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Chapter 17

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R.C. Hibbeler Engineering Mechanics Dynamics solution manual

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Chapter 17

1. 1. 91962_07_s17_p0641-0724 6/8/09 3:31 PM Page 641 © 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. •17–1. Determine the moment of inertia Iy for the slender z rod. The rod’s density r and cross-sectional area A are constant. Express the result in terms of the rod’s total mass m. y l Iy = x2 dm LM l A 2 x = x (r A dx) L0 1 = r A l3 3 m = rAl Thus, 1 Iy = m l2 Ans. 3 17–2. The right circular cone is formed by revolving the y shaded area around the x axis. Determine the moment of r inertia Ix and express the result in terms of the total mass m y ϭ –x h of the cone. The cone has a constant density r. r x 2 dm = r dV = r(p y dx) r(p) ¢ 2≤ x dx = rp ¢ 2 ≤ a bh3 = rp r2h h r2 2 r2 1 1 m = L0 h h 3 3 h 1 2 dIx = y dm 2 1 2 = y (rp y2 dx) 2 r4 r(p)a 4 b x4 dx 1 = 2 h h r4 r(p)a 4 bx4 dx = 1 1 Ix = rp r4 h L 2 0 h 10 Thus, 3 Ix = m r2 Ans. 10 641
2. 2. 91962_07_s17_p0641-0724 6/8/09 3:32 PM Page 642 © 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. 17–3. The paraboloid is formed by revolving the shaded y area around the x axis. Determine the radius of gyration kx. y2 ϭ 50x The density of the material is r = 5 Mg>m3. 100 mm x dm = r p y2 dx = r p (50x) dx 200 1 2 1 Ix = y dm = 50 x {p r (50x)} dx L 2 2L0 200 mm 502 1 3 200 = r pa bc x d 2 3 0 502 = rp a b (200)3 6 200 m = dm = p r (50x) dx L L0 200 = r p (50)c x2 d 1 2 0 = rp a b (200)2 50 2 Am A3 Ix 50 kx = = (200) = 57.7 mm Ans. *17–4. The frustum is formed by rotating the shaded area y around the x axis. Determine the moment of inertia Ix and b y ϭ –x ϩ b a express the result in terms of the total mass m of the frustum. The frustum has a constant density r. 2b b x z dm = r dV = rpy2 dx = rp A x + b2 B dx b2 2 2b2 2 x + a a a 1 1 dIx = dmy2 = rpy4 dx 2 2 rp A 4 x4 + 3 x3 + 2 x2 + x + b4 B dx 1 b4 4 b4 6 b4 4b4 dIx = 2 a a a a A 4 x4 + 3 x3 + 2 x2 + x + b4 B dx a 1 b4 4b4 6 b4 4 b4 Ix = dIx = rp L 2 L a0 a a a 31 = rpab4 10 A 2 x2 + x + b2 B dx = rpab2 a b2 2b2 7 m = dm = rp Lm L0 a a 3 93 2 Ix = mb Ans. 70 642
3. 3. 91962_07_s17_p0641-0724 6/8/09 3:32 PM Page 643 © 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. •17–5. The paraboloid is formed by revolving the shaded y 2 area around the x axis. Determine the moment of inertia y2 = a x – h about the x axis and express the result in terms of the total mass m of the paraboloid. The material has a constant a density r. x dm = r dV = r (p y2 dx) 1 1 h d Ix = dm y2 = r p y4 dx 2 2 h a4 r pa 2 bx2 dx 1 Ix = L 2 0 h 1 = p ra4 h 6 h a2 r pa bx dx 1 m = L 2 0 h 1 = r p a2 h 2 1 Ix = ma2 Ans. 3 17–6. The hemisphere is formed by rotating the shaded y area around the y axis. Determine the moment of inertia Iy and express the result in terms of the total mass m of the hemisphere. The material has a constant density r. x2 ϩ y2 ϭ r2 r r x m = r dV = r p x2 dy = rp (r2 - y2)dy LV L0 L0 r y d = rp r3 1 3 2 = rpc r2 y - 3 0 3 r r 1 r rp Iy = (dm) x2 = px4 dy = (r2 - y2)2 dy L 2 m 2 L0 2 L0 y5 r c r y - r2 y3 + d = rp 4 2 4rp 5 = r 2 3 5 0 15 Thus, 2 Iy = m r2 Ans. 5 643
