191prelim1 - I I“ I I I“ I I I 6.3 an...

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Unformatted text preview: I I“ I I I“ I I I 6.3: an z=a[fl—alnfl]=%%=n[l—mafl=k [fir =n'[1—2mas+m'sj andy=n{1—oca9]=e%= 6.1: 10 A(y) = 1;2{leg)(1eg) = 1;2[‘/1 _ 2 _ (_‘/1 _»_;,P]]2 = 1;2(24./1 11f)“ — 2(1 if)“: — 1,d— 1; V — Iffltyldy — ii. 2(1 may — 2L2; y3f3111 — 4::1 US) — 823 /}’£\ (cos 6, sin (“5 ‘5 r 2 m ‘77 77) (fl) 3%) 6-1= 44 Riv) = 2—29!“ and rty) = 1 e. v = fier([R(y]]’—[ (y )]=2)2y I01 2([2— W2] Mm. 22(4— 4yua+ yQIB _ 1)dy = «[3:9- _ 311;“!3 + 3y5’3f5fl = ”[3 _ 3 + 32(5)_ $1335 (-1,0) (1,0) 6.1: 46 a. r[y:] = 0 and R(y) = 1 _ W2 2;. V = I:fl([R(y)]2 _ [riyiiflidy = “I?“ —y,!2)9dy = «1.02“ _ y+ 742*: ,5,\ 92541112: “[y—yflfhyaflflg=ar(2—4;2+s;12)= 22!3 Z};— Eg/ b. PW)2 = 1 and R(y) = 2 — yf2 :32 V = I:WI[R(Q)]2 _ “mug”?! = ”1-:[9 _ Elia)? _ 1H}; = I? I— —J t' 13F? “In (4—2y+y2!4— 1de=2r[3y—y’+y3!12]%= (6-4+sh2) =1r(2+2,1’3) =sm 2’ 2 (0.4) 2' 2 3‘:be :a 6.1: 50 3.. A cross section has radius 1" = «\f2— and area. 'ir'r'2 = 22w. The volume is flr2qrydy = “[92]: = 2511. 3L‘E'..T.ff=|2:ssnm b. V( (is): J'A(h) ),d.h 50— =A(h)_ Therefore % = “3.“... .u. _A(h)__ BO % = filfii _ %_ For :§::£1:0 Ina: xdlx cntx c h = 4’ the area is 21mm): SGT so %_ — 8—“ 3%;- 3332 IN“ : .. 123W” ‘U 22= 22 L‘r- 21.. - ”tr—i1} =mi1Il—aJx-IJ= we. III-II“ Im *c '2 V = I: “in" WNW WMW=I§ HIE—1r} [#3— (4—21)] w= Emmi (i=— 2;) w: 33:33? I 1 ,1, = 552.4%- 42+?) as: 2.. [2:341 1. 41+ +1]: _ Mam 22rm— mamm— _ Dmu=56czx=l+tan2x 1+x2 31:5. Dsecx:senxt:nx 1 1 ix c 33:51, Jn2.xzmx=;tan ;+ .3.V:£2w[d1d1mdhu][alflllhaigfl]fir=l§2fl5_fl[l;_[j£_glfi]]dy=J-:2w[fi_m[f_$}w= :== xx 5 — 5 — — gu‘L/‘ffil WWW” gig‘rifii'"EF—f+¥iW—gwii—¥-$+fi]u—artwi’s—1mm_w¢+wm;_ DmI=—l 1— . 3353214.? 2 .. = + . ‘L V = 1: 22‘“ WHEN" 22222-22 =1: 22mm: [‘9'— - ("é— Hrl] w = Emma: [:9 — 2:} fir: Dsinhxzcns'hx 312111112: sechzx: 1, 12!]th Dsechx: —seclu|znhx Darchnhx: 11(1722] Dmhx: lief—1+? Darcsinhx: LIX/1+ +F 1d1x:x+C -23_%)¢=gw[r‘i-3;_%‘1+%]:=2w[1514—Mfm+aofm—mfim)= Hew— 1w. 3.2: an L v = J: w[R’|{y] 4593* = 11' “(igr— 1mm = w [—131 _ 21%]: = wu—ym— 11:3} _ (.133- 1313)] = 2;. b. V = fiafiahsn Indira][ahd] hdghfldz = Illflawzifd—{I—fle: = 2w I11;¢{=1Il—2:ifl = 2w [“If' _ '7'] 1 i = x : +C fnrnae—l lfi1_ I" x‘dix=1.n|x|+c a: n a m 2 E = —aint ma L? = 1+“: =:- {g} +{g} = ‘fE—ainij'+il+maij5 = m:- '° *6 mam—m .... 