Final - Name ID ME461 Mechanical Vibration 7 Term I Exam(Dr Zhu Fall 2010 —-—> Problem 1(351 kl XI A solid disk with radius r and mass m

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Unformatted text preview: Name ID# ME461 Mechanical Vibration 7 Term I Exam (Dr. Zhu) Fall 2010 —-—> Problem 1 (351 kl XI A solid disk, with radius r and mass m rolls without /\ i slipping and is restrained by a spring k]. a damper c. 7 and a massless wire connected to a bar with mass M — and length 2]. The other side of the bar is restrained ab by a spring k2. C X = :2 M9 = Q 95 No slipping W»— k: What is the equation of motion in terms ofr, m, kl, k3. c. l. M and unknowns 6). 9 and é ? kind}; Ema-66 2 _. ‘ Z ‘ 3. ‘ z T . :5- < ,45 (qe’mmb + :‘i m mg) + g (gmrl) 9 - J. M 2‘ ‘ ” :2<‘§)(~20 62 + —_;—(.g_m)rlaz: é(f%ir3 +5gnrl>82 _. .L a .2 n. U ~. K (9 .L x _. 1 a l“ > + 2K2 amfi‘a +_LC4KQV3)82 = éWW‘ , ‘Phob. ’1, New’i‘bn’s (Mome : F€>C>=t41, x ‘4 TCD 21'0un — TY‘ +Fr-Y‘ = J—mr’e RJ‘G 2 cré C?) E Page .. T~ Fb— KLrG—créz mré Fr (WP—>262) mr‘e +<1m9 +r<ma = 2—1— _%mm‘9 Oh m \‘ . ® 79 +CG +t<|9 —= QT/r. #50442. T ><=2r~8 X=Q‘9b v U M u Q 2‘33“ =~T~Q wax-fl: 73-May 8b=€Q'X 9b ' @ T: 495?:‘8 « %2r“8 Q @963 E \ ax +C®+K.© =~4~h18 ~39“ =~>E©M (3: +4g438 +CB+CR+4LQJBQ =0 Problem 3 (352 For the system shown, the damping coefficient is c = 600 (N-S/m); both springs are [(1 2 k2 = k = 3000 (N/m), and the solid disk radius is r =1 (m) with mass M = 40 (kg). The disk rolls without slipping on the surface. The horizontal coordinate x is associated with the center ofthe disk. X 1‘ b8- What is the equation ofmotion in terms of]: m. In. kg. 0, and unknowns xi A" and KI crew Haze ‘ ‘ \ 1 l ‘ 2. --— i.- ‘ 1 T :{MK + gezfirfle - 2(3—3x .. Q 2 U — igl‘KX +5:le ~g—LKfi-lQBKZ—4 fipuqxz KC 2 C i . g _ _ . — —- - ' v “a” a ' i i — ' . _ F50 Neflbnsfleory. X- r6 “ Ix @ZPQ—Cku+l<z)¥—C§(-FP=M¥ KY \\ l @ZT =R~r tbl'Mra‘S = Ell-Mirr‘ <____.<-‘-- <-Rv< @lr 4 Q3 ex 5M§<+Q5<+QKX=O Fr- .3 What is the natural frequency ( a)" ) of the system? m” : l 0 Fad/s Ql< F;— j4x 3000 a —- M :: ~——- = = l m 0‘3" " l 554/? SM 5x40 0 /5 What is the damping ratio (4’ ) ofthe system? J : 212* C C ngn :3 SM g S = -— 600 —. .L /1’ SMwn SXADXIO ~2 Name Name ID# What is the damped frequency (cad )‘? (ad : SB wa=u3nit~ s‘ = to xi“2\¥ = 55 What is the system response ifthe initial condition is .\‘(0) = 0.8011) and x'(0) :1 (m/s)? xc+)= e" 3®“t[c. coswat + cl snoth mm scon=5 wa=5J—S, °.