© 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 10–1 y Determine the moment of inertia about the x axis y ϭ bn xn a b x a Solution Differential Element Here x = a y n The area of the differential element parallel b a to the x axis shown shaded in Fig a is dA = (a - x)dy = aa - y n bdy bn n Moment of Inertia Perform the integration, Ix = LA y2dA = = L0 b L0 b y2aa aay2 - a n b a n b yn bdy yn + bdy a a n 3n + = c y3 - a ba by d3 3n + n n b = = b n ab - a bab3 3n + ab3 3(3n + 1) Ans Ans: Ix = 1010 ab3 3(3n + 1) © 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 10–2 y Determine the moment of inertia about the y axis y ϭ bn xn a b x a Solution Differential Element The area of the differential element parallel to the y axis b shown shaded in Fig a is dA = ydx = n xndx a Moment of Inertia Perform the integration, Iy = LA x2dA = L0 a x2 a b n x dxb an a = = = b n+2 dx nx L0 a a b b(xn + 3) ` na a n + a3b n + Ans Ans: Iy = 1011 a3b n + © 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 10–3 y Determine the moment of inertia for the shaded area about the x axis 100 mm 200 mm y ϭ x2 50 Solution Differential Element Here x = 250y2 The area of the differential element parallel to the x axis shown shaded in Fig a is dA = 2x dy = 2250y2dy x Moment of Inertia Perform the integration, Ix = LA y2dA = L0 200 mm = 2250 y2 c 2250y2dy d L0 200 mm = 2250 a y2 b y2dy 200 mm = 457.14(106) mm4 = 457(106) mm4 Ans Ans: Ix = 457 ( 106 ) mm4 1012 © 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 *10–4 y Determine the moment of inertia for the shaded area about the y axis 100 mm 200 mm y ϭ x2 50 Solution Differential Element Here x = 250 y2 The moment of inertia of the differential element parallel to x axis shown in Fig a about y axis is x 1 2 100250 (dy)(2x)3 = x3dy = ( 250y2 ) 3dy = y2dy 12 3 Moment of Inertia Perform the integration, dIy = Iy = L dIy = L0 200 mm 100150 y2dy 100250 a y2 b = = 53.33(106) mm4 = 53.3(106) mm4 200 mm Ans Ans: Iy = 53.3(106) mm4 1013 © 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 10–5 y Determine the moment of inertia for the shaded area about the x axis y ϭ x1/ 1m x 1m Solution Differential Element The area of the differential element parallel to the y axis shown shaded in Fig a is dA = ydx The moment of inertia of this element about the x axis is dIx = dIx′ + dA ∼ y2 = = = y (dx)y3 + ydx a b 12 y dx 1 (x2) dx 3 x2 dx Moment of Inertia Perform the integration = Ix = dIx = L L0 1m 53 x2 = 15 = x2dx 1m m = 0.133 m4 15 Ans Ans: Ix = 0.133 m4 1014 © 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 10–6 y Determine the moment of inertia for the shaded area about the y axis y ϭ x1/ 1m x 1m Solution Differential Element The area of the 1differential element parallel to the y axis shown shaded in Fig a is dA = ydx = x2dx Moment of Inertia Perform the integration, Iy = LA x2dA = L0 1m = x2 = x2 ( x2dx ) 1m m = 0.286 m4 Ans Ans: Iy = 0.286 m4 1015 © 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 10–7 y Determine the moment of inertia for the shaded area about the x axis y2 ϭ Ϫ 0.5x 1m x 2m Solution Differential Element Here x = 2(1 - y2) The area of the differential element parallel to the x axis shown shaded in Fig a is dA = xdy = 2(1 - y2)dy Moment of Inertia Perform the integration, Ix = LA y2dA = L0 = 1m L0 y2[2(1 - y2)dy] 1m (y2 - y4)dy y5 y3 = 2a - b 3 1m 4 = m = 0.267 m4 15 Ans Ans: Ix = 0.267 m4 1016 © 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 *10–8 y Determine the moment of inertia for the shaded area about the y axis y2 ϭ Ϫ 0.5x 1m x 2m Solution Differential Element Here x = 2(1 - y2) The moment of inertia of the differential element parallel to the x axis shown shaded in Fig a about the y axis is ∼2 dIy = dIy' + dAx = = = x (dy)x3 + xdya b 12 x dy 32(1 - y2) dy ( - y6 + 3y4 - 3y2 + 1)dy Moment of Inertia Perform the integration, = Iy = L dIy = = = L0 1m ( -y6 + 3y4 - 3y2 + 1)dy 1m y7 a - + y5 - y3 + yb ` 128 m = 1.22 m4 105 Ans Ans: Iy = 1.22 m4 1017 © 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 10–9 Determine the moment of inertia of the area about the x axis Solve the problem in two ways, using rectangular differential elements: (a) having a thickness dx and (b) having a thickness of dy y y = 2.5 – 0.1x2 2.5 ft ft x SOLUTION (a) Differential Element: The area of the differential element parallel to y axis is dA = ydx The moment of inertia of this element about x axis is ' dIx = dIx¿ + dAy = y 1dx2y + ydxa b 12 = 12.5 - 0.1x223 dx = 1 - 0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx Moment of Inertia: Performing the integration, we have ft Ix = L dIx = 1 -0.001x6 + 0.075x4 - 1.875x2 + 15.6252 dx L-5 ft = ft 0.075 1.875 0.001 ax + x x + 15.625xb ` -5 ft = 23.8 ft4 Ans (b) Differential Element: Here, x = 225 - 10y The area of the differential element parallel to x axis is dA = 2xdy = 225 - 10y dy Moment of Inertia: Applying Eq 10–1 and performing the integration, we have Ix = LA y2dA 2.5 ft = y2 225 - 10ydy L0 = 2c - 2.5 ft y2 2y 3 125 - 10y22 125 - 10y22 125 - 10y22 d ` 15 375 13125 = 23.8 ft4 Ans Ans: Ix = 23.8 ft 1018 © 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 10–10 Determine the moment of inertia for the shaded area about the x axis y y2 h2 —x b h SOLUTION d Ix = Ix = x b y dx L d Ix = b b y h2 3>2 3>2 dx = a b x dx L0 b L0 = h2 3>2 5>2 b a b a b x ]0 b = bh3 15 Ans Also, dA = (b - x) dy = (b Ix = L y2 dA h = L0 b y ) dy h2 y2 (b - b y ) dy h2 b b h = c y3 y d 5h2 = bh3 15 Ans Ans: Ix = 1019 bh3 15 © 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 10–95 The slender rods have a mass of kg>m Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A A 200 mm Solution 100 mm 100 mm Mass Moment of Inertia About An Axis Through A The mass of each segment is mi = (4 kg>m)(0.2 m) = 0.8 kg The mass moment inertia of each segment shown in Fig a about an axis through their center of mass can be determined using (IG)i = m l i2 12 i IA = Σ c (IG)i + mi d 2i d = c 1 (0.8) ( 0.22 ) + 0.8 ( 0.12 ) d + c (0.8) ( 0.22 ) + 0.8 ( 0.22 ) d 12 12 = 0.04533 kg # m2 = 0.0453 kg # m2 Ans Ans: IA = 0.0453 kg # m2 1108 © 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 *10–96 The pendulum consists of a 8-kg circular disk A, a 2-kg circular disk B, and a 4-kg slender rod Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O A O 0.4 m 1m 0.5 m B 0.2 m Solution Mass Moment of Inertia About An Axis Through O The mass moment of inertia of each rod segment and disk segment shown in Fig a about an axis passes through 1 their center of mass can be determined using (IG)i = m l i2 and (IG)i = mi r i2 12 i IO = Σ c (IG)i + mi d 2i d = c 1 (4) ( 1.52 ) + ( 0.252 ) d + c (2) ( 0.12 ) + ( 0.62 ) d 12 = 13.41 kg # m2 + c (8) ( 0.22 ) + ( 1.22 ) d The total mass is kg + kg + kg = 14 kg The radius of gyration is kO = 13.41 kg # m2 IO = = 0.9787 m = 0.979 m Am C 14 kg Ans Ans: kO = 0.979 m 1109 © 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 10–97 Determine the moment of inertia Iz of the frustum of the cone which has a conical depression The material has a density of 200 kg>m3 0.2 m z 0.8 m 0.6 m SOLUTION Iz = [ p (0.4)2(1.6)(200)](0.4)2 10 - [ p (0.2)2(0.8)(200)](0.2)2 10 - [ p (0.4)2(0.6)(200)](0.4)2 10 0.4 m Iz = 1.53 kg # m2 Ans Ans: Iz = 1.53 kg # m2 1110 © 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 10–98 The pendulum consists of the 3-kg slender rod and the 5-kg thin plate Determine the location y of the center of mass G of the pendulum; then find