hwsol8 - Homework 8 Constrained Rotation and Translation...

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Unformatted text preview: Homework 8 Constrained Rotation and Translation Part A: Assume that the functwn . x (t) r~ represents the length of x tape that has unwound as afunction of time. Find (J (t), the angle through which the drum will have rotated, as afunction of time. +A e... Circre O<i~ ro..n c..e 0 n tfJ., L X(I) .from e(t"::;.o) -/-0 (-t) e (t-).:::. pc{ot) ?<It)::. r' e /(If) W(... ? . -r her e.k, re- ~ f):=: (~{~ e r j Part B: The tape is now wound back into the drum at angular rate w t With what speed will the end of the tape move? Hno w it[ ;>( (n = r e (t)] =7 V(+) .=. rw (-t.J 1Aere{Or~e I --------, r W vCt)=- L-t"j Part C: Since r is a positive quantity, the answer you just obtained implies that v (t)will always have the same sign as w (t). If the tape is unwinding, both quantities' will be positive. If the tape is being wound back up, both quantities will be negative. Now find a (t), the linear acceleration of the end of the tape. Li ~ Q., +A ~ pre.. ~(tJ:: \I i 0 ()S cSfe p r ex. ct) Part D: Perhaps the trickiest aspect of working with constraint equations for rotational motion is determining the correct sign for the kinematic quantities. Consider a tire of radius T'rolling to the right, without slipping, with constant x velocity va:. Find w, the (constant) angular velocity of the tire. Be careful of the signs in your answer; recall that positive angular velocity corresponds to rotation in the counterclockwise direction. f;') . ./ Lx olire.e.+-/oVlJ o..l"1cA I'S OJ . ,~ j h The. clocK wjs~ -J he."" e. .po r e... n e,?> o..ri ve.. Vx IS j 1\ + h e.. fa 5~ rl \Ie.. x """ oJ.. ; r e.. G'h'o n . fhus w - v (-l:') r - Part E: Assume now that the angular velocity of the tire, which continues to roll without slipping, is not constant, but rather that the tire accelerates with constant angular acceleration a. Find (,I"" the linear acceleration of the tire. j J () c...e.w().s w WO-S n e.3 CA..:I-j \I e .) So , S 0 IS ~. V.x POSi+,ve,j D.-~ ;.5 posifl'lie. o../,sQ: a..~.:=. - 0< r y Introduction to Moments of Inertia r .J Part A: On which of the following does the moment of inertia of an object depend? A. linear speed B. linear acceleration C. angular speed D. angular acceleration total mass F. shape and density of the object G. location of the axis of rotation a x o r '!or W -rA E: f=: . -4r~b e hi om e..nt' tJf lnterf'o- is vnsl'fl.c;l--iecAj WI-rIt.OUt- ~ (\ o..)(,"..s. Part B: What is the moment of inertia I of particle a? m o.,s..s js on e. veLf' ia,b Ie for r-h e, m" fYI e t1 +- of +-1 ll1e. t" h\~. rA~ s ho.,p ol-e+er 1'11;1\e... e.. ().,nrA oLe-ns" ry f.o.r a,("e. ,. yYIpCJr to", a.....s -fA e y fro 1"'\ how ~ mo...ss e.,eme"'+ cA.M /J ("'0 0..)(.; fJre s. G: The. 10 co.