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)

12th Edition

ISBN: 9781259977305

Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.

Publisher: Mcgraw-hill Education,

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, Problem 18.156RP

To determine

The axis of precision.

The rate of precision.

The rate of spin.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 18, Problem 18.155RP

Chapter 18, Problem 18.157RP

Students have asked these similar questions

A rotating shaft carries four unbalanced masses 20 kg, 16 kg, 18 kg and 14 kg at radii 55 mm, 65 mm, 75 mm and 65 mm respectively. The 2nd, 3rd and 4th masses revolve in planes 80 mm, 160 mm and 280 mm respectively measured from the plane of the first mass and are angularly located at 65°, 135° and 270° respectively measured clockwise from the first mass.The shaft is dynamically balanced by two masses, both located at 55 mm radii and revolving in planes mid-way between those of 1st and 2nd masses and midway between those of 3rd and 4th masses. Determine, balancing mass by drawing couple polygon and their respective angular position graphically.

The space capsule has no angular velocity when the jet at A is activated for 1 s in a direction parallel to the axis. Knowing that the capsule has a mass of 1000 kg, that its radii of gyration are Kz=Ky =1.00m and Kz=1.25m A produces a thrust of 50 N, determine the axis of precession and the rates of precession and spin after the jet has stopped.

2 An object is made of two identical uniform 3m long rods connected as shown. The total mass of the system is 60kg. The object can rotate in the plane of the page about a fixed axis at Point A. There is a friction moment of 500 N m about point A. a. What is the mass moment of inertia of the object about Point A? b. If the object is released from rest in the orientation shown, what will be is initial angular acceleration? stationary Mf A = total mass= 60 kg 500 Nm 3 m 3 m g

Chapter 18 Solutions

Vector Mechanics For Engineers

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

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 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.