PROLEM 15.11 The 18-in.-rdius flywheel is rigidly ttched to 1.5-in.-rdius shft tht cn roll long prllel rils. Knowing tht t the instnt shown the center of the shft hs velocity of 1. in./s nd n ccelertion of 0.5 in./s, both directed down to the left, determine the ccelertion () of Point A, (b) of Point. Velocity nlysis. Let Point G be the center of the shft nd Point C be the point of contct with the rils. Point C is the instntneous center of the wheel nd shft since tht point does not slip on the rils. vg 1. v G = rω, ω = = = 0.8 rd/s r 1.5 Accelertion nlysis. Since the shft does not slip on the rils, C = C 0 Also, = [0.5 in./s 0 ] G [ C = + ( ) + ( ) C G C/ G t C/ G n 0 ] = [0.5 in./s 0 + ] [1.5α 0 + ] [1.5ω 0 ] Components 0 : 0.5 = 1.5α α = 0.33333 rd/s () Accelertion of Point A. (b) Accelertion of Point. = + ( ) + ( ) A G AG / t AG / n = [0.5 0 + ] [18α ] + [18ω ] = [0.4698 ] + [0.1710 ] + [6 ] + [11.5 ] = [6.4698 ] + [11.670 ] = + ( ) + ( ) G / G t / G n = [0.5 0 + ] [18α ] + [18ω ] = [0.4698 ] + [0.1710 ] + [6 ] + [11.5 ] = [5.530 ] + [11.349 ] A = 13.35 in./s = 1.6 in./s 61.0 64.0 PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 1151
PROLEM 15.11 Knowing tht crnk rottes bout Point A with constnt ngulr velocity of 900 rpm clockwise, determine the ccelertion of the piston P when θ = 10. Lw of sines. sin β sin10 =, β = 16.779 0.05 0.15 Velocity nlysis. ω = 900 rpm = 30 π rd/s v = 0.05ω = 1.5π m/s 60 v D = v D ω = ω vd / = 0.15ω v = v + v D D/ β [ v D ] = [1.5π 60 + ] [0.15ω β ] Components : 0= 1.5π cos60 0.15ω cos β Accelertion nlysis. α = 0 ω 1.5π cos 60 = = 16.4065 rd/s 0.15 cos β π = 0.05 ω = (0.05)(30 ) = 444.13 m/s 30 D = D α = α D / = [0.15α β ] + [0.15ω β ] = [6α β ] + [40.376 β ] D = + D/ Resolve into components. PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 1167
PROLEM 15.11 (Continued) : 0 = 444.13 cos 30 + 0.15α cos β + 40.376 sin β α = 597.0 rd/s : = 444.13 sin 30 (0.15)(597.0)sin β + 40.376 cos β D = 96 m/s P = D = 96 m/s P PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 1168
PROLEM 15.17 Knowing tht t the instnt shown rod hs constnt ngulr velocity of 6 rd/s clockwise, determine the ccelertion of Point D. Velocity nlysis. ω = 6 rd/s v = ( ) ω = (90)(6) = 540 mm/s v = v, v D = v D The instntneous center of br E lies t. Then ω = 0 nd v = v = 540 mm/s Accelertion nlysis. α = 0 D vd 540 ω = = = 3 rd/s 180 D = ( ) ω = [(90)(6) = [( ) α = [180α ] = 340 mm/s ] + [( ) ω ] = [180α ] + [160 mm/s ] D / = [90α ] + [5α ] + [5ω ] + [(180)(3) ] ] + [90ω ] = [90α ] + [5α ] = + D D / Resolve into components. : 160 = 340 + 5 α, α = 7. rd/s : 180α = 0 + (90)(7.), α = 3.6 rd/s D = [340 ] + [(90)(7.) ] + [(5)(7.) ] = [648 ] + [160 mm/s ] = 1745 mm/s 68. = 1.745 m/s 68. D PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 1179
