capitulo 15b mecanica vectorial para ingenieros Dinamica, Apuntes de Dinámica

¡Descarga capitulo 15b mecanica vectorial para ingenieros Dinamica y más Apuntes en PDF de Dinámica solo en Docsity! PROBLEM 15.153 Two rotating rods are connected by slider block P. The velocity vy of the slider block relative to the rod is constant and 15 directed outward, For the given data, determine the angular velocity of each rod in the position shown. b=8i0., 1 =9 ins. SOLUTION Dimensions: AP _ BP a sin20% sin60% — sin40 AP= 4,2567 in. BP=10.7784 in. Velocities, Note: P"= Pointof 4D coinciding with P. Vo E Vert Vara [y E 709] = [vpo 30% +19 in./5%. 609] PROPRIETARY MATERIAL. 3 2009 The MeGrew-Hil Companies, Inc. AU rights reserved, No part af ás Manual reproduced or distributed dx any forma or by any mecns, Prrap tan 400 _Dinds tan 40? ES =10.726 m/s _ 10.726 ins 4.2567 2.5197 rad/s Pra sin 40* _ 2 in,s sin40* in. 10 ¡p 2.52 vad/s ) 4 =14.002 in./s Pe Pap A BP 14.002 in.fs 10.7784 in. = 1.2900 rad's Op =1.299 radis ) 4 may be «displayed, wishatet the prior written permission of the publisher, or used beyond the limited distribution ta teachers and edacators permiica by MeGrav- Hill for ihoir individual conse preparation. lyow are a studentasina this Mariel, YO are asin ib ethont permission. 1153  PROBLEM 15.154 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in rod 8D and by the collar that slides on rod 4£. Knowing that at the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P. 044 radís, 0% =1.5 rad/s. SOLUTION 20 cos 30 00,4 = 4 ras ), gp =1.5 radís ) AB=2Min, AP=20tan30, BP= Let P” be the coinciding pointon 4E and «, be the outward velocity of the collar along the rod AE. Ve = Y p Y =APJO | Da +] Let P” be the coinciding point on BD and 4, be the outward spced along the slot in rod BD. Vo = jo + Vo = BP) 309] + [a 2 607] Equate the two expressions for v, and resolve into components, tr: -( 29 5 J0-5Xcos300) + cos 607 cos 30% or u =30+0,5u, (1) E (030394) = ll S)sin30% +1 sin 60% cos 30% in 60% sin From Eg. (1), 2 =304(0.51-33.333)=13.333 in./s Uy = [80 tan 30730 tan 30%] = 33.393 in./s Yo =[(20 tan 3004) |] + [13.333 —»] =[46.188 in.fs |] + [13.333 ins —-] y, =-/46,188* 413.339 =2.4037 m/s 46.188 t ap Mb PB=13.9 v»=348.1 in/5 73,99 « PROPRIETARY MATERIAL. 632009 Tho McGraw-Hill Companies, Inc. All rights rescrvod. Mo part of his Manual may be aisplayed. veproducod ar distributed in any foro ar by cry mien, without dhe prior wsriticn permission of the publisher, or used beyond (he Limited distibution to teachers ard educalors permitted by MeGrew-1HH for thelr individual conse preparation. Ifyow are a srudent using ais Marvel, poa are asta Evithort permission 1154  he] = a 30? z ” 27D B zo" PROBLEM 15.156 (Continued) (4) £=60% Geometry. Equilateral triangle: AH =BH =] Rod AE and block A: Y EV jo =[(d Ha 2 60%] +[Yypy 5309] Y = [fa GOA [Y pgs 0 309] (1) Rod 20 and block H: Va Vi o [BH 0% 60%+ [vs 230%] Yy = ea 60% + [vay 307] e) Equate eight-hand members of Eqs. (1) and (2): [co 60 THEY ya 309 = [leo 60%] + [ip 300] Vector diagram: — Yayo = Py 160 tan 307 = == 3 lar y + 530% 4 Y = MAR 5 Winap = e A PROPRIETARY MATERIAL. 40 2009 The MeGraw-HiH Companies, Inc, All riglus reservud. No part of this Máncal smay be displayed, reproduced or distributed de any farm o by emy means, voithout the prior written permission ale publisher. or used beyond ile Eiited distribatior so tecchers amd exfueators permitted by MeGrawEiK jor tir individaucl conexo preperatian, Ifvomr are e student using dis Manual. porare using owithoid permission. 1157  PROBLEM 15.157 Two rods 4£ and 2D pass through holes drilled into a hexagonal block, (The holes are deilled in different planes so thal the rods will not touch each other.) Knowing that at the instant considered rod AE rotates counterclockwise with a constant angular velocity 0, determine, for the given data, the relative velocity of the block with respect to each rod. O 450. SOLUTION A Angle between rods is constant, therefore | Y WD yz 0 pp 0 5 A Geometry: | PS A E Law of sines. 24 = Ea sin75% sinds? od sin 60? AM =1.1157 BH =0.8165) Rod 4£ and block A: Rod BD and block H: Yi Vr RV =[(4 HD 2457 + [vz 245%] Y =[1,115700.42 459] + [vg4g 33 459] (1 Pa War PV iran =[(80 075 +[v gap 215% Y =[0.816510005 75% +[Yym9p 415%) Equate right-hand members of Eqs. (1) and (2) [111520 245%] +[v)y 44 Do. 45% 3 ([0.8165/00%757+[v jpg 15%] a) PROPRIETARY MATERIAL. %> 2009 The MeGrsw-Hill Companies, Ino, All rights reserved. No part of this Manual may be displayes, reproduced or distributed in any fora or by any means, without fhe prior wrillen permission ef the publisher, er used beyond the limited distribution ta teachers and educators permitted by MoGraw-Hill foriheit individual conse preparation. If pow are a stedent using Hrs Mannc, gn ar tising dihont permission, 1158  PROBLEM 15.157 (Continued) Vector diagram. Equate components in direction parallel to Vyr ALAS? Equate components in direction parallel to. y ¿« 7 609 (08165100) cos 60%+v yy, cos300=1.115/00 Virap =+0.816/0 Vigo = 0-86 150 Equate components in direction perpendicular to Y yp« a. 450 (0,8 165/02)5in 60" v,ypy Sin 309 = Vip yz: (0.8165/03)sin 60*—(0.816/0sin30* = v4.1 Vopae =+0.299l00 Vriras = 02991007. 45% «q PROPRIETARY MATERIAL. 9 2009 The McGraw-Hill Compames, Inc, All rights seserved, No part of this Mannal may be displayed, reproduced or distributed im any form or by any means, withotl the prior written permission of the publisher, or ved beyond the mited distribution to teachers and vdncotors permitted by MoGrow-Elil for their máividuad course preparation If yor are a student using tés Mannal, ponare usiñy ibrithout permission. 1159  PROBLEM 15.160 At the instant shown the length of the boom 42 is being decreased al the constant rate of 0.2 nvsand the boom is being lowered at the constant rate of 0.08 radís, Determine (a) the velocity of Point E, (6) the acceleration of Point.B, SOLUTION Velocity af coinciding Point Bl on boom. Y y = 0 =(6)(0.08) = 0.48 més 5 60 Velocity 0f Point B relative to the boo, (a) Y amoo = 0.2 m/s 7 308 Velocity af Point B. Va = YY soon >: (1), =0,48c08 60" - 0.2c0530%=0.06680 m/s +. (11), =-0.48sin 60*—0.2sin30% =-0.51569 mis Ya = V0.066807 +0,51569* =0.520 mís _ 051569 tan f= % É 0.06680 PB=-82.6" Y y =0,.520 m/s 82.60 «€ Acceleration af coincidine Point Bon boom. Ay =re? =[6X0.08)? = 0.0384 mis? 9 300 Acceleration af B relative ta the boom. Boom = Ú Corialis acceleration. 20m =(2)(0.0810.2)= 0.032 mis? Eu. 602 PROPRIETARY MATERIAL. £5 2009 The Mobraw-Hi1 Companies, inc. All rights reserved. No part of is Munsal inuy be displayed. seproduced or distributed dm any form or by emy arcans, without the prior wise permission ef de publisher, or usod beyond the Hinited distibriion to teachers end cducators permitted by MeGronHRH for tele individacl comse preparation. JPyarcaro a stident using dis Marncal, poa are using ébowvithord permission. 1162  PROBLEM 15.160 (Continued) (b) Acceleration of Point B. Ay Ay +A room + 2000 +: (ap), =-0.0384c0830%+0-0.032c0560”=-0.04926 m/s? + (ap), =0.0384sin 3004 0+ 0.032 sin 60? = 0.008513 mis? ay =4/(0.04926)" +(0,008513)' =00.0500 m/s? _ 0.008513 tan P= p 0.04926 P=9.8 a, =S0.0 mms 29,8% « PROPRJETARY MATERITAL. 0 2009 The MeGraw-Hill Companies, Ino. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in emy form or by any means, withou! the prior writien permission of the publisher, or used heyond the limited distribution to teachers and educators permblted hy MeGraweHlill foraheir individual conse preparation. Afyos are a student using this Mansal, yomVJare using iwaihout permissior 1163  PROBLEM 15.161 At the instant shown, the length of the boom 48 is being increased at the constant rate of 0.2 nvs 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 oP Point 8, SOLUTION Veloció! of coinciding Pole Bloa boom. Y y = +60 = (0110.08) = 0.48 més 607 Velocity 0f Point B relative to the boom. Y aso = 0.2 m/s 7 30% (a) Velocityaf Point b. Va Vado +: (19), =0.48 cos 60" 40.2 cos 30% = 0.4132 mís +h (Yy), = 0.98 sin 6074 0.2 sin 30”=-0,3157 nfs vo = (0.4132) + (0.3157) = (0.520 m/s 0.3157 ta ==- MP ana Bara? V y = 0.520 mis 37.40 q Acceleration ef coinciding Point E" on boom. a =707 = (60.08) = 0.0384 m/s? 7300 Acceleration of B relative to the boom. Mimo = 9 Coriolis acceleration. 2 an =(2)(0.08)(2) = 0.032 m/s? 609 PROPRIETARY MATERIAL. 3 2009 The MeGraw-Hill Companies, Inc. AU rights reserved, No part of this Manual may be displayed. reproduced or distributed in any form or by any ercans, wilhent the prior written permission of the publisher, vr used beyond the Limited distribution to teacher end educators permitted by MeGraw-HBN fo their inclividual course preparation. Ifyou ave a student using this Marnnal, pos are usiag irrita permission. 1164  1 PROBLEM 15.163 a al ie %1% The sleeve BC is welded to-an arm that rotates about 4 with a constant angular velocity (2 In the position shown rod DF is being moved to the left al a constant speed a1=16 in./s relative to the slecve. For the given angular velocity to, determine the acceleration (a) of Point D, (6) of the point of rod DF that coincides with Polnt £. (1M=(3 rad/s)j. Vo — Vaio = 016 Mn ts)k; Aja =D AD=<5 0+(12 in.Jk Y y Fx AD =(3 ads ij [AS 1044 (12 mk] = (36 i0.45)i Uy A MAY py = (3 tad's)jx (36 in./s)i =-—(108 in.fs*)k AL = XV pj =(3 radés)j < (16 in./s)k = (96 in.fs?)I A) My 8 payo HR =-(108 in./s?)le+ 0 (96 in./s?)i Ap =[96 in.fs*)i— (108 in.isó)k 4 OLOF that coinmcides with dk. Var = Yao = [16 m.és)k, Ap =0 AE =-S ini vo =0X AE =(3 radís)ix (SS in./s)j=0 Ry MX WA =0 A ON jo = 28 rad/Sp<Cl6 in./s)k =(96 inJs*)i Ap = A PA o HA ap =0404(96 in./s*)i PI | PROPRIETARY MATERIAL. 