Final-Exam-Solution

Final-Exam-Solution - ME 274: Basic Mechanics II Fall 2011...

Info iconThis preview shows pages 1–18. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
Background image of page 11

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 12
Background image of page 13

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 14
Background image of page 15

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 16
Background image of page 17

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 18
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ME 274: Basic Mechanics II Fall 2011 D ecember 14, 2011 —-—--——-———-—————.—_.__._____________________ Instructions: Final Exam December 14, 2011 0 You have 120 minutes to complete this exam. 0 This is a closed—book, closed—notes exam. 0 You are allowed to use a calculator during the exam. 0 Usage of mobile phones and other electronic communication devices is NOT per~ mitted during the exam. Helpful Equations are on page two. Name: Problem Score Final Exam Page 1 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 —-———————_____________— Helpful Equations: Kinematics: 173 = 77A + (fig/A)”; + 03 X FB/A (EB : 5A + (EB/ALB, + 07 x FB/A + 20 X (fig/A)”; + Q X [LU >< FB/A] N ewton~Euler2 ZF’G = mag 21% : 1G5: Energy and Momentum: e : _{UB71,2"UA7L2 anl ’UAnl I12 2 AZAdt 2 Elm — [17/11 HA I [Gal‘l‘mT—‘lg/A X 27G Vibrations: _. 1g -: C 2 _ 2 £005) 2 e‘C*‘”"*t (S >!< sin(wdt) + C’ * c03(wdt)) Final Exam Page 2 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 M Problem 1 (20 points): Given: A pulley has a mass moment of inertia I0 and rotates around fixed point 0 located at the center of the pulle . The pulley is attached as shown to two springs y with stifiness [£1 and k2, respectively, and a hanging block of mass m. Neglect the effects of gravity. Find: Using the variables shown (7‘, k1, k2, m) derive the equation of motion in terms of 6 and determine the natural frequency for the system. Final Exam Page 3 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 W This page is for extra work related to Problem 1. “OX-TX 5 x (‘31?) 1 m 47‘) 74“ "‘T ‘J’ T ‘3 -" mbé 4F M Final Exam Page 4 0f 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 M Problem 2 (15 points): The motion of a system may be described by the following equation: (§M+%gi)zi+0d:+%Kx=0 The parameters of the system are given as M = 2 kg, [cm 2 4 kg/mz, R = 1 m, C: 3 N—s/m, K = 50 N/m Complete the following: a) Calculate the undamped natural frequency of the system b) Calculate the damping ratio 0) Using the initial conditions 30(0) = O and = 5/ 3, determine xc(t) d) At what value of C' Will the system stop exhibiting underdamped oscillations? 1 “J7. «5'1 K 4,". r t) J12. p E" - w: n; U» i (“ \‘ a: fin” \ g n A 'ifjm“‘ l f, 5, 1.1,» i ‘ 2; KM ‘0 l ,4 / O .A, a, a (v i ,7: ,. i7. \ fillings <9 flip/(,0) t L 7‘ " ( f. Final Exam N / , (M Page 5 of 17 ~ 1 a b x: f Mb :3 \ , DO \ 3 ‘7 ’\J i x N we 2 (00.64 «(Ickb : 6:- O,'@@t OK) “Iowa [30 or 7’2] fob; vyfi" mayday/arid) . _§~.. - [3‘05 ZW”? 0:0 N”: fez: Xfi—a Cst: 30%}. ,ygujl‘géJ ('iclfim‘la L3 F * {lgfi’dz “l L053 [W t: (O: i ’ ME 274: Basic Mechanics II Fall 2011 December 14, 2011 W Problem 3 (20 points): A given system has a forced—response equation of motion given by: (3M + 1135-) it + §Kx = Focos(wt) The parameters of the system are given as M = 2 kg, Jam 2 4 kg/mz, R = 1 m, K = 50 N/m Complete the following: a) Determine the particular solution, 331,05), to the given equation of motion, leaving F0 and w as variables. b) Using the initial conditions 55(0) 2 0 and = 10, write an expression describing the system general response, A6 m. laminar» \J Sights" (Ll >i<i & 2‘3; ‘x (“132301219 » .i ‘\ 7:1 1’ A _,; F: mi 2 i m <10.» M) f’ f ‘I, ' a :1 A .r ‘ ">- f—rr gu§:\\ “ch > W Rt.) QAUAHQ> ‘ N20 U)” ‘ J ‘ '\ f; a '\ 73"“ ‘ ‘1 I E‘s A (,7 (—1 (’I’fif'la 4'“? v I (“L/(3:? wq l‘tif‘bflmiul "i Lifimfluv‘z i WC} ' *3) ’ V f, i v. v v’ a \_.-‘ kn g o :1 A O t3 -‘ [if to? Page 6 of 17 ME 274: Basic Mechanics II Fall 2011 This page is for extra work related to Problem 3. L, \XCN} ,a v ‘9 x. I . Z W“ :-’ 2‘ g «H7; (from. bebum B “O dcxwlngifib : may ‘l' f: O VXCQ) .