hw7 - PROBLEM 15.114 The 6-in.-radius drum rolls without...

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Unformatted text preview: PROBLEM 15.114 The 6-in.-radius drum rolls without slipping on a belt that moves to the left with a constant velocity of 12 in./s. At an instant when the velocity and acceleration of the center D of the drum are as shown, determine the accelerations of points A, B, and C of the drum. [ SOLUTION Velocity analysis. v0 = 30 in./s VC 2 12 in./sa n—, w ) vC =vD+vOD or [12 ~]=[30 A» ]+[6ai w] —12:30—6a), ai=f163=7radls ) Acceleration analysis. an = 36 inls2 M», aC = ac [, a ) ac = aD + (ac/D) + (2.00)” [a t]=[36 m» ]+[6a +—1+[6n2 t] Components: 1;: 0 = 36 — 6a a = 6 rad/s2 ) +[z aC = (6)(7)2 = 294 inls2 ac = 24.5 ft/s,2 I 4 a}, : a0 + (am)! +91%)" :[36 ~—» [+[60l—~> [+[6co2 :[36 —4 ]+[36 » ]+[294 = [72 inls2 77» [+ [294 inls2 [ [ aA = 25.2 fi/s2 ‘C 762" < 38 : aD + (33,0)! + (38.0)” =[36—~l+[6all+l6wz 7”] =[36—»]+[36 i]+[294*”l = [258 inls2 «~—~ [ + [36 inls2 [ a3 = 21.7 ft/s2 7 79° 4 PROBLEM 15.118 Arm AB has a constant angular velocity of 16 rad/s counterclockwise. At the instant when 9 = 90°, determine the acceleration (a) of collar D, (b) of the midpoint G of bar 30. 1120111111 it) 111nm"? L < .775 r 1 1E1? SOLUTION Geometry and velocity analysis. 6 = 90° 60 sin 2 —— = 0.3, = l7.458° fl 200 fl VD and VB are parallel, thus the instantaneous center lies at so. wBD=0 Acceleration analysis. a“, = 0, (0‘43 2 16 rad/s, 603D = 0 a3 = [600.313 T]+ [6060313 1] = 0 + [(60)(16)2 l] : 15360 mm/sz 1 Point D moves on a straight line. a D = a0 —~ (amt 2 [60am «— ] + [200cosflaBD I] (aw) = [60ng l]+ [zoocosflwgD «- j = 0 11 3D = 33 + (up/B ){ + (aD,B)n. Resolve into components. a _ 15360 3D 200cosfl +1: 0 = —15360 + 200cosflaRD, = 80.508 lad/s2 ‘j (a) 3; (ID = 0 4 6003,”) + 0 = —(60)(80.508) = —4831mrn/s2 a0 = 4.83 m/s2 «— 4 (aG/Bl : [300530 +~—]+[100msfla3,) l]: [2415 +— ] +[7680 l] (ac/B)" = [300”; l ] + [100005 fiwfw <~v ] = 0 (b) 30 = ab, + (33/0)! + (fig/G)” [15360 l]+[2415 <~]+[76801]+0 I l [2415 mm/s2 «~— ] + [7680 mm/s2 l] ac = 8.05 m/s2 7 72.50 4 j i f 2011 mm E , 81.6.: i 7’ " 1001mm , 1 11m mm (1,, q P3601111“ 100 mm PROBLEM 1 5.125 Knowing that at the instant shown bar AB has a constant angular velocity of 19 rad/s clockwise, determine (a) the angular acceleration of bar BGD, (b) the angular acceleration of bar DE. SOLUTION Velocity analysis. wAB = 19 rad/s 3 VB : (AB)a)AB = (0.2)(19) = 3.8 m/s VB=VB “*2 VDZVDT Instantaneous center of bar BD lies at C. VB 3.8 \ a) = — : — 2 l9 rad/s 30 BC 0.2 / v0 = (CD)a)BD = (0.48)(19) = 9.12 m/s CODE — vi: = = 24 rad/s2 A) Acceleratzon analyszs. aAB : 0. a. {mm-.31] [( . W 1]= 722mm 30 : [(DE)aDE H+ [(015)ng M] = [0.38am l]+ [218.88 m/s2 ~—~ ] | I o N (20,8)! = [0.48am l]+ [0.20703 M] (aD,,,)n = [0.48wgD w ] +[0.2w§D t] = [173.28 m/s2 —-—> J + [72.2 m/s2 T] 3D = 33 + (aD,B )r + (CID/B )n Resolve into components. L: 218.88 = 0 + 0.2aBD + 173.28 (0) am 2 228 rad/s2 ‘3 4 +l: 0.38aDE = —72.2 + (0.48)(228) — 72.2, (b) aDE = —92 rad/s aDE = 92.0 rad/$2 } 4 PROBLEM 15.139 Rod AB moves over a small wheel at C While end A moves to the right with a constant velocity VA. Using the method of Sec. 15.9, derive expressions for the angular velocity and angular acceleration of the rod. SOLUTION SOLUTION P PROBLEM 15.151 Two rotating rods are connected by a slider block P. The velocity vO of the slider block relative to the rod on which it slides has a constant velocity of 30 in./s and is directed outward. Determine the angular velocity of each rod for the position shown. ,8 = 50° — 20° 2 30° AB 20 BP _ AP sinl30O sinfl = sin30° sin20° BP 2 136808 in., AP : 30.642 in. V}, = (BP)a)BP 7 50° v,,. = (AP)wAE 7 20° VP/F = V0 \i 20° VP = VP: + vP/F. Resolve into components. 1” ' (BP)a)BP cosSO° = (AP)coAEc0520° + v0 sin20° +1: (BP)a)BP sin50° = (AP)a)AE sin20° — v0 cosZO" Rearranging, (13.6808c0550°)w8p 30.64200520°a)AE = 30sin 20° (13.68083in50°)a)‘,flD — 30.642sin20°a)AE = —30cos20° Solving Eqs. (1) and (2), 6on : —4.3857 rad/s, wAE = —1.6958 rad/s 60/15 = 1.696 rad/s )4 (63;, = 4.39 rad/s ) 4 Note that instead of resolving into components, the triangle of vectors VP, VP, and vP/F can be constructed. Then, vP/F = 30 in./s, 1),. = i = 51.962 in./s tan [3 v = .30 = 60 in /s smfl — VP’ — 51962 =1.696rad/s, a)“: _ (AP) _ 30.642 _ vPr 60 a) _ : =4.39rad/S 3” (BF) 13.6808 F PROBLEM 15.161 At the instant shown the length of the boom AB is being increased at the constant rate of 0.6 1115 and the boom is being lowered at the constant rate of 0.08 rad/s. Determine (a) the velocity of point B, (b) the acceleration of point B. SOLUTION Velocity of coinciding point B' on boom. VB, = rm 2 (18)(0.08) = 1.44 ft/s T: 600 Velocity of pointB relative to the boom. vB/nboom = 0.6 fi/s z 30° (a) Velocity of point B. VB : VB, + vie/boom i; : (VB)X = 1.44cos60O + 0.600330O = 1.23962 ft/s +1 : (VB) : —1.44sin60° + 0.6sin30° = —0.94708 ft/s y v3 = (1.23962)2 + (0.94708)2 = 1.560 ft/s 0.94708 123962’ tanfl = — ,8 : —37.4° v3 2 1.560 ft/s T 374° 4 Acceleration of coinciding point B' on boom. a3. = rwz = (18)(0.08)2 = 0.1152 fi/s2 7 30° Acceleration of B relative to the boom. 33/1300,“ 2 0 Coriolis acceleration. 