63 2000 The MeGraweHill Companies, loc. Al rights reserved, No part of this: Manual mayo be displayed, repeodiecd vr distributed da aro Formar dy amy teca. wilhoat de prior weiflen permission uf de publisher, or used beyond the limited distriburion to teuchers and edicators permitied by MeGran Hi fia individual conse preparation. Ifyon are o student using Dis Mena! power sig ib withioal permission. 1167  PROBLEM 15.164 The cage oP a mine elevator moves downward at a constant speed of 40 ftís. Determine the magnitude and direction of the Coriolis acceleration of the cage ¡the elevator is located (a) al the equator, (4) at latitude 40% north, (c) at latitude 40% south. SOLUTION Earth makes one revolution (277 radians) in 23.933 4 (86,160 s). ; E 2. eo MZ =(72.926x10% rad la Velocity relative to ihe Barth at latitude angle q. Y picas = IO cos pi sin pa) Coriolis acceleration 2. MS TODA =(272.026:x10 ¡)x[40(=cos oi —sin q)] = (5.834 1107 cos )k (a) p=0% cosp=1.000 A =5.88:10 5 west «d (6) — p=40% cos p=0.76604 aj =447 10 5? west 4 (e) p=-40%, cos =0.76604 a =34.47x10 48? west «l PROPRIETARY MATERIAL, €: 2000 The Meúraw-Hill Companies, ac. All rights reserved, No part of iia Mantal muy be displayed reproduced or distributed in any fora or hy any meca, with the prior epitten permisxioo of ie publisher orouscd begond te limited divicibución to secehers and edncarors permitied by MoGrow-HlHL for tele ndividial course preparation. 1 yorcare a student using thés Mana. pam are tising twithont permission 1168  PROBLEM 15.153 Two rotating rods are connected by slider block P. The velocity vy of the slider block relative to the rod is constant and 15 directed outward, For the given data, determine the angular velocity of each rod in the position shown. b=8i0., 1 =9 ins. SOLUTION Dimensions: AP _ BP a sin20% sin60% — sin40 AP= 4,2567 in. BP=10.7784 in. Velocities, Note: P"= Pointof 4D coinciding with P. Vo E Vert Vara [y E 709] = [vpo 30% +19 in./5%. 609] PROPRIETARY MATERIAL. 3 2009 The MeGrew-Hil Companies, Inc. AU rights reserved, No part af ás Manual reproduced or distributed dx any forma or by any mecns, Prrap tan 400 _Dinds tan 40? ES =10.726 m/s _ 10.726 ins 4.2567 2.5197 rad/s Pra sin 40* _ 2 in,s sin40* in. 10 ¡p 2.52 vad/s ) 4 =14.002 in./s Pe Pap A BP 14.002 in.fs 10.7784 in. = 1.2900 rad's Op =1.299 radis ) 4 may be «displayed, wishatet the prior written permission of the publisher, or used beyond the limited distribution ta teachers and edacators permiica by MeGrav- Hill for ihoir individual conse preparation. lyow are a studentasina this Mariel, YO are asin ib ethont permission. 1153  PROBLEM 15.154 Pin P is attached to the collar shown; the motion of the pin is guided by a slot cut in rod 8D and by the collar that slides on rod 4£. Knowing that at the instant considered the rods rotate clockwise with constant angular velocities, determine for the given data the velocity of pin P. 044 radís, 0% =1.5 rad/s. SOLUTION 20 cos 30 00,4 = 4 ras ), gp =1.5 radís ) AB=2Min, AP=20tan30, BP= Let P” be the coinciding pointon 4E and «, be the outward velocity of the collar along the rod AE. Ve = Y p Y =APJO | Da +] Let P” be the coinciding point on BD and 4, be the outward spced along the slot in rod BD. Vo = jo + Vo = BP) 309] + [a 2 607] Equate the two expressions for v, and resolve into components, tr: -( 29 5 J0-5Xcos300) + cos 607 cos 30% or u =30+0,5u, (1) E (030394) = ll S)sin30% +1 sin 60% cos 30% in 60% sin From Eg. (1), 2 =304(0.51-33.333)=13.333 in./s Uy = [80 tan 30730 tan 30%] = 33.393 in./s Yo =[(20 tan 3004) |] + [13.333 —»] =[46.188 in.fs |] + [13.333 ins —-] y, =-/46,188* 413.339 =2.4037 m/s 46.188 t ap Mb PB=13.9 v»=348.1 in/5 73,99 « PROPRIETARY MATERIAL. 632009 Tho McGraw-Hill Companies, Inc. All rights rescrvod. Mo part of his Manual may be aisplayed. veproducod ar distributed in any foro ar by cry mien, without dhe prior wsriticn permission of the publisher, or used beyond (he Limited distibution to teachers ard educalors permitted by MeGrew-1HH for thelr individual conse preparation. Ifyow are a srudent using ais Marvel, poa are asta Evithort permission 1154  he] = a 30? z ” 27D B zo" PROBLEM 15.156 (Continued) (4) £=60% Geometry. Equilateral triangle: AH =BH =] Rod AE and block A: Y EV jo =[(d Ha 2 60%] +[Yypy 5309] Y = [fa GOA [Y pgs 0 309] (1) Rod 20 and block H: Va Vi o [BH 0% 60%+ [vs 230%] Yy = ea 60% + [vay 307] e) Equate eight-hand members of Eqs. (1) and (2): [co 60 THEY ya 309 = [leo 60%] + [ip 300] Vector diagram: — Yayo = Py 160 tan 307 = == 3 lar y + 530% 4 Y = MAR 5 Winap = e A PROPRIETARY MATERIAL. 40 2009 The MeGraw-HiH Companies, Inc, All riglus reservud. No part of this Máncal smay be displayed, reproduced or distributed de any farm o by emy means, voithout the prior written permission ale publisher. or used beyond ile Eiited distribatior so tecchers amd exfueators permitted by MeGrawEiK jor tir individaucl conexo preperatian, Ifvomr are e student using dis Manual. porare using owithoid permission. 1157  PROBLEM 15.157 Two rods 4£ and 2D pass through holes drilled into a hexagonal block, (The holes are deilled in different planes so thal the rods will not touch each other.) Knowing that at the instant considered rod AE rotates counterclockwise with a constant angular velocity 0, determine, for the given data, the relative velocity of the block with respect to each rod. O 450. SOLUTION A Angle between rods is constant, therefore | Y WD yz 0 pp 0 5 A Geometry: | PS A E Law of sines. 24 = Ea sin75% sinds? od sin 60? AM =1.1157 BH =0.8165) Rod 4£ and block A: Rod BD and block H: Yi Vr RV =[(4 HD 2457 + [vz 245%] Y =[1,115700.42 459] + [vg4g 33 459] (1 Pa War PV iran =[(80 075 +[v gap 215% Y =[0.816510005 75% +[Yym9p 415%) Equate right-hand members of Eqs. (1) and (2) [111520 245%] +[v)y 44 Do. 45% 3 ([0.8165/00%757+[v jpg 15%] a) PROPRIETARY MATERIAL. %> 2009 The MeGrsw-Hill Companies, Ino, All rights reserved. No part of this Manual may be displayes, reproduced or distributed in any fora or by any means, without fhe prior wrillen permission ef the publisher, er used beyond the limited distribution ta teachers and educators permitted by MoGraw-Hill foriheit individual conse preparation. If pow are a stedent using Hrs Mannc, gn ar tising dihont permission, 1158  PROBLEM 15.157 (Continued) Vector diagram. Equate components in direction parallel to Vyr ALAS? Equate components in direction parallel to. y ¿« 7 609 (08165100) cos 60%+v yy, cos300=1.115/00 Virap =+0.816/0 Vigo = 0-86 150 Equate components in direction perpendicular to Y yp« a. 450 (0,8 165/02)5in 60" v,ypy Sin 309 = Vip yz: (0.8165/03)sin 60*—(0.816/0sin30* = v4.1 Vopae =+0.299l00 Vriras = 02991007. 45% «q PROPRIETARY MATERIAL. 9 2009 The McGraw-Hill Compames, Inc, All rights seserved, No part of this Mannal may be displayed, reproduced or distributed im any form or by any means, withotl the prior written permission of the publisher, or ved beyond the mited distribution to teachers and vdncotors permitted by MoGrow-Elil for their máividuad course preparation If yor are a student using tés Mannal, ponare usiñy ibrithout permission. 1159  PROBLEM 15.160 At the instant shown the length of the boom 42 is being decreased al the constant rate of 0.2 nvsand the boom is being lowered at the constant rate of 0.08 radís, Determine (a) the velocity of Point E, (6) the acceleration of Point.B, SOLUTION Velocity af coinciding Point Bl on boom. Y y = 0 =(6)(0.08) = 0.48 més 5 60 Velocity 0f Point B relative to the boo, (a) Y amoo = 0.2 m/s 7 308 Velocity af Point B. Va = YY soon >: (1), =0,48c08 60" - 0.2c0530%=0.06680 m/s +. (11), =-0.48sin 60*—0.2sin30% =-0.51569 mis Ya = V0.066807 +0,51569* =0.520 mís _ 051569 tan f= % É 0.06680 PB=-82.6" Y y =0,.520 m/s 82.60 «€ Acceleration af coincidine Point Bon boom. Ay =re? =[6X0.08)? = 0.0384 mis? 9 300 Acceleration af B relative ta the boom. Boom = Ú Corialis acceleration. 20m =(2)(0.0810.2)= 0.032 mis? Eu. 602 PROPRIETARY MATERIAL. £5 2009 The Mobraw-Hi1 Companies, inc. All rights reserved. No part of is Munsal inuy be displayed. seproduced or distributed dm any form or by emy arcans, without the prior wise permission ef de publisher, or usod beyond the Hinited distibriion to teachers end cducators permitted by MeGronHRH for tele individacl comse preparation. JPyarcaro a stident using dis Marncal, poa are using ébowvithord permission. 1162  PROBLEM 15.160 (Continued) (b) Acceleration of Point B. Ay Ay +A room + 2000 +: (ap), =-0.0384c0830%+0-0.032c0560”=-0.04926 m/s? + (ap), =0.0384sin 3004 0+ 0.032 sin 60? = 0.008513 mis? ay =4/(0.04926)" +(0,008513)' =00.0500 m/s? _ 0.008513 tan P= p 0.04926 P=9.8 a, =S0.0 mms 29,8% « PROPRJETARY MATERITAL. 0 2009 The MeGraw-Hill Companies, Ino. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in emy form or by any means, withou! the prior writien permission of the publisher, or used heyond the limited distribution to teachers and educators permblted hy MeGraweHlill foraheir individual conse preparation. Afyos are a student using this Mansal, yomVJare using iwaihout permissior 1163  PROBLEM 15.161 At the instant shown, the length of the boom 48 is being increased at the constant rate of 0.2 nvs 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 oP Point 8, SOLUTION Veloció! of coinciding Pole Bloa boom. Y y = +60 = (0110.08) = 0.48 més 607 Velocity 0f Point B relative to the boom. Y aso = 0.2 m/s 7 30% (a) Velocityaf Point b. Va Vado +: (19), =0.48 cos 60" 40.2 cos 30% = 0.4132 mís +h (Yy), = 0.98 sin 6074 0.2 sin 30”=-0,3157 nfs vo = (0.4132) + (0.3157) = (0.520 m/s 0.3157 ta ==- MP ana Bara? V y = 0.520 mis 37.40 q Acceleration ef coinciding Point E" on boom. a =707 = (60.08) = 0.0384 m/s? 7300 Acceleration of B relative to the boom. Mimo = 9 Coriolis acceleration. 