Explanation Given: The region with moments of inertia is shown in the figure below: Also, the density of the lamina is assumed to be ρ = 1 gm cm 2 . Formula used: Consider ρ to be a continuous density function on the lamina of region R , the mass ( m ) of the lamina is determined by the formula: m = ∫ R ∫ ρ ( x , y ) d A The formula to calculate the moments of inertia about x and y axis of a lamina of variable density denoted by I x and I y is: I x = ∫ R ∫ ( y 2 ) ρ ( x , y ) d A And I y = ∫ R ∫ ( x 2 ) ρ ( x , y ) d A Where, y 2 is the square of distance to the x axis and x 2 is the square of distance to the y axis. The radius of gyration about y axis is determined by, x ¯ ¯ = I y m . The radius of gyration about x axis is determined by, y ¯ ¯ = I x m . The equation of a line in a two point form is given as: y − y 1 = y 2 − y 1 x 2 − x 1 ( x − x 1 ) ……(1) Where, ( x 1 , y 1 ) and ( x 2 , y 2 ) are the two points through which the line passes. Proof: Consider the provided figure of the right triangle, There are two points joining the hypotenuse that are ( b , 0 ) and ( 0 , h ) . Now, substitute these points in equation (1) such that the equation of the hypotenuse line is y − 0 = h − 0 0 − b ( x − b ) y = − h b ( x − b ) y = h − h x b Hence, bounds on y is 0 ≤ y ≤ ( h − h x b ) and bounds on x is 0 ≤ x ≤ b . The density of the lamina is provided as ρ = 1 gm cm 2 , therefore, use the formula of moments of inertia as follows, Moment of inertia about x axis is given as: I x = ∫ 0 b ∫ 0 ( h − h x b ) ( y 2 ) ( 1 ) d y d x Now, integrate the equation above with respect to y and assume x as a constant such that, I x = ∫ 0 b ∫ 0 ( h − h x b ) ( y 2 ) d y d x = ∫ 0 b [ y 3 3 ] 0 ( h − h x b ) d x = 1 3 ∫ 0 b [ ( ( h − h x b ) 3 ) − ( ( 0 ) 3 ) ] d x = 1 3 ∫ 0 b ( h 3 + 3 h 3 x 2 b 2 − 3 h 3 x b − h 3 x 3 b 3 ) d x Further, integrate it with respect to x as I x = 1 3 ∫ 0 b ( h 3 + 3 h 3 x 2 b 2 − 3 h 3 x b − h 3 x 3 b 3 ) d x = 1 3 [ h 3 x + 3 h 3 x 3 3 b 2 − 3 h 3 x 2 2 b − h 3 x 4 4 b 3 ] 0 b = 1 3 [ ( h 3 ( b ) + 3 h 3 ( b ) 3 3 b 2 − 3 h 3 ( b ) 2 2 b − h 3 ( b ) 4 4 b 3 ) − ( h 3 ( 0 ) + 3 h 3 ( 0 ) 3 3 b 2 − 3 h 3 ( 0 ) 2 2 b − h 3 ( 0 ) 4 4 b 3 ) ] = 1 12 h 3 b Similarly, the moment of inertia about y axis is given as: I y = ∫ 0 b

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ISBN: 9781337275347

Author: Ron Larson, Bruce H. Edwards

Publisher: Cengage Learning

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Chapter 14.4, Problem 30E

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To prove: The moments of inertia are Ix=112bh3 and Iy=112b3h for the provided right triangle and evaluate x¯¯ and y¯¯.

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Chapter 14.4, Problem 29E

Chapter 14.4, Problem 31E

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To prove: The moments of inertia are I x = 1 12 b h 3 and I y = 1 12 b 3 h for the provided right triangle and evaluate x ¯ ¯ and y ¯ ¯ .

Solution Summary: The author explains the formula to calculate the moments of inertia about x and y axis of a lamina of variable density denoted by I_x

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To prove: The moments of inertia are Ix=112bh3 and Iy=112b3h for the provided right triangle and evaluate x¯¯ and y¯¯.

Want to see the full answer?

Chapter 14 Solutions