— €710“: akQfi) + 3m (,%£) + Cwafitb .4 December 14, 2011 Ixct) : g ‘5lmé%‘t 'l' («55(0th ‘7‘ Cam W ugiwa IL :> 1L X? Gum/L3) . i O r 3 r ( 3(0): 0 : 3/4293 # Coy/(,0) 4 flaiflo> C): ~ 2:20: W V l T 31A 42 ‘ my 3% mafia) “row M A - . r, . h -><(o):10< %>;9Ao>* O C) 3v 9%003 fl (0 6,05 05 1 ,, E a (:g_; Cog ‘1 760% =- (05m ’31) + qw-z_,zg- v ; + c, «any Final Exam Page 7 0f 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 m Problem 4 (5 points): The kinematics of point P are described in polar coordinates using variables 7" and 6, as shown below. For a given position with 7' : 0.5 In and «9 = 2 radians, the velocity and acceleration for P are known to be: For this position (circle correct answer): the speed of P is increasing b) the speed of P is decreasing c) the speed of P is not changing Provide a justification for your answer, which may include words and/ or a sketch. “2 r3 WJ>, 5“ ,v) { P, Cc) L” F7 AL .0 K v J fl {/1 if J . V vs I- / i J i ’ i i v , \c l\ ‘ I: {may {I ’ ‘}\f1w «9x. // . Pr {I ,‘x A ‘ W A I all?) CC) 5“? ii? (“-1 {Vb i5 41d» i l 3" E’ ( .J ‘ ., _ ‘A ) $413145; .z/li i‘QGflpfl {1;} H.132 If a IWF': I V M \ A I W ‘ ‘ 4 A, .V .\ ‘ S J“ w u o i v r J , a—m P (NV ‘0 V I Final Exam Page 8 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 Problem 5 (10 points): Bar AB is released from rest with point A in contact with a smooth, horizontal surface. Describe in words or with a sketch the path of the center of mass G of the bar after the bar is released. You need to include an accurate free body diagram in order to receive full credit on this problem. Final Exam Page 9 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 Problem 6 (10 points): Particle P moves on the top horizontal surface of a rotating disk, which spins about its vertical axis at a rate of Q = 10 rad / s in the counterclockwise direction. The position of P is described in terms of the Cartesian coordinates (mp, yp) where the xyz axis is fixed to the disk at center 0. When (ring/p) = (8,6) ft, the absolute velocity of P (that is, the velocity as observed by a stationary observer) is known to be Up = (Mi—303') ft / sec. For this position, determine the velocity of P as seen by an observer standing at point 0. State your answer as a vector. . :1 4 0“"; t " TOP VIEW é \( / K c «< 1‘; " m 9" Li :4 :35 f m Ij\ 1“ a» semi at A -~‘ 3‘ i ‘ v‘ Page 10 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 201 1 Problem 7 (15 points): The following questions are conceptual. No partial credit will be given on these prob— lems. Particles A and B collide in the horizontal plane with initial velocities ml and um, as shown below at left. A A l‘ ‘21] B B y VB] x Circle ALL correct responses below. Some problems may have more than one correct response . Part A {5 pts) For system AB during impact: I, 51a linear momentum in the x—direction is conserved ) linear momentum in the y—direction is conserved | linear momentum in the n—direction is conserved linear momentum in the t—direction is conserved e) none of the above 1 i i l b C «.1 Part B ( 5 pts) For system A during impact: a) linear momentum in the :c—direction is conserved b) linear momentum in the y—direction is conserved c) linear momentum in the n—direction is conserved (@inear momentum in the t—direction is conserved e) none of the above Final Exam Page 11 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 Part C ( 5 pts) For system B during impact: a) linear momentum in the m—direction is conserved ) ‘ 3.x b) linear momentum in the y—direction is conserved aix‘lfi” {‘x f‘al (17’! ‘62, m “x c linear momentum in the n—direction is conserved ,. n » dljjinear momentum in the t—direction is conserved ‘l'i‘rfi/v [\2 ) e) none of the above Final Exam Page 12 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 M Problem 8 (25 points): The following questions are conceptual. No partial credit will be given on these prob— lems. Part A ( 5 pts) The mechanism shown below is positioned such that links AB and DE are vertical and ED is horizontal. If at the instant shown link DE is rotating counterclockwise: a link ED is rotating counterclockwise i, p b nk BD has zero angular speed at this