2= «2:: {1: :::]:2+mnw= v2; .12::.2=221: '31:. at [mmiaflom[fl,w]); [u=1—mat=%du=ahfit; t=fl=ku=fl,i=w=hu=2]—r figu‘lflm=‘f§[flu1fl]l=d. 5311235: #‘fi'=i‘[ii } =Iifli—2‘i‘f‘i=i‘5= In 1f1+tflfl 3+riflr= fiqfflifi+3+¥4i = 1 a cnsxdlx : sinx +C ainxmx=2cusx+c ifiWw=ifiw+¥dluw=i[; _ r1]:=% sec2xdlx=tnnx+c 3.56% fl_ 1 fl 3 nil 1 13:51” 1 1 _ I. T. — as y— =:- (i) = BE 3— =i- L ”.51...“ + iBfiE‘H— idfl= E,:glmfllfi|'— :Ii'l. _I_flaaaydy. 1 3‘ Y 65:23 I. [fire-0mmfigment-1593i“a351,.Thanx=—1fy+0md,ahme{fl,1]lmantha Awmfofizflxm m,C=1. Boy: -1—. h. (July-one. Wahwwflmmmafthaflmufimmdflmmdflumatmmdz. GE) nain3=.‘- [3)n=n'ain”#=i-L=L" {§j'+[{3)lds=_f;‘ flu[1_mflfl=nfi£,fi 145m” 2EL”|§II%|HB= staffing”: Aging]: =3a. fi.udlait1umdamnmgttas=5mdy=2fi.Thamgofmmufmmdiahmm[m Enmphl}.Thamdayatthtotmpdfln-mbmtadatthmmdthmdaat Wang,n]m[u,L].Tme= 5""—m— _=Ll.fiemay '—'“+"— I""=%=:|[Lffi,2£-f3]la thaamtmndmambmfim s.csuu=§z{2—!)u=j:{2z [22—4]:=s—1fi,rs= E=Hrr= El: lfl=[zs__': 415—13312)=Sfif31H=I;ig—%i”= [1f3+1f2]+{4—1}= [1;2+1}+[4—2}=7f2:-a=¥in=§. m: 12 Mu=L12[a+l)de+j:2uiz=jn1[:3+a]fl:e+f2mh= [‘Ti+’T“]:+[2"H= 2222.2: = 1.1m 1i22+1.'22==[e‘ +=]§+ [2212 = smany=sfl=ny=azles=i~en= .f{dz)'+[az'e:}2 =JH—Qs‘dz;u.=£julx“fH—uz‘u; [ar= 1+9ar‘=:‘du=Hizaifir=k fidu=arlihgar=fl=hu= 1, a: 1=-.-u=1+J]—+M.=£-r1“§u1i'eu= fiiill‘"ii"=fiil“‘"-ll- Find the center ofmass afawire ofcdnstam cbnsity :3 shaped like asernicircle nfradills u. Salutian We mndel the wire with the semicircle y = Va2 7 x2 [Figure 6.39). Tile dis- tribution afmass is symmetric about flleyaxis, so E = I].Tu find}, we imagine the wire divided into short segmens. The typical segment [Figure 6.39:1] has length: a‘: = adfl mass: din = B d; = Ba :39 distance ufc.rn. In mum's: ?‘ = a sin 6'. Masepcr umt leugib times leugih Hence, fjrdm flame-Sade fa»: I fa" 34: d3 7 Tllenemer Dfmass lies on the axis ofsymmeiry at the point (0, Zen/1r}, about terJu'rds of the way up from tile origin (Figure 6.3%). 842% —r:os 6']; Saw 2 "a. E: FIGURE 6.39 The semicirwls: wire in Example 6. (a)The dimensims and variables used in finding the center of mass. [13}1'he writer ofmas does not lie on the wire....
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