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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 Nonprogrammable 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 SON 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 GOOkg 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 Skg slender rod Shown in Figure Q5 (a) is suspended from the pin at A. If a Ikg 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 800kg cylinder shown in Figure Q5 (b) is attached to the lOOkg 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” = mmﬂrzH‘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 Cramdimres 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 ar 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):“ EMqJjgoror 2Mg='2‘.{grﬂk)p
i’rinciple of Waxk 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. (Fems 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
Principle0f 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.
 Spring '11
 IIT

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