Explanation Given information: The weight of the space probe ( w ′ ) is 3,000 lb. The radius of gyration along x axis ( k x ) is 1.375 ft. The radius of gyration along y axis ( k y ) is 1.425 ft. The radius of gyration along z axis ( k z ) is 1.250 ft. The weight of the meteorite ( w ) is 5 oz. The velocity ( v 0 ) is ( 2400ft/s ) i − ( 3000ft/s ) j + ( 3200ft/s ) k . The width of the side panel from center to point A ( b ) is 9 ft. The length of the panel from center to point A ( l ) is 0.75 ft. Calculation: Calculate the mass of the space probe ( m ′ ) using the formula: w ′ = m ′ g m ′ = w ′ g Here, g is the acceleration due to gravity. Substitute 32 .2ft/s 2 for g and 3000 lb for w ′ . m ′ = 3000 32.2 = 93.17 lb ⋅ s 2 /ft Calculate the mass of the meteorite ( m ) using the formula: w = m g m = w g Substitute 32 .2ft/s 2 for g and 5 oz for w . m = 5oz× 1 16 lb oz 32.2 = 0.009705 lb ⋅ s 2 /ft Write the relative position vector ( r A ) at the point ( A ) of impact as follows: r A = b i + l k Substitute 9 ft for b and 0.75 ft. r A = ( 9 ft ) i + ( 0.75 ft ) k Calculate the initial liner momentum of the meteorite using the relation: Initial linear momentum = m v 0 Substitute 0.009705 lb ⋅ s 2 /ft for m and ( 2400ft/s ) i − ( 3000ft/s ) j + ( 3200ft/s ) k for v 0 . m v 0 = ( 0.009705 lb ⋅ s 2 /ft ) × ( 2400ft/s ) i − ( 3000ft/s ) j + ( 3200ft/s ) k = 23.292 i − 29.115 j + 31.056 k Calculate the moment about origin ( M O ) using the relation: M O = r A × m v 0 Substitute 23.292 i − 29.115 j + 31.056 k for m v 0 and ( 9 ft ) i + ( 0.75 ft ) k for r A . M O = | i j k 9 0 0.75 23.292 − 29.115 31.056 | = i [ ( 0 ) − ( − 21.836 ) ] − j [ ( 279.504 − 17.469 ) ] + k [ − 262.04 − 0 ] = 21.836 i − 262.04 j − 262.04 k Calculate the final liner momentum of the meteorite using the relation: Final linear momentum = 0.8 m v 0 Substitute 0.009705 lb ⋅ s 2 /ft for m and ( 2400ft/s ) i − ( 3000ft/s ) j + ( 3200ft/s ) k for v 0 . 0.8 m v 0 = 0.8 ( 0.009705 lb ⋅ s 2 /ft ) × ( 2400ft/s ) i − ( 3000ft/s ) j + ( 3200ft/s ) k = 18.634 i − 23.292 j + 24.845 k Calculate the final linear momentum of meteorite and its moment about the origin using the relation: M = r A × ( 0.8 m v 0 ) Substitute 18.634 i − 23.292 j + 24.845 k for 0.8 m v 0 and ( 9 ft ) i + ( 0.75 ft ) k for r A . M = | i j k 9 0 0.75 18.634 − 23.292 24.845 | = i [ ( 0 ) − ( − 17.469 ) ] − j [ ( 223.604 − 13.9755 ) ] + k [ − 209.63 − 0 ] = 17.469 i − 209.63 j − 209.63 k Write the expression for the conservation of angular momentum about the origin for a system of particles consisting of the probe plus the meteorite as follows: M O = H A + M H A = M O − M Here, H A is the angular momentum of the probe

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

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Chapter 18.1, Problem 18.53P

To determine

The kinetic energy (T′) of the space probe in its motion about its mass center after its collision with the meteorite.

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Chapter 18.1, Problem 18.52P

Chapter 18.1, Problem 18.54P

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Chapter 18 Solutions

VEC MECH 180-DAT EBOOK ACCESS(STAT+DYNA)

Ch. 18.1 - A thin, homogeneous disk of mass m and radius r...Ch. 18.1 - Prob. 18.2P Ch. 18.1 - Prob. 18.3P Ch. 18.1 - A homogeneous disk of weight W = 6 lb rotates at...Ch. 18.1 - A homogeneous disk of mass m = 8 kg rotates at the...Ch. 18.1 - A solid rectangular parallelepiped of mass m has a...Ch. 18.1 - Prob. 18.8P Ch. 18.1 - Determine the angular momentum HD of the disk of...Ch. 18.1 - Prob. 18.10P Ch. 18.1 - Determine the angular momentum HO of the disk of...

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The kinetic energy ( T ′ ) of the space probe in its motion about its mass center after its collision with the meteorite.

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The kinetic energy (T′) of the space probe in its motion about its mass center after its collision with the meteorite.

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Chapter 18 Solutions