+-i 011 of +'he. ovci.s of roofiolt de,l-er~ i", es ((~'l Part C: Find the moment of inertia I:Zofparticle a with respect to the x axis (that is, if the x axis is the axis of rotation), the moment of inertia IlIofparticle a with respect to the y axis, and the moment of inertia I" of particle a with respect to the z axis (the axis that passes through the origin perpendicular to both the x and y axes). Wi+-}, f' r~s pe.c.:t +0 f"" x- o...x is : J'{:::m o.cJ.. ; US::. J I;<::. m r~7 = W,.th re,sp e.c,t -I-tJ '" y-o..xis T. >' r A. cJ.; lJ.s 3.... .:: ~o m (3"')'" : ['I fYI r'l / W j .,..h IS f e..:s P f-(.,t + tJ z.. 0.. Dw>r is T),~ oiis+-Gl.t'lce . f"'OM -the ~ Q)!ls ~ [1= j ii"a.+- J~" -= JiO r fz-;,omrj ,. / A ~ Z Part D: Find the total moment of inertia Jof the system of two particles shown in the diagram with respect to the y axis. 1he.. Co",+r;but-iof\ .(:.('011\ po.rt,'cle. CA.. : r 3r ."" .I = ()..t\ n" (3,.):1.; Cfmr'1.. r '}.. ef cA .f- rom fJ CA rh' c )e. b . , . 1=::. .. .? J"Y' ( 'I"" :a.) :::. ~ W\ ~ i he. +-or0...1 triO'" e"f o.p ;" g.=- ri 0.. loS: IIrnr1 Part E: Using the total moment of inertia Jof the system found in Part D,find the total kinetic energy K of the system. J~ ~ j YJ. 11~ I w J... n s e.r ti w ho.+ we.. jI..= -r 0 Utlo\ {Or' I: 1- fYl (rw)~ K= .!.mv2 Part F: Using the formula for kinetic energy of a moving particle of particle a and the kinetic energy Kbofparticle b. 2 K ,find the kinetic energy II r<. 0...::;' kb =- y~ r0.... (r w Y:). r"" ~ mf"W~ j;~ ;: (1" wJ?. = WI r 4W:l. Part G: Using the results for the kinetic energy of each particle, find the total kinetic energy system of particles. j( of the Io.. -r- Ib ~ " mr~w~ The Parallel Axis Theorem Part A: Using the definition of moment of inertia, calculate the moment of inertia about the center of mass, for this object. [a", c ,. m A J. rem. = yn r'" r c..1Y'I= -I- mr"J. J. m r '). r ,. Part B: Using the definition of moment of inertia, calculate IB, the moment of inertia about an axis through point B,for this object. Point B coincides with (the center of) one of the spheres (see the figure). T~= yY\ (O):l. +- m(a,.)~ LI U.s; I\~ -= lfmrJ Part C: Now calculate IBfor this object using the parallel-axis theorem. th (. fAro-II e,l-f>..xi.s fheOl'em I =- I c.,m r;: of- (m +- m) ,...:1 Jmr:l. + 1 m""'" = J:4mr9 o{= r~ Part D: Using the definition of moment of inertia, calculate lc, the moment of inertia about an axis through point C, for this object. Point C is located a distance rfrom the center of mass (see the figure). fhe. cl.i ~ I- a. n C. e.. he.tweef\ (). "Dl c C. q..{3 (MId- IS -rne J.~ b 0..1 { ol.::: r Ii. SO Ie.,.:::. m(rfi)""+- fY\Lf"fi)"1.::. rn,...~:t +-Q.fVlr" (-r = 'imr~J Part E: Now calculate ICfor this objeCitusing the parallel-axis theorem. rCo, =. I Com + (m r m) r ~ 'J.. Ie, = ~ 1r"2.t-;}mr L:r.c.