/*! 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 12 A copper wire of negligible mass... [FREE SOLUTION] | 91Ó°ÊÓ

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A copper wire of negligible mass, \(1 \mathrm{~m}\) length and cross-sectional area \(10^{-6} \mathrm{~m}^{2}\) is kept on a smooth horizontal table with one end fixed. A ball of mass \(1 \mathrm{~kg}\) is attached to the other end. The wire and the ball are rotating with an angular velocity of \(20 \mathrm{rad} / \mathrm{s}\). If the elongation in the wire is \(10^{-3} \mathrm{~m}\). The Young's modulus of the wire is (1) \(4.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) (2) \(2.5 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) (3) \(2.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\) (4) \(4.5 \times 10^{11} \mathrm{~N} / \mathrm{m}^{2}\)

Short Answer

Expert verified
The Young's modulus of the wire is \(4.0 \times 10^{11} \mathrm{~N} / \mathrm{m}^2\).

Step by step solution

01

Identify Given Variables

First, let's identify the variables provided in the problem statement:- Length of wire, \( L = 1 \text{ m} \)- Cross-sectional area, \( A = 10^{-6} \text{ m}^2 \)- Mass of the ball, \( m = 1 \text{ kg} \)- Angular velocity, \( \omega = 20 \text{ rad/s} \)- Elongation, \( \Delta L = 10^{-3} \text{ m} \)
02

Calculate Force Exerted by the Ball

The centripetal force acting on the ball is given by \( F = m \omega^2 r \). Since the wire is the radius through which the ball rotates, \( r = L + \Delta L \). Thus:\[ F = 1 \times 20^2 \times (1 + 0.001) \approx 400 \text{ N} \]
03

Use Hooke's Law

According to Hooke's Law, the stress developed in the wire is given by \( \frac{F}{A} \). The strain in the wire is \( \frac{\Delta L}{L} \). Therefore, Young's modulus \( Y \) is expressed as:\[ Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F}{A} \times \frac{L}{\Delta L} \]
04

Calculate Young's Modulus

Substitute the known values into the expression for Young's modulus:\[ Y = \left(\frac{400}{10^{-6}}\right) \times \frac{1}{0.001} = 400 \times 10^6 \times 1000 = 4.0 \times 10^{11} \text{ N/m}^2 \]

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Key Concepts

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

Centripetal Force
When an object moves in a circular path, like the ball attached to our wire, it experiences a force that keeps it moving along that curved path rather than shooting off in a straight line. This force is known as centripetal force. In simpler terms, centripetal force is what "pulls" the object toward the center of its circular path, ensuring constant circular motion rather than escape.

In the exercise, the ball rotates around a fixed point, exerting a centripetal force on the wire. This force is calculated using the formula \( F = m \omega^2 r \), where:
  • \( F \) is the centripetal force.
  • \( m \) represents the mass of the ball.
  • \( \omega \) is the angular velocity (rate of rotation).
  • \( r \) is the radius, which here is the length of the wire plus its elongation or stretch.
This force is central to ensuring the ball keeps looping around rather than flying off tangentially. Any increase in speed or mass, or decrease in the radius, affects the centripetal force dramatically, showcasing its crucial role in circular motion.
Hooke's Law
Hooke's Law is a principle found in physics that explains how elastic materials deform under stress. The theory holds that the strain (or deformation) of an elastic object or material is proportional to the stress applied to it, providing this stress does not exceed the material's elastic limit.

This law is expressed with the formula \( F = kx \), where:
  • \( F \) stands for the force applied.
  • \( k \) is the spring constant, unique to each material, showing how resistant it is to deformation.
  • \( x \) represents the change in length from its original state.
In our context, the copper wire acts like a spring, initially at rest but then stretched by the centripetal force exerted by the rotating ball. The stress of the wire equates to the force divided by the cross-sectional area. The strain is then the change in length divided by the original length. These ratios showcase how even minute changes in force or material properties can lead to visible differences in deformation.
Strain and Stress
Strain and stress are central concepts in understanding how materials react to forces. Stress is essentially the force experienced per unit area, and it quantifies how concentrated a force is on a material. Meanwhile, strain measures how much an object deforms due to applied stress, expressed as the percentage change in dimension.

In the problem, stress is evaluated as \( \frac{F}{A} \), where \( F \) is the centripetal force applied to the wire, and \( A \) is the cross-sectional area of the wire. On the other hand, strain equals \( \frac{\Delta L}{L} \), where \( \Delta L \) is the elongation, and \( L \) is the original length.

Young's Modulus connects these two by providing a measure of stiffness for materials, indicating how much they will deform under a given level of stress. It is described as the ratio of stress to strain, \( Y = \frac{\text{Stress}}{\text{Strain}} \). Understanding these concepts is vital in material science, as they help predict how materials will behave under various forces and conditions, ensuring they can safely hold loads or withstand environmental stressors.

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Most popular questions from this chapter

A piece of copper wire has twice the radius of a piece of steel wire. Young's modulus for steel is twice that of the copper. One end of the copper wire is joined to one end of the steel wire so that both can be subjected to the same longitudinal force. By what fraction of its length will the steel have stretched when the length of the copper has increased by \(1 \%\) ? (1) \(1 \%\) (2) \(2 \%\) (3) \(2.5 \%\) (4) \(3 \%\)

A sphere of radius \(0.1 \mathrm{~m}\) and mass \(8 \pi \mathrm{kg}\) is attached to the lower end of a steel wire of length \(5 \mathrm{~m}\) and diameter \(10^{-5} \mathrm{~m}\). The wire is suspended from \(5.22 \mathrm{~m}\) high ceiling of a room. When the sphere is made to swing as a simple pendulum, it just grazes the floor at its lowest point. Young's modulus of steel is \(2 \times 10^{11} \mathrm{Nm}^{-2}\). Find the velocity of the sphere at the lowest position in \(\mathrm{m} / \mathrm{s}\). (Given: \(g=10 \mathrm{~m} / \mathrm{s}^{2}\) )

An elastic material of Young's modulus \(Y\) is subjected to a stress \(S\). The elastic energy stored per unit volume of the material is (1) \(\frac{S Y}{2}\) (2) \(\frac{S^{2}}{2 Y}\) (3) \(\frac{S}{2 Y}\) (4) \(\frac{2 S}{Y}\)

A ball falling in a lake of depth \(200 \mathrm{~m}\) shows a decrease of \(0.1 \%\) in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take \(g=10 \mathrm{~ms}^{-2}\) ) (1) \(10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (2) \(2 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (3) \(3 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\) (4) \(4 \times 10^{9} \mathrm{~N} / \mathrm{m}^{2}\)

Bulk modulus for rubber is \(9.8 \times 10^{8} \mathrm{~N} \mathrm{~m}^{-2}\). To what depth (in \(\mathrm{m}\) ) should a rubber ball be taken in a lake so that its volume is decreased by \(0.1 \%\)

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