2 an =(2)(0.08)(2) = 0.032 m/s? 609 PROPRIETARY MATERIAL. 3 2009 The MeGraw-Hill Companies, Inc. AU rights reserved, No part of this Manual may be displayed. reproduced or distributed in any form or by any ercans, wilhent the prior written permission of the publisher, vr used beyond the Limited distribution to teacher end educators permitted by MeGraw-HBN fo their inclividual course preparation. Ifyou ave a student using this Marnnal, pos are usiag irrita permission. 1164  1 PROBLEM 15.163 a al ie %1% The sleeve BC is welded to-an arm that rotates about 4 with a constant angular velocity (2 In the position shown rod DF is being moved to the left al a constant speed a1=16 in./s relative to the slecve. For the given angular velocity to, determine the acceleration (a) of Point D, (6) of the point of rod DF that coincides with Polnt £. (1M=(3 rad/s)j. Vo — Vaio = 016 Mn ts)k; Aja =D AD=<5 0+(12 in.Jk Y y Fx AD =(3 ads ij [AS 1044 (12 mk] = (36 i0.45)i Uy A MAY py = (3 tad's)jx (36 in./s)i =-—(108 in.fs*)k AL = XV pj =(3 radés)j < (16 in./s)k = (96 in.fs?)I A) My 8 payo HR =-(108 in./s?)le+ 0 (96 in./s?)i Ap =[96 in.fs*)i— (108 in.isó)k 4 OLOF that coinmcides with dk. Var = Yao = [16 m.és)k, Ap =0 AE =-S ini vo =0X AE =(3 radís)ix (SS in./s)j=0 Ry MX WA =0 A ON jo = 28 rad/Sp<Cl6 in./s)k =(96 inJs*)i Ap = A PA o HA ap =0404(96 in./s*)i PI | PROPRIETARY MATERIAL. 63 2000 The MeGraweHill Companies, loc. Al rights reserved, No part of this: Manual mayo be displayed, repeodiecd vr distributed da aro Formar dy amy teca. wilhoat de prior weiflen permission uf de publisher, or used beyond the limited distriburion to teuchers and edicators permitied by MeGran Hi fia individual conse preparation. Ifyon are o student using Dis Mena! power sig ib withioal permission. 1167  PROBLEM 15.164 The cage oP a mine elevator moves downward at a constant speed of 40 ftís. Determine the magnitude and direction of the Coriolis acceleration of the cage ¡the elevator is located (a) al the equator, (4) at latitude 40% north, (c) at latitude 40% south. SOLUTION Earth makes one revolution (277 radians) in 23.933 4 (86,160 s). ; E 2. eo MZ =(72.926x10% rad la Velocity relative to ihe Barth at latitude angle q. Y picas = IO cos pi sin pa) Coriolis acceleration 2. MS TODA =(272.026:x10 ¡)x[40(=cos oi —sin q)] = (5.834 1107 cos )k (a) p=0% cosp=1.000 A =5.88:10 5 west «d (6) — p=40% cos p=0.76604 aj =447 10 5? west 4 (e) p=-40%, cos =0.76604 a =34.47x10 48? west «l PROPRIETARY MATERIAL, €: 2000 The Meúraw-Hill Companies, ac. All rights reserved, No part of iia Mantal muy be displayed reproduced or distributed in any fora or hy any meca, with the prior epitten permisxioo of ie publisher orouscd begond te limited divicibución to secehers and edncarors permitied by MoGrow-HlHL for tele ndividial course preparation. 1 yorcare a student using thés Mana. pam are tising twithont permission 1168  PROBLEM 15.165 A rocket sled is tested on a straight track that is built along a meridian. Knowing that the track is located at latitude 40* north, determine he Coriolis acceleration af the sled when it is moving noríh ata speed of 900 knvh. SOLUTION Earth makes one revolution (2 radians) in 23.933 h =86,160 s. Ze 86,160 =(72,926x10% radís)j Speed af sled. w= 900 km/h =250 mis Velocity of sled relative lo the Earth. Y picar = 250(sin pi4 cos pj) Coriolis acceleration. Mo = ZO Y ari a. =(20072.926x10* [250Gsin pi +eos pj)] =0.036463sin pk At latitude p=40%, a = 0036463 sin 40k =(0.0234 m/s? )k A =0.0234 m/s? west « PROPRIETARY MATERIAL, 05 2009 The MeGiraw-Hill Companies, hc. All rights reserved. No part of this Manual may be displayed, reproduced or distribiWed da ay jor or by any necas, widliont the prior written permission ef he publisher, ar used beyond the limited distribution fo teachers and educators permitted by MoGraw Hi for iheir individual cowse preparation. df yon are a stident astig this Manso! JONare ASÍaR Othon periastor. 1169  PROBLEM 15.167 Salve Problem 15.166, assumning that the direction of the relative velocity u is reversed so that portion 8D 1s being retracted. PROBLEM 15.166 The motion of nozzle D is controlled by arm AB. At the instant shown, the arm is rotating counterclockwise at the constant rate (4 =2.4 rad/s and portion £€ is being extended at the constant rate 1 =10án./s with respect to the arm. For each of the arrangements shown, determine the acceleration of the nozzle D. SOLUTION For each configuration, Pa = 01 in)i+(4 in7 Acceleration of coinciding Point D”. By OK a Pra ay =0- (2.4) (1i+4] =-(63.36 im./s* Ji—(23.04 in./s*)j Meceleration of Point D relative to arin AB. Born =D Length CD: ED=44 +4 =S in, Velocity of Point D relative to the arm AB, Case (a): Via = (10 in./s)i ) Case (b): Ya 3 Gi+ Aj) = (6 ini in] Coriolis acceleration. Case (a) Case (bj: Acceleration ef nozale D. (a) ZA Y (22.4k) (101) =-(48 in./s%)] AO x (6-8) = (08.4 in.) (28.8 i.%)j Ap = My Pa LOAN a, =-63.361-23.04/+48| =-(63.36 im.fs*)i (71.04 m.¿s*)j 2) =(6336 + (71.04) =95.2 ins? 71.04 . s ai P=483, Ap =95.2 in.ls? 48,3% « PROPRIETARY MATERIAL. € 2009 The Metirar-Hil Companies, Inc. All rights reserved. No part of this Mannal anay be displayed, reprodaced or distribirca in amv for or by an incas, withont (he prior vwólten periission of the publisher, or uscd beyond the limited distribution ta teachers end educotors permitted by MoGraw-Hi4 for beir individual conve preparation, Ifyorcarea student using ibis Mental, FOR are usinE dE ost pormissior un  (6) tan PROBLEM 15.167 (Continued) Ap = 63.361 23.044 38.41 28.8] =-(24.96 ins) (51.84 in.és?)j a)= 04.06 +(51. =57.52 ins? 18 O ao 404 dy =575 10.5 64,3% 4 PROPRIETARY MATERIAL, 0 2000 The McGraw-Hill Companies, hno. All rights reserved. No port of this Aarnial may be displayed, reproduced or distribircd di any Sora or dy any anecas, without de prior wrilien permission of the publisher, or used beyond ibie dinritod «disteiburion 10 4e0chers cmd ed Joa are asia withiand permiesion. 1173 sonorS permiticd by MeGraw-H4H jor theie tndidial cose preparation. Jfporare a students aj this dina,  PROBLEM 15.168 A chain is looped around two gears of radius 40 min that can rolaíe freely with respect to the 320-mn arm 48, The chain moves about arm 48 in a clockwise direction at the constant rate ol 80 mmés relalivo lo the arm. Knowing that in the position shown arm AB rotates clockwise about A at the constant rate (0=0,75 rad/s, determine the acceleration of each of the chain links indicated. Links ¿and 2. SOLUTION Let the arm 48 be a rotaling lrame ol reference. £2=0.75 rad/s =-—40.75 rad/s)k: Link de NA AO ml, y yg 7 = (80 mm/s)j ay r, =(0,75) (40) = (22.5 min/s)i a A 60 mms >= (160 mmis? Ji p 40 FOXY pg = (20.75) (801) = (120 mmn?s)i A = A] Hp RR pp = (302.5 mavsó)i a =303 mm/s. — 4 Link 2: r, =(160 mm)i+ (40 mun)]j il A a, = Ln, =-(0.75)*(1601+ 408) = (90 mms? )i (22.5 mm/s*)j Axiap =0 2 ar = (2H 0.75k (801) =-(120 mms?) A A A Vaj 901 22.5] 120] (090 min/só)i— (142.5 mmés?)j 00? +(142,5y H a = =168.5 mais? 3 - tan fs A B=si ye a, =168,5 mv 57,7% 4 PROPRIETARY MATERIAL, £3 2009 The MeGiriw-Hil Companies, luc. All rights reserved. No part of this Manual may be displayed, reproduced ar alstributed 1h any form or by oryomeans, without Hic prior written permission of de: publisher, ar used beyond Hue Jimited distribution to teachers and ecucators permitted hy MoGreneHiH for thenindividual conse preparation. [ver are a student asin dis Mama, Jomare usiag Fwiihont permission. 1174  PROBLEM 15.171 Rod 4B of length R=15 in. rotates about 4 with a constant elociwise angular velocity ax of 5 rad/s. At the same lime, cod 40 of lengih r=8in. rotates about B owith a constant counterclociónise angular —— velocity (da 0f3 tadís with respect to rod 48. Knowing e, that 6 =60%, determine for the position shown the Ri acceleralion of Point D. SOLUTION ata: R=15in, r=8in. 4=60* 0 =8k with a, =-S radís, 4 =0 00, =0k with (0 =3 radfs, 6, =0 LebF be a frame of reference rolating with the rod 4B. Then £L= «4 =4k Let DP” be the point of the frame coinciding with Point D, iy (A + rcos 8) + (esin 8) Corp = (reos B)i+(rsiné8)j Geometry: Velocity of Point D relative to the frame £, Wir = Oy ps n - (ak) x(rcos8i+rsin 85) = ariosin 0i+ cos 0j) Coriolis acceleration. MON y + a 0 [0,1 sin 8 1+ cos 8 j)] =-20 0r1cos 6 1+sin 05) Acceleration of Point D' in the frame. Bay = Vo =a UR +rcos O)i+rsin 8] Acceleration of Point D relative to the frame. , Mor Ep =-d% (reos Ai+rsin 0j) PROPRIETARY MATERIAL. €) 200% The MoGraw-Hill Companies, Inc. Alb rights tescrved. No part of His Manual may be displayed, reproduced or aisiributed in ary forn or by any means, wilhomd He prior weitica permission af the publisher, or used beyond the limited distribution to teachersand educotors prmitiod by MoGraw-HHU for Micir iodividial course preparation, 1 vow are e stdentasin dis Marat. por are usag ialhiod permission 1177  PROBLEM 15.171 (Continued) Acceleration ol Point O, By Ay PR po + ZO V p ay = Ria +200, +0 P (reos Bi+rsin 81) =[ a Ris (oy 10,) (reos 8 + rsin 05)] AS 15) 4543) (8c0s 60% +8sin 60%] =-(391 i./5%) (27.713 in.4s2)j Ap =392 int" 274.057 € PROPRIETARY MATERIAL, % 2009 The McGraw-Hill Compantes, lc. All rights reserved, Mo part ef is Menial may be displayed, reproduced or diswibrcd tn any Joricor dy any meats, without the prior written permission of ie publisher, or uscd beyond the drited aistriburion ta teachers and educators permined by MeGraw-HiH for drcir individral comrse preparation. Ifyomare astidentus iaa dis dlanal, Pon are rexirig dí avihoml perhriss ica, 117%  ¿ PROBLEM 15.172 The collar P slides oubward ata constant relative speed e along rod AB, . y which rotates counterclockwise with a constant angular velocity of e 20 rpm. Knowiog that r=250 mm when €=0 and that the collar E reaches E when € =907, determine the magnitude of the acceleration of the collar P justas it reaches B. SOLUTION > (20 ps L0RE) Le rad/s 60 3 a=0 0=90P.