TAM203_spring06_final_solutions - s o L o l/fO M3 TA name...

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Unformatted text preview: s o L o l/fO M3 TA name and section time: T&AM 203 Final Exam Friday May 19, 2006, 2-4:30 PM Draft May 15, 2006 5 problems, 25+ points each, and 90+ minutes. Please follow these directions to ease grading and to maximize your score. a) No calculators, books or notes allowed. A blank page for tentative scrap work is provided at the back. Ask for extra scrap paper if you need it. If you want to hand in extra sheets, put your name on each sheet and refer to that sheet in the problem book for the relevant problems. b) Full credit if x / . . o —>free body d1agrams<— are drawn whenever force, moment, llnear momentum, or angular mo- mentum balance are used; g 3 correct vector notation is used, when appropriate; T——> any dimensions, coordinates, variables and base vectors that you add are clearly defined; i all signs and directions are well defined with sketches and/ or words; —>| reasonable justification, enough to distinguish an informed answer from a guess, is given; you clearly state any reasonable assumptions if a problem seems Wig defined; a work is I. ) neat, II. ) clear, and III.) well organized; . your answers are TIDILY REDUCED (Don’t leave simplifiable algebraic expressions); El your answers are boxed in; and >> Matlab code, if asked for, is clear and correct. To ease grading and save space, your Matlab code can use shortcut notationlike “07 = 18” instead of, say, “theta‘ldot = 18”. You will be penalized, but not heavily, for minor syntax errors. 0) Substantial partial credit if your answer is in terms of well defined variables and you have not substi- tuted in the numerical values. Substantial partial credit if you reduce the problem to a clearly defined set of equations to solve. Problem 0: A Problem 1: [25 Problem is = EA +£ x EB/A +Xrez QB =21, +9.: X (n X rB/A) +Q >< rB/A +2a_2 >< Le, +amz Problem // 1 y E = W Problem 01 FA co to ’ [\D ‘3 Problem TOTAL: 0) -125 pt In order to not get -125 pomts you need to sign your name below. If you do not sign your name you get negative 125 points. Whether or not you sign, any violations of the pledge below will be fully prosecuted under the Cornell policies concerning academic integrity. I (Andy) have prosecuted many such cases and no student I have accused has ever been found innocent or had a decision reversed on appeal. ' Pledge: I realize that the regularly scheduled final might be identical to this test. No student taking the- late final should have any more foreknowledge of the test than have students taking this early final now. Between now and 3 PM Friday May 19 I promise not to discuss any aspect of this test with anyone, or within earshot of anyone, with the exception of TAM 203 staff and other TAM 203 students who also took this early test (assuming I know and recognize them and saw them taking this test). That is, there should be no posSible means by which any student in TAM 203 who is not taking the test with me now could learn by any direct or indirect way from me (for example though a third person overbearing me or reading my 7 email or through my parents talking to their friends etc) anything about this test. For example, and these are only examples, no-one will get in any direct or indirect way from me the answers to any of these questions: 2 ‘ 0 Did I think the test was easy or hard, fair or unfair? Was there a Matlab question on the test? ' Did the test have a statics problem, a problem from the lab, a problem involving pulleys, etc? How many questions were on the test? ' ' Were any formulas given on the test? Did the test include material from the final homework? How well did I think I did on the test? If anyone asks me any such questions or tries to get such information from me I will say that I am not allowed to even hint at the answers. If pressed further I will tell the person asking that such pressure is a violation of the rules of academic integrity. If pressed further I will tell 203 staff who was asking. If I know of any violations of this pledge I will promptly inform TAM 203 staff. By signing below I indicate that I understand and agree to the text above on this page. ‘ Signed (sign clearly and legibly) 1) (25 pt) A uniform square horizontal rigid plate ABCD has weight mg and is held in place by 6 negligible-mass rods. You need not write long vector formulas if you can confidently justify your answers without ‘ them. Find the tension in bar HD. ‘ filmy moweals glam/t 14C, Hie? Will-f twee Hml' doesn‘t {Moss Mme-oak f‘l': I3 the, Loom; alt/Lt, 4-6“,er M bar 6m“ slal'll- eclwrlflaf’rrt/M ZyMlc 5'. O 1:? ThalvlCI/Wiofln HD b6; O- 2) (25 pt) Make the usual assumptions about pulleys and the like. a) (20 pt) In terms of some or all of m and 9 find QB - j. That is, find the y component of the acceleration of point D. t b) (5 pt) Roughly speaking can you explain the answer to part (a). Hint: the answer to part (a) is a number multiplied by a symbol or symbols. That number is close to 2*" where n is an integer. Fbr example, if the answer to part (a) was 9m/g' (it isn’t) then we could say that answer was close to 23m/g and we would have n = 3. Use words and/or diagrams to rationalize the appropriate value of n from part (a). That is, somehow the mechanics has in it, approximately, n factors of two. Can you identify each one of these factors. [A very FED Fa “Messl‘” ” swam 15 /__,___._——-————-—~a 7252-411?