Explanation Given information: The weight of the space probe ( W ) is 3000 lb , the radius of gyration along x axis is 1.375 ft , the radius of gyration along y axis is 1.425 ft , the radius of gyration along z axis is 1.250 ft , the mass of meteorite ( m ) is 5 oz and the relative velocity of the meteorite ( v 0 ) is ( 2400 ft / s ) i + ( 3000 ft / s ) j + ( 3200 ft / s ) k and the position vector is ( 9 ft ) i + ( 0.75 ft ) k . Write the expression for the angular momentum of the meteorite about G . ( H G ) M = r A + m M v 0 ...... (I) Here, the position vector is r A . Write the expression for the angular momentum of the probe. ( H G ) P = I x ω x i + I y ω y j + I z ω z k ...... (II) Here, the angular velocity along the x axis is ω x , the angular velocity along the y axis is ω y , the angular velocity along the z axis is ω z , the moment of inertia about the x axis is I x , the moment of inertia about the y axis is I y and the moment of inertia about the z axis is I z . Write the expression for the moment of inertia along x axis. I x = m k x 2 ...... (III) Here, the mass of the probe is m and the radius of gyration along x axis is k x . Write the expression for the moment of inertia along y axis. I y = m k y 2 ...... (IV) Here, the radius of gyration along y axis is k y . Write the expression for the moment of inertia along z axis. I z = m k z 2 ...... (V) Here, the radius of gyration along z axis is k z . Substitute m k x 2 for I x , m k y 2 for I y and m k z 2 for I z in Equation (II). ( H G ) P = m k x 2 ω x i + m k y 2 ω y j + m k x 2 ω z k = m ( k x 2 ω x i + k y 2 ω y j + k x 2 ω z k ) ...... (VI) Write the expression relation between the angular momentum of the meteorite and the angular momentum of the probe. ( H G ) P = 0.20 ( H G ) W ...... (VII) Write the expression for the angular momentum of three-dimensional plane. ω = ω x i + ω y j + ω z k ...... (VIII) Calculation: Substitute ( 9 ft ) i + ( 0.75 ft ) k for r A , ( 2400 ft / s ) i + ( 3000 ft / s ) j + ( 3200 ft / s ) k for v 0 , 5 oz for m M in Equation (I). ( H G ) M = [ [ ( 9 ft ) i + ( 0.75 ft ) k ] × 5 oz 32.2 ft / s 2 [ ( 2400 ft / s ) i + ( 3000 ft / s ) j + ( 3200 ft / s ) k ] ] = [ [ ( 9 ft ) i + ( 0.75 ft ) k ] × 5 oz 32.2 ft / s 2 1 lb 16 oz [ ( 2400 ft / s ) i + ( 3000 ft / s ) j + ( 3200 ft / s ) k ] ] = [ [ 9.705 × 10 − 3 lb ⋅ s 2 / ft ] ( − 27 k − 28

Vector Mechanics for Engineers: Dynamics

11th Edition

ISBN: 9780077687342

Author: Ferdinand P. Beer, E. Russell Johnston Jr., Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 18.1, Problem 18.33P

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The angular velocity of the probe.

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Vector Mechanics for Engineers: Dynamics

Ch. 18.1 - Prob. 18.1P Ch. 18.1 - A thin rectangular plate of weight 15 Ib rotates...Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W=6 lb rotates at the...Ch. 18.1 - Prob. 18.5P Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Solve Prob. 18.6, assuming that the solid...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P

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The angular velocity of the probe.

Solution Summary: The author explains the angular velocity of the space probe and the position vector for the meteorite.

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The angular velocity of the probe.

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