**Physics Problem: Determining Angular Speed to Produce a Specific Strain****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 \(7.00 \times 10^{10} \, \text{N/m}^2\).)**Instructions:**1. **Sketch the Situation**: The object is moving in a circle, so the centripetal force is due to the horizontal component of the tension. 2. **Apply Newton’s Second Law**: Find an expression for the tension \(T\).3. **Apply Young's Modulus Equation**: Write the Young's modulus equation and solve for \(T\).4. **Equate Expressions**: Set your two expressions equal and solve for the angular speed \( \omega \) (in rad/s).**Example Calculation (Incorrect):**- Given answer: 7.42 (marked incorrect)---**Detailed Steps:**1. **Sketch the Situation:** - Draw a vertical wire of length 0.900 m. - Attach a mass of 2.00 kg at the bottom end of the wire. - The object is moving in a horizontal circle.2. **Newton's Second Law:** - The centripetal force equation, \(F = m \omega^2 r\), where \(m\) is the mass, \(\omega\) is the angular speed, and \(r\) is the radius (length of the wire).3. **Young's Modulus Equation:** - \( \text{Young's modulus}, E = \frac{\text{Stress}}{\text{Strain}}\) - Stress, \(\sigma = \frac{T}{A}\) where \(T\) is the tension and \(A\) is the cross-sectional area. - Strain, \(\varepsilon = \frac{\Delta L}{L} = 2.00 \times 10^{-2}\)4. **Area Calculation:** - Circular cross-section area

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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, if the wire is made from aluminum. (The Young's modulus for aluminum is 7.00 x 1010 N/m2.)

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, if the wire is made from aluminum. (The Young's modulus for aluminum is 7.00 x 1010 N/m2.)

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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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Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

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**Physics Problem: Determining Angular Speed to Produce a Specific Strain**

**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 \(7.00 \times 10^{10} \, \text{N/m}^2\).)

**Instructions:**
1. **Sketch the Situation**: The object is moving in a circle, so the centripetal force is due to the horizontal component of the tension.

2. **Apply Newton’s Second Law**: Find an expression for the tension \(T\).

3. **Apply Young's Modulus Equation**: Write the Young's modulus equation and solve for \(T\).

4. **Equate Expressions**: Set your two expressions equal and solve for the angular speed \( \omega \) (in rad/s).

**Example Calculation (Incorrect):**
- Given answer: 7.42 (marked incorrect)

---

**Detailed Steps:**
1. **Sketch the Situation:**
- Draw a vertical wire of length 0.900 m.
- Attach a mass of 2.00 kg at the bottom end of the wire.
- The object is moving in a horizontal circle.

2. **Newton's Second Law:**
- The centripetal force equation, \(F = m \omega^2 r\), where \(m\) is the mass, \(\omega\) is the angular speed, and \(r\) is the radius (length of the wire).

3. **Young's Modulus Equation:**
- \( \text{Young's modulus}, E = \frac{\text{Stress}}{\text{Strain}}\)
- Stress, \(\sigma = \frac{T}{A}\) where \(T\) is the tension and \(A\) is the cross-sectional area.
- Strain, \(\varepsilon = \frac{\Delta L}{L} = 2.00 \times 10^{-2}\)

4. **Area Calculation:**
- Circular cross-section area

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