“ *0, =>1Ts= 212 @FM‘,‘ 47}: mm L _ ‘Cmbmfms (D i@} -. I " !¢ l o MSCO$ €00, XDSQOS (70 \ Tho fi-wmrwar a? g.» is H»ch (gig/{Z— 6005:3430" :. ismzsoo ‘27, Tkm wc bind Hxa+ [fl=-é‘ . Th3 hv-w'Qbe/r4 '9‘” explmwi/r} ops {la/Haws: Frcrm Hm. kfmcmah’os we, llama Hnmt Maés :4 has ‘4 Hm ch awclmhw 0*? Mass 73> Qua/$5, A hash LG) +1‘Mos HM. lone/r431 i Hat/s aLm[na-F¢§I H~c Moth o'p’ MM 4:“ (Lafi- sys’rcM-x fie. tows-“sally we, Rave, H~c “(ciaka 0V4 D (w/ ngfllffiflrfit NQ$§>>PVHII¥¢3 ‘o-n‘ M‘bs A‘ 1.. arm/‘th +kc Wefflwr 0+ D is camm'ccl' by PL; rope bacaus¢ 0,? HM“ 6‘01)?“ g“ £00 7 . , =‘Vz. Z. Only MN:- HM3£ vie/{3144' cg}er by fave, ED ‘ bawv>¢n~e Hz] Futl'eY' ad» D. g. on“! ha”; Hui H/Nw‘mq/ rv'pi’é ED‘ is Mfrfod b7 r0190 AB \becmisc, 0—? [>01le ‘ V I so 90» vva MW 39 = tag/M :‘ egg/M; 28. How We, I Woe, anmhfais- FM hfncmec> 4N; End‘— u. Pom B m we m Mammy. a? mass/r um“; 0’9 FuHuj B- 5 Paer In»; a'nh.’ MU: Hur accdamhb’h 0’8 beeou’s‘: 0‘? ‘an FuHc1 D- 6. The; fi'éoMFO'nc/K‘F 'ogcoc‘ual—vbfln Mass D is hat? Hug, almS’gar-pc WIWPDEMV \o¢,ca\.v$( ¥kc le'Pe/l él‘n 30° :- 1/1. wa Hue, mace/lethm r9 " Ms D I”? “ Z‘ /\./ H‘mos bmmHe/Y‘ 'Hmrx fl' .wwlo‘ be. IX Viprec 1+; mohkltk’, smile/r I—Mm HWL since We map/am; Hg r‘ncrh‘bk 04" W95 D WP\<’-+¢«L\110\lmd HM»? wo—uld slow 4—H, syskflw Jo’wn mar—Pg. j V‘ 3) (25 pt) A person with mass m stands still at the back of a stationary boat with mass M. Then at t = 0 she walks the length L of the boat over time T according to the equation L(1 — cos(7rt/T)) HIP/b = ———2—— where mp/b is how far she has moved relative to the boat. Then for t > T she stands still in the front of the boat. a) b) 5 pt Make a plot of cc 5 vs t put t on the “3:” axis . Label key points on the “:13” and p/ “y” axes in terms of m7 M , Tand L. (10 pt) Make a plot of 921, vs 7:, labeling key points on the axis as for part (a). sub is the absolute position of the boat relative to a fixed reference frame. Assume the boat moves frictionlessly on the water. (5 pt) For parts (c & d) assume that the boat has friction with the water. The drag force is proportional to the boat speed: Fdrag = C'Ub- Eventually, as t —> 00, the boat speed tends to zero and the system comes to rest. What is the net impulse of the force of the water on the boat? That is, evaluate f0” deg dt (using basic mechanics principles this is a short calculation). (5 pt) What is {135(00)? That is, after all has come to rest how far will the boat have moVed? Show a lot of :3 vs .- x 4" P FED 44: SYQEH I . m MM.” is , filmeOWGT-‘t’di’. clear/y 4) (25 pt) A uniform ladder with mass m and length L slides on a slippery floor and against a. slippery wall. It is released from rest at angle 9. Immediately after release find the angular acceleration of the rod. Answer in terms of some or all of 0, g, L, m, i and j. Ifyou think you need Ia, I A or I 3 you can recall them or derive them or, for less credit, leave them in your final answer. H FED a? ladder AB ' LMg To Md ggA We use warmly 2 Hrs, 11m,» may 8,545 A i B Okra msHarthl 4123‘ Nova w‘ofia +i/xellr‘;"/r"esp echVe, \AMLUS- + 9‘- X—E-B/A + fixkg ><\ CB/A 1 g—XEB/A= Q l: 5: (Loose ‘ .k Lam-6f) : 4&9sz6 €+L5coaéf 'K'X (L we + Lsm ' : ' z A 9 case a v Lé-Zsme ? 2 /.\ l " '\ "2- I S‘b"> 0:" 0—,,- Layne—Le case => OLA = Lé’sine + Lézeose ‘E X \E 2< . l? i H r“ (B' 10 x (B. Thurs we, have; C. ,0 ‘ ‘1 ’.\‘ ’ M £61 : (Leéyne+ €103 e>b i" (’ l‘fst/bk 4— P‘J33m3 M5 baoLqMo AMB me gar -ML A 1" L A ~ 1 ’ _.—%’wbe k- : T'ZML IL 4' (im6L—'%Sln@f)x {W‘rgzgf 0A __ m 2" I \ “ NOTE: Th; him: 5 § k *5 " : T\',2:ML e L wife W'H‘ ézm‘te" 0u¥ Alive/r H"; ‘ cross— vadvo‘m E “:37 = ‘ ig- coae 35’) ’\ 2L :3) 3i '— ‘ {we “ 01C (/ho'r'QLi . Se+ T4 V = 5' (wash/Hr) and drefcrevx H“o»{~<.. w. r“ 4'. +UM&, ~ ’ J ' e.) (25 pt) A spool (like the mOVie’Heat Matment of Aluminum shoWn‘in lecture), without-lea: radius R rolls without slip, On a flat horizontal surface. The is at a. radius r and isrbeing pulled «filth a horizontal force F. At the moment in question the Velocity of the middle of the 'spool is vi. The mass of the spoolis m and its moment "ofzinertia about its center of mass'is Ia. What is the acceleration of point A on the spoolwhichis, at the instant in question, touching the ground. ‘ Answerinterm ofsomeorallofm,I¢,r,'R,g,vand F.V ‘ - ...
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TAM203_spring06_final_solutions - s o L o l/fO M3 TA name...

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