= E radians Uniform rotational motion. HE, + 001 Uniform motion along rod. FS hy oh rfi _ 050.25 _1 = == =-— ms, f 0.75 3 Vero = mis T Acceleration of coincidiag Point P?on the rod. (r=05m) ap = rar 2 2 = (0.51 =— oz) a == mis? l 9 =2.1932 m/s” Acceleration of collar P relative to the rod Aran =0 si 211 2 Coriolis acceteration. 20X Y py = 200 = (2) SEA 1.3963 m/s" Acceleration of collar P. Mp o py OA po a, =[2.1932 més” 1]+[1.3963 m/s? e] ap=2.60m/8) + 57,58 dp =2.60 m/s? 4 PROPRIETARY MATERIAL. 23 2009 The MeGcov-Hill Companies, lic. Ad rights reserved, No port of his Manaal may be displayed, reproduced or distributed 1 coty form or by any omcons, without the prior vrifiea permission of he publisher, or nsed beyond the limited disicibation to teachers and educators permitted hy MeGraw Hill forcthair individucal conse prepercrion. [vom area student using this Marnal, Jm are sing dtwvition! permission, 1179  PROBLEM 15.174 Pin £ slides in a circular slot cut in the plate shown ata constant relative specd a =90 mm/s, Knowing that at the instant shown the angular velocity «9 of the plate is 3 rad/s clockwise and is decreasing at the rate of 5 radís?, determine the acceleration of the pin i£itis located at (a) Point 4, (6) Point B, (e) Point €, SOLUTION lay PointA. (by) PoimB, Coriolis acceleration. Coriolis acceleration. =3 radís ), 00=5 radís >. 4 =%90 mm/s =0,09 més, 0=0 p =100 mm COS 2 60 31 mms" =0.081 m/s? Pp 100 a 2/2 ar =36 rad Is 201 += (23)00) = 540 mnvs* =0,54 m/s 1, =0 Y sa = 0.09 mís a =0 2 Ay — 7 =0.081 més" 7 p 2wu T=0.54 mis? T 2,¿=4,/+2 y +20 1] = 0.621 mus? T a,=0.621 m/s* T 4 Ty =0.14/2 m Ex 45% Y po = 0.09 més P A = or abr [0.2 M5) 57 45%] [(0:0. 1/2) e 45%] =[0.5/2 mis 7 45%]+[0.94/2 m/s 45%] 2 A dé 0.081 m/s? > Zu =0.54 ms > Ag Ag HA + ou] =[1.021 m/5* >]+[1.4 m4? 4] ay =1.733m/8 “2 53,9% PROPRIETARY MATERIAL, 33 2099 Vhe MeGraw-Hik Companies, Ine. All rights resceved. No part of ihiy Mental may be dispiaved, reproduced or distribuied in amy forma or by ay micas, wiihowt he prior oeritien permission 0f he publisher, er used bevond he distited distribidlon to deachers 06d educatora permitted hy Metiraw-HE for their individual course preparation. Jfyow are a student asia éhis Manual, Jon are sig ii withoul permission. 1182  PROBLEM 15.174 (Continued) (0) Point C. r.=0.2 mT Ver 0.09 m/s a =akxr, - 7 =[(0,25) 3-[nto.2 Ty] =[I mé? ]+[1.8 més? dy ae MS P = 0.081 ms?) Coriolis acceleration. 204 =0.53 mis? ) ASA Hay + end =[1 m/s? e-J+[2.421 mus? 4] a. =262wvs* 7 67.6% d PROPRIETARY MATERIAL, 0 2009 The McGraw-Hill Companies, Ine. All rights reserved. No part of this Manial may be elisplaned. reproduced or distribited in con form or hy amy means, witkowt the prior written permission of the publisher, or iped bevond ie Himited Aisiribatión to teachers end educator: permitled by MetirawHiH for their iudidial conse preparation. IyoVaro astudentusing dis Manual, Jon are minga tod perenissión. 1133  PROBLEM 15.175 Knowing that at ihe instant shown the rod attached at B rotates with a constant counterciockwise angular velocity (2, of 6 tad's, determine the angular velocity and angular acceleration of the rod attached al A. SOLUTION Liecmetry. AB=16 im, BD=16tan 30" in. AD=16se0307 in. Let the rod attached at 4 be a rotating frame of reference, Q=0, 3 Motion of coinciding Point D' on rod attached at A. Py =(AD)JO, =(L6s0c 30%), ¿2607 ay =[(4D)a, «60% + [ADS 5»309] =[16s0c30%, «2 607] + [ l6sec30%0% Ba 30] Motion of collar D relative to the frame. Vo un IO, Ape Fl AO Coriolis acceleration lau «00 Yo = Y y + Vr = 100.400 307)]00, 260%] + 5307] ap =[0.4s0c30%)0, 2 6091+[ (0.45c030)0, 230] +[0 30% + [260,0 42607] Rod BD: Yo = (BD), = (0.4 tan 30016) =[(2.4tan30%) in./s? =>] Rp = (BD ya = (0.4 tan 300X6) =[(14,4 tan 30%) in./s? T] Equate he 1wo expressions for vp and resolve into components. 60: (0.4sec 30%), =2.41an30%cos60*, wm, 51.500 radis Y 4 30% nm =2,4tan 30*c0530=1.2 mis PROPRIETARY MATERIAL, % 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of ihis Manual may be displayed, reproduced o distributed in any fora or by ony means, without the prior weliten permission of the publisher, ar used beyond the limited distribution to teachers and educators permitted by MoGran-HHE for theiriadividual comerse preparation. you are a student using this Manual, por are sing dllthont permission, 1184 PROBLEM 15.176 (Continued) Equate the two expressions for ay and resolve into components. +: 12m =(0,4scc 30%) er, cos 60" 0. 4sec 30% cos30* 0.4 ax, —12 Ea _—fóÉ£ó a Á 0.4 sec30%00s 60% (0.46 -(12)0.4) 0.4 sec30"cos 60? =-62.4 radis? 01, =624 radís” ) PROPRIETARY MATERIAL. 55 2009 The MeGraw-Hil Companies, Ine. All rights reserved. No par of dis Manual may be displayed, reproducca or distributed in auy forés or by any means, without the prior writer permisstom af he publisher, or used beyond the limited distriburion to renchers and educators permitted hy MoGraw-Hill for tele hidividua! course preparation. Ifyovwaore a studentusing this Mearnuol, Jairo iestag ibwitbiont permission. 1187  PROBLEM 15.177 AL lhe instant shown, bar BC has an angular velocity of 3 rad/s , . 7 : and an angular acceleration of 2 radfs”, both counterclockwise. Determine the angular acceleration of the plate, Li=4 q Hp im. ——. SOLUTION Relative position vectors, Gp = (Fin 413 in.)j Bye =-(6 indi+ (3 m)j Velocity analysis. Op. =3 radís 5 Bar 8£€ (Rotation about C): Mae =(3 radís)k Y = Ugo To = IAE 3]) =-—(U in.és)i—(18 in./s) Plate (Rotation about 1): ip = (Up k Let Point 8% be the peñot in the plate coinciding with A. Vi? = Dj Ip + pk ox (41 +3) =-Hi + Let plate be a rotating frame, Viur = Yiad Ya = Vit Y Jl + (40). + 4 )] Equate like components of Y. ii -I=30. 0/=(3 rad/s)k lo AED) AM Y =(30 ins) Acceleration analysis. Oy = 2 ruclis? y BaráC: ge = (2 rad/s*)k A Ur e re Pro = 2kx(61+31)- (3) 61431) =-12 5-61 +54 -27j=(48 in./s? Ji (39 in.s?)] a pa e Up pk Ay =D XT CO Yayo =0,kx(d+3)) y (Hi 43D =Jarpi+ de] 36127 PROPRIETARY MATERIAL. D 2000 The McGraw-Hill Companies, lbe. All nights aeserved. Nor part of his Marial may be displayed. reproduced or distributed o emy forncor by any encans, witlicut the prior wsrilien permission of the publisher, or swscel beyond the limited elistelbution to teuchers end educators permitted by Mera for their individual course preparation. (vom are a studen! esing this Mercal, arcano asias Frida PEereissio. 1188  PROBLEM 15.177 (Continued) Relativo to the frame (plate), Ihe acceleration of pin A is A 2 Vo. 30, Mai = lr), $ e = (re y =-(225 in./s ia (8,0), 5 Corialis acecleration, 20) XV payo a, =2(3k)x(-30)) =(180 in./s*)i Then Ay =Ap FA +2 04 =-(30) +36) + (da, —27)j-2251+(4,.,), 341801 a E CT Eguate like components ol a y. Lo 48=-Ba, +81) a) =-43 rad/s* ay =43.0 radis? ) 4 PROPRIETARY MATERIAL. €) 2009 The McGraw-Hill Companies, Inc. All rights reserved, No part ef dls Mannal may be displayed, reproduced sn distributed ln arg Jr or by anyomecss, velthont de prior written perarisción af me publisher. or ved beyond the limited distribution to teachers and educaters permitted by MoGrew-HHI for dicte individnad conse preparation. pom aro a student asin abris Monal, pamrare ustag Mvlibiosl perarission. 1189  PROBLEM 15.179 The Geneva mechanism shown is used to provide an Bo 1250. intermittent rotary motion of disk 5. Disk D rotates with a + constant counterclockwise angular velocity 6), of 8 radís, A pin P is attached to disk 0 and ean slide in one of (he six equally spaced slots cut in disk 5. lt is desirable that the angular velocity of disk 5 be zero as the pin enters and leaves each of the six slots; this will occur df the distance Disk 1 between the centers of the disks and the radii of the disks are wheng= 128% related as shown. Determine the angular velocity and angular acceleration of disk S at the instant when ¿4 =150% SOLUTION Geometry: Law of cosines. 1? =1.28% 42.507 — (21.25)(2.50)c0830* r=1.54914 in. P ad s a j L Law of sines. sn = 250,20 T E, 1,25 P 5 B 20 B B=23,794 2.50 in, Let disk $ be a rotating frame of reference. O=% SB Q-= a) Mation ef coinciding Poini Pon the disk. vpo =rex =1.549140% Y 8 Ar == kx Lp = Wo = 11.549 140, Y, 41+11.549 140 7 M Mation relative to the frame. Vas =4 =P ans =4 7 p Coriolis acceleration 2 idas N B Vo = Yao Vas =11.540 1400, Y, B]a Ju 7 8] Ap SA HA +2 NN =[1.549140% Y, B1411.549140 7 Plelo > 1+[ayu N Al Motion of disk D, (Rotation about 8) Y = (BP, =(01258)=10 11.5 27 30% dy =[(BP Ja 7 60% + [(BP 1 == 30] =04[(1.2518)" = 30%] =80in/s% 300 PROPRIETARY MATERIAL. % 2000 The McGraw-Hill Companies, Inc. All righús reserved, No pare of di Momia? may be displayed, reproduced or distribuied ia co: formoor by any neo viihowt the prior weritien permission of the publisher, ar ascd beyond the Limited distribution to teachers and educotors permitted by MeGraw HN for iheirindividaal conse preparation. Iyow cre csticden animes Manuel, Joare sing iwithont permission. 1192  PROBLEM 15.179 (Continued) Equate the bwo expressions for y» and resolve nto components. PB: 15491440, =10 cos(307+ 8) is 1eos53,7949 $ 154914 =3.8130 rad/s PEC sia 60 + 1 0Osin 53.794 = 8,0690 ins Equate the two expressions for a, and resolve into components. AB: 1.549 141%, — 2004 =8Osin (30%+ 45) ls " 1.549 14 =81.4 radis? _ SOsin53.794%+(2)(3.