91962_10_R2_p0827-0866 6/5/09 4:11 PM Page 827 © 2010 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. R2–1. An automobile transmission consists of the planetary gear system shown. If the ring gear R is held fixed so that vR = 0, and the shaft s and sun gear S, rotates at 20 rad>s, determine the angular velocity of each planet gear P and the angular velocity of the connecting rack D, which is free to rotate about the center shaft s. R P vR 2 in. S 20 rad/s 8 in. 4 in. For planet gear P: The velocity of point A is yA = vs rs = 20 a 4 b = 6.667 ft>s. 12 vB = vA + vB>A 0 = C 6.6 67 D + B vP a : ; + b a: 4 bR 12 4 0 = 6.667 - vP a b 12 vP = 20 rad>s Ans. For connecting rack D: vC = vA + vC>A c y C d = C 6.667 D + B 20 a : : ; + b a: 2 bR 12 2 yC = 6.667 - 20 a b 12 yC = 3.333 ft>s The rack is rotating about a fixed axis (shaft s). Hence, yC = vD rD 3.333 = vD a 6 b 12 vD = 6.67 rad>s Ans. 827 s D P 91962_10_R2_p0827-0866 6/5/09 4:11 PM Page 828 © 2010 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. R2–2. An automobile transmission consists of the planetary gear system shown. If the ring gear R rotates at vR = 2 rad>s, and the shaft s and sun gear S, rotates at 20 rad>s, determine the angular velocity of each planet gear P and the angular velocity of the connecting rack D, which is free to rotate about the center shaft s. R P vR 2 in. S 20 rad/s 8 in. 4 in. 4 For planet gear P: The velocity of points A and B are yA = vS rS = 20a b 12 8 =6.667 ft>s and yB = vB rB = 2 a b = 1.333 ft>s. 12 vB = vA + vB>A c1.333 d = c6.667 d + B vP a ; : ; + b a: -1.333 = 6.667 - vP a 4 bR 12 4 b 12 vP = 24 rad>s Ans. For connecting rack D: vC = vA + vC>A c y C d = c6.667 d + B 24 a : : ; + b a: yC = 6.667 - 24 a 2 b 12 2 bR 12 yC = 2.667 ft>s The rack is rotating about a fixed axis (shaft s). Hence, yC = vD rD 2.667 = vD a 6 b 12 vD = 5.33 rad>s Ans. 828 s D P 91962_10_R2_p0827-0866 6/5/09 4:11 PM Page 829 © 2010 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. R2–3. The 6-lb slender rod AB is released from rest when it is in the horizontal position so that it begins to rotate clockwise. A 1-lb ball is thrown at the rod with a velocity v = 50 ft>s. The ball strikes the rod at C at the instant the rod is in the vertical position as shown. Determine the angular velocity of the rod just after the impact. Take e = 0.7 and d = 2 ft. A d 3 ft C Datum at A: v ⫽ 50 ft/s T1 + V1 = T2 + V2 0 + 0 = 1 1 6 c a b(3)2 dv2 - 6(1.5) 2 3 32.2 B v = 5.675 rad>s a+ (HA)1 = (HA)2 6 1 1 6 1 1 (50)(2) - c a b (3)2 d(5.675) = c a b(3)2 dv2 + (v )(2) 32.2 3 32.2 3 32.2 32.2 BL e = 0.7 = vC - vBL 50 - [-5.675(2)] vC = 2v2 Solving, v2 = 3.81 rad>s Ans. vBL = -35.3 ft>s vC = 7.61 ft>s 829 91962_10_R2_p0827-0866 6/5/09 4:11 PM Page 830 © 2010 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. *R2–4. The 6-lb slender rod AB is originally at rest, suspended in the vertical position. A 1-lb ball is thrown at the rod with a velocity v = 50 ft>s and strikes the rod at C. Determine the angular velocity of the rod just after the impact. Take e = 0.7 and d = 2 ft. A d a + (HA)1 = (HA)2 a 3 ft C 1 6 1 1 b(50)(2) = c a b(3)2 dv2 + (v )(2) 32.2 3 32.2 32.2 BL v ⫽ 50 ft/s vC - vBL e = 0.7 = 50 - 0 B vC = 2v2 Thus, v2 = 7.73 rad>s Ans. vBL = -19.5 ft>s 830 91962_10_R2_p0827-0866 6/5/09 4:12 PM Page 831 © 2010 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. R2–5. The 6-lb slender rod is originally at rest, suspended in the vertical position. Determine the distance d where the 1-lb ball, traveling at v = 50 ft>s, should strike the rod so that it does not create a horizontal impulse at A. What is the rod’s angular velocity just after the impact? Take e = 0.5. A d Rod: a+ (HG)1 + © 0 + L L 0 + L MG dt = (HG)2 F dt (d - 1.5) = a m(vG)1 + © L v ⫽ 50 ft/s 1 (m)(3)2 bv 12 B F dt = m(vG)2 F dt = m(1.5v) Thus, m(1.5v)(d - 1.5) = 1 (m)(3)2 v 12 d = 2 ft Ans. This is called the center of percussion. See Example 19–5. a+ 3 ft C (HA)1 = (HA)2 1 6 1 1 (50)(2) = c a b(3)2 dv2 + (v )(2) 32.2 3 32.2 32.2 BL e = 0.5 = vC - vBL 50 - 0 vC = 2v2 Thus, Ans. v2 = 6.82 rad>s vBL = -11.4 ft>s 831 91962_10_R2_p0827-0866 6/5/09 4:12 PM Page 832 © 2010 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. R2–6. At a given instant, the wheel rotates with the angular motions shown. Determine the acceleration of the collar at A at this instant. A 60⬚ v ⫽ 8 rad/s a ⫽ 16 rad/s2 Using instantaneous center method: 8(0.15) yB vAB = = = 4.157 rad>s rB>IC 0.5 tan 30° aA = -aA cos 60°i + aA sin 60°j a = ak 30⬚ rB>A = {-0.5i} m aB = aA + a * rB>A - v2 rB>A 2.4i - 9.6j = (-aA cos 60°i + aA sin 60°j) + (ak) * ( -0.5i) - (4.157)2(-0.5i) 2.4i - 9.6j = (-aA cos 60° + 8.64)i + (-0.5a + aA sin 60°)j Equating the i and j components yields: 2.4 = -aA cos 60° + 8.64 -9.6 = -0.5a + (12.5) sin 60° 150 mm B aB = 16(0.15)i - 82 (0.15)j = {2.4i - 9.6j} m>s2 aA = 12.5 m>s2 ; Ans. a = 40.8 rad>s2 d 832 500 mm 91962_10_R2_p0827-0866 6/5/09 4:12 PM Page 833 © 2010 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. R2–7. The small gear which has a mass m can be treated as a uniform disk. If it is released from rest at u = 0°, and rolls along the fixed circular gear rack, determine the angular velocity of the radial line AB at the instant u = 90°. A u R Potential Energy: Datum is set at