4. 4. 91962_07_s17_p0641-0724 6/8/09 3:32 PM Page 644 © 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. 17–7. Determine the moment of inertia of the z homogeneous pyramid of mass m about the z axis. The density of the material is r. Suggestion: Use a rectangular plate element having a volume of dV = (2x)(2y)dz. h a – 2 y C (2y)2 + (2y)2 D = y2 dm Iz = 10 a dm 2 m 2 a – dIz = Ans. 2 12 3 a – a – dm = 4ry2 dz 2 2 a4 ry dz = r(h - z)4 a b dz 8 4 8 dIz = x 3 3 16h4 h r a4 r a4 a 4b (h4 - 4h3z + 6h2z2 - 4hz3 + z4)dz = a 4 b ch5 - 2h5 + 2h5 - h5 + h5 d 1 Iz = 6 h L 0 6 h 5 ra4 h = 30 h h a2 ra2 4r(h - z)2 a b dz = 2 2 m = (h - 2hz + z2)dz L h L 2 0 4h 0 ra2 ch3 - h3 + h d 1 3 = h2 3 ra2h = 3 Thus, 644
5. 5. 91962_07_s17_p0641-0724 6/8/09 3:32 PM Page 645 © 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. *17–8. Determine the mass moment of inertia Iz of the z cone formed by revolving the shaded area around the z axis. h z ϭ –– (r0 Ϫ y) The density of the material is r. Express the result in terms r0 of the mass m of the cone. h y dm = r dV = rpr2dz. Here, r = y = ro - z. Thus, dm = rp ¢ ro - z ≤ dz. The Differential Element: The mass of the disk element shown shaded in Fig. a is ro ro 2 x r0 h h mass moment of inertia of this element about the z axis is dmr2 = (rpr2dz)r2 = rpr4dz = rp ¢ ro - z ≤ dz 1 1 1 1 ro 4 dIz = 2 2 2 2 h Mass: The mass of the cone can be determined by integrating dm. Thus, rp ¢ ro - z ≤ dz h ro 2 m = dm = L L0 h = rpC aro - zb ¢ - ≤ S 3 h 1 ro 3 h 1 = rpro 2h 3 h ro 3 0 Mass Moment of Inertia: Integrating dIz, we obtain rp ¢ ro - z ≤ dz h 1 ro 4 Iz = dIz = L L 2 0 h = rpC aro - zb ¢ - ≤ S 3 h 1 1 ro 3 h 1 = rpro 4 h 2 5 h ro 10 0 From the result of the mass, we obtain rpro 2h = 3m. Thus, Iz can be written as A rpro 2h B ro 2 = (3m)ro 2 = mro 2 1 1 3 Iz = Ans. 10 10 10 645
6. 6. 91962_07_s17_p0641-0724 6/8/09 3:33 PM Page 646 © 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. •17–9. Determine the mass moment of inertia Iy of the z solid formed by revolving the shaded area around the y axis. The density of the material is r. Express the result in terms of the mass m of the solid. z ϭ 1 y2 4 1m y x Differential Element: The mass of the disk element shown shaded in Fig. a is 2m 2 dm = r dV = rpr2dy. Here, r = z = y2. Thus, dm = rpa y 2 b dy = 1 1 rp 4 y dy. 4 4 16 The mass moment of inertia of this element about the y axis is 4 dmr2 = (rpr2dy)r2 = rpr4dy = rp a y 2 b dy = 1 1 1 1 1 rp 8 dIy = y dy 2 2 2 2 4 512 Mass: The mass of the solid can be determined by integrating dm. Thus, ¢ ≤` 2m rp 4 rp y5 2 m 2 m = dm = y dy = = rp L L0 16 16 5 0 5 Mass Moment of Inertia: Integrating dIy, we obtain 2m rp 8 Iy = dIy = y dy L L 572 ¢ ≤` 0 9 2m rp y pr = = 512 9 0 9 5m From the result of the mass, we obtain pr = . Thus, Iy can be written as 2 a b = 1 5m 5 Iy = m Ans. 9 2 18 646
7. 7. 91962_07_s17_p0641-0724 6/8/09 3:33 PM Page 647 © 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. 17–10. Determine the mass moment of inertia Iy of the z solid formed by revolving the shaded area around the y axis. The density of the material is r. Express the result in a terms of the mass m of the semi-ellipsoid. y 2 z2 –– ϩ –– ϭ 1 2 b a b2 C C Differential Element: The mass of the disk element shown shaded in Fig. a is y . Thus, dm = rp £ b ≥ dz 2 2 2 y y dm = r dV = rpr2dy. Here, r = z = b 1 - 1 - a2 a2 = rpb2 ¢ 1 - 2≤ y2 x C dy. The mass moment of inertia of this element about the y axis is a dIy = dmr2 = (rpr2dy)r2 = rpr4dy = rp £ b 1 - 2 ≥ dy 4 1 1 1 1 y2 2 2 2 2 a rpb4 ¢ 1 - 2 ≤ dy = rpb4 ¢ 1 + 4 - 2 ≤ dy 1 y2 2 1 y4 2y2 = 2 a 2 a a Mass: The mass of the semi-ellipsoid can be determined by integrating dm. Thus, rpb2 ¢ 1 - 2 = 2 r p ab2 2≤ dy = rpb2 ¢ y - 2≤ a y2 y3 a m = dm = L L0 a 3a 0 3 Mass Moment of Inertia: Integrating dIy, we obtain rpb4 ¢ H 4 - 2 ≤ dy a 1 y4 2y2 Iy = dIy = L L 2 2 = 4 r p ab4 a a rpb4 ¢ y + 2≤ 0 1 y5 2y3 a = 4 - 2 5a 3a 0 15 3m From the result of the mass, we obtain rpab2 = . Thus, Iy can be written as 2 A rpab2 B b2 = a bb2 = mb2 4 4 3m 2 Iy = Ans. 15 15 2 5 17–11. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes O through the center of mass G. The material has a specific weight of g = 90 lb>ft3. 1 ft G ca b p(2.5)2(1) d(2.5)2 - c a bp(2)2(1) d(2)2 1 90 1 90 2 ft IG = 2 32.2 2 32.2 0.25 ft + ca b p(2)2(0.25) d(2)2 - c a bp(1)2(0.25) d(1)2 1 90 1 90 0.5 ft 2 32.2 2 32.2 1 ft = 118 slug # ft2 Ans. 647
