Exam 2 2008 - ES 223 — Rigid Body Dynamics Exam 1...

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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 Work-energy: T: 'l' Ul—)2 = T2 T1 +Vgl +Vel +U{_,2 =T2 +ng +Vez T=lww2 U =I5-d: 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 right-triangular 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 ¥\83-1W34§+ 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= na-aoveao"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 -37q2-314%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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