point A. When the gear is at its final position (u = 90°), its center of gravity is located (R - r) below the datum. Its gravitational potential energy at this position is -mg(R - r). Thus, the initial and final potential energies are V1 = 0 V2 = -mg(R - r) Kinetic Energy: When gear B is at its final position (u = 90°), the velocity of its yB mass center is yB = vg r or vg = since the gear rolls without slipping on the r fixed circular gear track. The mass moment of inertia of the gear about its mass 1 center is IB = mr2. Since the gear is at rest initially, the initial kinetic energy is 2 T1 = 0. The final kinetic energy is given by T2 = yB 2 1 1 1 1 1 3 my2B + IB v2g = my2B + a mr2 b a b = my2B r 2 2 2 2 2 4 Conservation of Energy: Applying Eq. 18–18, we have T1 + V1 = T2 + V2 0 + 0 = 3 my2B + [-mg(R - r)] 4 yB = 4g(R - r) B 3 Thus, the angular velocity of the radical line AB is given by vAB = 4g yB = R - r A 3(R - r) Ans. 833 r B 91962_10_R2_p0827-0866 6/5/09 4:12 PM Page 834 © 2010 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. *R2–8. The 50-kg cylinder has an angular velocity of 30 rad>s when it is brought into contact with the surface at C. If the coefficient of kinetic friction is mk = 0.2, determine how long it will take for the cylinder to stop spinning. What force is developed in link AB during this time? The axis of the cylinder is connected to two symmetrical links. (Only AB is shown.) For the computation, neglect the weight of the links. 500 mm A v ⫽ 30 rad/s B 200 mm C t2 (+ c ) m(yAy)1 + © Lt1 Fy dt = m(yAy)2 0 + NC (t) - 50(9.81)(t) = 0 + b a: NC = 490.5 N t2 m(yAx)1 + © Lt1 Fx dt = m(yAx)2 0 + 0.2(490.5)(t)—2FAB (t) = 0 FAB = 49.0 N Ans. t2 (a +) IB v1 + © Lt1 MB dt = IB v2 1 - c (50)(0.2)2 d(30) + 0.2(490.5)(0.2)(t) = 0 2 t = 1.53 s Ans. R2–9. The gear rack has a mass of 6 kg, and the gears each have a mass of 4 kg and a radius of gyration of k = 30 mm about their center. If the rack is originally moving downward at 2 m>s, when s = 0, determine the speed of the rack when s = 600 mm. The gears are free to rotate about their centers, A and B. A 50 mm B 50 mm s 2 = 40 rad>s. After 0.05 the rack has traveled s = 600 mm, both gears rotate with an angular velocity of y2 , where y2 is the speed of the rack at that moment. v2 = 0.05 Originally, both gears rotate with an angular velocity of vt = Put datum through points A and B. T1 + V1 = T2 + V2 y2 1 1 1 1 (6)(2)2 + b C 4(0.03)2 D (40)2 r + 0 = (6)y22 + 2 b C 4(0.03)2 D a b r -6(9.81)(0.6) 2 2 2 2 0.05 y2 = 3.46 m>s Ans. 834 91962_10_R2_p0827-0866 6/5/09 4:13 PM Page 835 © 2010 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. R2–10. The gear has a mass of 2 kg and a radius of gyration kA = 0.15 m. The connecting link AB (slender rod) and slider block at B have a mass of 4 kg and 1 kg, respectively. If the gear has an angular velocity v = 8 rad>s at the instant u = 45°, determine the gear’s angular velocity when u = 0°. v ⫽ 8 rad/s 0.2 m A At position 1: (vAB)1 = (yA)1 1.6 = = 2.6667 rad>s rA>IC 0.6 (yB)1 = 0 B (yAB)1 = (vAB)1 rG>IC = 2.6667(0.3) = 0.8 m>s 0.6 m 45⬚ At position 2: (vAB)2 = (yA)2 v2 (0.2) = = 0.2357v2 rA>IC 0.6 cos 45° (yB)2 = (vAB)2 rB>IC = 0.2357 v2(0.6) = 0.1414v2 (yAB)2 = (vAB)2 rG>IC = 0.2357 v2(0.6708) = 0.1581v2 T1 = 1 1 1 1 1 C (2)(0.15)2 D (8)2 + (2)(1.6)2 + (4)(0.8)2 + c (4)(0.6)2 d(2.6667)2 2 2 2 2 12 = 5.7067 J T2 = u 1 1 1 C (2)(0.15)2 D (v2)2 + (2)(0.2 v2)2 + (4)(0.1581v2)2 2 2 2 + 1 1 1 c (4)(0.6)2 d(0.2357v2)2 + (1)(0.1414v2)2 2 12 2 T2 = 0.1258 v22 Put datum through bar in position 2. V1 = 2(9.81)(0.6 sin 45°) + 4(9.81)(0.3 sin 45°) = 16.6481 J V2 = 0 T1 + V1 = T2 + V2 5.7067 + 16.6481 = 0.1258v22 + 0 Ans. v2 = 13.3 rad>s 835 91962_10_R2_p0827-0866 6/5/09 4:13 PM Page 836 © 2010 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. *R2–11. The operation of a doorbell requires the use of an electromagnet, that attracts the iron clapper AB that is pinned at end A and consists of a 0.2-kg slender rod to which is attached a 0.04-kg steel ball having a radius of 6 mm. If the attractive force of the magnet at C is 0.5 N when the switch is on, determine the initial angular acceleration of the clapper. The spring is originally stretched 20 mm. A k ⫽ 20 N/m 50 mm 40 mm C 44 mm B Equation of Motion: The spring force is given by Fsp = kx = 20(0.02) = 0.4 N. The 1 mass moment of inertia for the clapper AB is (IAB)A = (0.2) A 0.1342 B + 12 2 Applying 0.2 A 0.0672 B + (0.04) A 0.0062 B + 0.04 A 0.142 B = 1.9816 A 10 - 3 B kg # m2 . 5 Eq. 17–12, we have + ©MA = IA a; 0.4(0.05) - 0.5(0.09) = -1.9816 A 10 - 3 B a a = 12.6 rad>s2 Ans. *R2–12. The revolving door consists of four doors which are attached to an axle AB. Each door can be assumed to be a 50-lb thin plate. Friction at the axle contributes a moment of 2 lb # ft which resists the rotation of the doors. If a woman passes through one door by always pushing with a force P = 15 lb perpendicular to the plane of the door as shown, determine the door’s angular velocity after it has rotated 90°. The doors are originally at rest. 3 ft A 2.5 ft u P ⫽ 15 lb 7 ft Moment of inertia of the door about axle AB: IAB = 2c 1 100 a b(6)2 d = 18.6335 slug # ft2 12 32.2 B T1 + © U1-2 = T2 p p 1 0 + b 15(2.5)a b - 2 a b r = (18.6335) v2 2 2 2 v = 2.45 rad>s Ans. 836 91962_10_R2_p0827-0866 6/5/09 4:13 PM Page 837 © 2010 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. R2–13. The 10-lb cylinder rests on the 20-lb dolly. If the system is released from rest, determine the angular velocity of the cylinder in 2 s. The cylinder does not slip on the dolly. Neglect the mass of the wheels on the dolly. 