2a)“ : (2)(0'08)(0'6) : 0'096 W52 V: 600 (b) Acceleration of point B. 33 : 38, + amboom + 2am 2:. : (a3) = —0.1152cos30° + 009600560" = —0.051767 m3 4t : (513)], = —0.11525in30° — 0.096sin60° : —0.14074 m/s2 2 a3 = (0.051767)2 + (0.14074) 2 0.1500 ft/s2 0.14074 tan :—————, fl 0.051767 [3 = 69.80 aB = 0.1500 11/52 :7 69.80 4 PROBLEM 15.168 A chain is looped around two gears of radius 2 in. that can rotate freely with respect to the l6-in. arm AB. The chain moves about arm AB in a q clockwise direction at the constant rate of 4 in./s relative to the arm. i Knowing that in the position shown arm AB rotates clockwise about A at the constant rate a) : 0.75 rad/s, determine the acceleration of each of the chain links indicated. Links 1 and 2. SOLUTION Let the arm AB he a rotating frame ofreference. Q = 0.75 rad/s )= —(0.75 rad/s)k Link 1: r. = —(2 in.)i, vm = u l = (4 in./s)j a; = —er1 : 4(0.75)*(421) =(1.125m./s)1 31,113 2 “— : 42 z 8 111/53 —~+ = (81n./53)i 29 X VFW, = (2)(—0.75k) x (4j) = (6 in./s)i al : a; + am + 29 x VWB : (15.1251n./s3)i 31:15.13in/s2 -> 4 Lin/(2: r2 =(81n.)i+(21n.)j vMB : u r e = (4in./s)i a5 = 42%: : 7(0.75)2(8i + 2j) : —(4.5 in./sz)i - (1.125 in./s3)j 32/413 2 29 x vm :(2)(—0.75k)x(41): 7(6 in./s3)j a2 2 3'2 + a2,” + 29 x v3,” = ‘4-5i -1-1251' — 61' = —(4.51n./s1)1 — (7.125111./s3)j "' 7 32: (4.5) +(7.125) 28.43 111153 tanfl : i, fl = 57.7° 7.125 4.5 a2 = 8.43 111153 7 57.70 4 PROBLEM 15.176 Knowing that at the instant shown the rod attached at B rotates with a constant counterclockwise angular velocity (03 of 6 rad/s, determine the angular velocity and angular acceleration of the rod attached at A. SOLUTION Geometry. AB = 16 in., BD 2 l6tan 30° in. AD = 16sec 30° in. Let the rod attached at A be a rotating frame of reference. Q = (0/1 ‘3 Motion of coinciding point D' on rod attached at A. VD, = (AD)a)A = (l6sec30°)a)A 41600 an : [(Aoyxfl 4 60°] + [14ij tum] : [lésec30°a‘4 41600] + [165ec30°wfi 330°] iMotion 0f collar D relative to the frame, VD/F = u $1300, aD/F = Ll <<:300 Coriolis acceleration. 260/114 4&60O v0 = v0, + VD/F = [(1ésec30°)wl4 460°] + [u Vim] aD : [(16secaA) z{1600] +[(16sec30°)wfi $300] + [12 Tim] + [20,414 460°] RodBD. v0 = (152mm,, = (16tan30°)(6) = [(96tan300)in./s2 ——~ ] 30 = (BD)w§ : (l6tan30°)(6)2 = [(576tan30°)in./s2 I] Equate the two expressions for VD and resolve into components. 4:600; (16sec30°)(oA : 96tan30°00560°, (0A = 1.500 rad/s ‘34 “(30% u = 96tan 30°cos30° = 48 in./s Equate the two expressions for 3D and resolve into components. 460°: (l6sec30°)aA + 260/414 = (576tan30°)(sin60°) : 576sec30°sin60° — 2044;, _ 576tan30°sin60° — (2)(1.5)(48) a A l6sec30° 16sec30° = 7.79 rad/52 aA = 7.79 rad/s2 ‘34 ...
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This note was uploaded on 01/09/2009 for the course ENGINEERIN 440:222 taught by Professor - during the Summer '06 term at Rutgers.

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hw7 - PROBLEM 15.114 The 6-in.-radius drum rolls without...

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