Explanation Given information: The curve equation is x 2 a 2 + y 2 b 2 = 1 . Calculation: Sketch the horizontal strip shaded portion as shown in Figure 1. Write the curve Equation as follows: x 2 a 2 + y 2 b 2 = 1 (1) Modify Equation (1). x 2 a 2 + y 2 b 2 = 1 x = a 1 − ( y 2 b 2 ) Determine the area of the strip element d A as shown in below: d A = 2 x d y Determine the moment of inertia ( d I x ) with respect to x axis of a rectangular strip: d I x = y 2 d A (2) Substitute 2 x d y for d A in Equation (2). d I x = y 2 2 x d y I x = ∫ 0 b y 2 2 x d y = 2 ∫ 0 b y 2 x d y I x = 2 ∫ 0 b y 2 ( a 1 − ( y 2 b 2 ) ) d y (3) Consider y = b sin θ Differentiate both sides of the Equation. d y = b cos θ d θ Find the moment of inertia of the shaded area with respect to x axis by using direct integration. Substitute b sin θ for y and b cos θ d θ for d y and apply the limits in Equation (3). I x = 2 a ∫ 0 π 2 b 2 sin 2 θ ( a 1 − ( ( b sin θ ) 2 b 2 ) ) b cos θ d θ = 2 a b 3 ∫ 0 π 2 sin 2 θ cos 2 θ d θ = 2 a b 3 ∫ 0 π 2 1 4 sin 2 θ d θ = 1 2 a b 3 ∫ 0 π 2 1 2 ( 1 − cos 4 θ ) d θ = 1 4 a b 3

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Moment of Inertia

Inertia can be termed as a resistance offered by a body to change its state. A rigid body initially at rest will apply resistance when acted by an external force that tends to change its state of rest or velocity, or a body initially at motion will provi…

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Chapter 9.1, Problem 9.16P

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Find the moment of inertia and radius of gyration of the shaded area with respect to x axis by using direct integration.

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Chapter 9.1, Problem 9.15P

Chapter 9.1, Problem 9.17P

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Find the moment of inertia and radius of gyration of the shaded area with respect to x axis by using direct integration.

Solution Summary: The author calculates the moment of inertia and radius of gyration of the shaded area with respect to x axis by using direct integration.

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Find the moment of inertia and radius of gyration of the shaded area with respect to x axis by using direct integration.

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