0.5 ft For the cylinder, t2 (+R) m(yCx¿)1 + © Fx¿ dt = m(yCx¿)2 Lt1 30⬚ 10 0 + 10 sin 30°(2) - F(2) = a by 32.2 C (1) t2 (c +) IC v1 + © Lt1 MC dt = IC v2 0 + F(0.5)(2) = c 1 10 a b (0.5)2 dv 2 32.2 (2) For the dolly, t2 (+R) m(yDx¿)1 + © Lt1 Fx¿ dt = m(yDx¿)2 0 + F(2) + 20 sin 30°(2) = a 20 by 32.2 D (3) (+R) vD = vC + vD>C (4) vD = vC - 0.5v Solving Eqs. (1) to (4) yields: Ans. v = 0 vC = 32.2 ft>s vD = 32.2 ft>s F = 0 837 91962_10_R2_p0827-0866 6/5/09 4:13 PM Page 838 © 2010 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. R2–14. Solve Prob. R2–13 if the coefficients of static and kinetic friction between the cylinder and the dolly are ms = 0.3 and m = 0.2, respectively. 0.5 ft For the cylinder, t2 (+R) m(yCx¿)1 + © Lt1 Fx¿ dt = m(yCx¿)2 0 + 10 sin 30°(2) - F(2) = a 10 by 32.2 C (1) 30⬚ t2 (c +) IC v1 + © Lt1 MC dt = IC v2 0 + F(0.5)(2) = c 1 10 a b (0.5)2 dv 2 32.2 (2) For the dolly, t2 (+R) m(yDx¿)1 + © Lt1 Fx¿ dt = m(yDx¿)2 0 + F(2) + 20 sin 30°(2) = a 20 by 32.2 D (3) (+R) vD = vC + vD>C (4) vD = vC - 0.5v Solving Eqs. (1) to (4) yields: Ans. v = 0 vC = 32.2 ft>s vD = 32.2 ft>s F = 0 Note: No friction force develops. 838 91962_10_R2_p0827-0866 6/5/09 4:14 PM Page 839 © 2010 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. R2–15. Gears H and C each have a weight of 0.4 lb and a radius of gyration about their mass center of (kH)B = (kC)A = 2 in. Link AB has a weight of 0.2 lb and a radius of gyration of (kAB)A = 3 in., whereas link DE has a weight of 0.15 lb and a radius of gyration of (kDE)B = 4.5 in. If a couple moment of M = 3 lb # ft is applied to link AB and the assembly is originally at rest, determine the angular velocity of link DE when link AB has rotated 360°. Gear C is prevented from rotating, and motion occurs in the horizontal plane. Also, gear H and link DE rotate together about the same axle at B. E 3 in. H D For link AB, yB = vAB rAB = vAB a 6 b = 0.5vAB 12 For gear H, vDE = yB rB>IC = vAB = 0.5vAB = 2vAB 3>12 1 v 2 DE 1 6 = 0.25vDE yB = a vDE b 2 12 Principle of Work and Energy: For the system, T1 + ©U1-2 = T2 0 + 3(2p) = 2 1 0.2 3 2 1 1 0.4 2 2 ca b a b d a vDE b + c a b a b dv2DE 2 32.2 12 2 2 32.2 12 + 1 0.15 4.5 2 2 1 0.15 1 0.4 a b(0.25vDE)2 + c a ba b dvDE + a b(0.25vDE)2 2 32.2 2 32.2 12 2 32.2 Ans. vDE = 132 rad>s 839 3 in. B M A C 91962_10_R2_p0827-0866 6/5/09 4:14 PM Page 840 © 2010 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. *R2–16. The inner hub of the roller bearing rotates with an angular velocity of vi = 6 rad>s, while the outer hub rotates in the opposite direction at vo = 4 rad>s. Determine the angular velocity of each of the rollers if they roll on the hubs without slipping. 25 mm 50 mm vi ⫽ 6 rad/s Since the hub does not slip, yA = vi ri = 6(0.05) = 0.3 m>s and yB = vO rO = 4(0.1) = 0.4 m>s. vB = vA + vB>A c0. 4 d = c 0. 3 d + B v(0. 05) R T (+ T) c 0.4 = -0.3 + 0.05v T v = 14 rad>sb Ans. 840 vo ⫽ 4 rad/s 91962_10_R2_p0827-0866 6/5/09 4:14 PM Page 841 © 2010 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. R2–17. The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin v = 8 rad>s and its center has a velocity vG = 3 m>s as shown. If the coefficient of kinetic friction between the hoop and the plane is mk = 0.6, determine how long the hoop rolls before it stops slipping. (+a) mvy1 + © L v ⫽ 8 rad/s G vG ⫽ 3 m/s 0.5 m Fy dt = mvy2 30⬚ 0 + Nh (1) - 5(9.81)t cos 30° = 0 Nh = 42.479 N Fh = 0.6Nh = 0.6(42.479 N) = 25.487 N (b+) mvx1 + © L Fx dt = mvx2 5(3) + 5(9.181) sin 30°(t) - 25.487t = 5vG (a +) (HG)1 + © L MG dt = (HG)2 -5(0.5)2(8) + 25.487(0.5)(t) = 5(0.5)2 a vG b 0.5 Solving, vG = 2.75 m>s t = 1.32 s Ans. R2–18. The hoop (thin ring) has a mass of 5 kg and is released down the inclined plane such that it has a backspin v = 8 rad>s and its center has a velocity vG = 3 m>s as shown. If the coefficient of kinetic friction between the hoop and the plane is mk = 0.6, determine the hoop’s angular velocity 1 s after it is released. v ⫽ 8 rad/s G vG ⫽ 3 m/s 0.5 m See solution to Prob. R2–17. Since backspin will not stop in t = 1 s 6 1.32 s, then (+a) mvy1 + © L 30⬚ Fy dt = mvy2 0 + Nh (t) - 5(9.81)t cos 30° = 0 Nh = 42.479 N Fh = 0.6Nh = 0.6(42.479 N) = 25.487 N a+ (HG)1 + © L M dt = (HG)2 -5(0.5)2(8) + 25.487(0.5)(1) = -5(0.5)2 v v = 2.19 rad>s d Ans. 841 91962_10_R2_p0827-0866 6/5/09 4:14 PM Page 842 © 2010 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. R2–19. Determine the angular velocity of rod CD at the instant u = 30°. Rod AB moves to the left at a constant speed of vAB = 5 m>s. D vAB v CD u C x = 0.3 = 0.3 cot u tan u # # x = yAB = -0.3 csc2 uu # Here u = vCD, yAB = -5 m>s and u = 30°. -5 = -0.03 csc2 30°(vCD) vCD = 4.17 rad>s 842 Ans. A 0.3 m B 91962_10_R2_p0827-0866 6/5/09 4:15 PM Page 843 © 2010 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. *R2–20. Determine the angular acceleration of rod CD at the instant u = 30°. Rod AB has zero velocity, i.e., vAB = 0, and an acceleration of aAB = 2 m>s2 to the right when u = 30°. D vAB v CD u C x = $ x = aAB 0.3 = 0.3 cot u tan u # # x = yAB = -0.3 csc2 uu $ # # $ = -0.3 ccsc2 uu - 2 csc2 u cot uu2 d = 0.3 csc2 u a2 cot uu2 - u b # $ Here u = vCD, yAB = 0, aAB = 2 m>s2, u = aCD, and u = 30°. 