PROLEM 15.150 Pin P is ttched to the collr shown; the motion of the pin is guided by slot cut in rod nd by the collr tht slides on rod AE. Knowing tht t the instnt considered the rods rotte clockwise with constnt ngulr velocities, determine for the given dt the velocity of pin P. ω = 8 rd/s, ω = 3 rd/s AE 0.5 = 500 mm = 0.5 m, AP = 0.5tn 30, P = cos30 ω AE = 8 rd/s, ω = 3 rd/s Let P be the coinciding point on AE nd u 1 be the outwrd velocity of the collr long the rod AE. vp = vp + v P/ AE = [( AP) ωae ] + [u1 ] Let P be the coinciding point on nd u be the outwrd speed long the slot in rod. Equte the two expressions for vp = vp + v P/ = [( P) ω 30 ] + [u 60 ] v P nd resolve into components. 0.5 u = (3)(cos30 ) + u cos 60 cos30 : 1 or u1 1.5 0.5u = + (1) 0.5 (0.5tn30 )(8) = (3)sin 30 + u sin 60 cos30 : 1 u = [1.5tn 30 4 tn 30 ] = 1.66667 m/s sin 60 From (1), u 1 = 1.5 + (0.5)( 1.66667) = 0.66667 m/s v P = [(0.5tn 30 )(8) ] + [0.66667 ] = [.3094 m/s ] + [0.66667 m/s ] v =.3094 + 0.66667 =.4037 m/s P.3094 tn β = β = 73.9 0.66667 v =.40 m/s 73.9 P PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 11
PROLEM 15.163 The motion of blde D is controlled by the robot rm C. At the instnt shown, the rm is rotting clockwise t the constnt rte ω = 1.8 rd/s nd the length of portion C of the rm is being decresed t the constnt rte of 50 mm/s. Determine () the velocity of D, (b) the ccelertion of D. Unit vectors: i = 1, j = 1, k = Units: meters, m/s, m/s Motion of Point D of extended frme. Motion of Point D reltive to frme. r DA / = (0.3 m) i (0.4 m) j ω = (1.8 rd/s) k α = 0 vd = ω rd/ A = ( 1.8 k) (0.3i 0.4 j) = 0.43i 0.576j D = D/ A ω D/ A α r ( r ) = 0 (1.8) (0.3i 0.4 j) = 1.0368i+ 0.7776j v D / = 50 mm/s 5 = (0.5cos 5 ) i+ (0.5sin 5 ) j = 0.658i+ 0.10565j D / = 0 Coriolis ccelertion ω vd / = ()( 1.8 k) ( 0.658i+ 0.10565 j) = 0.38034i+ 0.81569j () Velocity of Point D. vd = vd + vd/ = 0.43i 0.576j 0.658i+ 0.10565j = 0.65858i 0.47035j v (0.659 m/s) i (0.470 m/s) j = 0.809 m/s 35.5 D = PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 17
PROLEM 15.163 (Continued) (b) Accelertion of Point D. = + + ω v D D D/ D/ = 1.0368i+ 0.7776j+ 0 + 0.38034i+ 0.81569j = 0.6565i+ 1.5933j (0.657 m/s ) i (1.593 m/s ) j 1.73 m/s D = + = 67.6 PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 18
PROLEM 15.168 A chin is looped round two gers of rdius 40 mm tht cn rotte freely with respect to the 30-mm rm. The chin moves bout rm in clockwise direction t the constnt rte of 80 mm/s reltive to the rm. Knowing tht in the position shown rm rottes clockwise bout A t the constnt rte ω = 0.75 rd/s, determine the ccelertion of ech of the chin links indicted. Links 1 nd. Let the rm be rotting frme of reference. Ω= 0.75 rd/s = (0.75 rd/s) k : Link 1: r1 = (40 mm) i, v1/ = u = (80 mm/s) j = Ω r = (0.75) ( 40) = (.5 mm/s) i 1/ P / 1 1 80 160 mm/s (160 mm/s ) u = = = ρ 40 Ω v = ()( 0.75 k) (80 j) = (10 mm/s) i = + + Ω v 1 1 1/ 1/ = (30.5 mm/s ) i Link : r = (160 mm) i+ (40 mm) j v = u = (80 mm/s) i / = Ω r / / = (0.75) (160i+ 40 j) = (90 mm/s ) i (.5 mm/s ) j = 0 Ω v = ()( 0.75 k) (80 i) = (10 mm/s ) j = + + Ω v / / = 90i.5j 10j = (90 mm/s ) i (14.5 mm/s ) j = (90) + (14.5) = 168.5 mm/s 14.5 tn β =, β = 57.7 90 i 1 = 303 mm/s = 168.5 mm/s 57.7 PROPRIETARY MATERIAL. 013 The McGrw-Hill Compnies, Inc. All rights reserved. No prt of this Mnul my be displyed, 135