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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

Expert verified
(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

01

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: \[ \theta = 3.34 \times \frac{\pi}{180} \approx 0.0583 \text{ radians} \]
02

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 \( \tau = \kappa \cdot \theta \).
03

Calculate the Torsion Constant

Using the relationship \( \tau = \kappa \cdot \theta \), solve for the torsion constant \( \kappa \): \[ \kappa = \frac{\tau}{\theta} = \frac{0.5076}{0.0583} \approx 8.70 \text{ Nm/rad} \]
04

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, \[ I = \frac{1}{2} \times 6.50 \times (0.12)^2 = 0.0468 \text{ kg m}^2 \]
05

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, \[ \omega = \sqrt{\frac{8.70}{0.0468}} \approx 13.6 \text{ rad/s} \]
06

Determine the Period of Oscillation

The period \( T \) of oscillation is given by \( T = \frac{2\pi}{\omega} \). Calculating, \[ T = \frac{2\pi}{13.6} \approx 0.462 \text{ seconds} \]
07

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: \[ \theta(t) = 0.0583 \cos(13.6t) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A 1.35-kg object is attached to a horizontal spring of force constant 2.5 N/cm. The object is started oscillating by pulling it 6.0 cm from its equilibrium position and releasing it so that it is free to oscillate on a frictionless horizontal air track. You observe that after eight cycles its maximum displacement from equilibrium is only 3.5 cm. (a) How much energy has this system lost to damping during these eight cycles? (b) Where did the "lost" energy go? Explain physically how the system could have lost energy.

A 175-g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 155 N/m. At the instant you make measurements on the glider, it is moving at 0.815 m/s and is 3.00 cm from its equilibrium point. Use \({energy\ conservation}\) to find (a) the amplitude of the motion and (b) the maximum speed of the glider. (c) What is the angular frequency of the oscillations?

A small block is attached to an ideal spring and is moving in SHM on a horizontal frictionless surface. The amplitude of the motion is 0.165 m. The maximum speed of the block is 3.90 m/s. What is the maximum magnitude of the acceleration of the block?

A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/m. At \(t\) = 0 the spring is neither stretched nor compressed and the block is moving in the negative direction at 12.0 m/s. Find (a) the amplitude and (b) the phase angle. (c) Write an equation for the position as a function of time.

An object with height \(h\), mass \(M\), and a uniform cross-sectional area \(A\) floats upright in a liquid with density \(\rho\). (a) Calculate the vertical distance from the surface of the liquid to the bottom of the floating object at equilibrium. (b) A downward force with magnitude \(F\) is applied to the top of the object. At the new equilibrium position, how much farther below the surface of the liquid is the bottom of the object than it was in part (a)? (Assume that some of the object remains above the surface of the liquid.) (c) Your result in part (b) shows that if the force is suddenly removed, the object will oscillate up and down in SHM. Calculate the period of this motion in terms of the density \(\rho\) of the liquid, the mass \(M\), and the cross- sectional area A of the object. You can ignore the damping due to fluid friction (see Section 14.7).

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