Dynamics 14th edition by r c hibbeler chapter 18

© 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–1 At a given instant the body of mass m has an angular velocity V and its mass center has a velocity vG Show that its kinetic energy can be represented as T = 12IICv2, where IIC is the moment of inertia of the body determined about the instantaneous axis of zero velocity, located a distance rG>IC from the mass center as shown IC rG/IC G vG SOLUTION T = 1 my2G + IG v2 2 = 1 m(vrG>IC)2 + IG v2 2 = A mr2G>IC + IG B v2 = I v2 IC where yG = vrG>IC However mr2G>IC + IG = IIC Q.E.D 912 V © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–2 The wheel is made from a 5-kg thin ring and two 2-kg slender rods If the torsional spring attached to the wheel’s center has a stiffness k = N # m>rad, and the wheel is rotated until the torque M = 25 N # m is developed, determine the maximum angular velocity of the wheel if it is released from rest 0.5 m O M SOLUTION Kinetic Energy and Work: The mass moment of inertia of the wheel about point O is IO = mRr + ¢ m l2 ≤ 12 r = 5(0.52) + c (2)(12) d 12 = 1.5833 kg # m2 Thus, the kinetic energy of the wheel is T = 1 I v2 = (1.5833) v2 = 0.79167 v2 O Since the wheel is released from rest, T1 = The torque developed is M = ku = 2u Here, the angle of rotation needed to develop a torque of M = 25 N # m is 2u = 25 u = 12.5 rad The wheel achieves its maximum angular velocity when the spacing is unwound that M is when the wheel has rotated u = 12.5 rad Thus, the work done by q is 12.5 rad UM = L Mdu = 2u du L0 12.5 rad = u † = 156.25 J Principle of Work and Energy: T1 + © u - = T2 + 156.25 = 0.79167 v2 Ans v = 14.0 rad/s Ans: v = 14.0 rad>s 913 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–3 The wheel is made from a 5-kg thin ring and two 2-kg slender rods If the torsional spring attached to the wheel’s center has a stiffness k = N # m>rad, so that the torque on the center of the wheel is M = 12u2 N # m, where u is in radians, determine the maximum angular velocity of the wheel if it is rotated two revolutions and then released from rest 0.5 m O M SOLUTION Io = c (2)(1)2 d + 5(0.5)2 = 1.583 12 T1 + ©U1 - = T2 4p + L0 2u du = (1.583) v2 (4p)2 = 0.7917v2 Ans v = 14.1 rad/s Ans: v = 14.1 rad>s 914 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–4 A force of P = 60 N is applied to the cable, which causes the 200-kg reel to turn since it is resting on the two rollers A and B of the dispenser Determine the angular velocity of the reel after it has made two revolutions starting from rest Neglect the mass of the rollers and the mass of the cable Assume the radius of gyration of the reel about its center axis remains constant at kO = 0.6 m P O 0.75 m 1m A B 0.6 m Solution Kinetic Energy Since the reel is at rest initially, T1 = The mass moment of inertia of the reel about its center O is I0 = mk 20 = 200 ( 0.62 ) = 72.0 kg # m2 Thus, T2 = I v = (72.0)v2 = 36.0 v2 20 Work Referring to the FBD of the reel, Fig a, only force P does positive work When the reel rotates revolution, force P displaces S = ur = 2(2p)(0.75) = 3p m Thus Up = Ps = 60(3p) = 180p J Principle of Work and Energy T1 + ΣU1 - = T2   0 + 180p = 36.0 v2 Ans v = 3.9633 rad>s = 3.96 rad>s Ans: v = 3.96 rad>s 915 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–5 