PROBLEM 15.112 The 18-in.-radius flywheel is rigidly attached to a 1.5-in. -radius shaft that can roll along parallel rails. Knowing that at the instant shown the center of the shaft has a velocity of 1.2 in./s and an acceleration of 0.5 in./s 2 , both directed down to the left, determine the acceleration (a) of Point A, (b) of Point B. SOLUTION Velocity analysis. Let Point G be the center of the shaft and Point C be the point of contact with the rails. Point C is the instantaneous center of the wheel and shaft since that point does not slip on the rails. vG = rω , ω = vG 1.2 = = 0.8 rad/s r 1.5 Acceleration analysis. Since the shaft does not slip on the rails, aC = aC 20° aG = [0.5 in./s 2 Also, 20°] aC = aG + (aC/G )t + (aC/G )n [aC Components (a) 20°] = [0.5 in./s 2 20°: 0.5 = −1.5α 20°] + [1.5α 20°] + [1.5ω 2 20°] α = 0.33333 rad/s 2 Acceleration of Point A. aA = aG + (aA/G )t + (aA/G ) n = [0.5 (b) 20°] + [18α ] + [18ω 2 ] = [0.4698 ] + [0.1710 ] + [6 = [6.4698 ] + [11.670 ] ] + [11.52 ] a A = 13.35 in./s 2 61.0° a B = 12.62 in./s 2 64.0° Acceleration of Point B. a B = aG + (a B/G )t + (a B/G ) n = [0.5 20°] + [18α ] + [18ω 2 ] = [0.4698 ] + [0.1710 ] + [6 = [5.5302 ] + [11.349 ] ] + [11.52 ] PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1151 PROBLEM 15.121 Knowing that crank AB rotates about Point A with a constant angular velocity of 900 rpm clockwise, determine the acceleration of the piston P when θ = 120°. SOLUTION sin β sin120° = , 0.05 0.15 Law of sines. β = 16.779° ω AB = 900 rpm = 30π rad/s Velocity analysis. v B = 0.05ω AB = 1.5π m/s v D = vD 60° ωBD = ωBD v D/B = 0.15ωBD β v D = v B + v D/B [vD ] = [1.5π Components : β] 0 = −1.5π cos 60° − 0.15ω BD cos β ωBD = − Acceleration analysis. 60°] + [0.15ωBD 1.5π cos 60° = 16.4065 rad/s 0.15 cos β α AB = 0 2 a B = 0.05ω AB = (0.05)(30π )2 = 444.13 m/s 2 a D = aD 30° α BD = α BD a D/B = [0.15α AB = [6α BD 2 β ] + [0.15ω BD β ] + [40.376 a D = a B + a D/B β] β] Resolve into components. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1167 PROBLEM 15.121 (Continued) : 0 = −444.13 cos 30° + 0.15α BD cos β + 40.376 sin β α BD = 2597.0 rad/s 2 : aD = −444.13 sin 30° − (0.15)(2597.0)sin β + 40.376 cos β = −296 m/s 2 aP = aD a P = 296 m/s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1168 PROBLEM 15.127 Knowing that at the instant shown rod AB has a constant angular velocity of 6 rad/s clockwise, determine the acceleration of Point D. SOLUTION ω AB = 6 rad/s Velocity analysis. v B = ( AB)ω AB = (90)(6) = 540 mm/s v B = vB , v D = vD The instantaneous center of bar BDE lies at ∞. ωBD = 0 and vD = vB = 540 mm/s Then ωCD = vD 540 = = 3 rad/s CD 180 α AB = 0 Acceleration analysis. 2 a B = ( AB)ω AB = [(90)(6)2 ] = 3240 mm/s 2 a D = [(CD)α CD 2 ] + [(CD )ωCD ] = [180 α CD = [180α CD ] + [1620 mm/s 2 ] a D/B = [90α BD 2 ] + [225α BD ] + [225ωBD = [90α BD Resolve into components. −1620 = − 3240 + 225α BD , : α BD = 7.2 rad/s 2 180α CD = 0 + (90)(7.2), a D = [3240 ] + [(90)(7.2) = [648 2 ] + [90 ωBD ] ] + [225α BD ] a D = a B + a D/B : ] + [(180)(3) 2 ] α CD = 3.6 rad/s 2 ] + [(225)(7.2) ] ] + [1620 mm/s 2 ] = 1745 mm/s 2 68.2° a D = 1.745 m/s 2 68.2° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1179 PROBLEM 15.150 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in rod BD and by the collar that slides on rod AE. Knowing that at the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P. ω AE = 8 rad/s, ωBD = 3 rad/s SOLUTION AB = 500 mm = 0.5 m, AP = 0.5 tan 