJ that ot:- o. Positions of 0 and Q are rccdrrocclt r=u6:(0+)/.(0+)- : l I sc(0+) +'r vy ,//\ Fi xe, 9.:.rsb J .J j ]J J -l "-l illPulse Consider a rigid body with fixed a-.iis of roLatiou at O along 5 (Fig-3'l9a)' Let an instantanebus ti"n the impulsi"Sffi and let 0 ati ta OCQ normal directiol = (impulsive force) /a(0) = Iegoact at Q in a relations: of momentum impul'+momeni ar O be -IO(0) = t.g. * IO9+. Impulse-momentum and angular l- J rJ (o) positive terms is zero: Each of the 4 terms on the l.h-s- of eq(a) should be zero since the suur of these u6(t) i= i-.I .J ,:- I (d) 14 IO9,p J J J U J j_l rJ '--I -J J J -J J il-J -J - jJ J J k, U '..r I U, 'lllhile working wur J[aitmci or hitting with a cric&ct bat' if the imp11 point is a centre of percussion w-r-t. the point wiere ih" h"rrdt" ilnlta, thi there b oo impubive reaction'(characteris[ic'stin8') oa hand' 3-s TMPACT OE RIGID IODIES 3-9.1 C'enerat Smooth lmpact of two {Jnconstrained Rigid Bodies at a Point point of body I having Consider smooth impact of [wo unconstrained riSid bodiej (Fig-3-20a) so that '4' plane at is called mass ml makes impaci rvith poiot B of body 2 having-mass ryz-The common taugen[ "t to the g, normal aL A direction The planegla.ae of .impacl. I*t.g, e1 be orthogonal urrii vecLors in this impagl b called ccatral impacl if lhc ccntt's ol mass G lie on ltrrr" of impact is ciled line of impact- The 'rol-cent6l impact.. The impac0 is modelled to occur in zero time' the liae of impact, othervrise it is called L,eL O be the point fixed in space which is coincident rvith / at impact' i just Let el and 96- = u;oe.n+uisq+rri'gr be the angular velocity aad velocity of cenlre of mass of bodv ui,S, * u.:.e5- \4re before the impact- Their cbrresponding values just after the impacr be ari' and li. = ui.g." + in direciions body of each momenta The set up 12 equations for the 12 scalar unknorvns irr: ,Al,C2,vb1,!.1,9, 9l are conserved, since the external impulse ou each ii only along e,", * TJ tJ u u u u t-. tJ LJ LJ LJ There is no external impulse I{o 1-.- *m2rt!2o = rnlutn * along q;, is conserved: tn?u?n. ; (3-366) ofeach body is conserved, since the external angular inrpulse on each about fi-ted point O is zero: v'on t- - a!^o (3-36d) = -e(o p, - t:*)' ttAa:9a -9, = &c', *g, xte6r)'ea, (3-36e) iu*n = tB:gq +9zx taCr)'9*={Slc: If body2isamassivebody.thenrn2=cp + 4=-;t, s'c.=9gr,t"go-ularand(3'36b)forthe sysiem of bodies and (3.36c) for body 2 do not yield aontrivial equatious. The slt scalar unkno*ns erl, 4r ] are determined frorn (3.36a), (3.36c) for body i and (3-36d): (g'gZ) u'1, = u1,, oir = ure , ELr+yrs,xwtv';;9- = qrr*!crox',r,t".:11 -:" -::" =-",(.q.ea -"x^)If body 2 is a massive body,,at' rest; 'then' the last of the'equations iu (3.3?) reduces tor,ufi,l.'= -eurn ' 3.9.2 Surooth lmpact of two IJnconstrained Rigid Bodies in plaue urotion at a Point rr'en=de-9^=$br+sal xL{cr}-g., lubo = ts - % = (!L2+dz x rac2) -9o, : rvith u u L. L. LJ LJ The six postimpact unknowns (Fig-3-20b) ,u'ro,llr,rtln,{2r,'-t'r,tt', are obtained using (3'36a,b):- ,i u'rr = ur,, ,'rr7tt2rr mro'r..t*rr!. =tlI1u1. (338o) *mzx?a and (3.36c,d) for the configuraLion ot'Ct,Czshorvn in Fig.3.20b: L- (3.386) + mlurnrl, €* rL - rn2v!2nr2 = &' rr- rnzu2n rt' obn-t7o=-e(oao-ue.), i.e., (ul, *u'2t2)-("t -arirr): -e(o2o *ur3r3)-(u1. -&r1rrl'(3'38c) since for the given configuration u,tn = uh -(rtrl, aBn=ttzo*ttzt'2. If body 2 is a massive body, then rr2 = oo * dz = t tz, *z: \2, utrn = !gn, and the t 1st e9ua!i11 in (338a) for the momentum of system of bodies aud, second equation in (3.38b) for body 2 do not yield nontrivial equatious. The 3 scalar unknowos ,1, l/1o, u'1, for the configuration of Cr, Cz shorvn in Fi8'3'20b' .io^\ /e 2aAr /a aa^\r-.^--:-^J r* ro t mZVz+ (3.38c)1,(3.386)1,(3.38c): --^ determined from are iSul, +m1u'1rr1 LJ L: LJ I L.. L. (3-36c) oir = impact and e is an experimentally deternriued constarrt, called the coe$cicnt of rc'stitution' Hence t-- L , ELr+rcrox ,ar/cr=E-cr*tcroX n1u6, + I{Lr*lcpx 'nlu'rng! =E-c'+bPx '}rtulnci' (3-36c) ttt?!'-c2+ EL,*lczox ''2''2,.e'=Ec,*lczox EL"+!c.or.*r*r=Ec2*lctox l"u"t'' where Wr)=[rc Jb,], Wrl-- tfctltgl'r], etc- Equations (3'36a.b,c) constitute 11 scalar equatious' Atr of crnpiricat ilalioais added to conrplete the set, viz-,e - trfto..vhere:u, is.the't'elocity'of separation just g. before along B g from .r{ along E just after impact, uo is the 'velocity' of approach of A tot'ards t-.- L: = ur, mrui. Lj L-. 0rr, 4. vr, , u'lo = v?a' = on the system of ts,o bodies, so their total monrentum u'r, tll - I$-. ,i' 1 7 ii i j'z LJI /be - -n F;3.3-zot) +8 --J{vza 'n. V- l I l r-I ,^-J i r..' (3.39) $u'r+m'o'vnr1 -- Su1+'rtlulnrtr trt')J' *uzrzl orirl) -,(". (u2- *cr2r2) - (oi, = -c(u2. to u'1. - atlr1 = -e(r,L -4111): reduces (3'3?) in equation last the If body 2 is a massive body at Eest, then Bodies at a Point 3.9.3 Smooth Central Impact of two Uncoustrained Rigid the angular velocities are unaltered and usiug Since the impulsive normal force passes through G, ,'r, -{3-36a-e), the postimpact rralucs are obtained from rr'r. ntoio *m2u'2o = mlula = or, u',, = uga, oL, = pzr. lie = Qt, *'rR21'3n' a'2o - ttln - -e{oz- - ur, )' (3-40) ,l.he velocity of the centre of mass c of the system of two bodies remaini constant since the orternal impulse system is only due to changr: in g' component on the system is zero. Ilence change in kinetic energf of the nr:) = (m1u'1, 1m2oiol(mr * rnl) l: .rf velocities of G relative to C I rca =(*rrro + rrf,-lrz)llU + Body 1: A?r = |nr1[ri; ]*,[',, - I!]+*#41' - Wl'- -1 :.}.#[,i._oi.},'.-(o1;.-,,.),]=,ffi(e2..tl)G'.._.t.), '- arr + alt: iffi(c2 3.:o< isGatled gcrfecttgclas{icimpactfiorurhichthereisoo =+ e3-1<O:+ O(c!f.T\case€=f pciccdlg ?lastic impact fot *Lich +he lcs of kinetic energlr (A?r - O) a$d ?, = ua- Tlre case,c = 0 is called loss of kinetic energy is ma-xirrrum and u' = [' and the consenratioa ofrmraentum For impact of body t with a massive body 2 (Fig320c)' *, = k, -\f <0 to in s,, does not yield a nontrivial equation' Uence (3'40) reduce : oz, u.L, = 'l'r tLn = u2,,' 92" to In particular if body 2 is fi:ied. then {4 =?.tr = O and {3'41) reduce ui, = ur." ,i, = ,ar, ,1, 0, l2o :'0, 4. = 0' u'r, = ur, u'tt : ut6, o!1, t)'rn = o"n = -e(vzo - q.r- (3-41) (3.42) -e7t1u' = 3.9.4 Other Cases impact A has zero velodtY just after If body 1 impacts a perfectl-v rough fixed body 2, then the point of of nronrentum about o: irnpact. The postinrpac0 angular velocity. gr! is decermined fronr cotrs,:rnation (3'43) rvhere l7bl = [ro][4]' Hb-,= E-c, + rsp X rnlucr 1 :: ttre rigid body I' :r: since just aflsiimpact !r+ = 9o: 0, i-e', O becomes a point of of each body shoul{tbe draws shorving FBD',s point, then lf one or fhe oiher body is constrained at some points' The postimpact data is obtained all the impulsive forces at the point of impact and the constrained of momentum eguations a'boilt centre of by writing impulse-momentum equations, angular impulse-moment and the coefficient of restitution equation' mass (or about a fixed point, if any) foo 3.1O GYROSCOPIC COUPLE "Jbody, inertia at o equal *' = 8'' 1. consider a rigid body with a fixed point o with two principal monrcnts of g. itself precess at coastant Y" fg and L,eC the body spin at constaat ratc sc3 about body-fixed a:<is 3 about O reeuirld, to maintain this about an orthogonal axis E fired in f (Fig.3.21a). The mome nL e-o acceleralion qif of the body ane motion is called ggttscopic o11plc. The angulapeio"ity 91 and angular @ i-e- orl - pE+ : = p, u2= 0, u3 = s, t.rl :.r3 se3 = Pg.s * se3, tir = P9t x sgs = -psg? /i cquations of motion: = 0' r'r2 = -ps' Applying Euler's 3 +fl rt<t ''J ';-l ,-J '.J I ,-J ,i'.J ,J .J ,J " - t)(u1" - t':o)"' :-J 99r >p J J L-J ,::J + t3J et_-J ri--J il -il '=:J ilr .J .:J J J :J il;J iJ J - )' o :t - l' \i E o directions' y = pE- s =sg. The gyroscopic couple is' surprisingly; normal to both precession and spin Hence 2- Co;ds the case of precession axis I at ao angle 0 l'o the spin axis 93 (Fig'3'21b)' = -pssia9s' ut = pS+ sg- p(sindq *cosaq3) * sse, ,L= pEx s*, = p(siuast *-:*':] 1"* eqgations: g--tt=tt*a$t-pssind- Applying Euler's i.e. <.r1 = psia0, on=0,@3 = s-+-p.cos tr,[o, = fr-t Gr- E)t :rs = 0, Mo2-- Ir9r.r--(€ - Ifl),.l3.^r1 = fflp2sin0cosd - 4(u+p"*O)psin6' 7[ot= Ea" .. (.[f, - I$)urp2 = a, (3.44) i-e-, Co = Mo: [Iflp?sin0cc0- /&(t*pcosd)psin0]g:' rni[r'q1z:-":l:' oI 3. Ler a Uody with ifl = I?ztoLale,abour fixed poinr O (Fig.3.22a)- We s''rsh to find the cos &in (3'44) , "+ r^r3 + g and = <,4 so . that preccssioa al cottslottl . is possi6la :Usiug Mo = -rrrgh sin ds2 l -s rvhere - Lr- L L.- flflrp?ccd- $r,,4p+mgtlsind - 0 + iL- > tj I L. tj LJ Ir t--- L-. LJ L-. L.. Pt; tz = .llllrngrlc()ss\l/zl ?r*Sh cos 0 ', r7: :--fr:-€'t ff".*t,t(r-:W tIg**:a It LJ Ll- F.8 Eg.=.z2cr- (") t FortIrecaseof-!>lt|,,nghcos0|(I$)2'tlresquarerooLternrisoipandedby.binomialtlreorerrr s9S retaining the predominant leading terrns, the precession rates are obtained as: ?E * A a-,'4 pt= n.sh/(f;33-il! ' p1= I$u4/(.Ifl1cos0), ,-NfY^ ^ - (, '' ./i,"";g The relations of tlr'rs sectioD are valid for a spizrrin g projcctile*itr, o "ra nrgl replaced by C and F.L, tvhere F is the aerodynanric force acling.opposite ut =' 1, -'i.r.zb :tr;:ll'xff:ffi;"j."ffiH"Jj.:"$;l'#i'.,o" :, r il3n+.rrse axis I The spinaing top shown in Fig3.23, spins at rate s about its axis es, its precess€s a! rate p abour the fixed vertical axis g.r ia the vettical plane through &, gs. The angular velocity r.r and trlo ate grve[ * &:i,1Y:::.1::Td;*:.,o.. The moment about O is Mo - -mghsin8qz. ?+ v = Since Ifl - I?r*d \\bAe*O kv tgr-' i!i., by z i I ! i ! ] OC'h j*gg. -"'{:3 t Ee3.r-3 t Mechanical energy is conserved since forces..".or15gF.ative: rll?L@2sinz0+d'?)+ Er3l *mglrcos0= constant= Mo3- 0, the third Euler's equaiion yields o3 8s constant: 74o"=I&,i,r-(IP, - $r)u,'.o2-_ 0 :+ cir3 LL-. l4nsht$lQrlccsl' _1 (3.46) that the sleepiag top Tire condition (3-a6) is delinitely sat.isfied if -3 > llnghlfr/IlrJ. It follorvs from conslant 0, is follo$rd (d = 0) is stable it t t! > l4rr.ghlfllf$!1. For a spinning top, the initial precession at at some staqe condition by rvobbling rnotion with nutation because as r.r3 decreases dire to inevitable friction, /f--\ rc i (3-46) get-. violated and precession at constany 0 is not possible' /", If condition (3-45) is'satisfied then the trvo possible precessioa rates, p1 (fast precession) and P: (slow pfecession), are given by the roots of (3'a5): l-- t-j (3'45) 0; or fflp?cos0-I$qp*mg'['-0- of the quadratic equation Thus tsoo r.alues (rools) p1, p2 of uniform precession.exist provided the discrinrinant (3-45) is positit'e, i.e- if z (3.46) u! > i.e., if 4(I?rcos0)msh 0 U|*i' - - sind = 0, i.e-, d = l =0 + qa3 Eo' t' ' =constant=os(0)' - (1) (2) t1 t a 50 I i; ,il. '-r 0' Eence che a:ds f4 is zero' sinle {4. fo = -rrrghi*e*'Es= The moment about tie fixed vertical 0 is constant' . abour rhe fi; verticlt axis E: through momenr of momenru*lg, E"l ,+ Iffpsin20*I&r.rgcos, = /fo.(o} {3) = consL. ffi@sind9'-tiq)+se9.l.(sin0q1tcos09") Eo.b= OF .A' RTGID BODY 3.12 TORQUE FR.EE MOTION priacipal axes at c' H (o) moment N-c =q. Let g'be the consider a rigid body rvith external l.,LioLh gc(o) = const' {i.e * E-c= .f1-1g1 * I&rrgr+ rfg!'atu = iavori<ibrcThe constant W =g vector [6 provides an izacn'a6le lite in :-q=F'2s=ig ,.!-l;rig,Il,: + f (Fig'3'2a)' T17=T-fc=const' + =;"; * w*= ;*;;A; \Hcucap = coust' pt.na * C ..-^. ct ucospA - cN : co''st' vector r'r drarrn from C lies in is-const'ant and the tip of the as rvell as directiou' lU' i"-a"t*"f ' g is variabte in magniiude invaiobleplane norma I to H.through i.e., the.rotational'kinetic.energr [1 3.13 ToRQUE PREE MoTIoN OF A *:- !. 3rd Euler's (1) + eq: BoDY \I/ITH I?' = $'f '&.- + Ec:/fiitte'at'e){r$"e"=consi'=[r,(a\ Y=g *--"'*=ttf'-('4i:id)+r$'al1=const' Ta =Mc-!*:o'" ' it + -3 =' co4st=0 l{ca: I}r-'-1'fi : '|)tt:'-=^' G) * 'i+'l=cor'xtii:Wftd+-3)itlg1.(,3=consr. the (l) Q\ (3) (4) (3).(4)1-2-=-?+-3+-i:GotEt.(5}o=arclan(ai+u$1ll?1-"1:const...(6) :corrst.+HgL)cos|=cotls!.(l),(5)+f=coost.i.l (2):+ZTa=fi.C-- (1)'+'t']tt=(J'rrsr*..'eesl*-'l -';--';'*"+*'3q =ff*ffi**' * : s: Per{* sea' Hence o rvith l ) u"t t?" : t':"t :':':!l.3' (6) iaclination trre body'fixed a-'ris g" and *rhose i*crinatioo a is a vector of constant rnagnitude. -rvith Equat'ion (8) shorss t-hat Irc nxed in i'ettial frame f are constant' o ' p with the invariable li*e es along boriS--fxed isintheplaoeofglyandg"andtheiirctraation0ofthebody-fixeda-tisesrsiththeinertialspace-fixedaxis ;:l,f-g"."" trtt motion is as if the ffi;;; shorro case rhe for constaut a'-xis e,.with { :inertial ss rernaias cone 2 of angle f on.* *ti, ."*tott **, a-,cis,qj, *ith a gtlRace-fixed cone I of angle about atis p * along s'itlr spin s of the bodl' The nrotioa consists of preeession of -ri= ", "ootlt initial dat'a for ..r as the determined abinitio from f... totque for -Jion "*1t The various e$tities rdb;:";;;il ffi **i*:"ljJi;: Ifi-,g, 'E-c + = I[6rsin0 = Ar'r.r1 +l + I$t.ts$ = Il6(sin 091 * H'scosl:= o,c" de3)' I{'ql$l"ua)' = 5d:Pcos0 = $ccosilI{s t,c - t{JIIccc0lI!" s=Ufi-$slnct*"':" regtlotprccessiott(Fig'3'26)' the body is said to have rctrosrcitcr''-""i;::;'';i; -t4,4, .r, 19 p..! tand = ar3 .r=<^r1/sin0=Hslll1, L] cos I?P* = Psing = Hc:ilnlll?r' , ForIfi >I83,s>0andthebodyissaidtohave + (s + = ig' u,fis,/'o Pto". er\ Wu 5t ("} - (6} ta tcoaod -^ <Is' -n--, ForIfi 'i if : ,:::.,":::,:,:t:"i:::,-1ff-n" = ,W,/ ":lli.. "f P- --. j ); {. ti r :r -.I --l J JJ JJ ,J ': I J JJ J J J J J J J ..J I ! J -J -J -J -J -J -J -J I -.J S.J ) 1.;"'1-,: r ) f-),J -,.:..: btirr:'ivnonr.r norerroN' or A RIGrD BoDY , ,,1i of insrtia r-t'-;'d' *" g where P::":Ptl,axes ,uo?, ,t= *.*,angular velocity ;rra o"art."a*a srndt distur-baa1 is giv6n :' warG to in cstigit€";hether *.u,,."io" bounded if We at C and Mc.= O tothebody.I,ettheingulatvelocitybegr(t)=.,,o9r+e;(0gidrrdng:,Y"*motion,thelalrreberng ;J;-i'6i;;;;;. the dbturbace r" 'ppii"a Jii " <r"o' I.ence s= €is:ur=trg*er' -@2:c2, &15=€3' til=i1' d?=i2t r!'s=€' . 3.14 -'-;" srABrjrri ,) I. L l, _-- -_t-lr-:.. Neglectingtermsofsecondordetofsmaltness'Euler,sequationsofnrotionatCyieId: ''. Mcz- 1,9$2-(rS-S.}15r,6=+ --$rir-(6-frlt"(t'te*er)=0' L. L rL Mcs-I{"ar-(Ilr-$.),tp2, ' r$'"-(fr- *' €'a-(€'- t?**zq *0' {2) 8rl<'1+'1)1'=0' $'a-tf'-r3pe2:0'(3) '"'r:' ' (1)+e13e1(0).Equaiions(2)and(3}arecoupleddifferentialequationsfore2,e3.lAreeliminate€3by forming (7I *d using (3): (4) * tqi', rg:Xr{, 8z€z- (rS - rf,),oa" iu _'g':'..'?'i'j:' eq(3) implies is harnrc rnic (bounded) [ *d hence :0. + i L G) If, * €r, $r * I&r, that ea The"solution of eq(4)'forez coe$cient of ezin'eq(a) is Posiiive' i'e" the is also boundedt if $r> I?" aad If, , ts IIence, tlre rotalion of a rigid body o6out rS, a : o- - or if rfi < r$ and : t, i:: < I&' ' (3.48) of itcrtia is {rtc lcrgesl or prittcipol azis fot ulrich lhe ttomellll slc6le(2) + (3(u)' Equation to:":'-::1.L:* e3 c3 = e3(0)' I$: Equati'on (3) + es = 0 :+ tr9, If, { &) .h^,,r. ,, = arcis et ls unscable' rotat'ron aboul e2 builds up with time and the e2 = e2(0) + (Is - Ifi)r.roes(0)t/s- rhus Equatious(1)-(3)* i;=0 + e6=e;(0) +thetotationaboutanva-xisisstable' (.) f;, =Ilr=I$: TLYING CIGAILS thc smallest'' is TO 3.15. I.LYING SAUCER.S AR.E STABLE COMPARED a-'ris 3 (Fig'3'2?a) ended uP b-v rotating An early cylindrical satcllite iaitie,lly spinning about lonSitudiual The the s'ell-established theory of Sec-3'14' about transverse axis i 6g.f-eZU1, "ont."ai"tinl a."*"ti"aly is body The rigid bodgliag. thc objcct wrongly as a e-xplanation of the actual obsertation lies in ,noacl the'rotatio[al'kinetic energy ?a is noc conserved deformable and though E-c iscooserved "irr"" ub = Q-, because of dissipation of energr rithin the body: (1) - E lz = flZ t2I& E-c :€".e : Ifru'e, "n(0) " constant, tlic body ends up turning about 1" 1*it ?.6 keeps decreasing due to d-rssipation aud since ffc is the dehomiuator of gxeressiol (l) for rrt -tl'"t for which ?p is minimum fot giveo I[6:, Hence it follows from ubich the pritcipat m:o:i: is flre marimurn' eventually the body would c'|d eV rctalittg obovt an n'i" 1o' By the sanle transverse atis I since lfiEence a cylindrical satellite ended up rotaiing about a -19 ' in a-tial rotation rvhereas siucer-shap4''bodi"t ; reasonr cigar-shqped bodies with I$ < ffi aru not stable with I$ > /fi are stable in axial rolation' p i- Tau' Sa.rCc< - Sr'.a?eJ I r'\ c () 5A i t o(ga *r* Ar4(n ;},rff -1 tr's'32s ':: :A.i'['ii* o)V' ,r i fixed rr.-r ilF' pu"sti .-o"iihroutr throug! 8nx"a that . MorroN ,ND"R .ENTR.aL 3.16 Eo,.cE tr T -.-L cetttrcl !o'ni [*"h Consider a body (Fig-3.2S) moving .rrra", "'*t".n al and 3-17' for of fot,,g' ln Secs' 3'16 p"i"t the,ccilrz' o i.*"0 point o ia ireriial frame r and F tt r". frame f' ire.'ity, rve write (')1y * ( ) and !L a are rv'r-t' inertial + Lcxtttgc=lcxii*=6@=9 (1) :+ bxgc=llbl*=b=constant' + is its value pet unit the mass is concentrate d * c and to all as if mornentum of monrent irthe rvhere E-b thc t*iectory O wlrich o::*"t * t-.-^1:"* Equation (1) + tc-L fu, i.e-, s6 is in a ptane through FIlcc :+ 1"xF=0 lcxnlc=constamt =Eb, with Choose tty-axes in this ptaae iu-oplane passiag tll,roug-h lhe ce-alre of fotceO' "u*" fu = lrse" origin at O- Equation (r) + rc x gc = rg. x (i9" +rC+) = nrass. of C is a plane "!e'= (3-49) : constant (3.4S) i t hol'"' v4 = hslr' 0 :* O =coustaut', i'e'' the trajectorY of C is aradial Ilence 6 retains the same sign. If ho:0, then e= r'ector of C fronr o is given by by the line- Thl rate at which creo ;i is srocpl tris'i'28) ' :*ttt:: (3'50) ;:,i$1t = h6l2d't - r"di,lz,' :+ ,"6 =roo = *- rouco = h6 rrnotion't,, is Kepler's secoud larf of Planetary Ilence the area is srvept at a constalrt rzrt'e:'This 3'ITMoTIoNUNDERIN\/ERSESQUAR3GRAVITATIoNALFoTicE 7- of F as ${ n I r2)g'" the potential srefSSI F: if it -(G For the- inverse square gravitati,onat central force telation for rsork-euergy the I{eace is at c is acting -cttmfr- c ' moving at speed o'becorrs: f"=lv._-r.,:+Tc+V-=const.+n,,i'{2-Gi{m|r=,aifi_cu,a1rg:Q-. e) +ZGM/r -ZGM/rc' body escapis to infinity if speed Do at occurs at r^6 and u-i. occurs at r-"r. The u? ='uE i ,orro*" that u-* 1= oo is ftence ) 0, i-e., if ,L :4 +2Gn{(L!a- U'al= ol-zlt{/'b > 0' i'e" n "oi {2Gltlrdrtz' giten bli inilependent of its dfuection''is the rnr'aimu m csca?c spccd t.to escape at rs. which is 1,=(2GMhiltt2' For circufaror6i{of radiusrs, i=0. f ,n(i- rods) = F, = - Equation of the Atajectory "3.17.1 of the -trajectory' in ;;;ion equation of motion using S= hol'?" ' moti'on =0' u=r{gd =uegoandcheequati'onof 1,. = 16| = {GM/4ltl2 ' -Glfmlri + (3.52) C r', f by solvirg the coordinates, for tlre ge*eral case' is'ottained, :='' =':.;r:f'- [ilt3 +GMl"'=o: '' - t#) = F = -GMtnlr?' Llr and using 6 = no1*: u The nonlinear equation (3) "* be sirnplified by the Crapsformation = m(f r=*(ij Substituting i from =-"+ =-)##-=-oo*,+ -lrfrus + GI*{uz =o d?u :+ G'M @+"=a?- Thecomplementary,particuIarandgeneralsolutionsll3,ltptuofeq(5}aregivenby u. = Acc{+Bsi'o= Dce(6-l?), vo=CMlhi, giverr where A, B. D,p are constants' Ilence the trajectory is -$3 (3) Lgt?: i=-ho##=-u""#' (l) in (3) yields a linear equation in u: -trer'# t3sr) by' tr = u.*up, =G^tltfl+ (4) (5) aco{d-fl' ,l ,J .T JJ -T JJ rI L/ -j $, q) -r* t- LJ I \w cent.o'a.l h /b a .\-. Af * : LJ L. f LJ L l- tJ lj (6) tw' L GM +Dccl(4-01' . n-*rt-al L=\E rh6 from the radial .o TIo ,r o! =0. Ilence d' is-the polar angte measured p,then eq(6)-+ positiooatr-;.(rig.3.29a).Forsimplicitnwedenotec.bydintheseguel.Thusuith$mco,s$'Elfmmlhe LeL6- -_ 6 _ ;;';;r;;"i "*,.,'*I IJ l-,j l-- IJ IJ L L t' L L. u L U, L,L rL "qrotioo of tLc ttaiu"f:ryj;"*r. , (353) ht, (3'53) w'r't" time an'f usrns "::::': The rodial speedlsobtained trv differentiat*:I !:'o'*-t f $ = ha:. (3.54) v'=i=lrsDsin{' + -il.=--DsinCi consf,. D from f-^. O) 6\ //frlist.ance (distance tof P from line .[) = haD|GM: (dbtance of P D fllll -rccsal =+ of fotce'direct'rix at distauce ll D trom (Fi8'3'29b) witlr /ocus-at thc ccttrc Hence the orbig is a conicsc(:,t('n :' (3'55) it and eccenlilcidg egiven [l5 e = ttiolGt{ - Eq(3-53) This is Kepler's first lErv of-plarrt.tarl" 1uotion' of .xl'its The on the value of Es nrui - e--trf,o1t;tl l oa > {2GMlro)rl2 e)l os= (2GMlrs)Ll2 e=l q <{2Gt{l1ttl2 e< I tao t- L l_ )u -- -- L- Ix os ;;t ---l rs Eo>0 escap€s Eo=0 Just escaPes 6o<0' Bo<o' = (GMlhlLP -GMnf Es = -Gt'lm/2o Eo= -G!!m,l! I L closed orbit closed orbit I i ; of senri-major a:cis a: ffi..ltip,r.icorbi[ ;:::;;:::'".r *, = *. :: ,'= o 1 ,:'=;"i" :;l T::::?:".;lx;;11;:;: n"llz flUg=r2U2:Ir9(o) Gl[ntr2' tnvtt - G l{ n lt1 = = ntfr/Z " nl'z + ul = "-:.:"'.:;;.:,--.r,ri,, = Pur u2 from (b) in (a): rnti/": -GMmf4 =n(tp1fr212/z-Gl't'', ^ fu= -GMm,/(rr + "z) = -GtIm/2o' D Put ul from (c) in (a) : E'o E6 (0) 2Gt{r2l;1(" +'z)'(c) /---fft7]>- vru'e *<-re gilru-rT t-;ttrs<u Orbit tbr Cltrs<d s.L7-2 3-L7-2 Time Period for i" \ -r --_: ...-i^i rnrl qer . .,,, Thetimeperiod?forelliptic.,rlrit(Fig.3.30),ofsemi-nrajorandsemi-minora-\esband6' I(3.56) 'l' - (Area of ellipse) / (Areal velocitl') = dabl(hs/2\ (r) r)c.sd = olffi*;ll i=#* ;e*o= rou + t*i" = (3.s6) & (2) i.e.,* 3.1?.3 =+ = Irfr' ' * ZiiI;' |('-"* * rmin) = e D1f.:1 (2) .'-' (3-5?) 2;.a3lz -4,a)t/'= - Lo 'D, -T T =Zroslzl(GMl'l', o3, which G Kepler's t 'a [-a(l- "'l'l', c?ltlz ttrid l'arv of planetary motion' of Initial Data Parameters of tlrtr Ortrit in Terms oc r=r(,;-,o)' - 2r;ab= 2rq2(L'l'= T Ftg-3.10 16' us' hs hiD , ,ttz_ \GMD, (3.58) os I TbeparameLercGM,ft6.I)arecomnrlledfromtheinitialvelocityugal,radiusroatanaag[ec6rvith position from tic minimunr (F€'3'31)' The inclination co of the initial radial due to gravity at the s*rface of isdetermined from the acceleration I radial poeition is an urrt no..'i- CM : nt!! * GM = 9-8?" Eeuation (3'a9) =+ earth having .rdiu. a, d Mtnl lt the circumferential directiorr so 5+ ! 116 = F4uations (3-53) and (3'54) for rs are solved = rouCo - l/rs = GMlhe+ Dcosdo, + uro = uositlco = IroDsiaC"' "=[(* -W)'*(H)']"" D=lu"o-GM/hal' (3'60) ,r#M' {s=ar*a. (3-5e) = rouocGco' simultatreously for D and $6: rP6 * uro=0' (3'60) * Eorlaunchingparalleltothesurfaceof theearth,o3=0i using previous results: once GM, Irs, D are ott4irred. then all required-entitiesarc-computed ho 116 -' Gi{ n I GM :,,1---- = t ' t rmin ha 'io'o oo3(l - e21u2 11no1z1' and if e < l' then ?' = 'obf A= '-t'*' * t-'')/2' e: ttf,OlGMt o= (r*"' cosd- ftlt-GMlhlllD' wittr u'=heDsin{' u4=h6lt, Atgivenr: fr(,m ,Irvz 12 - Gl{nfr = mufr - GM*lrs' alternatively, u! = * - ui, .r,here u2 is obtained- \. i earth iu a of:tbe velocitl: ui relative to the surface la:U:,tchirrgposifiorr is at the equator $'ith cr is the angular velocity of the'6arth' direction West to iEast so that,.us = ,,* a.,tol, *.,,lvhere is The &csl '*"';:j';';ili;;, . *".,.,". .'.to"itf{ior the area (Fig.3.30); the tim.-fes rraverse from B to D, since tle areal from t1a :: "i:1::thb'time'?5pO: ss'ept b;--the eosltion vector w't't- 'point .-1 : $reaOABil) = laobl4 - b(oellzl!(h612'l = lt - ze:!r!12;abl(l t!h:12\:."