A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn since it is resting on the two rollers A and B of the dispenser Determine the angular velocity of the reel after it has made two revolutions starting from rest Neglect the mass of the rollers and the mass of the cable The radius of gyration of the reel about its center axis is kG = 0.42 m P 30° 250 mm G 500 mm A B 400 mm SOLUTION T1 + ΣU1 - = T2 + 20(2)(2p)(0.250) = v = 2.02 rad>s 175(0.42)2 v2 Ans Ans: v = 2.02 rad>s 916 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–6 A force of P = 20 N is applied to the cable, which causes the 175-kg reel to turn without slipping on the two rollers A and B of the dispenser Determine the angular velocity of the reel after it has made two revolutions starting from rest Neglect the mass of the cable Each roller can be considered as an 18-kg cylinder, having a radius of 0.1 m The radius of gyration of the reel about its center axis is kG = 0.42 m P 30° 250 mm G 500 mm A B 400 mm SOLUTION System: T1 + ΣU1 - = T2 [0 + + 0] + 20(2)(2p)(0.250) = v = vr (0.1) = v(0.5) 1 175(0.42)2 v2 + 2c (18)(0.1)2 d v2r 2 vr = 5v Solving: Ans v = 1.78 rad>s Ans: v = 1.78 rad>s 917 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–7 The double pulley consists of two parts that are attached to one another It has a weight of 50 lb and a radius of gyration about its center of kO = 0.6 ft and is turning with an angular velocity of 20 rad>s clockwise Determine the kinetic energy of the system Assume that neither cable slips on the pulley v 20 rad/s 0.5ft ft O SOLUTION T = 1 I v2O + mA v2A + mB v2B O 2 T = 50 20 30 a (0.6)2 b(20)2 + a b C (20)(1) D + a b C (20)(0.5) D 2 32.2 32.2 32.2 = 283 ft # lb B 30 lb A 20 lb Ans Ans: T = 283 ft # lb 918 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–8 The double pulley consists of two parts that are attached to one another It has a weight of 50 lb and a centroidal radius of gyration of kO = 0.6 ft and is turning with an angular velocity of 20 rad> s clockwise Determine the angular velocity of the pulley at the instant the 20-lb weight moves ft downward v ϭ 20 rad/s 0.5 ft ft O SOLUTION Kinetic Energy and Work: Since the pulley rotates about a fixed axis, vA = vrA = v(1) and vB = vrB = v(0.5) The mass moment of inertia of the pulley about point O is IO = mkO = ¢ kinetic energy of the system is T = = 50 ≤ (0.62) = 0.5590 slug # ft2 Thus, the 32.2 B 30 lb A 20 lb 1 I v2 + mAvA2 + mBvB2 O 2 1 20 30 (0.5590)v2 + ¢ ≤ [v(1)]2 + ¢ ≤ [v(0.5)]2 2 32.2 32.2 = 0.7065v2 Thus, T1 = 0.7065(202) = 282.61 ft # lb Referring to the FBD of the system shown in Fig a, we notice that Ox, Oy, and Wp no work while WA does positive work and WB does negative work When A moves ft downward, the pulley rotates u = SA SB = rA rB SB = 0.5 SB = 2(0.5) = ft c Thus, the work of WA and WB are UWA = WA SA = 20(2) = 40 ft # lb UWB = - WB SB = - 30(1) = -30 ft # lb Principle of Work and Energy: T1 + U1 - = T2 282.61 + [40 + ( - 30)] = 0.7065 v2 Ans v = 20.4 rad>s Ans: v = 20.4 rad>s 919 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–9 The disk, which has a mass of 20 kg, is subjected to the couple moment of M = (2u + 4) N # m, where u is in radians If it starts from rest, determine its angular velocity when it has made two revolutions 300 mm M O Solution Kinetic Energy Since the disk starts from rest, T1 = The mass moment of inertia 1 of the disk about its center O is I0 = mr = ( 20 )( 0.32 ) = 0.9 kg # m2 Thus 2 T2 = 1 I v2 = (0.9) v2 = 0.45 v2 2 Work Referring to the FBD of the disk, Fig a, only couple moment M does work, which it is positive UM = L M du = L0 Principle of Work and Energy 2(2p) (2u + 4)du = u + 4u ` 4p = 208.18 J T1 + ΣU1 - = T2   + 208.18 = 0.45 v2 Ans   v = 21.51 rad>s = 21.5 rad>s Ans: v = 21.5 rad>s 920 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–10 The spool has a mass of 40 kg and a radius of gyration of kO = 0.3 m If the 10-kg block is released from rest, determine the distance the block must fall in order for the spool to have an angular velocity v = 15 rad>s Also, what is the tension in the cord while the block is in motion? Neglect the mass of the cord 300 mm 500 mm O Solution Kinetic Energy Since the system is released from rest, T1 = The final velocity of the block is vb = vr = 15(0.3) = 4.50 m>s The mass moment of inertia of the spool about O is I0 = mk 20 = 40 ( 0.32 ) = 3.60 Kg # m2 Thus T2 = = I v + mbv2b 20 1 (3.60) ( 152 ) + (10) ( 4.502 ) 2 = 506.25 J For the block, T1 = and T2 = 1 m v2 = ( 10 )( 4.502 ) = 101.25 J b b Work Referring to the FBD of the system Fig a, only Wb does work when the block displaces s vertically downward, which it is positive UWb = Wb s = 10(9.81)s = 98.1 s Referring to the FBD of the block, Fig b Wb does positive work while T does negative work UT = - Ts UWb = Wbs = 10(9.81)(s) = 98.1 s Principle of Work and Energy For the system, T1 + ΣU1 - = T2   + 98.1s = 506.25   Ans s = 5.1606 m = 5.16 m For the block using the result of s, T1 + ΣU1 - = T2 + 98.1(5.1606) - T(5.1606) = 101.25      T = 78.48 N = 78.5 N Ans Ans: s = 5.16 m T = 78.5 N 921 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–56 If the chain is released from rest from the position shown, determine the angular velocity of the pulley after the end B has risen ft The pulley has a weight of 50 lb and a radius of gyration of 0.375 ft about its axis The chain weighs lb/ft 0.5 ft ft SOLUTION Potential Energy: (yG1)1 = ft, (yG 2)1 = ft, (yG1)2 = ft, and (yG2)2 = ft With reference to the datum in Fig a, the gravitational potential energy of the chain at position and is V1 = (Vg)1 = W1(yG1)1 - W2(yG2)1 ft B A = -6(4)(2) - 6(6)(3) = - 156 ft # lb V2 = (Vg)2 = - W1(yG1)2 + W2(yG2)2 = -6(2)(1) - 6(8)(4) = - 204 ft # lb Kinetic Energy: Since the system is initially at rest, T1 = The pulley rotates about a fixed axis, thus, (VG1)2 = (VG2)2 = v2 r = v2(0.5) The mass moment of inertia of 50 the pulley about its axis is IO = mkO = (0.3752) = 0.2184 slug # ft2 Thus, the 32.2 final kinetic energy of the system is T = = 1 I v + m1(VG1)2 + m2 (VG2)2 2 O 2 1 6(2) 6(8) 6(0.5)(p) (0.2184)v2 + c d [v2(0.5)]2 + c d[v2(0.5)]2 + c d[v2(0.5)]2 2 32.2 32.2 32.2 = 0.3787v2 Conservation of Energy: T1 + V1 = T2 + V2 + (- 156) = 0.3787v22 = (- 204) v2 = 11.3 rad >s Ans Ans: v2 = 11.3 rad>s 970 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–57 If the gear is released from rest, determine its angular velocity after its center of gravity O has descended a distance of ft The gear has a weight of 100 lb and a radius of gyration about its center of gravity of k = 0.75 ft ft O SOLUTION Potential Energy: With reference to the datum in Fig a, the gravitational potential