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A 0.900 m-long metal wire has a circular cross section of diameter 0.950 mm. One end of the wire is fixed, and the other end has a 2.00 kg object attached to it. Determine the angular speed of the object (in rad/s) required to produce a strain of 2.00 x 10-2, if the wire is made from aluminum. (The Young's modulus for aluminum is 7.00 x 1010 N/m².) rad/s

A 0.900 m-long metal wire has a circular cross section of diameter 0.950 mm. One end of the wire is fixed, and the other end has a 2.00 kg object attached to it. Determine the angular speed of the object (in rad/s) required to produce a strain of 2.00 x 10-2, if the wire is made from aluminum. (The Young's modulus for aluminum is 7.00 x 1010 N/m².) rad/s

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:** A 0.900 m-long metal wire has a circular cross section of diameter 0.950 mm. One end of the wire is fixed, and the other end has a 2.00 kg object attached to it. Determine the angular speed of the object (in rad/s) required to produce a strain of \(2.00 \times 10^{-2}\), if the wire is made from aluminum. (The Young’s modulus for aluminum is \(7.00 \times 10^{10} \, \text{N/m}^2\)). **Given:** - Length of the wire, \(L = 0.900 \, \text{m}\) - Diameter of the wire, \(d = 0.950 \, \text{mm} = 0.950 \times 10^{-3} \, \text{m}\) - Mass of the object, \(m = 2.00 \, \text{kg}\) - Strain, \(\epsilon = 2.00 \times 10^{-2}\) - Young’s modulus for aluminum, \(E = 7.00 \times 10^{10} \, \text{N/m}^2\) **Required:** - Angular speed of the object, \(\omega\) in rad/s **Solution:** To calculate the angular speed of the object that will generate the required strain: 1. **Calculate the radius of the wire's cross-section:** \[ r = \frac{d}{2} = \frac{0.950 \times 10^{-3} \, \text{m}}{2} = 0.475 \times 10^{-3} \, \text{m} \] 2. **Calculate the cross-sectional area, \(A\), of the wire:** \[ A = \pi r^2 = \pi (0.475 \times 10^{-3} \, \text{m})^2 = \pi (0.225625 \times 10^{-6} \, \text{m}^2) = 0.707 \times 10^{-6} \, \text{m}^2 \] 3. **Using the definition of strain \(\epsilon\), which is the change in length per original length:** \
Transcribed Image Text:**Problem Statement:** A 0.900 m-long metal wire has a circular cross section of diameter 0.950 mm. One end of the wire is fixed, and the other end has a 2.00 kg object attached to it. Determine the angular speed of the object (in rad/s) required to produce a strain of \(2.00 \times 10^{-2}\), if the wire is made from aluminum. (The Young’s modulus for aluminum is \(7.00 \times 10^{10} \, \text{N/m}^2\)). **Given:** - Length of the wire, \(L = 0.900 \, \text{m}\) - Diameter of the wire, \(d = 0.950 \, \text{mm} = 0.950 \times 10^{-3} \, \text{m}\) - Mass of the object, \(m = 2.00 \, \text{kg}\) - Strain, \(\epsilon = 2.00 \times 10^{-2}\) - Young’s modulus for aluminum, \(E = 7.00 \times 10^{10} \, \text{N/m}^2\) **Required:** - Angular speed of the object, \(\omega\) in rad/s **Solution:** To calculate the angular speed of the object that will generate the required strain: 1. **Calculate the radius of the wire's cross-section:** \[ r = \frac{d}{2} = \frac{0.950 \times 10^{-3} \, \text{m}}{2} = 0.475 \times 10^{-3} \, \text{m} \] 2. **Calculate the cross-sectional area, \(A\), of the wire:** \[ A = \pi r^2 = \pi (0.475 \times 10^{-3} \, \text{m})^2 = \pi (0.225625 \times 10^{-6} \, \text{m}^2) = 0.707 \times 10^{-6} \, \text{m}^2 \] 3. **Using the definition of strain \(\epsilon\), which is the change in length per original length:** \
**Problem Description:** Two blocks are free to slide along the frictionless wooden track shown below. The block of mass \( m_1 = 5.02 \) kg is released from the position shown, at height \( h = 5.00 \) m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass \( m_2 = 9.20 \) kg, initially at rest. The two blocks never touch. Calculate the maximum height to which \( m_1 \) rises after the elastic collision. **Diagram Explanation:** The diagram illustrates the following: 1. The left side shows a vertical height \( h \) that is indicated to be 5 meters. 2. At the top of this height is positioned block \( m_1 \). 3. The track is curved downward and towards the right. 4. At the bottom right of the track is the block \( m_2 \), located on a flat surface extending horizontally from the curved section. 5. There are two blocks, \( m_1 \) and \( m_2 \), where \( m_1 \) is situated at the height and \( m_2 \) is on the flat part of the track. The objective is to find out how high \( m_1 \) will rise after an elastic collision with \( m_2 \).
Transcribed Image Text:**Problem Description:** Two blocks are free to slide along the frictionless wooden track shown below. The block of mass \( m_1 = 5.02 \) kg is released from the position shown, at height \( h = 5.00 \) m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass \( m_2 = 9.20 \) kg, initially at rest. The two blocks never touch. Calculate the maximum height to which \( m_1 \) rises after the elastic collision. **Diagram Explanation:** The diagram illustrates the following: 1. The left side shows a vertical height \( h \) that is indicated to be 5 meters. 2. At the top of this height is positioned block \( m_1 \). 3. The track is curved downward and towards the right. 4. At the bottom right of the track is the block \( m_2 \), located on a flat surface extending horizontally from the curved section. 5. There are two blocks, \( m_1 \) and \( m_2 \), where \( m_1 \) is situated at the height and \( m_2 \) is on the flat part of the track. The objective is to find out how high \( m_1 \) will rise after an elastic collision with \( m_2 \).
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