EGM1181 may2010 - INTI lNTERNATlONAL UNIVERSITY LAUREATE...

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Unformatted text preview: INTI lNTERNATlONAL UNIVERSITY LAUREATE INTERNATIONAL UNIVERSITIES FINAL Examination Paper (COVER PAGE) Session : May 20 i 0 Programme : Diploma in Mechanical Engineering Course : EGMllSl : Engineering Dynamics Date of Examination : July 15, 2010 Time : 11.00 am — 1.00 pm Reading Time 2 Nil Duration : 2 Hours Special Instructions This paper consists of SIX (6) questions. Answer any FOUR (4! questions in the booklet provided. All questions carry equal marks. Materials permitted Non-programmable Calculators Materials provided Nil Examiner(s) : Mr. Ragunathan Sivalingam Moderator : Associate Professor Dr Megat Mohd Hamdan bin Megat Ahmad T his paper consists of 7 printed pages, including the cover page. EGM1181(F)/Page1 of6 INTI INTERNATIONAL UNIVERSITY DIPLOMA IN MECHANICAL ENGINEERING PROGRAMME EGMI 181 : ENGINEERING DYNAMICS FINAL EXAMINATION : MAY 2010 SESSION Instructions: This paper consists of SIX (6) questions. Answer any FOUR (4) questions in the answer booklet provided. Question 1 (a) The tennis ball in Figure Ql(a) is struck with a horizontal velocity v,“ strikes the smooth ground at B, and bounces upward at 9 2 30°. Determine the initial velocity VA, the final velocity VB, and the coefficient of restitution between the ball and the ground. (15 marks) Figure Q1 (a) (b) A SO-N block is given an initial velocity of 3 m/s up a 45° smooth slope. Determine the time it will take to travel up the slope before it stops. (1 0 marks) EGM1181(F)/Page 2 of6 Question 2 (a) The disk illustrated in Figure Q2(a) has a mass of 20 kg and is originally spinning at the end of the strut with an angular velocity of co = 60 rad/s. If it is then placed against the wall, for which the coefficient of kinetic friction is pk = 0.3, determine the time required for the motion to stop. What is the force in strut BC during this time? (15 marks) Figure Q2(a) (b) The catapult in Figure Q2(b) is used to launch a ball such that it strikes the wall of the building at the maximum height of its trajectory. If it takes 1.5 s to travel from A to B, determine the velocity VA at which it was launched, the angle of release 6’, and the height h. ' (10 marks) Figure Q2(b) EGM1181 (F) I Page 3 of6 Question 3 (a) The skier shown in Figure Q3(a) starts from rest at A and travels down the ramp. If friction and air resistance can be neglected, determine his speed VB when he reaches Be Also, find the distance S to where he strikes the ground at C, if he makes the jump traveling horizontally at B. The skier has a mass of 70 kg. (12 marks) 30° Figure Q3(a) (b) The ski jump shown in Figure Q3 (b)has a trajectory approximated by the parabola given. For the given drawing, determine the normal force on the 600 N skier the instant he arrives at the end of the jump, point A, where his velocity is 9 mfs. Also, what is his acceleration at this point? (13 marks) Figure Q3(b) EGM1i81(F)/Page4 of6 Question 4 (a) At a given instant the train engine shown as E in Figure Q4(a) has a speed of 20 m/s and an acceleration of 14 ms2 acting in the direction shown. Determine the rate of increase in the train’s speed and the radius of curvature p of the path. (12 marks) Figure Q4(a) (b) The GOO-kg wrecking ball shown in Figure Q4(b) is suspended from the crane by a cable having a negiigibie mass. If the ball has a speed v = 8 m/s at the instant it is at its lowest point, 6 = 0°, determine the tension in the cable at this instant. Also, determine the angle 6 to which the bail swings before it stops. (13 marks) 12m Figure Q4(b) EGM1181(F)/Page5 of6 Question 5 (a) The S-kg slender rod Shown in Figure Q5 (a) is suspended from the pin at A. If a I-kg ball B is thrown at the rod and strikes its center with a horizontal velocity of 9 m/s, determine the angular velocity of the rod just after