£130)(8.0690) 01 =3.81 1ad/s ) 4 | PROPRIETARY MATERIAL, 12 2009 The MeGrowHll Companies, Inc. Al rights reservo, No port of his Manso! may be displayed, reproduced or distribmed in any fora or y ay mears, wiíhoul tre prior weilten permission of the publisher, or used beyond aho timited distribution to teachers avd edicion permitted by Meoliram-PRE jor Cheb incdbcidaad course preparation. lfyow area student idas this Manial. Jarare dsleg 4 ito permission, 3103  PROBLEM 15.180 Disk S Ry = By In Problem 15,179, determine the angular velocity and A angular acceleration of disk S at the instant when =135%, Ry=125, PROBLEM 15,179 The Geneva mechanism shown is ysed to provide ab intermiltent rotary motion of disk $, Disk Dl rotates with a constant counterclockwise angular velocity (0), Of 8 rad/s. A pin P is attached to disk D and can slide in one of the six equally spaced slots cut in disk 5 las desirable that the angular velocity of disk $ be zero us the pin Disk D enters and leaves each of the six slots; this will occur if the wbeng = 1207 distance between the centers of the disks and the radil of the disks are related as shown. Determine the angular velocity and angular aeceleration of disk S at the instan when A =150%. SOLUTION Geometry Law of cosines. 11.25 42.50% —(2X1.2512.50) 00945? 7 =1.84203 in. P n/P_ sin4s? Y £ Law uf sines. a . o % á B=28.675* Let disk 5 be a rotatina frame of reference. Q=a ), Q=a) Motion of coinciding Point Pl on the disk. y = 30, =1.842030,N 4 Ap = 0 K XT po — Po = 11842030 Y 44 11842030 7 81 Metion relative to the frame. Yo = 4 E Ma == 78 Coriolis acceleration. 2 yu N B Vo = Ye + Vos =11.8420300, Y, Bl+ la + 8] Ap Ap RA +2 NS =[1.842030%,, Bi+11.5420302 > B]+[4 7]+Baa XA PROPRIETARY MATERIAL. (0 2009 The MeGraw-HiH Companies, Ine. All tighis reserve. No port of iiy Mánual may De displayed. peproductd or distribied 1 any fora or by amy metas, without the prior weltien permission of he publisher, or ised beyond the limited distribution 10 teachers and educotors permitted by MoGrawabHI for their indibvidio! conse preparation. Ifyon area xtucen using this Monual, pan are using dl vlhaat permission 194  PROBLEM 15.181 (Continued) Coriolis acceleration. 209y pu =(212.400)(1,3416)=[6.4399 mís* > 8] Use A; =48 y 448 ypp + [20 yu A 1] and resolve into components. +27 fi: 0=-18c08 8 +0.2795 10%, +6.4399 yy = 34.56 radis? 4N A: 0=-18simB+1.61=0, 4=>6,4398 m/s" dp =[18—]+[(027951)364.56) 7% BI+[L61Y A] =[18—]4[9.6594 4]+[1.61 WAI 59,11 m/*3218,4* Summary: (d) 0% =240 radís ) yo =34.6 radis? ) 4 (6) — ve =1.342 m/s %. 63.47 ap =9.11 ms 18.49 4 PROPRIETARY MATERIAL. E 2000 The McGraw-Hill Companies, Inc. All tights reserved. No part af this Manuel may be displayed, reproduced or distributed ia any formoor by 0ny cars, withot the prior weillea permission of the publisher, ortesed beyond the finited alisiribution to teachers and educators permitted bp MeGreaw-HRl for dheir individual conese preparation. Ifvow re a stden sing this Manaol, jor are esingitwithornt permission. 1197  PROBLEM 15,182* Rod 48 passes through a collar which is welded to link PE Knowing that at the instant shown block 4 moves to the right at a constant speed of 75 in./s, determine (a) the angular velocity of rod AB, (5) the velocity relative to the collar of the point of the tod in contact with the collar, (c) the acceleration of the point of the rod in contact with the collar, (Mint: Rod 48 and lmk DE have Ihe same (9 and the same 0£.) SOLUTION Let 0=04) aud a=a”) be the angular velocity and angular acceleration of the Eink DE and collar rigid body, Let ¿” be a frame of reference moving with ihis body. The rod 42 slides in the collar relative to the frame of reference with relative velocity u= 4 230% and relative acceleration ú =% 2 30%. Note that this relative motion is a translation that applies to all points along the rod. Let Point 4 be moving with the end of the rod and 4 be moving with the frame. Point E is a fixed point. : 6 in. y Geometry. E 12in.— Velocity analysis. y =75imés y, =12w| Y = Y, +u Resolve into components. Has 7504 cos300 u=— 786.603 508 cos300 (a + Eo deT2o+usin3or 0 PO 6085 vals (a+ Angular velocity. 0=3.61 radís *) «4 (6) Welocity of rod 48 relative to the collar. u=86.6 in/s 30 4 (e) — Acceleration analysis. a¿=0 a.=1a | 120 ==120e) 4156.25 — Coriolis acceleration. a, = 260% 60% =625.01 in./s > 609 4¿4¿+ú+a, — Resolve into components. Ar 0156254 c0830% 625.01 c05600 1 =180.43 in./s? +: 0=-120+úsin30%4625.DIsin 600 a=52.624 radis* PROPRIETARY MATERIAL. 2 2009 Tho MeGiraw-Hill Companies, Inc, All rights reserved. No part of this Marnal may: be displayed, reproduced ur distributed in any form or by uy means, withont ihe priorowritien permission af de publisirer, or used beyond dae lianited distribution do leachers and educators permitted by MoGraw-HRH jor their individial cose preparation. dfyow are a studenttustag dis Mantal, yaware using doweiihoni permission. 1198  PROBLEM 15.182* (Continued) For rod AB, De =3.6085 radís *) Desa =52.624 radis? *y Let P be the point on 48 coinciding with collar D. Poy = 1200830 7 30 =10,392 111. 7 309, Ap =p), HA =0+[(10.392652.624) %e. 60%] +[(10.392)(3.6085)' 7730%] =[546.87 22.60%] + [135.32 730 =[390.63 — ]+ [405.94 | a, =56 ins 746,1 € A May ulso be determined from a, =41 + +4, using the rotating frame, The already calculated vectors 4 and a, also apply at Points P' and P. Ay =4 = 607 304 60 607 =[315.74 in./s037 30] +[78.13 in.1877 60%] Then ar =[315.74% 30%] +(78.13 60%] + (180.43 7 300] +[625.01% 609] =(135.31 27 30%]+ [546,88 22. 609] PROPRIETARY MATERIAL. € 200% The MeGraw-Hill Companies, inc. All rights rescrvcd, No pari of this Manual may be displayed, reproduced er «distribuled any farm or by any means, without Hhe prior written permission af the publisher, ar used beyond the limited distribition to teacher and eduecnlors permitlcd Dv MeGrow-HiHfor Meir individual conrse preparation. Ifpor are estudení axingy this Manzal, yon are 1eslag lp weltbont permission. 1199  PROBLEM 15.184 Plate ABD and rod OB are rigidly connected and rotate about the ball-and-socket joint O with an angular velocity 0=0 +0, j+0,k Knowing that v, =(340./5)i+ (14 in.fs)p+(v,),kand e, =1.5 rad/s, determine (e) the angular velocity of the assembly, (6) the velocity of Pomt 1. SOLUTION 0, =1.5rad/s (1.5 rad/s)i +00, + 0,k r,=-8 in )d+(6 in.)j+(4in.)k To =-H8 im.)1+(6 in. (4 m,)k (a) Y, =9XF, io jok =15 mM 0 -E +6 +4 yy = (de, 60, Ji 8, 6) j 419480, )k But we are given: Y =68 in.s)+(14 im./s)j+(v,),k (va): 40, —6m, =3 (1) (v,),; 80, —6=14— 0, =-2,5 rad/s (2) (1,) 9+80, =(%,), 56 Substitute (%, =-—2.5 radís mto Eg, (1): 40, 622.5) =3 6, =-3 rad/s Substilute e, =-—3 rad/s into Eq. (3) 9+8(3) = (11), (1,4). =-15 in4s We have: 0=(0.5 rad/s)i—(3 rad/s)—(2.5 rad/s)k «€ PROPRIETARY MATERIAL, 0 2009 The MeGraw-Hi1l Companics, Inc, All rights rescrvcd. No part ef this Mareal may be displayed, reproduced or distributed di cup forn or by cope mesias, withont Che prior weilten permission 0J dhe publisher, ar used beyond the lite distribution to teachers and edicarors permitted by MoGrinv-HdiH for drcir inclividacal course preparation. [fyouare a smuentusing dais Maria, Foiarcieing itwithont permission. 1202  PROBLEM 15.184 (Continued) (b) Velocity of D. Y S0XEp ij k =11.5 3 —25 +8 +6 4 =(12415)1+(-20+6)j+(9+24)k Y» =(27 in.sji (14 in/s)4+(33 in./s)k «€ PROPRIETARY MATERIAL, $3 2009 The MeGiraw-HTI! Companics, Inc. All rights reserved. No port ef dis Minmial may be displayed. reproduced ar distributed la any fora or dy any means, withowt ihe prior written permission of he publisher, or used beyond the limited distribudlon to teachers and educators permitted by MoGrane HH fe their tadididual comse preparation. IPyoware e studentising iris Merval. PUT Aena e eS doit Permission. 1203  PROBLEM 15.185 Solve Problem 15.184, assuming that (1, =-1.5 radís. PROBLEM 15.184 Plate 480 and vod 0% are rigidly connected and rotale about the bali-and-socket joint O with an angular — velocity o0=ea +0 j+0k. Knowing that y, =(3 inés)i+ (14 in.) +(0,),k and o, =1.5 radis, determine (0) the angular velocity of the assembly, (6) the velocity of Point D. SOLUTION 0, =1.5 radís 0=(15 radsji+0,j+0,k r, =-=(8 inJi+ (6 10.)j+(4 mk Tp =+H8 in) i+(6 in. —(4 in. )k (a) Y WT io jo k =-1.5 4, 0, SÁ +bo + y =(4e, 60, Ji+ (Sa, +6) j+ 69480, Jk But, we are given: Y, = (3 inósi4 CIA in.) j+ r,,),k Gdl 40, 60, =3 0) (1): Ba, +6=14 e. =-—| rad/s 2) 0): +88, = 0,4), (3) Substitute Y =-1 rad's into Eq. (1): 40, -6(-1)=3 (9, =-0.75 rad/s Substitute 2, =-0.75 rad/s into Eq. (3): -9+8( 0.75) =(v,), (v,), =-15 ins We have: 9 =(1.5 radisji-(0.75 radsj- Cl vadís)k 4 PROPRIETARY MATERIAL. £% 2009 The MeliravweHill Compáames, Inc. Allomghts reserved, No part of dis Menaol may be displaped, reproduced or alistribited in env form or by amomenns, without the prior wrilten permission of the publisher, or used beyond the limited distribution to teachers and educators permitled by MeGrHiH for their individual cose preparation. Uyow area studienasing ti Adariol, pom are using lowithowt permission. 1204  PROBLEM 15.186 (Continued) 1 From Eq. (2). 0, =—A40.3 0.3, rom Eq. (2) s 025 +) = 0.48 rad/s From Eq. (6), 0, “5 (0.640.250, ) =-1.6 radís ta) Angular velocity. 