!t Similarly,Tso=T{t+%!ella'TealTao--(a-2el1l{*+2e)<r' T.te to %l*i/a' GeostationarY satellite b'a circlc a[ou{ :\'-5 relative to the eart'h- llence iUs orUit Geoeta.tionary sateltitc has a &ted position tic cct,rl- of forcc trajectory of, a saGellite is coplatar uith azis ttt'uc*cl ot the ralcL.r of [he earth- But the cqtatoial ploac lhc i* r' r ortit ol ndias o at the eartb's centre. Eeqce geoetationary satellite has a circzla and moves ftom West to East at the rat'e r'r: km' r" J (Gl[lutz1rls - kfr luztl3:'{1000 + ac:,trtfc- (GMlr.)U? 3.17 4 3-1?-S alansfer Orbit should be requires change in Eo' The transfer Transfer from one type oforbit to another type oforbit for achieved *hLre ttre greatest increment in energt' AE'-is executed at the optinrum location in the orbit (i.e., for the same elxpenditure of fuel): rhe same change lAgl in the magnitude of velocity 5f,_(vatueof81inthefinal.orbit}...(laIueofEointhe|nitia|orbit) As)2.- 3.?1 ='-ft:4, $? + za g r&1 mu],G M m / ) - lm(fi -uf ) =, !rn(31' + r], i LI G (lmrfi l ! -( = a location u'tcrt 15 is maa-inzm' i'c:' ot the should be large, i.e.. Au shottld be alotg 4 and'al Au.gi Hence localiort of the minimum radius r*;t' a Satellite 3.17.6 Effect of Atmospheric Drag on the Orbit of 1\'e consider trvo cases. ! ---L -: rs- Hence a verl'short time $rhen r- 3 fmin 1. The sateuite is in clliptical or0it and is in the atmosphere for lsc applied ct r ='ro $'hich causes instantaneous the atmospheric drag is modelled as an i[slcn la7lcots imptt in t6e next orbiO (Auo is negative)' i change iu speed from us i' the initial orbig to uo *iAuo Es=\flir2ol2-GMn/rs- -GMnl2c+ <0' AEo=muoAos= Gttm6l(,l2o2 + 6ia'=2o2rta6oolcM tends to a by the above amount and tlie trajectory Hence the'semi-major axis of successive orbits decreases circulat orbit. 5q_ JJ J J JJ JJ J J J :.-I J J JJ ..!-'-- :,7.,;= <-;+-: : : :' ;,":ri: . :1 :l ,--i l'- tLT L_- L* LJ - :-: -ra. ' -€-i:!r:*' 11j-- foict F"i1 atmosphere undei-drag cir<rla.t otbit of radius r-in nibtly moves.in.a 2. a\esatellite ilr= -ZFlmo'/ -(t) o{*ll.'r2 =-F,-' ;;'-t-+;: speed tu= Eer'ce rdecreaS€s and p-Yiewa *lr - -2clm' cu' t'l potcatial drag, viscous the ""u For the case of linear ttis occurs-sine the decrease in atthoulh thete is ;;g';;t;' increases, (GMlirl2 a= by f' mot" thu" the negative work done * -GMm/2r,'4:r''+'ri' : ;; TO NON-INERTIAL FRAME M 3.18 MOTION RELATfVE of the centre of mass C of Coosideranoninertialframe(m)havingangularvelocitvurandangularacceleralion{w.r.t.iDertial TLe uot'Jo" oimotioo g,nu'('';'s.t2)' has accelera.tioa frame r and its point A a bodvot I',r[r^r, itx rct+'- x (sr x tce) +24'x ecw + ec-.l= tltg.clm = + Thus F -mgrl1t u: f- * T; - trt9_x'la^ - t,i+'-x [; ;:l gct*: {gx lr,i) -Ztw-x pseudo fforces' physical e-tternal fiotce rngclm equals the sum of actual orrrame, angurar accer.erarioa F and four pseudo (3.61) 'forces' rvhich are due to g-?i:-:t:Xil:::ffi:;l,JiTt:I":":ffi:JI accererarion . ' frame I attac.hed.:,9i.:the-ccntre to tlre:earth and 1n inertial attaclred (,o} ftame angular ihe at non-inertial frale m rot"t* Corrsider a stars €ig.3i3"iat under gravitational force "n" onry and with its axes to*rards of the earth to frarne velocity sg of the earth relative *d other external forces'l' Let r: ; ' "rn="a'g' Considei o' ooa' " a= ";; (3'61) becomes "-i g,;ln'.4m-- gcla- Equatiou 'uo' *= (U x m4^= L + z-- rnst;,ll - nry'x (el r) -2ruex9lm' gravitational fotce r = O'g1* = g' q' = 0' the (FiS'333b)' O o:Yo the at 1. For a stationary plumb.bob t2) eolrl= -ri9' + g<= !' bl' -mlc--= :+ L= nsal! -lE<o-'rrsold g= ho +'L-itnrd dLection general differs iu magnitude acoedration .1 g, *nhi.h in & where 4 is the effective gravitational from g-.At 2 Forsmallvaluesotl, w_e -"R)e':' x24)41 =1 ar!(<'r' & -t,'ott=d;a'1*i:T-A:itcetn=2rl(3600 obtain tl",rt" c"tJl"[s d;m in (1] and tt" "qu"to'] L-= -'n- 1--'.nt'f -- -@ compa."d neglect rhe centrifugal term f,.En = ?-+ mg--211'!4x then as given by the plump line, , - ..---^, only the coridis pseudo'force' of -Zttlt't x 9l-lo Ecst according to [Ut dit*tion ia North3lrcmisglerc and to we have{o iodude Eence, if rve use a tleoioles l- A. bodg ia free oerlicat fall ot the c*zator, arc dc!c*!-d t.o ;;; 4- The sinds btorving on the surface tf tbe lcftin t;c so,4bcnt hemiiohae:* tt"-;gu @ 3.33o. Hencet'he ' E-'tr.1+*3si',i r..o.a.o"u'*itt tr.",aitat ';''-2ry"* *q*lttd^itrii'-tcip do*n' &e rcsiolorl.o;-ftuout"' of the wini;;;;" cyctoacs formed due to motion ottttctclockuisc itt Notlhent bemisphctc-oo| "*"**c-ia'sordhcra I (3-62) 4il' a X / fr' =..=u trt-{r" , hcnisphcrc(Fig'3'35b)' i ' ,rrf\, 51?3-'ffi. \"i.4,^ 3' t3" 3 \ w f #r \ \} z' X&F 3'3so. :=:.{ i l; : Yn '% L I 3.3sb ',/ + irI !rJ . :-r_ 5. Approximatesolutiorr"fo{Jree fatl: E = : 0' Equation (3'62) reduces Lo (3) g=g--fu-xuon the r'h-s' o[ (3)' lt is solved b) successi*'e approximatiols, usiu! g frorn the previous approximation u- yt*la lstapproxinration: usingg-0orl r'h's' oi(3) :+ g= g- + :+ a = 9.-?4'x (g"t +go; 2nd approximation: (4).is used on the r'h's' of (3) g: go + yt-,!x (g.t2 +29r,t) r: ro + 9ol * lyt2-t' x (*g,t3 + !j2) (4) (5) (6) (7) to t he usual results for unilorm acceleration' The undcrlincd terms are the correcLions. due to earth's rotation. J -T J J =I J j J J J J -J iJ j J .J { J J jJ j :J J 97 *J -J J }-l F- ;i | ,', - i (' U c- lq - . - ';-'t -' . '' r t.i^!ze\a"e - Bodv 1: Cr + l- for point .t{ *'-r-t- a, y' z axes-for- th: systcm sholvn gia.tftu'd rtu'd &-ia.t 1yF$a E's' E's' ' Axes 1,2,3 are the principal a-tesof z-a-tis is principal anis at cr *'ith pt"ne ingi'^tt I;r;'rii,'$2^ rinllProne *t,:'#:,'-*= *.-:--' b- '.-----+'t F-*il.83' ,b o'" or mass f-" * t;;;;i"l;""' "** rllass synlmetrl'through inertiaatcl . Plane y-- is a planeof " :;,==0,'|u =o,'"' r,ul',;:;' *u':i,,=,,".;{:=:i"'0"'"' ")i sr=si,o' s?=-coso' sr=0' ""''uli"" ,1 =coSo' r12=:sillct' rl3=s' * :+ .i=g=sitrogl-cosoQ2' k=!!=cosagl-:-siuaq" - a" cos2 a)f 12 $; = tl"': Ifr'"1 * 8;ri+r$"3 = rn1(62sin2o* $; =t!;=lfi.r,i +t,g;ni+t"g;n"=rrr1(0?cos2o.+a3sinro)/12 rn1(6? - o?)si.ctcoso/12 8: - r'co' :,f,'"',.' + Ig"sznz* r$s3rr3 = rnas s-r'tnrnctr]' aL c2- Plarte r: is a glarre o[ Body 2: Axes I-.2-.3- are the pri'cipal a-xes of iuertia '; with through Cz + Y-a-\rs x PrinciPal a-tis at C2 l-IE--I l)3 . 1-.. l_.. f 'arious A'B'F'G'H and bodv 4 aL 3 aL A,B,F'G'H Kt i f] I *.t.i* l ' '*_ o'-.:r-'tri _T o, _ ,4 r E=-=,"'nr,'ft1,.,.11;tt;;".8,i;l;*L;*'" l-- I !- inertia llrff3;T" ilffi:U; ^ H f tl.:.e '/ 3 i is pi Nes,ecr:++:i:a:*"*:"f l-#*t'.,,':::'"lli,"jio*, 1 at ;::;'li.i:::T::; bodv ''l''J'x. bodv bodies at ttre rollorving poin*: iasfition'ror *:' n,D,P,Q,bodv ;':;;,D,P,Q,bodv "i'u i: E*AMPLES meoti of r,#:, HT I + L: I ,= rr.2='piTR2L, , 1;'r - r==='; 26.=csiuO, :l:-r.1t.;:'::'" yc?=0' *l'"",'n=r,tt:;l":i'f .- k=a'=sigggl+'ccoei, + t t: E t I = I?.',-- tg'=b"+ecos0' n!=0" ai=siud' ri;"i' + ri;"!' + rfi':? = n2l{R? f 4+ L" r' =':': ='ii"i":-f1:u::::::: Body 3: !, j, k are the principal a:ces Ll- t R2 sin'2 0121 I! :u:,,',,nrvith of inerLia ag f,3 z6'=b3 :;==':"::,:i;i-1;;';-,;:;"u"i^l'",,, tf; - p3a161c1(ci + oil1tz -Za.'t2c1l(r" /t'+ cilr2l + I?; = pso$ (ol + b"L) / L2 - 2,...,r2 cllt'' lz* p'l , tr l--lL=I lt2lcos'0 + L21'lz1sine'."0 tr L- rt]=cos0' 1cz- I!3^-=rfJri'+?ni'+i.?4'=n2l(R?/4+t21tz1sirr30+n2cos2o1z1 t fL=.. 1"1' rfi =',.2n=12 ''c1 ' J s8 P2l 'lt; = 'lr" = /'!'=o' !:' Body a: I i r ptiniipal axes of inertia are ;'he ac Cr witl::, ,,^ 8 = tf, = I!! =a, fi - zm+nils t!; = t!,- nu(3i?r/8)2 - nalzlils - (3nr/8)2J, If"' = E;' ma= pa(2etlfi\, ic. = yc = 0; _ 3c,=.b4+3ftt18' values of nr-o'i' abouL since for the hemisphere the a-tes i.e.,I!,= If;u= rfl = f[i(z'no]EiJ= im,&l' at .A for *,J The.elements of t#i,".ii. ,..,ir.i* "y"t",, ;;;;., of arc half o{ the values for a sphere x'!' z at E are finatly obtaled using parallel a-xes theorenrs: .4-* E.=Dti+rrt(y},+,3,)J,,r,=ir,i;+rn;(zi+'3,I..I*=EtL"..+",'.(4,+v},}l' t'=' . i=l 4'" - n;rs,vc,l, + = it'i' =flfi r=l i=l - r,;v6':6-J I*= ' * ftf; - 'n;'c""c'-l' Theprirrcipala..iesategir'errpoirrcofabodycanrsotrtetirne-<Ireobr.airredbytoolrirrgforaplaneofnras-. s1'rnmei;yoft1lebodypu.,inselrrouglrtlratpoirri.Thea-tisnoranal.l.osuclraplaueisaprirrcipala.lisat. trre above Procedure''ihen the third thar point. If trvo pri'cipal axes have been J".".*r1r"i ": i'Tt::a-.ies'-r' Trre ptirrcipal'a-'ces'a*'1'ariousip'f,i*ts of trre principar principal atis follorvs from *tutual: orthogoualiry parenthesis: rirass synrrnetry git-e* i* che of rhe bodies are listed rvith the planes of l- ,{: Bod-v 2- ,{ : Bodr-3-6: Bodl- 4- S: goas !(v:). J : j (z:), p : j(:::),n, i (c:), g : 9r,9:' ec (t2'23)' ei(?-3'}. i.i.!(:::.9.:), Ii: i'(v:)' Dt i:5'S' Ca (1'3"?'3-)" Q'.'4 t1'3-}' g= i,!,k(r:'v:)'ff' r', !,j.,k(r:.1[zv]. b(llrvl' i'J-'k (r:'Y':)' about a no[-principal atis mass C' is rotating Exatnple 3.2 A rigid. body, of nlass nl Nit'h ceutte 1f tlre bod-v has eleureu[s fior axes c'y': attached rc poiut O of inertia (Fig.E3.2a). The inertia nratrix at prof position' The-bearing ai B doeS not bearing reactions f", th";; the rind -.. *.Ign,.lr,t.r.inr. Consider i* particular tl::?: "i'oi"" tn't'* 'rj32b'f9z ,t ' 'pr&* J'ffi3o'' Lt{ks:,q, ;:-''-""fi,' ,ry#, 1"*.-<3o''lffir (G) :; (1' ide an-s axial *,9, --rtl*t;:l (bl Rrk, 8a = &!+ Esj', be the angular = ark, Ra ='R1i+'R?i+ The FBD Ot p*itio &a does not have axial comPonent' acceleration and bearing reactio*s at the gir.o ''is obtaiued fromgp Q using = the body is shown in Fig.E3-2c. The acceleration of C Solution ut = u\., oc = ei. I*t or sc = so+ dr x aC -t-' ^:-^^ ^ This can also be obiained directly since C ;;;,;&,8s + s-x (sx @) ntoves = in a circle of radius e ag rates point are determined using moment about ftxed M o =-@ E x (R, i + Rz i + Es b) * 6 L x (& i + n'si) + - (I?,6- rf;.r.r?)i+ U3,6 + t!'-aii + t!,ay q (l) -'f ei+i;ei i{ o L and E= r"'r'i' The 6 unknorvns tng6: + c ! x nrg(- sin 0 ! - cos 0j) I | -"- I --.)O F.',, L-- ,-l * L L l' '- '' , . [: j: '-" i>Rt- '/ t?12- t?,' 6'R4-onl=jrf;,a+r!''z D[s*= (2) i, + ;[;:a;#(Eri+;.;,.'f;r;;:"*:'1-"* "' ,=*j * i: :-l (5) Er*fie-nrgsino =-mtt2e L (6) .,;', f (?) :::I-mscosB=m'be .-a. .o 2 _..?onl.nna 7, ao=gfla+ tg,"-m:''r?ea+ntsasind)/r eorts),(;)'* R5=(I!'.,+I?"' +'t1,,ca4rngacos,)/L ^r=-Tr?,o+I!.,,,2+tnu?eb-nrs6si.0)/ R2:(I!,6-$,-.+mieb+*Ou*"rtlt't' * Eqs(2),(6) f I 1 i8) (9) ! o[c'? term and change their directions rotors ,lue to the presencespeed rrigrr foi rarge very are in grou.d refereuce These reactions respect to the fixed directions their consequently, --0"";-::;ith *'ear' The rotors nlus! be balanced to avoid this' as the axes rotaLevibratior,. forced ^r,.t r,ary rvith time and cause hea'y r LIL * i LJ - We consider some parLicular casesi'e'' lhe ceatre o[ mass C r' on the a-tis e =0' )'' erf 1 :, I i , I t. rotat'ton' i 'tt I t i ,i 2- I!" = L tf;"= 0' R7 L I l' i'e'' ac z-a-xis is a principal a'xis of irrert'ia o' L' = -(nu't2eb - ntsbsitt.)f fu - {-rw't"ea + ntqasin : /i3 = (,r-'e, R5 0)/L' = l ' ntgbcos,)/ L (rrtr'ea + ')rgo cos?)/ L R5=ntgocos,tL t-r.3.a=o,I!"=I?,__0,i.e-'tlrea.dsofrotationis:rlrrincipalaxiso[irierriaatCrvhiclrliesonit. Rv=ngbsin!/L' E2=ngbco,\fL ',l,=rrrgosi.0/L' L l- ;:':Y;',:;,T:"lTi::*1.,:"j'i':';ff:.:l;ffi:iii"i'i:; L] ; L F L-- ,) ) , Ul-r Lic l.=-l : lJi r-. lI ' cosoel lsinae2' "'= : - -'.:::" -ccSoQr' ';;,=i::i3"*;='f'"'"'*"rf,"":ir$s3rr3 ={I? -rfl)siiro'co ] ltj ljl * ;:*:;':;,=tu;'i;*1;:;:;;I."::;;1*'i= L= s= ' l- ' u [ " ?::;.;ti*"::"? :i**;;;;*i**:::|"'i= L er' e: aL o r+i'lrr s' a'l ',i.,e*ia j ispri.cipar axis a'L o ie Hence Eqs(8)-(9) *.here rg =+ R2 (I!'i l ntebcos|)l (10) L' = = -(I!,at2 - msbsinl)lL' : (-r.P + mgocos9)f L' n's = (-/fln 0)/L' E' fu = (8,-' * mgo sin t0)lL' R1 is given by eq(10). For rrre bodies,'a:r;,::1:, p2)/r2isinccoso _r,n(o2_ p21/12]sinoroso (l plare 1!.:7rns?r,z- nry2lr2)sinacoscr = [rn(q? (0 \:, ,"";r"gr;ar -q cos a lflsinocosa -1nR2 l+lsin r:-- 1!, to -:' r*eltt---- Misaligaed mff'12)sin c cos o = -pnP(2 --. thin disc 1'lrrz14 - rrrRzl2)sinccoso ' tL:(ii) (lQ Misatigped thin rod. E :(t- mtr2l12)sinccosc = -{r"h2112"""::,:.^ D.. 1!,_lm(R214+h2rr2')-,l,n2izisi,iacoso =m(h2!r2-ftzl4)sinocosc a ;;;;;;;*ru". .. : 6o , I) ,i t r\-l -'' note thiiihe > platc and if Ir : tfrR di'a'amic reactions ate :.r,ro it p = q for a rectangular lc ls interestingto a piinc-ipal axis of incrtiaO and the axis of rot'aLion is for a misaligned cylinder since for tke cas€s lf,, = (Fig'83'3a)- The several pulleys aad rotors Exaaple 3.3 (a) A shaft is rctatin' about i-axis carryi*g z;)' Find the reactious from the centre of nrass has coordinates (e;' y;' mass of a typical element is mi and torque is ?' Bearing at anSular velocity c'r and the driving the bearings at .,{ and B when the shaft rotates for A w'r't' body matrix The elen'rents of [5e inertia B does not provide any a-xial constraint to the shaftpllkular find Neslect sravi!.v' fixed rotating axes e, Y,z are !U) It P,r*er1+es|- 11= r, 1* . rF shaft in Fig.E3-3b' I:",C'--- t, f =lji3'lrf'ai;): j,mi ;,; --, i il -r i. 1 "?::i::Lt:XI""X;i: i'e,i 'ru,ru2 t ri o' a 0-l^ _- iE & tsJ "{ .i{ tb) (o) Di,r -:-p-r.,n)ThgFBDissholvn Solurign (a)Letthebearingreactrionsbe&,=:Eri*&j-r.,fe;nsi+&!+AoE *o*unt' eqlati-o'p:for:lbe 6xed in Fig-b3.3c. The 6 unknowns &-&r;fia,'r,fr, A, ;;a"r..i*'ifi -r __- !i .3, -- -., --. u-, lat.rU e{ - from t'lre - pointr4andf:frl.;gc;-Forptane*otlo,rnuitt4.'in,t'lrediregtionofuon-principalarrl"'E'at'-' ./ M-" - trk x (Br i+ R:i_) +r1= u:"a -Iid:)i+(Iin+ I:,u2],i+-tine' Rz--$*u'tf;'u,1lt {l) El = (/fl'r}+ t!'u21/t (2) _q ; =T/I!, + k: T = I!,n a:i t= Eri* 8zl+8r = )l*'i4g, + &,^. =Inu*.-'!'i. where tb,=t,kx (t lJurit--'ttri+vi!) = -(dryi+a'"'ili*(d':ri -'"vil! (3) ,.= + + :+ -i*r--i:,;-t-".,'''' LRt i rf;"; + t!.'r2 i: j: {{} {5} bearing point can also be talen in Iis3J' This appioach of taliug tbe moanent equation fot thd&'red O, :1 -O.lm, ml =f-5ks, r3=0' T-a2N-rir'y2=0$6rer' kg, (b) nq =2 rn3 = 4 kg, ,, I-0-O4mr 93:O, zs=O-{rn, z2=03rn L=O'5ln' 14:l00tad'fs' kgrp? I!, = -2x 0.05 x O-L - 0-5 x 0 x 0'3 -4x (-0-04) x 0'{ = 0'(}$4 lf;,= -2x 0 x 0-l -O-5 x 0:06 x O'3 -4 x 0 x 0'4: -0'009 kg'mi tg-ott r!, = 2x 0.1213 * 0-5 x 0.u3l3 + 4( oj€12 + 09421 = 6'o3s*62 ,, , o. :fi, - 0.2/c035467 = 5-639r rid/s? ' r ar- =.r?(0.04) i - 6(0'04)i ='t00 i - -r -r -'- ,. rc) (7) R2 r?^ : Rr .. \ : >-ir _{ (8] 0'2256j m/s3 << I _{ Usingdatafromeqs(6)-(8),eqs(r)'(2).(a)yield l' -q \' .-I !b, -' -..'?(0-05)i'+6t0:05)i --:1t^:::'::i::'::' (rr- 2-.?1o.oo15 tr;(0:06)[= -0'3383i-600!m/s?' +! :' x 1m2)/0.5 = 1080 N' * rm')/o.s = -oo'u'n,' ' -(0.054 x 5.639 + 0.009 2(-500 i + 0.2820 j) + 0.5(-0-3383 i-600j) + 4(400i-0'2256i)-1080 (-0.009 x 5.639 + 0-054 i+ 180'6 j = -{E0'2! - 1r9'? j N pass m is carocs' An autorrcbile of Baarnple 3.4 Bal'lking or srtperclcoation of highway au! milrcod proper rralue of 0 it its centre of mqss C travelling over a circular road 'banked' at ar1- angle 0. Figd the - <{ u't .i< 6t t I .,< -=- ilL- 'i- J-j * "o*,*iinH travers L: IJ 1, fr-a : -tt' tL sorution ;-*ffi-T (a) "' o",'*"' ll"ni tb> tty5:1,;T"i:"ffi'X3::ffi*:::':t;';:i': lixj::; sc - lo'l nls' - ("/Rx- cos 0 sin 0 cornpo.enrirof che sum of rrre trre represerr Fr Nr, Fig.E3.4b. in lormal,aul-1r^:r,.", The FBD is srro*'n f:, lJ . fif,ff,':i':l:*m;l:*iffi.::'"*:f;*:',i"":T,$::n:'L:i:J't:'::::ffi"fiilliliT; L r-. ?'f:["t':1"ffii[:::"'::':$:::[":"':;:::;:T: of inertia at C' ;i';':it::H:.:;i;':*il':::il is principal 0' as ! = t-L' = no'-principal w.r't' C equation for re3 u.e momerrt l_ L * L L tr '- -Ft - ir, "-,rilJ axis a ''"' jol';l;l';' ,l=;;'*f:-:i;'" ufi s"3"" '.,i:t,:::, tf, =-i' = 4i'"'= If'n1s. .4c')sin0cos0 -* = vietd "'**'+ + Fz -mgsin 0 = m(-uzcos,/E) F using eq(l)' rhe components of tr L :'=::;:, = itsc *t""'.'r*' :+ cos0/R-9sin0) ''v' + 1;2 = m(u2sin'/R+ecos,) Fr + F2 = rn(u2 (2) (3) (4) -$'"i Mc--gg,a-I3,"li-+(f,;!E'")i+11;t=inet*,(al.l2 (N2-/vr)6-L(F +F)=-1f,''2=u?'-i;;f 1 . .^r, U,1",;'i"f,l=,.il;,.X;;*;;:'i;:;i'::'ll^fi,r'!iil:?i::if);J"*'^'"' o?l R)2;!2b (7) - rfi1sin I+ l{ o'cos (ricz "il ) f 1. #;'J;::Ti,'fr::[:t:fffi;rion ; For no oduard slip, Fr * E 2- f eqs(3),(4) r- Y ;H:'i*ilil,:I"j# ;lx:1';q:;j:**:xtlxrf;:*:;''::'";:"'::: :,:::,H?J.':llJ;:::'::';' :;= i* :jrfli'tl{;t'",,",j,s': us (l) 9") e' + l- L , i"j333.,i'a1*:l:i:,:.9ff;"I"'.*r:Y$i,;:f#"iff.iffli -i 7ru,- f L-, ,:-, ; + Fornoinucrds,ip, Fz rhe optimum cnelc ror ba*kins: force shourd be zero. Eq(3) vierds s p(Nr * N2) where p - taro is the static coeffrcient of friction: -(" ca.lR- esin,) s P*(o2sin0/R'+ecos0) r;;::',1lf$l=ar='Rilll"l"'-o' tlll i II - 4-' ' =-"- 'For r,o'tippittc :+ ftocflii&,g),th:e snialler noraral + hsiallb)-(o'ln'i{hcq,0lb-sina +Uf" frlsinocosaTmDE}} } u2 Se.l?ltao o+b!h)ltl-(blhrtan0+U?z- f,1sine7lant'1 iv1= |m[e(cos '..J reaction /Vr must rcmaT positive, i'e', usiag eq(6) 0 JJ J J o (11) and for larger rralues Equation (1$ rerrcals that the car is more likely to overturn for smaller values of B'&.d of h,(Il2- Ift), r. lf (Ig - tfrl/r',,Al ( 1, then (11) simplilies to tl" S e8(tan e + b/hl lft - (blhltanil (12] p comparison of eqs(9) and (12) reveals that in this case tipping $'ould occur befote slippingif > blhyield (6)'(?) S---F-or-no 6ct*ing (d = 0)- The aris of rodation is a principal a-'ds- Equations I /vr,/v2 = lms(ll tw2.sba). J J JJ (13) Equations (9) and (11) yield the foll6rsing speeds u, and ur for the initiation of slipping and tipping: t! = psL-, (1't) * =(blhlsB are srnaller Hence tipping occurs before slippin-e, if r, > blh. The values.of r'! aud-urfor the unbanked road than {hose for thd:'bariked toad6- For a stmighl'Ian'ked rcad (R= co)- Equations (6) and {?} 1-ield Nr,iI? = |mg[cosf *(h/6)siag] than i.e-. in contrast to the circular case, the inner rvheels are subjected to greater normal reactiotr ;:*J:I'T:::::*f:::X*" vertical aris rrr ,reigrrr h a'd ,.id,h t ,^[*ll -J [er ^ a.'1i"8,L ,-r'i&t (Fig.E3.Sa). It.is rorating at. lT .T I 9'' lis hinged aL A and B abour a .,\. angular velocity a,o when it bangs against a stop D at the ground level'*fi-J. 1.t-- I -IThe coefficient, of testi[ution is 0-2- Find the irnpulsive forces on the t i':i I D i trt ;4 door from the hinges and the stop D during the instantaneous inrpact. <F4P6| Sotutioi The 1rBD for impulsive forces is shosm in Fig-Ilii5b- T'lre line of impact is along;i The aqgular velocity,beforeirnpscyisgr-:r.o\- {.ettbeirngularvelocityjustaftcrimpactbeC=o'L,andtheirnpulsive reactions from ttre bearings be . , &=Eri+ari+4"L, 8a=&!+a"i+nof The velocities of pointi 8, D just before and just after impacc are u'p'=0' tzr=-l.i,sb, o}"=0, ol=-''b' lLE=-uob\ gD=0, !f,=-a'$i'{a=0, = -0'2'o (1) Coefrcient.of restitution e'; o'g=-'v'4=-c(ue, -oo.l + -u'b=-0.2(-ra6!) *' ,,-! The fixed aris of.iotation is,not a principaL a:<is'of,inertia,at'tlre,fixed point r{ of the,d9o5* -m(Iix-dlz) = o; If. = -m\bt2l(-dl2l = mbdl4, I!, = Et =U!,L+ ti,i+ r*H-'q 6'H^= (rf,i+ i7;-+rLhx-'--; = -r:(*'aodj-+ 1= +' 'rrllb? fi2"+ l6ti)"1= 'rr.b" 13 |moj\)uo The angrlar impulsemoment of morneotum relation for the 6xed point .A: fang^ = A ff, q : t.I I I 63 .-J iJ J JJ J I rJ -J (2) \ (3) ..-J yields t-(id+c)k+pjl x ri-dkx (R.i+R6i+8.$ = -l-2(frnodj+ trn&?k),"6 :+ . [l: + Rs=$ dfr5 =0 :+ i' = 0.4mbzuolp k -g,F =-0.4m62r.rs i - (id + c)f - dfo :-o.3mDd.rE :+ [* = rn0[0.3 - 03(0.5 + / d)b/pl,.,q J (4) (5) (6) -J -J J :J J 4- ' Fo, or'tippittg (ooct'tiidag),tb:e smaller normal Jvr = + reaction rltr musi remain pcitive, i.e', using eq(6) |m[e(cos 0 + hsinllb, -$,2/R'fihcc,llb- sin0 + (Ifzu? (e.R(tau o+blh)lll-(blh)tane+u?z- If.)slngcosa/mDE]l > 0 ff|rine1*nn1 (11) Equation (11) reveals that the car is mote likely to overturn for smaller value of &e .d and for larger values of h,(If2- Ifr),r. lt U&- t?rrk"ru ( t, then (11) simplilies to u2 < el(tan e + blh) IIL - (blh)Lar.ol (12) Comparison of eqs(9) and (12) reveals that in this case tipping rtould occut before slippin g,if P 5.--Eor-r,o.ba;*ing (0:0). The aris of rodation is a principal a:ds- Equations (6)'(7) vield . /vr,/v2 (13) = |m9(1 1twz/gbill Equations (9) and (11) yield the folliwing speeds'u, and ,t - > b/h- u1 for the initiation of slipping and tippiug: ti pgR- (14) = $/hlsR Hence tipping occurs beforeslipping,if f > blh, Tlre values of, p' aud-u1for the unbanked road are smaller than ihose for the banked road6- For a stmight.banked rcad (R= co)- Equations (6) and (7) -vield ! Nr,ils = |mg[cosf +(tr/6]sin0l i.e-. in contrast to the circular ca-se, the inner g,heels are subjected to Sreater normal reacLiotr than rvheels foi the case of straiglrt banked road. ;:ffi. r::T::ilX'HT.;;*. 'r. t rreigrrt h a.d o'idtrr b6{ outer ^ fai.'tt-"&{ -,--irl&t is hinged at r{. and B abouc a vertical aris (Fig.E3-5a). It is rocating at angular velocity a,o when it bangs against a stop D ar the ground level-*l The coefficieni of testitution is 0-2- Find the irnpulsive forces ou rhe door from the hinges and the stop D during the instantaneous impact- <fqfa Solutioi The FBD for impulsive forces is shorvn in Eig-til5b- The line of impact is along,!- The '-gular velocityteforeimparcy is gf-={.rok- tet the angular velocity justaftcrimpact be +f = r.r'\., and the impulsive I & = E, i+ azi+ a"l 0, v',= -.a'bi 'gb = 0, og, = -usS, tfi" =0, t/6, = -u'b, o'D, = A, :+ -'a'b--0-2(-rob) +. 'ut' = -0'L*1-o (1) restitution ei u!a-vb,:-e(upr-rp.) CoeffEcient.of gD = The fixed anis of.rotation is,not a.principal anil oCinertia,at*the.fixed point A of the door* 4, = -qa11-itrz) = 0;' t{. tst= U!,1+ tf,i+ r* $q z+ : -tr(bl2')(-dt2) - mbdlt, I!, = m[b2 fi2 + l6/il21 :+ i: k: j: '- -0.4m62aro . -(id+ dF -dR.- -0.3m&d.rs PF = : :*, (3) d: Iary^ = A ffa yields t-(+d+c)k+pjJ x .ri-dkx (Ieri+R6i+8"U = . I.2(frnodj+ d&=0 + Es=0 JJ J I --J J \ -J JJ |rn6?k),.,6 F = 0.4nb2wslp f,a = m0[0.3-0-4(0.5 +qldlblpl.no (4) (5) (6) -J -J I rr J rJ -J :J i t-I I I - ,rrb"13 (2') ' a'Ht=(rf,i+ td,i+r*hx-'--o)=-1J(im0di+|rn0j$:uo The aagula,r impulsemoment of momeotum relation for the fixed point, iJ J &,=ni+a.i+nt The velocities of points E, D just befqe and just after impacc are ls - -uobl JJ n J J J J The velocity of u ul i: =* L ' lL :4'&.160 i = iaqla*-*"t*ttt* f.i+ erl+nuk+ n i+ iri+ nuf + F! = m(o'r + 05)&zoi El = rn6[0'3 - 0'4(0'5 - cld'lbfgl" A, + il+ F: O-6m&ao + O-f6GroL Using eqs(4!(6), ah." ) is gb = -0'Saro6 r"t"tioo ; f=,mLgr,' yields . ' . bcfore and just after the impact centE:kESi}io=t itz =o * j: .Rz*Es=0 itr - -8" + ftu+8"=o k: ,?3 : E6 = 0' provide a-tial constrail'' 63 and & cannot be e,raluated individually. If one hinge does not :n""and a thin disc o[ axial m-o'i' 11 Exarnple 3.6 The system shoru in Fig-E3-6a consists of an impeller mo hits tbe impeller' L_- tJ has an5ula velociiy ,.rqtust-beforc-a^-bird.of man The suppori at B does not provide any Model the bird as a mass-point and the impacg as instantaneous' 0 (a) if the coefficieut' of restitutio"^" : axial thrust to the shaft. Fiod the impulsive reactions of the support with a slight misalignment of 0- It LJ ffd;d;;;;;;il.#%" L-- r.tt-;5ef:=i n:,_ L_- LJ LJ LJ : - lfhe -(o) -^^ EBD for impusirc Solution i. ,W" sh6wn in Fig-Ei!'6b rvith forces on the systern of the bird a nd the ,otor is the impulsive reactions at A and -B being L* &a : &!