Question

The cranking device shown consists of a mass-spring system of stiffness \(k\) and mass \(m\) that is pin-connected to a massless rod which, in turn, is pin- connected to a wheel at radius \(R\), as indicated. If the mass moment of inertia of the wheel about an axis through the hub is \(I_{O}\), determine the natural frequency of the system. (The spring is unstretched when connecting pin is directly over hub ' \(O\) '.)

Short answer

Answer: The natural frequency (\(\omega_n\)) of the mass-spring system connected to a wheel is given by the equation: \(\omega_n = \sqrt{\frac{k}{m + \frac{I_{O}}{R^2}}}\) where, \(k\) is the stiffness of the spring, \(m\) is the mass, \(I_{O}\) is the mass moment of inertia of the wheel, and \(R\) is the radius.

Step-by-step solution

Formulate the Equation of Motion for the System

Begin by formulating the equation of motion for the mass-spring system by considering the forces acting on the system. For the mass-spring system, there is the spring force, given by Hooke's Law, and the gravitational force acting on the mass. For the wheel, there is the torque due to the spring force acting at radius \(R\). The equation of motion for the system can be represented as: \(ma = k(\Delta x) - mg\) Where \(a\) is the acceleration of the mass, \(\Delta x\) is the elongation of the spring, and \(g\) is the gravitational acceleration.

Determine the Torque on the Wheel

The torque on the wheel is related to the spring force acting at radius \(R\). To calculate the torque \(\tau\), we can use the equation: \(\tau = R \cdot k(\Delta x)\)

Relate the Torque to the Angular Acceleration of the Wheel

Next, relate the torque to the angular acceleration of the wheel, \(\alpha\). We can use the equation: \(\tau = I_{O}\alpha\) Which can also be represented as: \(R \cdot k(\Delta x) = I_{O}\alpha\)

Relate the Linear and Angular Acceleration

Since the mass and the wheel are connected, their accelerations are related through the radius \(R\). Express the linear acceleration \(a\) in terms of the angular acceleration \(\alpha\): \(a = R\alpha\)

Combine Equations and Eliminate Variables

Now, we can combine our equations to eliminate variables and solve for the natural frequency. Start by substituting the expression for \(a\) from Step 4 into the equation of motion: \(m(R\alpha) = k(\Delta x) - mg\) Next, substitute the expression for \(\alpha\) from Step 3 into the equation: \(m(R\frac{R \cdot k(\Delta x)}{I_{O}}) = k(\Delta x) - mg\) Now, solve for the elongation of the spring, \(\Delta x\): \(\Delta x = \frac{mR^2 g}{kR^2 + I_{O}}\)

Find the Natural Frequency

The natural frequency, \(\omega_n\), can be found using the elongation of the spring and the system parameters. We can use the equation: \(\omega_n = \sqrt{\frac{k}{m + \frac{I_{O}}{R^2}}}\) By plugging in the expression for \(\Delta x\), we can find the natural frequency of the system.

Final Answer

The natural frequency of the mass-spring system connected to the wheel is given by: \(\omega_n = \sqrt{\frac{k}{m + \frac{I_{O}}{R^2}}}\)

Key concepts

Equation of Motion

When dealing with mechanical systems, the equation of motion is pivotal. It describes how an object moves under the influence of various forces. For a mass-spring system, the forces considered are primarily the spring force and gravitational force.

According to the principles of mechanics, this can often be expressed as: \( ma = k(\Delta x) - mg \) where:

• \( m \) is the mass of the object.
• \( a \) is the linear acceleration.
• \( k \) is the spring stiffness.
• \( \Delta x \) is the change in spring length.
• \( g \) is the acceleration due to gravity.

Each of these components works in tandem to define the actual movement of the system. By understanding these forces and applying Newton's second law, we can predict and describe the behavior of the system.

Mass-Spring System

A mass-spring system consists of a mass attached to a spring, which exerts a force when compressed or stretched. It is a classic model used to demonstrate harmonic motion.

The behavior of the mass-spring system is governed by Hooke's Law, where the force exerted by the spring is directly proportional to its extension or compression. It manifests as: \( F_s = k \times \Delta x \) where \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( \Delta x \) is the displacement from the equilibrium position.

In our scenario, the system is part of a cranking device, coupling the spring's linear force with a rotational motion through a wheel mechanism. This connectivity allows for the transformation and transportation of mechanical energy.

Moment of Inertia

Moment of inertia is a measure of an object's resistance to changes in its rotation. It plays a similar role in rotational motion as mass does in linear motion. For a rotating wheel, it is crucial for understanding how the wheel responds to applied torques.

In the given system, the wheel's moment of inertia \( I_{O} \) about its axis significantly influences the dynamics of the mass-spring system. When torque is applied via the spring force: \( \tau = I_{O} \alpha \) where \( \tau \) is torque and \( \alpha \) is angular acceleration. The larger the moment of inertia, the more torque is required to achieve the same angular acceleration.

This principle is essential in solving for the natural frequency of the system, as it incorporates both linear and rotational dynamics to predict how the system will behave when set into motion.

Hooke's Law

Hooke's Law is fundamental in understanding how springs respond to forces. It states that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed, provided it doesn't exceed the spring's elastic limit.

The equation for Hooke's Law is \( F = -k \Delta x \) where \( F \) is the force applied in the direction of the spring's displacement, \( k \) is the spring constant, and \( \Delta x \) is the displacement.

In a mass-spring system, Hooke's Law helps us predict how much force the spring will exert back on the system when compressed or stretched. This law is pivotal when calculating the system's natural frequency, which involves evaluating how the spring's force contribution alters the motion of the mass.