PROBLEM 15.38 An automobile travels to the right at a constant speed of 48 mi/h. If the diameter of a wheel is 22 in., determine the velocities of Points B, C, D, and E on the rim of the wheel. SOLUTION vA = 48 mi/h = 70.4 ft/s d = 22 in. r = ω= vC = 0 d = 11 in. = 0.91667 ft 2 vA 70.4 = = 76.8 rad/s 0.91667 r vB/A = vD/A = vE/A = rω = (0.91667)(76.8) = 70.4 ft/s vB = vA + vB/A = [70.4 ft/s ] + [70.4 ft/s ] vB = 140.8 ft/s vD = vA + vD/A = [70.4 ft/s vE = vA + vE/A = [70.4 ft/s ] + [70.4 ft/s 30°] vD = 136.0 ft/s 15.0° vE = 99.6 ft/s 45.0° ] + [70.4 ft/s ] 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. 1052 PROBLEM 15.50 Arm AB rotates with an angular velocity of 20 rad/s counterclockwise. Knowing that the outer gear C is stationary, determine (a) the angular velocity of gear B, (b) the velocity of the gear tooth located at Point D. SOLUTION Arm AB: Gear B: (a) BE = 0.05 m: v B = vD + vB/E = 0 + ( BE )ωB 2.4 m/s = 0 + (0.05 m)ω B ωB = 48 rad/s (b) DE = (0.05 2): ω B = 48 rad/s vD = vE + vD/E = 0 + ( DE )ωB vD = 0 + (0.05 2)(48) vD = 3.39 m/s vD = 3.39 m/s 45° 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. 1067 PROBLEM 15.57 A straight rack rests on a gear of radius r and is attached to a block B as shown. Denoting by ωD the clockwise angular velocity of gear D and by θ the angle formed by the rack and the horizontal, derive expressions for the velocity of block B and the angular velocity of the rack in terms of r, θ , and ωD . SOLUTION Gear D: Rotation about D. Tooth E is in contact with rack AB. vE = rωD lEB = Rack AB: Plane motion = θ r tan θ + Translation with E vB = vE + vB/E [ vB Rotation about E. ] = [ vE θ ] + [vB/E θ ] Draw velocity vector diagram. vB = vE rωD = cos θ cos θ vB = rωD cos θ vB/E = vE tan θ = rωD tan θ ω AB = = vB/E lEB rω D tan θ r tan θ = ω D tan 2 θ ω AB = ωD tan 2 θ 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. 1074 PROBLEM 15.76 A 60-mm-radius drum is rigidly attached to a 100-mm-radius drum as shown. One of the drums rolls without sliding on the surface shown, and a cord is wound around the other drum. Knowing that end E of the cord is pulled to the left with a velocity of 120 mm/s, determine (a) the angular velocity of the drums, (b) the velocity of the center of the drums, (c) the length of cord wound or unwound per second. SOLUTION Since the drum rolls without sliding, its instantaneous center lies at D. v E = v B = 120 mm/s v A = v A /Dω , (a) ω= vB = rB/Dω vB 120 = = 3 rad/s vB/D 100 − 60 ω = 3.00 rad/s (b) v A = (100)(3) = 300 mm/s v A = 300 mm/s Since v A is greater than vB , cord is being wound. v A − vB = 300 − 120 = 180 mm/s Cord wound per second = 180.0 mm (c) 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. 1104 PROBLEM 15.83 Rod ABD is guided by wheels at A and B that roll in horizontal and vertical tracks. Knowing that at the instant shown β = 60° and the velocity of wheel B is 40 in./s downward, determine (a) the angular velocity of the rod, (b) the velocity of Point D. SOLUTION Rod ABD: We locate the instantaneous center by drawing lines perpendicular to vA and vD. (a) Angular velocity. vB = ( BC )ω 40 in./s = (12.99 in.)ω ω = 3.079 rad/s (b) ω = 3.08 rad/s Velocity of D: In ΔCDE: γ = tan −1 7.5 25.98 = 16.1°; CD = = 27.04 in. 25.98 cos γ vD = (CD )ω = (27.04 in.)(3.079 rad/s) = 83.3 in./s vD = 83.3 in./s vD = 83.3 in./s 16.1° 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. 1112 PROBLEM 15.88 Rod AB can slide freely along the floor and the inclined plane. Denoting by vA the velocity of Point A, derive an expression for (a) the angular velocity of the rod, (b) the velocity of end B. SOLUTION Locate the instantaneous center at intersection of lines drawn perpendicular to vA and vB . Law of sines. AC BC = sin[90° − ( β − θ )] sin(90° − θ ) l = sin β AC BC = cos( β − θ ) cos θ l = sin β cos( β − θ ) AC = l sin β cos θ BC = l sin β (a) Angular velocity: vA = ( AC )ω = l cos( β − θ ) ω sin β (b) Velocity of B: vB = ( BC )ω = l cos θ vθ sin β ⋅ ⋅ sin β l cos( β − θ ) ω= vA sin β ⋅ l cos( β − θ ) vB = vA cos θ cos( β − θ ) 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. 1118
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