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One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel (Y = 2.0x1011 N/m²). It has a radius of 0.68 mm and an unstrained length of 0.85 m. The radius of the tuning peg is 1.2 mm. Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions.

One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel (Y = 2.0x1011 N/m²). It has a radius of 0.68 mm and an unstrained length of 0.85 m. The radius of the tuning peg is 1.2 mm. Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions.

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 One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \( (Y = 2.0 \times 10^{11} \, \text{N/m}^2) \). It has a radius of 0.68 mm and an unstrained length of 0.85 m. The radius of the tuning peg is 1.2 mm. Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. ### Key Variables: - Young's modulus of steel, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \) - Radius of the piano wire, \( r = 0.68 \, \text{mm} \) - Unstrained length of the piano wire, \( L = 0.85 \, \text{m} \) - Radius of the tuning peg, \( R = 1.2 \, \text{mm} \) - Number of revolutions of the tuning peg, \( n = 2 \) ### Steps to Solve: 1. **Calculate the amount of elongation \( \Delta L \)**: - The circumference of the tuning peg \( C = 2 \pi R \). - Since the tuning peg is turned through two revolutions, the total length of wire wrapped around is \( \Delta L = 2 \times C = 2 \times 2 \pi R = 4 \pi R \). 2. **Substitute \( R = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)**: - \( \Delta L = 4 \pi \times 1.2 \times 10^{-3} \, \text{m} \). 3. **Calculate the tension \( F \) using Young's modulus formula**: - The formula for tension in the wire, \( F = Y \left( \frac{A \Delta L}{L} \right) \), where \( A \) is the cross-sectional area of the wire: \( A = \
Transcribed Image Text:## Problem Statement One end of a piano wire is wrapped around a cylindrical tuning peg and the other end is fixed in place. The tuning peg is turned so as to stretch the wire. The piano wire is made from steel \( (Y = 2.0 \times 10^{11} \, \text{N/m}^2) \). It has a radius of 0.68 mm and an unstrained length of 0.85 m. The radius of the tuning peg is 1.2 mm. Initially, there is no tension in the wire, but when the tuning peg is turned, tension develops. Find the tension in the wire when the tuning peg is turned through two revolutions. ### Key Variables: - Young's modulus of steel, \( Y = 2.0 \times 10^{11} \, \text{N/m}^2 \) - Radius of the piano wire, \( r = 0.68 \, \text{mm} \) - Unstrained length of the piano wire, \( L = 0.85 \, \text{m} \) - Radius of the tuning peg, \( R = 1.2 \, \text{mm} \) - Number of revolutions of the tuning peg, \( n = 2 \) ### Steps to Solve: 1. **Calculate the amount of elongation \( \Delta L \)**: - The circumference of the tuning peg \( C = 2 \pi R \). - Since the tuning peg is turned through two revolutions, the total length of wire wrapped around is \( \Delta L = 2 \times C = 2 \times 2 \pi R = 4 \pi R \). 2. **Substitute \( R = 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m} \)**: - \( \Delta L = 4 \pi \times 1.2 \times 10^{-3} \, \text{m} \). 3. **Calculate the tension \( F \) using Young's modulus formula**: - The formula for tension in the wire, \( F = Y \left( \frac{A \Delta L}{L} \right) \), where \( A \) is the cross-sectional area of the wire: \( A = \
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