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Exam 3 2007 - K ES 223 —— Rigid Body Dynamics Exam III...

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Unformatted text preview: K ES 223 —— Rigid Body Dynamics Exam III April 4, 2007 NAME/SECTION g 01 Ct 22/02/ 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. Equations: Uniformly accelerated rectilinear motion: Acceleration due to gravity: _ 2 v=v0+at g—9.81m/s 2 g = 32.2 ft/sec l s = so +vot+——2-at2 v2 = v02 + 2a(s — so) Normal/tangential (path) coordinates: K = Vét v 2 A . A (for circular motion) g_ = —— n + ve, , ,0 v = r9 2 V 2 v2 an=_‘=PIB =Vfl an=—=r€2=v0 p r a,=13=s a,=v=r6 2 2 a = an + a, Polar coordinates: Quadratic equation: ax2 +bx+c=0 = —biVb2 —4ac g=fér +r6é9 g: (F—réz)é, +(ré+2r'6l)é6 x 2a (for circular motion) Vr = 0 Power: v6, = r65 P ___ E '1’ a, = —r62 6 =Poulput a9 = ré. m ‘Pinpul Work-energy: Coefficient of restitution: Tl +U1—>2 = T2 Ti+Vg|+Ve|+Ul,—>2:T2+Vg2+Ve2 e:(v2)n-(v])n 1 (vl)n_(v2)n U=J£'d£ T=—-mv2 2 _ 1 2 _ 2 . U192 _ 2 k(x2 )5) (Spring) Selected integrals: 1 V = mgh V = —kx2 dx 1 g e 2 I =——ln(a+bx) a+bx b Imp l - : . 1 u se momentum Ism(a+bx)dx =—Zcos(a+bx) ’2 1 . £1 + IZEdt = Q2 Icos(a+bx)dx=;sm(a+bx) 1| Q. = mx '2 10. + [much = .1102 II 10 = 1X m2 Relative motion — translating axes: Rotation about a fixed point: v A = v B + v ’V '— — _ 3 v = r0) _ 2 9A — 93 +91% v an = r02 = — = va) r at = r0: . . . Relative motion — rotatmg axes: BA :23 +QXK+EreI v = a) x r . ‘ — ‘ gA =23+QXE+QX wx:)+22><2,e1+gm 2,. =QX(QX£) 9,, = -w2: Qt =QX£ M‘s—infifi Law of cosines: C a a _ b _ c sinA sinB sinC c2 =a2+b2—2abcosC b 1. (25) Link AB is rotating at 01,13 = 4 rad/sec in the direction shown. Determine the magnitude of the angular velocity of bar CD at the instant shown. Use the method of instantaneous center of zero velocity. Find: wCD using the IC method 31 Sum; 5m75‘ 55-. 10.92;? In Us: 55 ”ca .8. : :2— , :C3;2_Eié 5m”; 5”}60 We :6 U363 =EQ( LII) EC '- 97)“)? m C __E____________51 Then ‘ UzémeczQ we :- 22L—l L". z ( as m QZP- Q/Wa Fad/52L was 2% 5221:) Uc. = 9'7‘75’X3‘ 95) :- 2"52 m/Jec. ”CD: L251: 5‘38 _________________._¥ wCD:5'3? (flea/$6. 5 2. (30) For the instant represented, when crank 0A passes the horizontal position, determine the velocity vector of the center G of link AB. Use the relative motion (vector equations) solution method Find: 20 1 ‘ 5 1 Yi j<—77.<=vmm——>¢eeomw , l l S X % i 8 rad/s “ 1 l l 1 E 15V "via? ’\ A 'U’n : @Abxffio‘ —8,lzX(‘é0A ‘1 480? (EB/A - 993 x [3,; was Xx (—ISSflE- 903) A = #55,? a)“; 3‘1- 9011,93 L Jab 10+¢ Q) U54.“ - 480&-— [55/9 («JA334QowA32 591,14th 0'- 480 -/55 9 w 6.3 Qwfie: 3 08 rad/5 U65» W‘i’U‘G-lfi Ufi' H808 (EG’A: bagel)" «(g/R3 308€K£771940‘ ‘456‘3 7.ch = L180? {21403 +1343“, =9! '14)}: [38'62‘LWDJ’3-fll 3. (35) The quick-return mechanism consists of a crank AB, slider block B, and slotted link CD. If the crank AB has the motion shown, determine the angular acceleration of the slotted link at this instant. Express your answer as a vector. Additional information for the instant shown: map = 0.866 rad/s CCW and slider block B is moving toward point D with a speed of O. 15 m/s Find: gap . 0 95:9,“ 53‘0" 57‘ +91%" QwaP/L * Wallet + grel Q3 :- mafia,“ 4 fissure/A = -(3)"[/oo @5302 0051/3081 + cHzx {\DOCOSBO u- 4005:0306 ) = “779 2‘. 1» 4503137731‘ ”0‘— Qoyfpp- dwhrwoo - 300°\’c.o k wwxwmx Wile “(0.866) (3003.) - “9.152 agmxyull = ;zl(o.€66k)y( 150t)=15<t86 95ml; amt 5‘46 Into Q :390124’9996120+sao%o’3}~aasl +35q,8a"+4rd3 Equa+€ 5’ xaaq = 300 or“, +9593 (Xco: 3.23 :> logo: 3,233: mil/5L I 4- (5) State in words Newton’s First Law. + )l , {)/ Cl pantie/e remains an M5 or cm I was 4‘0 ”MM/e £01th uni-far‘m Vela“ ‘ \‘p 4km“; U ND unbalanced Vague filo/Hog on ff. 5. (5) The instantaneous center of zero velocity (IC) method is useful for finding the velocities of points on a rigid body undergoing general plane motion. Why must the instantaneous center of zero velocity method not be used to determine accelerations of points on the rigid body? The a66€l€raWL/W\J 0‘4 71/16 IC )5 Wed“ zero. ...
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