0 = -0.3 csc2 30°(vCD) 2 = 0.3 csc2 30° C 2 cot 30°(0)2 - aCD D vCD = 0 aCD = -1.67 rad>s2 843 Ans. A 0.3 m B 91962_10_R2_p0827-0866 6/5/09 4:15 PM Page 844 © 2010 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. R2–21. If the angular velocity of the drum is increased uniformly from 6 rad>s when t = 0 to 12 rad>s when t = 5 s, determine the magnitudes of the velocity and acceleration of points A and B on the belt when t = 1 s. At this instant the points are located as shown. 45⬚ B Angular Motion: The angular acceleration of drum must be determined first. Applying Eq. 16–5, we have v = v0 + ac t 12 = 6 + ac (5) ac = 1.20 rad>s2 The angular velocity of the drum at t = 1 s is given by v = v0 + ac t = 6 + 1.20(1) = 7.20 rad>s Motion of P: The magnitude of the velocity of points A and B can be determined using Eq. 16–8. yA = yB = vr = 7.20 a 4 b = 2.40 ft>s 12 Ans. Also, aA = (at)A = ac r = 1.20 a 4 b = 0.400 ft>s2 12 Ans. The tangential and normal components of the acceleration of points B can be determined using Eqs. 16–11 and 16–12, respectively. (at)B = ac r = 1.20a 4 b = 0.400 ft>s2 12 (an)B = v2 r = A 7.202 B a 4 b = 17.28 ft>s2 12 The magnitude of the acceleration of points B is aB = 2(at)2B + (an)2B = 20.4002 + 17.282 = 17.3 ft>s2 844 Ans. A 4 in. 91962_10_R2_p0827-0866 6/5/09 4:15 PM Page 845 © 2010 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. R2–22. Pulley A and the attached drum B have a weight of 20 lb and a radius of gyration of kB = 0.6 ft. If pulley P “rolls” downward on the cord without slipping, determine the speed of the 20-lb crate C at the instant s = 10 ft. Initially, the crate is released from rest when s = 5 ft. For the calculation, neglect the mass of pulley P and the cord. 0.4 ft A B 0.8 ft s 0.2 ft Kinematics: Since pulley A is rotating about a fixed point B and pulley P rolls down without slipping, the velocity of points D and E on the pulley P are given by yD = 0.4vA and yE = 0.8vA where vA is the angular velocity of pulley A. Thus, the instantaneous center of zero velocity can be located using similar triangles. x x + 0.4 = 0.4vA 0.8vA x = 0.4 ft Thus, the velocity of block C is given by yC 0.4vA = 0.6 0.4 yC = 0.6vA Potential Energy: Datumn is set at point B. When block C is at its initial and final position, its locations are 5 ft and 10 ft below the datum. Its initial and final gravitational potential energies are 20( -5) = -100 ft # lb and 20( -10) = -200 ft # lb, respectively. Thus, the initial and final potential energy are V1 = -100 ft # lb V2 = -200 ft # lb Kinetic Energy: The mass moment of inertia of pulley A about point B is 20 IB = mk2B = A 0.62 B = 0.2236 slug # ft2. Since the system is initially at rest, the 32.2 initial kinetic energy is T1 = 0. The final kinetic energy is given by T2 = = 1 1 mC y2C + IB v2A 2 2 1 20 1 a b (0.6vA)2 + (0.2236) v2A 2 32.2 2 = 0.2236v2A Conservation of Energy: Applying Eq. 18–19, we have T1 + V1 = T2 + V2 0 + -100 = 0.2236v2A + (-200) vA = 21.15 rad>s Thus, the speed of block C at the instant s = 10 ft is Ans. yC = 0.6vA = 0.6(21.15) = 12.7 ft>s 845 P C 91962_10_R2_p0827-0866 6/5/09 4:15 PM Page 846 © 2010 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. R2–23. By pressing down with the finger at B, a thin ring having a mass m is given an initial velocity v1 and a backspin v1 when the finger is released. If the coefficient of kinetic friction between the table and the ring is m, determine the distance the ring travels forward before the backspin stops. B v1 v1 r A Equations of Motion: The mass moment of inertia of the ring about its mass center is given by IG = mr2. Applying Eq. 17–16, we have + c ©Fy = m(aG)y ; N - mg = 0 N = mg + ©F = m(a ) ; : x G x mmg = maG aG = mg a + ©MG = IG a; mmgr = mr2 a a = mg r Kinematics: The time required for the ring to stop back spinning can be determined by applying Eq. 16–5. v = v0 + ac t 0 = v1 + a - (c +) t = mg bt r v1 r mg The distance traveled by the ring just before back spinning stops can be determine by applying Eq. 12–5. + B A; s = s0 + y0 t + = 0 + y1 a = 1 a t2 2 c v1 r v1r 2 1 b + (-mg) a b mg 2 mg v1 r (2y1 - v1r) 2mg Ans. 846 91962_10_R2_p0827-0866 6/5/09 4:16 PM Page 847 © 2010 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. *R2–24. The pavement roller is traveling down the incline at v1 = 5 ft>s when the motor is disengaged. Determine the speed of the roller when it has traveled 20 ft down the plane. The body of the roller, excluding the rollers, has a weight of 8000 lb and a center of gravity at G. Each of the two rear rollers weighs 400 lb and has a radius of gyration of kA = 3.3 ft. The front roller has a weight of 800 lb and a radius of gyration of kB = 1.8 ft. The rollers do not slip as they turn. 3.8 ft A G 4.5 ft 5 ft 2.2 ft B 10 ft The wheels roll without slipping, hence v = T1 = yG . r 30⬚ 1 8000 + 800 + 800 1 800 5 2 1 800 5 2 a b (5)2 + c a b(3.3)2 d a b + ca b(1.8)2 d a b 2 32.2 2 32.2 3.8 2 32.2 2.2 = 4168.81 ft # lb T2 = 1 8000 + 800 + 800 2 1 800 y 2 1 800 y 2 a by + c a b(3.3)2 d a b + ca b(1.8)2 d a b 2 32.2 2 32.2 3.8 2 32.2 2.2 = 166.753 y2 Put datum through the mass center of the wheels and body of the roller when it is in the initial position. V1 = 0 V2 = -800(20 sin 30°) - 8000(20 sin 30°) - 800(20 sin 30°) = -96000 ft # lb T1 + V1 = T2 + V2 4168.81 + 0 = 166.753y2 - 96000 y = 24.5 ft>s Ans. R2–25. The cylinder B rolls on the fixed cylinder A without slipping. If bar CD rotates with an angular velocity vCD = 5 rad>s, determine the angular velocity of cylinder B. Point C is a fixed point. 