Explanation Given: The region with the moment 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 given by the below-mentioned 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 polar moment of inertia is I o = I x + I y . The formula used to calculate the radius of gyration about y axis is x ¯ ¯ = I y m . The formula used to calculate the radius of gyration about x axis is y ¯ ¯ = I x m . Proof: Consider the provided figure of the semicircle with radius a , The polar coordinates are x = r sin θ y = r cos θ d A = r d r d θ Further, the bounds on r is 0 ≤ r ≤ a and bounds on θ is 0 ≤ θ ≤ π since semicircle is given. The density of the lamina is provided as ρ = 1 gm cm 2 , therefore with the help of the formula of moments of inertia, Moment of inertia about x axis is given as: I x = ∫ 0 π ∫ 0 a ( r 2 cos 2 θ ) ( 1 ) r d r d θ Now, integrate the equation above with respect to r and assume θ as a constant such that, I x = ∫ 0 π ∫ 0 a ( r 3 cos 2 θ ) d r d θ = ∫ 0 π cos 2 θ [ r 4 4 ] 0 a d θ = 1 4 ∫ 0 π cos 2 θ [ ( a ) 4 − ( 0 ) 4 ] d θ = 1 4 ∫ 0 π a 4 cos 2 θ d θ Further, integrate it with respect to θ as I x = 1 4 ∫ 0 π a 4 cos 2 θ d θ = a 4 4 [ cos θ sin θ + θ 2 ] 0 π = a 4 8 [ ( cos ( π ) sin ( π ) + ( π ) ) − ( cos ( 0 ) sin ( 0 ) + ( 0 ) ) ] = π a 4 8 Similarly, the moment of inertia about y axis is given as: I y = ∫ 0 π ∫ 0 a ( r 2 sin 2 θ ) ( 1 ) r d r d θ Integrate the equation above with respect to r and assume θ as a constant such that, I y = ∫ 0 π ∫ 0 a ( r 3 sin 2 θ ) d r d θ = ∫ 0 π sin 2 θ [ r 4 4 ] 0 a d θ = 1 4 ∫ 0 x

11th Edition

ISBN: 9781337275576

Author: Larson, Ron, Edwards, Bruce

Publisher: Cengage Learning,

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Chapter 14.4, Problem 32E

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To prove: The moment of inertia is I0=14πa4 for the provided circle and evaluate x¯¯ and y¯¯.

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Chapter 14 Solutions

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To prove: The moment of inertia is I 0 = 1 4 π a 4 for the provided circle and evaluate x ¯ ¯ and y ¯ ¯ .

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

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To prove: The moment of inertia is I0=14πa4 for the provided circle and evaluate x¯¯ and y¯¯.

Chapter 14 Solutions