+ E:i+ tratt' ftr = fQi+ 8si at d rvith angulat t'eloci6- c : :'16 ! just The rotor rotates aboul a non-ptincipal asis of inertia.of tire disc 1.2.3 are Lhe principal a''ces of inertia of the before impact and with a/ : rz[ k (say) just after impacg. A-tes [. LL- discarCwirh t1r=1]o6q,,'E:'r3= irorf1.I$=0si'ce!isaprincipaldirectionarCasi=-car!3.=-sirg, sl =sino.- s?:cos9k: n_cospe1 _singez i-"nosgr*.J"eg-.r,,=.*0, fleace for che rotor L I!, = If, - rn2(0)6 : O. .. 4,= tf;r-m2(0)b :€,=-[,co ='f1s1n1 *f2s3n3+I$s3't3 = lrn38isin0ccg' i1 = tr+ fr'? + f;,'{=+ rgnS = 11 * f rz2Rllsins 0 +2cos? al' H a = {I!, i+ rf i+ 11$t.l \I lI t- (.) The line of impact is .lo.,S The velocity of the bird just before impact: CE : :o( (l) (2) sil ct!)' :-ai=Henceitsvelocityjustafterimpact:i/e.=-usccoi*u'j-- The?unkaorvllsu"r^r{'8.r'l?2'83'8{'8sare o{ molltum telation for determined from the,'coe6cient of rcstitution relation, *gul". impulse'momenl and:bii.a: lte sdocity of D of the fuied poin[ d and the.impube-momenium relation for the system rotor ,-, i i.t I A. ! : justbeioreandjustafterirnpacti*gp=-uroBri,cb--u,Rt!.Ttlecoefficientof',.restitutioare}ation ,'r,-rb"--.c(tq-?oyl -" : Eence E-- {e = -vscosa!. o'-O=-0(-uosino-0) A{.e for rotor*bird yields i+ I* k)('' - tro) * AE x nrs(gi - * o'.:O' (3) Using eq(2}, I+ng^ = zbk x (A+ i + ) + *ri) : (If. i + ti gB) ! J L <- r.l '^fI -l L'- ! I -- J c O f +i: j: k: , '-"r t : c.a!t,l a .ii l' ".-:;--- .' h I The irgulse-rnomentum rq!.atioa L: it, i + a, j + :+ A?for rotor*bird E3 k + Fa t + yields a"i =''o(!! - !: .Er+&=0 l: k: A,+At=rlouosinc E"=0 : q'El = rnouo sin cj (7) (8) (e) il"=-0'5msussinc' R1 =h"= &:0' Thesolutionof eqs(a)-(9)yieldsa:r-rro, 1is=1-5m6u6sino, Ians, =afl..t and r=AP forrotor*bird' il;;;;;;;;;,ii;',ir,ir",q.E","udeterminedusins di{Iering from that of the rocor alone After impact rotor*bird form one rigid body with its inertia -"it Afu for rotor*bird vields conrriburio, o[mass-poinr rno ar'AD--Rrj+34k. !-ang^= ;;;;" zoL * x (n"i+ Euj) =[{ri,-mo(0)36}i+{ti,-rnol?r(30)}i+{I:,+ms(o?*E?)}\l'' (Er j + 36 !) x m6ue(-t(fi, i + ir1 i * I!, L)no+ cos di - siu ct i)] (i0) - I!.(u' -,o) - 3mouobsinc 2bR4 - lf;.{r,r' :yd- 3moEl0o" - 3.rn6usbcosc j: g I!fti -.o) * 'os&!"t' - m6uq81 cose E: yields Tbe iragulse;mometgum relation I = Lpfor rot'or*bird j)i R1 i - uot- cos o i - sin a E, i+ ar!+ Ito f + f-o i + Ar i =,ro(g! - ge ) = rns{-u' i Ar+4.-rne(uecosa-t'r''81) :+ !: j: ea+At=lttouosilro r?"=0 k: !: -2bfr!.s (ll) , ( 12) {13} ( t4) { l5) -r,E .j?s, Rr,FL",ff3 ".e computed successively usingeqs(12),(11),(t0),(13)'(14)'(15)is rotatirrg at, the rates shotrn relarive e,.'.6. a.z A body 2 consisting of a shafr B D zrtd a rcd G E(Fig'FII'?a)' The rod.is of length 'L and to platform l totaiiog about a fixed t'ertical atis at the given rates the a:iial testraiut- '4't' lrut' prof ide an'r' ma.s ra- Neglect the inertia of the shaft- The bea.ing ot B does (b) trvisting the SD' shaft *r"rifinil (a) the bearing re4ctions at B and D and the couple applied to ;i"", ;;;;;;;;;;;r""t t; -a force and shear force at section :lu"r tt'" slender rod' 'orSnal ;[i!,: ort,t,, atI iro! l,lr'ldWt-* ik*,,, 7FE e,f . --l+ {t La.J**,rrit-u*{; $:, (at Solutioa G) f*t t!r,4tand sh. Q2 be the angular velocities and acceleraLions 4r : gfr =.i'r i L 93 : t4r f cr3i =rr1 i+-ri ,izz = ri;.r+.i'z i + tlr x t tz L=tirr i * q;r 'z I J f, ,L leJ- sEa ! ;. ! lg- - *il* k-Lle -*i (() of bodies l 2 t'r'r'!- ground: ' (u rile j* rer1o2 k' (2) the momeo' of body 2 is sborvn in Fig-83.7b, where C is its ce-nme of mass and lf is Tllil '-" of motion of body 2: equations the 6 fronr 31re determined M Fz, F3, E:, Fr, 81, unknowns The 6 the s[a[t- Tte FBD 6s _.. JJ JJ J J ''l-J .-l J J ]J :J :J U r*J U r'l ,tl -J J -lJ J -J -'J J J k: E= trru"oa a eo;pS\tiatlo-qp [- f . lb = so * -' L- lJ [J' E' H H' H L L L 1-. L l- tJ lJ t. . l ]JJ L) r u lj,-, l- 2 remains fixed' foa C-'-foi-* O on the dgid cxtension o-!!oar cz-x 9C + * " !*x gC) c,,i(D + dlli + &ar(6 + d) - i'{;'"I L Mc= - (-683 + 6r.3) i+ The principal x (&i+B3H+ ?iLl+ei) x (rri+rzi+rsE) (M + iLr.g+ !z&)!+ (6& - it& - 6rr) k ii+eir'i-Di) axes of body 2 at c are parallel tb z,g,z axes l;:' ll,'. (4) witlr I!, = o. tc = Ig' = mL2 lt2' Forbody2eqs(1)-(2):+t')s:Qt'tdv-Lt2''tu'=O'&t=i1'j"=it2'"t)'=t't1<'t2 body 2 for C yietd: The Euler's equations of motion f-br + Mc, = 19,a" - Uf" - I!.\,,,r,,t. Mcy - Suau -99"- €,\,.-' Iy[c,=t,6,-(C,- $"\-'uv. = + (Fa-8")A=O (5) u + \t(Rs+ Fa) = nL2&2f L2 6(Er - Fi - iLF? = mL2sYu2t6 (6) (7) F'--nec+(fta+rr-,r,e1ii&!+(Ra+ra)E='n[-]r'!f,!+{"tp*-410+d)}i+{t'rr(6+d)-}62'}LI Er * .Fr = m(g'Vtit'Itt ' B',+Fr-m(g'Pt|tlzl' + : ' R" + Fs : m[dr1(6 + dl {') .. Ft-ttt-tllttzL-u1(b+d)l *;lLlzl t'O' . - tilr(6 + dll d;lzbl' 8s = f3 = ]rn[ti1(0 + d) - ''"Llzl R1= \m[g -.!r1z l 2',,p2l.2lcb -ctiL(b + dlfztJ; f' = molp:f, -:'11(6+ d)| Ft= i*lc -.trLlz-:,"'tru2Lz/ra+tillo * The solution of eqs(5)-(10) vields M = lmLl?fuf,l3 when ri2 = 0' Notc that a torque M is needed even rvith centre of orlss of mass rod of.the (b) The FBD of part AE 'o(L -r)/tr (d 6)i + !(l + "1i i-ointr'O, f are on bodv 2 with aC' -- + C' is shor*rn i1 p;g-Ii3'7c' 4 t OC' +r'z x (g& x OC'l + il luz(L+')]k =l\.trtr * .li+ lrrrr(t+') -'i1o +d)lj + ['".r(6 W.= CR-rG -.')!xEn gc. = =" go+ Thegrincipalaxesofbody AEatc.".:p1.ll"ltot,r,:-oowith r'rt'dl"" The Euler's equations of motion for body A'E fo' C.';ilh l;c" (11) (12) t =0'Ifr' ='nfi'=m(L'-tl3lt?,L' as forsodf"2- yield: "t-t - t.a, - ({, - Ili)Pt""t, =o rfa"llzL Mc.y - €, a" - g9, - lllY',', = m(L M"., = ry; a, - t*; - $")n,', = n{L - rlsuYo2l6L Mc., + W. = rrr(L -.)"(d'zi+ r-[m(I-"]lLlsc' :+ ?*tPzL),ll2L Ll9x.' -[*(, - t)l LlgL+4* = Pn(L - 'll l--,1' lj.-1 (3) - - ir; r, i + l,-t1tt2L - G (13) I t -;= (r5) (tuapt(2L-,)-&'?(6+d))LI/6r Mc.+i@-dix& rn(f,-r)2ttr?(2.t;;)-&,1(ii;)$ is the normal e the ! comPonent of !P is in the -idirection' Henc at cross-section the io Ient beins .he i '{' The normal tht ihear force- The twisting mot together-consiitt'to j components and ! of ea together force and the sirice tlrc i ana'b ilponents n,ment b;;di'u tht is z,eto. Qailself componeni of Qp is moment' bending the .""",o"r" vertical shafr, at O (Fig-E:t-Sa). mass at C is hitr; a a axis at a pxa-rple 3.8 'A body of mass rn *ith centre of roaft is- rlated about the fixed'verticat ti",-i?r,.i* ,r" o ilr,".-,h *o*"rrr"ii tts principal ". "rrig to tlte vertical' Find oc has a consta"i il'.ri""aun J..uvv such that consranr rate c.r os-" = "- i., ,K,rry-=iffi;:" """'ffii;;;; +=I' N"L(-l .,,.n "-"LNi: ';,{o'L (b) . ,' to) L ^ . rls equ lt -: ti tio*s for Solutionsincethe.bodyhasplanemotion'thesolutio"-t:t::O*tnedusingl'Euler moiion fcr non-principal a-'ies' equation for plane tuomenf, in the direc{'ion rrirrs"joir* does nor,r-tert any Fig.ffi;;--,*,. i. sfterrn is body the of tlre fixed point O' FBD Trre rfrom 3 Euler's Jquations ior att"'*iita L" Cr,Cs'g 3-unkno.'os The axisof the hinge *"utut"i-"i-J'ut'o"at tt' The angurar velocity r,r and angular principai axes or Z' -o*""t trJ3=-L)cosA"av=:-'sin0' tt=t9=-{^rcos0i+r'rsirroj' u:g+ Mo =Gti+ Crl-nrghsindk d:=0' ut:it":rit:O(t) (2) TlreEuler,sequat,ionsofmotionforthebodyforo.usingeq{l).(2}.r,ield: Mo, = I?,6, - (19, - $rl,,-o 0, ,0=0, or sin0 = i-e.., or 0=t, . Cz=O - rrrg& si t0 = {€, coso or. {3} ft.=S: r+ + * Mo,=19i,-(8-*.Yu-. l{s, = I?uau - 1t!, - f,Yo'-' iff }-: sin 0 cos 0 #) ,a \j t5) -L' = *st41;?r- {?,1'" 0 =cos-![rrrgt71 €r' 8)t"! a'? Tlre third solution exists provided -L' < > WgnlT?v - 19"|l' ! imshlt'fi.- !?1"-l: 1]: 'if oierr *r m"-*,"1r in .be directiirn r,ils.ioinr does not z- The FBD of the body is showu io Fig.Hi"-aretu" equations for the 6-xed po'int determined from 3 moment Mt,Ms,e unkno*ns 3 The a:ris. hinge the of O- The axis of rotation is the noa-principal . oc'rs 9"=g=-cc9i+singi = g= -sin0!- c1s0! 8r=O 9,: L + sc - ry^ = 'nt=-cosd' a'=sin0'^ nt=0 0 s' = -sin0' s' = -cosO; s' = since k is principal a'tis at O' + * t!"-* ,i'. = *rq, - + I?r""n" + f;'s'n' mglrsin o4* Megs 67 =le..- -l rfl)sinOcose ' I t E- I.:, Mt=: -+ Mor=f;r&-$rz- 'r3* 1y{oz = tSa + 8*" i. * '+ Mo"= t&' MoL ) --) ) Equarion (?) ror 0 is - ,-,, -r',, -row2r*g.*e $yf sir-e"s'e (?) (7) raghsino ==(e,- (s) Ms=A rheT-".::(:l-.y';::::;::":;:f::;,3;H';r:l*-issupporred * Td teagth tr' (a) Find uniro,m srenire*oi oi-* force"shear the norrnal 1u1 ri*a on ff;3,',::r. i"*t "tfi$;.lffiilffi;" ex.rtea;;;t load au" |oom' -theof thJb.om-dt-d' rhe rension i' tt " c.biJ "oa t+rytion f *"-""'';[i beading ana momenr twisring ,2t4t . lef l_(J20,'T ( force, , i offi,*, j\eft;i.il" _ffio, ;W t o$r-. '.4:8N7" '.-=Q+t I '-rt-rr to ; .Q,E C\' '" with the a-tis so rhat a-.ris y is coplanar aL o are choseu z'y,z inertil of axes ( its centre of is principal T rvhere 'rs Solution (a) Tlre boom shorrn in Fig'83'9b' au, The axis"iri" ,,, trre directio* of trre hinge axis- Tbe 6 unknorvns of the rod and the vertical *"*.", any exert uot and 3 Euler's mass. The hinge joint does i-notio. of the-boom: L=',.gc n"r",n"'l]j";;of of the boom are R1,R2,R3.,C1,Cz,Tareietcrmined augular acceleraLion r': *;"0 *ili, TLe o. para noa the equations for ""*,* L l-. L. L. L. r-- L L L L L .rtbl I tal (1) u' =osin0' t'tt =0' <" --!^'e:c"cc0i*trsin0j + "t'=t'tcos0-' -'' =arsiu0' t'r' =0' t't:i9=ricosg!+cisin0j * i':ocosl' I ) ly 2 sit'h gI.=|li' tb: k) + (2) usine eqs(L) aqd (2)' ,o-x &+ ex {gx QC) ' '' + r"2 cos0i- - !) = llsin o(-"'! sin0 i rng(lrsin0)lL b= G i*cai +ffLcosl - (r) (4) C : qithl;o'=0,lf;,=I!;:nlz|3and.ec(4},tlreEulecsequatiorrrofrrrotlotlfiortlreb:omforoyield: I 11i[or=8",b"-(g-$1-'t'' 8t)", Mo. = I+'., -{*- + + 'g;'=''!ml1;"itta TLasa- lmgr'sin0 ro [:'mgs + - lmf,lt^'zsro e*el i (6)' (?) ' (8) j;nrlsip0(+rzsin0!+r12c0!-t'r!) ;,tiiri* R8k+r(-si.0i+cm0i-)*rng(-cos0i-"']'aq: 0+31Di-lmf,sino&! sirio} !+! m sin 01c,! f, cm + & = Rr i+a,i+Rs k : !n9(1+cm2 o)lz.oe 0-}m t*12 the tringe joint on the boour' where B is the totat reactiolr force from 6g E I (b) The FBD of pait'-r{8 of the-rod of mass m(L - {l L *lith centre of inass D is shown io Fig-E}$c. i"i"o o.D ateoothu'EooH;ith O2.= +(t+r)i, -i.,,. .,. b,:9o,1 ex'OD* otx(gxgP) ''i': M^o The principat axes of body : (s) *U,*a)sind(-ru2sin0!+r.,2cos0j-&!i'' An + \fg,- a)coork- *tZ -z)1 x {a AB aLD are parallet to c,y,: aries (10) n'ith I!. = O, 1f,, = I!,: m(L - zl3f t2L. F=rn(l -dL)eD : = ln(L-z)?sind(-,.,?sin li+uzcali-Au The Euler's equations of motion for body AB tor D, with Mo, ..;,s,..;.z,. . . same as (11) for the b6om, yield: - t?*" - U?" - I!),*t"u, =a l{o, = t"Drio" -(e. - I!'b,-. = n(L- r)3asinp/tzt 0' ca M o, =T!.b, - (I!. - tfll-,., - tu(L - ,c)3,.r3 | t2L "in M-o=.i'n(f, -zfsiao(<irj +o2 cos? \)/\2L= ea * lqt -r)cosd!- j(tr -,a)i x fo (12) F'n is obtained frori eq(f l) and thea C* is obtained from eq(12). Ttre rrormal to the cross-secti6o at .e is in the - ! direction. Hence the ! component of {p is the normal force aod thp i *d k components together ' + 0 . constitute the shear force. The tsisting nromeut being the ! conrponent of AR is zero. Eence Ga itsetf is the bending monentAs in Example 3-8, this problem cal also be solved using the nroment gquations for plane nrotion- 'Blairrple 3.10 L gytoscogic gtir,dcr (Fig.E:I.10a) consists of a'uruller'or grinder in the form of, a disc of radius r and mass zn, which is rnade to rotate abouc a fixed vertical a-xis ai a constaot rate f)- The disc rolls wit^hout slip. Neglect frictiooal force on it iu the radial directiou. The disc can rotate freely on the rod about its a:ds. Find ihe force exerted by the grinder on the pan- Find also the reaction of the pirr aL O on the l;shi rcd-OA- d in tF y't,i s qfu.zEg ;' *K1rYK''-*K# d;ffi-r.r\'.; i.toVr. -t*W6,! c, Y, \]_ d/ . \\J f -'-. <h"' s .f,- -tgt //,> ! 7T t-i"sg Inl f3 (dr . i\A - Solution The axis e of the disc is a principal axis at point O on its rigid eiGnsion] O remains fixed iri the ground reference. At the instant shorvn (Fig.E:|.10b), the principal a:iis y is cliosen in the vertical plarie through e-axis. The FBD's of the disc, the rod and the rod+d'rsc, are shown in Figs.ES-10c-e- The radiat friction force F1 - O. Cn has no component along the axis of'the 4;rsc.- Ma has no component aloag the axis of tlre pin. The 12 unknorvns Mr,ll[z,Ct,Cz,rV,r'3,f'e(3),8(3) ate determined from 12 equatioas of motion of the rod (6) and tod+disc (6). The angular velocity of ttre disc 2 relative to body I is aloog i say r.r1 !- Let ur, % and rir1, ri2 be the angular velocities ald accelerations of bodiei 1, -2 rv.r.t. gro,'nS- \trIith O B = 5i - ri, and g,a - 0 (uo slip), ir 9r = Og: Q(-cosdi+sin0i), (1) cnil t-J 'aU 'at ':I J j J JJ J J -I J J J JJ JJ J Jrl ..-l JJ J JJ ,J JJ J E L. ....,'j-'* t"---.. 1 -1 "..-, ' lj, lj, l-'. L IJ TJ L t_ IJ IJ L L L 1_ Il.L IJ u tIJ L L L l- L I :. '= (.rr ,..o, - Eqs(2)'(3) ':+ :+ ,t:''ro+ r.rr gtz - x@'= -(arl-oi#"blr*6Osiu{l!= + =A(-ug -bsiallr) L 1- (3) =0 ;:e:;*;l;;:;,-:fi::;;l;T""ff:-eanolr)! - m*/2, 4 : *, = m(?14+62) and eqs(4), (5) vield htc - -69sing/r, r.r, - Osin6, d. = 0, 6, = 0. 6" =0, r;r, - For disc For light rcdOA: -O2sin6(cosp - 5sing/r) (5) (6) . =+ M.a=qE+0ix(,tr!*Ezi+ae!i) W=Ma-en+&ix(-$-9 :+ Mi* Mzi.=(Cz - 6Es)i+ (Cs * eBz)k :+ Itfi = O; Mz -- Cz- 6Ea, 6s + 6Rz - 0 . (4) Z, I!, (7') For the rod*disc, Euler's equations foi trted'poini O are applicatle dnce rod is inertialeS: ' Wl{6" x fiEitr' -:rce0)Irtk +'M:i+'([i- "j] - -rfsi+ (ll:-[fali+ [N(6sing -'rcos0) - mgDsind]! -n96sin0'k'+ : f,.u, -(r?r- I?,furur2 + (OsinA -r.F3 = :+ 0 &=0 (8) i (e) + !t6,.=I!,b,-(*-(;l-;,-.t, + JV ir'(6sino-rcoe p)-mg0sin0 = mO?sind[]6rsin0-(62 + = fm96sin0+mfl2si!0{}&rsin8-(63+ Irct the normal reactioa :+ r1I for thestatic JYlnrsr : case (A: f plwO'117(0sin0-rcosO} 0) be IVs1. r2/4)cf] (10) Eg(10):+ I{"g: mgDsinA/(6sin0-rcos0). I + [sind -$l2b+2bltlex0$gr!2s (ll) .'. For 0 = g0r,cq(11) =+ IVflrI"g- 1+ Qz{2g,Ttrusa'largevalueof Oimpiesaverlrtarge valEcof AI aud the crushing action of tbe SdDder b better. ,t mo.,"s in a circle of tadius esiaA 1 rcte fl. Ilence q1 = -fF6sia0l.'Fot t'he rcrd*disc, F= mqa + &+ (lV -mglJ =-O2&sin,! +. .& = -O26sitr0f +(m9-Ar) J N is given by e,q(U).: Equatirins (?) qnd (9) imply It{ p = l{1i+ ff3i = 9' p,1a-ple 3.11 (a): A uaiform bloctr of rnss m - 120 kg is supportcd at D b: a b:rll qrd soctet joint (Fig-1p.lf a). It is subjected to aa instantaneous inipulse with P = 2'10 Ns. Find the, impulsive reaction at iUe-support Eind the velocity of C and the aagular velocity of t\,block ju1 after-t-he applicatlduof the n'here , instantaneous impulse. (b) Rcwork part a if the bQc\.is lupportea ii D by aierticat incxtcnsible cable for -(i) ncw-oik nail b if the block is supported at D by a spring of t*,o cases l. p = 240 N.s, 2- F = -2{0 Ns. jost before stiffness 120 kN/m. (d) The bloc! is not supprjitid'biiU b trEuslating vertiiilty{orru at 2 m/s just impact with a corner (rcqr ncar it) of a emooth horizontal table-at point .8. -Find its lagular vetocitl' poiot t tras is rough.a^od p*J3,if the after impact. The coefEcbat of rcstitution c = O.6. (e) Rcrrork ,1a"!t9 joint is hit at B by a zero velocity just after impact. (O The blodi support€d At D b, ; 6all and'socket L1 L', L." ri1 Q It €*5 rf' 1.> in -6q , ur o' '.:g- +rtr' '*._, i.r99'' -s '.5 31,/.*P* I &l*,n) o'rj /+_2t GI t9 !-he *: TrW.,ffi-,ffi;'G" t1'u t {' t*} (t) (z) .lo*o J Sot.rtioa (a)Let&betteimpulsivereactionatD(Fig.E3.llb)and u=u'ilu.vj+u,kbetheangular t ,n, velocityjustafterapplicationof theinstantaneousimpulse. LHo=Ia,.E, yields ar, llc=oD*oxDC yields uc aud mAg6 = f yiel& !. r. y, t a.e the principal aries of inettia at fixed point D with = 120(0.52+0-3?)/12+O-252]= 10.9 kg.*r, 1f;, =120(0.12+t?11t2= rf, = 120(12 + o-52tll2+ 0.2s2] - 20 kg.m?. larg,o= DAx (-60i- 240j + 120\) = (0-5i+0.15$ x {-60i:240i + 120E) = 36!-69!- 12OL Nm-s, Il, 10.9 kg-m?, (1) (2) The angular impulse-moment of momeutum relation for fixed point D.yields /.g.o: f.ogo +. I!,.,L+ Iflutr!+ I!,tr' L= 10.9.2'!* l0.g,y j + 2&', E = 36i- 69j - 12t)& + .,. - 36/103 = 3'303 rad/s. ov = -69/10'9 = -6330 rad/s' ot' = -120i20: -6 rad/s =+ r.r - 3.303!-6.330j - 6k rad/s + oc: b +u x DC= 1a-SOai- 6.330j :6!) x (-O.i6j) = -1.5i- 0-8258k m/s -! + lfrya Je -OOi-240j+ 120k :+ Ij,,= 120(0.52 +032)/12=3.4kg.m,, I!, = 1fr(12 + o -52\I 12 = l2-5 kg-*'. "ogUl"t If,, = 120(0-e3 +f2;7tZ= 103 kg-nf, (-60i-l'i*r20$ - (30+0.15P)i-69i+(15-05P)L, impulse-moment of mornentum -J :-l --l :.l .-J --J J .J the instaotaoeous impulse applied by the cable be ' i (!I. 2 O).The FBD is shown in Fig-E3-llc' The principal axes at C are e,y, s axes with C,{ = 0-5i+ 0.25j + 0.15! m, ite - -.1 A= -r20i+ 24Oi- 219-1k N-s -(b) kt Larrs.-'GAx -'l -l The impulse-momentum relation yields n&pa, ! relation A fl+ = Irrrr. rJ -/ Lr (3) (4) for centre.of mass Cryields - €.-,L+ Ifru"!+ r9,.,k - 3-4r, i+ 10.9r.r, j * 12.&r, k - (30 + 0-15P) i - 69i + (15 - 0-5P) k -' :+ o. - (30 + 0.18P)/g.4 rad/s, o, = -69/10.9 = -6.330 rad/s, u,, = (15 - O.5F\l:r2-5 rad/s (5) =+ r^, = (30 * 0.15P)/3.41i - 6-330 j + (15 - 0.5P)/12.5lk radls (6) lb = I!m= [-60i+ (f - Pli+ 120\]/120 m/s (7) uDy = acy = (f - F'11LZO 9-e =yr+ glx CD = tb + sx 0.25j =) : I0 and eq(7) yields f = F - 240 N-s > 0u5r, = then il^ taut, than thai +1,the ^-Lrcable ---.i-o remains iarrl l. F =240 Ns. r{ssumc rl.-i m/"' Eeuce the assumption is v'alid and eqs(S), (6) yield gr = 19.41 [- 6.330i- 8-4 [. rad/s, g" = -0'5 i-t- L 2. F =-240N.s.r{orr-.thatthecableremainstaut,thenup, - 0 andeq(?)yields f = F'= -24$ Ns / 0. Eerice the assumption is not valid. Ttus the cable becomes slacL. with T = 0. Equations (5)' (6) yield g=_ -1-?65i- 6.330i+ 10.8k rad/s, sc = -0.5!* 2i+ t m/s. 7t -l l il r -J -J -J J J J J -_-J r -J sJ I since its stiffncss is.8'mte- Henee l.j t. IJ L:, l: t_. IJ IJ L (.),,.Tb9.:PIl$.G,:.*-qbl:.r*rO;;#'Provide anv impulsive reaction #ti""'.ui-!qs(s),191 *ltu f = 0' ii'.1;J;ir;; "H:T": **1ol lrorizinfr .Atil ::',t"": t1 .,''T:l'"J" er 'rt'Po (d, l're [n€ :ffi ff j::1t':', being gt,cb, gq1. The FBDf is sLorvn in (,- .. f,ranaa l,o r,t, be .the values just begre 1:-yr*1 T" Y: m, oc 05 * 0'25 j 0' 15 L m' Hence ;;';;fu ;; 'l *-;:: ! = ?;:;r: .i;,- @= -o.si- 0'2sj + 0'15 k e i m/s, m /s, ar = 0; 0. * = o!"i' d =' ,LL+ ursj + ,1 g t^' gc = !E= -2j {p = * + d x @=(o-r*rl +0.25e,)ia (oi -O't*n,-0'5ur1}i+(O'5-i -0'2&.,1)k conservation of nroment of momentum The 4 unknorr ns u',,u!r,4' u| are obtained using coefficienl.of resritution *::-t_ abour n*"1ryr"i d""iraaorr wirh o, and the m5) + + llio (- (8) (9) Ha*OC x _ OC x rr.elt =0 *OC x rQc Eb = Itc + €.t'.i+ tfirt'ri+ IgyiL+ (0.5!+0.25j - 0.15h} x 120(-2j) 3.4u,,i+l0.9oii+ 12.5a,: k + (0.5i+ oj'i] 0.15$ x 120uij = (10) ruf = -10'59 - 529au!" :+ !: i-*'.* o"; : -36 u|=o ':+ 10-9.ri=O i: (11) a'o--o'Lfu''-0'5tI=-0'6(-2X13) obr=-0'6ue, + (12) + o'' .stibsrituting q,,,d,'fromeqe(10), (12) in eq(13) vields ui = -1-2X7 "tlt-a1^::"(q' 3'662k tad/s'radls, tl,= -i-UU, ."alt' U"o* lc-= -L'237i */t' i' = -4'0'tli-4-041 point 0 is consen'ed s'ith 0 about fised (") The FBD is shown ia FigB3-lre- The momenL of ulomentunr (0'5'0'25' jt'"t dter impact' since g! = g = sb- Usiog EC = being a poiqt of the rigid Utai .l oa*' ]t'' . 13.6 kg-,.u, Ifl = A- 120(0'5X025) = -15 +0-1521 ; e. uL-0='e(ue.-i) = Ijt, = 12d(0:S2+0J?)/12+O2sa l2(0'?5)l It=0ks'*'' rfl = 120[(0 .32 +Lzllr2+Gls"+0-52] ={3'6 :ill':4'5'kssr2 9 &gro" If; = lz}f(f+ 0.52)/12+0-52 +0,5?l = 50 kg'nrz' t?' =0 - 120(-0'15X05) =: 120k, [r€}[!4J : ilo =9.+ QC y mgc: (0.5i+ 0.25j - o.15$ x 120{_2i1 -36i-t+l - [i: *j * ][;il =[-il] + gr= -3'6't3i+0'3?63j =+ Note thae this (f) =[riff] 2-I3h md/s does not comespond to the case of e = 0 for thesysteaof masspointand btocl(m*m1) isslacrsg in case TheFBDof - lfi] ofg! - I whi& only the-cornpone:t "'-""^: 9' Fig'E3'llf' DB = 0'5i-0'3i+0'1L' * = +uif+'i-tl _,'. =Q, e8 =-zi+i-: L m./s, s't - !8 - 6 * ! * ry:(-l i+'i'i+ oi !) x'(osi - 0'3i+ $' uif c8 I '':10:.;r;'+o-&al)i+(0'5r.t'}-'0:la';li-(0'34+0'&ailE , The mogrcnt of momentum g'o= gD of the systcm of mass point and ' (14) -(15): the block is conserved: ED + t*4i+1*r'r!+I!,u',L+Dg x nir* = 0+LExmttb =+ro.erli.+r0.cr;L+2erl&+10S1-03j+0'1Llx2(uli+ui!+ui$ !: 1 f.0.9ar! : !: ., - O2tti - 0-6o! : 0-'{ ,::ro.*.r! +o2o'r- "1 = tu" + 0.6u! + d, = 4-2 LL ;...r:;1 ixi.: i tj .Lil;.'.] l+I,i -' =+ =+ + x 2(:2!+ i- k) = (0-5i- o'3j +0'1k) u{ = 0.03669? + 0.018349ui + O'OSSOapu! . -'" u!, .\.r,'' 1'-' =0-9-t5046 - 0'0183't9u! * = -0-01 - 0.03"1- 0'05u' 0'091?a3o! (16) (1?) (18) , =... 1. Thelindof impactisidong!e.,---[. Eence ot,=t,'g','='-1.i-u..gt=uli+i-k ttecoefficieotof restitution relation Yields :+ a'k't'r+0'&'1 -l/'--0'6(0+2) (19) ob,-'6. = -e(aar-'*) (20) eqs(16)-(rS) .+ uL=O, @, =-0.036697-0-01gea9uf, o! =-0-06-0-03ui .using eq(20) in (19) : i, - 1.165? m/s, eq(20) =+ ari = -0.05809 rad/s, ar! = -0-09497 czd/s + y' = -O:05809i-0.09497! radls 2- Eor sticliing together v', * = = O-Lt {s- Eguation (15} + t'o* o,*":, io =o-*'t',- 0-tari' Substituting from eq(21) in first part ofeqs(16)-(18) yields + l-rt.t 0.3 -0.r1 [r.,'-l I-o-sl rdtil LrlJ= L_tl *r Lj;il- y' --0-03455i+0.05168j 1"' (21) = -0'3''t -A'lt', - l.',1 = [0'03455 I l:iJ L;nu:.1 0'00965{f, rad/s (g) , The FBD's of the block, rna.ss-point, and mass-poirrt*block ..g 5[esn in Fig-E3-1I5' ,Bet P be the fi.ted point, in space coincident rrrith B- ' CB=0.5i-0.05i+0.1!m,rc.=-0.5i*0.05j-0.l!nr gc=98 =3!-2j m/s, ga=O lb= -2i+ !- ! m/s' 'i ='',L+t'ri+t'ri k Forblock Ep- I{p + l- The liue of imPact g' =i 8".!i+tfr-'ri+/9"-'.1+ PC-xmls-g+PC xrnk {22) e3) :+ $=o.!-2jm/s, Q4l *=u',i+j-km/s- The b unknowrrs or,o'ri*t'.rtt'r,ul ate deterrnined f,rom eq(23), cou*rrration 'of notuentum of a-dfuection and the coefrcient of rcstitution relation. Usingeqs{22} and {24) in eq(23! yields tr + r'zr ifl 3.Ut t+lo.qrik+12.er1h+(-o-si+0-05j-0.1t')x120(u"i-2i)=(-0.5i+0.05i-0-tux120(3i-2i) . !: j: ' k; p,for m*mr: 3.aw!-24=-24 = !")'_ 10.gr; :+ :+ :+ *:'r: -12a,--36 12.*ri+120-6u"=102 .L20a,+2o',- 120(3) +2(-2) + d. x cB =.o,i- 2i+ (-li+.i,i+-lH x (0-5iu" * 0.1tri, + 0-05(,1 = t-13409.u, - 030228 'oBz = F (25) =0 -3.3028 + t;', = -L.44 * uf = 60u, 178 - l-1009u, 0-48u; 0-05j + 0-1k) la:* :+ :+ (1.13409o, -0.'10228) - (I78-60u") - -0'6(3+2) ob.-oL, - -c(as, -re,) e: :+ 'tr' a- = 2.8691 m/s and eqs(26),(2?) yield u| = -0.L442 cadls, u!,= -0'06283 rad/s : -O-1442j - 0.06283k rad/s + 2- t€t gL - uri+lryi+urL. The 6 unknowns ufrru'r,tt'r,us,uyru: conservration of momentum g for m * nr1. Equation (23) yields {27) (28) (2e)' are determioed from eq(23) and ?.*,f.L+LA-9uiL+tZ.&,t' k+(-O.Si+0-05j-0.1$x 120(u" !+u, j-+u. k) = (-0.5i+0.05 -I3 (26) j-0-l $x120(3i-2j) J .JJ ,l J J J J J J rJ J JJ J J JJ I k, - LJ ' : lr. ,n ';*turr* 12'"---24 + --t' ,--_l j L : B: * U u L t- = = (tr, * 0.1c.{ + 0'0&.,:) " i+ {o' - 0'tr'{ + o'st"')i+ of momentunr g 1 Usi:rg eq(33), the conservation t* ' . li: - (32) o's'"i) L + 1 + L tJ L l--l--. L L. for rn * 120u' k: 120o.