= ljTY) r'l B c o E Consider an irregular object of mass m. Its moment of inertia measured with respect to axis A (parallel to the plane of the page), which passes through the center of mass (see the second diagram), is fA ::: O.64mr2 given by . Axes B, C, D, and E are parallel to axis A; their separations from axis A are shown in the diagram. r : r I r I r Part F: Which moment of inertia is the smallest? XA ~ r i~ Slnd".1 the.re... f()re.:fA is smo-lIe-.s-f. Part G: Which moment of inertia is the largest? IE":::; r is t 11e. I ~ (' fj ~ sr. Part H: Which moments of inertia are equal? Ie CJ" IG Co re x r 8 e..c.a..VS t. th t. 1)'\ 2 Part I: Which moment of inertia equals 4.64mr ? ~.s~ 15 i r (' e5 v loJ"'.J ~ ~ yY\ r".J -1-0 t,Llr tXCA.C nor /--1 Y 1 0;: M (~ r J :a.. ::;::... '-{ m yo ':a. -7 w),,, c It is c los ~ 'I. 6lfmr~ Ix Part J: Axis X, not shown in the diagram, is parallel to the axes shown. It is known that Which of the following is a possible location for axis X? = 6mr2 . To Tc "-" IV 4 YYI r 'J.. /V ,r.- 'i tnI' thus ~ I')t rno..y be <t' b e;rwe<n ::Co ke Problem 9.22 10-7 Compact Disc. A compact disc (CD) stores music in a coded pattern of tiny pits m deep. The pits are arranged in a track that spirals outward toward the rim of the disc; the inner and outer radii of this spiral are 25.0 mm and 58.0 mm, respectively. As the disc spins inside a CD player, the track is scanned at a constant linear speed of 1.25 m/s. Part A: What is the angular speed of the CD when scanning the innermost part of the track? An.fju/().r or /ine.o..r v::. ve.loct'l-y are.. re./o.tuJ.-: fYl r ()..J r::: O. OJ.S V:::. I. :1.5 Mis /)J=" I. ~ S ""/.5 O. o~5h1 Part B: What is the angular speed of the CD when scanning the outermost part of the track? Gte 0...; f\ v~ r w v:::. t. J. 5 ~h r.=. f.AJ 0.05 g yY"\ - I.:>' 5 "Ys O.058'rn - .2 J. 55 Part C: The maximum playing time of a CD is 74.0 min. What would be the length of the track on such a maximum-duration CD if it were stretched out in a straight line? It we, UY\WOLJrtcL the.. frO-c-t' Q.;r 1e.Y\f)J, I..l ~x. r he... t rOLe.f\ 1Aetime.. ,-S Sca.n n e.cA is 5 ""/s 74r1'* ~os_-=YYl T\ 'fL/lf(JS f<. e..Gall ; vt ~ x ~ I. l5 mJS ~ 'I if 'i0 oS AX::=' L:i x = .5 5 5 0 fY\ :;. 5. 55 K m Part D: What is the average angular acceleration of a maximum-duration playing time? Take the direction of rotation of the disc to be positive. CD during its 74.0-min re mem be.,. +he. -For m IJ JO-: lies 0-= \Iof - v.' a...,dl ~t ~c c eleY"Ovr"()" +he... ~v.me. o..fJfJ veJoc,i7y : 0(; for a"3 v b..,. (J.)t- - Wi 6t - SO r~ec. - ~ J. 65 r~ 'I'-I'/Os F. L( IX IO-=-.s ] Problem 9.46 A light, flexible rope is wrapped several times around a hollow cylinder with a weight of 40.0 N and a radius of 0.25 m, that rotates without friction about a fixed horizontal axis. The cylinder is attached to the axle by spokes of a negligible moment of inertia. The cylinder is initially at rest. The free end of the rope is pulled with a constant force P for a distance of 5.00 m, at which point the end of the rope is moving at 6.00 m/s. If the rope does not slip on the cylinder, what is the value of P? ,here.. ()..re. sever,,--I t-nJ."~.s of fo cons;O<e/'": Who.+ IS the for 0-- Y'fJomen-/ lflUfia. of bo...rrel ') f-lle hollow cy/;tlcJ..er: -;;~ m ('::::. whCkt ~ "-0 N ~ (, ~ 5 m ) 2 :::. 