instant c) link ED is rotating clockwise Part B (5 pts) A homogeneous disk of mass m is acted upon by a torque T at its center 0. Assume that the disk rolls without slipping on a horizontal surface. Circle the answer below that most accurately describes the friction force acting on the disk by the surface on which it rolls. (fijhe friction force acts to the right b) The friction force acts to the left 0) The friction force is zero d) The direction of the friction force cannot be determined without knowing the coefficient of friction. Final Exam Page 13 of 17 ME 274: Basic Mechanics II Fall 2011 December 14, 2011 Part C {5 pts) Shown below left is a thin ring (of mass m, outer radius R, and center A) that is released from rest on a rough incline. Shown below right is a homogeneous disk (of mass m, outer radius R, and center B) that is released from rest on a rough incline. Both ring and disk roll without slipping on their respective inclines. Let 0A2 and U32 represent the speeds of A and B, respectively, after both A and B have dropped through a vertical distance of H. Circle the answer below that most accurately describes the relative magnitudes Of ’UAg and U322 a) UA2 > UB2 b UA2 = UB2 "C /A2 < 7132 / ‘ 7: , ,, \ t0 f” (int?) {\e V}! p 7/ » 2 ‘ 5' VI ‘3" 3;"; kl. (7 . Ci“: 'Ci"i{.M/‘\ k)“: r A“% l A i 044“ (A) J/ i i i i 4"‘4 4.,- x ~ -» x I» "(- r ‘ ‘ ‘ ‘fk, ’ i ; Y“; (151‘ (JR; flair 2,5, (1)7«igfiwgfii‘rngi {ah MW if}? «if {‘15 \li ) s c Q , H K . us A l"; ‘ l I i t , ” n.1,, LI“, . e I , in {Iii l’lg'lfi: if" {,2 M (Wang r ‘ ({‘fi:,'>jf;\w~i:\~) yxAGQ; Hm my 1:; Tm“ ' 4 l 4' '\ ’ (A “)11‘” K an” Vlr‘t""‘:?g 1‘ la; w 111;. 5 {.1 lack LS»: Final Exam Page 14 of 17 3, .l/ ME 274: Basic Mechanics 11 Fall 2011 December 14, 2011 Part D {5 pts) A particle connected to ground with a massless, flexible cord at A is released from rest at position 1. At position 2, the cord has contacted a fixed peg at B and swung to an angle of 900 > 6 > 0 from the vertical with the particle still moving at this position. Answer the following TRUE / FALSE questions about the motion of the particle in going from position 1 to position 2. rm ! liga’fifl Energy for the particle is conserved T UE o Linear momentum in the vertical direction for the par— / ‘ ticle is conserved TRUE 0r Linear momentum in the horizontal direction for the ; particle is conserved 5 r.@ Angular momentum about A for the particle is con— " served The \Cimr.(>thi‘”i"‘:’> «57‘s. monarch *‘t'lAn, ‘A\<:£\ (4“ 4 MM r. a" l i I 3 i4 R l QC UL? H‘- R ‘1 1) 1‘. I r \s’ o W mi, v ‘ g “i Era “A f T we" (:“Qi/kfd‘ - ' ‘ i l e x w -. rm imam , A \‘i I, , i 2 . l \r 1‘ x/jk ) g l ‘a/lfiwz - \ i k, u 31ciotme, ill/iii til/Lair :5 l ~’ ~“’ ' ~ 5 3... “I . __,. -v_—-s (‘2 . i < 4 ‘ I v i C, ‘\ f, [N FVHgé/Q Vt”? ,7.) a , ,, r p v \ A, _ r x V ‘4‘ an [3 M(_)[email protected]<iw - I , \J [\afigflmé‘ivfigiy» (Lg \i\~ a» {AA (3 ,, t n‘ (1' if . ‘ . w: x ‘ '3 m‘injf 13%;) Ff VaFi um for {we}; jimg firs i. 1:5, (5mg QM (a; a 2 Oct “ M / \ Final Exam age 15 of 17 ‘7 ME 274: Basic Mechanics 11 Fall 2011 December 14, 2011 Part E (2.5 points) Given: A block slides down a rough incline With a speed of no. Gravity acts on the block in the vertical direction, as shown. Consider the work, U Jam-c, done by the friction force on the block. Circle the item below that most closely describes U Jam. :ilfric < 0 Ufric : 0 C) U fric > 0 d) More information is needed to determine the Sign of U We. 3 Final Exam Page 16 of 17 ME 274: Basic Mechanics ll Fall 2011 December 14, 2011 Part F {2.5 points} Given: A disk rolls Without slipping down an incline with a speed of 00. Gravity acts on the disk in the vertical direction, as shown. Consider the work, Ufm-C, done by the friction force on the disk. Circle the item below that most closely describes U We. a) Ufm'c < 0 fric : O C) U fm'c > 0 d) More information is needed to determine the sign of U We. g Final Exam Page 17 of 17 ...
View Full Document

This document was uploaded on 12/25/2011.

Page1 / 18

Final-Exam-Solution - ME 274: Basic Mechanics II Fall 2011...

This preview shows document pages 1 - 18. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online