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Chapter 2: Problem 21

The system shown consists of a rigid rod, a flywheel of radius \(R\) and mass \(m\), and an elastic belt of stiffness \(k\). Determine the natural frequency of the system. The belt is unstretched when \(\theta=0\).

Short Answer

Expert verified
Answer: The natural frequency of the system, denoted as \(\omega_n\), is given by \(\omega_n = \sqrt{\frac{2k}{m}}\), where \(k\) is the spring constant of the elastic belt and \(m\) is the mass of the flywheel.

Step by step solution

01

1. Analyze the Forces and Moments Acting on the System

First, we need to identify the forces acting on the flywheel and the belt: a gravitational force is acting on the flywheel, and an elastic force is acting on the belt. These forces will create a torque (moment) whose axis is through the center of the flywheel.
02

2. Define a force balance equation for the torque acting on the flywheel

We write the torque balance equation considering the gravitational and elastic forces: \(T_G = I \alpha \) Where: - \(T_G\) is the torque due to gravitational and elastic forces, - \(I\) is the moment of inertia of the flywheel: \(I = \frac{1}{2} m R^2\) - \(\alpha\) is the angular acceleration of the flywheel. For the gravity force, the torque is given by \(T_g = m R sin(\theta)\) while for the elastic force, we have \(T_e = k R (R\theta)\). Therefore, the total torque is \(T_G = m R \sin(\theta) - k R^2 \theta\).
03

3. Set up the torque balance equation and find the equation of motion

Replacing the total torque and the moment of inertia in the torque balance equation, we get: \(m R \sin(\theta) - k R^2 \theta = \frac{1}{2} m R^2 \alpha\) We can simplify this equation by dividing both sides by \(\frac{1}{2} m R^2\), resulting in: \(2 \sin(\theta) - 2kR \theta = \alpha\) Now, we can linearize this equation by assuming small angle \(\theta\) so that \(\sin(\theta) \approx \theta\): \(2 \theta - 2kR \theta = \alpha\)
04

4. Convert the equation of motion into a standard form

In order to find the natural frequency of the system, we will convert the equation of motion into a standard form: \(m\ddot{x} + kx = 0\) To do this, we will assume a one-to-one relationship between the displacement \(x\) and the angle \(\theta\): \(R\theta = x\) Which means: \(\alpha = \frac{d^2x}{dt^2}\) Replacing this into the linearized equation of motion, we get: \(2 \frac{x}{R} - 2kR \frac{x}{R} = \frac{d^2x}{dt^2}\) Simplifying the equation and rearranging, we get the standard form: \(m\frac{d^2x}{dt^2} + 2kx = 0\)
05

5. Solve the standard form equation for the natural frequency

The solution to the standard form equation has the form \(x(t) = A\cos(\omega_n t + \phi)\), where \(\omega_n\) is the natural frequency of the system. To find \(\omega_n\), we compare the coefficients of the standard form equation: \(\frac{m}{2} \omega_n^2 = k\) Now, we can solve for the natural frequency: \(\omega_n = \sqrt{\frac{2k}{m}}\) This gives us the natural frequency of the system.

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Most popular questions from this chapter

A railroad car of mass \(m\) is attached to a stop in a railroad yard. The stop consists of four identical metal rods of length \(L\), radius \(R\) and elastic modulus \(E\) that are arranged symmetrically and are fixed to a rigid wall at one end and welded to a rigid plate at the other. The plate is hooked to the stationary railroad car as shown. In a docking maneuver, a second car of mass \(m\) approaches the first at speed \(v_{1}\). If the second car locks onto the first upon contact, determine the response of the two car system after docking.

A \(12 \mathrm{~kg}\) spool that is \(1 \mathrm{~m}\) in radius is pinned to a viscoelastic rod of negligible mass with effective properties \(k=10 \mathrm{~N} / \mathrm{m}\) and \(c=8 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). The end of the rod is attached to a rigid support as shown. Determine the natural frequency of the system if the spool rolls without slipping.

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\) and a viscous damper whose coefficient is \(1 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (a) Determine the response of the horizontally configured system if the mass is displaced 2 meters to the right and released with a velocity of 4 \(\mathrm{m} / \mathrm{sec}\). Plot and label the response history of the system. (b) Determine the response and plot its history if the damping coefficient is \(5 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (c) Determine the response and plot its history if the damping coefficient is \(10 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\).

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\). If the coefficients of static and kinetic friction between the block and the surface it moves on are respectively \(\mu_{s}=\) \(0.12\) and \(\mu_{k}=0.10\), determine the drop in amplitude between successive periods during free vibration. What is the frequency of the oscillations?

Determine the natural period of a typical ice cube floating in water. Measure the dimensions of a typical cube from your refrigerator and calculate its natural frequency in water. (The dimensions may vary depending on your particular ice tray.) Confirm your "experiment." Place an ice cube in water, displace it slightly and release it. Make an approximate measure of the period of an oscillation with your wrist watch, or a stop watch if available. Repeat this operation several times and compare the average measured
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