30°, BP = ω AE = 8 rad/s 0.5 cos 30° ω BD = 3 rad/s , Let P′ be the coinciding point on AE and u1 be the outward velocity of the collar along the rod AE. v P = v P′ + v P/ AE = [( AP)ω AE ] + [u1 ] Let P′′ be the coinciding point on BD and u2 be the outward speed along the slot in rod BD. v P = v P′′ + v P/BD = [( BP)ωBD 30°] + [u2 60°] Equate the two expressions for v P and resolve into components. : u1 = 1.5 + 0.5u2 or : u2 = From (1), 0.5 u1 = (3)(cos30°) + u2 cos 60° cos30° (1) 0.5 −(0.5 tan 30°)(8) = − (3)sin 30° + u2 sin 60° cos30° 1 [1.5 tan 30° − 4 tan 30°] = −1.66667 m/s sin 60° u1 = 1.5 + (0.5)(−1.66667) = 0.66667 m/s v P = [(0.5 tan 30°)(8) ] + [0.66667 ] = [2.3094 m/s ] + [0.66667 m/s ] vP = − 2.30942 + 0.66667 2 = 2.4037 m/s tanβ = 2.3094 0.66667 β = 73.9° v P = 2.40 m/s 73.9° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1212 PROBLEM 15.163 The motion of blade D is controlled by the robot arm ABC. At the instant shown, the arm is rotating clockwise at the constant rate ω = 1.8 rad/s and the length of portion BC of the arm is being decreased at the constant rate of 250 mm/s. Determine (a) the velocity of D, (b) the acceleration of D. SOLUTION i =1 Unit vectors: Units: meters, m/s, m/s , j =1 , k= 2 rD /A = (0.32 m)i − (0.24 m) j Motion of Point D′ of extended frame AB. ω = −(1.8 rad/s)k α=0 v D′ = ω × rD /A = ( −1.8k ) × (0.32i − 0.24 j) = −0.432i − 0.576 j a D′ = α × rD /A − ω 2 (rD /A ) = 0 − (1.8)2 (0.32i − 0.24 j) = −1.0368i + 0.7776 j Motion of Point D relative to frame AB. v D /AB = 250 mm/s 25° = −(0.25cos 25°)i + (0.25sin 25°) j = −0.22658i + 0.10565 j a D /AB = 0 Coriolis acceleration 2ω × v D /AB = (2)(−1.8k ) × (−0.22658i + 0.10565 j) = 0.38034i + 0.81569 j (a) Velocity of Point D. v D = v D′ + v D /AB = −0.432i − 0.576 j − 0.22658i + 0.10565 j = −0.65858i − 0.47035 j v D = (0.659 m/s)i − (0.470 m/s) j = 0.809 m/s 35.5° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1227 PROBLEM 15.163 (Continued) (b) Acceleration of Point D. a D = a D′ + a D /AB + 2ω × v D /AB = −1.0368i + 0.7776 j + 0 + 0.38034i + 0.81569 j = −0.6565i + 1.5933 j a D = −(0.657 m/s 2 )i + (1.593 m/s 2 ) j = 1.723 m/s 2 67.6° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1228 PROBLEM 15.168 A chain is looped around two gears of radius 40 mm that can rotate freely with respect to the 320-mm arm AB. The chain moves about arm AB in a clockwise direction at the constant rate of 80 mm/s relative to the arm. Knowing that in the position shown arm AB rotates clockwise about A at the constant rate ω = 0.75 rad/s, determine the acceleration of each of the chain links indicated. Links 1 and 2. SOLUTION Let the arm AB be a rotating frame of reference. Link 1: Ω = 0.75 rad/s = −(0.75 rad/s)k : r1 = −(40 mm)i, v1/AB = u ↑ = (80 mm/s) j a1′ = −Ω 2 r1 = −(0.75)2 ( −40) = (22.5 mm/s)i 802 160 mm/s → = (160 mm/s 2 )i 40 ρ = (2)( −0.75k ) × (80 j) a1/AB = 2Ω × v P /AB u2 = = (120 mm/s)i a1 = a1′ + a1/AB + 2Ω × v1/AB = (302.5 mm/s 2 )i Link 2: a1 = 303 mm/s 2 r2 = (160 mm)i + (40 mm) j v 2/AB = u → = (80 mm/s)i a′2 = −Ω 2 r2 = −(0.75)2 (160i + 40 j) a 2/AB = −(90 mm/s 2 )i − (22.5 mm/s 2 ) j =0 2Ω × v 2/AB = (2)(−0.75k ) × (80i) = −(120 mm/s 2 ) j a 2 = a′2 + a 2/AB + 2Ω × v 2/AB = −90i − 22.5 j − 120 j = −(90 mm/s 2 )i − (142.5 mm/s 2 ) j a 2 = (90) 2 + (142.5) 2 = 168.5 mm/s 2 tan β = 142.5 , β = 57.7° 90 a 2 = 168.5 mm/s 2 57.7° PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 1235

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