8. 8. 91962_07_s17_p0641-0724 6/8/09 3:33 PM Page 648 © 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. *17–12. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes O through point O. The material has a specific weight of g = 90 lb>ft3. 1 ft G 2 ft 0.25 ft ca bp(2.5)2(1) d(2.5)2 - c a bp(2)2(1) d(2)2 1 90 1 90 IG = 0.5 ft 2 32.2 2 32.2 1 ft ca bp(2)2(0.25) d(2)2 - c a bp(1)2(0.25) d(1)2 1 90 1 90 + 2 32.2 2 32.2 = 117.72 slug # ft2 IO = IG + md2 m = a bp(2 2 - 12)(0.25) + a bp(2.52 - 2 2)(1) = 26.343 slug 90 90 32.2 32.2 IO = 117.72 + 26.343(2.5)2 = 282 slug # ft2 Ans. •17–13. If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the 4 ft mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A. 1 ft O Composite Parts: The wheel can be subdivided into the segments shown in Fig. a. The spokes which have a length of (4 - 1) = 3 ft and a center of mass located at a distance of a1 + b ft = 2.5 ft from point O can be grouped as segment (2). 3 A 2 Mass Moment of Inertia: First, we will compute the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O. IO = a b (4 2) + 8 c a b (32) + a b(2.52) d + a b(12) 100 1 20 20 15 32.2 12 32.2 32.2 32.2 = 84.94 slug # ft2 The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A can be found using the parallel-axis theorem b + 100 20 15 IA = IO + md2, where m = + 8a = 8.5404 slug and d = 4 ft. 32.2 32.2 32.2 Thus, IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2 Ans. 648
10. 10. 91962_07_s17_p0641-0724 6/8/09 3:34 PM Page 650 © 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. 17–18. Determine the moment of inertia of the center 0.5 in. 1 in. 0.5 in. 1 in. crank about the x axis. The material is steel having a specific weight of gst = 490 lb>ft3. 1 in. 0.5 in. 490 p (0.25)2(1) ms = a b = 0.0017291 slug 4 in. 32.2 (12)3 0.5 in. 0.5 in. a b = 0.02642 slug 490 (6)(1)(0.5) mp = x 32.2 (12)3 1 in. Ix = 2 c (0.02642) A (1)2 + (6)2 B + (0.02642)(2)2 d 1 1 in. 1 in. 12 + 2 c (0.0017291)(0.25)2 d + (0.0017291)(0.25)2 + (0.0017291)(4)2 d 1 1 2 2 = 0.402 slug # in2 Ans. 17–19. Determine the moment of inertia of the overhung 20 mm crank about the x axis. The material is steel for which the 30 mm density is r = 7.85 Mg>m3. 90 mm 50 mm mc = 7.85 A 103 B A (0.05)p(0.01)2 B = 0.1233 kg x 180 mm 20 mm mp = 7.85 A 103 B ((0.03)(0.180)(0.02)) = 0.8478 kg x¿ 30 mm Ix = 2 c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 1 20 mm 30 mm 2 50 mm + c (0.8478) A (0.03)2 + (0.180)2 B d 1 12 = 0.00325 kg # m2 = 3.25 g # m2 Ans. *17–20. Determine the moment of inertia of the overhung 20 mm crank about the x¿ axis. The material is steel for which the 30 mm density is r = 7.85 Mg>m3. 90 mm 50 mm mc = 7.85 A 10 3 B A (0.05)p(0.01) B = 0.1233 kg 2 x 180 mm 20 mm mp = 7.85 A 103 B ((0.03)(0.180)(0.02)) = 0.8478 kg x¿ 30 mm Ix = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 1 1 20 mm 30 mm 50 mm 2 2 + c (0.8478) A (0.03)2 + (0.180)2 B + (0.8478)(0.06)2 d 1 12 = 0.00719 kg # m2 = 7.19 g # m2 Ans. 650