6 . ;t55 (fl'J. f\~ 's P: , -rh e. \Nor K o..one.. ()n t-tdS ;\1\0.1 cy uY\J.e .:: r l.s : w ~ ~ 1'\ ;:. Y;I. r w ~ -7 we,; p . .5 V'V"J Fo t'\ l' 0# rc e... cL i S 1-0.." e,.,L r~ O.~55 r w 1~'6~~ b l. fo r 0 V :::. C, Mis =- ,...t) 1-d; e.. fAe, r() p e- t:o..lts ff. w.=. ~ = :J '11'()..ts J". IE = 1 (o. d.. J. 5 5 '"" '"' ') (;). '1 r't".t) ~::c p. 5 m e.te.rS ~~ ~ '1.7 '3 'cUl~)..I OLoO )J.l1 - 1"' dh}jJ1 f -I- /"I ~ 7f =r I -::::c. lI'tJI(S r d J- ?J'a\( ds ?:. ~ r L 't V JJh "t ~/7 -:r: ~ -e'tJ"w :::-Il? '1s r 'tJ P. h d-::: .., L(A t =I ..J {! 0 i 7' 0 1111 '{ ~"'O 0 J 'l: >J (~~ JL *17) % : MtI:r I cI + =11 F'tJ..JJ- ~ : 1J/ifllo/l d-:w >s""O \AI s;;,~ f- A-f .' SilO) 10 Sf ! S' I 1""",NI "d 'f f O "0 '1 J 1: Ct \/ Mw' : 6)J'f) C( Lfr olf.../. r 0 cis 1'!f S -() J () f ' . , {OJ dO ,rays ~ Lf..J- Sf'1 d y;ooM f 0 l'{'J "I" ~,,, ,J Q "'b~-{.n\I J 7Q .j."iJW oW -,oT'ro o-} 1""'D U "iJ hi Problem 9.77 It has been argued that power plants should make use of off-peak hours (such as late at night) to generate mechanical energy and store it until it is needed during peak load times, such as the middle of the day. One suggestion has been to store the energy in large flywheels spinning on nearly frictionless ballbearings. Consider a flywheel made of iron, with a density of fI, in the shape of a uniform disk with a thickness of L. Part A: What would the diameter of such a disk need to be if it is to store an amount of kinetic energy of Ewhen spinning at an angular velocity of '-'-'aboutan axis perpendicular to the disk at its center? For ~ Solie). cy/incl..er rotCA.ti~ a..loour its o.xis: I :: M. ({.:J. Y:l aut \,Wh a.+- '05 J1. ~ " !>j ec,t for ll. Jolio. 11:: [J V for 0.- C Y lin c! e,.r v: c3) In L- - &~~(L) +h fj"f\ (,1\ M:=. j:J L 7ft{ ~ oJ ly f uttit\<j t-h is b o-.c. K we n~ fbY'I: r. ~ la. (TTfJ ~ LA~) R-:l. M wJ.~Y\ ,'f- F'il'ttA.fly) .sol v e a--\- f6r the.. ene-If of the.. fly wA~~( f s ro r~y; w. fl. f< ." ) ~ 1-\ E =. Solve 1 I. w 1.::. y,,- (t f' rr L . f\: 1'J~ ("IE " I'L.Tfw:A. +or " )''f :r:. ~ 1 lJ f' vV e.. \IV tMlt ~.. ell (1.'" e t e. r ! D::::J. '" ::: l.. trw:t) Part B: What would be the centripetal acceleration of a point on its rim when spinning at this rate? f3 -r l(E. Yy Q c..::' ::!-'J., f\ but v:: RW -;> \i 2..:=. R ~w '). frf!) Fe>-+" f hVJ ()..c. == R w 'l. D..c.::; w~ "'\ li11 0 w R yY\ + A; ('1 pLITWa . E ) '!'1 Lv "' ? aM h /a -I "'W ~ .. filIAl ~Ir 'f ~UI ";' Gw 't q/\ -=;1 ( ~IU ~..f + 'lIM 'Cir ) q/l := If (0 '1 W Gv.rf\ 1 0~ ...JOt 't 'II ?uJ h~1-+ q" "t' dI tAl ht:.,. to ql\ q \M '(/. - u I - ~lr G q, V\.I II 'C1 d v.J ~;( -I- 't: q1\ q \AI f1 = It G et UJ ~!t,1 .;fi\ : u 1M uy.? 't?J JlYl -f! ?w ~ -:J'#'f"~1 ,(JI n y. "J I'll d 1M <:";' := 1.'#11 (IJ :L Torgue and Angular Acceleration Part A: The seesaw is pivoted in the middle, and the mass of the swing bar is negligible. Find the angular acceleration a'of the seesaw. W-e. 11\1\ 0 vv' r == I. '" oS 01 ve... .f 0 r Sol vt... .:r:. : r. .I.=.