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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

Expert verified
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

01

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.
02

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: \(\tau = R \cdot k(\Delta x)\)
03

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: \(R \cdot k(\Delta x) = I_{O}\alpha\)
04

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 \(\alpha\): \(a = R\alpha\)
05

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, \(\Delta x\): \(\Delta x = \frac{mR^2 g}{kR^2 + I_{O}}\)
06

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.
07

Final Answer

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

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