0.3 m D 0.1 m C A vCD ⫽ 5 rad/s vD = 5(0.4) = 2 m>s B 2 vB = = 6.67 rad>s 0.3 Ans. 847 91962_10_R2_p0827-0866 6/5/09 4:16 PM Page 848 © 2010 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. R2–26. The disk has a mass M and a radius R. If a block of mass m is attached to the cord, determine the angular acceleration of the disk when the block is released from rest. Also, what is the distance the block falls from rest in the time t? I0 = c + ©MO = ©tMk)0; mgR = a = R 1 MR2 2 1 MR2 (a) + m(aR)R 2 2mg R(M + 2m) The displacement h = Ru, hence u = Ans. h R u - u0 + v0 t + 1 a t2 2 c 2mg h 1 = 0 + 0 + a bt2 R 2 R(M + 2m) h = mg t2 M + 2m Ans. 848 91962_10_R2_p0827-0866 6/5/09 4:16 PM Page 849 © 2010 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. R2–27. The tub of the mixer has a weight of 70 lb and a radius of gyration kG = 1.3 ft about its center of gravity G. If a constant torque M = 60 lb # ft is applied to the dumping wheel, determine the angular velocity of the tub when it has rotated u = 90°. Originally the tub is at rest when u = 0°. Neglect the mass of the wheel. u 0.8 ft T1 + ©U1 - 2 = T2 G 1 70 1 70 p b(1.3)2 d(v)2 + c d(0.8v)2 0 + 60 a b - 70(0.8) = c a 2 2 32.2 2 32.2 v = 3.89 rad>s M Ans. *R2–28. Solve Prob. R2–27 if the applied torque is M = (50u) lb # ft, where u is in radians. u 0.8 ft T1 + ©U1 - 2 = T2 p>2 0 + L0 50u du - 70(0.8) = 1 70 1 70 ca b (1.3)2 dv2 + c d(0.8v)2 2 32.2 2 32.2 v = 1.50 rad>s Ans. G M 849 91962_10_R2_p0827-0866 6/5/09 4:16 PM Page 850 © 2010 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. R2–29. The spool has a weight of 30 lb and a radius of gyration kO = 0.45 ft. A cord is wrapped around the spool’s inner hub and its end subjected to a horizontal force P = 5 lb. Determine the spool’s angular velocity in 4 s starting from rest. Assume the spool rolls without slipping. 0.9 ft O P ⫽ 5 lb 0.3 ft A t2 (+ b) IA v1 + © Lt1 MA dt = IAv2 0 + 5(0.6)(4) = c a 30 30 b (0.45)2 + a b(0.9)2 dv2 32.2 32.2 v2 = 12.7 rad>s Ans. R2–30. The 75-kg man and 40-kg boy sit on the horizontal seesaw, which has negligible mass. At the instant the man lifts his feet from the ground, determine their accelerations if each sits upright, i.e., they do not rotate. The centers of mass of the man and boy are at Gm and Gb, respectively. a + ©MA = ©(Mk)A ; 2m Gb 40(9.81)(2) - 75(9.81)(1.5) [1] Since the seesaw is rotating about point A, then ab am = 2 1.5 or am = 0.75ab [2] Solving Eqs. (1) and (2) yields: am = 1.45 m>s2 Gm A = -40ab (2) - 75am (1.5) a = 1.5 m ab = 1.94 m>s2 Ans. 850 91962_10_R2_p0827-0866 6/5/09 4:17 PM Page 851 © 2010 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. R2–31. A sphere and cylinder are released from rest on the ramp at t = 0. If each has a mass m and a radius r, determine their angular velocities at time t. Assume no slipping occurs. u Principle of Impulse and Momentum: For the sphere, t2 (+ a) IA v1 + © MA dt = IAv2 Lt1 2 0 + mg sin u(r)(t) = c mr2 + mr2 d(vS)2 5 (vS)2 = 5g sin u t 7r Ans. Principle of Impulse and Momentum: For the cyclinder, t2 (+ a) IA v1 + © Lt1 MA dt = IAv2 1 0 + mg sin u(r)(t) = c mr2 + mr2 d(vC)2 2 2g sin u (vC)2 = t 3r Ans. 851 91962_10_R2_p0827-0866 6/5/09 4:17 PM Page 852 © 2010 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. *R2–32. At a given instant, link AB has an angular acceleration aAB = 12 rad>s2 and an angular velocity vAB = 4 rad>s. Determine the angular velocity and angular acceleration of link CD at this instant. B vAB aAB vC = vB + vC>B 45⬚ A C nC S = C 30°d 10 45°b S + B 2vBC R T + ) (; vC cos 30° = 10 cos 45° + 0 (+ T) vC sin 30° = - 10 sin 45° + 2vBC vBC = 5.58 rad>s vC = 8.16 ft>s vCD = 8.16 = 5.44 rad>s 1.5 Ans. aC = aB + aC>B C 44.44 c60° S + C (aC)t 30°d S = C 30 S + C 45°b 40 45°d S + c2(5.5 8)2 d + B ; 2aBC R T + ) (; -44.44 cos 60° + (aC)t cos 30° = 30 cos 45° + 40 cos 45° + 62.21 (+ T) 44.44 sin 60° + (aC)t sin 30° = -30 sin 45° + 40 sin 45° + 2aBC (aC)t = 155 ft>s2 aCD = 2.5 ft aBC = 54.4 rad>s2 155 = 103 rad>s2 d 1.5 Ans. 852 2 ft C 60⬚ 1.5 ft vC D aCD D 91962_10_R2_p0827-0866 6/5/09 4:18 PM Page 853 © 2010 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. R2–33. At a given instant, link CD has an angular acceleration aCD = 5 rad>s2 and an angular velocity vCD = 2 rad>s. Determine the angular velocity and angular acceleration of link AB at this instant. rIC - C 2 = sin 45° sin 75° rIC - C = 1.464 ft rIC - B 2 = sin 60° sin 75° rIC - B = 1.793 ft B vAB aAB 2.5 ft 45⬚ A 3 = 2.0490 rad>s 1.464 vBC = vB = 2.0490(1.793) = 3.6742 ft>s 3.6742 = 1.47 rad>sd 2.5 vAB = (aB)n = Ans. (3.6742)2 v2B = = 5.4000 ft>s2 rBA 2.5 (aC)n = (3)2 v2C = = 6 ft>s2 rCD 1.5 (aC)t = aCD(rCD) = 5(1.5) = 7.5 ft>s2 aB = aC + a * rB>C - v2 rB>C -5.400 cos 45°i - 5.400 sin 45°j - (aB)t cos 45°i + (aB)t