impact. The coefficient of restitution is e = 0.4. (12 marks) Figure Q5 (3) (b) The 800-kg cylinder shown in Figure Q5 (b) is attached to the lOO-kg slender rod which is pinned from point A. At the instant 6 z 30" it has an angular velocity (u; = 1 rad/s as shown. Determine the largest angle 6 to which the rod swings before it momentarily stops. (13 marks) Figure Q5 (b) EGM1181(F)/Page 6 of6 Question 6 (a) Determine the velocity of the slider block shown in Figure Q6 (a) at C at the instant 6 z 45", if link AB is rotating at 4 rad/s. (I 5 marks) Figure Q6 (a) (b) A flywheel has its angular speed increased uniformly from 15 rad/s to 60 rad/s in 80 s. If the diameter of the wheel is 0.6 m, determine the magnitudes of the normal and tangential components of acceleration of a point on the rim of the wheel when r= 80 s, and the total distance the point traveis during the time period. (10 marks) - THE END - EGM] 18! 09/May20} 0(s)/Ragimarhan/0406201 0 EGM 1181 Engineering Dynamics Final Examination Data sheet Center of Gravity and Mass Moment of Inertia of Homogeneous Solids Cylindex I =1”. ‘= I'll-(33+ h!) 1'fl=%mr3 5‘ Hemisphbm' _ 3 2. . 1 2 :1” =9259W2 :2 = gm: In =1” = mmflrz-H‘a ) In: filnr ' X Thin Circutar disk F1, =1” = 34W; In}: %mr2 Ifl- = gm? I” 3: é ml)? 1?3 = firm! In -'=' {7th b?) Slender Rod gag”: 115mg? 15.: 13y: 3m” [flea Fundamental Equatic‘nns of Dynamics KINEMATICS Particle Rccfilincnr Motion Variable a Cuna'lanf a : an (It! , ., ' d5 _ _ 1 .- yF—«Eg STSP‘Pvu't'i'aavl)’ ‘ada‘ 2 lid?) 2? -—'- ‘33 + 21145:“.9‘3) Particle Curvilinem- Motion H )5 35,2: Coordinates. r, 9, .z- Cram-dimres 15:1: £1,525? 3-,: ?- q,r=iE-—ré2 v";):r aim}? vamrfl a9=r54 2139 vz=z (12:: 132:2 «5:73 1:, l, b Coordinmes v .i r = i, pd” a = x ‘ =-_— \ ' (1.9 a 1’1 [I + (dy/dxizlm ‘ = ‘-" p = . . r I? it? 44113 Relating Marion v” = TA '17 53M avg = 3'}; 33M Rigid Body Motion About a Fixed Axis Variabia a Comm»! a -=- ac dun “=7: w=wo+gll 3 9 = '90 + we! + in}: {ogita : ogdfi af- = of; + 20:49. - 99} For Pain! P s = Gr 1) = (or mt a-r a4, = a?! Relative Genemi Plane MuiiuIi—Ti'anslaling Axes Vs 7' VA + Via/Mm“) 3}; x A + BBiAgpin) Reiafive Genera] I’lane Motion—Tim and Rut. AXES W = VA 'i‘ a”, X 1'3“ X (valryz, 83 *“a 3A * 9-“ {RM +9 X (93x, *3/45 + 2‘0 X (YB/Ahyz'x (QB/Ahy: KINETICS Mass Moment of Inertia I =' J 72 dm. .i'graitelusixis Theorem I 2' IG 1‘ mi2 . _ . _ g Radm: of 63mm»: k m \“n- EGM 1181 Engineering Dynamics Final Examination Data sheet Equations of Motinn Pam‘eie 2F = ma Rigid Body EFX=MWGL (Plane, Mode?!) 2153,: 212(06):“ EMq-Jjgoror 2Mg='2‘.{grflk)p i’rinciple of Wax-k and Energy T: '1’ U1—2 = T2 Kinetic Energy 1 Particle 'I‘ = 3:12:24 Rigid Brady _ : (Plane Motion) T Z £11193 + [brag Work .- Variabie farce UP =1 F cos 9 d3 Comma: fame. (Fe-ms 9) As ii’e'igfrt U“, = — W is): Spring U£= ~62 ks: — k5?) Cot'tple meme”? [13—1 M :19. l’flweégfid Efficiency P U. m un a , ___ -au't.-= em P d: Fng E 7 Pin: “in (huservaiion of Energy Theorem T1+VizTg+ifz Potential [Margy y' = v.8. + Vawhere Vg = iWszc' = +% ksg Principle-0f Linear Impulse um] quenlmn Parade my] + 2er d1 = my; Rigid Body 'mfiéfi + ZIina‘ = MUG): Cbnservntlun of Linear Mamentum ‘ 23(53'52. mifh = 27(sy5tg, 171v}; - .w .' 5. Coefficient of Rasfinltionr e- " 7 — (“Aiihivsii Principle of Angular lmpuise and Momentum. {He};- + BiMo 5” = 910); wherello '= '(riszr) (Ha): + :fmc d: =- {Hal when:qu é low (115i +L2J‘Madr = (no): where H0 = {0(1) Particle Rigid Bod y (Plugs pituir'on) Conservation of Angular Momentum E(syst. H)l =. E(sy'st. H)2 ...
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This note was uploaded on 05/02/2011 for the course EGM 1181 taught by Professor Iit during the Spring '11 term at American Internation College.

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EGM1181 may2010 - INTI lNTERNATlONAL UNIVERSITY LAUREATE...

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