ENB211-T1-S1-2010

ENB211-T1-S1-2010 - GUT I|II||||||||I|I Surname Given...

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Unformatted text preview: GUT I|II||||||||I|I Surname Given Name/s III-I..- Examination Paper SEMESTER: FIRST SEMESTER EXAMINATIONS 2010 UNIT: EN8211 DYNAMICS — THEORY 1 DURATION OF (EXAMINATION: PERUSAL: 10 MINUTES WORKING: 2HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS EQUATION SHEETS - THREE (3) PAGES ATTACHED GRAPH PAPER MM - ONE (1) PER STUDENT EXAMINATION MATERIAL SUPPLIED BY THE STUDENT: WRITING IMPLEMENTS CALCULATORS - NON PROGRAMMABLE INSTRUCTIONS TO STUDENTS: Students are prohibited from having mobile phones or any other device capable of communicating information (either verbal or written) in their possession during the examination NO NOTES MAY BE MADE DURING PERUSAL TIME FIVE (5) QUESTIONS ONLY ARE TO BE ATTEMPTED ALL QUESTIONS ARE OF EQUAL VALUE Queensland University of Technology GUT Gardens Point QUESTION 1 The vertical slotted guide is moving with a constant velocity 5c 2 150mm/ S during the interval of motion from x = —80mm to x = 80mm. The equation of the fixed parabolic slot is 2 y=100[l— x 10000 calculate the n- and t- components of acceleration of the pin P, which is confined to move in the parabolic slot. From these results, determine the radius of curvature p of the path at this position. J, in which the units of x and y are all mm. For the instant when x = 60 mm, (20 marks) QUESTION 2 The 2-kg sphere is projected horizontally with a velocity of 10 m/s against the 10-kg carriage which is backed up by the spring with stiffness if 1600 N/m. The carriage is initially at rest with the spring uncompressed. If the coefficient of restitution is 0.6, the rebound angle (9: 86 °, calculate: 1) the rebound velocity v'; 2) the maxium travel 5 of the carriage after impact. (20 marks) )< \mfl e 5 k = 1600 NI‘IIL \ .“ if: -.\y cont/... ENBZI 1T1.101 QUESTION 3 The collar has a mass of 2-kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50-N force. Calculate the velocity v of the collar as it passes position B. (20 marks) 1.5 m QUESTION 4 The pin A of the hinged link AC is confined to move in the rotating slot of link 0D. The angular velocity of 0D is w = 2 rad/s clockwise and is constant for the interval of motion concerned. For the position where 6 = 45° with AC horizontal, determine: 1) angular velocity of A C, wAC; and 2) the velocity of pin A. (20 marks) cont/... ENB211T1.101 QUESTION 5 The bicyclist applies the brakes as he descends the 10° incline. The combined centre of mass of the rider and bicycle is at G. A critical deceleration ac will cause the dangerous condition of tipping about the front wheel A, which means the friction and normal forces at point B are zero. Calculate: 1) the critical deceleration, ac ; and 2) the fiiction force at point A in this critical situation. (20 marks) QUESTION 6 A 90-kg man stands at the end of a diving board and causes a vertical oscillation which is observed to have a period of 0.6 s. If the mass of the board can be neglected, calculate 1) the static deflection 6“ at the end of the board; and 2) the frequency for the vertical oscillation, if the stiffness of the diving board is reduced to the half and other conditions keep the same. (20 marks) END OF PAPER ENBleT1.lOl ENB211 Dynamics Equation Sheet Rectilinear Motion (displ S, vel v, accn a) ds v — — = s dt dv . d 2s .. a :: —— = v = —— = S dt dtz vdv = ads or sds = S'ds (area under vel vs time curve) 12 AV 2 v2 —v1 = Jadt fl (area under accn vs time curve) v2 32 Ivdv = lads = %(v22 ~ v12) v1 31 (area under accn vs displ curve) Equations of Motion For constant accn. .. v=v0+at v2 ~v02 = 2a(s —s0) s = s0 +v0t+%at2 For accn as a fn of time a=f(t). .. v=v0 + Mad: f s 2 s0 + Ivdt 0 For accn as a fn ofvel a=f(v)... v vdv s=%+; V0 f(v) For accn and vel as a