Explanation Given: The radius of the shaded area is 4 ft . The height of the shaded area is 4 ft . Show the intersection of shaded area at the arbitrary point ( x , y , z ) as in Figure (1). Conclusion: Calculate the differential mass of the shaded area. d m = ρ d V (I) Here, density of material is ρ , and the differential volume of the shaded area is V . Substitute π x 2 d y for d V in Equation (I). d m = ρ π x 2 d y (II) Substitute 1 y for x in Equation (II). d m = ρ π [ 1 y ] 2 d y = ρ π d y y 2 Compute the inertia of differential element with respect to the y axes. d I y = 1 2 d m x 2 (III) Substitute ρ π d y y 2 for d m and 1 y for x in Equation (III). d I y = 1 2 [ ρ π d y y 2 ] [ 1 y ] 2 = 1 2 ρ π d y y 4 Compute the inertia of the shaded area with respect to the y axes. I y = ∫ 0.25 4 d I y (IV) Substitute 1 2 ρ π d y y 4 for d I y in Equation (IV). I y = ∫ 0.25 4 1 2 ρ π d y y 4 = 1 2 ρ π ∫ 0.25 4 d y y 4 + 1 2 [ ρ π ( h ) 2 ( 0.25 ) ] ( r ) 2 (V) Substitute 4 ft for r and h respectively in Equation (V). I y = 1 2 ρ π ∫ 0.25 4 d y y 4 + 1 2 [ ρ π ( 4 ) 2 ( 0.25 ) ] ( 4 ) 2 = 1 2 ρ π ∫ 0.25 4 y − 4 d y + 100.5 ρ = 1 2 ρ π { [ − 1 3 y 3 ] | 0.25 4 + 100.5 ρ } = 134.03 ρ Express the mass of the material. m = ∫ 0.25 4 d m (VI) Substitute ρ π d y y 2 for d m in Equation (VI). m = ∫ 0.25 4 ρ π d y y 2 = ρ π ∫ 0.25 4 d y y 2 + ρ [ π ( 4 ) 2 ( 0.25 ) ] = ρ π { [ − 1 y ] | 0.25 4 + 12.566 ρ } = 24.35 ρ Express the radii of gyration k y for the solid

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Author: HIBBELER

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Moment of Inertia

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Chapter 21, Problem 10P

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The radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis:

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Chapter 21, Problem 9P

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The radii of gyration k x and k y for the solid formed by revolving the shaded area about the y axis:

Solution Summary: The author shows the intersection of shaded area at the arbitrary point (x,y,z).

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The radii of gyration kx and ky for the solid formed by revolving the shaded area about the y axis:

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