the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through G O y SOLUTION y = G ©ym 1(3) + 2.25(5) = = 1.781 m = 1.78 m ©m + 0.5 m Ans 1m IG = ©IG¿ + md2 = 2m 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12 = 4.45 kg # m2 Ans Ans: y = 1.78 m IG = 4.45 kg # m2 1111 © 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 10–99 Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O The material has a mass per unit area of 20 kg>m2 O 50 mm 150 mm 50 mm SOLUTION 150 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) This segment should be considered as a negative part 400 mm 150 mm 150 mm 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 IC = 1 (3.2)(0.4 + 0.4 2) - c (0.05p)(0.052) + 0.05p(0.152) d 12 = 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 Ans: IO = 0.276 kg # m2 1112 © 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 *10–100 The pendulum consists of a plate having a weight of 12 lb and a slender rod having a weight of lb Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O O ft ft ft ft SOLUTION IO = ©IG + md2 = 4 12 12 a b (5)2 + a b (0.5)2 + a b (12 + 12) + a b(3.5)2 12 32.2 32.2 12 32.2 32.2 = 4.917 slug # ft2 m = a kO = 12 b + a b = 0.4969 slug 32.2 32.2 IO 4.917 = = 3.15 ft Am A 0.4969 Ans Ans: kO = 3.15 ft 1113 © 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 10–101 If the large ring, small ring and each of the spokes weigh 100 lb, 15 lb, and 20 lb, respectively, determine the mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point A ft ft O SOLUTION Composite Parts: The wheel can be subdivided into the segments shown in Fig a The spokes which have a length of (4 - 1) = 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) A 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 = ¢ 100 20 20 15 ≤ (42) + B ¢ ≤ (32) + ¢ ≤ (2.52) R + ¢ ≤ (12) 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 IA = IO + md2, 100 15 20 where m = + 8¢ = 8.5404 slug and d = ft Thus, ≤ + 32.2 32.2 32.2 IA = 84.94 + 8.5404(42) = 221.58 slug # ft2 = 222 slug # ft2 Ans Ans: IA = 222 slug # ft 1114 © 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 10–102 z Determine the mass moment of inertia of the assembly about the z axis The density of the material is 7.85 Mg> m3 100 mm SOLUTION Composite Parts: The assembly can be subdivided into two circular cone segments (1) and (3) and a hemispherical segment (2) as shown in Fig a Since segment (3) is a hole, it should be considered as a negative part From the similar triangles, we obtain z 0.1 = 0.45 + z 0.3 450 mm 300 mm z = 0.225m Mass: The mass of each segment is calculated as 300 mm 1 m1 = rV1 = r a pr2h b = 7.85(103) c p(0.32)(0.675) d = 158.9625p kg 3 x y 2 m2 = rV2 = r a pr3 b = 7.85(103) c p(0.33) d = 141.3p kg 3 1 m3 = rV3 = r a pr2h b = 7.85(103) c p(0.12)(0.225) d = 5.8875p kg 3 Mass Moment of Inertia: Since the z axis is parallel to the axis of the cone and the hemisphere and passes through their center of mass, the mass moment of inertia can be 3 computed from (Iz)1 = m r12, (Iz)2 = m2r22, and m r32 Thus, 10 10 Iz = ©(Iz)i = 3 (158.9625p)(0.32) + (141.3p)(0.32) (5.8875p)(0.12) 10 10 = 29.4 kg # m2 Ans Ans: Iz = 29.4 kg # m2 1115 © 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 10–103 Each of the three slender rods has a mass m Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center point O a a O SOLUTION IO = B a sin 60° ma2 + m ¢ ≤ R = ma2 12 Ans a Ans: IO = 1116 ma2 © 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 *10–104 The thin plate has a mass per unit area of 10 kg>m2 Determine its mass moment of inertia about the y axis z 200 mm 100 mm 200 mm 200 mm SOLUTION 100 mm Composite Parts: The thin plate can be subdivided into segments as shown in Fig a Since the segments labeled (2) are both