11. 11. 91962_07_s17_p0641-0724 6/8/09 3:34 PM Page 651 © 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. •17–21. Determine the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through point O. The slender rod has a mass of 10 kg O and the sphere has a mass of 15 kg. 450 mm A Composite Parts: The pendulum can be subdivided into two segments as shown in Fig. a. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated. 100 mm B Moment of Inertia: The moment of inertia of the slender rod segment (1) and the sphere segment (2) about the axis passing through their center of mass can be 1 2 computed from (IG)1 = ml2 and (IG)2 = mr2. The mass moment of inertia of 12 5 each segment about an axis passing through point O can be determined using the parallel-axis theorem. IO = ©IG + md2 = c (10)(0.452) + 10(0.2252) d + c (15)(0.12) + 15(0.552) d 1 2 12 5 = 5.27 kg # m2 Ans. 651
12. 12. 91962_07_s17_p0641-0724 6/8/09 3:35 PM Page 652 © 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. 17–22. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing O through point O. The material has a mass per unit area of 20 kg>m2. 50 mm 150 mm 50 mm 150 mm 400 mm 400 mm Composite Parts: The plate can be subdivided into the segments shown in Fig. a. Here, the four similar holes of which the perpendicular distances measured from their centers of mass to point C are the same and can be grouped as segment (2). 150 mm 150 mm This segment should be considered as a negative part. Mass Moment of Inertia: The mass of segments (1) and (2) are m1 = (0.4)(0.4)(20) = 3.2 kg and m2 = p(0.052)(20) = 0.05p kg, respectively. The mass moment of inertia of the plate about an axis perpendicular to the page and passing through point C is (3.2)(0.4 2 + 0.4 2) - 4 c (0.05p)(0.052) + 0.05p(0.152) d 1 1 IC = 12 2 = 0.07041 kg # m2 The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O can be determined using the parallel-axis theorem IO = IC + md2, where m = m1 - m2 = 3.2 - 4(0.05p) = 2.5717 kg and d = 0.4 sin 45°m. Thus, IO = 0.07041 + 2.5717(0.4 sin 45°)2 = 0.276 kg # m2 Ans. 652
13. 13. 91962_07_s17_p0641-0724 6/8/09 3:35 PM Page 653 © 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. 17–23. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing O 200 mm through point O. The material has a mass per unit area of 200 mm 20 kg>m2. 200 mm Composite Parts: The plate can be subdivided into two segments as shown in Fig. a. Since segment (2) is a hole, it should be considered as a negative part. The perpendicular distances measured from the center of mass of each segment to the point O are also indicated. Mass Moment of Inertia: The moment of inertia of segments (1) and (2) are computed as m1 = p(0.2 2)(20) = 0.8p kg and m2 = (0.2)(0.2)(20) = 0.8 kg. The moment of inertia of the plate about an axis perpendicular to the page and passing through point O for each segment can be determined using the parallel-axis theorem. IO = ©IG + md2 = c (0.8p)(0.22) + 0.8p(0.22) d - c (0.8)(0.22 + 0.22) + 0.8(0.22) d 1 1 2 12 = 0.113 kg # m2 Ans. 653
14. 14. 91962_07_s17_p0641-0724 6/8/09 3:35 PM Page 654 © 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. *17–24. The 4-Mg uniform canister contains nuclear waste a material encased in concrete. If the mass of the spreader beam BD is 50 kg, determine the force in each of the links F H AB, CD, EF, and GH when the system is lifted with an 30Њ 30Њ acceleration of a = 2 m>s2 for a short period of time. B D E G 0.3 m 0.4 m 0.3 m Canister: A C + c ©Fy = m(aG)y ; 2T - 4 A 10 B (9.81) = 4 A 10 B (2) 3 3 TAB = TCD = T = 23.6 kN Ans. System: + c ©Fy = m(aG)y ; 2T¿ cos 30° - 4050(9.81) = 4050(2) TEF = TGH = T¿ = 27.6 kN Ans. 654