* 2(2-Ow3-0'55045o'+0'l?645u',+2'8405u')=.i ;*, o'352s p'i+ mr yretds u' B + * 2(4'?859 ,li,]'l [Il 125'68 J lu' :=:ffi."l:::,t?l:[i' 3.12 *m15 j: [[-i.roos Exampte rny4 i' ,i,#:^ - m4 + rnl* = 24 = 120(3i- 2j) + 2(-2i+ i- $ 356 * o'24rtu - 0'55046u') = l20u' *'to'*,li'J*1'13409o' 120(u'i* L k + 0-240" = 0'1764?u') = -238 + 3il5Z9o'+ tr_ilfkl rad/s'r'r' =o'r881 ,,*::^ l [[l= [;li:;:,J rad/s'' = ;o'"t1*::t:t n" ope,,. oPsl'-l1i.ig.Ea.rz.) *o na,. one face lralf ace' has oue A cuboidal bo:t, out. in space. J; *"^'i-"t""t intothe c'on$guration i[ .,o-rhereisnoexternarmoment"::lI'*i*ft=i*:il::1,"",$i::*;::" ;tffi':i::;::ii:T::T":'T;i:il;l"J*veroci'lva'clhisi:mii IJ l' l" L IJ L rJ Fug :a-O e conhgurar*'ra in the: .oloig,r."aior Ut the centre of mass 0 tEcJ = trn![el = trc[o r*t o "olr + surface atea of ttre bethe mass per unit box z (cl zo' (e\ Zo- Solution fr"t'? ,- rrc': 7 & l_- : Es uox'-1 = *:::*nea lOo?c :*lu.t . c"' = lz(ir+q(u')Jo = mr (r) F . - tG - : r rG,.rl ='i:""::]::*1.*f,Lt u1 = aao11;ms 9 : i. j " deletion) 2- : ; eatdAEKF '?t3 =o26 - :1Ut .T',:"i-"E " * planeyzisaplaneofmassWlmltrratc ant iirapriacipalasisofincrtiaitG lo*:1o' nara{H l*- ) ) Ir--i ) L-' o 'J Ir.\ l- (31) =(0.07772*r-r340eo'*a-24a,-o-s5o@;r$:r*l!;i,,Xi}iir.f;T:11"(33) L_' L- (30) -l!1647o'; "'d''=8-16+o'48u'+43ur L2.5u!.-60or-6u'=102 *'" ('Lii (0'5i- 0'05i+ 0'1E) -;i+'t'^u)x r+ o'i+ x vL + d. CE=o'!+ ls = '3'5294u' , +., l, . !1,,=-',*'111'1009u'-5'5046u' -39 10sc.r!+60ur-tfi.: o::-i:::: ;ffiHJ{r,*.;;"i* GCt=..(i- $o/zo ' lf;, =-mr(ec0 (Gct)' u = - 10o2o(- o | zo\(a | cc'=u'l*'719(iJ =9(i-!)a/20' fl decornposition Gcz=bz-'rc - *'(@"@'- 2ol- ( + *s(Gc")' ";(6t-*l'ol .r+ - I!'=0' : tG c"l' o" o(tg oI nl { - 1s4/20} Sol2o' = z,"l c I to :' -i4 rf ilf ;Y -\rlA: t.:-d jv ,;Full box: paoel ABDE: panel ADKF: _!{ t- aa oPen box: ec(1) + ;; t- L- il,- o-a.tr. vr just after closur' eof Let t t: ar" i* u, !* urk be the angular velocity rvith G mass of centre principal a-tes of inertia at its $, -- {'' + A, :3o4o * l0o2o[02 + (al2}l"] = l2laaat4l f;. - -azalr.zllz+(9al2Ql2l= -343aaof l2A0 S. : a2 ol(az + 01 ! 12 + (tsalror?l= t283o{a/1200 E; : (l2lI 40 - 343! l2O0 + r2$/ 1200)oro = 457 o / L20 Hc = (8?j * 457\'loaousll20 lid. For aclosed box, Ql z,y,: -,. : 0, tny --'8t"tol740, $' - . :3n'o (3) -\ i a{ J-\ _.r_ eqs(2) and {3) Sield *'. = 45?/360vo, :? t= . < -.-.t gr=Il"ut,1i--f;lrrj+I.c.r.,.k=I7aac(u,i+-rj)/6*laaou'k II.c = fl;' llelce are the I If,o =2l2a2o{a2+(2a)2}ll2l+2[2o2o{(2o1?ltz+(alz)?}}+z[axa{a21tz+o271=37oaa/6' Since tlre'eriternal momenf, is zero, qq iEToto/i40)i + ({57@s1360) ! -c \ ,' ,r}j;xc, A spherical sateltitc of radius 8.translates rvitir \ i velociry uoi in a rarefied atmosphere having rr molecules per uuit' i.;f. f;q I - ;1>/) l{!(Jg3\(fi1'=u/g volume, each of mass m- Find the drag force on the satellite. \r-= '\------l th+g restitutron e. o[ restitution 1: T- - - -- -fY€ Assume frictionless collisions rvith coefficient of sphere aucl tlte [o frame i[ertial an solution r4re neglect chauge in speed o[ sate.llite due to irnpact. Atiach satellite in the by srrept zrR2rts volutne consider velocities rv-r.i. this frame- The motecules in the cylindrical of the collisions Pet unit trme' unit tirne make impact with it (Fi8-E:!-13). The drag fotce equals the impulse The impulse l, on it is expressed in L{olecule 1 moving with to maties impact rsith line of irnpact along'g,. terrns of its velqcity componeots just before ap-d just after irnpact: Blarnple 3.13 u'r. = -eugcc9. t'1. = rt1, = euosin0 uoccr! f1 rn(ul- -urn)e, = -(I*c)rnuscos0e,. = ula.= t'_'a- _r-<l '.::.-,'__._ 1. ,| .",--r l-.ti ' of these impulses ori the The opposite impulse !2on the sphere is Iz = (1+c)mr5cos0g,"- The componeuts in the ditection i add sphere in directions oorrrrol to ! due to all impacts cancels out. leheri'as€oulPo[ents up. Tlre component I, of !2 in direction i'is I" = (I*c)rnue.*AJ(-cos0)= -(l Let JV be +rlnruecos?d the number of molecules hitting_ the sphere per unit titne ar angle 0, ,lV betrveen angles 0 and' 0 + d0' in unit time) = a x (volume swept b1r the arc between 0 atd 0 * d0 cos 0(ftd0) and length us) = n x (volume of cylindtical ih"ll of .udius Esin d, thicliuess a R2 o,sin I es 0 d0 n[2r(E sin 0) (coe 0 A dlr]ool = 1 .!- !1 c€ 0d0' The impulse per pnit time due to the i*p""t oi'rV *"f"1ks- is .:\'I" = -2r(l + clmnffvisin? by git'eu rl2, is from 0 Lo Hence the drag force D, equal to the total impulie per uuit tinre for 0 ranging *12 ' :' .',. ' ' D.:-Zr(t+c)1a1n113l"'"tr.*. iA;,-2"q1+c)mnp?r,i1-|cci{g}fl/?=-}:r(1+e)ma.Ezu! - : ' +*r rD :! - I Ll. tj l;, "t''1'; 6i.fo ,i,Y ht slarnpre 3.14iDisc-rof iass and its centre has " ""t;"'L!il1;nt-*:"Yn z attached to a sPrins Ij lj l-, rJ IJ I*L The line oi tttz=05 L rj L L. L L I-- olr^,rttzo are i*n"o is along kg, .u!,r=t'?r eq"trr,(ir-+,:; =;; -5€060 . tal and conservation of'fo- rnornentum: - , r/- ;r"'1;1_:;,j.-1;,1', on disc 2 is Jv'q' lnstantaneous impulse The mor.ion of disc 2 aftcr lnpact lltt" r ,. .., . .. talies place under a t" iffi; - ;;:':1;;" f6lss 3,s rvell --,:" force which is a couservativeenergv of disc "ott"n";:;;; * ianical sr.:*:ff.;;;**"n'ls'Ius'ia*erimpac'l''Lrre 0s r:*::l*"**:;,[T.*TIffi 1 *. ..u'rJilffi ;-;;; = aa+ l-o.z * tq ".a spring extcnsion,6, 0'2 = -l'l:h: - -1'129' *E'16go -si*og) =4(-cos0g nr/s' *t*'*" st2:'12g,,.49.'=?'2(-sinde' ' + ,i= -t'izm/s' u! = 8-16 m/s' a r1 r = 0'{ - 0't = 0-3 m a,.nd spring extensio* - e" = B' spring' tlrc of p*el ' t'ol' In the maxirnum compressed t'aoalv' i" this pcition be *t""i"'j'"t'*ro""ti"t rn' 51 = -0.1 * t* ur. = ' 'Q'= ;) u-l u,, t'--\ 'T+v m/s I'6'32 rn1 = lo'rz O'6(8'16)103 0'6(8'16)/0J-= :fr+V1=f*V,:.1*,1o,i4,1+ite1=!::,,:+.,i1+}t8, ..;,,. .'.,.. }(0.6X1632f1}t(0.r)"=.}(06)tq.ro?.+r..rzzJ+}[(0.?)? . &=39?0 N/m:3'9?0kN/m ,r,"l-onr exteosioa u" b -rtt " '- t'elocitv uc: .1.= 0 and circumfe*ntial =.q'j}[oit-t + 6) 6(8'16)/(0'{ + 6) = 4s96/(0'4 "':::1 -"1t0-ulto:+esg6'?/io'4+5H+ Ho,=@*o=,.*'{dn r , ??" -:: * uto) ,L^-r'ro= -e{u2* .', mroio*m*Ln=rlr1o1o.**'-:''+l2u"o+0'6r'!*=l'2(6)+0-6(--3):5'a(2) =+ IJ 0't determrt * ! t': and' cos ' .'-' *+ IIo' =ttr2rluct = ta2tlo!4 i 9*B * d;;';*) are gi'en by titu i""t t* t-'-:t just U'*" veiocities of centrq *a 'ogt'lo or' j l0sing=6 m/s' 'dl="t1 =2taAfs L0co0=*nll"' u'r. = =urr ttL=1'2kg, =-3 m/s' u\ =t'4 =0 rtt m/s; =,-5sin0 : = -rl Solutioa L f- tZry -trt'*1 {(;:;)j:-o \-1ff1; )"1^ i 1*;Tf:#: V \|{ B GFIC-it Hfji"i'3i"n"oir'iiilillii;il*i.* i ill*":: lHilj;II:'lJ:"J': .":[ r"-t ''JY"nu"-"lrrl' and b transr,tioe *ith'Jffi; 'il is Gv'€r ' *t:'ffi*'.xi'i'::5'lu.";fii:i:-':ii:iff";il:: T'"n'l;,. I\y - /'"t surface. rhe impact I ;of 1[":J:::'";;;;; i,t,, ;t Y"H:t compression = ?U 0'?' 1'hema-'rimum '. smooth with coefficient "it*"*tt"" *-'t.dY.,: -fr lult, i1^e"- /% thespringin subsequen' *o":n :t' :::I},'#jfl:tjilff*: and the """rsrduringirnpacr,instantaneous ;5T*-#r"*1r::H:X.*"*t qrnL o",tiilri; , ou disc ";;i;;;ngjnrubsequeor impurse on -:t:..':y;i;1ff.'P'ins motionvelocities -;"::"",, :+ sino" = -::6- The of-tbe-spnng----]--*r-r* o" r4b) and cos60;a/5 = 0'8 = --rnaximum extension aogular velocitv =f +v * i(3{0}61=' =lionltt,iu'+t't221+}lrszoyo:)2(4) 66J +nsa-ze: - 1s9.9, -.--::-; + 39?05{+317 * (6+0-1)[39?A5t +r,tg52+ 15?'86 --tz,s..fl =-o 1?5i,9--=,8'" =+ 39?06!'+ 2ng62+ 15?'86-- "= "ifil t'i17.538 =o (5) , - ''i' r'Jg "i : ; : '":83'):: ']16"+i -'=:t .'' ' al I -r ll fi'< compressed position 5 = -0'1 is a toot where (5 + 0.1) has bei& fgtofed out from eq(4) since the ext_:me Cardau's method,' is 6 = 0'2005' which is of eq(a)- The only real rqot of the cubic *"oiio" (5), obtained by ,-< g' (b) ti'ewotk part a for the [he plate at angle coefficient of restitution e = 0'4 and 2' if the rod imbeds in the motion of the rod just afLer impact plate rvith ''---- the same motion just before irnpact- (c) Find uvv r'e'v vr free L@E case of fs ;nrooih rvith c = 0'4' ()Jt in part a if the plate is massivc, i-e., m2 ) mr and ll1: i*O1rct-F.t \-+v^ - { < 2-tls Ai" r-il;.- , ,i L lJ*a i"l oo-:-ri'es, f-Y1-o{ :' (cl Solution (") t * t* 0'24 m' u'rr: it. - -5sind = *3 m/s. t'la = -icos0 : -{ m/s' ri = 0'3cc's0 = :O-38 m, ''' = 0-4 rn' oL = -2 radfs, o.,: = 10 radr/s, dr'= 0.16 m, @ 0-2? + 0-05?J - 0'016 ks'nr? r9; = 0.1(0.6)3/12 = o.oo3 kg-rn" 19- = 02[0.6? + 0-f ]/12 + [{o' for rod L' ttb' - Ho' for rod+plate and The 3 unknos'ns ut'1,t,\,u". are deternrined using f/i- = grouud coirrcident. q{th '{' coefficient of restitution relation, r,ihere D is a fixed point of tr":ri- rnlu!,r11 = f9't, -trl1tr1'11' HD,' :+ :+ a.,i = 8ui* +30 0.003o', - 0.lui.(0-2a) = 0.003(-2) - 0'l(-4X0'24) :+ I!,r;'2 + r9.'"ri + rrrlr:'1.d1 - *ur+ I?:-, + r7r1u1'd1 For rod+platc . IIb, - Ho,, , :+ For rod IIb, == :+ 0.016a,1+o.OOi1tu',* + .u', = * 30) + (t) 0.1ui-0.r6) = oI)16(10) + 0'003{-2) + 0'r(-4xo'16) {2} -2-btt'1n U'Bn-u'en= + ui. = -e(uso-tA;) -0.9?143 m/s, and :+ -!'ro - eq(Z) + l2-92oi^ *72) = :: - {3) Jt + (-2X0-24) = -4'48 rn/s = ola + (-lrr a'tn =u'r* +r.ri11 = .r'ro * (Sui" +30)(O.il;: 2'92u'ri+7'2 t)'8,, ='-tt t'z = 0-a(-25u!* ) =,-,oti 'n P. = .n2rr':0i4 x'10"-'4 m,/S. tu/.l ei : 14\ < -0'4(4:+ 4'48) .a!r=2-429 rzd'ls. O for rod+plate from Hb,: Hs. for rod+plate' Let m'o'i' about z-axis at kg'*'' Ifl+[I9,. +m1(di +a{f --0.016+0.003+0.t(0-163 +0'38?)=0'036 2. -i (fig.Sf.rsc) is obtained be ro- P = :+ lo,.t'o = I?r-z+ I?.'., + m1o1od1 Hb, = Ho. + 0.03&ri = 0-016(10) + 0'003(-2) +o'l(-{XO'16) - o'l(-3X038) =+ (b) 1- The FBD's are shown in Fig'Eil'l5d' For rod+plate: + m1u1rd2 ca! = 5.667 rad/s e.:, = ao +.r2 k x OCz = l0 L x (0'2 i + 0'OSi) = -O'5 i + li ^1" ,i = orr='-0.5 m/s, u2, = 2 m/s. I9,' = 0'2(0'61 + o'l211tz = 0'00?5 k8l-" 71" t *: : : f^,, f.: r_ L L L L L LL IJ ;' fc tod*plate *a For plate 2, + :+ tri : + :,.t1: ::,':l '-l =8ui' + 30 t::yi-rn2u!.(0'2) = *-' (5) - m2o2'(02) (6) :+ rnlolrr + m2t/2n = ntlvto * rn2t'3o p!. = -0.5u,1' 0.lo!o *o2l/,, = o.l,(-4} + 0.2(2) + tL - 'S^^ - -0-4(u3o - urn) p.'r: p, (7) - 0-13334) - (2.9,2ol' +7'21 ---0'4({+4'48} 2-066?of. - 2.92oi,: 3-9413 * ,i" = -o.oozo *7" [2-066?(-0.5) - 2-92]u'1" - 3.9413 ,4:0-{985 mfs,' . qrl'= 1'992'rad/s (2-06674 usingeq(7) ,Eqs(?),(6) Yield The 3 unkuorvns u!,d2.,Q Gig,E3-l5e) are determined using P', = rod+plate, where .B is a point fixed in Sround and coincident, rvith c:- 2- + .:--- 0-00?*.,1-o-2ui.(0.2) =0.00?5(10)-0.2(2X0'2) 5.3333uin -0-6666? + ollo : ?2o +{.r?(0-2) =2 + 10(0.2) = 4 rnfs, e + :+ u (8) P" l'" = pr| H'8. - E2' uc. = -b.i - 4j m/s, !c, = -0-5i+ 2!ni/s, {c"= 'i'i+ 4vi (ui" *, = *,+-lk x CzCr:4-i+ u1, i+ -lL x (-0-04i + 0.33j) = (u!, - 0'33r{)i4 u1r: -3 m/s, ul, -4 m/s, p2. = -0'5 m/s. tr1, = 2 m/s' o'r"=4,-0.33o!, *: n -o-oa<.r! - tot O-O{:ri)j (e) " pL=?, + :+ plv = ps + l--_ IJ ]J l- L-.'i'' Ll) 0-1(4= -O-33&{} +O2aL,=0-l(-3}*02t-0'5) itlTtlr* tIr=4, = tnrrrr; + n!3u2e b.r("r, 0-04-i) + 0.2ui, = 0.1(-4) + 0.2(2) + u!r-zS36 L_--l , + uj,.. =- 1.3331i + O-ir-i1f = 0.01333&-.ri rn2sc. :: 0'22',i)l O; (tl) l rad/s. its 1nglrlar velocity temains unaltered- Ileuce ot' = o9" 7 O '4(10) -'A;m/s' relatiirn' where The 2 unknowns ,1,{to are determined ftom H'9, = HD, and the coellicient of restitution D is a point fixed ia grouid ard coiocident uith ,{. . (c) Since the platc is massive Usine eqs(3),(4) 13 r,',". 0.0071,i + 0-00iu'l + 0.r1'g 0a,-t'02666?t^r!) - 0'33(-l-3333'0-33(-3)l = 0.0075(10) + 0.003(-2) + 0'lt-0'04(-4) - As in eq(l), forrod Lj-i : Lj"l ; + *, : (-1-3333 - O-22oili - 0.02666?q'i j I?: riL + tf,',ri k + czc r x *&,1 = A:.; k + tr,? " t f ]$ + Hb.= Ee, * Eqs(e)-([l) t'[J:- mrui-+roa{, --riirur: *lztzttzt - + IJ ; Eb. :: :+ :+ : ED, E'p.: Ee. 1 as in eq(f), for rod Using eqs(3)da) and l{ IJ L- , -..1.i.:! ': For rod+plate L I- f platr-2.t'o= po CiaetAmined uiing f6.-= Eo. Yrod l. rr;' = HD. fot with :{' co'incident groirnd of point *"fiA"i:&o=t6otloo'retat1cn, whlie D'.o afixed ate4ualruow*ri:*li}fi 1 IIb. - If o, + = 8ui' * urn) r.r', *1 " _l*,:]1..= -0.4(us, ' +- .i - { -.'(2.92ut;'+?.2) = -0.4(4 +4-48) Eq(12) =+ r./.= 30-526 rad/s ,.. -[g (12) 30 u'1, = 0.06575 m.fs < to a thin disc of mass lringed B1-mple 3.16 A'iqd. Q4 pf y.nass 0-6 kg is hinged aL O t: ptates at 40 rad/s' the sijiona*,.lhldisc hetd is rod position, the given the In .I.t'" hinge at is frictionless 0-f kg at / (Fig.E3.16a). asupport and .'j 'A spring is unstretcbJ' spring has a compression of 40 mnr and the rotational Find (a) released' is rod The the rod from hinge at oon N-m of 0.1 torque frictional constant is there and velocity of the from o just after ,.1"".., (b) the.angular the angular acceleration o[the rod and the reactio. t"' from o:'n:i_':::::i,:.t:::::o;,'$":to rod, the angular accelerarion of rhe rod and the reaction \' R, ioi,,_ lot o{r=4d h9v/-s -l:Tl Ad#,rF:t++T_-{f e, Y ,ph"..ffi ,,,{T>€i, 33i ;!'rtn e+l ;(,{=o*6, (b) tbl V6= V^ o'Bwr = o.BCr and the disc o.l"Lo*uo in Solution The FBD's of the roi+disc ;ffiH+, :T :-q(.-N ,,; / t'tt ttll - T:t-"3, t', :' Fig'83-l6b' Tt Let ,1,0;1,It,"ii;,i, the rn? : ;, *r, ,*. = 1:1.(u1),/2 = i;i;;i";;i;l ;.;l;,,; L')r'.l,2'i'o'tttug"1"*f poriitionofthecenlreofnrass'4ofdisc2' a,,gurar 0.0005 kg t'elocities'oftherodandthediscbe t' Ma=! + t-'2=0 + I,o.u'='1{a-=0 * (li u3=co8scant-40rad/s the ntoutettt' nr and g^ - 0-&'rt go' Fo: the rod+disc' The e-xtension of the spring is 0 : 0.2 siu 0 - 0.O4 L" [or tlre ai o, kiDetic *"'g'-1'and potential elrerg}' of mornentu m Ho,abotrt the fi-xed axis of rotation general position 0, are given b1' EoT : r!,,,t1+ gf;u2* 0.8nr2ua) 0'o2 O'OSe'rf + 0'4 +(0.128)ri + lto.r(0.s,.'r)" + oioom(o1:1 = - 0-128*rr + 0-0005(40) + o-sto'l)(o'o.11": : l€,4 + l!m2vi+ !41-;1 = o':11' + (2) (3) y - lkraz+|r(o-zsina-0.041?-n.r9(0.4sin0)-*zg(0.8sin0) =92+50i0-2sin0-0'0{)?-0'32ssin0(a} ro' = No, gields ri1 ' ur = 0 arrd rRr ', R: (a) The FBD of rod+disc is shorvrr in Fig.E3.l6c. using "q1Z;' are obtairred using f - I*iec- and kinematics' I{6'=!t,fg,+0.192&t=_0.1+{(0.2)+0.6g(0.a}+o.1g(0.8}:}4.,1=20$0rad/s?(5) t6! m/sz' * = -ul1a.a)i+6r(0-a)i= Sitnfs?, ee = -"'i1o't)i+i1(0'8)i= + Eri+ Eri+ ('l+0'6g+0'1e)!= o'6(8i)+0'l{l6j} F= Imicc' + ltr = 0, '.R2 = -4'467 N' yields:'r1' configrrra*ion 2 f9r 0 =3}9'='irf6 (b) Work-enerpr relation from the configuration f ,for.0 ] 0-P putl on The and kinernatics' ,"i"g l:imtg6, Using.q(2), H6. =':Ms,yields dr1. &,Roare obtai*ed dre spring for 0 = 30'is 100(0'2sin30o - 0'04) = 6 N' + Vr) - Wn"r-2 + 50(0'2sin 30" t(o-09&/? + 0.4) + {(r/o)'? * (Tz+vz) * r.rr -("r - 0'04)2 - 032esin30"ll - (0.{} + 50(-0'04)?l = :3.451 rad/s frs. Ms, -O-192ir = -0.1- + 100(0.2sin30" :0.04X02cos30o) *0'69(0'4cos3f )+0'1g(0'8cas30o) + &1:2.772 rad/sz a = -uzr(O.a)e. +6r(0-41e4= -4'?638g. * 1"10889 m/s?' ?1 -0'lr/6 (6) '2(z.16) (7) (8) :- C yo_ ',,,.,...,.. ='', ' " ': '" :--; r'-; : t' --' - j05(:4.?638)+o.l(-g.ozis) E=Err,.g"r. :+ - ^: 4:'L45N -'(9) fr'+ &+(0.6g+0-1g-6)sin30o : * (10) s. + 6) cos30" = OO1'''0"1 O'1(2J1?6) O'ls'(0-6g * + R+ :.0't* : er-* general configurati'on' ri1 can also t" coovealent oipression with system d.oJ..1," one {f ia f. For this . o 1 -) @A=-r.r?CI.g)e''+€i(o-s)+-+-9'52?5p;+2,21?6qrtils2' - 1: " 0.096(2trrrirr)+[2'+50x2(0'2sin00'0a)(0'2cos0]-0329cos0\'t1=-0'1tr1 (ll) cos0] = -0'r + so x 2(0'2sin0 - o'oaX0'21T0) - 0'32s + 0-096(2;t) [zo + eq(7) obtained earlier from Equation (11) for f, = 30o -is: identicat to o1factor cornmoD write the quations of on caocelling for rod+disc' $'e catl also *"*"oilequation th. writing of lto..,Instead t!. ilo, = to obtain e[minat,e the r:11,ioFs 8s"fi4 t.he disc separatery and rod tlre of ",,a motions is shown in Fig-El}'t6d- N' (c) The FBD of part og of the rod having miss rns = 03 kg .rjng i f' itf, are obtained aB: 3,ri(o-21 :+ N - s-osg'N 8'+JV-+(0'3g-6)sih30o + . f." =m3aDi Fe=m3aso + E +S+(03g-6)cos$!r=::t(o'l ^=.,',i:=2'814N equutions of motion of /t{p,=I!,61:+ilf-0.1-2('/6}+(03g-6)(0.2cos30.}+S10.+1,=.tcg(o.al./31-t = 0'5954 N-m N,F'Mcatral.obeobtainedusiagequatiotrsofrrrotiorrof(partBr{)+uisc. + ,1I 3-1? A 4-bar rinkase -*"Y ': ::::::1 ::i" ::T:i:T::'i.:':;:::':J r"Hji'x'I'l the glr'en instant '::.t_i:j", i;;:iT:""'.":J; ; i;;;.;;.ir.). Find rhcsupporr reacLions Iat Exampre i 1) B ,in Solutiorr \\t first exPres &r t'tz,t">3,A2'@3 -{Y ?,1t Bia*- 9' the connectcd tinkaggs' iu ternrs of arl'&l usiug kinenralics o[ Pointso.r{onbodyl;.r{,B,Coabody2andD,Eonbody3havetlresanreplaneofnrotioa.Heuce rlB=0'4jnr' 9B= 1-2!m !r3=r,3k' !4=!'p=9'' m' OA: tn 95 =t"'3k, go=uD=-ql. -t4e =0'2j gz:-rL, ir =.rrL, 0-4j -'''t'(t'2)i g8 = ee +;'i x AB = os(DB)! + a^ =u1(OA)j-- 2 * lj = 2j' "*.1:':.: t1&-+ a"! x'48 = -n{(pB)i+a{DB)i ,^ = -u!(o.A)i+6r(oA)i: -4!+ alj-, ea = etj + -4i+ar j +ti2 k x 0.4j = -1'666??(l'2)i+'3(l'2) +!:-4_o.4uz=_3-3333'j1w1=|.2t;4+til2=-1.666?rad/sz.til3_o.&83&.lr(2) (3) sc = sA-uiec+-zkx Ac= -4i+ti'tj+ti'zLx0-2j=-3'6667i+d'rj 9 (3 x 3) g,r=2krad/s, gta:-2k, The 9 unkuowns 8r' The FBD,s of 3 links are shown io Fig-E3.17b. il ;l"TI"" il::r,"",' "f' ,J I I I I I U U _) 24 - a1e t i t I :_ ; I determined'using + 0-3333&;" [1(1)2/3]ar + & = -te',ss R6=-10.886-o.*ilt =l|.2(L.2)2|3]ri3.:+ o.$s+ F4 = link3: Ittp,=IDt,t3 + -6_r.2g(0.6)-1.2R6 & - Et -O'1, Fy - m2aai + link 2 : "''R5'61 1 (4) (5) = 0'4cir 8o t ... : : - o.&r + , 6t = 3-ZS1 Bd/s2 tis = 3.151 rzdfez, fufi5 = -12.40 N E€(2),(4),(5) :+, -17-83 N, :+ :+ Bs +.8" : Q-A4444 Iink 2 1 Mc, = 19.&, -0.2(r?" t ng) = ll.4(0.4r2 /121(-1-6667) + -10.886 +o..E., : (-19.095 + 0.3333ur) -0.4c .F, = n2aca + Es(8) and (9) liukl: + -&- 0.4(-3.6667) 8s :0-7555 N = -0-7111 N, l?r * Es = t[-&,i(o-S)] = -2?(0.5) Rz * R<- s = l[.ir(0.5)] = 3-781(0.5) =+ Iiak 3 -- r. = m1agr" 1 = tn3acai + - Fy = rn3aca- :+ ft: J?5 =-35 - Rs = 1.2[-&13(0.5]] L-2g = 1-2[d,3(0.6)] = : f (7) (8) :- (e) = -l-4667 Es Fr=mracrz Fy R5 (6) r-2[-1-666??(0-6)J -2{3-l5l(0-6)J - {'i (10) :+ -.Er = -2-756 N + E2 = 29-53 N :+ l?r - -2.711 N + Es: t-641 N I -r '\_ g1:mFle 3.18 Each of the trvo rvheels in ttre mechanism of Fig.E3.16a has mass m1 and a-dal radius of gyration h- Each link OB has mass rn2 and is modelled as a thin rod of lenglh f,. The collar of mass ,n'3 slides on the fixed vertical shaft rvith i constant friction force .F- The spring has stiffness t and is contacted by the collars rvhen the links rcach the horizon[al position- The s.vstem is released from rest at position 0 : 01- Assume that the friction is suflicient, to prevent the rvheels from slipping. Find (a) the acceleration of the tolt". the.instant of release, (b) the velocity arrd acceteration of the collar *'tren d - d2 > 0, (c) "t the r-elocity and acceleration of the collar rvhen 0 = 0 and (d) The maximum compression of the spring. (e) \l:ork ouL parts a to d if the rollers are smoolll. T. : ._ :'._r l ._q ,s b J\ :-\ \a/ i-_j- v _i9 L :.fl+r---" > : l*r dotqm (di it tbi Solution The work<nergy pdnciple is conveuient to use for the conrpletc s]'stem of I bars. 2 rvheels and a collar, since this s1'stem has only one degree of freedom and the expressions of.T,Y and tft6 can be s,ritteu conveniently for the general position of the system. B moves verticatly down and O moves horizontaily. Eence the inst,antaaeous centre of rotatircn of link OB is at the point of interseciion 12 of [he norruat to uo - -to t e O and the normal to og = -u j at B- llence the angular velocity r.r2 of link 2 aod os,uo can be related to u using IzB -- LcqO,IzC = Ll2, I-tO = f,sin 0: f-ul = ol(I2B) ': __rf :i -j ,l : a .!_ L os . trr=uo/f =utatilr. ?and'[/ are expressed iU ternrs:'of,trand 0: 7.- lrm3tf +2flmpf;+ ]msefrolJ +2{lm2y1+ +(m"Lzl0)tl,3l (3) = i[*. * 2rn1(l + k]o/':')tarr2 0 +lm2*c2 elvz (41' V = tn3gLsir.fl +2nzs( |f sinOl + +&42 =(rr.1--l- nr3)glsin e * l}d,2 the t+rm |ta2 Ueing included only if thespring compresses by A- Wac= -Fu. Work-energr relation is ."_ ,a 1 F- .,| Body 1 rolls withsut slip =+ applied in iategrated fiorm to get u and in rate fiorm to t+'t = Wnc :+ Tt ri in the configuration of iuterest; ,Using eqs(l)-(a) ii 't /(I2Ol : vc/{IzCl, :+ lal=o*.€{L.. ry = }usec0, uo = uean0 (l} (21 $:-tsec0lL., .' Ya = Lsift?i {;l,:r' ii=,Lce00..-.:-a . + - _.,=r lF ,i ;'-r -i._E ' r\s-{l i< i !!; r-- :! ;: _< r .< ',' E [-"+2m1(1 +k|lrzltzrnze+!rn2seczrJlon+llhrnr(l+t3/r2)randsec2d+$m2secz 0tart0lv20 *(m2 +m3)grcos 06 ,,81 = -Fo (5) ,--- -'iii6.ri=;1qt Substiiuting 0 from eq 2 "*a ll'!' ."*.ayl,i iyyu,".F** ti1:," r:"" iiigl*lo" [m3+2mr(1+r3t")t"n'zai3"Iirf;.itt : ::, : ^',,,,,., ,s<;c,'*$mziec'dt"n'lo2.*otL"'. (a)Attheins0aniofrelease,0=0larrdu=C.I{eoceeq(6}yields^e.r,. 0t1 gr rl / [''e + 2*'(.L+ *3/t') ttf + ]m3sei 1r = (ra2 + 'o3)9 0 :82 and using zt 0 =01 to configurat'iou 2 aL c< >nfiguration ! from relaiion Applying rvork-energy (b) eqs(3).(a): 1zi + va) - (]I'r + V1l = Wsa'-' * -sin82) orlo'+(mz*mslgL('sin92-singr) - -Ff'(sin-gi k,k.rt,"..'2e.+$*r*"; (7} 0'}l'/' + it*r+2rnr(t+ {ms+2tnr(1+e3lf )tt"t0e }"'2sec? +. o =12{(*z* rns)9 - F}L(sin,, -"r"ili I 03 for d and u fromeq(7)' from eq(6) oir substitucing obtained then is The acceleration'u i *i:"tttt* o: ,Pt(*r'imr)g (.) For 02 = n- [(mz+'ta)s - F]/(ttt3* 5tn2; Fop'6,- 0, eq(6) vields rrheels have zero vltocitl' Iz-is at () and hence tlre configuratioa' this ia since *-hen the m1 does not appear in eq(8) 3 of the syst'em (Fi€1'E3'f8e) configuration in ;;tt* of at0"ction co.figurat'ion I and Lhe {d} t et 5 be the t-*i*t'* ."Liioo u.t-*gr;rhe initial verocitn 4ppryi*g';;.;;;.gy zero wrrore slstem.has yields .."fi*.ran 3 and using eqs(3) and (4) (T3+Ve)-(fi+%)-Wnct-s 0l= -F{Lsindt +6) rneo6 ][62J-[0+(m:+ma]glsin (e) *ti;t::'ff [0-2ra,2sf6l2)+ + rlLsin,1 = 0 (*' + (}, eqfT) 1;; * (E) j'Fl'f,ntn"'/t* : Yields - - "r3)e - '16 of the spring' is the nraximum de0ection ;1"'f; :";;;:1",":':#itH*"il'H.:#':'3;*1""..x.*:::,T;*:ll:*""=""t'1" l-€', tJ1 = ur' rrErrvv *'flo;l' of mass is zero, enrain unaltered arrd (9) r'.fi:"::L::: external forces about its centre to o1' Inciderrtally eqs(S) at'":.:'.:"::,ti::,Xl:l:;ir11::i;; "*ta.r."l since tlr-rs"t".'n terms involving t6 sec to zero. pcitions because u.r1 = 0 in these ttuo "ti*t t*"" fot it'" case of rolling rvithout slip' ^ ffi n *-------I r"' ;;:-,;;;";'';".'::' ]iilfffi "i"T:;':i;;.';.1*' ;n"-*o*'o'k :Y'::ll'":lili;'Jllii -{[-:',-- lFil tle:::::1.:l::: lffie,'*a'&[ -P lfQ{-i'" ;"";;;;*i'ir'"i' cen*es'r t-) e i*:' Tn',*:::-:t *Y"'l::: svstem 1L3 'fP 'cthe --ir orrorarion is 16rnrfr. The ffi:,ffiJffi:;:;""tsrops \o) \o) !.r^.1-< :. ,. ,. - -c,r.^ brocris*-v--Hq^i -.tr' rt= v-*Q-, s1.s2 preventing sudinsor rhe inertia or f::[[fJ:ii. i,i;;r. tt*ou"a' 6)-ttt io =-1'7'n1t" ind angulat acceleratioir of Utt CD' velocity . :'tl The stops $1,s2 it€ simultaneousl' find angular velociry - CD' the ttnsion in the clle-an:d, -l-t'd -fHL*l^ otthe block"a when so- thal th1,1a"timu1 r'rs be should What (b) | the axis of rotation' uro so ".*'fr ;t l*01*"*"*'oi uu"r A is rs? (c) What should'be ;;; .,{ rq? Ilst9f i|]fil Jieiust b""o-o slack when r{ has moved.out by bar and acceleration of .A relative to il:;ffi;;;;-cD TJq; ' l'u+.' .l::'" '- --9 ;T'?-t'")l"a ^r fi it: "t'",::: R,C ibl 8a ^ 1 Hl: ':**,': "1:::.3.":::::fJ"T: ;::i':"ti;::'": T;'n:::i*:k;*;::;T:',ri::Tri!; :,::?":JI"T:1Ti;i*il.ff ":i'ill.ff Hi:.,T*f sol.tiorr The momeirlofdheextern' -tl'-' :::::.JH"fi ff;i,ff ffi :*1;tt;f"';;i::"'",:X::"* leb) f SmT#;'TlllT:T;T;:"J#'tr:{;"%;;::,iil:-"P'Qrravemoveddown f"i:$;i ilT:lffi:::;;;;;;""i1' 'l';|2=: i|2(rig n{52ti +t}y" = E3 -- Ho. =16rnrfro +2tntlc't+ m(52r1+2rz),'t = rn(52rfr a 2rfr)r"s ^ 1r. . a -.-.-2r-.- 2(2m)(3rs\2t't coostant (r) 54mr3';o = (2) . ,--.^:..-rl =+ Ho,=m(52rfi+2r21-+4rnrir^r=0 (ii)')i + 2[]r(r -'o)'i -4mg$.