/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 38 A uniform, solid metal disk of m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A uniform, solid metal disk of mass \(6.50 \mathrm{~kg}\) and diameter \(24.0 \mathrm{~cm}\) hangs in a horizontal plane, supported at its center by a vertical metal wire. You find that it requires a horizontal force of \(4.23 \mathrm{~N}\) 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
The torsion constant for the metal wire is 8.71 N.m/rad. The frequency and period of the torsional oscillations of the disk are respectively 13.92 rad/s and 0.450 s. The equation of motion for \( \theta(t) \) for the disk is \( \theta(t) = 0.0583 rad * cos(13.92t) \).

Step by step solution

01

Calculate the Torsion Constant

First, we need to calculate the torque which is given by \( Torque = F * r \), where F is the force and r is the radius of the disk. The radius (r) can be calculated from the diameter which is given as 24 cm. So, \( r = \frac{24}{2} = 12 cm = 0.12 m\). The force (F) is given as 4.23 N. So, \( Torque = 4.23 N * 0.12 m = 0.5076 N.m \). The angle is given as 3.34 degrees, we need to convert it to radians, where we know that \( Rad = \frac{degree * \pi}{180} \). So, \( \theta = \frac{3.34 * \pi}{180} = 0.0583 rad \). Now, we can calculate the torsion constant (k) using \( Torque = k * \theta \). From that, \( k = \frac{Torque}{\theta} = \frac{0.5076 N.m}{0.0583 rad} = 8.71 N.m/rad \).
02

Calculate the Frequency and Period of the Disk

The moment of inertia (I) of the disk is found with \( I = \frac{1}{2} m r^2 \), where m is the mass and r is the radius. m is given as 6.5 kg, and r we've calculated as 0.12 m. So, \( I = \frac{1}{2} * 6.5 kg * (0.12 m)^2 = 0.0468 kg.m^2 \). The angular frequency ( \( \omega \) ) is calculated using \( \omega = \sqrt{\frac{k}{I}} \). So, \( \omega = \sqrt{\frac{8.71 N.m/rad}{0.0468 kg.m^2}} = 13.92 rad/s \). Now we calculate the period (T) using \( T = \frac{2 \pi}{\omega} \). Thus, \( T = \frac{2\pi}{13.92 rad/s} = 0.450 s \).
03

Write the Equation of Motion for the Disk

The equation of motion for \( \theta(t) \) is given by \( \theta(t) = \theta_{max} cos(\omega t + \phi) \), where \( \theta_{max} \) is the maximum displacement (initial angle), \( \omega \) is the angular frequency and \( \phi \) is the phase constant. The angle is not given so it can be considered as zero and if the disc is released from rest, \( \phi \) will be \( \phi = 0 \). So, the equation is \( \theta(t) = 0.0583 rad * cos(13.92t) \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torsion Constant
When we think about the torsion constant, consider it as a measure of stiffness for rotational movement. The torsion constant, often denoted as ul
  • (k)
    • quantifies how resistant a wire is to twisting. The formula to calculate it is quite straightforward:
      Torque \(\ \text{(in N.m)} = k * \theta \text{(in radians)}\)
      where the torque is the product of the force applied tangentially to a point at a certain radius (like the rim of a disk) and the radius itself. Converting the given angle from degrees to radians is a standard step since mathematical operations involving oscillations often use radians.

      For example, if a disk requires a certain amount of force at its rim to be twisted by a small angle, by knowing the applied torque and the angle, you directly compute the torsion constant using
      k = \frac{Torque}{\theta}
      . This can help in engineering applications where understanding material deformation properties under twisting forces is crucial.
    Moment of Inertia
    The concept of moment of inertia is essential in rotational dynamics, akin to mass in linear motion. For a rotating object, it provides a measure of how difficult it is to change the object's rotational motion. The moment of inertia, denoted as
    • (I)
    ,relies on two factors: the mass of the object and how this mass is distributed relative to the axis of rotation. For simple geometrical objects like disks, formulas are derived to make this calculation straightforward; for a disk, it is given by
    I = \frac{1}{2} m•r^2
    . In this equation,
    • \(m\)
    is the mass, while
    • \(r\)
    is the radius of the disk.

    In practical terms, this means that as the mass and its distribution away from the center increase, it becomes harder to start or stop the rotation. In the context of our disk: the moment of inertia directly influences the nature of its oscillations when joined with a torsional wire.
    Angular Frequency
    Angular frequency is a concept that helps describe how fast something rotates or oscillates. It is an important concept in physics, especially when discussing oscillations.

