Finding Tension in a Piano Wire Wrapped Around a Tuning Peg

[jmccray]

A piano wire (Y=2.0*10^10 N/m^2) has a radius of 0.80 mm and a length of 0.76 m. One end of the wire is wrapped around a tuning peg whose radius is 1.8 mm. The other end is fixed in place. Initially, there is no tension in the wire. Find the tension when the peg is turned through 2 revolutions.

 

[siddharth]

What are your thoughts/ideas on this problem? You need to show your work to get help.

 

[kyle8921]

The tension in the piano wire can be calculated using the formula T = (Y * A * ∆L) / L, where T is the tension, Y is the Young's modulus, A is the cross-sectional area of the wire, ∆L is the change in length, and L is the original length of the wire.

In this scenario, the cross-sectional area of the wire can be calculated using the formula A = π * r^2, where π is the constant pi and r is the radius of the wire. Substituting the given values, we get A = 3.14 * (0.0008 m)^2 = 5.024 * 10^-7 m^2.

Since the wire is wrapped around the tuning peg, the change in length (∆L) can be calculated by multiplying the circumference of the peg by the number of revolutions it has turned. The circumference of the peg can be calculated using the formula C = 2 * π * r, where r is the radius of the peg. Substituting the given values, we get C = 2 * 3.14 * 0.0018 m = 0.0113 m. Therefore, after 2 revolutions, ∆L = 2 * 0.0113 m = 0.0226 m.

Substituting all the values in the formula for tension, we get T = (2.0 * 10^10 N/m^2) * (5.024 * 10^-7 m^2) * (0.0226 m) / (0.76 m) = 1.49 * 10^7 N. Therefore, the tension in the piano wire when the peg is turned through 2 revolutions is approximately 14.9 million Newtons.

It is important to note that this calculation assumes that the wire is perfectly elastic and does not take into account factors such as friction and other external forces. Nevertheless, this calculation provides a good estimate of the tension in the piano wire and can be used for further analysis and experiments.