00 =(0.480 rad/s)i — (1.600 rad's)] + (0.600 rad/s)k « From Eq. (1), (14), =-0.25m, =(0.400 mís From Eq. (3), 4 =-030, 0.480 m/s (5) — Velocity of Point A, Y y = (0,400 m/s) +(0.300 m/s) + (0,480 ns)k or Y =(400 mms)i + (300 mms) j +(480 menYs)k € PROPRIETARY MATERIAL. £% 2009 The MeGrew-Hill Companies, Inc. All rights reserved. No pari of rie Mano! mayo be displayed, reproduced or distribisted in any formoor by eme meons, velthont the prior writien permission of the publisher, or used beyond de limited distribution to teachers and educators permitted by MoGraw-HRU for thieirindividual conse preparation, If yow are astudent sing tes Manual, Jaware using iówithont permission 1207  PROBLEM 15.187 At the instant considered, the radar antenna shown rotates about the origin of coordinates with an angular velocity o=0 +0 j+0k. Knowing that (v,), =100 mms, (v4), =-90 mms, and (1), 120 mnvs, determine (a) the angular velocity oF the antenna, (6) the velocity of Point 4. 3 0.25 1 SOLUTION E, =(0.3 m)i-0.25 m)k y =(0.1 mésji — (0.09 m/s) +(v,),k ij k vy=0Xxr 4 0. 1i—-0,097+(v/),k=[0, 0, 0, 03 0 0,25 0.110.095 +(v,),k=-0.250,1+ (0.300, 40.250, )j—0.309,k h 0.1=-0.250, (0) E —0.00=030m +0.230, 0 ko (v,), =-0.30, 6) Eg =(0.3 mil (0.25 m)j Vo =(Yg),1+ (17), 3+(0.12 m/5)k i ik Y¿=0XF (vo), + (14), +40.12k= 0, 0, a, 0.3 -0.25 0 (1), 14 (17), +0, 12k =0.2520,14+0.300,+(0.250, + 0.3)0, k E 0), =025, E) do (y), =030, O ko 0.12=-0.250, 0.30, (6) S ; 0.1 ; From Eg. (1), 9, = ET =-0.4 radés PROPRIETARY MATERIAL, € 2009 The MeGraw-8l Companies, Jae. All rights reserved, No part of dis Manual may he displayed, reproduced or distributed in cuy form or hy any means, withomi the prior written permission of the publisher, or used beyond the limited distribution te teachers and edecators permitted by McGraw TH for iheir éndovidual course preparation. [fyom are a stidend msing dis denisccol, pon are using dowébar permission. 1208  PROBLEM 15.187 (Continued) a l From Eq. (6 =-——(0.12 +0. rom Eq. (6), 0, a 2+0,30,) = | From Ey. (2), (0.094 0.2 rom Eg. (2) (a, META + 0.250, ) =-00.36 radís From Eq. (0), (1), = 10.3-0.4) =0,12 m/s (a) Angular velocity. 0 =-—0.400 rad's)j—(0.360 rad/s)k «€ (b) — Nelocity of Point A. Y, =(0.1 més)i — (0.09 més)j 0, 12 més)k or Y =(100 mmnís)i— (90 mms)] + (120 mm/s)k «€ PROPRIETARY MATERIAL. 13 2009 The MeGraw-Htll Compantes, Inc. All rights reserved. No port Of is Manuel mej: be displayed, reproduced or distributed inca form or by cas mears, who ihe prior written permission of the publisher, or used beyond the Liniied distribatión to dcachers and edacators permitted hy Moliran HH for dere iadindwal comerse preparcition, divo are a student using ais Mami, yecarecnisina dello permission. 1209  PROBLEM 15.190 In the system shown, disk 4 is free to rotate about tho horizontal rod OA. Assuming that disk E is stationary (0, =0) and that shaft OC rotates with a constant angular velocity ay, determine (a) the angular velocity of disk A, (6) the angular acceleration of disk A. SOLUTION Disk 4 (In rotation aboat O): Since 0, =%, 0, = 0, + a+ 014 Point D is point of contact of wheel and disk. a Yo 04 To io j ok =4, 4 0 =— —R Vo = (Ra +raJi+ Ra j-r0,k Since (4 =0, 1, =0, Each component of y, 15 zero. (y), =50,=0, m1 =0 (Yp), =-Ra) +02, =0; (a) Angular velocity. Da <a1+ [Ear A E (5) — Angular acceleration. Disk A rotates about y ancis at rate (4. ai a dd meca lec Uy 0 00 A G+ PROPRIETARY MATERIAL. 22009 The MeGrawoHill Companies, inc. Adl rights reserved. No part af dis Manual may be displayed, reproduced or distributed in any forn or by any medas, withont the prior weitlen: permission of he publisher, or used beyond the binited distriluetion to teachers and educators perniticd by MoGraw-H1H1 for their individual course preporction, por are stadent sing Mis Mancl, Ja ire esing dwiahont permission. 1212  PROBLEM 15.191 In the sysiem shoven, disk 4 is free to rotate about 1he horizontal rod GA. Assuming that shaft OC and disk B rotate with constant angular velocitics 44 and (0, respectively, both counterclockwise, determine (a) the angular velocity of disk 4, (6) the angular acecleration of disk A. SOLUTION Disk 4 (in rotation about 0): Since 0%, =(%. 0, =0,4+6j+0,k Point D is point of contact of wheel and disk, Disk 8: From Eqs. band 2: Foro ==14= Rk ojo k Yy 04 o O, A 0 Dor <R Yy = Ray +raJit Ro, 1ra,k (1 0, = 0] Vo E Oy AF = or] — RK)=-Réni 2 Vy= Y: Era ra it Ro j-ro k=-—Resi Coetficients of k: 0, =4 a =0 Cocfficients ok EFRa+ro,)=-Ra,; 0, = =ca —=,) (a) Angular velocity. m,=aj4+ Pta ak 4 (9) — Angular accelcration. Disk A rotates about y axis at rate e). asoma, bo 19h 0H] 0, Cato api 4 PROPRIETARY MATERIAL. 33 2009 The MeCiraw-HiH Companies, Jue. All rights reserved. No partof this Manual may be Lispleyed, reproduced or distribuidas any formar by any iicans, avtthioml the prior veriiten permission of the publisher, or ved beyond tire limited disiribirion to tecchers and educators permitred by MeGraw-T1 for their individual cose preparation, Ef yo area siuder using this Mensal, yow reina if withont permission 1113  156 ram PROBLEM 15.192 The L-shaped arm 8C0 rotates about the z axis wih a constant angular velocity €, cl 5 rad/s. Knowing that he 1SOonmeradius disk rotates aboul BC with a constant angular velocity (0, of 4 rad/s, determine the angular acceleration of the disk. y le 10 , SOLUTION Total angular velocity, £0= 4 + ak Angular acceleration, 20=(4 rad/s)j4(5 rad/s)k Frame Oxpz is rotating with angular velocity (= ak. a= = Dgo FAX =0+0kx(0,j+09k) == 04 = ASMA ==20i 0=-(20.0 radis "ji PROPRIETARY MATERIAL. 03 2009 The MeGraw-Hill Companies, nc. All rights sesorved. No part af Mis Manial may be clisplayed. reprodueccd or distributed in any fora or by cnomicans, withiont cie pri wriren permission of! he publisher, ar uscd beyond the Hinaired distribution to teachers and cducotors pernitted by McGrow-AHiifor their individial conrse preparation. If you are a student using dis Manual, pon are rising hwillroul permission. 1114  PROBLEM 15.194 (Continued) (6) 8 =90%. Acceleration at Point P, Tr, =(0,25 M)j Wp = 0X Fo =(5i+4k)x0.25 =-(1.25 fUs)i+ (1 1Us)] Up EXE FIX VA =-20/0.25j4+ (514 4k) (1,2514 j) =0+0-625]-4j+0 =-10,25j ap =-(10.25 Us) 4 cl PROPRIETARY MATERIAL, > 2009 The MeCirioHil Companies, Inc. All iglhts reserved. No par ef its Monnol my be Siplegned repreduccd or distributed 16 eny form or hy quy means, aio the prior written permission ef the publisher, or used Bevond the lisnited distribution fo wachers and educators permitted by MeGrear TT for ibehe individual cose preparation, Ifyon area stdeni using this Mental, Joicare sing iticithont permission, 1217  PROBLEM 15.195 A 3-in.-radius disk spins at the constant rate «9, =4 radís about an axis held by a housing attached to a horizontal rad that rotates at the constant rate «o, =5 rad's, Knowing that 8=30", determine the acceleration of Point P on the rim of the disk. SOLUTION Angular velocity, 0=a+0k 0 ($ rad/s)i + (4 rad/s)k Angular acceleration, Frame Oxpz is rotating with angular velocity £2= 4 i, AR O =0+1x(091+0,k) =-a0! =-(M(5)j= 205 ¿=-(20.0 radis*)j EOMEtrY. 8=30" To = (3 in.cos30% + sin 30%) =(0.25 ficos 30% +sin30*j) Velocity of Point P. Vp = WXYp i j k = 5 0 4 0.25cos30% 0.25sin30% 0 —(0.5 ft'5)i+(0.86603 ft's)j+(0,625 fi/s)k Acceleration ef Point P. Ap = XT p AX VA i i kl |i ¡ k 0 -20 0p+| 5 0 4 0.25cos30% 0,25sin30% 0| |-0.5 —0.86604 0.625 4.3301k -3.46411 -5.125]+4 3301k ap =-(3.46 fs )i (5.13 fs? )j+(8.66 fis" )k « PROPRIETARY MATERIAL. € 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No par of ihis Manual may be displayed, reproduced or distributed ln any for ar by any meca, without ¡he prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGram Hi for their individual comese preparation. Ifyow area studentusing this Meal, porare using itwithont permission. 1218  PROBLEM 15.196 Á gun barrel of length OP=4m is mounted on a turret as shown, To keep the gun aimed at a moving target, the azimuth angle fis being increased at the rate el Blcdt =30%s and the elevation angle y is being increased at the rate dp ldr = 10%. For the position B=90 and y=30%, determine (a) the angular velocity of ihe barrel, (6) the angular acceleration ol the barrel, (c) he velocity and acceleration of Point P. SOLUTION df, 307 , A ) Let =-Lj=- 2 ads (¡=-| — rad? e o ao E ra >) E ng dy. (10% , r =P E dis li== E i 10 Eb ls Ea sj E rad) (2) Angular velocity 0= 0 +0, 9 =—(0.1745 rad's)i (0.524 radfs)j «4 (6) Angular acceleration. Frame Oxyz is votatinig wilh angular velocity Q =0, (E dy = ore FDA, =0+00, <((0, +0) Ek m=-(0.0914 rad/s”)k « le) Felocity and acceleration of Point P. For 4=%90" and y 30%, Ep = (4 mijsin 30%] 4 c0830%k) eo (2) (E) o 0 Asin30% deos30 =-1.81381 +0,6046]-3,4907k Yo =-(1.818 mís)i +(0.605 m/s)p- (3.49 nvs)k 4 PROPRIETARY MATERIAL. 0 2009 The MeGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displaved, repraduced or distribided da any forn or by any means, witlreia the prior written permission af De publisher, or used beyond the Hinited distribution to teachers and educators permilted by MeGraw HU for hebindividual course preparation. you ave a student msn this Montal, Jo are ing itwithonr permission. A 1221 pan 2 PROBLEM 15,197 (Continued) Y = 0xr io jok o, m0 4 25 0 = (2.50, — 4a, Jk Recall from Eq. (2) that v, =S0k, and write 50=2,50, 40, (4) Add Eqs. (3) and (4): 440=15,56, (0, =28.387 radís Eg. (4): 50 =2.5(28,387) - 400, 0, =35.242 rad/s (a) Common velocity ofunit 48. w=(28.387 rad/s)i+(5.242 rad's)] 0 =(28,4 radís)t+(5.24 rad/s)j 4 (6) — Angular velocity of shafl FA. (See figure in text.) Point N is at nut, which is a part of unit 48 and also is a part of shaft GH, 1 La En TS Xp dt a) pj i+v,] 2 za Yu) Nut Was a partofunit dB: (1 (28.387 rad/s)i + (5.242 rad's)j VW, = XT, =(28.3871+ 52420) | > 5 + va) y, =(28,387y, —2.621y, )k =+(25,766y, Jk (5) NutÑN as a partof shalt FH. Oy, = Ojgy1 Y y Oj FE, 1 = (00% 1) q + y) = Oy Y, Ke (6) Equating expressions for vy. from Eqs. (3) and (6), +(25.766 y, )k = Oj 1, k Mew =(25.766 rad/s)i Wen =(25.8 rad/s)i 4 PROPRIETARY MATERIAL. 