sin 45°j = 6 sin 30°i - 6 cos 30°j -7.5 cos 30°i - 7.5 sin 30°j + (ak) * (-2i) - (2.0490)2(-2i) -3.818 - (aB)t(0.7071) = 3 - 6.495 + 8.3971 -3.818 + (aB)t(0.7071) = -5.1962 - 3.75 - 2a (aB)t = -12.332 ft>s2 a = 1.80 rad>s2 aAB = 12.332 = 4.93 rad>s2 2.5 b Ans. 853 2 ft C 60⬚ 1.5 ft vC D aCD D 91962_10_R2_p0827-0866 6/5/09 4:18 PM Page 854 © 2010 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. R2–34. The spool and the wire wrapped around its core have a mass of 50 kg and a centroidal radius of gyration of kG = 235 mm. If the coefficient of kinetic friction at the surface is mk = 0.15, determine the angular acceleration of the spool after it is released from rest. IG = mk2G = 500(0.235)2 = 2.76125 kg # m2 +b©Fx¿ = m(aG)x¿ ; 50(9.81) sin 45° - T - 0.15NB = 50aG + a©Fy¿ = m(aG)y¿ ; c + ©MG = IG a; G NB - 50(9.81) cos 45° = 0 (1) B (3) The spool does not slip at point A, therefore 45⬚ aG = 0.1a (4) Solving Eqs. (1) to (4) yields: T = 281.5 N 0.4 m (2) T(0.1) - 0.15NB(0.4) = 2.76125a NB = 346.8 N 0.1 m aG = 0.2659 m>s2 a = 2.66 rad>s2b Ans. 854 91962_10_R2_p0827-0866 6/5/09 4:18 PM Page 855 © 2010 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. R2–35. The bar is confined to move along the vertical and inclined planes. If the velocity of the roller at A is vA = 6 ft>s when u = 45°, determine the bar’s angular velocity and the velocity of B at this instant. vA A sB cos 30° = 5 sin u u # sB = 5.774 sin u 5 ft # $ sB = 5.774 cos uu (1) 5 cos u = sA + sB sin 30° $ # # -5 sin u u = sA + sB sin 30° (2) 30⬚ B vB Combine Eqs. (1) and (2): # # -5 sin u u = -6 + 5.774 cos u(u)(sin 30°) # # -3.536u = -6 + 2.041u v = u = 1.08 rad>s Ans. vB = sB = 5.774 cos 45°(1.076) = 4.39 ft>s Ans. From Eq. (1), *R2–36. The bar is confined to move along the vertical and inclined planes. If the roller at A has a constant velocity of vA = 6 ft>s, determine the bar’s angular acceleration and the acceleration of B when u = 45°. vA A See solution to Prob. R2–35. u 5 ft Taking the time derivatives of Eqs. (1) and (2) yields: # $ $ aB = sB = -5.774 sin u(u)2 + 5.774 cos u(u) # $ $ $ -5 cos u u2 - 5 sin u(u) = sA + sB sin 30° 30⬚ B Substitute the data: vB $ aB = -5.774 sin 45°(1.076) + 5.774 cos 45°(u) $ -5 cos 45°(1.076)2 - 5 sin 45°(u) = 0 + aB sin 30° $ aB = -4.726 + 4.083 u $ aB = -8.185 - 7.071 u 2 Solving: # u = -0.310 rad>s2 Ans. aB = -5.99 ft>s2 Ans. 855 91962_10_R2_p0827-0866 6/5/09 4:19 PM Page 856 © 2010 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. R2–37. The uniform girder AB has a mass of 8 Mg. Determine the internal axial force, shear, and bending moment at the center of the girder if a crane gives it an upward acceleration of 3 m>s2. 3 m/s 2 C A Equations of Motion: By considering the entire beam [FBD(a)], we have + c ©Fy = may ; 2T sin 60° - 8000(9.81) = 8000(3) T = 59166.86 N From the FBD(b) (beam segment), a + ©MO = ©(Mk)O ; M + 4000(9.81)(1) -59166.86 sin 60°(2) = -4000(3)(1) M = 51240 N # m = 51.2 kN # m + ©F = m(a ) ; : x G x 59166.86 cos 60° + N = 0 N = -29583.43 N = -29.6 kN + c ©Fy = m(aG)y ; Ans. Ans. 59166.86 sin 60° - 4000(9.81) - V = 4000(3) V = 0 Ans. 856 60⬚ 4m 60⬚ B 91962_10_R2_p0827-0866 6/5/09 4:19 PM Page 857 © 2010 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. R2–38. Each gear has a mass of 2 kg and a radius of gyration about its pinned mass centers A and B of kg = 40 mm. Each link has a mass of 2 kg and a radius of gyration about its pinned ends A and B of kl = 50 mm. If originally the spring is unstretched when the couple moment M = 20 N # m is applied to link AC, determine the angular velocities of the links at the instant link AC rotates u = 45°. Each gear and link is connected together and rotates in the horizontal plane about the fixed pins A and B. 200 mm 50 mm M A C k ⫽ 200 N/m 50 mm B D Consider the system of both gears and the links. The spring stretches s = 2(0.2 sin 45°) = 0.2828 m. T1 + ©U1 - 2 = T2 1 1 p 0 + b 20 a b - (200)(0.2828)2 r = 2 b C (2)(0.05)2 + (2)(0.04)2 D v2 r 4 2 2 v = 30.7 rad>s Ans. Note that work is done by the tangential force between the gears since each move. For the system, though, this force is equal but opposite and the work cancels. R2–39. The 5-lb rod AB supports the 3-lb disk at its end A. If the disk is given an angular velocity vD = 8 rad>s while the rod is held stationary and then released, determine the angular velocity of the rod after the disk has stopped spinning relative to the rod due to frictional resistance at the bearing A. Motion is in the horizontal plane. Neglect friction at the fixed bearing B. 3 ft vD A 0.5 ft Conservation of Momentum: c + ©(HB)1 = ©(HB)2 c 1 3 5 1 a b(0.5)2 d(8) + 0 = c a b(3)2 dv 2 32.2 3 32.2 +c 3 3 1 a b (0.5)2 dv + a b(3v)(3) 2 32.2 32.2 v = 0.0708 rad>s Ans. 857 B 91962_10_R2_p0827-0866 6/5/09 4:19 PM Page 858 © 2010 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. *R2–40. A cord is wrapped around the rim of each 10-lb disk. If disk B is released from rest, determine the angular velocity of disk A in 2 s. Neglect the mass of the cord. A Principle of Impulse and Momentum: The mass moment inertia of disk A about 1 10 point O is IO = a b A 0.52 B = 0.03882 slug # ft2. Applying Eq. 19–14 to disk A 2 32.2 [FBD(a)], we have t2 IO v1 + © ( +) MO dt = IO v2 Lt1 (1) 0 - [T(2)](0.5) = -0.03882vA 1 