Answer to Question #99500 in Mechanics | Relativity for Aled

Question

A steel axle carries two railway wheels at a distance 1.448 m apart at 20oC. Calculate the spacing at 2.3oC and 32.3oC and determine the difference in spacing of the wheels at the two different temperatures (spacing at 32.3oC less spacing at 2.3oC )

(The thermal expansion coefficient of steel is (12 x 10-6) oC-1

Give your answer in mm to 2 decimal place.

Answer 1

Length according to temperature

$\Delta\,L = \alpha_L\,L\,\Delta\,T$

or

$L_{final} - L_{initial} = \alpha_L L_{initial}(T_{fianl} - T_{initial})$

or

$L_{final} = L_{initial} + \Delta\,L$

At $32.3^{\circ}C$

$\quad \Delta\,L(32.3) = 12*10^{-6}*1448*(32.3 - 20) \approx0.21\space(mm) \\ \quad L(32.3) \approx 1448 + 0.21 \approx 1448.21\space(mm)$

At $2.3^{\circ}C$

$\quad \Delta\,L(2.3) = 12*10^{-6}*1448*(20 - 2.3) \approx -0.31\space(mm) \\ \quad L(2.3) \approx 1448 - 0.31 \approx 1447.69\space(mm)$

Difference in spacing

$L(32.3) - L(2.3) = \Delta\,L_(32.3) - \Delta\,L_(2.3) \approx 0.21 - (- 0.31) \approx 0.52\space(mm)$