-ro)12 2 + ;'11-*'41(zm)fl&ardt + (3) +21,1'ml$'t')' T +v= !(romrfr)c, 2m9(r- ''o) 2-54ry(r' + 1'5mi? 'ol2l'a - rn(26rfr + ,'),''f radial velocit'y of A be ur' for r'= 216 be srr'g,r and acceleration Ho' = 0 [eq(2l' (a) r-et the angular velocitr--and -"Y y =*n.o.]'i 1*i"t "*''. ;ili;, f,lo, from =consramr arl,11,61 are obtained _.0 (4] 0'9r';E - 0'9(g/re)ti' 54mrfrc^rs :- , :.:= Eq(l) ? m[s2rfr +2(2ro)i]-r = i =:n(26rfr+rl)t.,fr Usiogeq(3),f+,Il=constant+,,4zo,fr+1z,.o1?1..,f+r:6rrruf+2.5,agnfr|tg-2ttt.gti * ffi'lifi'ir-4nr(2r.s)u1&,1 =0 l*r, * p for-trris position "ru "r.or.,r. The FBD,s of btocks ,{ a,,d. btock P ; block -.t : Using ' {G} :a' = -8,1ro1l60rs = -0;r'*53(c/ro) block P of in Fig.H}-lgb- The equation 'rotio. 'of 2T *2rrrg :+ + F, =2*up' {8} T -2futttg = 'r{i-r'i(2rs}j ri'l mG f" = F6-m,,r$+z;61+At1=rn(2r6i'1-l2r;;-1)(9) (tO) 1"-rng"j0 + F-:mi rz?=0Sirng' Atl=f i39mg' N2:1tr,!' i=0'089' ))vields eqs(?)-(lt of eqs(4)-(6), thesolution i 8-O8g ia tlrc arerrl = L211(grs)U3 a*d = CD tod to relati*e .,{ blocli and acceleraiion of The veloci* outward radial direction' *!.1"':: ,*:y:::.:::"::1,.f,: ilY.Tl'ilH"ffi,". eq(3)]' ::;[j, :."."**i*tl)l a'd r +v =coustart [usins Eq(l) + *L52rt+2(2toi21.,n :!4tnilato using.eq(3) , T* v= (.) e con-staru.i;: ''n nor which yietds t^r6 = 0 J2,t s af€ obtainCd a ..+,,, ,n[2ffi + (2r:d;?Ic3 +i.5rIg1fr/ro - Irom eq(11) in eq(12) i (11) 2m9re = nr( 2orfr +.ipl (12) = 0'4303(cl"l'1" - r- FBD's of blocks the ror r*Lu3be 19'J:'::LH:.1"ff?r"."=;""."'t ["q(r)], l:'Y"" ",,go1".,"locG * in Fig'E3"19b with ? =!'.4r3'are'r al and P are the ""*" ra\for block A and F' ='Zmap" for block P: ' Substituting A t?) =zrn(-ilzl = ca2 :='* rhe -.it - Eq(l) =+ (13) m[52r[ + 2(2ro)2]t'a = S4mrfrars ; F, =2tnop, btock/ : F.=*(i-tC2) block P o3 = 0'9rs + (14) :+ i =2g + -2mg =2n{-il2l (15) + -2-5mg=m[i-r'f(2rs)] * uo=l'666?(9/r3]t/z Exalople3.20Acylinderofmassraisstation..y.,ll:.a.brllkofnrassmllringedtoafixedsupportai is 0'2 and for contact of'the of brock rvirrr trre cylinder o (Fig.E3.20a). The coefficient of friction i., "".r*a 83 -. }.. tr U u klj rJ ) ..\:'a :tr: '. .! f'-' r ' L tJ L lJ L 1_IJ l.j l--tJ u IJ lJ t-" u L tl-. u[., [-=,, [j] 1-- L. .t: -il 'i _j.ri crylinder.rrith tlre gtglit+5p-+ -A constaat forcc P i1.1qnti9{i"?:.th9 c1!inf1 at a cggstant height of L2R -'ra1"9 -6f tf,i'cyUnaer fot trvo cases: 1- ra1 : 3m, of P for impendift'uiiiii6,ii aboye the gound. (a), Find tUi 2- tal: 9.6m. (b) For mr = 9.6m, find the instantineoG'iititei"tio"rof C and the angulat acceleration of tbe cylinder if the applied forcc P is greater than the .v{u9, gbtained in part a- (c) If P - 2-3rng for at bottr rrl1 :9-6mg, 6nd thc velocit| and position of C w[en ,tip S--.-^ ;+-ti +r r ,- lj IJ L- /- llrl ii .":-.::'- 'r "i*i" Y8'3 .se ff'8 ilffii-t.,,t'o#1 ',;ffh: 3.(ffi1,,T-{ h)x tt'.(.ril}* Nr -*e$ f"1-,?, "PPE -.1 ^ ur roltiuf s'ithout:slip at either -A or B- The FBD's (a) The cylinder o,iit-tlve ir"piiaiiig;niriiJn "f of the block and the cylinder are shorvn in Fig.E3-20b. The $ un[llqrvns P,/{r,Fr,N2,f2 are deterrnined' usiag M6, = 0 for block, 3 equations of equilibrium for the cylinder and the condition of slip at B or A. Sottrtion Mo. = rVr($.66)'- m1g(O.it) = :+ O /r! = 5m1gl,6 (1) ltr6asslracimpendingtollingrvithoutslipatSandwitlrslipat,riwithfr=O:2/Vroppositetothe directiron of impending slip (Fig.E3.20c). Mg.=0 .Fr=$ .+ &=o 1- F.or m1 - - :+ P-A.ZNL-Fz-o :+ A'2-rng-/Vr=0 3m, eqs(1)-(E) yield rVr whether the assumption is correct: #zNz:0-1(3.5nr9) - + 2-5ng, P = 0.8333nrg, - 0.35m9 + P=NJ3 =i 0-2N1(2.e)-L.2RP=0 f f? = F2 F2l = O.3333m9 S 0.1333Nt = 0-3333mg, (2) (3) At = 3'Irng' \\'e clecl' PzNz. Eeoce the assumption is correct- =+ lmpending rolling without slip at B occurs for P = 0-8333mg- [t should be noted. that slip does npt necessatily occur at [he surface where the coefficient of friction is smaller2- Fot m1 = 9.6m, eqs(l)-(a) yietd N1 - 8rn9, P = 2.667mg, Fz i 1.06?m9, JtI2 = 9rn9' \1'e clcct whether the assumption is correct pzNi:0.1(9rn9) = 0.9m9 :+ lFrl= 1.067rn9 f.PzNz Eence the assumption is not correct. :+ Assume that the impending moiion is rolling witt-igut slip at .A and wit.hslipat8withFz-0.1/Vzoppositetotheditectionofimpendingslip(Fig.E:l.2M). fr=O M* =S .. Fr=O :+ _, Nr-*g:-.9*C:.Q :+ 0.84P-0.lJVr(28) =0,.,.' : :r::' 'ii'.ri P:0.1rV2 -'Fr = 0 =+ , . :. Ifih cfcd:whether:the assumption is corect: PrNr=0.2(8m9)-l.6rn9 Eeoe the assumption is correct. + : =) i:: :+ * 'r:+ JVz P ' f1 = =225mg !'t'35mg :::.: .. . - |Frl=.1.35mg S PrJVr Impending rolling without slip qt .A occurs for P 8!+, , 9m9 = 2-25m9- : (5) (6) (?) 'I ';'. ") .t. . :, : .,1'. : i I ,]-. 'i t' ' ' ., ,,'.,t,' ^.-- .,' rolls'---lvrthout slip at A with ri = ft) \-, For P > 225mg,'tf,E cflintter i, , R o are oLtained using 3 gqua'tio9L:. of motion of the cylinderl , F,=0 * ,A'r':tb:8mg=0 t,'l (8) .t- "=Y^^ Ft=P-0'9rr.g-ma (9) * o=t)R=(1-6^P- 3'6ng)l3n (1O) F.rE-O'1r1Iz R-O2PR= lrrr&z(alRl + F1 = 0-3rng +O'4667P :. o or*2-7861119' O'2(S'ng) i'e''7f 2'25ng < P S is valid if lprl ( ,.rgr, ,i.e-. if O.3rng + 0'4667P S F,=ma * t{6.: I!,,) + Eqs(9) and (10) + This soluiion * , alL(Fig-E3.20e)' The 3 unliaovns P-O'1JV?-F1 =ntz ' (Fig'E3Jof)' Ttre 3 unknos'ns ,Fof P >2-Ts6mg,ttre cylinderrolls with slip at A and B A'2'6'c cl4inder: obtained using 3 equations of motion of tlrc + Fr=O F, = qa + !{ct = I!,'b + are (l t) =+ Atz-9mg (r2) =+ o=(P -2'5mg)/n * ' it= (1'4mgE -}'4P)lnR" {13} N3-nr9-8nry=9 P - 0'1N? - 1-6n:9 t'6m9.R-O-lJt!:R- OSPR= '"'"t''-- (c)Since2.25mg<P=2.3m9.<2.?86ryc'tl,uirri+ialrnotionisrolliugrvithoutsliparS{EigrE3.2og}. For block ? I{o,-:Y1r-9-6rnd0'50] = 0 : geuerat position The equations of motioa <[ the c-5rlinder for + &:o 0-f iVt Tlrus slip s.arts at both contac* fior z ,--,!:= +u? + (t-0e333 - = o-722gb. ( 1-{} = 4.8m90/r ': 1\': = (1* 4.80/rlm9 *ma =+ Fr =(2-2-}-4ilblfimg (1-09333 - 0-64612)s g't = iilR"(al Rl a, a =uR= 0,160/zlrng < 0'2(4'8rn96/r) i'e'' \tlocity u1 o[c (r5) {16) (1?) if z ( a'7n9b' iu rh'rs position is obtained using eq(l?)" 0:41')'g d' l:,'?"e333 I+ u1 = 0'1738(sb)'tt' -0'6461n(0-7229/A'(|;ls ossl)g * = [1-09333(0'72296 -0-66) 1 - P, = ina F"=ma :+ t{s, = f,a * & B: o-lNzE - 0-2R(2'3rn .Fi = (1'106? * 0'16b/c)rne Eqs(16) and (i?) :+ (l:106? * Tlris solurion is valid if lfllf ar1^r1, i-e-, if + r yield At2-rng-I[s:O 2.3tng :\ fo""'d'= the rvheels is rn and tlte centrc ] massof aself-propelled vehicle (Fig.E3.2la) irrctuding radius of g1'ratio. of t1 ' l-3 ' wheel assemblies have uIESS rr1 , rn2 and a-tal of mass is at C. The rear and front driven car' ttrs,rvheels:.arerp, ild pt' (1) A rear'-rvheel The static and dynamic coefficients of friction 1si u is f,he torque M1 ' The rvind resistauce is cu2:'where starting from rest, m-oves without slip uader a drivi.g the t a*d after travelling distance z' (b) Discuss velocity of the car- Find its acceleration and speed .i ri*" marcimum the (c) Find slip of a rear-rvlteel driveo car' basic method of analysrs' for motion with possible achie=e and the driving corque required to possible acceleration for a rear-wheel dri,ven car without slipping driven car x'ithour the maximum possible acceleration for a fiour'rvheel it- Neglect rvind resistance. (d) Find 'Mr.Mzrequired on [he rear and front wbeels to-ltrie;'e it' Neglect wind slipping and the driving torques driving *" car and the angular accelerat'ious of its u''eels if the resistance. (e) Find ;"";":i;J;; "i wind resistance' torques are Mt,l{z and the wheels slip' Neglect 8s ,i : : J JJ J JJ J J J J J J J J J J J J J J J iJ J J J J J J J J .+v re$ -vrQ t-1 T*. ql lNz i,tq*, F, rPe- .-.-o of rouitg uirhort sfip. Irence *ork-eaergy 'o' ,.r*t".?0"."". esuation' ir" mot or the car' rront *'heers gou"oios rorce .TlT;?. ':? ,]*;:;":"il:tlt:ffJt'-'t'"* oo rriction force rt.^ r*ant ..hels and the friction iaput torque to- rrre rront rvheers and-the f::::til*t::ffi;'J;ffi;;"fr:"i"* of t5e car' In ttre FBD che desir-ed angular acceleration' on these is in the backward directio' * u.o''i" provide forrvard accereration- The rsork trrc forward direction to in is wheers the rear on force frictiouar the the verocities of poirrcs o[ zero (such as frictiou fiorces since citrrer is ' forces internar and external doae by the and the driving tU" wtlf.aone by the rvind fotc* ctt2 Lri .. (such zero pair-rvise contact are zero) or a,.d angular ".""0i of the car' and the aogular velocitl' torque M1 onthe rear whee\. I*t, o,t t,itu" 'r'" """ulrati'on car and rate of & = alR- The kinetic energy of the' * ,i* . implies slip No = wheels. its of acceleration given by by all internal and external forcqs are ;;;a;. + tnr2&!(u/R)?] + 1 = |(m - m1 - r",r)r'+ []r,1u1 + !nr'*!(ola]"t []'zau? (u m2killttllu2 (mr&? + + '' (2) . = if* g{ - Mp+cuz(-u) =(M1lR-c'z)o qnd' cancetrut. cancelling rcomr1roa factor n: t{ QII.' T= tiz' in ratc foun ?= -- -r- --..-'d ' ahe acceletation is obtained using worh-cner'r.retatioi (Mtf R- cu?)u [* * (*.ei + 'r'r3]/E2l oit = 'a=(M1lR-"'?)1@*(*tt-? +n+ki)l?2l :+ tb obtain u(t) [o. coEstant Mr' iategratc ;;;;4ri lr" ro.'"o*tant Mr.integralc A#=ry .- I t lm +(-"t? + (3) wariables' the cq(3) w.r.t- t by writiag a = dvldlaad separating auyanaad sepaqatini t&e tariables. cq(3) rv'r'L z bv wtiting oI - u "r?ti)/E'zl' l r' ffidv =t / [m + (m1!] + m,Li) I R21' .: obtained usiqS'the'work'e'aergy For eoastaot Ml, o(t)caq also be directly being Mt(xlB}. work done by Mrin-distance e (rotatioa z/E) ., tz-Tt=Wt-z .3 |t',"+{1rri For.Mr(t), o(t) is obtatuied by intcgrating eq(4) ,= [/ Ho, -(m1tlo1 * ir1uh1) =MplB' rv"r't' timc u,t t] dtl I (b) Tte FBD's are showu in'Fig'Eii.2lc' +r'f.2k1.llR2lu2 relition in'inteigated form' the RVn+ (m1ef + $"k11I Rzl' coincident with A. Let O be a futed point of the grcund f (mzti rz + *rYlrr] +#,1X13 = 86 mplr + mrti w1 * mzkloz- (5) lr ,ll ,<- -,1 .t . 1'q' be deterrematic viriables a'61'&2' can O:' foidei;\il1;IViifr,Pz''& 8 unkaowns, 11 The "iU.tbodn front and rear wheel assemblies of car) 'Rz'fu'\ i"di* (main the 4 rEised usi.g I (3 x 3) eqiraiions of motioa "i'g we shali writc ? equations iavolving-ouly and 2 conditions of slip or uo slip tt e "rra i.-g;;et*, For the whole For rear - Ho, Fc=o car: rrheels: nrgb2- Mu- . Fr=ma Msr. = $,'itv euzd=lrtoh+m*lua+mzkZ6' 'vl6me-ilr-Nz-t + p, _._ 1-- _ 1:l - trr - &R = m1klbl .I] ContactcoaditionatB:ifnoslipthenul:olI-'ifslipthenFl=peNv(11) or if stip thin if no slip then u2 = a! R' Contact coudition at z{: For tLe case of impending slip .,..< ::i (8) &- F2-ctt2=ttto Mt < i,,).. F2 = pvN2- -{ \ (12} assumption is made regarding condition replaces p1 iu eqs(1I) a,nd (12)' An ofsliporuoslip.at"orit,i.t"landB--/\rr,Nz,fr,Fsaree.teyledinterrns'ofo,6l,b2usingsuccessively equations of (11) andldTz) La 5'ield - <r 3 appro'lxiate ir eqs(6),(?),(g),(10). These are substirured made 'about the "-tii""o are computed' A thecl is finatly forces and'the solved are These o,riliria. for of the equations slip and rvhether the value of the mignitudes ' p.op". directions of fiiction for-ces at the contact(s) rvith frictioaforcesatcontact(s)withnosliparelesst'hanorequalto[lrerespective.O.t"dl"Oofsiaticcoefficien[ rsit'h other satisfied then the problem is rervorke<i these ;;;" If reaction. normal and friction of "ot por is gi'en b.v 'q(a)' i* Fig.E3.21o *,y:l_lrl",:-'i* force. no:l*,." the front rvheels rrill lose amount since at some st'age either florvener, a cannot be increased to unlimited can be determined b1r Grst Thu *"r.i*om.pcsible aceeleration stipping.l start wheels rear the or contact 1olR u1 = i2 - al R' i,ll = findiug tbe forces as in part b. For no slip, "'2 +'rl2k3)/R|v (13) Ho, =(m1l;fr.l1*m1uJr1} *$n2k!u2}n2llh2) }m3uh3 = [,nn+(m1[i -_ < r! il"H:"rril,:15;::*": Forthe6uaknown'qJVr'IVa,fr,fz'd.'M1'rveusethefollowing6equattons: NiO - mg0r = {mlr +{rar*? +'rn2*?2ttlRlo M6. : iIs. For the .rhole car: ra9-lVr-A'u=0 Fr=0 :+ F, = Msr, = I!,"t1 tna For rear wheels: tu) (15) (16) -F2-no Mr - FrR= *1*11alL F1 aA (18) - wheels: i;;;g ,u" *o1r.-"nergv retatien in rate form' i'e'' eq(a)J: . +rr.zkl\lR]rollb' N1:lrnsb2*{mlr+(mlef +n2ktr')/Rla)lb, N2=[msD1-{*h+(m1&! ,/r.2hlllrzlaa f'r': (m + tuzk:'/Rz)o, M1 = fm+ (mlei + Ft = n2+],alY\ , Substituting these in eqs(19) yields no'slip "i-.reil wheels if and no E$-off of hont wheels :+ I- The ma:cimum o if 1 o1= p,9bz/6[1 + m2kl/rr.}} a S az = s\flh+ (m1ef acceleration': a = min(c1'42)' - 1 ^-{ ,- i- i! lt (17) FzR = r.2klol R M6r, = I!,'b2 nr, > g- (tg) w\ee[s: and condition of uo lift.off of frogt 1I4,Nr' Ft aL B: slip no of condition and in terms of o using successively eqs(14)'(15)'(18)'(16) The forces lvr,Nz,fz,Fr and M1 areexpressed t For front i! (20) (21) P,{h+ (m1ef + m2L1lll'rl.Bll' - ._ = + n2*lltmBl' < g'r : q 'r!'' of wheel gom eq(21)' -The liftpff c- --';*"* 'tto(t' is obfned tot' ac&lerarion this g3ea .t Atrf :"* T::..;:;;aed ii due being stry a beias of c * the vaue or :": ao' to *qd;;;;";i* *ohuo ti" teo-sheel a 9$-'-+i; '" "*t". thi direction of frictioo force on analvsisfor;i;"!*5r-**"i:"l;:fff ris-E3-Jld.. No'ice u = w viaa". ":":,'::' "' -' i) requir"d Mr of Tbe value . ""'i'," lll,:f #:; :::il:;::**:"T"'T'T:;; (mrt? + {i[1 + Mz)/t*+ tR LL.r [rn (M.1 + Mz)ol^ * + (m1tf + mzkzz\l?2]uu = a = r/.zk:2)l P.1P. (22) - ,R:. ,... h,E - ;3 *#-ry d=# -E-- ' it-t LJ LJ rvheels: For ftont wheels , For rear L-. LJ u * IJ -i"-i--ii)"r", ] (23). ' (24) [:l *[z- FtR= m2klolB =(rr+fz) l,m<P'{NttiI:}= jol"- ' ?hema:crmumacceteration " l{t eqs(20) from eqs(26} arrd (2?} ujing . Mzarre obiained Mt = $.el&R{b rcl Ho: The FBD's are shor'vn in Fig-E3'21t'- car: \) Fq front .and (2s} for IrIr , J{z , Fr , F2,.o| Jqb F, =ma -. a rtr'PtNz=mo ttllrlr+ IrErrI' wheels: o"r; t''ioz + : t2..Mr-ltNrJl= mrti&r m*?6t \'-- ^**'R= :) j a1s (31) (32) t33) \oo' i (35) Snbetituting&y&2,oftomeq9(33).to-.(35)ineq(30)arrdusingNr*JV:.=rngfromed3t)yietds ' tv'L---Mzllb- (36) E)) tttl-- Mt Pt(tr JV2 'r [rng{61-Ft(h = 8]] + Mt* Mzllb' fV1=tmgt6z+pt([- rl f (29) . - atg mg-ltlr-N:=0 i{o' Fr=o-, ! _) ",- ^^.tit**t ' F'=o'N,- 5 is given by eq(t')' The -unkno,sns"I{r'JV3'a'ur1'<'r2 Mo' = . y'{rr"dim=pts ri/E} + * v,lrrruit+ (*tei + mzti)}/l '"i :'^: For the whgle l.: J IJ 1-, IV'D-nrgbz=' ''' mg-Nr-J\r'=0 Ft* Ft =trt? Mt - Ftf ='rni-{c/R + f. = t'td Mct.= I3-"i" h{cr, : €:a,' '-. l- u L ': brirrnno*r,, Ai..lv,;Fr,F3,o,Id1'It[2: EguationsCI3)and{2a)yi,eldlVr.Hzgivenbyeq(20).Itfollort-sfromeqs(25),(28}and(24)that t_- l-r.- _*;.*.,,,,..# q'h$';ii""f;tt*::.i,::,,it; frovidedthereil'"m*e "r'ttie ' 1t6,, = trur For the shole car: lt u tu 1" u L f--- "##*,t,iT-,\ or-Tt, .. (d). aa [t ^'--l ,j-J from eq(36) eqs(33) and (34) usin5 lv1'il3 ,,,t.rizare theu all the wheels "o*p.aiii;m the rear *i*r" tral;i'r.and,with Tbe analyses of the retardation oi. ,"r,i"u with only reveals that a (35) and equations (29) ":J f:,J Compirisoa of braked, are sinilar to those presented for accilerttion' slip on in such *f1jt that there is nq " gr€t€r retardaiion of lr,gis obtained if all the wheels are brakedthere is actual jlip on the ground yielding are locked, then the ground. On the other hand if all the rvheels para*er km rvith a verocirv orsoboo km/h is raunched ar an arrirude aod its minimum and 'to the earth,s surface. Find 1- the minirnum and maximum altitudes of the satetlite tlles of travel from the otUit and its time period' 3' the maximum velocities, 2. the eccentricity oi itt" velocity components and 4'.-theradialanrl cirlumferential perigee-and apogee to the end of the minor a-tis and (ii) after traversing altitude of 1600 km for the first time' and the speed when (i) the satellite reaches an to a malfunction o""* ?*i,--"' (b) Work out part a il 1ue the satellite in a polar angle of 60o ar earth's centre rro* ti* (c) tlom s" * iL" circumfere*tial directio'' the vdocity at launch is at an outrvard "rrgi. or the optimunr location in Find of-eccenLricity 1-2' trajectory a llur" to launch"d be is to probe part b, a space the objective' (d) in its *elocity at launch to'achieve the orbit where i[ should be launched and the "rr"r,5. be launched from it so t'hat U'-aA oUlivation module is to At the maximum altitude of the sarcllite in Part ;Xj:T:;t'i.T;::;erire "ry. in its subsequentorbit its minimum ai"'"""eno*theearth'siu:-.::i::i:, taunchiirg' viug cases: 1- circumfettntial i:"""*il?, rockets) &r'the-'follot required at launch:(imparted by appropriate .f'z t. of the earth as 6400 2- radial launching frorl tlre satgltite. Take tlre radius,E |^,I, ,<u nn;.: ilit'*itlffi il;'ffi determine the sofntion 3'*"1*, i \- ro - = 6400 +650 =?050 cMlh|= rmin = km)? sR2 =9.s1+{6400 1-16416 km, . . -, = 19245 km/h, E,:,650 hr.n,. omin = ftolrr,mrx ,.rrffi(64o0 x'10-4 km-r ?050 1fi<fLl!"| ,-7 t,r km), =520?56 x l0r2 km3/lf km/h' L6 =16o6o = 211-5 x 106 km3r/h = omax = - ft=aigg t*' :*1 Irglrmin 30000 km/h'- e:=':y':*:^0.::*' km = *:' =':ly,lz): [ = c(l -_'?l't" = (rmin +rim.,g)/Z = 9020 : 1t' 1 1 the mindr alris (Fig'E3 '22a) a.d of end zt'the to r*in Q at perigee P Let ?rbe the time to travel from be tle time to travei from.apogee '4 at rm'a;g to 8' !i = (area OPQ\ I $ol2) = 0'5078 h' area'O PQ = tobf 4 - 6(oe)/2' ?i (area OAB\ | (hol2) = 0'6?17'h' b(oe)12' J .c : ',: I T 'J J J J 8802 OAB 81---. "--f J ,lJ I U -T2 rabl4+ = T-he launching pcitiou turus out to be at rn in' Llr'= GMlhf,+ DcosC :+ e =70'27o ' At r = 6400 + 1600 = 8000 km : tim./h' + e = (ui + o?)rl2 =ZbStZ tmTl' v6 = lrsDsind - 5062 km/h. 04 = hglr=261137 At {:600 : km'/h' + u = (uf + v1)u2 = tl7o5 km/lt' v, = hgDsind = 4658 km/h, vo = l:,ab = 2?311 area J Sl*t1 Br.. oo:ooo:30&}0 km-! D = lllts - GM ft:|il - o'25428x l0-{ imax = $Mfte- D)-r = 10990 km' Lma:r -J *,' \ the given dita for ihis porallel } basic constauts Glr{,ho,D from o km' Kl|zth"+ D)-r = [*ir, = rmin - = 1 1b; laraching. GM '- (ffi"9*K# J-I sY. J ;q k-r;=--+{YF:"r -t r ..-J J J J "rJ ,J -J JJ J !': \ --/ L l-S f--. '-'t '' The speeds -w" u could'il;* first determine i" "ro 6400 = ui,= t- : I - uscc3o cM/h t - = 116, 299589 =1.16236 using energr conserva[ion: km/h' -' x t0-{ km-l ;j*' E tr =.roud'o (Fig.H]'22b) =2LL'2LX 106 km2/h D=l(tl\-GM/hil2+(r'"/ho)?lt/2 - -* o'zot8ss x lo-{ km-r **' ;: = umax = l'o/"min = "'J^:;'='*'-;,'j:* km/h' e=^io1cu =o,243r 6-a(r-"')'t=6?90-6km' T=rob/(hs/2)=2j59h 30186 il':'":,:'"',::::[::;';,:]#::$.",:,11,';]ili']';1i:l''ii"iH:[HffiS:::T::i r',=..gi#jqx"1;=-. At 0 = "; =';'"1"; :,ffi:}f] l-_ The fact that the values of u at IJ il,:r:::::L'ffi:y;e f ,1$:;,,'_:?km,h ;, = o"',-, |-ffir,?'-:'';='liij,i ),r -2a{ss km/,r r = 6000 km are the same in parts a and b follorvs directly from the equation be launched ia the circumferobjective with the minimum Au, trre probe should :if"*:l':X#i:l[*in*;i:;i:l':]",',1'u,;i'.1.H"15:,H;S,I;'*"* D1/r'2 =) =L2. + l/rmin = GMlhl+ DL + D1 = 0.77g555 x 10{ km-r' + ur = ucr = h1/r*11-401164 km/h, er:h2rDr..Gi't , L] I -' Gr{lh?= l/69e? - (I + tlt'2\Dl h1= (L.2Ct{ Au = ur :+ tr, L I :':,';::xlT;:., ..., km/h, + :+ lJ ur r- 263'129 x 106 km2/h I'02?8 km/h' 1/11044 -=:::'":;,'o.*''^ [r, = ul - 19124 = -154 ,' km&' 1i:::il111[*'.]i11ilm:':II[ Lffi';*,:lt,'r;*mlm'*T1;.Tffi: is giveo bv. 12 = 6700 km of the module Conservation of energy l-, L' fj' Dr\'t'= themoduleare l1044kmand67fr) km' = ucr = h6lLL044= rgszo o2 t- l -r.,1.*=40464-30186- (d) l.-r"rnt"...",ttu--u*imumandminimum,radiigr!.treorbitof Li r,o, D refer to the trajectory of the probe (Fig'83'22d)" t L = io.4 /cllnltgi ^::Z:o,=::';*i!,,",',1,i,', ;=[X Zi^o,Y,',i',!,;]r:]:::: 16'49o bndo : (""//'o) I $/'o -GMth?)'=0'2960?' {s = lJ L" L 'ro o-(tr,',ir,*rmax)12=9020'5km f tr L L I F . = no/'*o*- 19124 km/h. ";; \ma2.-GMmf t '- '-*-- ;' *ft' uo = 30000 k,.r/h' uro = uo sin 3o = 15?0'08 ,-, "urained D foFttris ao,.pamllclloonchiog kg1, zo50 ==:::';::;, ;:: lJ I ' + 060 ui; Y : -- ooz = holrz = 11044 lm(ul+rsrzl2l-G x 19124/6700 MmlLL*a4 = q,hich'ts very large compared to laulobtained in part 1' ea = 31523 km/h' \'tttti-GMmf tz + u' : 4(F6 km/h **--) r,-*l ;;-I aJ exc-luding the fiork' is mr with ceotre of mass I}xarnple 3.23 The mass-pf -a fork-lift tru& (FigE333a), at Q' The fork is supported by a U at Ct. The mass of the fork "od t5" o"te i" -, ii "*:"."f .-4* and vertical reactions' The inertia Ugttt nofzontal smooth roller at A and a connection tt .B *hfu-;upport-s just su6cient at A, B , E,F if the fork is given of the wheels of the truck is neglegible- (a) Find the'reaction! (b) Find the ma:rimum deceleratioa the one of the ground reactions tP *to' ..'iJ t.'iJ t:J upward acceleration to reduce slip nor nei[her slips nor tips and the rvheels neither truck can have with four-wheel brakes so ttrat th" "."t" the centre of its centre of mass is at height h3 above lift-off the ground. The mass of the crate is m3 and ils the fcrk is f:- (t) the ground is p1 aud for the craie and base- The coefficient of friction for the wheels and at tlre isstant the fork has uprvard acceleration Find the ground reactions for a rear-r,rheer driven fork-iructi /t/r' A;ume o1 relative to the truck and th-e.-drivt'r t"'o'l t' tn::' **""FffTfjpfiI* ,,J -J :1i#-i u l fffffiFffi*#,ffi"ffi 'ffiffi"H$qf*4"Hffi#j to, ;f 1 JOo, ,;a.i.qffi 'z J *-fN, f:*S'{'h_-_Wffi tf 'i{r. Na +*/ ,J , ^a: :"F.?('ft8'i.. - e. Nr _!-,}oJ' }#, i#,:t, -}_u" solved N1 iv'l'n.."tr'oa''iit s.r.,tio;'dlhis probrem,t:,i,:", ="...3,J,:i,,#ra;.1.",' for individ"'i'ot* '"oich invoh'et:T;:"'r:ltl;l"ll1r";:::"J.,lll= Mc==00forindrvroua for dre rttsc, Ma Using F = lixed points t=r:lil 1. 1- usins =rttg<.t = io, forconvenient . ir":1i.:_=*5::*xT:HTl"ii:-Ti:J:rternar-ro,ces ' follos'ed io 8x321' on trrc svsreor appear ia 'lrre " tt i. "pp'o"tt' was o.vsram cao be ren'tit as F-ngcattanslatingsystemcaobere$'titten 3. Theequationsof motion,F:tng6, rn-=nof act'ing at c constitute a null g, i.e., the external-forces *a",,il. ruirh * 'itertia lin'-**" g-, Mc: using F = Q' [ - O for this augmented force systenr-and with rvorking ^ in consist's E thac method systern- The adt'antage'over the first trvo methods ttt" poiit' to"'""itttt any is A where this null force system, is zero' Method 2 rvould require Proper ;;'ii11i::i.:::* sv"t'm."b;;; force augmented the of 3. momeut equations- computation of H6,,though be the same as in me|lrod the resulting moment equation rr,ould acting on the Weillustrat"*u,hod"',and3inthesolution.\1,efirs[.aPpl1,metlrod3. er<ternal and inertia fcirces -rn;gc: The c' be fiork the of (a) t et the upwaril acceleration sysLems' Hence in Fi5'83'23b' These are null force t'" wheels (l) and.the fork+crate truck, "ho*o Fr=O Pront$rheels: MP,=O Me.. = 0 B.ear wheels 'Whclettuck: MF =0 ; =; Fork+crate:. MB,=g F'=0 Fr=O (2) F'=.9 ' :+ JVr=[m1963+m2(9+a)(61+62*6r)l/(bz+6:) ;; *; .-'*'(o+ a)Drl / (62 + re) '' N =-.z(g+a)bqld Er=[ = ft2=m2(g*c) Jj J 'J J ,! i ir J -J ,J ! :: i -_.1 ,-:{ (3) :- l/,\ (4) ,(\ (5) < (6) -q (7) -i:t'-- si .v-.< +r S -! ttI ,.. ' t* . -:i.i.1; !.;; .; ^:f Ii r.-i.-:::-r'=.':-: 'r,ar!{-i ' r.' i. :. -1, tot a '= (a1 tu.t /v1 increas€ P,9 trr lecrlases with a' 4rz becomes ryxo (?}to eqs(3) usirg fvi;vr1n Ef;e;"- be determined {mg7/m2b1- 1)s. lyith this-viluc-ot i, **o1. *d iuertia foi:ccs -m"rc- a :ting on the (b) Lct the backward accelera6on of the. trudkll"'t,-fr" lThese are null force systems' Eence : ' truck, crate and the wheels ate sihown in i'ig-ea-23c. (8) N2=liig6,2-mz9br-(m1alr1*m2olr2! /$z+b) -' Ms,=A Fortruck : lr is observed oo*aug6-iq{ l^ -.