/*! 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 42 A thin metal disk with mass \(2.... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 14: Problem 42

A thin metal disk with mass \(2.00 \times 10^{-3} \mathrm{kg}\) and radius 2.20 \(\mathrm{cm}\) is attached at its center to a long fiber (Fig. E14.42). The disk, when twisted and released, oscillates with a period of 1.00 s. Find the torsion constant of the fiber.

Short Answer

Expert verified
The torsion constant is approximately \(2.14 \times 10^{-6} \mathrm{N}\cdot\mathrm{m}/\mathrm{rad}\).

Step by step solution

01

Understand the Physical System

We have a metal disk attached to a fiber which acts like a torsion pendulum. When twisted, it oscillates with a known period. Our task is to find the torsion constant of the fiber, which quantifies its stiffness or resistance to twisting.
02

Identify the Key Equation

For a torsion pendulum, the period of oscillation \(T\) is related to the moment of inertia \(I\) and the torsion constant \(\kappa\) by the equation:\[T = 2\pi \sqrt{\frac{I}{\kappa}}.\] We rearrange this equation to solve for \(\kappa\): \[\kappa = \frac{4\pi^2 I}{T^2}.\]
03

Calculate the Moment of Inertia

The moment of inertia \(I\) for a thin disk about an axis through its center perpendicular to its plane is given by:\[I = \frac{1}{2} m r^2,\] where \(m = 2.00 \times 10^{-3} \mathrm{kg}\) is the mass and \(r = 0.022 \mathrm{m}\) is the radius. Substituting these values:\[I = \frac{1}{2} \times 2.00 \times 10^{-3} \times (0.022)^2 \approx 5.41 \times 10^{-8} \mathrm{kg}\cdot\mathrm{m}^2.\]
04

Compute the Torsion Constant

Using the calculated moment of inertia and given period \(T = 1.00 \mathrm{s}\), substitute these into the equation for \(\kappa\):\[\kappa = \frac{4\pi^2 \times 5.41 \times 10^{-8}}{(1.00)^2}.\]Calculating this gives:\[\kappa \approx 2.14 \times 10^{-6} \mathrm{N}\cdot\mathrm{m}/\mathrm{rad}.\]

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 91影视!

Key Concepts

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

Torsion Constant
In the context of a torsion pendulum, the torsion constant is an important concept to grasp. It represents the stiffness or resistance a material has to twisting. Imagine twisting a fiber or a rubber band; the torsion constant measures how quickly it tries to return to its original state.

The greater the torsion constant, the stiffer the fiber and the harder it is to twist. For a pendulum, this impacts how quickly or slowly it will oscillate when released.
  • A high torsion constant indicates less flexibility.
  • A low torsion constant means the material bends or twists more easily.
Understanding the torsion constant helps in designing materials and systems where controlled twisting is essential. You calculate the torsion constant using the formula:\[\kappa = \frac{4\pi^2 I}{T^2},\]where \(I\) is the moment of inertia and \(T\) is the period of oscillation. This formula shows the relationship between how the mass distribution (moment of inertia) and the period of oscillation contribute to the torsion constant.
Moment of Inertia
Moment of inertia is a fundamental concept in physics, especially in rotational dynamics. It tells us how the mass of an object is distributed with respect to a given axis. Think of it as the rotational equivalent of mass in linear motion. For the disk in this example, its moment of inertia determines how easy or hard it is for the disk to start rotating.

More mass farther from the axis means a larger moment of inertia, making the object harder to spin. For a thin disk like the one in our problem, the formula to calculate moment of inertia is:\[I = \frac{1}{2} m r^2,\]where \(m\) is the mass of the disk and \(r\) is the radius.
  • Use this formula to find \(I\) by plugging in the known values.
  • In our exercise, \(m = 2.00 \times 10^{-3} \) kg and \(r = 0.022 \) m, resulting in \(I \approx 5.41 \times 10^{-8} \) kg路m虏.
The concept is crucial in understanding how various forces and torques affect rotating systems.
Oscillation Period
The oscillation period, represented as \(T\), is the time it takes for one complete cycle of the pendulum's motion. In other words, it's the time from one peak of the swing to the next. For a torsion pendulum, the period depends on both the moment of inertia and the torsion constant.

Calculating the period helps to understand how changes in the moment of inertia or torsion constant alter the motion. The equation used is:\[T = 2\pi \sqrt{\frac{I}{\kappa}},\]allowing us to see how these variables play off each other.
  • If the torsion constant increases, making the fiber stiffer, the period decreases. This means faster oscillations.
  • On the other hand, increasing the moment of inertia increases the period, leading to slower oscillations.
The oscillation period provides critical insight into the dynamics of a torsion pendulum and helps in designing systems with precise oscillatory behaviors.

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

CP A rifle bullet with mass 8.00 \(\mathrm{g}\) and initial horizontal velocity 280 \(\mathrm{m} / \mathrm{s}\) strikes and embeds itself in a block with mass 0.992 \(\mathrm{kg}\) that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 18.0 \(\mathrm{cm}\) . After the impact, the block moves in SHM. Calculate the period of this motion.

CP SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a 0.200 -kg pan is suspended from its lower end. A butcher drops a 2.2 -kg steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

A 0.0200 -kg bolt moves with SHM that has an amplitude of 0.240 \(\mathrm{m}\) and a period of 1.500 \(\mathrm{s}\) . The displacement of the bolt is \(+0.240 \mathrm{m}\) when \(t=0 .\) Compute (a) the displacement of the bolt when \(t=0.500 \mathrm{s} ;(\mathrm{b})\) the magnitude and direction of the force acting on the bolt when \(t=0.500 \mathrm{s} ;\) (c) the minimum time required for the bolt to move from its initial position to the point where \(x=-0.180 \mathrm{m} ;\) (d) the speed of the bolt when \(x=-0.180 \mathrm{m}\) .

The tip of a tuning fork goes through 440 complete vibrations in 0.500 s. Find the angular frequency and the period of the motion.

BID (a) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note that is sung. If someone sings the note \(B\) flat, which has a fre- quency of 466 Hz, how much time does it take the person's vocal cords to vibrate through one complete cycle, and what is the angu- lar frequency of the cords? (b) Hearing. When sound waves strike the eardrum, this membrane vibrates with the same frequency as the sound. The highest pitch that typical humans can hear has a period of 50.0\(\mu \mathrm{s} .\) What are the frequency and angular frequency of the vibrating eardrum for this sound? (c) Vision. When light having vibrations with angular frequency ranging from \(2.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) to \(4.7 \times 10^{15} \mathrm{rad} / \mathrm{s}\) strikes the retina of the eye, it stimulates the receptor cells there and is perceived as visible light. What are the limits of the period and frequency of this light? (d) Ultrasound. High-frequency sound waves (ultrasound) are used to probe the interior of the body, much as x rays do. To detect small objects such as tumors, a frequency of around 5.0 \(\mathrm{MHz}\) is used. What are the period and angular frequency of the molecular vibrations caused by this pulse of sound?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Modern Physics

Read Explanation

Medical Physics

Read Explanation

Electricity

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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.