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EM 311M - Practice Final Exam

EM 311M - Practice Final Exam - 5c éufl‘obzx m-WM‘MQ...

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Unformatted text preview: 5c: éufl‘obzx m-WM‘MQ N w 1.22:!”va EM311M - Dynamics Final Exam PHR 2.220 Friday, December 14, 2007, 9-12 1. The homogeneous thin plate has mass m and dimensions b and h. Determine the product of inertia [W in terms of m, b, h. (The origin of the system is located at the center of mass.) (5 points) 55370 /. Chg-Mfwcfé I [y] W...“ ....... o<x,<,¢. 0<yl < b- 5f}: 2: .‘ attic Wheat/c6 Axes 7‘5? corcm eel» A -3. :AMHL L~f -m——-- émék amé/«L $36 2. Derive the principle of angular momentum and impulse for a an arbitrary system of particles With respect to a fixed point. (5 points) HO * \ go: 3. Compute the work of force field F = ($2, y) [N] along the curve shown below. (5 points) “Panama f'rzlzah kw. : .__—-—- mm...” / / UM = ffmf’ ' flight? = /(5f§Z/~/4zz)%5 A o / .s 4 / féfizfiMw 317425-14 = fwf=§faz 0 4. Show that the force from the previous problem is conservative. Determine the corre- sponding potential energy and use it to compute the work again. (5 points) ' \7 2, _?~_ ’2, . XN Qx/ <2}, ’2?) VXF=~6 ~z) f4? “5(sz 9,] o) Wmfl‘rc, (0,, 0/ a } (EEMWML L: S “l7 V X3 52f=_;; =4; :7 Vaw—HE'fiCQ) (2.1/— — 5... 4-: CI:A¢+C ’33. _F# > ’z’ 3 I’, 319 /l/= —:§- - A +c/ @ c a). mwmw—m [m4 ~ .. = J- L 5‘ __ Um'TQVb 5+2x¢ZAVMJ 5. How many degrees of freedom does the system depicted below have ? Explain. (5 points) # do’L fmivl‘ B '3 QJ f WSW!}A.0J .' Z/(f‘fl'h‘icf/wnf) + / (464/8):wa J a 3 # do; Mud : 52 5 =— 2, ’7' p 6. The metal plate is attached to a fixed ball—and—socket support at O. The pin A slides in a slot in the plate. At the instant shown 33A 2 1m, itA = 2 m/s. The plate’s angular velocity vector is w = (0, 0, 2) rad/s. Compute the velocity of pin A with respect to the fixed frame. (5 points) y‘f \ l . /x l f X l \ - «KR/M + 93] ‘25? y=0.25x2m 3 7L @ Cocufufmw «9% Its/m (wwmmc/ M0519“ J 9: 0,25xz. ‘ X=/ =5 51 :026“ 41] 3:05“. i=2, ->;'—~/'[2"~‘] Ade/M = (kg) = (1/, / )[55/ 9g [0, o, 2, ) X {(l) 0.15/0) All/k:(2/ //7‘ “0.5sz 7. A rigid body is rotating about a fixed point 0 with an angular velocity vector to = (1,0, 1) [rad / s] with the components specified in a system of coordinates centered at the point. The corresponding moments and products of inertia are Ix 2 lg = 10, I z = 20,1953, 2 [W = 0,112 : —10 [kg m2]. Compute the angular momentum of the body with respect to point 0. (5 points) (/30/ fiat—£9}: O 0 [£7 £7 ,0 l-O/i: (.30/ @ 8. A flat square plate with mass m and dimension (1 in the position shown below, is under— going a planar motion with vo(vx, 11y, 0) and w(0, 0, w). Compute the angular momentum of the plate with respect to a fixed point A. (5 points) / 9 L 95.1 z,‘ +/_/. 4' 6504 6/144 AA? 10/ 5a, 0) +{0f01 1:5!) 2 infirffir «3.); (OI OI Z4/IL Jake/not) 3/010; mm (393-34) +ELM0‘36‘)?’ @ 9. The mass of the slender bar is m and the mass of the homogeneous disk is 4m. The system is released from rest in the position shown. If the disk rolls and the friction between the bar and the horizontal surface is negligible, compute the disk’s angular acceleration in terms of g and R. (25 points) V v 3 1"] l L 9 P—9-1 o c .-‘ \ \\ Q/ ”’9 I‘fp cl.o./ Karla/"wafi‘ca .' ya 6/0,0,a() 373 = 3a + 5X§f3 3542/3 :- (—42] 0,01% Eflffl 0,10] (63:12,,0/ = (’dkl woe, o) ,0 3A = 9.5 7‘ 3‘59,- xfiA- 7&435- (ééaf/O, 0/ 0(6ij 53,4” /~/v£/ '26, 0.) (Ra/Aar/ ’széar / 0} Q/A : 0 :> /' V,_,.,,....__...~mwm. MW ..,... key “If " xZ/—— 3°05” : O :> “be" :- __ Z1 : (‘XR/ “dz/0) 1L J’A/fmb(zf ca‘ao/om .' ,MM_...,______. 4 _. ,9&$k: 1C: 24/1123 ‘ @ bar: "65/ mee= [imez' ,x'léVB \J/ 3 H8) H3 Lyyta [1/ ———-—> ‘ ——’6 4M 00?, / T 15' 2, /,// V 2L” 2 2% 2.4 m2 ac Q , B R \\ / TEN/é \\_’__? H “ D ,L VA T Lina”? ’{Vb zéarvt dim Mpco @ _. V3/91/+, m3 3%1jm/‘x/Q 13,73 —-2—f-—Mo</——§-:-+ gM'QQ’ _. 2LMR/- 4Mx/B/R =C7 _ (2/:me +émg) +— .ij —-me, ~§er£ +24 ”1"“? .L ..J__ ,6" 5...]..— _,= ,__L (4/ I 1772/3431 1/)Kk4 2,? ' 74' M00" — 3%»44 / Mswzzj: : 86.?6[shy] 10. A 2800—lb car skidding on ice strikes a concrete abutment at 3 mi/ h. The car’s moment of inertia about its center of mass is 1800 slug—ft? Assume that the impacting surfaces are smooth and parallel to the y axis and that the coefficient of restitution of the impact is e = 0.8. What are the angular velocity of the car and the velocity of its center of mass after the impact ? (25 points) O [lapel/217° (”I d '—’—’+ widen?" 7‘mza Conant/aha: 9/ larva? momcafzem I'M J—cén'rfi'o“ , 0 S m “a; fin _-_~> ija/xi; 0 <5) Cauéou/Qfiba o/ defame fowmfiu.“ @354“ 7:50 QC,{-d,z,0) x0c[—c(,,20 % X €6fl/‘M4, o, 0) 3. £6 96/1/3470, 0.] +[0, 0/ Moog/I “HGT—5:}; (0, 0/ - /73.7 V934”) Cocflz‘ag‘mf o/ 1csfi‘ficr/foq 3 'a/M, mix“ 43/. l/ W”. s a [I] (9 4.4 ,erzemqua _‘ cat/n “HZ“ Af/w 4/41 ><(0’ 0/ C" / W93~ch911k§v0=(v€:%//OO/fw(cé “/2 0/ “#:wa W , m 1: 0) —— /75, 9 - 7/3 W, /" -—3,J‘J'.. / — 4/49 :05; fl? 11. The inertia matrix of the 2.4 kg plate in terms of the given coordinate system is shown. At t = 0, the plate is stationary and it is subjected to a force F = (0,0, —10) N at point A(220,0,0) mm. No other forces nor couples act on the plate. Determine the acceleration of point A. (25 points) Cauflm'd A e éWJzoL—Mrprso -—- 30575 [1:ka :5 = 4'22” 4A Ir‘zzaz— 2y. 1470.5‘0 if 3/7 4 l :- 336/3933 PM“? i S f Xe: '22-: //0[mmJ 50 mm — _ _ .220 / 2 1;, IX .1” 0.0318 0.0219 0 MW”, __ - . , , [—5, 1y: —I,z]=[—0.0219 0.0357 0 ]kg—m2. 3): 5 5/7 4’ fl" 220 7r ”—0 I‘D ~11, —1zy 122 0 0 0.0674 3 r. ,4 Jf€$73 3 [mm J 5x % = 7: ’ W” 5‘“ “J IX— _.. I)...“ 6": 0.03/2P— 2.9 0,0736: 0,0049? [49“? fl: .ZZ’_. MXCQJC 0.03:7— 2‘4 0_ ”a : 0.0067 [WWI] 12’ I? ” M (”‘3"ch ~‘ 0'0574/v 2,4(0,0¢?2+o.//‘j= a, mygzkjmi/ (-7 aw?) Ix! :- rxj -—M,)CC¢/C = 0,02/7 — 2,1, (c.0990. ”j ._ 0.0040 [fly/”2.7 M ’ 0100?? 0,009 0 1 Z 21 - 0.009 @0067 a [65m] 0 0 0,0/5-9' Wax/1. S m f7? «4! mo 7‘7 'cm, 2.7 cgor .1— O 9 413x =2 O ,2 9 (LC ~ 0 —== Q =~ O @ (k > ‘2} & 2’? 4cz_ = “/0 3‘) at? ‘7‘43/7 [ J2. %M‘0Kfifl run/3’0“, N :: - «G K ( 0 / 0/ /0) (440/ /./ / o / ’ .49 0,00?! 