fns of displ a=f(s), v=g(s)... ENB211T1.101 Plane Curvilinear Motion (Position 5, vel y, accn g) Rectangular (x—y) Coordinate Systems K = xi + yi’ r=' r=xi+y1 v = M = 1/562 +332 = w[12x2 +vy2 (magnitude of vel vector) v 1 tan 6 = —) VX (angle of vel vector) 2 = ‘2 = Z = 561' + H a = |c_z| = 4562 +)'}2 = 1(a: +ay2 (magnitude of accn vector) av tana = —' a X (angle of accn vector) Projectile Motion ax = O av = g = 9.81 m.s'2 vx =vx 0 =vcost9 vy =vy‘0 +gt = vsinH—gt x=x0+vx 0t 2 y=y0 +vy‘Ot—égt VZ—V y yo 2 -2g(y-yo} Normal and Tangential Coordinates 2 = v9, = pfie, (tangential vel) Normal and Tangential Coordinates (cont...) :2, = fig. 2 V a=—e _n + (accn vector) 2 v an 2 — : : Vfl p (normal accn) a, = 1'1 = s“ (tangential accn) 2 2 a = la] 2 11a" + a, (magnitude of accn) Circular motion in n-t coordinates... v = r0 2 v . . a" =—=r62 =vt9 r at = 1'1 = r6 Polar Coordinates E = V9 .e:r : egg £0 = _6€r r = r'er + r923 g = (i; — r192 _,. + (re + 2mg Circular motion in polar coordinates... v, = 0 v0 = r6 or = —r492 a = r49 HEhQ Relative Motion EA = KB + XA/B 2A = 9.3 +QA/B Newton ’s 2"" Law F = mg iF=mg Rectangular coOrdinates EFX =max ENB211T1.101 (ii) Newton ’s 2"" Law (cont. . .) Z v = may 2 2 2 a=1lax +ay +az - — 2 F2+2F2 _ 2: +2 y z n-t coords. .. E] : ma” 2F, = mar Polar coords. .. EFF =ma,. 2F67 = mag Work & Energy U = Lid: Vg = mgh (close to earth) 2 V = — ng (in orbit) g r Ve = jkxdx = §kx2 0 UI_2 =AT+AVg +AVe U,_2 +Tl +Vg1+Ve1= T2 +ng +Ve2 Impulse and Momentum Linear... Q = my ’2 25:9 [Eggnog —G, =AG ‘1 Conservation of linear momentum... Ag = 0 or Q = 92 Angular... £0 = r X my Hx = m(vzy—vyz), Hy = m(vxz—vzx) (iii) Impulse and Momentum (cont...) (Motion Relative to Rotating Axes) Conservation of angular momentum... 1‘ = (0 xi = a) x j Ago :0 or £01 2&02 _ — _ EA _ EB +er+yrel Impact = + a) X V mlv1+ mzvz = mlvll +m2v2' t XY t xy v;_vl' . 9A =eB+Q><r+Q>< QX£)+2Q><2,.ez+g,.ez e = ‘ = (we; separat10n)+(v,.e/ approach) V1”V2 QA =91) +2QX21‘el +9413] Constant Velocity Translating Systems Plane Kinetics of Rigid Bodies 2E : "I’lgrel Urel = ATrel = mg ZMG = Ia 25 = Qt, 2MP = la + mad =Agrel MB zflBrel 2MP =ipg+8xmgP I = mk2 (radius of gyration) 1 = §mr2 (solid circular disk) E = mg ‘ 2 ( 1 b ) 1 2 (ml )/ 12 solid rectangu ar ar r G U12 +T1 +Vg1+Ve1 =T2 +ng +Ve2 2 - _1. L . . . T ‘ 2 mv + 2 2m! Bil Vibration and T lme Response Q = my 2 fl Free vibration without damping: 56 + mix = 0 M0 = £0 MG = flG where (on = 1lk/m ; solution: flp :flc‘l’gme x=Acosoant+Bsinwnt 2%!) = G + p x ma Free vibration with damping: _ 5€+2§wnic+wfx=0 where g =c/(2mcon); . . o o . - < : : _(w"' . Plane Kinematics of Rigid Bodies SOlutlon When 4’ 1 x ce “(wit H”) or d6 . dw x = {A3 cos wdt + A4 sin codt} 64”“ where a) = — : ’ a : —-— : dt dt cod = a)" 1—? mdw = “d6 v Undamped forced vibration under harmonic v=a)r, v=co><r . . .. , F. - 2 — * — excztatton: x + (0pc = —°—sm cut ; solutlon: _ 2 _ V _ m an ‘60 r—Ta 9n ‘QX 0X5) x=xc+xp where xc=Asinwnt+Bcoswnt and = _ . . . F0 /k at ar Qr—QXE x =Xsmwt,1nwhlchX=—————, ” 1— (w/w")’ (Relative Motion: nonrotating refl axes) Damped forced vibration under harmonic = . . ’1 F . 2A 33 +151”? exatation X+2§wnx+ng=—°smwt; _ _ m vA/B_wr YA/B ‘QXK solution when (< 1: x = x6 +xp where QA 2 QB +QA/B x[ 2 Ce““’"’ sin(a)dt +911) and x = Xsin(a)t—¢) , 9A :93 +(flA/B)» +(aA/B)t ij 0 in which X = m) 2 ___—____ (“A/B)" = vA/B /V = 602” {[1—(w/wn)2]2 +[24’w/wn]2}l ‘ (EA/3),, =Qx (OX5) 2420/0)" (CIA/13);: {)A/B :6” (9/1/19), =QXE 1—(co/con)2 ENB211T1.101 ...
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This note was uploaded on 11/03/2011 for the course EN 40 taught by Professor Mcgregor during the Three '10 term at Queensland Tech.

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ENB211-T1-S1-2010 - GUT I|II||||||||I|I Surname Given...

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