Explanation Given information: The radius ( r ) of quarter circle is 15 in . . The specific weight of γ aluminum is 170 lb / ft 3 . The uniform thickness ( t ) of aluminum plate is 0.05 in . Calculation: Divided the component into three geometric portions 1, 2, and 3. Find the mass of portion 1 in the x − y plane as shown below: m = ρ V = ρ [ 1 4 π r 2 ] t (1) Here, ρ is density of aluminum. Substitute 15 in . for r and 0.05 in . for t . m = ρ [ 1 4 π × ( 15 in . ( 1 ft 12 in . ) ) 2 × 0.05 in . ( 1 ft 12 in . ) ] = ρ [ 1 4 π × 1.5625 × 4.1666 × 10 − 3 ] = 5.1133 × 10 − 3 ρ Find the moment of inertia about x and z axis of portion 1 in the x − y plane as shown below: I x = I z = 1 4 m r 2 (2) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (2). I x = I z = 1 4 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 4 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 1.99738 × 10 − 3 ρ Find the moment of inertia about y axis of portion 1 in the x − y plane as shown below: I y = 1 4 m r 2 (3) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (3). I y = 1 4 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 4 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 1.99738 × 10 − 3 ρ Find the moment of inertia about x and z axis of portion 2 in the y − z plane as shown below: Here, mass is same for portions 2. I x = I z = 1 2 m r 2 (4) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (4). I x = I z = 1 2 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 2 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 3.9948 × 10 − 3 ρ Find the moment of inertia about y axis of portion 2 in the x − y plane as shown below: I y = 1 2 m r 2 (5) Substitute 5.1133 × 10 − 3 ρ for m and 15 in . for r in Equation (5). I y = 1 2 ( 5.1133 × 10 − 3 ρ ) ( 15 in . ( 1 ft 12 in . ) 2 ) = 1 2 ( 5.1133 × 10 − 3 ρ ) 1.5625 = 3.9948 × 10 − 3 ρ Find the mass of portion 3 in the x − y plane as shown below: m = ρ V = ρ [ 1 4 r 2 ] t (6) Here, ρ is density of aluminum. Substitute 15 in . for r and 0.05 in . for t in Equation (6). m = ρ [ 1 2 × ( 15 in . ( 1 ft 12 in . ) ) 2 × 0.05 in . ( 1 ft 12 in . ) ] = ρ [ 1 2 × 1.5625 × 4.1666 × 10 − 3 ] = 3.2552 × 10 − 3 ρ Find the mass moment of inertia ( I mass ) as shown below: ( I x ,mass ) = ρ t I area = ρ t ( 1 12 r 4 ) = ρ ( 1 2 r 2 t ) ( 1 6 r 2 ) = 1 6 m r 2

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ISBN: 9781260536225

Author: BEER

Publisher: MCG

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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 9.5, Problem 9.139P

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Find the mass moment of inertia with respect to x,y,z coordinate axes.

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Chapter 9.5, Problem 9.138P

Chapter 9.5, Problem 9.140P

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Find the mass moment of inertia with respect to x , y , z coordinate axes.

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