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Unformatted text preview: GUT Student Number Surname Given Name/s Examination Paper SEMESTER: FIRST SEMESTER EXAMINATIONS 2007 UNIT: EN8211 DYNAMICS — THEORY 1 DURATION OF EXAMINATION: PERUSAL: 10 MINUTES
WORKING: 3HOURS EXAMINATION MATERIAL SUPPLIED BY THE UNIVERSITY: EXAMINATION BOOKLETS
GRAPH PAPER MM  ONE (1) PER STUDENT
EQUATION SHEETS — THREE (3) PAGES ATTACHED 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 NOTES MAY BE MADE QNLX ON THE EXAMINATION PAPER DURING PERUSAL TIME EIGHT (8) QUESTIONS ONLY ARE TO BE ATTEMPTED ALL QUESTIONS ARE OF EQUAL VALUE Queensland University of Technology GUT GUT GUT GUT Gardens Point Kelvin Grove Carseldine Caboolture QUESTION 1 Magnetic tape is being transferred from reel A to reel B and passes around idler
pulleys C and D. At a certain instant, point P; on the tape is in contact with pulley C
and point P 2 is in contact with pulley D. Pulley C has a rotating speed of 20 rad/s and
the tangential component of acceleration of P3 is 30 rn/s2 at this instant. Determine a) the speed of the tape, b) the magnitude and direction of the total acceleration of P1, and c) the magnitude of total acceleration at point P3 on pulley A . 13,, 50 mm a . {4; 5 mm 6) l ,_
:10 C \{3 _. //6\
a: (O Q
Q
B / .'
100 mm QUESTION 2 The springs are undeformed in the position shown. Ifthe 6 kg collar is released from
rest in the position where the lower spring is compressed 200 mm, determine a) the velocity of the collar just before compressing the upper spring, and b) the maximum compression of the upper spring there after. 250 2"“ mm It B : l750 Nr‘m
\ 500 l _ 3 6kg 150111111 “If“ g _ i9: 250 mm A31 — 8400 me i—_ ENB211T1.071 c0nt/... QUESTION 3 The ball strikes the heavy steel plate with a velocity of v0 = 30 m/s at an angle of
60 deg with the horizontal and the plate is initially stationary. Ifthe coefﬁcient of
restitution is e = 0.7 and the spring stiffness is k = 155 N/m, determine a) the ﬁnal velocities of both masses immediately after impact, and b) the deﬂection of the spring. QUESTION 4 The two Vpulleys form an integral unit and rotate about the ﬁxed axis at 0. At a
certain instant, point A on the belt of the smaller puller has a velocity of VA = 1.75 m/s, and the bigger pulley has an angular acceleration of on; = 112.5 rad/52‘ For the instant
shown determine, a) the magnitude of the acceleration ac of point C and sketch the vector of ac,
and b) the velocity of VA, if the acceleration as were to double. 800 139
mm ’le/ (V f I i Tags i ,_ ‘ l .C/f’q" ill at l 15 ll Xx ’/ ‘1‘! 70”)!” ENB211T1.071 cont/... QUESTION 5 For the 4bar mechanism shown, link BC has a clockwise angular velocity of (035 =
3 .5 rad/s. For the position shown, determine (a) the angular velocity of the triangle
plate A81), and (b) the magnitude and direction of velocity at point D. 500 mm ;3 a i
he  100 mm WWW—a»: QUESTION 6 The three identical spheres, each of mass m, are supported in the vertical plane on the
30° incline. The spheres are welded to the two connecting rods of negligible mass.
The upper rod, also of negligible mass, is pivoted freely to the upper sphere and to the
bracket at A. If the stop at B is suddenly released, determine the velocity v, with which
the upper sphere hits the incline. Note that the corresponding velocity of the middle
sphere is 12/2. Draw a diagram of the system at the time of release and alongside draw
a diagram of the system at the instant the upper sphere hits the incline. Show the
velocity vectors on the diagram. ENB211T1.071 cont/... QUESTION 7 If the frictional moment at the pivot 0 of the grooved drum is 2.5 Nm. The drum has a
mass of 8 kg and a radius of gyration k0 = 225mm. Determine, a) the angular acceleration of the drum and
b) the acceleration of the masses. (Note: moment of inertia Io = m(k0)2 ) . in» H513: 300 mm l
l 200 mm l l ' 7 kg 12 kg QUESTION 8 The cylinder with mass m as shown, is displaced by an amount yo: 0.1 m downward
from its equilibrium position and is released at time I = 0. Determine the displacement
y, and the velocity v when t = 3 3. What is the maximum acceleration? Plot the
position, velocity and acceleration of the mass for 1 cycle on graph paper supplied. Equilibrmm
position ‘“ I " m = 4 kg ~‘l ENB211T1.071 C0nt/... QUESTION 9 The cart is moving down the incline with a velocity v0 = 20 m/s at t = O, at which time
the force begins to act as shown. After 5 seconds the force continues on at the SO—N level. Determine the velocity of the cart at time t = 7 seconds and then calculate the
time t at which the cart velocity is zero RN 1
3
l 50 .. . .t 5 Linear END OF PAPER ENB211T1.071 (i) ENBle Dynamics Equation Sheet Rectilinear Motion
(displ 3, vol v, accn a) ds .
v=——=s dt dv . dzs ,.
az—zv: ’1 :5
dt dt‘ va'v : ads or add : Eds t2 As=s2—sl=Tds=J‘vdt 21
(area under vel vs time curve) ti.
