Explanation Given information: The weight of geostationary satellite ( W ) is 800 lb. The angular velocity of the satellite ( ω 0 ) is ( 1.5 rad/s ) j . The weight of a meteorite W ′ is 6 oz. The travelling velocity of the meteorite v 0 is − ( 1 , 600 ft/s ) i + ( 1 , 300 ft/s ) j + ( 4 , 000 ft/s ) k . The distance value b is 20 in.. The radii of gyration of the satellite along x , y , and z direction ( k ¯ x , k ¯ y , and k ¯ z ) is 28.8 in., 32.4 in., and 28.8 in. respectively. Calculation: Determine the mass of the satellite ( m ). m = W g Here, g is the acceleration due to gravity. Substitute 800 lb for W and 32.2 ft/s 2 for g . m = 800 32.2 = 24.845 lb ⋅ s 2 /ft Determine the principal moment of inertia along x axis. I ¯ x = m k ¯ x 2 Substitute 24.845 lb ⋅ s 2 /ft for m and 28.8 in. for k ¯ x . I ¯ x = 24.845 × ( 28.8 in . × 1 ft 12 in . ) 2 = 143.11 lb ⋅ s 2 ⋅ ft Determine the principal moment of inertia along y axis. I ¯ y = m k ¯ y 2 Substitute 24.845 lb ⋅ s 2 /ft for m and 32.4 in. for k ¯ y . I ¯ y = 24.845 × ( 32.4 in . × 1 ft 12 in . ) 2 = 181.12 lb ⋅ s 2 ⋅ ft Determine the principal moment of inertia along z axis. I ¯ z = m k ¯ z 2 Substitute 24.845 lb ⋅ s 2 /ft for m and 28.8 in. for k ¯ z . I ¯ z = 24.845 × ( 28.8 in . × 1 ft 12 in . ) 2 = 143.11 lb ⋅ s 2 ⋅ ft Determine the mass of the meteorite m ′ . m ′ = W ′ g Substitute 6 oz for W ′ and 32.2 ft/s 2 for g . m = 6 oz × 0.0625 lb 1 oz 32.2 = 0.0116 lb ⋅ s 2 /ft Determine the initial moment of meteorite ( p 0 ) . p 0 = m ′ v 0 Substitute 0.0116 lb ⋅ s 2 /ft for m ′ and − ( 1 , 600 ft/s ) i + ( 1 , 300 ft/s ) j + ( 4 , 000 ft/s ) k for v 0 . p 0 = 0.01164 × ( − ( 1 , 600 ft/s ) i + ( 1 , 300 ft/s ) j + ( 4 , 000 ft/s ) k ) = ( − 18.62 lb ⋅ s ) i + ( 15.13 lb ⋅ s ) j + ( 46.56 lb ⋅ s ) k Consider that the position of the satellite mass center plus the meteorite is essentially that of the satellite alone. Determine the position of the point A relative to the mass center. r A / G = x i Here, x is the horizontal distance from the satellite center. Substitute 42 in. for x . r A / G = ( 42 in . × 1 ft 12 in . ) i = ( 3 .5 ft ) i The angular velocity of satellite before impact ( ω 0 ) is ( 1.5 rad/s ) j . The angular velocity of satellite before impact along x , y , and z axis is ( ω 0 ) x = 0 , ( ω 0 ) y = 1.5 rad/s , and ( ω 0 ) z = 0 . Determine the angular momentum of satellite–meteorite system before impact ( H G ) 0 . ( H G ) 0 = I ¯ y ω 0 j + r B / G × p 0 Substitute 181.12 lb ⋅ s 2 ⋅ ft for I ¯ y , 1.5 rad/s for ω 0 , ( 3 .5 ft ) i + ( 1.667 ft ) j for r B / G , and ( − 18.62 lb ⋅ s ) i + ( 15.13 lb ⋅ s ) j + ( 46.56 lb ⋅ s ) k for p 0 . ( H G ) 0 = ( 181.12 × 1.5 ) j + | i j k 3.5 0 0 − 18.62 15.13 46.56 | = { 271.68 j + i ( 0 − 0 ) − j ( 3.5 × 46.56 − 0 ) + k ( 3.5 × 15.13 − 0 ) } = 271.68 j − 162.96 j + 52.96 k = ( 108.72 j + 52.96 k ) lb ⋅ s ⋅ ft Principle of impulse and momentum for satellite–meteorite system: The value of moments about G is ( H G ) 1 = ( H G ) 0 = H G . The expression for H G is H G = I ¯ x ω x i + I ¯ y ω y j + I ¯ z ω z k . Determine the angular velocity about x axis. ω x = ( H G ) x I x Substitute 0 for ( H G ) x and 143.11 lb ⋅ s 2 ⋅ ft for I x . ω x = 0 143.11 = 0 Determine the angular velocity about y axis. ω y = ( H G ) y I y Substitute 108.72 lb ⋅ s ⋅ ft for ( H G ) y and 181

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ISBN: 9781259633133

Author: BEER

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Chapter 18.3, Problem 18.130P

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Chapter 18.3, Problem 18.129P

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The precession axis, rates of precession, and the spin rate of the satellite after the impact.

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