~, (%) 1...,.. ~;l. (f) L2= 1.:::. (J'Y1I-t""~Jt:l. 4 for r,=(~,~)(}) r ,.. (1Y'A'){~) r~ (m, -m~) Solve. tor- -r or. :::. ~ :r: =- (-m ~ ~ , rm - 1) ..t '~ ("". > ,. =: I j" (M, -rn~) (I'Y\, -r-maJ --jl <- .J~ Part B: In what direction will the seesaw rotate, and ~1zat wilr the sign of the angular acceleration be? The rotation is in the counterclockwise direction and the angular acceleration is positive. S r l' C e. D( m I ;> m:a.) rot (). (J 1\ t. whe.n S C 0U JI\ te."" c lo c.. i'\ w; s(. c()vnfe..rclocf\w/se- is Fos;t-;\I'!... V'oro.t-ioY'\ i..s ol.re c t't<:>n. Part C: Now consider a similar situation, except that now the swing bar itself has mass Find the angular acceleration ofthe seesaw. A~Cl'Y\ T.::Io{./ tnt]{lt. bv-r we. ( ')~ mvSt" re..e.xo..~;ne. bo...r ~ I 'C 'l-L eY\t" I =.. M, ( .L) ~ =l +- m:1;:- +- I J-;: ty\ :t. ~ ft'\ 0 tr1 o.f. I',.e.,.f,'~ o{ (>.. b 0,'" t>Jo~v+ ;r.s cera+-~r. r::. Yn, f m,; ~ t =~ 1 ~ Lf1 """"~ do.. rhes e. f-e:t'm s ca."" eel ,s"1'\ce.. bOwr ,'5 bcd~eAt So Iv,'n5 0( Por 0(: =- .IT. (m, -m~) .::!:-,. ( +-} t'1 bOwf' ) (m I -YV\ a.l '-f mit- m,., (tn, -rm~ +);3 ~ b~,.)J... --. ':1~ Part D: In what direction will the seesaw rotate and what will the sign of the angular acceleration be? The rotation is in the counterclockwise direction and the angular acceleration is positive. e5In c..- m I :> m;). J -th e.. .s e e. ,sG\.w' wi /I positi" 1'0 y.a..+O-r ~ ttccelero.r"ollt . Co ur\+ercJocf( v.II',Se, wjt~ e a..1\~vlo."" Part E: This time, the swing bar of mass fH1wis pivoted at a different point, as shown in the figure. Find the magnitude of the angular acceleration of the swing bar. Be sure to use the absolute value function in your answer, since no comparison of Tn" m2, and mbt,rhas been made. Wl.. now nt.e..c.l ro re.ca1crJ Io.+e botA. I ca... r:. ::.r:. In I (!)~+ 1"1,."3) ~~ T (2ll c.....) 3 ~ 'J. ~ Z\ T.fl -I- mp&4.~ (}) '-----?> I ~m I Use. rara.[{ el o...x/s -tAeort,tn /6 c.~/crJ10-""e q 1." +- ~J.. ~ ~ 0( fYlhtA.r l~ - q T~ /lI,,~ () +- ~ "'bo-r ~) ( ~) (t ttl bt>/ 3) (t) - tn" ( 2 (~) I I, 11\ ~ - f-=- jL (3_ 3 fI':;)... -.1m 3 I .r J-WI , ) bo.r ~r - ::.r ~1. .L 1.. ( '3 J1 fil, ~, M_, .,..~ mI r Lf rr) ~ bt>.t' ) Part F: Ifm_1 is 24 kilograms, m_2 is 12 kilograms, and m_bar is 10 kilograms, what is the direction of rotation and the sign of the angular acceleration? Re.c..~fl 0 v(' eXf re.ss ic n fo r >VI? fOP-QvE: r.,. [ :J. +- ~ fYl b oJ' - -I: rn b ",I" mI J IS .s,. fI ce (p:f'ter 1'1 VB fj inj n vm&a ~~ P.i;r.t sj't:J..e. LeFt f') in) side. R .. ~t ~ YII siol 0{ e > leff- sl'd..e r A er e-for"e r 0 ro.:!- "0 is c 10 c /-<. e. a.t)oA. ,'s l1ej 0.+/v e ! Problem 10.4 Forces F1and F2are applied tangentially to a wheel with radius 0.330 m, as shown in the figure. What is the net torque on the wheel due to these two forces for an axis perpendicular to the wheel and passing through its center? "I "2 fhe... +0 ('~ u e... olu~ T I =--F1 r oI.u e. to ~ J. 1-0 P,: (he- -1-0 r~ Ue. F).: r J..::::' r [he- to +0.. 