One end of a piano wire is wrapped around a cylindrical tuning peg and the other

Question

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 0.75  mm and an unstrained length of 0.62  m. The radius of the tuning peg is 1.4  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.

Explanation / Answer

Circumference of the peg = 2 * pi * (radius of peg)

= 2 *pi * 1.4 mm

= 8.796 * 10^-3 m

When the peg is turned through two revolutions, then the the length of the wire changes by

2 * 8.796 * 10^-3 m = 1.76 * 10^-2 m

For the wire:

Initial length Lo = 0.62 m

Change in length deltaL = 1.76 * 10^-2 m

Cross sectional area A = pi R^2 = pi * (0.75*10^-3)^2 = 1.76 * 10^-6m^2

Y = (T/A)/(deltaL/Lo)

T = Y * A * deltaL/Lo

= 2.0 * 10^11 * 8.17 * 10^-7 * 1.76 * 10^-2/0.62

= 4638.45N

Ans: 4638.45N

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