The propeller of an airplane is rotating at a constant speed ω x i , while the plane is undergoing a turn at a constant rate ω t . Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω t k , and (b) the turn is vertical, downward, i.e., ω t j . Prob. 20-1

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Answer 1

Textbook Question

Chapter 20, Problem 1P

The propeller of an airplane is rotating at a constant speed ω_xi, while the plane is undergoing a turn at a constant rate ω_t. Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω_t k, and (b) the turn is vertical, downward, i.e., ω_t j.

Chapter 20, Problem 1P, The propeller of an airplane is rotating at a constant speed xi, while the plane is undergoing a

Prob. 20-1

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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Answer 2

Textbook Question

Chapter 20, Problem 1P

The propeller of an airplane is rotating at a constant speed ω_xi, while the plane is undergoing a turn at a constant rate ω_t. Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω_t k, and (b) the turn is vertical, downward, i.e., ω_t j.

Chapter 20, Problem 1P, The propeller of an airplane is rotating at a constant speed xi, while the plane is undergoing a

Prob. 20-1

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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Answer 3

Textbook Question

Chapter 20, Problem 1P

The propeller of an airplane is rotating at a constant speed ω_xi, while the plane is undergoing a turn at a constant rate ω_t. Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω_t k, and (b) the turn is vertical, downward, i.e., ω_t j.

Chapter 20, Problem 1P, The propeller of an airplane is rotating at a constant speed xi, while the plane is undergoing a

Prob. 20-1

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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Answer 4

Textbook Question

Chapter 20, Problem 1P

The propeller of an airplane is rotating at a constant speed ω_xi, while the plane is undergoing a turn at a constant rate ω_t. Determine the angular acceleration of the propeller. If (a) the turn is horizontal, i.e., ω_t k, and (b) the turn is vertical, downward, i.e., ω_t j.

Chapter 20, Problem 1P, The propeller of an airplane is rotating at a constant speed xi, while the plane is undergoing a

Prob. 20-1

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

Answer 8

Expert Solution

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

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:

(a) 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) 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 turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

(a) The angular acceleration of the turn is horizontal ωtk→ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ .

Answer 13

Answer to Problem 1P

(a) The angular acceleration of the turn is horizontal ωtk→ is α=ωxωtj→_ .
(b) The angular acceleration of the turn is vertical, downward ωtj→ is α=−ωxωtk→_ .

Answer 14

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.