0 2009 The MeGrow-Hill Companies, Inc. All rights reserved, No part of dis Marual may be displayed, reproduced er distributed im any forn or by cry mens, without the prior wrilten permission of the publisher, or used bevond the limited distribution to teachers and educators permitted by MeGraw-HiILfor their individual conse preparation. [fyon are a student using this Manual, yor are sine ibwithanl permission. 122 PROBLEM 15.198 A 30 mme-radius wheel is mounted on an axle OB of length 100 mm. The wheel rolls without sliding on the horizontal floor, and the axle is perpendicular to the plane of the wheel. Knowing that the system rotates about the y axis at a constant rate «1 =2.4 radís, determine (4) the angular velocity of he wheel. (6) the angular acceleration of the wheel, (c) the acceleration of Point C located at the highest point on the rim of the wheel. SOLUTION e Geometry, [=100 mmn=0.1m y b=30 mm = 0.03 m + g 2 tn A=7=03 ¿El i Sh — P=16.6992 A o Ty =-d sec Bi fp =—1 cos Bi+ bcos Bj (a) Angular velocities. For the system, L=aj=(24 radts)j For the wheel, 0=0,4+0,j+0,k v¿=0Xxr, =(0,+0,j+0,k)x(4sec Pi) = 0 le, sec B)]— (len, see B)k=0 aa, = 0, 4 =0 o =e1 Vy =0XF =4,4x(-]cos Bi+ bcos Bj) =(0,bc0s 4)Kk For the system, Vy =12Xrp =04jX(cos Bi+bcos Pj) = (ex! cos 0)k Matching the two expressions for Y, 0.bcos 8 == 1008 1 or 10) e rad/s (1 =(8.00 radis)i € PROPRIETARY MATERIAL, 5 2009 The McGraw-Hill Companies, Inc. All rights reserved. No part of dhic Manual moy be displayed, reproduced or disteliwted in any fora or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to deachers and educalors permitted hy McGraw-Hill for their indiridual course preparation. dfyo area student ves ias tiis Mantal, From are asing iwcithori permisxion. amm AE PROBLEM 15.198 ; (Contínued) (b) Angular acceleration. a= se Digo FAO =(042,4j)85 =-(19.2 rad/s? )k =-(19,20 rad/s*)k « le) Conditions at Pol E ¿ =-( cos PB —bsin P)i+ 2bcos Pj = (95,872 mm)i4 (57.47 mea)j Y == MXTFo =81x (958721457, =(459.76 mm/s)k Me E XT AAN =-19,2k x(-95,87214 57.471) +81x 459,76 =(-1103.4 mus? pi (2004.6 munds”)] e =-(1.103 vs) (2,005 m/s) j PROPRIETARF MATERIAL. € 2009 The McGraw-Hil Companies, Tac, AR dighas reserved. No part of dis Monual may be displayed reproduced or diseibured in any form or by any acans, wéhour the prior writien permission 0f the publisher, or wsed bevond re disrited distribution lo teschers and edacators permiticd lo) MoGrowAH8l for their individual come preparation, Ifyow are a student asag is Manual, Sonar using iwlthont permission. A: 1226 PROBLEM 15.200 In Problem 15.199, the speed of Point 8 is known to be constant. For the position shown, determine (a) the angular acceleration of the guide arm, (6) the acceleration of Port E, PROBLEM 15,199 Several rods are brazed together to form the robotic guide arm shown, which is attached to a ball-and-socket joiot at O, Rod OA slides in a straight inclined slot while rod GE slides in a slot parallel to the z axis. Knowing that at the jostant shown ve=(180 mm/s)k, determine (a) the angular velocity of the guide arm, (6) the velocity of Point A, (c) the velocity of Point C. Dimensions in mn SOLUTION y leo Dimensions 1h mm. IL y=U44 200 40 ¿ 1 1 e : * Z09 : Yo Ha Since rod at D slides in slot which is of 1:2, (Yp), =-2(vp),, and 0 =Ar o), Angular velocity. o=4 +0, j+0,k Y, =0Xr,: (180 mmés)k = (0,1+ 0, ]+,k)(240 mm)j 180k =2400,k - 24001 1 Coefficients of ki 180=24062, 00, =0,75 radis Coefficients of k 0=-2400, («=0 Y ¡=0XFy4 y, =(0,751+ 0,1) (200k) VJdr0), +0) k=-1503+2000,1 PROPRIETARY MATERIAL, € 2009 The McGraw-Hill Companics, Inc. All rights reserved. No part of this Masuof may be displayed, reproduced or distributed in any form or by cuy means, without the prior writien permission Of the publisher, or used beyond the limited distribution to teachers and educators permined by McGraw-HiHior their individual conse preparatioa. Ifyow are a sirident using tris Manel, yon are using ibveithont permission. 127 PROBLEM 15.200 (Continued) Coefficients of j: (v,), =-150 Coefficients ofi: (4), = 2000, Coeficiente of k: (1,),=0 Recall the Eq: Vd. = 2000), 2000, =-2(-150) «), =1.5rad/s and (v,), =300 mos 0=(0.75 rad's)i+(1.5 rad/s)] Velocity od: Y = (600 mms) (150 movs)j Velocity oF C: ro = 10014 80j4 40k Y = XT io jo K =1075 15 0 100 $0 40 =601 —30j+(60-150)k vo =(60 mmn/s)i— (30 mm/s)j — (90 mm/s)k Ay = 0% Ey +0 (xr) = XT XV y i ¡ok i ik 2a=]x, e, ,|+/075 15 0 0 240 0 0d 0. 180 2400 142400 k +2701 1353 1 =(270-240er, Ji 1351424002, k m E » ' (a), =270-240, =0 e =1.125 radís” (24), =-135 (ay), =-135 minis? (aj). =240%, 0 a, =0 Ay S0Xxt,+0x(0Xxr/) OT HAN ij k i ik 10 e, L125|+ 075 15 0 0 0 200 300 -—150 0 2008, 14 (-112.5--ASO)k a, = 2000, 1-562.5k 1 1" A PROPRIETARY MATERIAL. 0 2400 The MeGiraws-Hill Companies, Inc. All cighús reserved. No part of tHils Masia inay: be displayed, reproduced or distributed in any fer or dy axyomeans, without De prior writlas permisión of] de publisher, or used beyond ile limited distrib to teachers and cducólors permitica by MeGrawe DB for bete Individual corso preporetias, Iva are cstudert sta dbris Minmiol oscar estaa diia purmbision  1228 PROBLEM 15.200 (Continued) Thus, (0), =2006r, (0,),=0 (0/), =-562,5 muvs” But (4), =-Aa4), =0 Therefore, (0,),=200%, =0 e, =0 (a) — Angular pcceleration: 0.=(1.125 rad/s? )k < (6) — Acceleration of €: A = DT + O MAT.) = DEA XV Te = (100 mo)1 +(80 mm)j+ (40 mm)k ojo k ojo ok a=| 0 O 1.125/4+1075 15 o 100 80 40 60 30 -90 =-9014-112.55-13514-67.5j4 (22.5 -90)k 1, =-(225 mmist Ni +(180 mas )j- (112.5 mnvs*)k 4 PROPRIETARY MATERIAL. 0 2000 The Meticov-Húl Compañies, Inc. AD mghts reserved. No part af his Mental may be displayed, reproduced or distributed tn cayo fora er by any imcans, vitlioul the prior weritíén permission 0f Ihe publisher, ot usd beyond the limited alistribucion to teachers and educators permiitod by MeGraweHill for dreir individol come preparation. df yow are o stden using bis Moral, poare asin dovithiornr permission. 1631 PROBLEM 15.202 Rod 4B of length 275 wm is connected by ball-and-socket joints to collers 4 and 4, which slide along the two rods shown. Knowing that collar A moves toward the origin O ata constant specd of 180 mms, delermine the velocity of collar 4 when e=175 mm, SOLUTION Cicometry. c=0173m, [4 =0273m Dip = 0 ROL1SO RD or 0.2758 0.175 40.157 44 b=0 15m Fun =-£0.175 m4 +(0,15 m)j+(0,15 m)k Velocity of colar B. Ya ==(180 mmés)i =-(0.18 m/s)i Velocity of collar 4. yv, =vw,k Va Vs EV where Va 0 Eo Noling that v y Is perpendicular to Ey, We gel Fay Y 0 Forming F yg o Y y. We get Tr Ya E EV E Y) SW a Ya or Doo Ma = Ep Ve m From Eq. (10), (-0.1751+0,153+0.15k)-(v,L)=(-0.175140.153+0.15k) -(-0,18b) 0.15v, =(-0,175)1(-0,18) or y, =0.2H m/s Y, = (210 mms)k «€ PROPRIETARY MATERIAL. *: 3009 The MoGeaw-Hil Companies, Inc. All rights reserved. No port of His Manual may be displayed, reproduced vr distribated Da omy form or hy any means, awvilliout the prior written permission of the publisher, or vcd bevond the limited distribution to teachers and educalors permitted dy MoGraw-Hdl for their individual conrse preparation. vorare a student asing this Manual, Parrartcnzing Howvithoml permission,  Lima _ 7 e _ 1232 PROBLEM 15.203 Rod AR of length 275 mm is connected by ball-and-socket joints to collars 4 and E, which slide along the bwo rods shown. Knowing thal collar £ moves toward the origin O ala constant speed of 180 mmés, determine Uhe velocity of collar A when c=50 mm. SOLUTION Geometry. Velocity of collar £. Velocity of collar 4. where e=0,05 m, Lap =0.275 1m Pp=P+0.1500487 or 0278 =008 40.140) b=0225m Ly = (0.05 m)4+ (0.15. m)j4(0,225 mjk Y =(180 mirvs)i =—(0.18 m/s)i Var Was Lara Noting trat Y yy ds perpendicular to Y yy. WE BE Py Vasa = 0. Forming E yy + Y, We gel ol From Eq. 0), or Cano Ya Va (Va EV) Eve Va PL Va Fur Va Tae cp mm (-0.051+-0,15j + 0.225k)-(1,k) = (0.051 +0.15]+0,225k)- (0.181) 0,225v, =(—0,05K-0.18) v, =0.04 mís y, = (40.0 mm/s)k « PROPRIETARY MATERIAL. 6 2009 Vhe MeGraow-Hdl Companies, Ine. All vights reserved. No part ef tés Manual may be displayed, reproduced or distribucd in ar: for or by anys s9ms, wwithicuel the prior weitren permission ef the publisher, or used beyond the binmited disiriburion ro roachers and cbicators permitcd by Motirave: HI for ibelr individual cose preparation. Pyon ave a vtadenl sing ii Metal, ari d irollhcal PEradasIOn 1233 PROBLEM 15.204 Rod AB is connected by ball-and-socket joints to collar A and to the 16-in.-diameter disk C. Knowing that disk € rotates counterclockwise at ihe constant rate (9, =3 radís in the zx plane, determine ihe velocity of collar 4 for the position shown. 25 in, SOLUTION Geometry. Fic = E in.Jk Tay =-(25 m))i +(20 i)j—(8 in)k Velocity at B. Va = Ai Tao =3jx(-8k) =-—(24 in./s 4 Velocity of collar A. Y ¿=v,] Va= Va Yan where Varo Di gg Noting that v yy ls perpendicular o Eg, We gel Pay Vga 0. Forming Y yp Y y, We get Vairo Ya ET [Y E Ya) Faro Vo Pa o 4% Tao Ya Fea Wa 0) From Eq. (1), (251 + 20|-8k)(v3) = (251 +20j-8k)- (241) 20v, =-—600 or v,=-30 in./s Y, =-(30.0 in./s)j «€ PROPRIETARY MATERIAL, £% 2009 The McGraw A5H Companies, Inc, All rights reserved. Ne part of tés Manual may be displaved, reproduced or «istribued in any form or by any oneans, without the prior vwritien permission of the publisher, ur used beyond the Limited distribnaion to teachers and edacators permiticd by MeGiravaFHl for their individial course preparation. If por ore a studentusing this Menual, youre sing diowithomt permission,  1236 PROBLEM 15.206 (Continued) Felocitv of collar B. a len y, SONOVi—180K) . 