holes, the y should be considered as negative parts 200 mm 200 mm x 200 mm 200 mm 200 mm Mass moment of Inertia: The mass of segments (1) and (2) are m1 = 0.4(0.4)(10) = 1.6 kg and m2 = p(0.12)(10) = 0.1p kg The perpendicular distances measured from the centroid of each segment to the y axis are indicated in Fig a The mass moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem Iy = © A Iy B G + md2 = 2c 1 (1.6)(0.42) + 1.6(0.22) d - c (0.1p)(0.12) + 0.1p(0.22) d 12 = 0.144 kg # m2 Ans Ans: Iy = 0.144 kg # m2 1117 y © 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 10–105 The thin plate has a mass per unit area of 10 kg>m2 Determine its mass moment of inertia about the z axis z 200 mm 100 mm 200 mm 200 mm SOLUTION 100 mm Composite Parts: The thin plate can be subdivided into four segments as shown in Fig a Since segments (3) and (4) are both holes, the y should be considered as negative parts 200 mm 200 mm x 200 mm 200 mm 200 mm Mass moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) are m1 = m2 = 0.4(0.4)(10) = 1.6 kg and m3 = m4 = p(0.12)(10) = 0.1p kg The mass moment of inertia of each segment about the z axis can be determined using the parallel-axis theorem Iz = © A Iz B G + md2 = 1 1 (1.6)(0.42) + c (1.6)(0.42 + 0.42) + 1.6(0.22) d - (0.1p)(0.12) - c (0.1p)(0.12) + 0.1p(0.22) d 12 12 = 0.113 kg # m2 Ans Ans: Iz = 0.113 kg # m2 1118 y © 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 10–106 Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center of mass G The material has a specific weight of g = 90 lb>ft3 O ft G ft 0.25 ft SOLUTION IG = 0.5 ft ft 90 90 ca b p(2.5)2(1) d (2.5)2 - c a b p(2)2(1) d(2)2 32.2 32.2 + 90 90 ca b p(2)2(0.25) d (2)2 - c a b p(1)2(0.25) d(1)2 32.2 32.2 = 118 slug # ft2 Ans Ans: IG = 118 slug # ft 1119 © 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 10–107 Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through point O The material has a specific weight of g = 90 lb>ft3 O ft G ft 0.25 ft SOLUTION IG = 0.5 ft ft 90 90 ca b p(2.5)2(1) d (2.5)2 - c a b p(2)2(1) d(2)2 32.2 32.2 + 90 90 ca b p(2)2(0.25) d (2)2 - c a b p(1)2(0.25) d(1)2 32.2 32.2 = 117.72 slug # ft2 IO = IG + md2 m = a 90 90 b p(2 - 12)(0.25) + a b p(2.52 - 2)(1) = 26.343 slug 32.2 32.2 IO = 117.72 + 26.343(2.5)2 = 282 slug # ft2 Ans Ans: IO = 282 slug # ft 1120 © 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 *10–108 The pendulum consists of two slender rods AB and OC which have a mass of kg>m The thin plate has a mass of 12 kg>m2 Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G 0.4 m 0.4 m A B O _ y G 1.5 m SOLUTION y = 1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8) 0.1 m Ans = 0.8878 m = 0.888 m IG = C 1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3) 0.3 m (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 + (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 - [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 IG = 5.61 kg # m2 Ans Ans: y = 0.888 m IG = 5.61 kg # m2 1121 © 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 10–109 Determine the moment of inertia Iz of the frustrum of the cone which has a conical depression The material has a density of 200 kg>m3 z 200 mm 600 mm SOLUTION 400 mm z + z = , Mass Moment of Inertia About z Axis: From similar triangles, 0.2 0.8 z = 0.333 m The mass moment of inertia of each cone about z axis can be determined using Iz = mr2 10 Iz = ©1Iz2i = 800 mm p c 10.82211.333212002 d10.822 10 - p c 10.2 2210.333212002 d10.2 22 10 - p c 10.2 2210.6212002 d10.2 22 10 = 34.2 kg # m2 Ans Ans: Iz = 34.2 kg # m2 1122 ... 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. .. 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... 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