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

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

12th Edition

ISBN: 9781260916942

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Conservation of Energy

In physics, the rule of the conservation of energy states that energy doesn’t disappear from the atmosphere; instead, it changes its form from one type to another. Energy is one of the most important physical quantities in nature, and it cannot be create…

Videos

Question

Chapter 18.3, Problem 18.130P

To determine

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 18.3, Problem 18.129P

Chapter 18.3, Problem 18.131P

Students have asked these similar questions

14 in. 18 in. VB 1500 ft/s -18 in. A 0.05-lb bullet is fired into the side of a 20-lb square panel which is initially at rest. Determine a) the angular velocity of the panel immediately after the bullet becomes embedded and b) the impulsive reaction at A, assuming that the bullet becomes embedded in 0.0006 s. B

Please asap

The greater angle of the launched projectile creates the greater horizontal range.” Do you agree with the statement above? Justify your answer. Use some calculation tosupport your justification.

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...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.

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

Solution Summary: The author explains the precession axis, rates of gyration, and spin rate of the satellite after the impact.

Concept explainers

Videos

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

Want to see the full answer?

Chapter 18 Solutions