/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 113 A motor of mass \(M\) is support... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A motor of mass \(M\) is supported by springs with an equivalent spring constant \(k .\) The unbalance of its rotor is equivalent to a mass \(m\) located at a distance \(r\) from the axis of rotation. Show that when the angular velocity of the motor is \(\omega_{f}\), the amplitude \(x_{m}\) of the motion of the motor is $$ x_{m}=\frac{r(m / M)\left(\omega_{f} / \omega_{n}\right)^{2}}{1-\left(\omega_{f} / \omega_{n}\right)^{2}} $$ where \(\omega_{n}=\sqrt{k / M}\)

Short Answer

Expert verified
The amplitude of motion is: \[x_{m} = \frac{r(m / M)(\omega_{f} / \omega_{n})^{2}}{1 - (\omega_{f} / \omega_{n})^{2}}\]

Step by step solution

01

Identify Natural Frequency

Determine the natural frequency \( \omega_n \) of the system. This is given by \( \omega_n = \sqrt{\frac{k}{M}} \). This expression represents the frequency at which the system would naturally oscillate without any external forces.
02

Define Displacement Due to Unbalance

Consider the unbalanced rotor with mass \( m \) at a distance \( r \) from the axis. The centrifugal force due to this unbalance is \( F = m r \omega_{f}^{2} \). This force results in a displacement of the motor, leading to a motion amplitude \( x_m \).
03

Determine Motion Equation

The system can be described by the equation of motion for forced vibrations: \( M a = F - k x \), where \( a \) is the acceleration, \( F \) is the force (centrifugal in this case), and \( x \) is the displacement.
04

Solve for Amplitude Using Steady-State Condition

In steady state, the acceleration \( a = \omega_{f}^{2} x \). Substitute the force and rearrange: \[ M \omega_{f}^{2} x = m r \omega_{f}^{2} - k x \]Rearrange to find \( x \):\[ x_{m} = \frac{m r \omega_{f}^{2}}{k - M \omega_{f}^{2}} \]
05

Substitute Natural Frequency

Replace \( k \) with \( M \omega_{n}^{2} \) (since \( \omega_{n} = \sqrt{\frac{k}{M}} \)), yielding:\[ x_{m} = \frac{m r \omega_{f}^{2}}{M \omega_{n}^{2} - M \omega_{f}^{2}} \]Simplify further to obtain:\[ x_{m} = \frac{m r \omega_{f}^{2}}{M (\omega_{n}^{2} - \omega_{f}^{2})} \]
06

Simplify Final Expression

Finally, factor out \( M \) and simplify:\[ x_{m} = r \frac{(m / M) \omega_{f}^{2}}{\omega_{n}^{2} - \omega_{f}^{2}} \]This simplifies to the desired expression:\[ x_{m} = \frac{r (m / M) (\omega_{f} / \omega_{n})^{2}}{1 - (\omega_{f} / \omega_{n})^{2}} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Natural Frequency
In mechanical systems, the natural frequency refers to the frequency at which a system tends to oscillate when disturbed and then allowed to move without any further external forces. For the setup described in the problem, the natural frequency \( \omega_n \) is a key parameter. It is determined by the properties of the motor and the supporting springs. Mathematically, the natural frequency is given by the formula: \[ \omega_n = \sqrt{\frac{k}{M}} \] Here, \(k\) is the spring constant, which measures the stiffness of the springs, and \(M\) is the mass of the motor. This expression tells us that the natural frequency depends on how easily the system springs back to equilibrium when displaced. When oscillating at this frequency, a system experiences its maximum amplitude of vibration if not damped. Understanding the natural frequency is crucial for evaluating how the system will respond to external forces such as those induced by an unbalanced motor.
Centrifugal Force
Centrifugal force comes into play in systems involving rotational motion. In the scenario given, the motor's rotor is slightly unbalanced, meaning there is a mass \(m\) at a perpendicular distance \(r\) from the axis of rotation. This unbalance causes what is known as centrifugal force, calculated with the formula: \[ F = m r \omega_{f}^{2} \] Where \( \omega_f \) is the angular velocity of the motor. This force acts outward from the axis of rotation, as if trying to "flee" from the center. It can cause parts of the machine, like the motor housing, to vibrate unnaturally. Understanding centrifugal force is pivotal in predicting these kinds of forced vibrations and determining their effects on system motion.
Steady-State Vibrations
Steady-state vibrations occur when a system oscillates at a constant amplitude over time, under continued external excitation. In the context of the motor problem, after the initial transient responses have died out, the system reaches a steady state, dictated by the balance of centrifugal force and spring restoring forces. The amplitude \( x_m \) then becomes a function of both the exciting frequency \( \omega_f \) and the natural frequency \( \omega_n \): \[ x_{m} = \frac{r (m / M) (\omega_{f} / \omega_{n})^{2}}{1 - (\omega_{f} / \omega_{n})^{2}} \] This equation demonstrates how the amplitude of vibrations is influenced by the relationship between the forcing frequency and the natural frequency. If \( \omega_f \) is close to \( \omega_n \), the system can reach very large amplitudes. This scenario is referred to as resonance and can potentially lead to mechanical failure if not controlled or damped. Thus, understanding steady-state vibrations helps in designing systems to avoid detrimental resonance effects.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A slender \(5-\mathrm{kg}\) bar \(A B\) with a length of \(l=0.6 \mathrm{m}\) is connected to two collars, each of mass \(2.5 \mathrm{kg}\). Collar \(A\) is attached to a spring with a constant of \(k=1.5 \mathrm{kN} / \mathrm{m}\) and can slide on a horizontal rod, while collar \(B\) can slide freely on a vertical rod. Knowing that the system is in equilibrium when bar \(A B\) is vertical and that collar \(A\) is given a small displacement and released, determine the period of the resulting vibrations.

The 8 -kg uniform bar \(A B\) is hinged at \(C\) and is attached at \(A\) to a spring of constant \(k=500 \mathrm{N} / \mathrm{m}\). If end \(A\) is given a small displacement and released, determine \((a)\) the frequency of small oscillations, (b) the smallest value of the spring constant \(k\) for which oscillations will occur.

An 18 -lb block \(A\) slides in a vertical frictionless slot and is connected to a moving support \(B\) by means of a spring \(A B\) of constant \(k=8\) lb/ft. Knowing that the acceleration of the support is \(a=a_{m} \sin \omega_{j} t,\) where \(a_{m}=5\) \( \mathrm{ft} / \mathrm{s}^{2}\) and \(\omega_{f}=6\) rad/s, determine (a) the maximum displacement of block \(A,(b)\) the amplitude of the fluctuating force exerted by the spring on the block.

A 4 -kg collar can slide on a frictionless horizontal rod and is attached to a spring with constant \(k\). It is acted upon by a periodic force of magnitude \(P=P_{m} \sin \omega_{f} t,\) where \(P_{m}=9 \mathrm{N}\) and \(\omega_{f}=5 \mathrm{rad} / \mathrm{s}\). Determine the value of the spring constant \(k\) knowing that the motion of the collar has an amplitude of \(150 \mathrm{mm}\) and is \((a)\) in phase with the applied force, \((b)\) out of phase with the applied force.

Determine the amplitude and maximum acceleration of a particle that moves in simple harmonic motion with a maximum velocity of \(4 \mathrm{ft} / \mathrm{s}\) and a frequency of \(6 \mathrm{Hz}\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.