0:009 ”(k ,_ /7 03004 0,0047 ‘ 0W M (ac/c? 4’} 2-0 _—_-> K’Lro / 4.49 0.009 0.06/26 ‘ 0( : /‘/ . 0:0667 I: . :- /72€/~a1n]1( X 0.0059? 0,009 0'00994’296 .rz /0-001/ 0.0067 9.09M a? — 0.02621” ”1,! 9, /, (x .2 / on? / S 5—. 6197/sz J moaooeme /\ ' 04/26/4437, 0-; , " er g%=gcffXEA:/0,0l'fil7j7‘ (0.0 /—g,7p/ 0-) r ( 0/ 0/ W522.) : (0' 0/ ”315) [AA/1;, M/ n 12. The radius of the pulley is R 2 1000 mm and its moment of inertia is I = 0.1 kg—mz. The mass m = 5 kg, and the spring constant is k = 135 N /m. The cable does not slip relative to the pulley. The coordinate x measures the displacement of the mass relative to the position in which the spring is unstretched. Determine a: as a function of time if c = 60 N—s/m and the system is released from rest with ac = 0. (25 points) Kine/Ma flag ,' l 1 7’ ., ; U l l — 70? =7 X“ A 5: i T' 7‘ Z 2 M? m)? fifl W‘s/Won“ “077161...v‘°1%e4°/0/ m£=m5vT#C); 2) T: m —,c‘x'»mx ,a,.,........... _. r Id=(kx~7j? I[-—-/é;‘_ :- kx -—. (raj—(ng—smgj (Mfg?) £4— cxyl—bx :Mf 5/ )ECai-éOx'f-WL‘wT/C =- 47.0J/ @ @mh‘caflm d'oll/z (efaAZKb/zau /)O€"%b“J X r 0, 353 [W] Gram,“ £9&L%:Gu 0% ‘/&€ Aoma7€atoead (7&1 ,‘ 2, ,7 +- //,’7(/? /— .86'4‘7-“0 “fliggf V 32,477 A : 52,4/7 942‘ 2, 7 SM c-z 0U- fica/ Q/P ‘ 6! 22,2 }/:’§.72é, gar—3.05 - 3.72.4 4' ,_ 3,034— X[H= 6/6 +666, . + 0.363 35$ 3:35“meg.“it???“ if” 7% 13155fo . , £7266- _ aosb X: —<’?.725 C/ e — 3036267 Im‘fi'a/ wuoé'fi baa ,' x(o) :0 :> 0/ +61] +0,3g3 =0 22(0) :0 => - 8.7269 P505 (”2/ r 0 @ ~0,363 / / / 0 *3,05 /.0/ —£’.726 ’303/ / / ~0 357 ‘5-726 0 3 /67 C = “W...“ = 45 :- —o,5“s”é ’ 4 4.676 5' 76 __ 47,7266“ _ 3,031- X=(+J= 0/77 a —— 0 we a 74 0353 13 (bonus problem). The slender bar of length l and mass m is pinned to the L—shaped bar at O. The L—shaped bar rotates about the vertical axis with a constant angular velocity we. Determine the value of we necessary for the bar to remain at a constant angle B relative to the vertical axis. (25 points) val/3%, .2.— VfiL’Lec X/SRM of 900106075253: (3, {S’fah’oua’g/ *1 2' W“‘""“’_\fi,g) Watch“ *0 +gc wank/7231 Vflf’kd'é 1 r \/ ska/F; Kg”?! ( at» [as/Wt Skew-a, ‘ h‘ ' a *2? ’3' *10/ '7, J >/g, airman W/ a—égwc«a( MM ”7%: Ear. ' .90; £- 77'61444‘ farv‘ha figs“, ”um/7Q; 7/001,“ >35];- 7‘0 kg 2.5 / 0 0 7 0 344% —- mfl 0 Wfl pint/5 Cage/Cart- 1/6/0619 Vt'ci‘m. ,‘ a) : (0, O, we) 4;: )99, $5/ A‘I? #3135 émr’ géafffitfl .' -—-—-‘--.... msfiWn- Ho / 4,. , v0 [ w Mabel? '1'; 8160/6 [baa/fez? techie/”4,6 a: (a? ”La/7’0““ . ' 4 _. (f. , x 2 MM" 6‘] _ Mal) / 4“ $73M. X, 3/ ,1} 0 4?, =W§fé +fm/3) a‘X/ : Q12 :0 W‘5 12(72&CA ,‘ (/0 f: ’14? ff _. a A - 0 - ”154/0 (6 7",g Jag/.24) Wm (M ”721) We came 7‘0 Wé Ab; x79. Axum,“ ~73“: (Le, Vfipc mam!“ 8757-2“; 740 £Va£¢¢afi quzfuk cf «(Eugefi 22L H, ./‘j (52% )v‘fxew WC 1/ ‘Lmle’o I OJ = 0 W0 ~C «f M 0 {3’ [Elwin/e) Sue/:QGJO fénzl £51kfl610 X (0 / “Cooflédo I SILL/3 do "- cJ éc CA4 ( 0 / 0 ) /2,"L[ Emfl / M .1. 2. (_ 2‘451‘01/5 [/0 +é-co’31/3‘2HU , O / 0) :- —/é'fl(6 +/§/C;3/5 ( %2[e+f—zh/3J) “W‘WW 0 H0 at _Z/-% r “ figm/Efftrl’é + é%fl(b+2€ Shufl) @ ...
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