[. t-_ L-t (s) - FY=0 t_ '"E --0"-t" (10) o=(rt +Fz)l\qt*mz) ord N1 increases by the same arrcunt' " : (11) o1o1=(m162-*rtrl!!-Vrlrr*rnah:) Fornolift-offofrearwheels,usiugeq(g): JVz20 = (Lz) pr$ir+N:r)l'ftar+"tz) noslip of wheels, using eq"(rg),(9i,o - (f'r+ Fzll(mr*mz) S "Ls LJ Equations (S) and (9) imply l that 'L-r JVz ^,^ decreases ^":-'' with in ease For L* t-. For fr = Q cra[e: fv=0 :+ M9, =A L= l.l- (13) II3 ='fi13a = m3! 0 Ar3D - nr3sd7l2*n3efu- (i4) JV3 =+ b = d1l2- ahslg (15) msa.f l.?@3g :+ ' a-S g,zg : Fs ( ,raNs !i:l (17) a{:gd1l2hs + b--'|1./2'-"'A"ry',0 F6r'notippingof,crate, usingeq(15):' c:miu(o1' Prl' lt29' gd1/Zfu\' Eence the maximum retardatiou : wheels are shosrn in F(-83'23d' (.) The external and inertia forces -m;lLc;acting ori the truck and lhe Fot no slip of crate, using eqs(13).(14) LJ t, L_- These are null force systems' Eeace L-. L LL-. L- LJ lt L__ u t-t. L. LJ L LJ L*. L-._ _i Frontrvheels : Mp. =0 Rear wheels : Mq, =0 (18) Fr:0 r;- LhlR 'Il9l t20) :+ a=MJR(m1 +mz}wholetrucli: {'=0 + (mr +&z)a=fr.*Fz mzahzll (bz+fu] (21) nrt = [mrgts*mz(g+atX0r*6a*&3)- rzlolrl MF. -0 (ba *03) Q2\ JV! = [m1s6 z- rrlz(9*ar)6r * mralrr +m2oh2ll frr (22) is given bv eq(20)' ' where o iu eqs(21) and"!. (3)'(5)dB)'(15) aad (21)' we illustrate method 2 for obtaining rroment equations Fig'El]'23e' consider points o and (") lrt u be the upward vetocity of the [crk- The FBO'" are showtr in the given instant' S n*"a to ground and coincideat with .F. and B at :+ ec(3) Ho,=-m2u(61+02+h) + Mo'=U:'-m2o(61*6:*6s) Ttucti: :+ ' eq(5): +': Mi;'='H5t = -'n2o6{ ' :'''l : Hst J'-r,..2u6l' fork*crate , Consider (b) Leruberheforwardvelocityof theforkwith.i='-a-TheFBD:ar:.Yn,rnBie:EXlt?3ft ' given instant* ai the : points G and K fixed to ground a.od coincident ncith E,and,D Ttu&:HG,=m1tlIr1*tn,thz:+Mc,=iIn,=-m1aIr1-m2ah2eqr6}.: tc(15]":' + ' Crate: U*"=fisah3Y*r=i{x'=-rn3o[3 velocity of the fork- Thc EBp' i'' (") lrt u be the forward. velocity of the tructi and q be the uprsardwiih F' Iet' the location of Ci from and coincident shown iu Fig83,23g. Cor,"iao P*t'g fixed to grotind pr.and iih".g,r.* instant c = 01 *62*0s:9 = lr2. For.the t14', 0 be giveu by e,y, whe re i:,,,i,:= i Ho,Jmsulrl{t.r.zoy-rr.2v1, :+ - -rr. Mo'=Eo,=m1oh1}m2cr!*nzvil--r/r2a1a-m2v1i=m1clr1+m2ay+?,/l210u|-t,12.,|Z-T,.:I..:i ' ':+ eq(2t) i: . - nt1ah1 *mzohz - rr.2r.v(b1* 62 * De) . L-.) .; .- -- ::r=g:ri : €-|t+'':: ii l:. ..::, " -' - - -::l:i":"t':':-; -. -::;=".r,1:: ' L_- LJ L L- ga l:IJ Qr I Ct .: lj. lj . ..... a. tdrrarroNAr, MEcrrANrcs aND EQUTLTBRIUM lj tj tj t q is called uirlual displacentent 6q(t\: 6q(t) = q-(i) pathofmoiion.Notation'6O'differsfromthenotation'dO''LetaforceF;(f)actatthematerialpoint the virtual 1*O]"t-":lt"t.": i. The positiou .,r".to. 1;(tiof point i cau be expressed in [erms 9f S(t)' Let in thG rvork done by F The '-(t) poinr i due to virtual displacement 6q(t) of the sistem be 5r;' ;;";r*; F'(t)'tirt' t' is time value at at its virtual displacement iu zero time, keeping the value of the force conslan[ external and i*ternal forces acti'g the all b-v done rvork uirlzal urortdone by F;- The virtual i;;;ilahe denoted b1'- 6l'{r(t)' oo th" system in virtual displacemenc 6q(r) of the system is the path q(t) is catled [he tarialiarr The diflerence in tl're value of an entity for tlre path q'(t) and lor r-alues of the entity is called the f rsl of the entity and is denoted bv 6( )- The first order diflerence in the Hanriltoo's Pn-nciple: uaiationof the entity and is denoted by 6(r)1 ). \'Ve norv state tlte .A geometrically admissible motion c(r) of a syslenr betrveen prescribed confiSuratiorls {1 at tl and ri2 indicator (V-I') vanishes: at t2 is the actual moiion of the system if the follorving variat'ional 1-. u [J t: tl-. l- v.r. t-. l- = J['i r,' ;.rr, * 6w)dt =.o in tlre srnall /,rbitt1rry varied nrotion of the q=i.r,. s'(r) = c(t) + 5q(t) frame' inertial called valid is is refereace frame in rvhich Eamilton's Principle neighbourhood 42 LAGrtANGElq EQUATIONS FOIL DISCRETE SYSTEM coordinaies Consider a discrete .system o[ a degrees of freedom rvith geueralised u l_ 91 , " 6w= I8i6c; and sumrnation L l- : Q;6q;, where q, =T L,'H' (4-3) ,. generalised coordirrate is called the generalis cd forccfior the ith *rr"ntior. nl" u""r. used- e; ? is obtained as follorvs- Let 6g(t) = e4(t)' ie" rvith {rr) = r1(t3) = 0, and e ( c'(t)=q(t)+ea(t), 1' (") a function of e, and expanded by Taylor's T.he kinetic enerst function ?' for the vatied path is expressed as series about e = 0, i.e-, about the original path q(t): * #1,=o{ i#l.=rt? + ':'' 6rtrtd'(r), q'(t),4-"[.i(t], c(r),4 -"(.) -rl.=o = #1.=o' * i#1,=;2+-"' ?\.' : i I a-. g;. The first variation of 1* t a i tj lj t ' ' !In, rvhich are all in terms of 59: lJ 1 r-l of g(t)'- l'he independentofeachother. Leiq=lgr gt ...- {n-l tld f bethethecolumnvectorofthegeaemlised actjns zlr-i =I;({r'"''9'}' can be coordina[es. The virtual work do[e lry internal and external [orce-'Fi ercpressed (-; (4.21 for t- cg (4.1) -o(i), ,virtual, displacement since ar actual infinitesinral displacement dq(t) occurs in infinitesinral This is called a in zero time and is not along an-r actual time along a paih of rnotion, ivhereas virlual displacement 6q(t) is t-- L t- :: t t7 4.1 EAMTLToN's PRTNCTPLE '".' .: FiA'('l move system a qLet be Let the ge'eralised coordinatrs to, d""".ibu the configuration of a system t2- Consider a geometrically admissible frc:n the configuration g-1 at time t1 to the configuration f2 at time tir, C({z) = 'i2 (Fig'4'1)' Consider motion q(f) of the system between these configurations such that S(tr) = these confiSuratious such that g- (' I ) = another geometrically admissibte motion g'(t) ;f the system bets'een q(t)' Tlre iafnitesimal qr, q-{lz\- f2, which is an aroi{raryratied rnotion in the smallneighbourhoodof tinre I ,displacement, from ar.r admissible configuration q(t) to a neighbouring configuration q'(f) at the sanre t-: t_ tjt_j tj Q. : - ?!i'({), c'(t),fi} = r[d + €q, q+er;, {] = ?(e} ='[.=o E3 ,".i' '*'-'' chiio,rokj of airjr"T i f1'1]sfrief ';"li;'$9i€i.--.. , ; o,ui"-- #1ffir'#Tr' ; Yrpil} = i# *'+ a\e 6rst o.a"..L,,sos:.te-:l-.tlt- tj it, SuLir'*1-r.g(43);(4'4)in(4'2)andusing6s;=er;;vields'-'r'--"'.'1 *ffo'*Q;t;l,tte u tj zf,e srsr t_ LJ ( tj t-. tj ]j -ide 4f,r by parts term in the integrand or "q(i) is integrated g\r€'f,ogrcag c's egtalidrs for discrctesystcnr Y'!U-rU Qi=-ast strbat'iunigtr - lJ t'I ,' t-. lj --, I H H L- a factor of r1 as in the othir two ternrs: trtence' the inCegrand in t:: zero as : : { i =r,.. ., n. O':*U aud AV l-]. get (6) (4.s) aotdinates itilepettent of each and nonconservativq par ts Qr' other' a Qi" 3 #i:r*t'"'"u"*o*ve T:ffi###";ill :H"ff .j6w=Qi5qt:(Qi+Q?.}6qt=j?+Qt.6ci=?x+Qi.)6c; rrie &hoe u tj tn (a)' ni(tl 0T _ n. aldi,) - 6fi= v; lJ ]j : tfitffi-#,-Qllrdt=8' ' .':"]'., r ,, (c) should be -ts zero since 6(t1) = 6(t2) = 0 by O"<rjniegral term o,o be valid for arbitrarv TlreSe ': =o' l,l't#* tffia,(t)l[ - /:' 'ii"ihn ff'ti' t_: LJ tj . Qi rromeq:ti;: Q; = Q?' _'av -:i':':i, i= _ u; '' + ;(ad,) - aq, 6*= ^.. the Logra*gian fittctiott tr o[ rhe "y"i"* *=#_X woRK *If'u +.2 a*d(46) r....,n. bv ,(,i, g' t) = ?(4' c' t) * (d) - (4.6) y(c)' llence *rkt_#-_o,., i=,,..,n I.TLINCIPLE oF \/IRTUAL in equilibriutn, ie-, t-trnsider a system rvhich remains c(t) = {-r.= iu: * d o * T = 0- (4i) For rhe ^ of the first order aud hence the generalised velociiy i'(t) = 6GI the 6q(r). + Vqriedparh C.(r) = C(t) " (4'2) reduces a [1"6W at = order, * OtUf = 0. Hamilto"titin"ipt" is of ilr" ?*, e knetr?energy "".o,..1 6w ."'f q(,r ro(,,) = 6q(t3) is initiatly in equilibrium and 6l{'=0 for all virtual displacements 5g' thetl to the t\>nr.ersely itthesystem generalised coordinates q(t) corresponding are zero. Choosing t'he origin of the forces of the generalised solution rlr 0' Hence the initiat .oJiriont are g(0) = 0' d(0) = ttre configuration, equilibrium :o:ttal and' the sy:tem remains in equilibrium' of motion for this case is c(t} = 0 uork lben the hsms{eneorS.L"g...rg"{"quations '.;.16o{ of virtual work: If a sistc* is ia cguili6ritm' principle the proved have s.e gcoilticolly a fhos in orbittary uirlual dkploccmcnt b zctt' Conrterselg' clone. fC thc intcmolatd. exlental lotqcs j - 0 +. =0' uJrta<sieleconfiguvtiott,ofosyslenlls.cllequiliDriunrcottfgalrrtiott.!!"*,:,::"::o,""bythcexlental is iai{iolly i* eqrilibrium:.. loirtlrat displacemed' ptooided fi]e systea .rr.A inlc*o I lorces is zero for arbitmry (4'8) . 6W=0 v 6q' Foradiscretead.o.f.(degreesoffreedonr)systemrvithiidependentgi,eq({.8)+1 :+ Qi-O' V 6q; 6l4r=Q;6qi=0 lhe generuEsed [orlr,s ate zc,to' ip.rforc discrele system in eguilibium' all 94 i=l'"''a (4.e) t - ,HE . ll'Ki' ilD.J \, )'-- -- H-i A porENTrAL ENERGy srATroNany 4.4 pRrNcrpLE ot' -. J.{. -.-- -:.--' - ---l-ret 6W. and 6[7," be the virtud ivorF doae by the conservative and nonconservative forces, then eq(&8) 6W = 6W. * 6Wn. = -iV + tVn. - A, =) If alt internal and external ficrces.are conservative, then eq(4.8) 6w = -fi :0, + V 5q- 6Wn": 5V .+ v 6q. til = o I j']- e =+ (4f0) lq 11 r' ({-11) - i.e-, cbe first variation of potentiat energl is z,ero. Thus the potential energl/ has a stationary value (an extremum) at an eguilibrium configuration. Thus priuciple of stationary potential cnerry follows I/ a systcm srbjected.onlylocotsert;alioeforccs rsincguiliDnunt,thet6V=OY6q. Conuersely,if 6V -OV5q and if lhe system is iniliatty in cqailibiam, lhen it remains in equilibriumFor an n d.o.f- conservative system, eq(4-tl) .+ :+ V v 5qi, 5q,, :+ ilr=Y&,=o =fr;6qi=0 i= I,---'n- Y=o U*=O 6V ;-t r< G-rz) - ir Equations (4.i2) are Lhe equalions of eqailibrium of on n d.o.f. conserttaliue system. 4.5 PELINCTPLE : Or RATE OF \rfRfUAt .wortK :- Consider a sei of adnrissible virtual velocities ri .at an adnrissible equilibrium configuration- The rate rt. rateo[sork done s,ith these virtual velocities- Since ri.an as re{l be of virtuat'work 6liz i, a"nn"a ". v i ,r,ir-o (4 - =- \ 13) i.e., for a syslem in equilibriurt, lhe tole of llrrtual work dol.e is zero for orbittory viriual oelocitiesNOTE: If the system is constrained and the forces of constraint (internal or exCernal) together do zero *ork in virtual displacement consistent rsith the constrain6, then in eqs(a.8)-(4.13) the virtual rvork is only due to . the active internal and external forces. If all the internal forces together are *'orkless, then in eqs(4-8Ha-f3) the virtual work is only due to the e-xternal forces on the s-vs!el'rlDircct Proof ol Prittciple of Vrtua! firo* for a rigid body. l{ for a rigid body can be ecpressed as _ lA = F -ya + W.u, + d.W = F -gcdt+ Mc.g_dt - F.drc + Mc.d4where d0 :,"tdt- Hence,..* (4-14) ! 6w- = F -6u + y" A rigid body is in equilibrium iff f, =!,M.c- Q, provided it is initially iu equilibrium. Ilence eq(4J$=+a ,{h rigid body is in equilibriuro iff5t4z = 0, V virtual displacements 5t, 54. provided it is initially inequilibtiu.m - 44 - 4.6 srABrLrry oF EeurLrBruuM .=.-! -E : lj,' :- !s. coNFrcuILATroN W 1Ji r; - An equilibrium configuration of a system is said to be stable if ony smcll initial disraficncc (displace*ar/li \ of,f,o and / r,elocity) results in a motion rvhich b timired b a small teighbotrhood of the equilibrium configuration :& W The equilibrium position o[ rod I in Fig-42 is stable but those of rods 2 and 3 are utrstable. Rod 3 is unstable U,rl since a small iuitial angular velbcity causes large displacement from the equilibrium position-. F,3.q.z Cotditioos of stobitity o/equilibrium configuration of a corserualiue system of connectcil'igid bodi* arc:. (r)- If the potential,energy of a conservative system of counected rigid bodies has o lbcol''iniaimua at a position of equilibriurn, then that equilibrium configuration is stable. \ Proof,. Choose the datlrm of V aL the equilibrium configuration 96, i.e-, 7(96) = 0. Since V has . minimum at qo, for posiiion g in the neighbourhood of gs, Y(c) > 7(q6) = 0. Let the iaitial displacement and velociiy imparted to the sysi.em be q' and ri'. Then ' =,n (1) (1) r(,i)= E(c)-V(cl Q) +V(c)=?(,i') +V(c')-E(c')>0 =) "(,i) E(q') has a small positive vatue since > 0 and tz(q') > 0 by eq(l). Equatiou (l) + there exbt aset "(,i') ofg,int,heneighbourhoodofg9sottrat7(q,)=a(g.)andeqs(2);(3)+ T(i') =.8(s') - V(s') = 0, 9s + i' =A. \^ . , : .{ \< - i -'l -t > __ -t _J .,-] -I l -J S/ I t- Ll- l. L L a- ffiIt*Xy":I;1":#rI"1["#;;H.ffi;E kt-_ l- equiribriirm .""rir,i.."rti* {s and th't6anm"m t gs' IIence V has a local minimum d go' where g' is in the ngighbou+;d of is given by with coordinate g, the equilibrium configuration 96 For one d-o-f. systern LJ dvl dq lq=qo L-- tJ L: orr dq' lc-co (4-15) =0F-tli .t2v #v *'o ir ffro, #:#='^"=';#10,."'q=qo aY ^.r.._.-bY ci^ rs grr€I Civenr b-t . ---€-.-alinn -. -^-a. dt - the equilibrium configuration {r3'92s For trvo d.o.f. system rvith coordinates 91, 92. 0t' 0. AV * qr =916, {? = 920. - LL. ac, --n" The equilibrium confiEurt tJU L l_ IJ L 1- 0q, i'"" i. stable if 1z tras ('t-ro) (4'17) derivat ,ives at 96 satisfy: a local minimum at {o' i'e'' if the 4.T,,NUMBERoFINDEPENDENTEQUATIoNSoE.EQUILIBRIUMoEAftIGIDUoD: g. Altlrouglr Mg = g,it does rroc -.-ield bod.v are: f-= o L{.:= a rigld o[ equilibriunr of Tlre equatiotrs .-\ anJzmoreequationsindependentoftlrep,"'io,'ones.T-trecumberofequationsofequitibriunr.rr'ritterrin .olporr".rt form, for different force s1'stenrs are as follows: : 1.General,4=9,1'11=0(6)-Parallet'll:-a-xistli=0'Mt'--O'M^v-0(3)' 0 (3)' 4' Coplanarineyplane "F'=0'F"=O"{f;' = (3)' 3- Concurrenl;F=0 F' =0 (2)' 6- dpl"n".concurrentinr-yplane:I."=o' 5- collinear ll r-axis:& =0(1)O (2)' ?- Coplanar parallel, ll z-a'<is in zy plane : I1 = 0' I6a' = 0 for '4 on lg- Alt forces intersecring the same line r (d , F = 9, M t:Q (5) since M-^'9= SYSTEM INDETERMINATE ,"O,ICALLV DETEI,JVIINATE AND STATICALLY XN." ".^'='" I reaction forcbs and,couples atthesupports ,---- ' U u IJ IJ l-. I ,, t-. l- - . if the aud the A system is said to be statically dete -rinate 9'' Llt(f ) = using only the equations of equilibrium [E(P) = internal force resultants can be determined b-v fiorcc''dcforrnation case' p; otherwise it is ca.Ued .ror,.oit, itilclcnninate' For the laiter Ql for all its parts relations are needed for compete soluLion' be dererr.ined lry the teacrtions if supports cannoc if itdeteminatc is'cxlentally.slatically A system intcrual force system is iateraally stoticollg ;il,c1e|{.inale ifi|the using only the equaiions.of eqtlilibrium- A systens tqt'"tions o.!'g'uilibriunr' Stiticalll|.tinf.:,t:1"*e resuliints cannot be detetminea UV ,r.iog oniv'ittt timperature' during assembly and due to rise in are subjected to forces due to initial mismatch in in Pig'a'3 is preseated the fotlorving'l'al>lc: The information for statical determiT'Jc1i;";";:*t Remarks : force resultants No' of Eqs' No- o[ gq' f Rractions 1 Rr, Rz,R: (3) Rt, Rz,Rr' R< (a) 2 beam arch3 Rr-&(,%,86)(6) beam beam 4 Rt, Rz,c, (3) 5 Rt, Rz, r?e, Cr (4) r---^ A *hf''3!t] P- P- p' 13\ . 3 3 3+3 - I3 3 N' "t' n4' s (3) 3 N'M'S(3) 3 ili a'r' s 'v: /v' M' S'T {4} staticallv deterrrrinate externally stat' indet' staticallvdet'emrinate 3 ::*'SLi".;=J:;tr; r-1"".l{a ]J Jrtrft*iu { i i NHH{L +=# t ",fr' ft-,\ f?u*t{F :TiTk ^"a""'si "# IJ "* l-.. ' ".,rX":P.** r-- I lj LL. beam I i r\grr'\Yr-- L-_ u l.J *" n -,, 96 3 - int'ernally srat' indet' (.)€ Ll -S''. wwi{,e < . )€ -)f:: < ,,. Alinternalcross-sectionofabeamwhmeoutrvard(fromthebeam)ao-rmalisdirectedril-1:::::::: is is directed io tf. -y-.*'-o-':,::,1'rection same morneDt resulLants are assigoed the and force called a -ve face. The cornponenG of the internal in directed and face components actiag on a +ve siga if their effect on the beam is similar- Thus ttle in direcLed and -ve and those acting on a -ve face +se coordinate direction are assigaed a positive sign and corrrponents acLirrg on a +ve face .*:J[:":';fi":;;;;.;;;;';;;;;' ditecad in-ve<oordinate-direction "irr sign' in +re coordindtc direction ate also assiSned a negative ,-< e-axis along irs discretc and distributed loads' cl"9* C.oosider a straighr;; JjJ;+t*. ; of tlre components plane of toading (Eig-{-'lb)' Let ihe ceuuoidat a,xis and y-*ds transrffse'to i[ in rhe distributed the and q(a) N/m' r be n(z) N/m and distributed force in the e and y directions at location N(r)' oiequiliblium for lhe axial normal force equations tl,e momcat be m(r) N.m/m- ll'e.raot to establi.L i from gf bear' a element ru(r)- ihe FBD of an rhe ransverse strear force S(:) and ihe bendiog;;;* are eouilibriul of in riq.:a'sa- Its equat'iqns ta t*.Lx$,hich.does ror carry any discreb load, is slrown .= ' :. .i{ f" = S(:r+32)-S('}+g"*Ac =r'0' F,= JY{e* az) - N(r}+n-At = 0'. ta} (1"Ar}atr'e=0' A[t; = it{$ +Ar}- Jr/(r)* rnry&*S(t:+A'r)Ae+ . l' Dir'iding load goAz from '{ and thetefiore rsLete 1Ae is tlre distance of tbe resultant of the transverse " inFolvitg tcrm the r' at' values average vdlues becorrr the cqs{a} by Az and taking the timit as Ar - 0' the q drops out and rt'e get , dY nt {4-19) dA' ,=-.rr, E- -;$-nrE=-C, - t'' := . d2.u ds dn E=-E-E o! -r <!q (4-20) E=c-6- Eqaerions (4.19) are Lhe diffcrc.*tiol cqvalions f-ir! coplanat load- For cqnilibrivmof a straight beam under f =-r. '#--,,a dtn #= -r, #--'' __ (:r'2t) -- CoasidcrthefB{}daoirrtrnitesimalelement(Eig.4SbJofthebeanatorrgd'o'$'he.eadbcreteforce of eguililrrium, [ = o' & = f-.g+ Frj_ aad a discele couple ifok acts *j;:;,rr" b"..*-'Tlr.the.qo.tio*s ol quilibittttt "t jump coaditians 0, lf;, = 0, of this elernerri io tt Emit as ., - o, "i"ra il __. " ---< i N(rf,}=N(af,)=-F..'s("f)_5(16-)=-Fyo,if(:f,}-l{(rf,}_-l{o.(i1.21) ' / . '' Tteiumps in ir-, S. M at a6.equal negative of f'o' Fvo' Ms' tespective$'- ,+ Geosctrical'Itcz,Vltctiot of ncrlt.Its.' E{uationsi (4'19) -!{ fl(,,}.=rv(e1).i_|1,ou..-.-(areautrder-a.ediagram) l8z - S(41) = : < g.r diagram) -(area undl:t - J,.'oo *= M (rz) -rlf(21) =' j:: t * !,,' ^ S(az) i< 5-z and ra-e diagrams) -(sum of areas under - [=]Sif.nr=ol. (slqcof&r-edurve)=-n, (slopeofs-rcurve}=-g. (slopeofi/-l.curve)=-(S+m), a.part ort'e bearn considering bv obtaincd r". aiil-.. roaded seg;en* "iu".- are si;';i; with rf' s' tf -:TA a = 0 to z in Fig'(4'4b) is considgr-ed to *. side of the sectioo at, r. If the b""", *;;;;f-. * Irc.e. ri - !e g -(ortrer rorces in a dir-I, = -(other.**:::,ri.-r' 1; -,::1':-::f"::H -r .v :-{ ii:t*'' i{ i ,,1) *r4, tx+ &>( _{ _{ The cootdinate direc[ion are atso '..i8;ned a.pcitive "ign lEig'a'a1' thce acting on a -ve face and direct-ed and i.g",ii" . are assigned d, *M _{ I oE STRATGHT BEAM EQUIiBRIUM equertoNs t; li tol 4-s ,18 .J )-' l -{ -{ -q : 9't :. /jt il 'i i i i a' t i L-.- *gn*"t are computcd and the N-t, S-z' M-z The end values and the extrgmal values if any within "".h results pt""""t"a eatlier is helpful for this' diagrams are drarvn- The geometrical interpretation of the and the constant's of intcgration If q(r) is not a simple function, theo eqs(4-19) are directly intcgrate<i de[ermined from the end conditions' L ., L L t L. L.- -L, ) L- - Lt, L.-- , iiqfliry:ts; I , : \-- t- ., ) [, t-..q t- _) ) L..t- ) t--- L L ) _) tI *l Li lr _) LJ. ) L_ 4.10 EQUTLTBRIUM OE TRUSS so that their centroidal a'ies are A framervork of tiangnlarsf,apes formed by joining bars'(members) lruss' The joints ate idealised concilt:rcnt at the joiis a'rd the apptied looils ect otly-aurlcseToirls-is-calleda joints for space truss' The members are idealised as as lringed joints for plane truss and as balt aul soctet joink- Thus alt nrembers of the truss are two force nrembers wcighllesscompared * 1L. t6rads applied at the o[ connection to the joints' t'et Fi be ttith the forces necessarily acting along the tin" 5oining their points of a joint' the forces due to members are shown the tensile force on the ith member at its ends. i" " fiO r:alue for Il'' equations of tire joints yields a negative as pulls on the joints. lf the solution of the eguilibrium then the actual force in the ith member is compressive' ' j joinG alld having c -roun{atLn consttaints' 'Tn" c,onsider a frame made of, 6 rigid bars, connected at j joiuts is 3j (2j for a plane fran're)' The tcrms appearing nuqrber ofcoordinaies needed to specify position of t.t'""' The number of constrain[ equations specifying iu ihe parenthesis in this section refer to a plane f.*r,',u 7 z^- ,,B')" = L2al'and the number of geometrical the fixed length of 6 bars is 6, e'g'' (ca - 16)? *(y^ -y'f +( O".tn: d'o'f- o{the c' e'g', constraiat equations duc to c foundatiou constraints is '-e =,0'-UC:0'^Tt: n=3j -b-c(n=2j -6-")' [ngeneral' lhe frame- lfallthecousfraintequatio*s are independent, then Hence n Z 3i -b-c(n>2j -6-")numberof independent constraint equarions is S(0f c)a mecicaism in which some individual member(s) is it then If a - 0, then it is a rigidJrcme and if a > o, still of sonre nrenrber or some founda0ion coustraint catr move as rigid body. A franre is over-rigid if renror"al frame is foundation constraint is calted rcilvndattL A malies it a rigid frame rvith a : O- Such a member or constrainr nrakest' " "t:tlTl:T fisr-igidif removal of any member or any foundation of forces at j joints is 3j (2j) and the number concurrent The number of equations of equilibrium for unknown fcrces,in 0 members amd from " ? condition for a' rigid frame is that n : 0' i'r ,er of unknown forces is less than or equal numb the taat lly dctctminc{c'is condition for frame to be statico " equatiotls are arranged in a matrlr to tlre nunrber of equilibrium equations, i-e., 6+c <3j. If the equilibrium the unknotl'n force vector [flte+"1 and form: [n][Ft = [P], where [A]1e;]xtc+c) is the coeffiiient matrix of nccessary rank [A, P] :=16 * c, nhich is therefore the i"f,"r, I ii" U.ir""to.. Irs solution exists, if rank [A] = ooi snffcicot condition forlhc frome to bc stat.iullg dctcnninatc' and su-ccessive addition of A simplc spaliol (ptator) lruss is formed Ly a basic tetrahedron (triangli) so formed is just-rigid' joinG- The framework' 3 non-coplanar (2 non-collinear) members to form other are independent in this ese' tf the The constraint equations provided by the fixed length of members motion o[the trussas ":i9:j:e91t:j: fotrndati,on constraints are also independent, then [hese Prevent 3j:6+6(2j:6+3)' aodc=6(c=3). Eence,thed-o-f--ofsuch.asimpletrussarea=3j-6-6-0 =+ of ttre whole ttuss' equilibrium of The 6 (3) foundation reactions are dltcrmined from the 6 (3) equations considered nerd,' The forces Tte jcrint which is the last to be formed during the fotmation of the "lls:-: (2) equations of equilibrium for in ohly 3,(2) members meeti'g-at this joint Je'determined frbm the 3 and solne its 3 (2) fi" next clnsider- the last but oni joint that was formedjoint' the concurrerrt for"" "yrt"**!ltT:*d that ite procedure equations of eqtrilibrium fior the unknorvn forcesin tie3(z]membe* which qeated that is repeated for the last but two joinL ttiat was fiormed and the fiorces.in Q-*"*O:rs ' all memberc' Eence' joint are determined. This sequential procedure leads to the determina:io1 o[ forces.in yr.l rT:_i?TIr. asimplctntss-roi{[6(3}ilrdcpcndcltfovndatio*co*sltlirttsisstoticallgdclctmilatc...'