Question

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

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

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.

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 \(T_G = m R \sin(\theta) - k R^2 \theta\).

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 \(\sin(\theta) \approx \theta\): \(2 \theta - 2kR \theta = \alpha\)

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: \(m\frac{d^2x}{dt^2} + 2kx = 0\)

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.

Key concepts

Rigid Body Dynamics

Rigid body dynamics is the study of the motion of solid objects that do not deform during their movement. When we talk about a rigid body, we assume that the internal structure remains constant no matter how fast it moves or what forces it experiences. This makes it easier to predict and analyze the movement since we only need to consider the motion of the entire body rather than its individual parts.
In the context of the given problem, we have a rigid rod and a flywheel. Both of these are assumed to move as single units without bending or twisting.
This simplification allows us to focus on the key forces and torques acting on the system to determine how it behaves when disturbed from rest.

Flywheel Dynamics

Flywheels are fascinating components, often used to store rotational energy. Practically, they are circular, heavy objects that work as energy reservoirs to maintain the angular momentum of a system. The dynamics of a flywheel revolves around its ability to regulate the speed of machinery by a balanced outflow and inflow of energy, often through accelerating and decelerating.In our exercise, the flywheel is an integral part of the system, influencing how the system responds to force application. Key factors contributing to flywheel dynamics include:

• Moment of Inertia: A measure of how much torque is needed for a desired angular acceleration about a rotation axis. For our flywheel, this is calculated using the formula: \(I = \frac{1}{2}mR^2\)
• Angular Acceleration: The rate of change of angular velocity, influenced by how the torque is applied.
• Energy Transfer: Stored or released by the flywheel to keep the mechanical system stable and functional.

Understanding these dynamics can help in predicting the motion under varied mechanical constraints and inputs.

Torque Balance

Torque is quite similar to force but it's specifically tailored to rotational systems. It's a measure of the force that would cause an object to rotate around an axis. To maintain equilibrium in rotational systems like the flywheel setup, a torque balance must be achieved.
In our setup, torque arises mainly from two sources:

• Gravitational Torque: This arises due to the weight of the flywheel acting at a distance from its pivot.
• Elastic Torque: This comes from the tension in the elastic belt wound around the flywheel.

The essence of the torque balance is ensuring the sum of all torques equals the rotational inertia times the angular acceleration: \(T_G = I \alpha\).
This balance helps us determine how these torques influence the overall motion, making it possible to predict how the system's state changes under various conditions.

Natural Frequency Calculation

Every mechanical system with a certain arrangement has a natural frequency which defines how it oscillates when disturbed. This frequency is intrinsic to the system's properties like mass distribution, stiffness, and geometry.
To determine the natural frequency, we turn to the standard form of the equation of motion. In our rigid body and flywheel setup, this translates to:\[ m \frac{d^2x}{dt^2} + 2kx = 0 \]Here, \(k\) represents the stiffness of the setup and \(m\) is related to the mass effects. To solve for the natural frequency \(\omega_n\), we use:\[ \omega_n = \sqrt{\frac{2k}{m}} \]This equation provides insight into how quickly the system will oscillate if perturbed. Understanding this concept is crucial, as it helps to design systems that either avoid resonance with external frequencies or leverage it to enhance performance.