    The angular frequency, often symbolized by
    • \(\omega\)
    , instructs us about how many cycles or rotations take place per unit of time. For torsional oscillations, it is computed using this fundamental formula:
    \omega = \sqrt{\frac{k}{I}}
    ,where
    • \(k\)
    is the torsion constant, and
    • \(I\)
    is the moment of inertia.

    This relationship is critical since it links the inherent properties of the system—the stiffness of the wire and the rotational inertia of the disk—to the oscillatory motion's quickness. By determining the angular frequency, one can also find the period of oscillation, as
    T = \frac{2\pi}{\omega}
    . Overall, angular frequency provides insights into the dynamic behavior of oscillating systems, showcasing how the system's physical characteristics underpin its motion.

    One App. One Place for Learning.

    All the tools & learning materials you need for study success - in one app.

    Get started for free

    Most popular questions from this chapter

    Quantum mechanics is used to describe the vibrational motion of molecules, but analysis using classical physics gives some useful insight. In a classical model the vibrational motion can be treated as SHM of the atoms connected by a spring. The two atoms in a diatomic molecule vibrate about their center of mass, but in the molecule HI, where one atom is much more massive than the other, we can treat the hydrogen atom as oscillating in SHM while the iodine atom remains at rest. (a) A classical estimate of the vibrational frequency is \(f=7 \times 10^{13} \mathrm{~Hz}\). The mass of a hydrogen atom differs little from the mass of a proton. If the HI molecule is modeled as two atoms connected by a spring, what is the force constant of the spring? (b) The vibrational energy of the molecule is measured to be about \(5 \times 10^{-20} \mathrm{~J}\). In the classical model, what is the maximum speed of the H atom during its SHM? (c) What is the amplitude of the vibrational motion? How does your result compare to the equilibrium distance between the two atoms in the HI molecule, which is about \(1.6 \times 10^{-10} \mathrm{~m} ?\)

    An unhappy \(0.300 \mathrm{~kg}\) rodent, moving on the end of a spring with force constant \(k=2.50 \mathrm{~N} / \mathrm{m},\) is acted on by a damping force \(F_{x}=-b v_{x}\). (a) If the constant \(b\) has the value \(0.900 \mathrm{~kg} / \mathrm{s},\) what is the frequency of oscillation of the rodent? (b) For what value of the constant \(b\) will the motion be critically damped?

    A \(5.00 \mathrm{~kg}\) partridge is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is pulled down \(0.100 \mathrm{~m}\) below its equilibrium position and released, it vibrates with a period of \(4.20 \mathrm{~s}\). (a) What is its speed as it passes through the equilibrium position? (b) What is its acceleration when it is \(0.050 \mathrm{~m}\) above the equilibrium position? (c) When it is moving upward, how much time is required for it to move from a point \(0.050 \mathrm{~m}\) below its equilibrium position to a point \(0.050 \mathrm{~m}\) above it? (d) The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?

    Consider the system of two blocks and a spring shown in Fig. \(\mathrm{P} 14.66 .\) The horizontal surface is friction less, but there is static friction between the two blocks. The spring has force constant \(k=150 \mathrm{~N} / \mathrm{m} .\) The masses of the two blocks are \(m=0.500 \mathrm{~kg}\) and \(M=4.00 \mathrm{~kg} .\) You set the blocks into motion by releasing block \(M\) with the spring stretched a distance \(d\) from equilibrium. You start with small values of \(d,\) and then repeat with successively larger values. For small values of \(d,\) the blocks move together in SHM. But for larger values of \(d\) the top block slips relative to the bottom block when the bottom block is released. (a) What is the period of the motion of the two blocks when \(d\) is small enough to have no slipping? (b) The largest value \(d\) can have and there be no slipping is \(d=8.8 \mathrm{~cm} .\) What is the coefficient of static friction \(\mu_{\mathrm{s}}\) between the surfaces of the two blocks?

    BIO Weighing Astronauts. This procedure has been used to "weigh" astronauts in space: A \(42.5 \mathrm{~kg}\) chair is attached to a spring and allowed to oscillate. When it is empty, the chair takes \(1.30 \mathrm{~s}\) to make one complete vibration. But with an astronaut sitting in it, with her feet off the floor, the chair takes \(2.54 \mathrm{~s}\) for one cycle. What is the mass of the astronaut?

    See all solutions

    Recommended explanations on Physics Textbooks

    View all explanations

    What do you think about this solution?

    We value your feedback to improve our textbook solutions.

    Study anywhere. Anytime. Across all devices.