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Unformatted text preview: ES 223 — Rigid Body Dynamics
Exam 1] March 26, 2008 NAME/SECTION / Closed book and closed notes. Show all work and enclose your ﬁnal answer with a box.
Problems will be graded for partial credit only if the work is neat and all intermediate steps are clearly shown and presented in an easy to follow format. For
problems 2 and 3 you must show a coordinate system. Equations:
Uniformly accelerated rectilinear motion: Acceleration due to gravity:
v=v0+m g=9.81m/s22
1 g = 32.2 ft/sec
s = s0 + vot + Eat2 v2 = v02 + 2a(s — so) Normal/tangential (path) coordinates: gzvé,
v2 A A (for circular motion)
g=—— n +ve, ‘
P v v=r49
2
V '2 ' v2 . .
an=_:pﬂ =Vﬂ an=—=r62=v6
,0 r
at at:\}:r6
2 2
a: a +a n 1 Polar coordinates: Quadratic equation: A ~: 2 _
Z=rer+r9e9 ax +bx+c—0 g=(r'—r92)é, +(ré+2fé)é9 =—bi\/b2*4ac. x
2a
(for circular motion)
V, = 0 Power:
V0 2 r6 P =E'E
._ .2
a7 — —r6 e _ Poulpul agzrél m P Workenergy:
T: 'l' Ul—)2 = T2
T1 +Vgl +Vel +U{_,2 =T2 +ng +Vez T=lww2 U =I5d: 2 l .
U142 = ‘Ekoczz _ x12) (Sprlng) n=1m2 P;=ngh 2 Impulse—momentum: 2 V
an=rco =—=va)
r
a,—ra
2=er
[email protected]
_ 2
£n_—a) E
g,=gx: Law of sines: C a a _ b
sinA sin B c
sin C Coefﬁcient of restitution: :ML—ML
om—m» e Selected integrals: I dx =—1—ln(a+bx)
a+bx Isin(a + bx)dx = — % cos(a + bx) Icos(a + bx)dx = g—sinw + bx) Relative motion — translating axes: 2,4 :23 +2% £A=gg+g% Relative motion — rotating axes: BA =ZB+QXE+V _ rel gA :gB +QXE+QX(QXE)+2QXVI‘€[ +grel Law of cosines: 02 =612 +b2 —2abcosC 1. (25) At a certain instant, vertex B of the righttriangular plate has a velocity of
200 mm/s in the direction shown. If the instantaneous center of zero
velocity for the plate is 40 mm from point B and if the angular velocity of
the plate is clockwise, determine the velocity of point D. Indicate the
direction of the velocity of point D with an arrow on the ﬁgure. Use the
instantaneous center of zero velocity method to solve the problem. Find: the magnitude of VD and show the direction of VD on the
ﬁgure using the IC method 7/8: mo 9 aoottww 03 Ewe/s D 2. (30) At the position shown in the ﬁgure below, link 0A has a constant
counterclockwise angular velocity of 3 rad/s. For the position shown,
determine the angular acceleration of AB. Apply the relative motion solution
method using vector equations. Express your answer as a vector. A A. ,\
Find: QgAB 1:909} " 3 k ) @hg‘B‘h; (£33433 )3. fall/6
9308 : D ) 9(5'3: doshiﬁ’lbc.‘ 0(a): kind/1‘
Additional information: At the instant shown, wAB = 3 rad/s clockwise and .5 (03¢ = 3 rad/s clockwise. (D QA=96*9‘*I&
_._.— » —  g‘" N
Qei‘53/o*9£x£”/o B
= qmnrxai‘: Mls‘ « «
0V5 : " (Srﬁlg‘l‘ SBLXIQIL : ‘q{'22'331) +q‘5‘ 11‘ tad“ Q ‘. \8’2a ¥\831W34§+ 1%42 “V51 A
ngy w1ram+ €330} Saw q li’élw‘c’t'aekxpﬁl
I" /\
: ~\8A‘D\O(M5L
i r 0)
NM Subdvlm’e back "7 ‘34 4 a“ Z
48;; =\8’L+\8% we(g}+§%w "8% _ 0“
Emma. o= naaoveao"8 ‘9 0W“
Equq+ezf~l8 = 18+Aﬁi§crél°fw =35 Weeﬂg 4 3. (35) The air transport B is ﬂying with a constant speed of 480 mi/hr in a
horizontal arc of 9 mile radius. When B reaches the position shown,
aircraft A, ﬂying southwest at a constant speed of 360 mi/hr, crosses the
radial line from B to the center of curvature C of its path. Write the vector
expression for the velocity of A as measured by an observer in and tuming with B. Ezra! Find: velocity of A as measured by an observer in and turning with B. Be sure to express your answer as a vector. A
tgbzelxoxm) at
Y ft/sec=(m1/hr) x 1.47 ft=m1x5280 ___ ,1 ob$++ISec N W :ptowm)
= 529 ¥+ISeL 'th = 5m £05 452 — Em gas 45":
é___u;sm a En; “37% 4:7ng 9/329 37q2314%v706%+3qB§ —. 37th ~ @873 \ (grep Sig/33L 5 (lg/rd 3 4. (5) State in words Newton’s First Law. ‘ 5. (5) n the rotating reference me equat1on for acceleration, there IS the term 2va +a _ rel —rel Explain what this term represents. Please be speciﬁc as to what the term represents and do not merely repeat the equation. Use words. Do not use
symbols. ...
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 Spring '07
 MORRISON
 Acceleration, Angular Momentum, Angular velocity, Velocity, Polar coordinate system

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