Question

A steel piano wire \(\left(Y=2.0 \times 10^{11} \mathrm{Pa}\right)\) has a diameter of \(0.80 \mathrm{mm} .\) At one end it is wrapped around a tuning pin of diameter \(8.0 \mathrm{mm}\). The length of the wire (not including the wire wrapped around the tuning pin) is \(66 \mathrm{cm}\). Initially, the tension in the wire is \(381 \mathrm{N}\). To tune the wire, the tension must be increased to \(402 \mathrm{N}\). Through what angle must the tuning pin be turned?

Short answer

Answer: The tuning pin must be turned by approximately 1.98°.

Step-by-step solution

Calculate the initial length of the wire wrapped around the tuning pin

To find the initial length of the wire wrapped around the tuning pin, we'll first find the circumference of the tuning pin and then multiply it by the initial tension divided by the Young's modulus, Y. Formula for the circumference of a circle is: \(C = 2\pi r\) Where, \(C\) is the circumference and r is the radius Given, diameter of the tuning pin, \(d_{pin} = 8.0\,\mathrm{mm} = 0.008\,\mathrm{m}\) So, radius will be, \(r_{pin} =\frac{d_{pin}}{2} = 4.0\,\mathrm{mm} = 0.004\,\mathrm{m}\) Now let's calculate the circumference of the tuning pin: \(C_{pin} = 2\pi (0.004\,\mathrm{m}) = 0.0251\,\mathrm{m}\)

Calculate the change in length of the wire wrapped around the tuning pin

To find the change in length, we'll use the formula for the elongation of the wire, which is given by: \(\Delta L = \frac{L \cdot \Delta T}{Y\cdot A} \) Where \(\Delta L\) is the change in length, \(L\) is the initial length, \(\Delta T\) is the change in tension, \(Y\) is the Young's modulus, and \(A\) is the cross-sectional area of the wire. Given diameter of the wire, \(d_{wire} = 0.80\,\mathrm{mm} = 0.0008\,\mathrm{m}\) The radius of the wire, \(r_{wire} = \frac{d_{wire}}{2} = 0.0004\,\mathrm{m}\) Area of the wire's cross-section, \(A = \pi r_{wire}^2 = \pi (0.0004\,\mathrm{m})^2 = 5.0265 \times 10^{-7}\,\mathrm{m^2}\) Given initial tension, \(T_{i} = 381\,\mathrm{N}\) Final tension, \(T_{f} = 402\,\mathrm{N}\) Change in tension, \(\Delta T = T_{f} - T_{i} = 402\,\mathrm{N} - 381\,\mathrm{N} = 21\,\mathrm{N}\) Using the elongation formula, we get: \(\Delta L = \frac{L \cdot \Delta T}{Y\cdot A} = \frac{(0.66\,\mathrm{m})(21\,\mathrm{N})}{(2.0 \times 10^{11}\, \mathrm{Pa})(5.0265 \times 10^{-7}\,\mathrm{m^2})} = 1.3821\times10^{-4}\,\mathrm{m}\)

Calculate the angle the tuning pin must be turned

Now we'll use the formula for the arc length of a circle: \(Arc Length = \theta \cdot r\) Where \(\theta\) is the angle in radians and \(r\) is the radius of the tuning pin. We want to find \(\theta\), so rearrange the formula: \(\theta = \frac{Arc Length}{r}\) Given, The change in length of the wire wrapped, \(\Delta L = 1.3821\times10^{-4}\,\mathrm{m}\) So, the angle is: \(\theta = \frac{1.3821\times10^{-4}\,\mathrm{m}}{0.004\,\mathrm{m}} \approx 0.03455\,\mathrm{radians}\) Now let's convert radians to degrees: \(\theta_{degrees} = \theta\left(\frac{180}{\pi}\right) \approx 0.03455\,\mathrm{radians}\left(\frac{180}{\pi}\right) \approx 1.98^\circ\)

Final result

The tuning pin must be turned by approximately \(1.98^\circ\) to increase the tension in the wire from \(381\,\mathrm{N}\) to \(402\,\mathrm{N}\).