10 a b A 0.52 B = 2 32.2 0.03882 slug # ft2. Applying Eq. 19–14 to disk B [FBD(b)], we have The mass moment inertia of disk B about its mass center is IG = m A yGy B 1 + © (+ c ) Fy dt = m A yGy B 1 t2 Lt1 0 + T(2) - 10(2) = - a 10 by 32.2 G (2) t2 IG v1 + © (a +) Lt1 MG dt = IG v2 (3) 0 - [T(2)](0.5) = -0.03882vB Kinematics: The speed of point C on disk B is yC = vA rA = 0.5vA. Here, yC>G = vB rC>G = 0.5vB which is directed vertically upward. Applying Eq. 16–15, we have vC = vG + vC>G C 0.5vA S = B yG R + C 0.5vB S T (+ c ) T c (4) -0.5vA = -yG + 0.5vB Solving Eqs. (1), (2), (3), and (4) yields: Ans. vA = 51.5 rad>s vB = 51.52 rad>s yG = 51.52 ft>s T = 2.00 lb 858 0.5 ft O B 0.5 ft 91962_10_R2_p0827-0866 6/5/09 4:20 PM Page 859 © 2010 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. R2–41. A cord is wrapped around the rim of each 10-lb disk. If disk B is released from rest, determine how much time t is required before A attains an angular velocity vA = 5 rad>s. A Principle of Impulse and Momentum: The mass moment inertia of disk A about 1 10 point O is IO = a b A 0.52 B = 0.03882 slug # ft2. Applying Eq. 19–14 to disk A 2 32.2 [FBD(a)], we have t2 IO v1 + © ( +) MO dt = IO v2 Lt1 (1) 0 - [T(t)](0.5) = -0.03882(5) The mass moment inertia of disk B about its mass center is IG = A 0.52 B = 0.03882 slug # ft2. Applying Eq. 19–14 to disk B [FBD(b)], 1 10 a b 2 32.2 we have m A yGy B 1 + © (+ c ) Fy dt = m A yGy B 1 t2 Lt1 0 + T(t) - 10(t) = - a 10 by 32.2 G (2) t2 IG v1 + © (a +) Lt1 MG dt = IG v2 (3) 0 - [T(t)](0.5) = -0.03882vB Kinematics: The speed of point C on disk B is yC = vA rA = 0.5(5) = 2.50 ft>s. Here, yC>G = vB rC>G = 0.5 vB which is directed vertically upward. Applying Eq. 16–15, we have vC = vG + vC>G B 2.50 R = B yG R + C 0.5 vB S T (+ c) T c [4] -2.50 = -yG + 0.5 vB Solving Eqs. (1), (2), (3), and (4) yields: t = 0.194 s vB = 5.00 rad>s yG = 5.00 ft>s Ans. T = 2.00 lb 859 0.5 ft O B 0.5 ft 91962_10_R2_p0827-0866 6/5/09 4:20 PM Page 860 © 2010 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. R2–42. The 15-kg disk is pinned at O and is initially at rest. If a 10-g bullet is fired into the disk with a velocity of 200 m>s, as shown, determine the maximum angle u to which the disk swings. The bullet becomes embedded in the disk. O u 30⬚ 200 m/s a +(HO)1 = (HO)2 1 0.01(200 cos 30°)(0.15) = c C (15)(0.15)2 + 15(0.15)2 d v 2 v = 0.5132 rad>s T1 + V1 = T2 + V2 1 1 c (15)(0.15)2 + 15(0.15)2 d (0.5132)2 + 0 = 0 + 15(9.81)(0.15)(1 - cos u) 2 2 u = 4.45° Ans. Note that the calculation neglects the small mass of the bullet after it becomes embedded in the plate, since its position in the plate is not specified. 860 0.15 m 91962_10_R2_p0827-0866 6/5/09 4:21 PM Page 861 © 2010 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. R2–43. The disk rotates at a constant rate of 4 rad>s as it falls freely so that its center G has an acceleration of 32.2 ft>s2. Determine the accelerations of points A and B on the rim of the disk at the instant shown. A aA = aG + (aA>G)t + (aA>G)n G B ( a A)x R + C (aA)y S = c32.2 d + 0 + B (4)2(1.5) R : + b a: (+ c ) c T v ⫽ 4 rad/s T (aA)x = 0 (aA)y = -32.2 - (4)2 (1.5) = -56.2 ft>s2 = 56.2 ft>s2 T aA = (aA)y = 56.2 ft>s2 T Ans. aB = aG + (aB>G)t + (aB>G)n c( a B)x d + B (aB)y R = c 32.2 d + 0 + c(4)2 ( 1.5) d : + b a: A+cB c ; T (aB)x = -(4)2(1.5) = -24 ft>s2 = 24 ft>s2 ; (aB)y = -32.2 ft>s2 = 32.2 ft>s2 T aB = 2(aB)2x + (aB)2y = 2242 + 32.22 = 40.2 ft>s2 u = tan - 1 ¢ (aB)y (aB)x ≤ = tan - 1 a Ans. 32.2 b = 53.3° d 24 Ans. 861 1.5 ft B 91962_10_R2_p0827-0866 6/5/09 4:21 PM Page 862 © 2010 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. *R2–44. The operation of “reverse” for a three-speed automotive transmission is illustrated schematically in the figure. If the shaft G is turning with an angular velocity of vG = 60 rad>s, determine the angular velocity of the drive shaft H. Each of the gears rotates about a fixed axis. Note that gears A and B, C and D, E and F are in mesh. The radius of each of these gears is reported in the figure. A vG ⫽ 60 rad/s vH G H F C E vC = vB = rA 90 v = (60) = 180 rad>s rB G 30 vE = vD = rC 30 v = (180) = 108 rad>s rD C 50 vH = B D rA ⫽ 90 mm rB ⫽ rC ⫽ 30 mm rD ⫽ 50 mm rE ⫽ 70 mm rF ⫽ 60 mm rE 70 v = (108) = 126 rad>s rF E 60 Ans. R2–45. Shown is the internal gearing of a “spinner” used for drilling wells. With constant angular acceleration, the motor M rotates the shaft S to 100 rev>min in t = 2 s starting from rest. Determine the angular acceleration of the drill-pipe connection D and the number of revolutions it makes during the 2-s startup. D 150 mm 60 mm For shaft S, S v = v0 + ac t 100(2p) = 0 + aS (2) 60 M aS = 5.236 rad>s2 1 u = u0 + v0 t+ ac t2 2 uS = 0 + 0 + 1 (5.236)(2)2 = 10.472 rad 2 For connection D, aD = uD = rS 60 a = (5.236) = 2.09 rad>s2 rD S 150 Ans. rS 60 u = (10.472) = 4.19 rad = 0.667 rev rD S 150 Ans. 862 91962_10_R2_p0827-0866 6/5/09 4:22 PM Page 863 © 2010 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. R2–46. Gear A has a mass of 0.5 kg and a radius of gyration of kA = 40 mm, and gear B has a mass of 0.8 kg and a radius of gyration of kB = 55 mm. The link is pinned at C and has a mass of 0.35 kg. If the link can be treated as a slender rod, determine the angular velocity of the link after