Ch16_Sec5-6_Ex-54-61-71-90-95-98

Ch16_Sec5-6_Ex-54-61-71-90-95-98 - 16-54 e gear rests in a...

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Unformatted text preview: 16-54. e gear rests in a fixed horizontal rack. A cord is wrappe around the inner core of the gear so that it remains ho izontaily tangent to the inner core at A. If the cord is pu ed to the right with a constant velocity of V 2 ft/s, dete mine the velocity of the center of the gear, C. 1/ i i i a) vc=l.33ftls “- Am . WWI: p 15:0 16-51. The piston P is moving upward with a velocity of 300 in./s at the instant shown. Determine the angular velocity of the crankshaft AB at this instant. I From the geometry : l . 415 sin 30° v, =t {300j 1- his v. = —u, cos 30°i+ u, sin30°j aa= —m.pk rpm i= {-S(:0581.66°i+ 55in81.66°j} in. v, =v. + mx rm 300j = (-1)” c0530°i+ v. sin30°j)+ (—wgpk) x (—5c0581.66°i+ Ssin81.66°j) 3001‘=(—1)”C0530°+5$in8L66°mnrH+(D. sin30°+ 5cosBl.66°aap)j Equiting the i and 1 components yields : 0 a -v. c0530" + Ssin81.66“wgp 300‘: v5 sin30°+ 5cos8l.66°a).p Soltiin; Eq:i.(l) and (2) yields: \ m: 83.77 rad/s u. = 473.53 ian For bmkshaft AB : Crankshaft AB rotates about the fixed point A. Hence Us 3“ 0mm": 478 53: w45(1.45) a)“ = 330 rad/s '7 An: 387 ‘ 16.71. lf’the hub gear H and ring gear R have angular velocities i0” := 5 rad/s and'wR = 20 rad/s, respectively, determine’thq angular velocity (as of the spur gear S and the angul r velocity of its attached arm 0A. j ‘ Kinematic 113:]: ram .- Sincelink 0A is rotating about fixed point 0. men .A IS always (1' and perpendicular to link 0A and its magnitude is vA = mot. r‘M = 0.20;“ (3:: as shown). Also. poimB and C are moving in the wnh , speed of u,” = ”r” = 5(o.15) = 0.750 m/s and U; = mm = 20(025) = 5.00 m/s(di.rec as shown) respectively. 1 Velacily Edna! an : Here. um = Q’s-rm, ‘= 0.1w5and is directed as F shown. Applying Eq.116— l5. We have l : I 3 VC=V5+ch [5.00] = [0.750]+[o.1m,] i +)i 5.00 = —0.750+ 0.1a); H i a), = 57.5 rad/s An; Here. um 4; 05+“, = 57.5(0.05) = 2.875 m/s and is directed as shown. Applying ELL/16¢ 15. we have "A = V: +VAIB l [0.2mm ] = [0.750] + [2.7875] l l ‘ i +)§ 0.20)“ = —o.7so+ 2.875 1' i mm = l0.6 rad/s Ans 1 16-90. lf link GD has an angular velocity of we]; = 6 rad/s, determine the velocity of point E on link BC and the angular velocity of link AB at the instant shown. Vc = Q'DUCD) = (6)(0.SD =J.60 m/s ”6 _ 3-“ rc‘uc 0.6 an3fi1 \ mac= = 10.39 rad/s )= 7.20 m/s l’ : “A3=l=—wJ—46mdls‘) Ans l i ‘ 0.6 = = 10 39( V5 mac's/1c ( k )‘COSSO" V: = wacrmc = 10.191/(00 un30°)2 + (0.3)2 = 4.76 m/s Ans 0.3 6: "l __._)= . ti 5 m" (0.6 misc" 4° 9 “‘5 l . - 16- . . If the hub gear H has an angular velocity w” = 5 ra¢/s,:determine the angular velocity of the ring gear R = so that the arm 0A attached to the spur gear S remains statienall'y (mm = 0). What is the angular velocity of the spun gear? / l a The 161i up. 0. TQM/s of 5 z 7;. (”5:0057El5‘0md/s Ans )0; 1 S~ mo =0-7Sm6 m=aé%=3.00md/s Ans ( ) f ‘ velocity pf S rad/s. l l l l . 1 l l l l 3 v. =o.2(5)l= l m/a: » l Membernq: Val; = 0.4899 m ’51“: 0.4 3 - : —..._ sin75° sin45°i ram: = 0. 5464": l l ‘ eke-m: .830rad/s V: = 0.4899(1 336) = 0.897 m/s / Am ...
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