Explanation Given: The rectangular brick has dimensioned as a, b and c . which is length, breadth and height respectively. The mass is M and the center is situated at origin and also parallel to the coordinate axes. Calculation: The edges of the brick parallel to the coordinate axes and also center is at the origin. Therefore x varies from − a 2 to a 2 , y varies from − b 2 to b 2 and z varies from − c 2 to c 2 . density= mass volume = M a b c = k Initially, calculate the moment of inertia over the x -axis. Therefore, the integration will be over ( x 2 + y 2 ) ⋅ ρ ( x , y , z ) but here the density is constant. The moment of inertia over the x -axis is, I x = ∭ E ( y 2 + z 2 ) ρ ( x , y , z ) d V I x = k ∫ − c 2 c 2 ∫ − b 2 b 2 ∫ − a 2 a 2 ( y 2 + z 2 ) d x d y d z Integrate the above integral with respect to x and apply the limit of it. I x = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ y 2 x + z 2 x ] − a 2 a 2 d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ ( y 2 ( a 2 ) + z 2 ( a 2 ) ) − ( y 2 ( − a 2 ) + z 2 ( − a 2 ) ) ] d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ y 2 ( a 2 ) + z 2 ( a 2 ) + y 2 ( a 2 ) + z 2 ( a 2 ) ] d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ 2 y 2 ( a 2 ) + 2 z 2 ( a 2 ) ] d y d z On further simplification the above integral becomes, I x = k a ∫ − c 2 c 2 ∫ − b 2 b 2 ( y 2 + z 2 ) d y d z Integrate the above integral with respect to y and apply the limit of it. I x = k a ∫ − c 2 c 2 [ y 3 3 + z 2 y ] − b 2 b 2 d z = k a ∫ − c 2 c 2 [ ( ( b 2 ) 3 3 + z 2 ( b 2 ) ) − ( ( − b 2 ) 3 3 + z 2 ( − b 2 ) ) ] d z = k a ∫ − c 2 c 2 [ ( b 2 ) 3 3 + z 2 ( b 2 ) + ( b 2 ) 3 3 + z 2 ( b 2 ) ] d z = k a ∫ − c 2 c 2 [ 2 ( b 2 ) 3 3 + 2 z 2 ( b 2 ) ] d z On further simplification the above integral becomes, I x = k a ∫ − c 2 c 2 [ b 3 12 + z 2 b ] d z Integrate the above integral with respect to z and apply the limit of it. I x = k a [ 1 12 b 3 ⋅ z + b z 3 3 ] − c 2 c 2 = k a [ ( 1 12 b 3 ⋅ ( c 2 ) + b ( c 2 ) 3 3 ) − ( 1 12 b 3 ⋅ ( − c 2 ) + b ( − c 2 ) 3 3 ) ] = k a [ ( 1 24 b 3 c + 1 24 b c 3 ) − ( − 1 24 b 3 c − 1 24 b c 3 ) ] = k a [ 1 24 b 3 c + 1 24 b c 3 + 1 24 b 3 c + 1 24 b c 3 ] On further simplification the above integral becomes, I x = k a [ 2 24 b 3 c + 2 24 b c 3 ] = k a b c 12 ( b 2 + c 2 ) Thus, the moment of inertia of the rectangular brick is a b c k 12 Thus, the moment of inertia of the cylindrical brick about the x -axis is k a b c 12 ( b 2 + c 2 ) . Calculate the moment of inertia over the y -axis. Therefore, the integration will be over ( z 2 + x 2 ) ⋅ ρ ( x , y , z ) but here the density is constant. The moment of inertia over they y -axis is, I y = ∭ E ( z 2 + x 2 ) ρ ( x , y , z ) d V I y = k ∫ − c 2 c 2 ∫ − b 2 b 2 ∫ − a 2 a 2 ( z 2 + x 2 ) d x d y d z Integrate the above integral with respect to x and apply the limit of it. I y = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ z 2 x + x 3 3 ] − a 2 a 2 d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ ( z 2 ( a 2 ) + ( a 2 ) 3 3 ) − ( z 2 ( − a 2 ) + ( − a 2 ) 3 3 ) ] d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ ( z 2 ( a 2 ) + ( a 2 ) 3 3 ) + z 2 ( a 2 ) − ( a 2 ) 3 3 ] d y d z = k ∫ − c 2 c 2 ∫ − b 2 b 2 [ a z 2 + a 3 12 ] d y d z Integrate the above integral with respect to y and apply the limit of it. I y = k ∫ − c 2 c 2 [ a z 2 ⋅ y + a 3 12 ⋅ y ] − b 2 b 2 d z = k ∫ − c 2 c 2 [ ( a z 2 ⋅ ( b 2 ) + a 3 12 ⋅ ( b 2 ) ) − ( a z 2 ⋅ ( − b 2 ) + a 3 12 ⋅ ( − b 2 ) ) ] d z = k ∫ − c 2 c 2 [ a z 2 ⋅ ( b 2 ) + a 3 12 ⋅ ( b 2 ) + a z 2 ⋅ ( b 2 ) + a 3 12 ⋅ ( b 2 ) ] d z = k ∫ − c 2 c 2 [ a b z 2 2 + a 3 b 24 + a b z 2 2 + a 3 b 24 ] d z On further simplification the above integral becomes, I y = k ∫ − c 2 c 2 [ 2 ⋅ a b z 2 2 + 2 ⋅ a 3 b 24 ] d z = k ∫ − c 2 c 2 [ a b z 2 + a 3 b 12 ] d z Integrate the above integral with respect to z- and apply the limit of it

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ISBN: 9781305266643

Author: James Stewart

Publisher: Cengage Learning

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Chapter 15.6, Problem 44E

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To find: The moment of inertia of the rectangular brick if the solid has constant density k.

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Chapter 15.6, Problem 43E

Chapter 15.6, Problem 45E

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The moment of inertia of the rectangular brick if the solid has constant density k .

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