energy of the gear at position and is V1 = (Vg)1 = W(y0)1 = 100(0) = V2 = (Vg)2 = - W1(y0)2 = - 100(4) = - 400 ft # lb Kinetic Energy: Referring to Fig b, we obtain vO = vrO / IC = v(1).The mass moment of inertia of the gear about its mass center is IO = mkO = Thus, T = = 100 (0.752) = 1.7469 kg # m2 32.2 1 mvO2 + IOv2 2 100 a b [v (1)]2 + (1.7469)v2 32.2 = 2.4262v2 Since the gear is initially at rest, T1 = Conservation of Energy: T1 + V1 = T2 + V2 + = 2.4262v2 - 400 Ans v = 12.8 rad>s Ans: v = 12.8 rad>s 971 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–58 The slender 6-kg bar AB is horizontal and at rest and the spring is unstretched Determine the stiffness k of the spring so that the motion of the bar is momentarily stopped when it has rotated clockwise 90° after being released C k A Solution Kinetic Energy The mass moment of inertia of the bar about A is IA = (6) ( 22 ) + ( 12 ) = 8.00 kg # m2 Then 12 1 T = IA v2 = (8.00) v2 = 4.00 v2 2 1.5 m B 2m Since the bar is at rest initially and required to stop finally, T1 = T2 = Potential Energy With reference to the datum set in Fig a, the gravitational potential energies of the bar when it is at positions ➀ and ➁ are (Vg)1 = mgy1 = (Vg)2 = mgy2 = 6(9.81)( - 1) = -58.86 J The stretch of the spring when the bar is at position ➁ is x2 = 222 + 3.52 - 1.5 = 2.5311 m Thus, the initial and final elastic potential energy of the spring are (Ve)1 = kx1 = (Ve)2 = k ( 2.53112 ) = 3.2033k Conservation of Energy T1 + V1 = T2 + V2 + (0 + 0) = + ( - 58.86) + 3.2033k k = 18.3748 N>m = 18.4 N>m Ans Ans: k = 18.4 N>m 972 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–59 The slender 6-kg bar AB is horizontal and at rest and the spring is unstretched Determine the angular velocity of the bar when it has rotated clockwise 45° after being released The spring has a stiffness of k = 12 N>m C k A Solution Kinetic Energy The mass moment of inertia of the bar about A is IA = (6) ( 22 ) + ( 12 ) = 8.00 kg # m2 Then 12 T = 1.5 m B 2m 1 I v2 = (8.00) v2 = 4.00 v2 2A Since the bar is at rest initially, T1 = Potential Energy with reference to the datum set in Fig a, the gravitational potential energies of the bar when it is at positions ① and ② are (Vg)1 = mgy1 = (Vg)2 = mgy2 = 6(9.81)( - sin 45°) = - 41.62 J From the geometry shown in Fig a, -1 a = 222 + 1.52 = 2.5 m  f = tan a Then, using cosine law, 1.5 b = 36.87° l = 22.52 + 22 - 2(2.5)(2) cos (45° + 36.87°) = 2.9725 m Thus, the stretch of the spring when the bar is at position ② is x2 = 2.9725 - 1.5 = 1.4725 m Thus, the initial and final elastic potential energies of the spring are (Ve)1 = kx = (Ve)2 = (12) ( 1.47252 ) = 13.01 J Conservation of Energy T1 + V1 = T2 + V2 + (0 + 0) = 4.00 v2 + ( - 41.62) + 13.01 v = 2.6744 rad>s = 2.67 rad>s  Ans Ans: v = 2.67 rad>s 973 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–60 The pendulum consists of a 6-kg slender rod fixed to a 15-kg disk If the spring has an unstretched length of 0.2  m, determine the angular velocity of the pendulum when it is released from rest and rotates clockwise 90° from the position shown The roller at C allows the spring to always remain vertical C 0.5 m k ϭ 200 N/m A B 0.5 m Solution 0.5 m D 0.3 m Kinetic Energy The mass moment of inertia of the pendulum about B is 1 IB = c (6) ( 12 ) + ( 0.52 ) d + c (15) ( 0.32 ) + 15 ( 1.32 ) d = 28.025 kg # m2 Thus 12 T = 1 I v2 = (28.025) v2 = 14.0125 v2 2B Since the pendulum is released from rest, T1 = Potential Energy with reference to the datum set in Fig a, the gravitational potential energies of the pendulum when it is at positions ① and ② are (Vg)1 = mrg(yr)1 + mdg(yd)1 = (Vg)2 = mrg(yr)2 + mdg(yd)2 = 6(9.81)( - 0.5) + 15(9.81)( - 1.3) = -220.725 J The stretch of the spring when the pendulum is at positions ① and ② are x1 = 0.5 - 0.2 = 0.3 m x2 = - 0.2 = 0.8 m Thus, the initial and final elastic potential energies of the spring are (Ve)1 = kx = (200) ( 0.32 ) = 9.00 J 2 (Ve)2 = kx = (200) ( 0.82 ) = 64.0 J 2 Conservation of Energy T1 + V1 = T2 + V2 + (0 + 9.00) = 14.0125v2 + ( - 220.725) + 64.0 v = 3.4390 rad>s = 3.44 rad>s Ans Ans: v = 3.44 rad>s 974 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–61 The 500-g rod AB rests along the smooth inner surface of a hemispherical bowl If the rod is released from rest from the position shown, determine its angular velocity at the instant it swings downward and becomes horizontal 200 mm B SOLUTION A 200 mm Select datum at the bottom of the bowl u = sin - a 0.1 b = 30° 0.2 h = 0.1 sin 30° = 0.05 CE = 2(0.2)2 - (0.1)2 = 0.1732 m ED = 0.2 - 0.1732 = 0.02679 T1 + V1 = T2 + V2 + (0.5)(9.81)(0.05) = 1 c (0.5)(0.2)2 d v2AB + (0.5)(vG)2 + (0.5)(9.81)(0.02679) 12 Since vG = 0.1732vAB Ans vAB = 3.70 rad>s Ans: vAB = 3.70 rad>s 975 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–62 The 50-lb wheel has a radius of gyration about its center of gravity G of kG = 0.7 ft If it rolls without slipping, determine its angular velocity when it has rotated clockwise 90° from the position shown The spring AB has a stiffness k = 1.20 lb/ft and an unstretched length of 0.5 ft The wheel is released from rest ft 0.5 ft B 0.5 ft G k = 1.20 lb/ft A ft SOLUTION T1 + V1 = T2 + V2 + 1 50 50 (1.20)[2(3)2 + (0.5)2 - 0.5]2 = [ (0.7)2]v2 + ( )(1v)2 2 32.2 32.2 + (1.20)(0.9292 - 0.5)2 Ans v = 1.80 rad s Ans: v = 1.80 rad>s 976 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–63 The system consists of 60-lb and 20-lb blocks A and B, respectively, and 5-lb pulleys C and D that can be treated as thin disks Determine the speed of block A after block B has risen ft, starting from rest Assume that the cord does not slip on the pulleys, and neglect the mass of the cord C 0.5 ft A SOLUTION 0.5 ft D Kinematics: The speed of block A and B can be related using the position coordinate equation (1) sA + 2sB = l ¢sA + 2¢sB = ¢sA + 2(5) = B ¢sA = -10 ft = 10 ftT Taking time derivative of Eq (1), we have yA + 2yB = yB = - 0.5yA Potential Energy: Datum is set at fixed pulley C.When blocks A and B (pulley D) are at their initial position, their centers of gravity are located at sA and sB Their initial gravitational potential energies are - 60sA, - 20sB, and -5sB When block B (pulley D) rises ft, block A decends 10 ft Thus, the final position of blocks A and B (pulley D) are (sA + 10) ft and (sB - 5) ft below datum Hence, their respective final gravitational potential energy are -60(sA + 10), -20(sB - 5), and -5(sB - 5) Thus, the initial and final potential energy are V1 = - 60sA - 20sB - 5sB = -60sA - 25sB V2 = -60(sA + 10) - 20(sB - 5) - 5(sB - 5) = - 60sA - 25sB - 475 Kinetic Energy: The mass moment inertia of the pulley about its mass