/*! 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 25 A long wire hangs vertically wit... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 25

A long wire hangs vertically with its upper end clamped. A torque of \(8 \mathrm{Nm}\) applied to the free end twists it through \(45^{\circ}\). The potential energy of the twisted wire is (1) \(\pi \mathrm{J}\) (2) \(\frac{\pi}{2} \mathrm{~J}\) (3) \(\frac{\pi}{4} \mathrm{~J}\) (4) \(\frac{\pi}{8} \mathrm{~J}\)

Short Answer

Expert verified
The potential energy of the twisted wire is \( \frac{\pi}{2} \mathrm{~J} \), option (2).

Step by step solution

01

Understanding the Problem

We are given a wire that, when a torque of 8 Nm is applied, twists through 45 degrees. We need to find the potential energy stored in the wire due to this twist.
02

Convert Angles to Radians

The angular twist is provided as 45 degrees. First, we convert this angle into radians since the potential energy formula involves radians. To convert degrees to radians, use the formula: \[ \theta = \frac{\pi}{180} \times \text{degrees} \]Thus, \[ \theta = \frac{\pi}{180} \times 45 = \frac{\pi}{4} \text{ radians} \]
03

Use Potential Energy Formula

The potential energy \( U \) stored in a twisted wire is given by the formula: \[ U = \frac{1}{2} \tau \theta \]where \( \tau \) is the torque and \( \theta \) is the angular deformation in radians.
04

Plug Values into the Formula

Substitute \( \tau = 8 \text{ Nm} \) and \( \theta = \frac{\pi}{4} \text{ radians} \) into the potential energy formula: \[ U = \frac{1}{2} \times 8 \times \frac{\pi}{4} \]This simplifies to \[ U = \frac{1}{2} \times 8 \times \frac{\pi}{4} = 2 \times \frac{\pi}{4} = \frac{\pi}{2} \mathrm{~J} \]
05

Compare with Options

The computed potential energy is \( \frac{\pi}{2} \) Joules which matches with option (2).

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.

Torque
Torque is a measure of the rotational force applied to an object, leading to its twisting or turning. It's similar to how regular force causes linear acceleration. Torque depends on both the magnitude of the force and the distance from the pivot point (or axis of rotation). When you apply a torque of 8 Nm on a wire, it induces a rotational effect, essentially twisting the wire.
The formula for torque \( \tau \) is given by:
  • \( \tau = r \times F \times \sin(\theta) \)
where \( r \) is the distance from the pivot (or the radius of rotation), \( F \) is the force applied, and \( \theta \) is the angle between the force and the lever arm.In the context of a twisting wire, torque is that twisting force applied at one end, while the other is held fixed. You can visualize it like turning a screwdriver; the handle applies torque to turn the screw. Understanding this concept helps explain the relationship between torque and potential energy stored in the twisted wire.
Angular Deformation
Angular deformation refers to how much an object has rotated or twisted due to an applied torque. This deformation is measured in angular units such as degrees or radians. When a torque is applied, the wire twists and its shape deforms angularly, a concept known as angular deformation.
The equation for angular deformation due to torque in a wire is:
  • \( \theta = \frac{\tau}{J} \)
where \( J \) is the torsional rigidity of the wire, which is not required to solve the problem but crucial in practical applications.
Angular deformation is crucial since it helps us determine how much the object has twisted, which is essential for understanding the energy stored in the wire. It's like measuring how much a spring stretches when pulled. In the example problem, an angular deformation of \(45^{\circ}\) indicates the extent of the wire鈥檚 twist.
Radians Conversion
Radians are the standard unit of angular measure used in many areas of mathematics. Converting angles from degrees to radians is vital, as radians are necessary for most physics equations involving rotation or circular motion.
To convert degrees to radians, use the following conversion formula:
  • \( \theta = \frac{\pi}{180} \times \text{degrees} \)
For example, the 45-degree twist in the problem becomes \(\frac{\pi}{4}\) radians when converting from degrees.
Understanding radians is critical for computing physical quantities using formulas like those for potential energy and torque. Radians provide a direct and naturally scaled measure of angular displacements since they relate directly to the arc length subtended by the angle on a unit circle. Always remember:
  • 360 degrees equals \(2\pi\) radians
In any physics problem involving angles and rotation, radians allow you to utilize mathematical and physical formulas more effectively.

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

A thick rope of density \(\rho\) and length \(L\) is hung from a rigid support. The increase in length of the rope due to its own weight is ( \(Y\) is the Young's modulus) (1) \(\frac{0.1}{4 Y} \rho L^{2} g\) (2) \(\frac{1}{2 Y} \rho L^{2} g\) (3) \(\frac{\rho L^{2} g}{Y}\) (4) \(\frac{\rho L g}{Y}\)

Two wires of the same material and same mass are stretched by the same force. Their lengths are in the ratio \(2: 3\). Their elongations are in the ratio (1) \(3: 2\) (2) \(2: 3\) (3) \(4: 9\) (4) \(9: 4\)

A copper wire and a steen wire of the same cross-sectional grea and length are joined end-to-end (at one end). Equal gind opposite longitudinal forces are applied to the free ends giving a total elongation of \(l .\) Then the two wires will have (l) same stress and same strain (2) same stress and different strains (3) different stresses and same strain (4) different stresses and different strains

A wire is suspended vertically from a rigid support. When loaded with a steel weight in air, the wire extends by \(16 \mathrm{~cm}\). When the weight is completely immersed in water, the extension is reduced to \(14 \mathrm{~cm}\). The relative density of the material of the weight is (1) \(2 \mathrm{~g} / \mathrm{cm}^{3}\) (2) \(6 \mathrm{~g} / \mathrm{cm}^{3}\) (3) \(8 \mathrm{~g} / \mathrm{cm}^{3}\) (4) \(16 \mathrm{~g} / \mathrm{cm}^{3}\)

On applying a stress of \(x \mathrm{~N} / \mathrm{m}^{2}\), the length of wire of some material gets doubled. Value of Young's modulus for the material of wire in \(\mathrm{N} / \mathrm{m}^{2}\), is (assume Hooke's law to be valid and go for approx. results) (1) \(x\) (2) \(2 x\) (3) \(x / 2\) (4) Insufficient information
See all solutions

Recommended explanations on Physics Textbooks

Electric Charge Field and Potential

Read Explanation

Medical Physics

Read Explanation

Electricity

Read Explanation

Oscillations

Read Explanation

Torque and Rotational Motion

Read Explanation

Radiation

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.