201.246 =(22.3607 munís)i (44.72.14 mm/s)k Velocity of'collar A. Y Fa] VE VA Van where Va = Ops Noting that v yg is perpendicular to E yy, WE Bet Fijo Vas 50, Forning Y ya + Wy, We get Lon Va = Fa AV do) Lo Ya FE Yava ur Pomo Va = Lg Va m From Eq. (1), (7401 + 280]-100k) -(v,1) = (401 +280]—100k)-(22.36071-44.7214k) 280%, = (-40122.3607) + (100K-44.7214) or v, =12.7775 mm/s Y, =012.78 mms) j PROPRIETARY MATERIAL. % 2000 The Metirao-Hill Companies, Inc. All rights reserved, No port of this Manual mav be displayed, reproduced vr distribiaded de ny forma ce dy any means, withonki the prior written permission of the publisher, or used beyond the limited distribution to teachers cul educators permitted hy MeGraw-HiN for thacio individual comrse preparation. [fyoi are e student ring this Mentel, Fon erre sin dt withiat permission.  EM 1137 PROBLEM 15,207 Rod AB of length 300 mm is connected by ball-and-socket joints to collars A and B, which slide along the two rods shown. Knowing, that collar A moves toward Point Data constant speed of 50 mms, determine the velocity of collar 4 when c=120 mm. E, - 150 na - SOLUTION Geometry. T=5i, Tr, = (90 mm)i T¿ = (180 mm)k Fox: > Tp Te =(4.5inJi—(9in.)k dep =a (90) + 180Y =201.246 crac) Pi == BE 1 $0 _ 120(001 — 180k) 180 =(60 mi — (120 mm)k Ty = Et Poe =180k + 601 120k = (60 mm)i+(60 mmk Fai Ta Eg =-—60i+ pj —60k Le o E + DE 300 =60" + y +60 »=287.75 mm, Ey = (260 mai + (287.75 mm) — (60 mm)k PROPRIETARY MATERIAL, 5 2009 The MeGro Hill Companies, lac, All rights reserved, No part af ds Minual may be displayed, reproduced or distribiled da amy forn or by auy incans, without the prior writien permission of the publisher, or used beyond the limited «istiburion to teachers and educatora permitted by MeGiraw-P0Ljor their individual course preparatton. If pow are a student using this Manual, omar aia diario perniissión. Velocity of callar B. Velocity of collar A. where Forming F yy > Y¿, We get or or 1238 PROBLEM 15.207 (Continued) y ey Cp (S0K901 —180k) A a 201.246 = (22,3607 mn/9)i (44,72 14 mimn/s)k Yi =v,] Ya Va + Von Vai = Dn ag Notinig that vyg is perpendicular to E yy, WE gel Ej Ya = 0 Pago YA E E Vara) a Y Ps a Cae" Ya Ta Va 0 From Eg. (1), (-604+287.75j-—60k) -(v,) = (-601 + 287,75]-60k)-(1.118031—2.23607j) 287.75, = -0002.3607) + (6044.72 14) v, = 4.6626 mm/s v, =(4,66 mnvs)¡ « PROPRIETARY MATERIAL, 0 2009 The McGraw-Hill Companies, Ino. All rights reserved, Ne port of dis Manual may De displayed. reproduced or distribited in any forn or by ans means, elibont the prior written permission 0f the publisher, or used beyond de Limited distribution fo toachers and edieators permitted ly MoGrawePiR for thotr individual coneve preparation. if par are e stuelontiesing this Manual. parar nzing itwithont permission. 1241 y PROBLEM 15.210 Two shafis AC and EG, which lie in the vertical pz plane, are connected by a universal joint at D. Shaft 4C rotates with a constan! angular velocity (7 as shown, At a time when the arm of the erosspiece attached to shafl AC is verlical, determine the angular velocity of shalt EG. SOLUTION Angular velocity of shafi AC. Dio = Mk Let en] be the angular velocity of body DD relative to shaft 4D. Angular velocity of bed» D. Oy >0k+0,j Angular velocity ofshaft EG. (My = (0,(cos25*k —sin 25?j) Let en i be the angular velocity of body D relative to shaft EG. Angular velocitv of body D. 60), = (cos 25% —sin 25%) + 09,1 Equate the two expressions for 6, and resolve into components. ii 0=0, mM o m3-ajsin25" (2 k: a =0,c00825% (3) From Eq. (3), a y == osin 250 + 005 25%) 4 cos25* cos25 PROPRIETARY MATERIAL. 9 2009 The MeGrw-Hi1l Companies, Ine. All rights reserved. No part of His Manual may de displaved, veproduced or distributed ln any Joc or by gay mear, vlibiost this prior veiltten permission of the publisher, or used beyond he limited distribution to teachers and educators permitted hy MoGreow-HHE for their individual course preparation, [fyom area studentasing this Manual, vor are usitg owithort permisxion. 1242 . PROBLEM 15.211 Solve Problem 15.210, assuming that the arm of the crosspiece attached to shaft 4€ is horizontal. SOLUTION Angular velocity 0] shaft AC. (0, =9k Let cai be the angular velocity oF body 1 relative to shaft 4D, Angular velocity of body D. 0 =4k + 034 Angular velocity of shafi EG. 056 =0% (cos25*k —sin 25%) Let c% be the angular velocity of body D relative to shaft EG. Where A is a unit vector along the axis, the clevis axle is attached to shaft EG, y A=c0s 25" + sin 25%k » (0,4 = e, cos25*j+ «0, sin 25% É 25 Angular velocity of body DD, My 0 A 6, = (ea, cos 23% —, sin 25%) +(ar, sin 25% +4, cos 25” )k Equate the two expressions for €, and resolve into components. 1 0a=0(1) p D= a, cos 25” e, sin 25% (2) k: a =esin25"4+0 00525" (3) From Egs. (2) and (3), (0, = 4 cos 257 Wi = 0 Cos 25 (sin 25% + 00825") «€ PROPRIETARY MATERIAL. %5 2009 The MeGraw-Hill Compantes, 1ne. All rights reserved. Mo part of is Mainal may: de displayed. reproduced or distribmied in aay form or by any arcars, withont the prior weltien permissión ej tae publisher, or used beyond the line distribution lo teachers and educolors permitied by MoGraw Hill for their individual conse preparation, [fyon are a stidentrsing is Mental, yo are sig ¡Dovidionl permission 1243 PROBLEM 15.212 ln Problem 15,203, the balland-socket joint between the rod and collar A is replaced by the elevis shown. Determine (e) the angular velocity of the rod, (6) the velocity of collar 4, SOLUTION Geometry. e=0.05 m, Lin =0.275 m Es=cA401SO07+ A or 0,275 =0.05 40,157 + p? b=0.223m E ¿y =—(0.05 mi +(0.15 m)j+(0.225 m)k Velocity ol collar 8, Ya =- (130 mms =—(0.18 més)i Velocity of collar A. y, =1,k Angular velocity of collar 4. (1, =40k The axle of the cfevis at Bis perpendicular to both ihe z axis and the rod AB.A vector P along this axle is P=k xr y =0.05j+0.151 P= ¿(os +(0.15) =0,158114 Unit vector A along the axte: A= z =0,0486814 0,31623j Let 60, be he angular velocity of rod 48 relative to collar a, 0, =04=0.316230,1+0.94868m,¡ Angular velocity of rod AB. 0D yy = 00, + 0d, (y =0.948680,1+0.316230,j+ak (1) WO A i ¡ k v¿k =-0.181+|0.048684, ¡Gli a 0,05 0.15 0.2258 Resolving into componenis, ii: 0=-0.18-0.154 +0.071150, (2 f 0= 0.0544 0.2 13450, 3) ki y= 0.1581 la, (4) PROPRIETARY MATERIAL. 4 2000 The McGraw JH Companies, Inc. All rights reserved. No port ef is Meal may be displayed, reproduced or distributed in cue foemor by omyomcoas, withewl fe prior written permission of he publisher, or used beyond the Heited alistribution to teachers and educators permitted by MeGrewHil for their idivideal course preparation. If yowere a student using sis Marwal, von are using owithionl permission.  PROBLEM 15.213 (Continued) Solving Egs. (2), (3) and (4) simultancously, Y, =-30 in.fs, (0, =0.27867 rad?s, (1, 1.1429 rad/s (er) — Angular velocity of rod 48. Erom Eq. (1) 012 =(0,30478)01,1429)1 + 0.27867j+(0.95242)(1.1429)k 0 y = 10.348 rad/sM + (0.279 rad/s)] + (1.089 rad/s)k < tb) Nelocity of 4, Y, =-(30.0 in./s)j «€ PROPRIETARY MATERIAL, 1 2009 The MoGraw-Hill Companies, Inc. A rights reserved. No: part ef ahís Menical may be displayed, reproduced ur distribielted in ex for or dy any nrcans, witbord the prior writica permission of the publisher, or nxed bevonid the limited distribution to teachers and educarors permitted hy MoGraw- bd for their individual course preparation. Jfyom are a student using ds Mannal, por are sta Fl velthowl permission. 1247 PROBLEM 15.214 For the mechanism of the problem indicated, determine the , acceleration of collar A, Mechanism of Problem 15.202. Á HN PROBLEM 15.202 Rod 48 of length 275 mm is connected 150 q 27, by ball-and-socket joints to collars 4 and B, which slide a along the two rods shown. Knowing that collar B moves - * toward the origin O at a constant speed of 180 mmís, ? determine the velocity of collar A when c= 175 mm. SOLUTION Geometry. e=0.175m, ly =0.275m Pp=0+01500+b or 02758 =0.1758 40.15 +1? b=0.15m E yo =(0.175 m4+ (0.15 m)j+(0.15 m)k Velocity of collar 8, v¿ =- (180 mms)t =-(0.18 ms) Velocity of collar A. where Va 0 ¿ppp Noting that Y yy is perpendicular to yy, We get Pyoy Via = 0. Forming £ yy * Y ¿> We get Lao Ya = Tap [Na E ap) Era Va Pas Yara or Tar Va Eur Vo (1 From Eq. (1), (-0.1751+0.15j+0.15k) (,k) =(-0.175140.15j+0.15k)-(—0.18%) 0.15v, =(-0.175)-0.18) or v,¿=0.21 m/s y, =(0.21 ms)k elative velocity. Vas = Va — Va =(0.18 mís)i+(0.21 m/s)k Van" Yan = (0.18 mis)? +(0,21 m/s)? = 0,0765 m/s? lez PROPRIETARY MATERIAL. 0 2009 The MeGraw-Hil Companics, Inc, All rights reserved. No part af his Manual may be displayed, reproduced or distributed in any form or by any means, without fhe prior writien permisston of the publisher, or wsed beyond fe dimited distribution to teachers and educators permilled by MeGraw-Hi8l for their individual course preparation. [fyoware a student rsing this Mental, pom are stas ¡wirhoni permission. 1248 PROBLEM 15.214 (Continued) Acceleration of collar 8. ay=0 Acceleration of collar 4, a, =0,k ¿25 PA where Mg O q E yg 0 yg Va Noting that (¿gr yyy is perpendicular LO Y; y, We get Y yyy + 0L yy XE yyy = 0 We note also that Pi Da AV a Vat La O EV an o Vai o 2 = AY gp ) 7 a Then Pain A gy 0 (Y gp) 2 =—V ay) Fonming Y yyy +9, Wwe get TI] ap a or Eg Ey VA — (a Y (2) From Eg. (2) (-0.17514-0.15)+0.45k) (0,¿k)=0—0,0765 0.1544 =-0.0765 dy =0.51 mis? Az =-(510 mms )k PROPRIETARY MATERIAL. £3 2009 The MoGras-HHL Companies, Inc. All rights reserved. No port of His Mernial may be displaved reproduccd or «distribrled da np Jorcor hy any neans, without the prior written permission of the publisher, pr used beyond ¿he limited distribition do teachers and educators permined dy MeG Hill for their mdtvidual conse preparation, Ifyow ore a studentusing dis Marne, Por dare using dl wiihicid permission.