Question

A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C, the outer diameter of the shaft is 8.70 cm and the diameter of the central hole in the wheel is 8.69 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range : α\(_{steel}\)= 1.20 × 10\(^{-5}\) K\(^{-1}\).

Answer 1

Given

• Initial temperature: \( T = 27^\circ\text{C} \Rightarrow 300\,\text{K} \)
• Outer diameter of shaft at 27°C: \( d_1 = 8.70 \,\text{cm} \)
• Diameter of wheel’s central hole at 27°C: \( d_2 = 8.69 \,\text{cm} \)
• Coefficient of linear expansion of steel: \( \alpha = 1.20 \times 10^{-5}\,\text{K}^{-1} \)

Concept

• Both shaft and wheel are steel; when the shaft is cooled, its diameter decreases.
• The wheel will slip on the shaft when the shaft’s diameter shrinks from 8.70 cm to 8.69 cm.

1. Required change in diameter

\( \Delta d = d_{\text{final}} - d_{\text{initial}} = 8.69 - 8.70 = -0.01 \,\text{cm} \)

2. Linear expansion (contraction) relation

For small temperature changes:

\( \Delta d = d_1 \, \alpha \, (T_1 - T) \)

Here, \( T_1 \) is the (unknown) lower temperature of the shaft.

3. Solve for \( T_1 \)

\( -0.01 = 8.70 \times (1.20 \times 10^{-5}) \times (T_1 - 300) \) \( \Rightarrow T_1 - 300 = \dfrac{-0.01}{8.70 \times 1.20 \times 10^{-5}} \)

Compute the denominator:

\( 8.70 \times 1.20 \times 10^{-5} = 10.44 \times 10^{-5} = 1.044 \times 10^{-4} \)

So

\( T_1 - 300 = \dfrac{-0.01}{1.044 \times 10^{-4}} \approx -95.8 \,\text{K} \) \( \Rightarrow T_1 \approx 300 - 95.8 = 204.2 \,\text{K} \)

Convert back to Celsius:

\( T_1 \approx 204.2 - 273.2 \approx -69^\circ\text{C} \)

The wheel will slip onto the shaft when the shaft is cooled to about \(-69^\circ\text{C}\).