center is IG = a b A 0.52 B = 0.01941 slug # ft2 Since pulley D rolls without slipping, 32.2 yB yB vD = = = 2yB = 2( - 0.5yA) = - yA Pulley C rotates about the fixed point rD 0.5 yA yA hence vC = = = 2yA Since the system is at initially rest, the initial kinectic rC 0.5 energy is T1 = The final kinetic energy is given by T2 = 1 1 m y2A + mB y2B + mD y2B + IGv2D + IG v2C A 2 2 = 60 20 a b y2A + a b ( -0.5yA)2 + a b( -0.5yA)2 32.2 32.2 32.2 + 1 (0.01941)( -yA)2 + (0.01941)(2yA)2 2 = 1.0773y2A Conservation of Energy: Applying Eq 18–19, we have T1 + V1 = T2 + V2 + (-60sA - 25sB) = 1.0773y2A + (- 60sA - 25sB - 475) Ans y4 = 21.0 ft>s 977 Ans: vA = 21.0 ft>s © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–64 The door is made from one piece, whose ends move along the horizontal and vertical tracks If the door is in the open position, u = 0°, and then released, determine the speed at which its end A strikes the stop at C Assume the door is a 180-lb thin plate having a width of 10 ft A C ft Solution ft u B T1 + V1 = T2 + V2 + = 1 180 180 c a b(8)2 d v2 + a b(1v)2 - 180(4) 12 32.2 32.2 v = 6.3776 rad>s vc = v(5) = 6.3776(5) = 31.9 m>s Ans Ans: vc = 31.9 m>s 978 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–65 The door is made from one piece, whose ends move along the horizontal and vertical tracks If the door is in the open position, u = 0°, and then released, determine its angular velocity at the instant u = 30°  Assume the door is a 180-lb thin plate having a width of 10 ft A C ft Solution ft u B T1 + V1 = T2 + V2 + = 1 180 180 c a b(8)2 d v2 + a bv - 180(4 sin 30°) 12 32.2 32.2 G (1) rIC - G = 2(1)2 + (4.3301)2 - 2(1)(4.3301) cos 30° rIC - G = 3.50 m Thus, vG = 3.50 v Substitute into Eq (1) and solving, Ans v = 2.71 rad>s Ans: v = 2.71 rad>s 979 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–66 The end A of the garage door AB travels along the horizontal track, and the end of member BC is attached to a spring at C If the spring is originally unstretched, determine the stiffness k so that when the door falls downward from rest in the position shown, it will have zero angular velocity the moment it closes, i.e., when it and BC become vertical Neglect the mass of member BC and assume the door is a thin plate having a weight of 200 lb and a width and height of 12 ft There is a similar connection and spring on the other side of the door 12 ft A B 15 ft C ft SOLUTION D ft ft (2)2 = (6)2 + (CD)2 - 2(6)(CD) cos 15° CD2 - 11.591CD + 32 = Selecting the smaller root: CD = 4.5352 ft T1 + V1 = T2 + V2 + = + c (k)(8 - 4.5352)2 d - 200(6) Ans k = 100 lb/ft Ans: k = 100 lb>ft 980 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–67 The system consists of a 30-kg disk, 12-kg slender rod BA, and a 5-kg smooth collar A If the disk rolls without slipping, determine the velocity of the collar at the instant u = 0° The system is released from rest when u = 45° A 2m Solution Kinetic Energy Since the system is released from rest, T1 = Referring to the kinematics diagram of the rod at its final position, Fig a, we found that IC is located at B Thus, (vB)2 = Also (vA)2 = (vr)2rA>IC;  (vA)2 = (vr)2(2)  (vr)2 = (vA)2 u B 30Њ 0.5 m C Then (vGr)2 = (vr)2(rGr>IC);  (vGr)2 = (vA)2 (1) = (vA)2 For the disk, since the velocity of its center (vB)2 = 0, (vd)2 = Thus, T2 = = 1 m (v )2 + IGr(vr)22 + mc(vA)22 r Gr 2 (vA)2 (vA)2 1 1 (12) c d + c (12) ( 22 ) d c d + (5)(vA)22 2 12 2 = 4.50(vA)22 Potential Energy Datum is set as shown in Fig a Here, SB = - cos 45° = 0.5858 m Then (yd)1 = 0.5858 sin 30° = 0.2929 m (yr)1 = 0.5858 sin 30° + sin 75° = 1.2588 m (yr)2 = sin 30° = 0.5 m (yc)1 = 0.5858 sin 30° + sin 75° = 2.2247 m (yc)2 = sin 30° = 1.00 m Thus, the gravitational potential energies of the disk, rod and collar at the initial and final positions are (Vd)1 = md g(yd)1 = 30(9.81)(0.2929) = 86.20 J (Vd)2 = md g(yd)2 = (Vr)1 = mr g(yr)1 = 12(9.81)(1.2588) = 148.19 J (Vr)2 = mr g(yr)2 = 12(9.81)(0.5) = 58.86 J (Vc)1 = mc g(yc)1 = 5(9.81)(2.2247) = 109.12 J (Vc)2 = mc g(yc)2 = 5(9.81)(1.00) = 49.05 J 981 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher 18–67.  Continued Conservation of Energy T1 + V1 = T2 + V2 + (86.20 + 148.19 + 109.12) = 4.50(vA)22 + (0 + 58.86 + 49.05) (vA)2 = 7.2357 m>s = 7.24 m>s Ans Ans: ( vA ) = 7.24 m>s 982 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–68 The system consists of a 30-kg disk A, 12-kg slender rod BA, and a 5-kg smooth collar A If the disk rolls without slipping, determine the velocity of the collar at the instant u = 30° The system is released from rest when u = 45° A 2m Solution u Kinetic Energy Since the system is released from rest, T1 = Referring to the kinematics diagram of the rod at final position with IC so located, Fig a, rA>IC = cos 30° = 1.7321 m  rB>IC = cos 60° = 1.00 m rGr>IC = 212 + 1.002 - 2(1)(1.00) cos 60° = 1.00 m B 30Њ 0.5 m Then C (vA)2 = (vr)2(rA>IC);  (vA)2 = (vr)2(1.7321)  (vr)2 = 0.5774(vA)2 (vB)2 = (vr)2(rB>IC);  (vB)2 = [0.5774(vA)2](1.00) = 0.5774(vA)2 (vGr)2 = (vr)2(rGr>IC);  (vGr)2 = [0.5774(vA)2](1.00) = 0.5774(vA)2 Since the disk rolls without slipping, (vB)2 = vdrd;  0.5774(vA)2 = (vd)2(0.5) (vd)2 = 1.1547(vA)2 Thus, the kinetic energy of the system at final position is T2 = = 1 1 m (v )2 + IGr(vr)22 + md(vB)22 + IB(vd)22 + mc(vA)22 r Gr 2 2 1 (12)[0.5774(vA)2]2 + c (12) ( 22 ) d [0.5774(vA)2]2 2 12 + 1 (3.0)[0.5774(vA)2]2 + c (30) ( 0.52 ) d [1.1547(vA)2]2 2 + (5)(vA)22 = 12.6667(vA)22 Potential Energy Datum is set as shown in Fig a Here, SB = cos 30° - cos 45° = 0.3178 m Then (yd)1 = 0.3178 sin 30° = 0.1589 m (yr)1 = 0.3178 sin 30° + sin 75° = 1.1248 m (yr)2 = sin 60° = 0.8660 m (yc)1 = 0.3178 sin 30° + sin 75° = 2.0908 m (yc)2 = sin 60° = 1.7321 m 983 © 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher *18–68.  Continued Thus, the gravitational potential energies of the disk, rod and collar at initial and final position are (Vd)1 = mdg(yd)1 = 30(9.81)(0.1589) = 46.77 J (Vd)2 = mdg(yd)2 = (Vr)1 = mrg(yr)1 = 12(9.81)(1.1248) = 132.42 J (Vr)2 = mrg(yr)2 = 12(9.81)(0.8660) = 101.95 J (Vc)1 = mcg(yc)1 = 5(9.81)(2.0908) = 102.55 J (Vc)2 = mcg(yc)2 = 5(9.81)(1.7321) = 84.96 J Conservation of Energy T1 + V1 = T2 + V2 + (46.77 + 132.42 + 102.55) = 12.6667(vA)22 + (0 + 101.95 + 84.96) Ans (vA)2 = 2.7362 m>s = 2.74 m>s Ans: (vA)2 = 2.74 m>s 984 ... 2016 Pearson Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced,... Education, Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form... Inc., Upper Saddle River, NJ All rights reserved This material is protected under all copyright laws as they currently exist No portion of this material may be reproduced, in any form or by any