Explanation Given information: The curve equation is y = k x 2 3 . Calculation: Sketch the vertical strip shaded portion as shown in Figure 1. Write the curve Equation as follows: y = k x 2 3 (1) Refer to Figure 1. Consider a differential element to y axis width x shown in Figure. At x = a , y = b . Find the constant k using the relation: Substitute a for x and b for y in Equation (1). b = k a 2 3 k = b a 2 3 Substitute b a 2 3 for k in Equation (1). y = ( b a 2 3 ) x 2 3 Determine the moment of inertia ( d I x ) with respect to x axis of a rectangular strip: Refer Equation 9.2 in section 9.1B determining the moment of inertia of an area by integration: d I x = 1 3 y 3 d x (2) Substitute ( b a 2 3 ) x 2 3 for y in Equation (2). d I x = 1 3 ( ( b a 2 3 ) x 2 3 ) 3 d x Integrate equation (2) with respect to x . I x = ∫ d I x = ∫ 0 a 1 3 ( ( b a 2 3 ) x 2 3 ) 3 d x = b 3 3 a 2 ∫ 0 a x 2 d x = b 3 3 a 2 [ x 3 3 ] 0 a = b 3 3 a 2 [ a 3 3 ] = b 3 3 a 2 a 3 3 I x = 1 9 a b 3 Thus, the moment of inertia of the shaded area with respect to x axis is 1 9 a b 3 _ Find the area ( A ) of the shaded area

Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977206

Author: BEER, Ferdinand P., Johnston Jr., E. Russell, Mazurek, David, Cornwell, Phillip J., SELF, Brian

Publisher: McGraw-Hill Education

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Moment of Inertia

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Chapter 9.1, Problem 9.15P

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Find the moment of inertia and radius of gyration of the shaded area.

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Chapter 9.1, Problem 9.14P

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Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

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Find the moment of inertia and radius of gyration of the shaded area.

Solution Summary: The author calculates the curve equation as follows: y=kx23.

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Find the moment of inertia and radius of gyration of the shaded area.

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