1 tor~ue, 15: r-:;. ~- F, ) r (F Problem 10.19 A string is wrapped several times around the rim of a small hoop with radius rand mass m. The free end of the string is held in place and the hoop is released from rest. Part A: Calculate the tension in the string while the hoop descends as the string unwinds. Dro..w 0- fret..-hocfy c11().JrClr)1 Por -t-he- t-A~ h()c>~. -rn~ is ~. \..Il\l>.f Yle.t -Forc..e. CA..G1-~n!J on hOop ~~-r :: m Ctem :;:::. j - T J11 01." w <!. r= I A K f\ 0 W a.h 0" tX .j--0 I" ~ tJ e. ? 1 ;;:.M fI, a~'!! rn ~ (..,.. oSi 0 Yl e.,1\ ;;:. ro..e>V /).s) oJ.u. -r = r:0<.. fA v 5 Yl'I'j - I e).,t") :: > T ()..1\c{ 0 -r R => T =- f{ 0... em T == :Fi fla./Iy Part B: Calculate the time it takes the hoop to descend a distance h. l! == ~ ) ----J of' ~a.ss 0 Yl W J,~t is t~ e ()..c.c.el e.rcd-; 0/\ P -f}, e c etl1- ef ex~ TesS; ~ y:1. 0. -{ ~ ::::.) o...c,m .::::. ~ U.s; "3 r- h e Ii1}1em at,'Co X"== h ~ Y~{-f) So / ve- 10 r + _. IN F-r-;;-;:---l t~ v 3 ~ t~ Part C: Calculate the angular speed of the rotating hoop after it has descended a distance II we.. H Y'O We.. J"O'Ve.. v'~m W Cl.;e f'~/~""e.ot by: w~ for ~1h vc~ R Vc."" usin '.1. :::: J -rAe Min em ().,f,'c e.xtre.:SS.'on v'c'm = v';l: ~o..~ ..1. (~) h -> J;h f- t. Us ,~; {'tJ z...R Problem 10.26 A Ball Rolling Uphill. A bowling ball rolls without slipping up a ramp that slopes upward at an angle the horizontal. Treat the ball as a uniform, solid sphere, ignoring the finger holes. Pto Part A: What is the acceleration of the center of mass of the ball? St,J- Up {)... tre,e..-booty C-.F ~ M)S(t'\fl lorCfJUe. o(.;~5"'a.m - f:::. Mo..cfYl 1:~ r f\ ~ r. 0{ ;: (i- f'{f{") (~) f /<=- ~ P1V3 O-c.m K~Sj M{iCA-cm {-;;. }'S M O-cm f-hi.s bo..cl{ i1\l-" rAe. rYl Forc.e, e,u,,-hon 1-0 So)v~ f-o" n;3 Jy5 M o...c :=. 11().. em ~ oS i "/!>:::. ~ O-c. m (O-e'" "'-y <;jsin;.3/ Part B: What minimum coefficient of static friction is needed to prevent slipping? We.. /I Xnow I;;. ':np~ J"t3Si nf3 foz )/5 fit ~CIV\ rn==t1~COSj3 f jA-z::,_ Y? Jk5 s; f\j3 ~ F;, %Cosj3 7 frAy!> Problem 10.61 The mechanism shown in the figure is used to raise a crate of supplies from a ship's hold. The crate has total mass 46.0 kg;. A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius 0.250 T1I and a moment of inertia' = 2.10 about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius 0.120 1lI, the cylinder turns, and the crate is raised. b I 'I k(, . m~ What magnitude of the force F applied tangentially to the rotating crank is required to raise the crate with an acceleration of 0.850 ? (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.) illSj .) We, 1.-\Yl 0 \AI r-:;;. r )( F b V +So. r:; (},I 0.. w CJ- ot i 0... ~ yO- Y)'\ f- 0 .s e ~ the. o f-h er fore e.s (Ac.+it\~ I. 0<. -r m{jro.-)A A 0(::. ~ puHiY\j + h is oJ' (0. IJ. -1-0 Be, f n er ... Vh(~t-()o.) yY\) ,:.:: I.~ R ~olve + ~ (a.rj) ~ We h ()..V e. -ro lor m ~) ( F) .. , So w~ p/Uj ( 'f b JI\ J .n n IJ tyIb e/'s (O./JY1\)F::. ( :J.. . 10 JA, ~ 0 . ~ 5 mIs ~ to. F=J.Otf :t5/'h) r - 3) (0.8" 5 '"Is:l r 9.8) ( . ~ 5M) - ...
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