Av : v2 — v1 = jadt (area under accn vs time curve) (area under accn vs displ curve) Equations of Motion
For constant accn. ..
.— —0v + at v —v0 2:2a(s—so) _ 1 2
s «so +vot+3at For accn as a fn of time a=f(t). .. v=v0 +jf(t)dt s 2 s0 +jvdt For accn as a fn of vel a=f(t9...
s:%+ywh fa) For accn and vel as a fns of displ a=f(S), v=g(s)... Plane Curvilinear Motion
(Position :, vel y, accn a) = S (magnitude of vel) Rectangular (x—y) Coordinate Systems I r=xi+vj v=r_‘= Sci+yj v:[v,: W/x2 +y2 *1/12 +12},2 (magnitude of vel vector)
‘j tanl9 = y vx
(angle of veI vector)
a: 2f = x'z_' + yj a=a= ‘ljc +y 231/61 +ay2 (magnitude of accn vector)
a V mud1*"—
a .‘C (angle of accn vector) Projectile Motion vy 2 vylo +gz‘ =vsmt9—gt x=xO +vx0 y=yo+vy0
v 0 2g(y~yo) Y F Normal and Tangential Coordinates
E = vet = 10,82: .
(tangential vel) (ii) Newton ’5 2’” Law (cont...) Normal and Tangential Coordinates (cont...) (accn vector) '.’ an :sz62 :]}ﬂ
,0 (normal accn) at = 1'! = [9' (tangential accn) 2 2
a : gl : Jan +al (magnitude of accn) Circular motion in 111 coordinates... v=r6
v2 .7 .
an=7=r9‘=V9 Work&Energy
at=ﬁ=ré UZIE'dZ.
ngmv2 Polar Coordinates Vg = mgh (close to earth) r 2 re 2
_ r , m R . .
Q, = 63539 l/g : — 8; (1n orblt)
9:9 : —6€r . . V3 = kadx = also
‘1: r2. ”96.3.9 0 “
g=(F—r6lz _, +(ré+2r’é)ga U,_2 =AT+AKg +AVe U1—2+E+Vg1+Vel 2]; +Vg2 +Ve2 Circular motion in polar coordinates...
vr = 0 Impulse and Mromentum
Linear. . . v8 =r6
a, = —r6“
a6 =r6 Relative Motion v. = v + v_ . .
‘A “B ““3 Conservation of lmear momentum... AQ=00rQ1=Q2 “4 2%; +QA/B Newton ’s 2'"i Law Angular...
H0 =1”?! E=mg ZFzmg Rectangular coords. ..
ZF x 2 max Ex 2 m(vzy — vyzf)
H y = m(vxz — vzx) H, = m(vyx — vxy) Impulse and Momentum (cont.. .)
2M —H'_O ——0 Conservation of angular momentum... Ago I 001’ 1—101 2 E02 r
mlvl +1112V2 = mlvl ’lI’l’izv2 I I I — v .
e = —‘——1 = (we; separation)+(vrel approach)
v1 — 172 Kinetics of Systems of Particles ZE=mg U112 + 2;. + Vigil + £21 : 7:12 + Vrgl + 1/er l4 :1 +3.4 VABZQW 34BZQXC
9.4 :39 +433 EA 2.513 +a(. 4/3)" +a( aA B) (“A/B)" :VHJS /r—(‘) r (“A B): :vA/B ‘5”, (ﬂA/B)» :QX(QXF) (ﬁg/B): ZQXL’ Motion Relative to Rotating Axes) izwxi 1‘:ng (1A zaB +a§xr+a_)x(wxr)+2(_0xvn,, +are,
A :aP +2(_e_)_><v +a _ rel __ rel 2MP :}a+m2id
2MP :ng'i‘meQP I = mk2 (radius of gyration)
I = 3m)" 2 (solid circular disk) Vibration and Time Response
& w§x= 0 where wf = a/k/m has the
solution x= Acos wnt+ Bsinwnt Given initial conditions x0: Aand x32: Bw,
the displacement function is Xi? x = x0 coswnt + *5111 wnt . WW The static deﬂection is D ST = 331g: 3.
W k ...
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