. @ -ro fi:::*j:I i 1 : 1 I i ! i t l I i I I i i i 1 I itre Es \t, ry' . -\ .r-{r. ,-F* '...r f op-y nn Fq u! :5' - - sls.q.b --Cig.k.6 F; is negative' if *: *l:T": of Ia the jo,:nhwise'deresrinarion of meniber forces descr'rbed .{qve,joints in whic'h it appears -anv subsequently' of equilibriumoi then this negitiru rrdu.l"l"Urri,r,.a in all equations plane tbe special case of joints 'A and I of a consider pnlts Ft. oaly The FBD's of all joints should shorv joint r{ yields Fz - 0 and applied lload' f' = 0 for truss, shorvn in Fig.4.6, rvhich are not subjecteJto 4 7' 8 are zcrc-totxe members F! = 0, & = 0 for joint B yield Ft = 0, F.l = 0' Thus members ,r,cmbers of a plottc {rzss' For this purpose' o Sometimes we are interested in littding forces irt iusl lcro it into tro parts by cutting a member of interest a section is taken through the truss so that it divides aodgnfenblyoilyluoothermemberswhichott,nolconctrt./rlltuithil'TheFBD.ofthetrusstooneside sectioaed' using the equations of equilibriunr of this sge.lion is dra*'n shorving pulls on all the meiubers often convenient to select A at the point of Ml:0, F" = 0, .F, = 0, lve can n,.a ro..o i,. l **b.o. lt is above FBD' then these ficundation reactioos aPPear in the concurrence of some unknorvn membe.r forces. [f l1 l- ...:- t'i --. , p,lr directed oppcite to the impending r I; ( prJv- For impending slipping rvithout tipping of a bod1" '=' .---.^^4 impending 1.. simultaneous imoendinq For -i*rrtranonrrs othe area or conract- D^".,"r.o. tipping aod slipping, F = y,N aud .tr[ acts at a corner' load is int:teased, tne assume 2- Ia ordet to decide rvhether tipping or slipping occurs first as 1he applied is 'that it tips 6rst and find.4 N and .r,".t ,ui.tt* tat 'S P,/V- If the c'hecli holds' then the assumption of slipfing-takingPlace first' cortect- Else the assumption. is *.rong and rve ret'ork on the-.basis a prescribed load P' we find the values P1 order to decide $'hether the systen'r is in equilibrium undet -1* I* tso directions' If P lies outside the range of and Pz of the load for s,hich there is impending motion in the ittremains in equilibrium' Pz. then the systern does not remain in equitibriurn, btherrcisi Pr "nd and no slip at another' one -A:"*: 4- C,onsider a body'rvith trvo cont-acts $'iih the po+sibility of slip at opposite to impending'stii: direction] and impending slip at contact 1 (say) [i-e., /V1 and Fr = P' r Nr directed check wlether'the a-ssumption equilibri"no slip ar iontact 2 [i-e., /Vi and Fe]- Solve th" "qrriions of "n9 a-ssumption i'S correct and the then the of no slip holds, i.e., check whether l&l I FrrN2. lf Lhe check holds, with the assumption of no olution b over. Otherrvise, the assumption i *.ong and the problem is tervorked ' pe,N2 directed opposite to impendiirg slip at I [i.e., /Vr and F1] and impending slip at 2 [i-e'' N: and F2 = slip ditectionl, and finally check that lftl S P,1Nl' including atl the fric[ion 5- For a.system of bbdies with multiple contacts, count, the number of uuknorvnsslip.equals the difference ' fotces as'the unknorvns- The number of contacts at wtrich there is impending of all the bodies' rhe number o[ independent equilibrium equations of the number:of ;;;J;;*;;;;; stip, and employ at these contacts & = r'jffi Assume the specific:1oJ",., .f these contacts rvith impending .; t*n"nutn, slip as obtained from assumed kinematics' Solve the f,quations dtuect€d opposite a il;;;; at the contacts rvhere of equilibtiu,m and these kinetic conditions ani thun check whether lql< t'jtntY: has beun obtained, else r.ercork wiih no slip is assumed at,the beginning- If the check holds then the solution ;;;;;;";;. --< '...,,< :'r< !\ .- I Thus1/rcteaclionPoothejottntalisla*gctttloacitc,leofrcdilsr=.Rsinc. 4'7 il INYvrOf,\/rING FRICTION 4.12 rIrNTS FOR EQI'ILIBRTUM PROBLEMS - t r--^^ -thenormalreactionNisatacornerandt,hefrictional'::""\.[A '{:r- For tipping rvithout slipping of a body' i,i#I;,iJ'#T',:[ I -1 4.11 N.EACTION AT A JOUILNAL BEARING journal beli-ng of the same radius (Fig'4'7)' The inevitable consider a shafitof radius R at rlst in a short (radial) foice,v and atangent'iatfrictilonal f11ce 1'N clearance results in a li,e contact *'ith tle no.rrr", to Nn *'hcre Luthe bearing' The tot{ reaction P is at an angler'c relative to slip rera,Ye about to sllp is abouf, rrhen the journal ts 'Tlre .^m tha ..'is Tlre a-xisthe r tr from distance ,ut at sorne some dBlance 1P p act radially but o _ t*_ip, is theranSle of friction- Thus does not j-' b1./mourerrt M of the reaction about the a-xis is given " 6 :- ^ ;t {,;Itf I r=(P'N/P)'R=Bsina -i+)l; \ -i. -l 1 of the whole truss' If more than 3 members als-o used in the solutio'' = : ,lI equilibrium should be determined beforehand b1r considering the of equiribrium are equations irr"i. made.nd are are cut then additional sections ;;=;.^;:;'.o 1< :\ 1-1 .: i + i. i 1-rrr I .-r : ; ,i1- l ?-)' -..< .,< .jt < : :-a .a 99 \l , r'' .;:--'' - "'1 ... ' l ,. lr :'i'" Exaraple 4-1 .A .0iuck with e-ear wheel drivte:;'- i'"; (Fig.Bt-l) has a'winch-mounted al its back: 'Thii's'; winchi!: 't-'i."':' driving torques on the rear wheels and on the ti-"1 :' are ?1 and ?2. Assume no slip for the rvheels"itte' drag force coefiEcient of friction for block 5 is p- The c1u4 and wind' are on che truck and block 5, due to ct'c2 are c2us where r{'u5 are their velocities and system constants. Derive equations of motion of this l_- rj rJ { using c aud l{ l-..I{ t{ IJ Solution f:. (+r*q E)(AMPLES.4 ,',.1- .ri, l.-. i<;\: as generalised coordinates 3,a>f) G.vll F;3- e q-l rotationd-R of-.he-r*'heels of the truck a,d tt u;.ot"ti*7 or the rvincrr cause ilR and the velocity of block 5 of block 5- The angular velocity of tlrervheel s k The di"ptac"nr.nt r and displacement (z * r{) is (i + ra)- l{re take the initial configuration and potettizl energy are given b-v o? tnt =ytt"* tt tlre datu.r for pote.tial energv Y- The ki'etic l.d]= 6d2]+ ]mo;' *.l*'tt + + I2(i/l?)?J+ jt"rii + r1(i/a]2] ][,nzi2 ![mfi|+ ,*+''ut16'+ msri'i'',;"' m?:+:m3 ] ma'*rn5J {t' i lepa"li?'+}(rr * =![rnr 0 T= / -v' ' '::i'l : (ti1 'r ,n2 + .f,3'1'' nta)g'z iin O 'l- fisgr(z * rd) sitr (1) (2) '..' 5 moves up by (6r -i- 166]- The rvheels rotate b-v 6tlr-and block rhe 5r;6o, displaceme[t virtual In a do virtual rt'ork on the rear The driving torques T1 and 12 : prn5gcos,. pN is s block on frictional force do no I'irtual rt'ork sinpe it' torques acting on chassis 4 opposite Tlre respectively-rvheels and the wincrr frictio'ar forces on the trlo.r 5 are c1i a.d c2(i + rp)- Trre tra'srates- The drag forces on trre truck "rrd Hence of points of con0act is zero for no slip' rc-heels do no *,ork since the velocitl.riXor + t tdl. es 0 (62 * t 66) ; c1i6c' "z{i+ . -, 6iryo. = Tt(6r l. Rl * Tz66 - pmsg 1611166 Pmsgr cos 0 - t:(i + + ri)l6z [?r ct(i+ ci cGd - = W I R -pm5r s Qi" * Ut + I)lR?li + 'ns"d' ff =yl.r* rn: * m3 + ',,{ * ms = (rnr{ rn2 * rn5)g sin d' dv 0z =lT2 - =vrl R- pra59co6a : c;i:'c2(i+td['.'ar ,rr3 + "r.{ + .# fi pm3'orcosf, =(ra *msr3)d L L [rn1 { rn5ri' =,r,sgrsino, *(#\-T|-X=QZ'' *rn2*me*m4*ms+(Ir d h)lr1li*msr6, ^ * (ror * arz * rns * rq * ms)g sin 0 = T l-R' Prr,sgcos 0 T:o. ctz T-s. A.E, ': The Lagrange's equations' of motion are: *ff)-**T=e2", -c3(i *te)tl t3) - i .:.:: c1i - c2{i + -d)' (Ia+m5r2)6+m5ri+m59rsin0_Tz.pm5!!tcos0.""(i+'d)" U U L L. r, L : - A m1 with a thin bar of mass m2 consists of a thin disc of mass Example 4.2 The System shorvn ia Fig.E4.2 by a linear sring of,s;tiffnes a horizontal track and is restrained on slip rvithout rotls disc The it. to pinni:d the d'sc' The springs arc uodeforned -rs connected bet*reen the rod a$d E- A torsional spring of stiffness !r motion' rvhen 0 = 0, a = 0' Derive the equations of foo l, The frictionat fegge -on ihe rotor does no rvork since the I'elocity of the point of contact is zerq. The displacement z of the <iisc cenf,1e aud rotation Solution d of the bar, imply rotation 6 - z/f-of thedisc. The angular'velociiyul of the disc and aagular velocity ar2 of the bar are given by u1 = if R, u2 = 0 and zA = u* f,sind. The telative twist of the torsional spring'ts $ + x/R)' The virtual displacement 6:c, 60, + 6xa = 6x * Lccsl 60, and 6Wn' : P 6z-a : P[62 + Lcos| ei" = P, qt'= 601 = Q'fu 5). I + Qi"60 JJ J + J J J Eis.€tr.z PLcosA- The velocity of C, the kinetic energ,' and the potential energy rvith datum at O are given by ec=lLa .dk * Be=;!-akx (I/2)(sindi+cosdi) - (i+ ILitcosl)t-*Lisinei T=llllrrrgzltilRl' +rrr1izl+ |nr2[(i +*Li)cosllz +(*Lisin0]'1+*(*rtz1tz'1P = l(3m1/21- n2\i2 + {.n2l2l6)it2 + lntzLii} cos? V - +k* + +tt(d * rl R)2 - f,m2sLcoso = 6*rt2rrr.z\i+ ff lrr.:Idcosd, { = ff X=lr1k{o+x/R)/R rnrLzlft+ l,n2Liccso, = t ,te + x/ R)+ *rnssl The Lagrange's equations of motiorr are: d r?Tt 0T aU d OT AT o0 AV' a J : = -f,m2liflsinfl. (l} 0 = PLcs0' i AI (2) 4 ,Exarnple 4.3 The system shown in Fig-D4.3 consists of a rotor n'ith axial nroment of inertia / and two bars.pinned to it. Each bar has mass ,,1 aud the mornenLs of inertia about principal a-xes'g.at the centre of rnass C ate'. I1.,lzz,Ias. The bars >i \ i\ I/ ' vA+ s2'x Ag -- (u a u, x o A) * ttz x &-= 6 tx = -(B + csin 0)d & + aa cosdi+ o2s = (E + osino)z$z + o202. o0sin gi I tol Ri * (Oi + o g x o(sin d [ - cc0!) 9L d are restrained by two torsional and one linear spring. The springs are uudelorrned for 0 = 0. The torque applied to the rotor is fi and an internat mechanistn ipplies lorque 7i to each bar to raise.them. (a) Derive the equations o[nrotion and shorv thaL these are directly related to the.p.rinciples.,of, moment of momentum and sork-energy. (b) Obtain the first integrals o[ these equations. (c) Find the angular velocity d of t\1 rotor for which the bars make constant angle d6 to [[e vertical for T1 = ?i = 0- (d) iConsider the case of.free roiation of the rotor, i'e', Tr = 0' Initially q = uf il = 0' way thai a such iu bar to each torque applies t/2. mechanism The internal rvhen 0 the bars are lorvered ro t,1r" ,"stical configuration and 0 = 0 at 0 = 0. Find the angular iD velocity of.the rotor rvhen 0 = 0. and the work done by the internal mechanism to change the configuration ofthe bars from horizontal to vertical- . Solution Tle angrlar vetocities of rofir 1 and bar 2 are g1=, jj- and gz = 6i+0L- Hence gc' is tiven by gi = J J + {3m112*mr)i+'}rn3l18ce 0-i2$n?l**a* k.(0 +z/R)lR= P ,n2L2612+ rlrn2tr(i cosd -. irisin g) + !r*2Li0 sin 0 * I;r(f + zl Rl * ]rnssDsin n t26p*ImzLiced+ k40 +zlR)*|m2gtrsin0 - PLcqA' F.quatlons (1) and (2) are the equations of motion of the system- JJ -0, ;lril-*+fu=Q';"' a\7*)-fr+*:Qi" :+ r?Tt AT sin 0, AT J .J { ( a-. .\ JJ J J JJ J J J J J J JJ J J --.1 J { r 1 The exteasion "q*i*p!ne*,j-!$0: The ki19!ic.+ergv and potcntial ' '' energr : G;G;ilr.,iaa"i"il;i.'d':*..9^n,'i1x...,.:;.-.i1:..,.'.'... t = iI i'+ 2(i)ft{(r? + ;i'* ai.i'+ a2021 + '.!' .f" o+ rzed? sin2 e + tr.itz1 ='Vli)+rrrt'o+I:zsiq?0-+m(R+asin8l16'+lls+'rar.2102 acoso * 2,L'sin? o * +z(+kcazl= -2mg = -2tngacosO* |qzrsinal? f (1) ktoz- (2) reversea gotqYo by the torques ?i on the 01" O.:t tlrl In a vittual displacemen t 6l,6l,r'irtual :*.1 i" done virtuat rotation in direction 1i actidg on the rotor do no virtuil work since the rotor has no . 6Wn' :Tt6i 1 Q{' 46dilSt6O-+Q!:6e-+-- l- Hence Qt' =zTz' =Tl l : qa0 = 2lI 12* rrr cc? I {ae =ilfu+ * I32sif d + rn(E + osin 0)?]<i *OQ A - 2*orsin0* a0 --0, 5=0. o0 2kL"sio?i +2:k.0 =o'"' + *(#)-#-#=Qi"' *(#)-'**# - $avn* r:, *', * ,-y1- Zlle*tt*210 -Kt"r- siuof )il (3) =A 11".::,": :rr:::::^-a,,1r t r11)$in,cos 0 +Znlft*asin8)acosa]t'3+?rnsasin a +2kL2sin2o azb,o:trz(41 the **T* about the fixed a'tis Equation (3) is the moment of nomentum equation fot at 'r1' Equation (4) is ttre moment of, rnomentum equation'of:::t'ty::':'axb fo.* ih" 6 fottowing equation ftom I t2* 11 1 cosz 0 + *, tt r-'- r, [Le 12{ '- 1 * ,i The Lagrange's equations of motion:-' (") - mazli Izzsinz .in o + D0' i{st = i{6,'' : cqs(3],({i: eq(3)C+eq(l)0 0+ m(E + csin0)2}d 2',.(R +a sin 0 )a cos alit - 1U* +,,nailii +2k':1!0=T'6+tTzi -(Ir"-Ir1{sinpcos.0'+2n(R+osin0lgccd}tid?*[2mgcsind+2&f,?sinzd trtrlr* rr; i";r,r +,r;,sins? +daii"ioon+6qg6,-.1;;fcisosinaiz-€ru#tsin8)acodltii? i"'i' si"? + 2;u 6z'1'26;1 zff iil6' = t'Q + ?ui +2[t$ * i'l i :+ fiUrn+frrcos2 0*I22si,:0*rn(8+asir,efl6' ' rhich is the -- -1to +'ro91162 - work-energy relatiou for the whole systcm have been writtcn utr*ti ;; ; *"-"otoirro-*t =T'i ECuat-roT^(3) rm equati'on and wort-energl syst€$r. lntegrating eqs(3] a,od (5]'oace, we get .. '':: '. ' ;-:. t I i zmgo ccO +2&f,2 sins 0 + *ta1 in the rate form' t ' adt{5) t"luitT for i +tFzit 1 fould the shole , a(r'tV) =Wo' l i (6) i L,- ) L.-) L.h:+ L- r) 'l i I I t a .lo5l:-, f L-- i,' t) _(c) Tr=Tz=O,0=dq'"]\ Jf, - i .: i = 0, lzmsasina +2bL2 sin20 ;-r t--.d t = a, and eqs(3),(i) yield d - constanl = fs (say] and -- *2k.01/ IUzz- r11;fi,racosa +2nr(E+asind)aerdl (d) LeL u2 be the angular velocity of the displacement from 0 = r/2 b e -0 to get I rotor rvhen the bars are vertical. We apply eCs(6) for the -.< i 1 + Iu + m,B2)utt - Z{I 12 * Izz * m(R + a\2)u1 t t2 = II /2 * 122 *,n(.R + a)!q | (I /2+ f,, + mE2) t['"c - (T + Vb _ (,l+ rz1, J^Ut lz* r- + m(E + o),|-i + zkL2 + k.(r /2)?l + 2(I /2 =+ (7) /z* lU r11 * ma\,.ti -znzsal - (8) The rvork done by the internal mecha.ism is given by eq(g), *,lrere a,3 is given by eq(7). a : Exarnple 4-4 Atl instrument to monitor conditions iu hazardou, (t-S ,r,2 B l* "ntironmentispositioncdb1,alazy.tongtypeder,iceslrorr,ninFig.E4.{a,lr-vappl.ving"M.).jlE1 torque.tly' to the handte of the bar zlB. The mass of ihe instument i",r. und mass of rods is I *:::,^:.j::Y ";;.* *;, *;;;;;t;;. kg/rn. The pirch o;,h" t'::'"",::t:,Y"*'*T:tion-,Sor.tilrheslfl ihu*t one tlrey-coordinatesofthesepointsofapplicationare.< y3 = (@ + 2b +c) =Zac-rx,l :+ 6(A8r= Applyiug the principle of virl,ual c.r./:\ c,*/'.. ' rvork c ' i\ ;1Zi sin -2asiad60 6W - b3, considering - p6gl2r :+ 6d -t'.\ /S:--\ -\m *k#." }t.' = _4rcsinl60lp- =f,Frr6y;* l{6ti =0 + + 4\bg 6y2 * 4\cg 6% + 4\dg 5y4 + rtug 6ys + M6O O. . = [Ig{o' + 4b(o+6) + 4i(a +2h + cl+4d(a+26+ 2c+ d.ll +ory(o * 2b * 2c -t- 2d)] cos 0 - Qtralp)Atsir 2\ag + i. :, \:H $ir'a'{,,:li{-u "::.:::::i'!;^ -48 i. X\, 1 0, \Zzt e+(o* 2b+2c*2d)sinr?. 7\ or u oo; j:" =^," *"1 : :': ::"'' dya =(a +2b+2c+d)cosdd0. -: d3r5 =(o *2b+2c+2dlcosl69. f-;lf-\,'-.' -or -, '--*Y.'"! slr5 -tst.vt.cizarcosttcv. y, = |a sin 0, yz = (a * 6) sirr d, y. = (o* 2b+2c*d)sing, Us = The rotatiou 6<lof the screw ca' be ecpressed in terms of 69 :+ .-'i' ;il ,i,i lras degtee of freedorn- Let d be the ge*eralised coordinate. Forces *,hicrr ,ro nork act in the y-direction at poinG I to 5 displacenrent sd, f,d, {Fig-Ea.4b}- For viriual displacearent + t)fo A 6y1 M = p*10$s{o' ++6(o +0) + 4c(a * 2b,q sy The torque llf decreases as 0 increasesExam.ple 4.5 Derive the er2ression of the forie P aod the moment M required to maintain the equilbrium of the pendulums shown in Fig-Bl-ba. Tha springs are undeformed for 01 - 02:0-.The pendulurns are modellcd as uniform bars-) ,. ,., r * =0 4d(o*2b + 2c+ d)} +mg(o+ 3_r Mi 0t 60 V.)6e 2b *2c+?i)l / 4ra *[* {2-,,r, s:.q-{r trrI i k.tez-e,) K<,,, '+",1*f,{le'-o') Solution llre shall present three solurions usiug 6W - 0, 6W"c - 6V - 0, and 0Vl0q;: O: i l- The extension of the spring is 0sinds. The rela[ive rotation betw'een the ends of the torsional spring connected to the pendulums is (gz-0)- The forces acting on the trvo bars are shown in Fig.Bl-Sb- Note that : the'pair of moments due to torsiooat spring does some net work on the system during a virtual iisplacement -: J ) l -J ! I !*o_3 J J - , k.'\r tj_ lj-1-^ t. l' u tJ 0sin01' y1 1* L2sirt6z, 6gy= L1cm0r60r * Lzccgz60z, = lrr.in e, ce, - rlz;,.02602, 13 - iLrcc d1' 6e3 - -iL'sin ='r'r-Jr,* +rr.o. rr, ,r; -- ,,. n',- r.rdro, fotces on the bars The virtual work of ext'ernal and internal 0)150r+ [-e{02 (-I;Dsin01)6vr + M 6A11 F (0' 5W : P 6st* m2s 6x7* r..ts 6zs+ Ae' q + t[ + r'V, -. m2)s L1 sia0' - &0' .!.;\ '- "'' ":i ::" = fP Lr cose L - (m 12 + Y 6i0t and V60: k'(i02-'A)1682:O 0z\rrr2g/;2si *lP,1co;021 Hence, tbe coefficients o[ A r (n1 12 tr i d1 601' I O'}FA' (l) - ie-' * m'lg Lrsin 01 ."u""r"n"r.i:i 7:Hltx"lT:,f:[is:.fu as.-,v 'using Y: sprins and sravilv forces is orrtained * |f,2 cos02) 8r)' - \*'gL'cos01 - mz9(Ltcos01 = !t(6sin9r)' + i&r(0, 6tl:&6"sin01cos01 60t+kr(6t-rdtX6gz-rr,l*trnrg.LrsilrLrlar-rn2$(-f,1sin04-|L2sin036d3{4) rz f,ence 6W = 6Wo" -6V - P 6w.+ M 60r-6Y70d{ibetl 6att from eq(4t'the eq(t) given vidds' after substiruaion calculation' "*ti' . ..earlier- The test of the procedure is same of Yirtual as constant f6r the PurPose 1o1k zretre.ied p M moment and force 3- The Siven In thai sease as tgitlr P and M ate trea'ted as consert"ative * f,tz'G'azl l*'gL'"T": *2s(L1ccfi c661 - *rn:g& coo: *gi;1e* - e'1'- (t'" + 'n:)eLr v = -pyt- araflrto"indl)3 +!E (az-l-rlz - IJ L Lr cu momegt 1L l_ L- should separately be zero' t; i.t II, ,t I (2) &r(0r - 01} = 0 - &62 sin 0r cos h * i{ * (3) 01) = 0 P Lac<x,l2 * !m29 I'zsin01 - k'(Az the eqs(J) and (f)r are*respectivel-v' ** is lf then and P r*-rrorc' (3) ef the vietds Equatioa :o('| rnomenr equ.ibriu'riqulq -obtained ^I:0l::lt"' J th<j, equiribrium equati,on.for lr. z P L L l__ L: 501' and' 502 - il :i ;::'f$:;";::':::':XiltT"iilX"i::T:S'ffi1':f;::i:';:""':i'i'1d 6yt.= 6cosd1 6d1' tJ L IJ L L l-. L L L IE 4' whereas TJ lJ llJ IE ,ti L tIJ .t 1 - = -P(.L1sin,r * ,r ,, *Tf ** + *t(osla "iJ arc co.oscraatioc, theait is most corwcnieat to asc !e E Fauations (6),(?) agree with eqs(2) and (3)' titted up by tbe force trL from Exaroptre 4-6 The platform h Fi8-E4'6 is are light' Find the rralue of P the pistoa of the hy&iulic cylinder' Tde hnks for given (5) lhc c4uilibn.,- eq.,alio,.,x ovldqi = 0v pLrcos6l*f62sin01 cos01 - kr(02-01)+(|rn1 *m)gLtsino1 =0 =-M 00t 40 aV ' = a -PL2c<x,01t t'(gz -g.)+ *mtgi2sin02 do2= ' - f,e sindz) 0 (6) (i) c t^t i jr-OH=O gi:c'oA=AG for 0 - *14' Mt, Mz, Mg' Tte springs are undeformed as resoh'ed is E treai A,Ur,Ur,U, as constants' Force rralues of R. - Solutioa [cos C,sin We R= -RcosCi+ Rsindj *ith (a sin 0 - c)l/t(d C) - [(d - a cos 0), a cos 0)2 * (a sin I - wich datum'at The potential eoergy of a constant force F 'l]" cl"lt 12 "t:Ot.:ft," rhe *r.;;fi;:';acts ar, G- ff" to'r{r iswlt2and *.H;5:;1ff:A;.:fi"rffffi;.I";J'i.;-"..*aie t64 .. I .d f)l .:s.- ',i, :ir-:. -.... , :. potential energf of a moint6trl/, aqtiog on body f in.positive of body i. For the given one d.o-f- system. V is given by ; direction is -M;0; where d; b the rctation ii ,, {iiliJ v = -(-W)va - (-P)" E - ?Q * Wtl2];v{- (-M'X" - 0) - (-M'zXo) - (-MeXd) i l-l ++t ,P'- ol4l2 -(-Ecos{)eg - (Esin{}v6 =-(-w\(zlsind+6)-(-P)(c+2acq0) -?a.yw1l2)(4osiud)-(-Mrl(". '*0)-(-M2Xo) Rsin{(osin8} - (*M';1{e)* }&r,.[ a] - ..l2l2 + ][,,[0- ol4l2 - (-BcosdXacos0) + 1e',[. ' : 2A - The equilibrium equation ry: a0 i p -| *1212 - qF- &L ".." *Y ,! in Fig'81'7a' The unstretched if the handle-srltfjo'*ll1 bv equal forces 'uhe nrechatristn st:orvn * | r .. r I Yt "/r;ffi'\ o,tr c.sc tIet Solution obtained by U,ty a1c -- u2b '+ ''t't2 = u{lb The componen0 of the velocity of poiut D in the direction of force P is virtual rate of rvork Yields 6W ; F'ri+'t-(a - Ln)u,6 - Pugafb = Q uza = uPofb. The PricciPle of P:1ri+*lL- .t I -t -..J ir --i !-J I j I *riting the r"elocity of l{ in trvo rvays: . _i ,-J Che the clamping 0 as fixed, a rircual angular velocitl'r'r1 is be most.onu"ni"nt..Iteeping to in this problem, since it happens of the forces are shown giten to link 1- T11e virtual velocitl- ccmponencs of points C and r1 it rhe direction instantaneous ceni're I: of in Fig-pl.?b. From sytnmetry, the virtual velocity of poini B is along OB' T'he angular velocrry o2 virtual link 2 is obtained at the point of intcrsection of the normals to 3L and 5' The 'rs I ,l |<- a .af--.-1q|i principlb of virtuat rate ot *-** force,be P- The application of --l--o --'r-1 I t A-'( a ---k- c ->t+- b t.1 I .Sl-- ta I I .""1!, n.Jj,',1"","',-' on cjre,obiecl I-.J I ,_l i T::l;:',;j'f;:";,1^r;j is .Lo. Find the clampin* tength of the :plrrc friction. Neglect F(o) of link 2 i ZWoas|-2Pasin 0 +4(Q-W1/2)acos0- lt1* t43 (l) t2t,,[ 20 -x/21+t,,10-n141-l?ncospsing-Rasin{cosd=0 Example 4.2 Inacessible objects are clamped by rf- l: TV/Oq;:0 yields P: 2wocosd + 4(8 l-, L. .L, Ldlblac U, \l -a -t .l I J -,1 l !/ -t --l \i--j --J tl J \l J ,I tl J-t *l ! u I ,] curvature of the peripheqr at of the top cylinder above the point of cootact is h. Let Ot,Oz be the cenires of lo5 I _j -.j ..J ..J J Ll\l. 1'\- .: . -;::-.. rt. ::-. / l=- ) L l- L_ t- L--- LlLIt L-- l'l- 1- L--- LL. ILJ tj LJ IJ l.J LJ LJ L-. TJ LJ u rj t-u t-: u *ui"T, P<il{4?'c. rhc point of "r"1""i tL"4qp-:?*4;i*-ti""."f small rotatior (Fig'E4'8b) when -ry..v,-?,.{l*3 :i; cont"cii" D with z3oti=e-T.u"olup "*aitiolimji[es $7fualR2 BD=DA :+ -''R2Q=:Ro'tr'7n * verticat' 3nd ihe tine Ab2Q is at *"81e In the displaced position, O1,D,O2are collineat at 4gle 0.with fhe with datum at Or is 0 + d wit; the vertical- The,potestial eaergy Ir of body 2 of mass'rn (I) v(0\=mg(orol)ced+(o26Jcc(0+cI = ms[(fl1+'Rz)cc0+(lr-82]cos(L+ hf F.:)al v,(0D=#=_*onEr*&}siad1(Il:a?X1+Rr/8:)sin(L+h|.R2}0| (3) + RylR)z cc(t+ Rtlb)el (r.r v,(0) * =-me(Er *.82)c0 + -.R?Xr .-.0 as the egui[brium configuratioa' which is using eq(2), the equilibrium cmditioa v' = 0, yields 0 y"(0) otherwise obvious- This equi[brium pcition is stable ) 0' i-e-' using eq(3), if ntlez|z]= '#(Er +Ez)2[RrEz if v"(0)=-mc(Er aRzl+(r.-R Xt+ i.e.'ifn<Rt&ltRr*Ez)]=ll$lR|+l/Ed;....Ki /(Rt+Ez]-al decided by eraluating higher deriratives-liace .For rhe casetr =.E1Br/(Er+aa), tue statrility is 0- For this case, eq(3) +* /?:){cos d cos (t + ful R:.)9l Vt'(0) = -rng(Es Vo'(g)- me(Er + fi2)[siuf \' L-. L-- (ric'F:'8a)- Mark th" on body 2. @nsider LJ L_ LJ LJ i]}=*: .*--4qonom!&tio ,0;0 rhe points - - (1+ E1/82)sin(1 + v"'(0):o yrttt(g) : *c(11r + E2)[cos0 - (l + V'* (Al = ,lr.s{R1+ 8z)[l - (l + hl > 0 Y"(0) = (5) (6) &/Riel (7) Ri&zl2 cos(l * (8) P'i.lR2]01 -1s-) Rzltl < 0 a local minimurn at 0 = 0' alld therefore follows from eqs(7), (9) that'tb,e potential encr8y v does not have the equilibrium configuration is onstable for this case' equilibrium configuratioa seduces to For the particular'cases given in Fig3l-&, the conditio"jn) :t:t*le (]'is belorv the centre of curvature OaI . Br : oo: siable if h < Rz,Le-. if the ceuire of mass 2. Ez = oo, stable if L < Er3. Rr = -R: stable if h < Rzl(l - RzlR). It * 4. R2 = -k stable if h.< Erl(l - &lR). 'trr4 " stableifll(0,i.e.,iftheceltreofmassisbelos'thepaecteoint' 5. 8r=0orr?z-0orEr=Ez:0: lJ 11 ! ? (fiS'S+,S1], ; at A suppo* the Exarnple 4.9 'The bar AB is piaaed to c A tu :.*,-,"';;%:;:Tfi;":is Ij:.jl,Ll.i at u:J:;"*;la;;]1(' -&a 5 is ^u^mY, D to bar BC at B. The bar BC which pioned l'o a roller and guided in a vertical slot- The bars are coastrained by five springs' Find i ".