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Exam3 - EM311M Dynamics Exam 3 Monday Dec 4 2006 6:00-8:00...

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Unformatted text preview: EM311M - Dynamics Exam 3 Monday, Dec 4, 2006, 6:00-8:00 p.m., WEL 1.316 1. Use polar coordinates to compute the moment of inertia for a homogeneous quadrant of a Circle with respect to axis x. (5 points) (Fara mcfu ‘Za/I'ou g 2<= I‘C/u’.) Q 1) < r’< 73/ :7 c r' 51‘” C? 0< Q < L7“ 7?“ L a. 2—; Q 2, .— 2 2 fly — f f /‘ 24/1 6 rc/rC/Q . a O 0 am 29 :/-Z%tr72(9~ f' _. 31'an c ”66939: = f j I‘396r 1:er are: {I O f) .24— z, , 4 , / , . 2,? . , . f8n9~JJQ ’jgfitze; =3? (Ziggwékuk: Z. 0 § ’L l 33 L : €772, _ f 77'»? i r: , h” @ /6 ‘7 " 4‘ 2. Determine the transformation matrix from system Omyz to system O’x’y'z’. (5 points) 2/ - / 0 0 ’0 “/ (7 3. Define the inertia tensor (matrix) and explain its relation to the angular momentum of a rigid body. (5 points) fffasré’) —f§xj —f§xa— I» 13% I”). ’1‘ *- _ r.” 3 71.0 * flit} ff“ O N 4: J? éfmw jEQew) 4 The objects consist of identical 3— it 10— lb bars welded togther. If they are released from rest in the position shown, which one will havea greater angular acceleration ? (5 points) imzmx—wi.‘ 1 CL fcdrHc @ f’hamfulzl qt I‘Lzfitflh .1; WC 657/(97ruf— ‘ fl #1:: +1429 Shut/24mg, mcwdl'fii/ ,éJ ”(at V5? d’flfi“fi‘91 01 CflF/fi-i Q/ “raw ) noanwuf He 0/ Alf/{far W 44-505 (fa/cc 0/ run)»: /: waraf 243:0 fa '6’; 6764’5’1Nc" . ‘ / V‘tcaf/ly‘jfi X = O 4'“ 604& 604647 . 5. Derive the equation of angular motion for a rigid body undergoing an arbitrary motion, in the body-fitted system of coordinates (5 points) m=ge . C“ Fifxfgcdif/C 0% GAE/fU-Qfll— Z l/uf/yag /JC/ W/ MOr’L/CLC7ZK:(M . éc flc A <f£cw)=r49c Tk=émézsé'j' 5— 5/5176ng 6. The masses of the slender bar and the crate are 9 kg and 36 kg, respectively. The crate rests on a smooth horizontal surface. If the system is stationary at the instant shown and a counterclockwise couple M = 300 N—m is applied to the bar, What is the resulting acceleration of the crate ? (25 points) rol- U ; ... 2,14: we 25—mwa5fi 50.3/4 5w!) sh‘fuxh'aj "LO/o (a J /50( : — 46.15—[6‘04/00212/ +800 /23 0< = 255,695 :5) m = 2,047 [325:] 48X :~2,0£[%] w Ic = [If/mug = if) z 4 :05g7fkjmaj 7. The l—kg sphere A is moving at 2 m/s when it strikes the end of the 2-kg stationary slender bar B. If the velocity of the sphere after the impact is 0.8 m/s to the right, what is the coefficient of restitution? (25 points) Commz/arhbn, 2g ZC'may 2/??OIl/ffn/2LM 414 X;- C/(mg-fzcmfii “Age Sysficqfl ,_ ,) Mmmfl~wwnmwwemg ~«,v......-m “gymbmh , , €245,492.» 4%“ 15:.“ ”or? +Z‘éx => “93 W [s] @ COW/7 «2/ Am We y—i‘m 7‘01 W [W t‘vawm.mm-q—tm . . [2+0 l\ _ ». “.7, “was...” ”N...“ . W, "m one-,1. {swanawa—M—wm‘ . MW 1% 4/5?!» 