Explanation Given information: The time is 1 s , the mass of the capsule is 1000 kg , the radius of gyration in x axis is 1 m , the radius of gyration in y axis is 1 m , the radius of gyration in z axis is 1.25 m and the produced thrust is 50 N . Write the expression of angular momentum about point G . H G = I x ω x i + I y ω y j + I z ω z k ....... (I) Here, the moment of inertia about x axis is I x , the angular velocity in x direction is ω x , the unit vector in x direction is i , the moment of inertia about y axis is I y , the angular velocity in y direction is ω y , the unit vector in y direction is j , the moment of inertia about z axis is I z , the angular velocity in z direction is ω z , and the unit vector in z direction is k . Write the expression of angular momentum about the mass centre. H G = ( H G ) x i + ( H G ) y j + ( H G ) z k ....... (II) Here, the angular momentum along x axis is ( H G ) x , the angular momentum along y axis is ( H G ) y and the angular momentum along z axis is ( H G ) z . Write the expression moment about the mass centre. M G = r A / G × F A × Δ t i ....... (III) Here, the distance of A with respect to the mass centre is r A / G , the thrust is F A and the time period is Δ t . Write the expression of the distance of A with respect to the mass centre. r A / G = y 1 j + z 1 k ...... (IV) Here, the distance of A from y axis is y 1 and the distance from z axis is z 1 . Write the expression of principle of impulse and momentum. H G = M G ...... (V) The below figure represent the angular momentum. Figure-(1) Write the expression of angle of axis of precision as shown in Figure-1. tan θ = ( H G ) y ( H G ) z ...... (VI) Substitute ( H G ) x i + ( H G ) y j + ( H G ) z k for H G in Equation (I). ( H G ) x i + ( H G ) y j + ( H G ) z k = I x ω x i + I y ω y j + I z ω z k ...... (VII) Compare the unit vector i on both sides of Equation (IX). ( H G ) x = I x ω x ...... (VIII) Compare the unit vector j on both sides of Equation (IX). ( H G ) y = I y ω y ...... (IX) Compare the unit vector k on both sides of Equation (IX). ( H G ) k = I k ω k ....... (X) Write the expression of moment of inertia about x axis. I x = m k x 2 ....... (XI) Here, the radius of gyration in x direction is k x . Write the expression of moment of inertia about y axis. I y = m k y 2 ...... (XII) Here, the radius of gyration in x direction is k y 2 . Write the expression of moment of inertia about z axis. I z = m k z 2 ...... (XIII) Here, the radius of gyration in x direction is k z . Write the expression of resultant angular velocity. ω = ω x 2 + ω y 2 + ω z 2 ...... (XIV) The below figure represent the angular velocity component. Figure-(2) Write the expression of angle between angular velocity components as shown in Figure-(2). tan γ = ω y ω z ...... (XV) The below figure represent the diagram of rate of precision and spin. Figure-(3) Write the expression of lamis theorem applied on Figure-(3). ϕ ˙ sin γ = ψ ˙ sin ( γ − θ ) ϕ ˙ sin γ = ω sin θ ...... (XVI) Here, the rate of precision is ϕ ˙ and the rate of spin is ψ ˙ . Calculation: Substitute 2 m for y 1 and − 1.25 m for z 1 in Equation (IV). r A / G = ( 2 m ) j − ( 1.25 m ) k Substitute ( 2 m ) j − ( 1.25 m ) k for r A / G , 50 N for F A and 1 s for Δ t in Equation (III). M G = { ( 2 m ) j − ( 1.25 m ) k } × ( 50 N ) i × 1 s = { ( 2 m ) j − ( 1.25 m ) k } × ( 50 N ⋅ s ) i ...... (XVII) The Equation (IV) is a cross product of two vectors, now change Equation (XVIII) in determinant form. M G = | i j k 0 2 m − 1.25 m 50 N ⋅ s 0 0 | = ( 0 − 0 ) i − { 0 − ( − 62.5 N ⋅ m ⋅ s ) } j − { 0 − ( 100 N ⋅ m ⋅ s ) } k = − ( 62.5 N ⋅ m ⋅ s ) j − ( 100 N ⋅ m ⋅ s ) k ..... (XIX) Substitute − ( 62.5 N ⋅ m ⋅ s ) for ( H G ) y and − ( 100 N ⋅ m ⋅ s ) for ( H G ) z in Equation (VI). tan θ = − ( 62.5 N ⋅ m ⋅ s ) − ( 100 N ⋅ m ⋅ s ) tan θ = 62.5 100 θ = tan − 1 ( 0.32625 ) θ = 32 ° So, the axis of precision is at 32 ° . Substitute − ( 62.5 N ⋅ m ⋅ s ) j − ( 100 N ⋅ m ⋅ s ) k for M G and ( H G ) x i + ( H G ) y j + ( H G ) z k for H G in Equation (V)

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ISBN: 9781260265521

Author: BEER

Publisher: MCG

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Chapter 18, Problem 18.156RP

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The axis of precision.

The rate of precision.

The rate of spin.

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Chapter 18, Problem 18.155RP

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Chapter 18 Solutions

VECTOR MECH...,DYNAMICS(LOOSE)-W/ACCESS

Ch. 18.1 - Prob. 18.1P 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 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 axis of precision. The rate of precision. The rate of spin.

Solution Summary: The author explains the axis of precision, the rate of spin, and the expression of angular momentum about point G.