EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.
EP ENGR.MECH.:DYNAMICS-REV.MOD.MAS.ACC.
14th Edition
ISBN: 9780133976588
Author: HIBBELER
Publisher: PEARSON CO
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Chapter 20.3, Problem 1P

(a)

To determine

The angular acceleration of the propeller when the turn is horizontal.

(a)

Expert Solution
Check Mark

Answer to Problem 1P

The angular acceleration of the propeller when the turn is horizontal is α=ωxωtj_.

Explanation of Solution

Write the expression of angular acceleration at constant speed.

    (ω˙x)XYZ=(ω˙x)xyz+Ω×ωx        (I)

Write the expression of angular acceleration turning at constant rate.

    (ω˙t)XYZ=(ω˙t)xyz+Ω×ωt        (II)

Here, ω˙ for the angular acceleration, x,y,z for the translating-rotating frame of reference, X,Y,Z for the fixed frame of reference, and Ω for angular velocity.

Write the expression of angular acceleration.

    α=(ω˙x)XYZ+(ω˙t)XYZ        (III)

Conclusion:

Substitute ωtk for Ω, 0 for (ω˙x)xyz, and ω˙si for ω˙s in Equation (I).

    (ω˙x)XYZ=0+(ω˙tk)×(ω˙xi)=ωxωtj

Substitute 0 for Ω, and 0 for (ω˙x)xyz in Equation (II).

    (ω˙t)XYZ=0+0=0

Substitute ωxωtj for (ω˙x)XYZ and 0 for (ω˙t)XYZ in Equation (III).

    α=ωxωtj+0=ωxωtj

Thus, the angular acceleration of the turn is horizontal ωtk is α=ωxωtj_.

(b)

To determine

The angular acceleration of the propeller when the turn is vertical, downward.

(b)

Expert Solution
Check Mark

Answer to Problem 1P

The angular acceleration of the propeller when the turn is vertical, downward is α=ωxωtk_.

Explanation of Solution

Substitute ωtj for Ω, 0 for (ω˙x)xyz, and ω˙si for ω˙s in Equation (I).

    (ω˙x)XYZ=0+(ω˙tj)×(ω˙xi)=ωxωtk

Substitute 0 for Ω, and 0 for (ω˙x)xyz in Equation (II).

    (ω˙t)XYZ=0+0=0

Substitute ωxωtk for (ω˙x)XYZ and 0 for (ω˙t)XYZ in Equation (III).

    α=ωxωtk+0=ωxωtk

Thus, the angular acceleration of the propeller when the turn is vertical, downward is α=ωxωtk_.

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