Explanation Given: The width of the shaded area is a . The height of the shaded area is 2 h . The density of the material is ρ . Show the intersection of solid formed by revolving the shaded area at the arbitrary point ( x , y ) as in Figure 1. Conclusion: From Figure 1, Calculate the differential mass element. d m = ρ d V (I) Here, the density of material is ρ , and the volume of the paraboloid is V . Substitute π y 2 d x for d V in Equation (I). d m = ρ π y 2 d x (II) Here, the centroid of the differential element along the y - axis is y 2 . Calculate the centroid of the differential element along the y - axis. y n = h n a x y = h a 1 n x 1 n Substitute h a 1 n x 1 n for y in Equation (II). d m = ρ π [ h a 1 n x 1 n ] 2 d x = ρ π h 2 a 2 n x 2 n d x Compute the moment of inertia of the differential element that is parallel to the y -axis and about the x -axis. d I x = 1 2 d m y 2 (III) Substitute ρ π h 2 a 2 n x 2 n d x for d m and h a 1 n x 1 n for y in Equation (III). d I x = 1 2 [ ρ π h 2 a 2 n x 2 n d x ] [ h a 1 n x 1 n ] 2 = ρ π h 4 2 a 4 n x 4 n d x Calculate the entire mass of the solid formed by revolving the shaded area. m = ∫ 0 a d m (IV) Substitute ρ π h 2 a 2 n x 2 n d x for d m in Equation (IV). m = ∫ 0 a ρ π h 2 2 a 2 n x 2 n d x = ρ π h 2 a 2 n ∫ 0 a x 2 n d x = ρ π h 2 a 2 n ∫ 0 a x 2 n + 1 2 n + 1 d x = ρ π h 2 a 2 n [ n x 2 + n n 2 + n ] | 0 a = ρ π h 2 a 2 n [ n a 2 + n n 2 + n ] | = [ n n + 2 ] ρ π a h 2 Express the moment of inertia of the solid that is formed by revolving the shaded area about the x -axis

Engineering Mechanics: Statics and Modified Mastering Engineering with eText and Access Card (14th Edition)

14th Edition

ISBN: 9780134229287

Author: Russell C. Hibbeler

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 10.8, Problem 90P

To determine

The radius of gyration of the solid that is formed by revolving the shaded area about the x -axis.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 10.8, Problem 89P

Chapter 10.8, Problem 91P

Students have asked these similar questions

I do not understand how to approach this question. I tried to answer it but I kept getting it incorrect.

(read image)

Chapter 10 Solutions

Engineering Mechanics: Statics and Modified Mastering Engineering with eText and Access Card (14th Edition)

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

The radius of gyration of the solid that is formed by revolving the shaded area about the x -axis.

Solution Summary: The author calculates the radius of gyration of the solid formed by revolving the shaded area about the x -axis.

Videos

The radius of gyration of the solid that is formed by revolving the shaded area about the x -axis.

Want to see the full answer?

Chapter 10 Solutions