the assembly is released from rest when u = 0° and falls to u = 90°. A yD 0.25 vCD = = 5vCD rD>IC 0.05 yE = vA rE>IC = 5vCD (0.1) = 0.5 vCD The velocity of the mass center of gear B is yF = 0.125vCD. The location of the instantaneous center of zero velocity is as shown. Thus, vB = yE 0.5 vCD = = 5vCD rE>(IC)1 0.1 Potential Energy: Datum is set at point C. When gears A, B and link AC are at their initial position (u = 0°), their centers of gravity are located 0.25 m, 0.125 m, and 0.125 m above the datum, respectively. The total gravitational potential energy when they are at these positions is 0.5(9.81)(0.25) + 0.8(9.81)(0.125) + 0.35(9.81)(0.125) = 2.636 N # m. Thus, the initial and final potential energy is V1 = 2.636 N # m V2 = 0 Kinetic Energy: The mass moment of inertia of gears A and B about their mass center is ID = 0.5 A 0.042 B = 0.8 A 10-3 B kg # m2 and IF = 0.8 A 0.0552 B = 2.42 A 10-3 B kg # m2. 1 The mass moment of inertia of link CD about point C is (ICD)C = (0.35) A 0.252 B + 12 0.35 A 0.1252 B = 7.292 A 10 -3 B kg # m2. Since the system is at rest initially, the initial kinetic energy is T1 = 0.The final kinetic energy is given by T2 = = 1 1 1 1 1 m y2 + ID v2A + mB y2F + IFv2B + (ICD)C v2CD 2 A D 2 2 2 2 1 1 1 (0.5)(0.25 vCD)2 + C 0.8 A 10-3 B D (5vCD)2 + (0.8)(0.125 vCD)2 2 2 2 + B u Kinematics: The velocity of the mass center of gear A is yD = 0.25 vCD, and since is rolls without slipping on the fixed circular gear track, the location of the instantaneous center of zero velocity is as shown. Thus, vA = 125 mm 50 mm 1 1 C 2.42 A 10-3 B D (5 vCD)2 + C 7.292 A 10-3 B D A v2CD B 2 2 = 0.06577 v2CD Conservation of Energy: Applying Eq. 18–19, we have T1 + V1 = T2 + V2 0 + 2.636 = 0.06577 v2CD Ans. vCD = 6.33 rad>s 863 75 mm C 125 mm 91962_10_R2_p0827-0866 6/5/09 4:22 PM Page 864 © 2010 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. F⫽6N R2–47. The 15-kg cylinder rotates with an angular velocity of v = 40 rad>s. If a force F = 6 N is applied to bar AB, as shown, determine the time needed to stop the rotation. The coefficient of kinetic friction between AB and the cylinder is mk = 0.4. Neglect the thickness of the bar. 400 mm A For link AB, 500 mm B 150 mm C a + ©MB = 0; 6(0.9) - NE(0.5) = 0 IC = a + ©MC = IC a; v NE = 10.8 N 1 1 mr2 = (15)(0.15)2 = 0.16875 kg # m2 2 2 -0.4(10.8)(0.15) = 0.16875(a) a = -3.84 rad>s2 a +v = v0 + at 0 = 40 + (-3.84) t t = 10.4 s Ans. *R2–48. If link AB rotates at vAB = 6 rad>s, determine the angular velocities of links BC and CD at the instant shown. A 250 mm 30⬚ B Link AB rotates about the fixed point A. Hence, vAB ⫽ 6 rad/s C 300 mm yB = vAB rAB = 6(0.25) = 1.5 m>s 400 mm For link BC, rB>IC = 0.3 cos 30° = 0.2598 m vBC = yB rB>IC = rC>IC = 0.3 cos 60° = 0.15 m 60⬚ 1.5 = 5.77 rad>s 0.2598 Ans. yC = vBC rC>IC = 5.77(0.15) = 0.8660 m>s Link CD rotates about the fixed point D. Hence, yC = vCD rCD 0.8660 = vCD (0.4) vCD = 2.17 rad>s Ans. 864 D 91962_10_R2_p0827-0866 6/5/09 4:23 PM Page 865 © 2010 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. R2–49. If the thin hoop has a weight W and radius r and is thrown onto a rough surface with a velocity vG parallel to the surface, determine the backspin, V , it must be given so that it stops spinning at the same instant that its forward velocity is zero. It is not necessary to know the coefficient of kinetic friction at A for the calculation. v vG Equations of Motion: The mass moment of inertia of the hoop about its mass center W 2 is given by IG = mr2 = r . Applying Eq. 17–16, we have g + c ©Fy = m(aG)y ; + ©F = m (a ) ; : x G x N - W = 0 N = W W a g G aG = mg mW = c + ©MG = IG a; mWr = W 2 r a g a = mg r v = v0 + a t1 0 = v + at1 = mg b t1 r vr mg The time required for the hoop to stop can be determined by applying Eq. 12–4. + B A; y = y0 + a t2 0 = yG + (-mg) t2 t2 = yG mg It is required that t1 = t2. Thus, yG vr = mg mg v = r A Kinematics: The time required for the hoop to stop back spinning can be determined by applying Eq. 16–5. (c +) G yG r Ans. 865 91962_10_R2_p0827-0866 6/5/09 4:23 PM Page 866 © 2010 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. R2–50. The wheel has a mass of 50 kg and a radius of gyration kG = 0.4 m. If it rolls without slipping down the inclined plank, determine the horizontal and vertical components of reaction at A, and the normal reaction at the smooth support B at the instant the wheel is located at the midpoint of the plank. The plank has negligible thickness and has a mass of 20 kg. B G 0.6 m 2m A Equations of Motion: Since the tire rolls down the plane without slipping, then aG = ar = 0.6a.The mass moment of inertia of the tire about its mass center is given by IG = mk2G = 50 A 0.42 B = 8.00 kg # m2. Applying Eq. 17–16 to [FBD(a)], we have a+ ©Fy¿ = m(aG)y¿ ; N - 50(9.81) cos 30° = 0 b+ ©Fx¿ = m(aG)x¿ ; N = 424.79 N 50(9.81) sin 30° - Ff = 50(0.6a) a + ©MG = IG a; Ff(0.6) = 8.00a (1) (2) Solving Eqs. [1] and [2] yields Ff = 75.46 N a = 5.660 rad>s2 Equations of Equilibrium: From FBD(b). a + ©MA = 0; NB (4) - 20(9.81) cos 30°(2) - 424.79(2) = 0 NB = 297.35 N = 297 N + c ©Fy = 0; Ans. Ay + 297.35 cos 30° - 20(9.81) - 424.79 cos 30° - 75.46 sin 30° = 0 Ay = 344 N + ©F = 0; : x Ans. Ax + 424.79 sin 30° - 75.46 cos 30° - 297.35 sin 30° = 0 Ax = 1.63 N Ans. 866 30⬚ 2m