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

Question

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Answer 1

Question

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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

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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Answer 2

Question

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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

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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Answer 3

Question

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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

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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Answer 4

Question

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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

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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Answer 5

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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

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→_.

Answer 6

Chapter 20.3, Problem 1P

(a)

To determine

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

(a)

Expert Solution

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→_.

Answer 7

Chapter 20.3, Problem 1P

(a)

To determine

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

Answer 8

(a)

Expert Solution

Answer to Problem 1P

The angular acceleration of the propeller when the turn is horizontal is α=ωxωtj→_.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

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→_.

Answer 13

(b)

To determine

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

(b)

Expert Solution

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→_.

Answer 14

(b)

To determine

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

Answer 15

(b)

Expert Solution

Answer to Problem 1P

The angular acceleration of the propeller when the turn is vertical, downward is α=−ωxωtk→_.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

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→_.

Answer 20

Ch. 20.3 - Prob. 1P Ch. 20.3 - Prob. 2P Ch. 20.3 - Prob. 3P Ch. 20.3 - Prob. 4P Ch. 20.3 - Prob. 5P Ch. 20.3 - Prob. 6P Ch. 20.3 - Prob. 7P Ch. 20.3 - The disk rotates about the shaft S, while the...Ch. 20.3 - The electric fan is mounted on a swivel support...Ch. 20.3 - Prob. 11P

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

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

Answer to Problem 1P

Explanation of Solution

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

Answer to Problem 1P

Explanation of Solution

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