=Xo=0~%,ca=élco%w2 3 Kat): -51: J“. ‘ : I+5xo. _. e «454.5. -yfi 0“ %(+) = A9: " C CostcoA~<123) I A =,iC;’1-c2’ = 0.987 ‘St g 2 hon—K c‘le) 2: +ah4<ég73~ 10,635 =7 xcc) = 0.9379. costsrgt_ 0,525). m . What is the system logarithmic decrement ( in )" ln = :23“ XE . X2 fig x. __ Q“ 2 ‘2“3 —_._ JE— _ 12.3. i072: = SODA-C - Swnwmfiz-Sz. TIT? E — fig" :. be:ka $596? a” 4150 BE use} To FIND E'E-JILUMW S.- (SME' 51W) —.!‘|.. '7. J... __L 11— TL _1 1._ Ema-t 4» am“): +£3.19 _ mex i=f‘é é," Eéir-EL T = firm->1“ avmziu a;er :mw .. ii1[m1+mz*% séétmg .. T O" _ —m'x: (“M 35%“ mrfix ms; = Mur“ flfz + F TL “UT HERE!) W F -- 1 X. =._.. Xirre 6:“ gr 9 (2:, Flt-ht): I '- r-. xlfi £19 x3 = %‘(—FF): ‘2pr 3352(3); tjfifiom 1M T‘EEMS O-F X. z z w t .. _ j“. n xl=£a x'fifi/fihéfi; n-fix x1 = {19‘ ab: 1;: ‘ %L(rr)bifcfifi 31-. sue. @953 [um G.) ‘ H w *Egfi.._mfl-. Fp(m"+ H") + £|(W“-*"‘X“m*%[fi 3 la "2‘ “3 F31) (renews mum} fi" ,0 (fli)+flm¢(%“)+ft{3ix)+J—M “1&5? ff“ mrykafixlr rm: r, r ,x . z FF 2 ? kin -‘ _- fill '. hH-n _ 3'90?) .. [m _ :9 runs In ,9: . .. . - Elxtfqr)-EIX.P+T¢=305 (1) mi‘ + __-——+ _=__.L film I: m if} + I :9 [ P Fe rp*‘zrf*r;*€fi¥[”v‘—$X _ airline/fix} - Fk.><._ fink: Tait?) MUMT '3‘" T we W955; 3171* ftm‘i4r r‘k.x+ kgrikgx:o LC'EE‘ + (twig '* (f‘kfi- Ric-1.711613% :Q-‘l ; ._fls:aumE Emu-19 - pawn-IE .9 rs magma?- EMT Hum 5T. ta. I] J33)— j;____._._-—-—-—-— All in: in? TIGL‘W/ 71,5139 _ T5 a)“ KR kas T5 jean v cg/§J-kx¢{§)= fileé 1 a." 1- 1 TEWE ‘5 “iii—a + K§a=o gram CWSTIL 69.1mm! tnfiish aims +gfb=o _ ci‘ 1 '5qu: T i JIGW'JO-Cé) Flu-D: STEMT mm; 265mg;er 5W (%;M)£“é + = £Es-uwt '5 {(-3 Ma“ (30; 4 Shh-Jflq‘: 53.333 Ira-v3 X15 " {9550 Makes; + .5‘,‘) ' {-512.5— M-m @9450 EESPOLHE {5 5mm gr am: X mm) : __.§:'"—. {J X J't/i‘ll F=E‘%= H.341; (we a) L)”: j—E-fii-‘SIWB mus 1}}1fi3'n‘ E1: = m: Hit-g 55'» ‘_ val = F it) In. F 51 1-53!— : Lil—- F 0.32 mi may u-n (Aufivmfi 056m» A 9) {Ma X = Hi :. 0-31 Pad- 3 ‘ F" 3. film: = "(100599? 965): —8_s?xm‘¥ 5m (1041.12 t) ...
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This note was uploaded on 01/29/2012 for the course ME 444 taught by Professor Staff during the Fall '08 term at Michigan State University.

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Final - Name ID ME461 Mechanical Vibration 7 Term I Exam(Dr Zhu Fall 2010 —-—> Problem 1(351 kl XI A solid disk with radius r and mass m

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