+6f' *. \" I "!L1 X" ;;;; ; ;'il';'";;;; ;*.iia :cquiriirium coonsurarion fl_*aeg, -r^/ i4}u- @u " of thesystembecomesunstablellki'B\p1'1lt<rf| \1,-. Solution Inthedisplacedpositiono€thesystemGig'Ea':lb)'anglesa '" f,I 1 {y \i* r and { are related by oW, (.) -r ,#' 6 A ,, zl'i -$=a|tb 8E=Dsind:csing- :+ uetg2ld.ordertcrmsip+ug *"'*' &3 Lquals {t + 0 * C) - r = O*The telative rotation betwecn euas of.the-tprsioaal ' -.' I : i : s"H: potlntial energy I/ of this conseratirc slqtem,,accurate.upto 2nd order terms. is given by v(0) = ikLoz +2[]e:(asi!0)1+ ].rs(a + d)2 + *br|" + P[c cc0 + (0 + c) cosdJ : + Pfacxx? +(6+dcos(crela) = tr*re' +-ezicsino)2 + |r"1e +allblz +^if7"otu12 . = i[*, + [.(r + o/o)' + e$2 fi1e2 + &242 sin2 d * P[a cos I + (6 + 4 "*Gflf)J -'' - v,(01=[Br+ts(l +albl"+k*2lb?\rt-rtzc2sin2d-P[csill0+(0+c)(c/o)sin(adlbrl v"(0)=[&r+ts(t *alb)z+k*2lbzl*tkza2cs20-'Pfoccxl+(0+c)(c/0)2cos(c0/!)l;.'":-t*:-i:... -: =. 19_6 .. ,l -t ..I ,tl' 'ing*ition'f,=o' For the eguilibrium !o, ,I.,*'..r..,:: ,:.- ,"toii foi* *"(, 0 The equilibrium configuration = v'(0);0 (o'k) ald (, '1 t,r : \ ,: pla*(0+cXglq2l...,,. * alb)z. * *oo21b2l *2kza2 -_ O is stable rf y"(0) > 0' i-e'' t ti ( (r, if l: J J P<tte,+&:(1+,,lbl''v*4o",1021+2k2a2lll4+(6+4(a/D)']'(1) load Po: the minimum vatue of p at which the equilibrium is cated the critical positiou becomes unstabre eq(u:+=[{&r+!s(l+o|b\z+kqaz|b2l+2k2a2|lfo+(b+cXc/6)2J.(2) If I(a = 0, = lli;;.ff; I &1 u,:,',J.-T:::::Jiff-:r'#"T-ril1t';t;l'ilT:"'* orthe reg is stabre ir " =:' '= :L-,,,"";'i*l;,1;;;A);;'d;;;;";i*ongu'"iioo I ,/i11 -- t, ry Zkla i'e'' if k > mgo/Z' &2 = 0, te = i: &'- ' ns < }k, rnsisting of 6 li(s equrlru'ruru t::"f":l:l::::"i.L1':;ler f,ne equilib.rium !-rnd the mas\ Exarnple Exarrrple 4.10 Find ""-'r.'"i"it ity. Each smaller links has masl (Fig'E4'10) and discuss-thethe lengthof free ot""Ja*ther All links are uniform and th<: lit;;;;;4m' bigger each rn, and Siil:;:Tl ,ri'iltlit*ition of the springs I;1 d of trre svsrem, the renscrr *, thisconserl,ati,esystemwiththedaturnforgravitationalpot.errtialenergy,atoisI (1toe 8e - ccc0)2t -2m1g(acos0) + Y(d) = 2[]81(2ocos 0 -2l.ceLs]? ]tu 0) - v@ 2ar19(3ocos 0l =;11;;';' - r'"f = {a| =2(4t1 * *2}sio" 0 =O + (t). Thus the positions @ at 'o' Sf J + coso[4(2m1 sino=0 (ii)of equitibr-ium are at J J J : * m2 (2) t']"'("*o -cc0e)l a- 4tn)ga- 2({&l + &r}a2(cos 9 - cos (a 06}| +*:)trGos0-cosoo]l:O [4(2m1 *rr.z*4to\sa-'(n: : V, JJ J d) V'{0} --2sin0(4&r *}2)ai(cosO sinf,[4(2m5l-ta?+ 4ta]so-2(4k1i f, J - 2',.9(0'541e -^2m9(3'5ocsg):2(anr)gto*" ,r, * mz * 4m\sacso - *:: - 1(2*''+4(2mr t mi* 4rn)gasino cos09) * m2si(aacoso - ' 1' d = 0' 2' 0 = a' J JJ J is '\r :'# ? ; [#"::i:;x::r;x;l;I,'iln a;*;1',:!:.'; J (4) aud 2rnr 3-0,='at= cos-rfi2*r*mz*{rn}s/(at1+t2)o+9os0ol l. 1t cosOo) < 0 4(2llnt* mz*am)g - 2(4&r + [z)a(l ,-.., ,, * m: i-4rn)9 / (aEi +i&)c*cc : 0o < 1 (5) positions' these equilibrium '' consider the stability of cosg3 if only exists position this o"ios"o(1}*" condude that since 1;W" ' "oa poqition' v,,(0) = 4(2,o1* mz 1- 0 _ e = o l"'to unstable equilibrium ao th."'ir'(;i position' oito, e, posiiion "t a- if equitibrium o = 0 is a stabte equilibrium ,-o ;;;"(o) n& 3d ' oI<0' 0=risan b. if equilibdum posiriou & $od "r"r, *r*)r;:';d;;;itJr; iL "-;;i o: 2. 0 - *: v"(r)=:[n(2!ry11*'+t#fid-);;;'(; equilibdum Position' 3. 0 =0zz ' using eqs(3) and @- qs1l* v,,{0sl=44rr+tztrfin1 * (4)' t^\nnn-t(4t.*&r art+(z*r +mz*4m)s o-2l4kt*kt)c2(ce t 03-cos gol} '' = 2t(4el+&2fsin'es> 0 and heoee 0'= 0s b a statle equilibriur-n Pqition' lol :.\{ , \ I - a L-. 9^1 <. ladqlYy:ttre'-zgrro force memb!* in the trusses showa in ./ Fig.Bl.ll. (c) 2 L J tt B 6 I !- A L Solution (.) By successively drawiqg the FBD of joints B, 7, 5,- 16, 12, 14 are zero force members. {b) BysuccessivelydrawingtheFBDofjoints K, D, E, J, F we conclude that members 3, B,C,G,F,D,E,F, I,if,.C,we.concludethatmembers 3, 5, 10,8,7, 13, 12, 15, 27,26 are zero force rnembers. (.) By successively drawing the EBD of joints B , C, E, G, H, I , s,e conctude that rnembers 14, I [, ?, lg, L 18,3 are zero ficrce members. : p,lsample 4-12 Find the forces in aII members of the bridge truss shown in Pig.tsl-CIa. ? R3 \-\-- E EA <-vL / ', 8c .4-- ---4 ,L. t-=Lrz 0 9 lr, :{K' F,3 Q: ':.7ffiil- Solrrtiorn The FBD of the whole truss is shown in Fig.&t.l2b- Tlre 3 reactions from the support deibr:miaed from the 3 equilibrium equations. for the -r :+ !1 o.l (o) 'z I50 l{s, = l6rtr -4(20) - S(50) - 12(80) = 0, l?r=90kN, 8z=60kN. Es=0 *hJ t..r"" under coplanar F" = J?s = 0. Fc = Rrt are loads: Ez -20-50-80 =0 The forces in the membexi are determined by coasidering the equilibrium equations ofjoinG in lhesequence A,B,C,D,E,F,b,G so that at eveqr stageonly two unknown rhembfr forces have tobe determined. cc0= 4l@'+42)tl2:0.8, o-24?s 0. F- = -Fr- Frcos€ --0 :+ Fr = ISS kN, F1= -\10 kN., Joiot B F,= Fz-fr=0. fc:-80-&-0 + Fa--F2--l20kN, Fs=-80kN-," foint C lf; = F1.cos0 - JIs cc 0 I fu Ft = Fs* .Fr siu d * fs sin 0 - fe sin c = 0 "*o<,, :+ 0.8Fi.+0.9?01.F5 t!9, 0.6Fi - 0.9?01& - -f0 =+ fs = 25 kN, f5 = 103-l kN = Joint D lI' = (&-&)*u,i = 0, Fr = Fz*(Fe *fa)sinc = 0 =+ f6=F6=103.1 kN, fe = -50 kN Joint.E: F, = -so F7-(& +rs)sina= 0, f, = Fr Fro * (F5- Fe)coed =Q I &=-25 kN; F'ls=-80 kN =+ Joiat F: f'=F.o-frs=0, 4:-2A-Frr =0 + Frs=-80kN, Frr=-20kN Joint O: & = &s -l Frs * Ftzc<x,l = 0, =) Frz = 100 kN, Fy : Rz- Fl2sind 0, = Joiat r{: : coso- 41,'2+42lrt2=0.g701 + 'sin0=Q.g, sinc= & = Rr- F1siu0 = : I a j I :: I ).-rr tcB 1 I It -'i-.,a= ;; r"i;l :t' :"';)' I 1: '-:"'- .' '' I i Joint G : F*= f'scosc*(6t- F$cos010 '+ ' 0'01?3 = 0 2ud check eq!"t]:" J I : arithmetic rather if rve had used exac[ The 2nd and 3rd check equations wootd have been exactly satisfied .ftaificant dElts. Tlrcre r*'ill always be 3 equilibrium equalions *'hich tlran roundi[g-off the numbers,to 4 since the forces in these members ate pro*ide a check on the solution. Members 1,5,6,i, 12 are in tension the forces in these members are negative' posiiive- Members 2,3,4,7,g,t0, f f,'fa are in compression since 15'of tLe K-truss 2 4.13 Find-the forces in rrembers 5 and 9 of truss 1 and in members 12, 14, Exaraple o\i ^g*ffiffii'Eff; ,Tffi, ," "' ,,,t1ur?ffi#,'*"i ,o lo 30 /--K--J|A}\*.r*btt 3s*X9. ;i ffi; ="", *:7(!\;l'7(_ -r"j 1 + A/i\l \l/:\ o--'-=7,\i*.i /l\:[';fr I.j*o,', -Bi-( -nl - iJ YZo..P ..S'R *ffiiii uKffi hffi :;Wr', ^:ffi" *i *, ;ffi.'SKSFYy^,'r -:, i. t;l;f" %-,i?,+"" q -#\ t"--t -{f- 'i rs i"';i''= :ifi,f1TTf^ -r rlnk- ! "t]t"-rtj['l. J - $5 $:. equations Ttre reactions are computed firsr from the equilibrium l'[s:12R.r -20 x4- t0x 10+10x3-30 x 3=0 =+ + + F,=R3t30-10=0 \ts rd; (FS-E{-r3c): of the truss =20 kN' .R3=-20 kN, lrcs| =2l(22 11z1r|z: OS944' 8F5 -20'x 8-30 x3*Ez x {= t I i , I t kN i I oblained py'findi$6"the''point of Thus member 5 is';having a'tensile:,{orceio126,25'kN. ,f5 can also be intersettion O of A;,Fe *a'*i.,g Ma =O: 8Fs- 160-90+10 x4=0':+ Fi=26'25 kN It[s = i I Rz= 10 kN' sin? =0-'4472' F, = R1 + Rz -20 - 10 =0 the is better to take appropriate sections thtoughSince fotces in a ferv membes are to be determined, it the part of the equilibrium equaiioqs of tie truss cutting Lhe.rnernbers of intercst- 'f5 is determined from Lruss to the left of section PP (Fig-Bl'13d): :Fo x'1+30x'1*8r x4-'R: x4:0 + 4f'6+30-20x4- f0x4-0 :+ f6-22-5 f! - f<cc0*Re*Fo*30-0'=+ 0-8944& -2A+22'5+30=O * It--36'34tli'' :+ f5:26'?5tN F, = Flsin0 * Rz-20+Fs=0 :+ -36'34 x0'4472+10-20r'Fs=0 iI\ I &1 A{p \ t I I 0 + I 4 members' Eence we have to firstfld A section ihrough member 9. sucL as sections QQ and,Rn', cuts and using Mc = 0' or find fto by the force in one of these members- \ff,e can hnd Fz by taking section SS s'e can find r'g by considedng the taking a section ?ll" and using M6: :0- Once Fz or Fto is determined, Consider the equilibrium o[ part of equilibrium of joirrt D, rvhere now ooly two unknown forces are acting' the point of intersection ou the right of section Uf (Fig-81.13e). lVe fiud F6 by tlking ntomeut about t t truss C of the unknorvn forces Fu,Frz: = R1x 4* 10 x 3 - 10 x 2+ Fro(cos0 x 3 *sind x 2) = 0 20x4+30-20+&o(0-894{x3+0'4472x2)=0 + F16=-25'16kN + Considet equilibriurn of joint D (Fig-81.13f): Fe = 25'16 x 0h472--ll'25 kN Fy =-Frosrod - f.s = 0 + Ms : : : .: l ro3 -l L ]J, L L L L IJ t' L 1: IJ 2. * For the l-.. l_- IJ IJ L z.di1s --''413 e.{ (rfg-Bn'rs8) + sia, = 0-8. cosd = 0.6.'A"".tioo'ke 'Tlre The -- --'*--f --not.conce*tent' a're -^1 i'ti'i;Jiand anv three orthem . - ro 22, oo 23. o1 bei.s L-i-d amarrrant coocutrent 14, r',z2.z3 nith members r8, ;H"[};";ffi ]'[ ";; ;;"*; ffi'"i'""II"tlitffiil:H*-#:;il . ? ,t .-..-^ .^ -iohi of af section wtion XX XX MB, = 0 of parr of the- truss to risht ;ii.*ffi;;;il; "oo".io. ":$"?riff t': _ f11=40kN-' Me"-18Fu*7)x 18-15x 12-15x24-15x36=0 + forces find forces in inner membqrs' we first fiod This is'the procedute for outer members of K-truss. To of the joini wh"t"-.':T meet- A section Y\' in adjacent outer members and thea consider the equilibrium 0 at G' C'onsider equilibrium equation M4:' = cuts members 8, 12, 16, 1? of which 12. 16, l? are coucurrent yyle:g-a('rg0): of part of the truss to the right of section i, - 18f's*20 x 18- !M< oonsider the equilibriup of TJ L 1-- *-i;;i;, 15 joint g1E:3-€h.r3[): fitFrz'= 37.5 x 0-6 - 5 - 1?'5 klt 4.14 Draw the axial firrce,llril,' B-M' diagrarns:::* o:::t-* = F6- Fr+ -.Frssin0 = 0 + F,=&zt.Prscos0*5=0 =+ F, g:6arnple F6-l0kN + xL2-L5x24=0 "+frt.." If-rat:-'' t!rJ L t-ffi IJ L L IJ tok.U-rSs +lL-J rok*l^+lm _> R,,, 10-40-0-8Frs =0 * -31"5rkN 'r, in Fig'81-1aa- NB., r-bl -'6[-.. f'1 r (cl .,, l-fl-\-,.-4 i<-------. x j., =r,, ,--6;i:, I i-- 1;^, .6-1 -,'-.-.1o ..,., . ----....,--J €=_ -.-...=:-:-u_.a-_,_.__;'', Uott' sides- the FBD of the supiorted-oi Solution The reactions are detetmined firsi "fi.e the bEam is :';', beam is shown ia Fig.Bl-t4b- The equations of equilibrium'yigld L L I II ; -i t{s-108s-rt0=(10x4)x4+2Ox8=0 :+ ]?s=4kN' Fr=R1+R3+20-(10x4)=0:+ {r:16kN' F,=Rz*15=0 + ft2--lSkN, r' For the first 3 ranges we draw FBD We derive expressions of JV,S,i/ ficr five differcnt ranges of trvo ranges (Fig.B[.lab) of the part to tbe teft of the sectiori and for the last to the right of the sectioir- we dtaw FBD of the part i O<z<1: .IV=-ff =-'Rz:15kN, S=-Fi - -Rz=-16.kN' l{=-Me:Etz-162' M(0+) = 0, a-('-l = 16 kN'm:+ S:-Fi --Rz=-[6kN' IV=-ff =-Ri:15kNi L<r12: -..- ilo 1 l -r- -.t.t f ) (,, A. tt 'l I 2<r <6: + -J -J =i .-J r^l kN-m' t+1 =56kN'm' M(2-): ?2 74 =.-lta-gE1:f 40= fPc+{0' kN:' -i" :: Irr= -F1..--Rz:l.5kN!,, . S=:S=-B:+t0(z-J)=tot':36 l ,S(2+) = -16 tlr' 5(6-) = 24 kN I i a\ i Mhasertremalvaluewhereschangessigninthissegmentfors=toz-36-0'atz=3'6m40'- $(z -2t212' M - -M,+= Rt:.+ 40 - 10(c - 2\2]/2- 16c+ N=Fi=15kN' M = Mt - E3(r0 - --J + 20(8 - r) = 200 -242'I) M(6+) = 56 kN-m, I4(S-) = 8 kN'm' ?.J M(8+) = 8 kN.m, Ir'l(10-) = --J N=fI =0, 8<e(10: -l M(6-):56 kN'rn' M(Xf| = 84'8 kN'm' S=^s=fts*20=24kN' M(2+) = 72kN:m. * 6<z<8: -J S=Fi =E3=4kN, y' i{'1=E3(10-:r)=40-42' 0' of tY' S' if for diFetent i B'Ir{' D rr diagrams rr-zr'nlc are ate drarvtt d in Fig'Bt'14c otttrg the expressions and S.F' force, a-tial The -^.,^- - rierrar.c force in u-direct-ron, or l J :h:]:Hll;l;ill?;T;,;'i:,;;J;;*"::=:::::t:'",.:'",:,":r:H,'[i::ii"l#FI];ll 'i:r:::,::;i::"1;T:il#i*i#H;;il""F;:.-=iyr::.::':'t:*T,"ffi rarving :l*.T; varr;s.euadrarilart*:with , rtrte( diagranr the segment tfitl't g =cons[atl[' ) varrcs l#'":'ffa::Tf,:;'=x#:fl:::*;;;;;;;","*;: ..o,nu -r,e rrf undegi:S-E. of the area undes'i:S'E' and A'l/ ctluals -te -ve of the load on ihat portion acts' ;;;;t coupre couple act's' dis.ere discrete a cidrr 6r -S- As o.vhere s "r,",,so'=isn, r f;?t;i;i:#"":::J::ff";;;;.;;";" iFia F/ tS; Cpse' -- -r-^..,* urhere .4-h -+1 ;- Ji-i. _-J '.,o1< equ;als \ r*-)' '1" ..J l l< z -'4l'.5 4 t.4 If {okx , lot<* -d-51<- -.1 -- --*;>*."*1.},r - i) i:- 4+- 6<==..^.=b=56{ N l" z'--->1-t'5 AL tt. {<-- z,1 - rr N ,-_ ai1 (c') . i_ 'ii I . 1_I +to L+l -i90 r?.s k-z__L'_ q - ro-G7 ., l,i i <{l A 4jL |.-q , .-/tit ,E ec ==$fl=[ztx4) | f,*r,*J.3,9ffi+ ,,<-x-c':rf . .-___-__F1 bv its-equivalent at point -\ i!< --, ii'!r << BcDisreplaced (Fis'pt'lsb) ror""lvst*..ri's;;o;; !ft.- -40 kN-m' t* " *o"'"nt f,' = 10x0'5-90x0'5 expressious of D: forces 90 kN and 10 kN in negative " ""d5;"it"o lone side' We derive beam is supported on determined .rn" The reactions need not be N,S,M for five different ranges of r " "irr""-tt trsing FBD's in Fig'84'15c' lv! :-Mt=-42-2'212' S=-fi =4*2t' /f=-Fi=0, 0' M(2-\ = -12 kN'm' 5(0+) = 4 kN. S(2-) = 8 kN M(0+) = :+ M = -*[t= -4x-2*/2+10(z-2) N =-F"o - -?.5 kN; s = -Fi = 4*2c-lo' 2<z <6: ' S(2+) = -2 kN, 5(6-) - 6 kN + 10 = 0 at r = 3 rn- ' 4*2cfor S segment [his = in sign changes e-ttremal value where S O< z 12: ! I 1i<< }<< .-..< -r'- l ilr M has ilI -< .-j .u/-di: li ii ./ .l r{ ';,'< il i: - - '-2(!kN'm, tt(l)= -11 kNrn' 4+2x6-10=6kN' 6<s<?: N ---F|=-zJix, -'s=-ry= io1'- 2) = -6r * 16 (c ird+ i--] t-.\ [_, *" 8': <:c < LJ I L- ir,-.-+:{(6-) -i^ : -42- (2 x 6) x -3) + kN'm ,1/(6+) - -z}kNrn, M(7-\: -26 10- 10: -4 kN' lf = *4l = -?-5*90 = 825 kN' S= -Fi =4*2x62) + 10 x (' - ?) + 40 = 4z - L4 M'= -Mtr = -42- (2 * 6) x (z - 3) + 10(z - 7a ? -. = 1(z-8)= -Fi - 4+2x 6- 10- 10+ 10s(8+) = 6 kN' s(11-): -6 kN S LJ LJ LJ LJ L_- L-. t_ t' L tu U 1. *. .&f has extremal value rvhere .t"a = 5 m, 10 i v - z (' - 8)'' 14-s- (r-8)2 = 0 =: 10 rn is admissible' r^ - f--?I+40-l0f: +(z-s)212 M - -t'{a: 4t-(2x's.l-(",1}]10(:.-?f10x(r-?)+40-r0(z-8} s.ii{s -:l'l::5 8)/31 = -62 * uu ,: +*8(z 1l: -slJ(z - 8)t(' M(10) 1057 kN-m"'''' rn ofwhich only ilffiH" r-3'5 kNlm &'ig-Et-16a. ;' 'z z^-i*,^4,%€ '-x k}{'m' = s'.,lf for difierent solution We consider the equilibrium of Me=q? + (2i- 2i- 3 k) - (2i en = 4!* r2j + 12k kN'm ?lhe normal to the section - thJeam e + 4$ + (2i- &=4i-si+8LkN 2i- B x 6j * (2! - i) x (-+ll+'(!) in -i negat'ive face)' Hencc x (-4!) = Q the comPonentQ------t i coorpinate directions are positive: dit' = 8 kN, &=4i*8t'kN' =? S.I1=rest'of thgcomP- of =+ Mr= ComP. of eR io -i = -12 af B.M.:rest of the comp- en= 4i+ 12L kN'm' M, = -4 kN.m, M, = -12 kN, M = (42 +l22ltl2 = s, = jkN, *tli'+)l tbr (Pig.Bi-16b): to the right of the section ai E (a at E io Fig.Bt.l6b is in -ve 3r-direction poiniing in lhe ;rrc ComP. of F,e B 4-X}ryt fW "lV Ytrf1,*ty ^fi ^l*o* :t - F=&*(2i-4\)+61-4L-'4i-0 IV - -.-rt, -6*rz1\ e1ft*/m zk|\ (c) ;tC":e "- 14 of N' l.-.- ' t: Li l. tj L s= - the oipressions are drarvn in Fig.Bt-r4d using Tbe a:rial force, s.F. and B-M. diagrams rvith varying slope of'-g and hnearly, s *"lo q,aJraticatty segments- For the segment in rvhich c :ari; E 26 kN'm altd occurs rvhere The ma:simum magniLude of M &f rraries cubically with varying slope of -5' a d.iscretc couPle acts' shown in 'Hod and B'[{' zt-E for the bent beam the axial force' s'F'' trvisting momeot . * t_:, changessign in thissegment for lf1a*1=lt nnot*' M(rl-'1= 3 L L,-, tJ tj s Llz(' -sI(" -8) & = -8 kN, 's = (4'+82)r/2 = 8'944 kN kN'rn' dit' 12'65 kN'm ' , : .' rla t I =:Po* *,u, -- t,^=r,Yffi!,f; 4. tu=0-1;Ir=0-53m' tJ ih-op *^\ 2.17t**o ]-. u Y? ^\.S(r-r *:v fcv-43,"#q1,H', "}(.S"tLr +"/ (o-) u N l$Nczu* 6x 1$)--,.\{6nlx J}\<,** tr eAS(:: rkE*-'#K';'*M-s, '6+{; +o, H ,",3i?-\:;,".::r-jr"* * "*, ;:';;:T."iH;;:;;.;;;"or,,,i.i..,t ];1 f : orequiribrium .,r.*.iro. are srrown "q,"rio.," ",o.r,lJi".to derermine the I unknorvns: 'rv'1.F1,N2'Fz'ivs'Fs'e'P' in seaerar' i'l. ifi;:::"J::t,1il:::f,:*t;*Xil';f:"H,:l'lJ::"J:"fr:::.'.:::' '" ;i.*'xi-#**xl:tg:;x:"':tr:':i::ir';:l;I:"ff"loiorounacr f --l I ira Ms=A2N7x0'4-F3x0-4:0 :+ It- N2-240-Fs-0 Fv:r\rs-320- t 02N2= we check rchether the assumptio-n E [ 1f F3:O'2N2 :+ N2-240-0'24r2=0 + i;'rX:=lil;l];ff. no sliq at B is corlect' + N2=300N' F3:60N' A'3=360 N' .mption is correct. The ;;;#i4.,:tl'*;i*i*;:i;;'"]:r*I,T:Ii1fi,;.:TL;11i.iffii,';; lJ **Xi:'ir:,Xh:*:l*.**;:?:,'T:tilY,i#'::;"[ff':I'i'lapeadiit N' Ar = - t6o*Go = o * F,=P-02Nr-120-300=0 MB=L60x02*120x0'6*3tl0x0'4-Px[-Nre=0 + Fy E I . I =Nr 1oo P-4{0N' s-2'24-4'4h tr fff;j.":1,?$*;n*jj;,X;::nxHi;;,:::Ti#i:;;::": 1-I and tr ] H,i L, l-. r "** (21 m''fi'n r*:i**n*rii-*;:**,m;::n:Hr"::l:::::Tril;:';ffi5T:t tr j 1-= (i) & is arbitrary- Equatioas of equilibrium orrrnvnnJ-annY u;==;:l::::ljJ'*300x04-Px053=' F=P-F'-120-300=0 u3 = P=422'6N' I *],oo*, =+ F1=2'6'N- I ... ' 1 02' 1lhus motion impentls for P The aisumption of rio's{ip- is o+. siuce l&l/X1 = 0'026p1 slie ai 8,_ *itr, tippiru of the blcck abour E and rolling of the cylindet wi!hou_! : {if2'6 N Cascsgand{llFsllx:=0.15?9}t,s=0.landhencethqTul?'igoofnoslipatEiswrong.Eencethe tbc eylinder Kinema[ics implics 94 = 9.D = g,t and hence cylinder has impendingllp "t B and no slip at .4. dorawards is shorvn in Fig'Bt'l?f rvith Fe = lrIVs = 0'1/vs has impending translation: upwar&. The FBD and Fr arbitrarY' .TJf :lii .,I'J }J iJ - 0-1AIs x 0'4 = 0 + Fr - O.lNs * r\! = 355.6 N, Fz = 3556 N, :+ .lt/3 - 320 - 0.1Ns = 0 &=Na-320- F2=0 * 5, - 275.6 N' F'=Nz -240-0.1'rY.=0 =+ N2-240 - 0.1 x 355'6 = 0 i'e'' the assumption of no slip at r4 is \!e checL that l'F2l/N2 = 35'56/275'6 = 0'129 I pz = 0'2 is o'k'' will have 3 equilibrium equations imply that it Mo = Fz x 0.4 ! i rI t I I and only correcL. The 4 unknorvns, M1 ,Fr,r,P for the btock E, rvith the 4t'h equation provided by this coodition' point either impending slip or impending tipping about rvith Fr = lrJVr - 0'2Nl wc assttmc impenditrg slip uithout tipping. The FBD 15 5|16\\'n in Fig'Rt'l7g t) dow'orvard and 160 -J J I 'l J I fv = Nr - 1.90 + 35.56'= 0 2?5'6'='0 , F, = P= 0.?/Yl - 120 ir's = J x G2* 120 x 0-6+2?5'6'x'0''4'- p:'v l1:- lllr =0 * P i 420:5:I*li + z, = L.722 -J i,$360h J tl (3) -J i.e', $'hether \ve check rvhether the assunrption of no tipping is correct, motion the assumptiou of no tipping is correct' Thus casc 3: lr = 0.5, eq(3) + r = 0-032 at slip without '{' and traoslation of the cylinder impends for P = 420.5 N with sliding of the block > 0. H"n." &=Nr-160+as.5o=0 ., F,=P-fi-120-2?5-6=0 '1 = 0-069 < i, 0 + * P=404-23N' JV1 = 124-{ N' + f1=8'03 N' for P =0-2. Thus motion impends : W,I I i _-l -< -- 4M'?3 N -< rvithout slip at '4' rvith tipping of the block about E and translation of the cylinder roiatlcn irrespective of the va]ue of torque Example 4.18 A disc 1, mounted on a shaf[, is prevented from iu contact rvith it' Firid the condition M applied in the direction shorrn in Fig.E4.18a by placing'a disc 2 Neslect weight-of disc 2' so that disc 2 acts * u rri"tion lock for disc I' ;";:;,;;;r;& ,-*-)N--,1 W\ffi": I L Cascl:}=0'53,eq(3)=+r=_0.0694(0.Hencetheassumptiooofnotippingiswrong.Tbustlre with Nr' ]?t stipping' The FBD is sho$'n in Fig'E4'l7h blocL has impending tipping about point E rvithout at .8 and F1 is arbitrary. Equations of equilibrium Ms = 160 x 0.2*120 x 0'6+275'6 x 0'4- P x 0'53 = I -< < .ac=.1R -< to only trvo reartions at A and B' Eeoce the Disc 2 is a 2-force member since it is subjlcied tt. roration of disc 1 is prev-eutcd if disc 2 two forces must ac! along the line ,4B as showu in ris.E4.rai, e(tan-rtsr andaitt"-tp2' FronrFig'E4'18c as20=(d-r)/('R+r)' .{ Eence the reguired condition is --< .< solution ><i-a doesnotslipate"na61]i"u.,if 0 = !co6-r(d-r)/(E+r)J ( min(tan-r !'t,tan-t ltz) :+ (d-fi1(R+r) > 2cos[min(tan-r p1'taa-lp2)J I l :- fitr ;i - a :r{ ! Lj :. LJ LJ i LJ LJ ! ! : t- L. Lr, L IJ iN N/(b) (t') I -t- r =' '\t't' SolutionThenormalreaction.lsw|LN/m.Assumeihattheimpend"ingsli':li:1?::.1'=::"'::: ihis assumlfioo ls not i"-; *'"uD is as shown in Fis,E4.leb- Eowever, ;::X;";;;.*;; bar ls rn rne s'u,c urtw n. - F, r^L Tbe -rB-- fi<fion f-:+:^f^..force ls not satisfied by the forces in Fig.Bl-l9b0g;not Ma equation the equilibrium uus sYu.^urru*' = since slrlce correct correc. :atrcr,Ieo :I ""t 1L --- :- impending :-^--r;-rarrrinn totation on- Hence there is partly ' - F:-D, ro-.-;1L has to be distributed partly iu idirection aad i"- idk*lt .ninorvn. - an =-,,-i-nn.n io Fig.Rl.lec wirh c shoron * rso; :r-;: |j.T:T ;ffi;J;*,il';";;";ih. -Eguations of equilibrium: (u +P=(L-2alpwlL s=P-(pwlL)(L-o)+(vrflLlo=o 2a2 -4La+ L? =0 a: (1- Lld,'L "+ p =-1'fZ root less than -' is listed- Eencq eq(U * .'l.o*.;,, with a i'ertical st'ep' is 9?r, with its front.wheels in contact weight of trolley wheel A four 4-20 Exampte P rgquired of tle trotler' Find tbe:i:rirnr 'iu -fioice pulred by a force p (Fig-Bt-20a)- .C is tl" .",it " or mass all surfaces t"t:T,::it-h the 8!9'1rn4 do rnount the step if tlre coefficiert of static frictionlat "f "5 ;;r. t , *" ,**tble lj u L y#'# ;TY#+, &r x% c IJ IJ L lj IJ L L IJ r-. L. L. -,, -,, A-- fU. k-a* -4-r:r,o -*4 F?i.r ''- -}'t<-15rc---+{ 1_ t, L ,i^=1u*1L\(iL-o)'12- (pwlL)o(L-a+o/2\=0':* IJ J t4 to) u L I =-,,*.r: o thibrizontal valle of ,,t9a). Find the value ^.e tH;lar'r"stn,9,1a holzgntal,table (Fig-Bt'l9a)' 4-19 Exanple - be uiformlv Assu,,," the *ormal t"-T::"" from'the force p for which,o,* * t"-i*;".ion. T: (cl is p' i*i.ti""U. The weight of the bat is W. ail' the coefficient of static frictionl<-o-ri/t- z 1 0V;/ *lI,* --- L-.* (b) i q%.= lr:,,., atEat E since tt trJn.yi-t3t*-"tt' tft" luip by losing contact Solution along " -TT:: it is a 2-force member with the tbrces actiag is sub-iected to only two forces through ,4.and D. Eence fu: "l ? *u @ acting along AD asshorsn in Fig.E4-20b- Similatly, \rheel 2 is a2-force member rvith the l{ence the FBD of the trolley is a-s showl BG u:shosrn in Fig-Bl-20c rvith sin0 :32A.,40O: 0.8,ccsB': 0'6: There is no reactioo in Fig-pt.2m. Notice the in[eresLiug fact that the.friction forces this coplanar system Yield I?1, .R2' P: Fr=P-Ezcos0=0 & = & +.&zsin0 -W =O + :+ + are zero--The'3 equations of equilibrium of R2=$PlT R1'=W -4P13 a 1500rv-3000(14/ -4Pl3l-400P=0 Me = 1500W - 3000Er - 400P :0 kN is hinged Exaurple 4.21 A uniform rectangular'plare ABCE of weight 6 The abour its ed6e AB and supporred by a light flexible cable cD.(Fig.E4-21a). ai hinge the but 'A axis hinge at I can only support a force normal to.the hinge the and B and can atso support axial load- Find the reactions at the hinges / ' * P:5W/t2 tension in the cable. of The FBD of the plate' is shown in Fi5'81'21b' The coordinates The (-l'2'5)' the midpoint G of the diagonal AC ate (0,0,6) + (-2,4,4)ll2 = reactions and the tension are given by Solution t9 t_ B, 1i-o where t = TI(CO) arid r?o ' AB -- 0 since &B L AB' The 7 unknorvns Position vecr?r, -. - , R5,t are obtbined from- Ma = g, {. = g.- &s ' AB = 0' & = Rr i+ Rz j_+fis ki 4, = I?a !+ fts!+ fio k, T=rwJ tors are obtaiaed by subtracting the coordinates "i icml .i - i' l,t5. u- iJz-l G-O)]= lrom those of its tip, s; r-F tt 5tI Lb) lI{ t !o 4 .1 -'. "- .t' AG*-:-i+Zi- E: I &,q,+ Es.+ L{a- ?+6k-q 6k+lC x T.+ SB x L2-24t*4&+2.R5:e, A.G ,- krn,'e:-2i+di-ZE-,,' &" - AB = (Eni+ /?"i+R.t) -(4i- ..is .< x 2L) ra:nj-rt* : =a&._ 2Ra = a. li i a"=l_l i l0 0 (i) :. (2) EI .,1* 3,_il-li 4i -21:O 85 l?l k _r 6 i: j: 5-L2t-2&=0, -':----l k: -4&=0, '=+ &-0, l=0-5, .' 2R6 -F .l?5 -.0 ' Eq"(t)and(3) :+ .Es-.116-0 :+ &a=0 and ?: t(2!- 4i:48: i-2j- Z1 :+ Eq(z) :+ Ee : -Ee --I- 6li = *i + 2j - ,{t kN- :. :. (3) T= (L2 *22 12zyr12 : - 3 kN .Er! :7-_ .f R5 :-. -r..'E Ma-(fi2)q-p.&=0 The equitibrium eguaiion Me . a: '. tte ?-* =0 for the whole rig yields 1g1: - ftr}- wa- pR=o :+ The equillbrium equation M, r ! 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