Question

A uniform, solid metal disk of mass 6.50 kg and diameter 24.0 cm hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of 4.23 N tangent to the rim of the disk to turn it by 3.34\(^\circ\), thus twisting the wire. You now remove this force and release the disk from rest. (a) What is the torsion constant for the metal wire? (b) What are the frequency and period of the torsional oscillations of the disk? (c) Write the equation of motion for \(\theta(t)\) for the disk.

Short answer

(a) \(\kappa = 8.70\, \text{Nm/rad}\), (b) \(f \approx 2.16\, \text{Hz}, T \approx 0.462\, \text{s}\), (c) \(\theta(t) = 0.0583\cos(13.6t)\)

Step-by-step solution

Calculate the Angular Displacement in Radians

To find the angular displacement, first convert the angle from degrees to radians. The conversion factor is \( \pi/180 \). Thus, the angular displacement \( \theta \) in radians is given by: \[ \theta = 3.34 \times \frac{\pi}{180} \approx 0.0583 \text{ radians} \]

Define the Torque and Torsion Constant Relationship

The torque \( \tau \) required to twist the wire is the product of the force \( F \) and the radius \( r \) of the disk. Since the diameter of the disk is 24.0 cm, the radius is 0.12 m. So, \( \tau = F \cdot r = 4.23 \times 0.12 = 0.5076 \text{ Nm} \). The relationship between torque and torsion constant \( \kappa \) is \( \tau = \kappa \cdot \theta \).

Calculate the Torsion Constant

Using the relationship \( \tau = \kappa \cdot \theta \), solve for the torsion constant \( \kappa \): \[ \kappa = \frac{\tau}{\theta} = \frac{0.5076}{0.0583} \approx 8.70 \text{ Nm/rad} \]

Find the Moment of Inertia of the Disk

For a solid disk, the moment of inertia \( I \) about its center is given by the formula \( I = \frac{1}{2} m r^2 \). Here, \( m = 6.50 \text{ kg} \) and \( r = 0.12 \text{ m} \). So, \[ I = \frac{1}{2} \times 6.50 \times (0.12)^2 = 0.0468 \text{ kg m}^2 \]

Calculate the Natural Frequency of Oscillation

The angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{\kappa}{I}} \). From previous steps, \( \kappa = 8.70 \text{ Nm/rad} \) and \( I = 0.0468 \text{ kg m}^2 \). So, \[ \omega = \sqrt{\frac{8.70}{0.0468}} \approx 13.6 \text{ rad/s} \]

Determine the Period of Oscillation

The period \( T \) of oscillation is given by \( T = \frac{2\pi}{\omega} \). Calculating, \[ T = \frac{2\pi}{13.6} \approx 0.462 \text{ seconds} \]

Write the Equation of Motion for \(\theta(t)\)

The equation of motion for torsional oscillations can be described by \( \theta(t) = \theta_0 \cos(\omega t + \phi) \). With \( \theta_0 \) as the maximum initial angle, \( 0.0583 \) rad, and considering initial release from rest, \( \phi = 0 \). Thus, the equation is: \[ \theta(t) = 0.0583 \cos(13.6t) \]

Key concepts

Moment of Inertia

Moment of Inertia refers to the resistance of a body to angular acceleration, similar to how mass represents resistance to linear acceleration. It's crucial in determining how a rotating body behaves under various forces. For a solid disk like the one in the exercise, the moment of inertia (\(I\)) is calculated using the formula:
\[I = \frac{1}{2} m r^2\]where \(m\) is the mass of the disk, and \(r\) is its radius. In this case, given the mass is 6.50 kg and the radius is 0.12 m, the moment of inertia was calculated to be 0.0468 kg m². This value tells us how difficult it is to twist the disk and, thus, influences the frequency and amplitude of oscillations.

Torsion Constant

The Torsion Constant (\(\kappa\)) is a measure of the stiffness of a wire in resisting twists. It's similar to a spring constant in linear motion, but applies to rotational motion. The torsion constant defines the relationship between torque (\(\tau\)) and angular displacement (\(\theta\)) as follows:
\[\tau = \kappa \cdot \theta\]In the exercise, we calculated the torque using the force necessary to turn the disk tangent to its edge and the radius:
\[\tau = 4.23 \times 0.12 = 0.5076 \, \text{Nm}\]By dividing the torque by the angular displacement in radians, the torsion constant was found to be approximately 8.70 Nm/rad. A larger torsion constant indicates a more rigid wire or material.

Equation of Motion

The motion of the disk in the exercise is described using the rotational analogy to Hooke's Law, resulting in an equation of motion for torsional oscillation:
\[\theta(t) = \theta_0 \cos(\omega t + \phi)\]In this expression, \(\theta_0\) represents the maximum angular displacement, \(\omega\) is the angular frequency, and \(\phi\) represents the phase constant. Given the disk starts from rest, \(\phi\) is zero. Therefore, the equation simplifies to:
\[\theta(t) = 0.0583 \cos(13.6t)\]This equation characterizes how the angle of the disk changes over time, exhibiting periodic motion analogous to a spring-mass system.

Angular Frequency

Angular Frequency (\(\omega\)) is an essential parameter in oscillatory motion, representing how rapidly an object oscillates. It's directly linked to the torsion constant and moment of inertia, using the formula:
\[\omega = \sqrt{\frac{\kappa}{I}}\]Substituting the calculated torsion constant (8.70 Nm/rad) and moment of inertia (0.0468 kg m²) from earlier, the angular frequency was found to be approximately 13.6 rad/s. This figure tells us how fast the disk undergoes torsional oscillations, affecting both the speed and energy of the motion.

Period of Oscillation

The Period of Oscillation (\(T\)) is the time taken for one complete cycle of motion. It's inversely related to angular frequency and provides insight into the cycle's duration. The period can be calculated using:
\[T = \frac{2\pi}{\omega}\]Applying this formula with the angular frequency previously found (13.6 rad/s), the period of oscillation is approximately 0.462 seconds. This short interval indicates a rapid oscillation, showing the wire-disk system returns to its starting position within less than half a second.