Explanation Given: Unbalanced weight of tail rotor is 1 lb . The distance of unbalance weight to axis of rotation is 6 in . Equivalent stiffness of tail and tail rotor is 8800 lb/ft . Equivalent weight of tail and tail rotor is 165 kg . Operating speed of rotor is 500 rpm . Concept used: Write the expression for natural frequency of system. ω n = k M ...... (1) Here, ω n is the natural frequency of system, k is the equivalent stiffness of tail and tail rotor and M is the equivalent mass of tail and tail rotor. Write the expression for amplitude of vibration of motor. x m = e ( m / M ) ( ω f / ω n ) 2 1 − ( ω f / ω n ) 2 ..... (2) Here, x m is the amplitude of vibration of motor, e is the distance between mass center of rotor and its axis of rotation, m is the mass of rotor, M is the mass of tail and tail rotor and ω f is the forced frequency of rotor Write the expression for the forced response of tail. x p ( t ) = x m sin ( ω t − ϕ ) ...... (3) Here, x p ( t ) is the forced response of tail section and ϕ is the phase difference. Calculation: Substitute 8800 lb/ft for k and 165 lb 32.2 ft/s 2 for M in equation (1). ω n = 8800 ( 165 32.2 ) = 41.44 rad/s Substitute 41.44 rad/s for ω n , 500 rpm for ω f , 1 lb for m , 6 in for e and 165 lb for M in equation (2)

12th Edition

ISBN: 9781259977237

Author: BEER

Publisher: MCG

See similar textbooks

Concept explainers

Free Damped Vibrations

Free vibration is a type of vibration in which the external force is removed after giving the initial displacement to a body/system. In other words, in this type of vibration, force is applied at the initial displacement of the system, and after the init…

Question

Chapter 19.4, Problem 19.120P

To determine

The forced response of the tail section.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 19.4, Problem 19.119P

Chapter 19.4, Problem 19.121P

Students have asked these similar questions

Q.1 The system shown below consists of a block of mass m and a cylinder of mass 2m. The block and the cylinder are joined by pulley arrangement as show. The coordinates x and y are dependent in which x = 2y. Use m= 2 kg. The angle 0 is any value between 0 and 90°. If the cylinder is pushed downward 20 mm and released a) find the subsequent response of the block, x(1) b) find the displacement and velocity of the block at t = 1 s. The specific value of k is given in separate excel sheet. K=1985 2m IT, x = 2y

A large industrial cooling system of mass 160 kg is supported by six springsarranged in parallel. Two of the springs have equal stiffness of 30 kN/m, twohave stiffness of 15 kN/m, and the remaining two have stiffness of 10 kN/m.Determine:i) the total stiffness of the spring system;ii) the natural frequency of vibration of the motor;iii) the periodic time for the vibration.

Problem 5.1 A rotational spring-mass system is shown to the right. The slender rigid bar has a mass of 18 kg and a length of { = 1.5 m. The springs have stiffness k = 360 N/m. The bar has a displacement of –0.06 rad and a speed of –1.1 rad/s when a harmonic moment M (t) = (80 N·m)cos 16t is applied. Find the response of the system. Assume small displacements. You will need to solve the differential equation because the results from lecture and in the book are for M(t) = M sin wt, not M (t) = Mo cos wt. You do not have to derive the homogeneous solution from scratch; you may use equations from Module 3, but the coefficients will be different. k - 3 Мо cos ot Answer: 0 (t) = {0.0540 cos 10t – 0.110 sin 10t 0.114 cos 16t} rad k

Chapter 19 Solutions

Vector Mechanics For Engineers

Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - A particle moves in simple harmonic motion....Ch. 19.1 - Prob. 19.3P Ch. 19.1 - Prob. 19.4P Ch. 19.1 - Prob. 19.5P Ch. 19.1 - A 20-lb block is initially held so that the...Ch. 19.1 - Prob. 19.7P Ch. 19.1 - A simple pendulum consisting of a bob attached to...Ch. 19.1 - Prob. 19.9P Ch. 19.1 - A 5-kg fragile glass vase is surrounded by packing...

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 forced response of the tail section.

Solution Summary: The author explains the force response of the tail section.

Concept explainers

The forced response of the tail section.

Want to see the full answer?

Chapter 19 Solutions