0 a .2 we”, “ ~=> V57 ==- 0 @ comm-Mm. 4?! age/m ’”_O“?f€:flfféi:..m"a 1L 5" @“f “waif mmflffioaw‘iéé é‘wfi’zfi‘ 450/7» ' a/Im. Q/ffib afl‘n O: .225‘) ._ mué).0,g :: 0.4:»,6764) '—0,7(/ ___:> mitt/{m /, 08 z:— mqu @ .~‘ warm-«3‘1 L— s {0, 01/09) . 56‘ ., r- .— mwsa a - 55+ 93 we WW (Mm; :(Ag4cp/o/0) (0169:3550) COL” [ghee-Ken}; O “76555 fififiéfl’“ I“ @ k §= 71,9- ’3 [7:37 8. The mass of the Z—shape homogeneous slender bar shown below is 12 kg. 61) 0 Show that the center of mass of the bar coincides with the origin of the system of coordinates shown. é) 0 Determine the full tensor of inertia of the bar in system xyz (watch for symmetries!) c) o A single force F = (20, 0, 40)[N] is applied at point (1,1,0)[m]. Determine the accel— eration of the center of mass and the angular acceleration with respect to an inertial frame. 4) 0 Determine acceleration of point (—1, —1,0) with respect to the same frame of reference. (25 points) x (C 00) ' fic=€*F = (20,0 (/02 (‘40, ~40/ ~40} [WV/4t] VU‘ .r. «a? 7 ‘2’ [— 3%] éa '3 40 J7 Sw—/ZO 200 M a z , - ’5 <19 2 90 ,H 9‘ ,— /“3 ‘5’0/“ —?0+/2/0 ___ 10 :2— 57/1 44] 9 ” 7 7 M30 = f /z,{ 40‘ aw, 46%):[115 0, (/0) 6* ~ m @ Q r 7 l w QC}: 355 .,0 d) QA:§’L +o<xCAv46dxqui4 9 (bonus). A satellite can be modeled as an 800—kg cylinder 4 In in length and 2 m in diameter. If the nutation angle is 6 = 20° and the spin rate n3 is one revolution per second, What is the satellite’s precession rate «,0 in revolutions per second ? (25 points) g3}, : (0, 0/ Y) “1 X12": l’j’ £3 an a 0 — 2:12. 6 o n m, a)“ /L\u Va 9- Cd, 5 0 I 0' q :3 O 94 ' M 91L. 6 0 an a 17,, m 5, 7; A; f‘O‘FQ'ffou O; x .7 .3; MIL A I 3/ ?.) $ 5; ) . "< M 6, ow f ’5‘! an 9 0 , >1“ 9 x 7 X 77’an 4 {o 'LMQIVIOfl Wk 5: 0 / O ”C ¥,3,?,-—-aX//9« 61'46‘ a 603:9) yilfujnfa-L Vt (om/{J 071 ‘Xy ?— 4L k7}- 3:? g g ' r? .7: (Q); 7‘6le :7: /‘ 5’?” 67 ’5‘! / 0/ C203 7/ ) afigpmhm g/C (am AZ 0/ ‘MC JakKZLVLC, 4'14. 5’7 9" e=§§+ M , _‘,———~ /L7j 2 (—e'nfiyfi/o/ mfifvéfolj fin , 0/ Saw/K: éivwn/ m 075/014,: 1* _... . ,(y :géc% +£M§gj =9 0 O c» Q? xH W a N Q? Q 29 W. V, x + CD n $1 ‘9 x \g. Q RH %. 35 V. Z X (0 ”IX m6;-zl.,e9 7/ 2"+ 1% 60398221 6 7; + @5126 y? / 5” :’ O @ Ca,,/m/4/7m 07; :2; x: rune 0<4~<R Ar ‘7 FSt‘u'e 0<6<2//—- g 2- & 0<&-<..A, 27"", 5g 11/ ’ :2;=:f6/i22) = 5 of / M—zam ~22) New-4M9— 0 o Z7'OC2./ ‘9‘? v” =W/ a q R, . 3 ; _/_‘___ ?_ 519(29- o 0 f) o _ 77’2ch 600w: #24],th 9 3731:6“5 /"003w' 2/ K217i» b 2,". L ’/ ,4, f/ f2zpo£rc€9a¢+ = 2—:- 9/ 35/ 00 a a o 5 o : 772243 3 (w. 5‘ ”72%, 2 L FQW+ $22267_ M 334—151 13‘ =pey/ ‘1 3, - 4 5 fl /6 I; = 3’00 27+ ‘5“) = 9467fl5¢m17 Ia = 35%?” = $00001W= 700114375413 Momtuf- ml" at)”: K figural? -/&w¢vlh. C I; = 7457‘“ 8300-42,: [867 [fig/Hi] “r51 ...
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