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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**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
Transcribed Image Text:**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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