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

The given temperature, T = 27°C can be written in Kelvin as:

27 + 273 = 300 K

Outer diameter of the steel shaft at T, d₁ = 8.70 cm

Diameter of the central hole in the wheel at T, d₂ = 8.69 cm

Coefficient of linear expansion of steel, α_steel = 1.20 × 10^–5 K^–1

After the shaft is cooled using ‘dry ice’, its temperature becomes T₁.

The wheel will slip on the shaft, if the change in diameter, Δd = 8.69 – 8.70

= – 0.01 cm

Temperature T₁, can be calculated from the relation:

Δd = d₁α_steel (T₁ – T)

0.01 = 8.70 × 1.20 × 10^–5 (T₁ – 300)

(T₁ – 300) = 95.78

∴T₁= 204.21 K

= 204.21 – 273.16

= –68.95°C

Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.

Answer 2

Here at temperature ${\displaystyle T_1={27}^{\circ }C}$, diameter of shaft ${\displaystyle D_1=8.70cm}$

Let temperature ${\displaystyle T_2}$ the diameter of shaft changes to ${\displaystyle D_2}$= 8.69 cm for steel

${\displaystyle \alpha =1.20\times {10}^{-5}{K}^{-1}=1.20\times {10}^{-5}{}^{\circ }{C}^{-1}}$

∵ Change in diameter ${\displaystyle △D =D_2-D_1= D_1\times \alpha \times (T_2-T_1)}$

${\displaystyle \therefore 8.69-8.70=8.70\times 1.20\times {10}^{-5}\times (T_2-27)}$

${\displaystyle \Rightarrow T_2 =27-\frac{0.01}{8.70\times 1.20\times {10}^{-5}}=27-95.8=-68.8{}^{\circ }C}$ or ${\displaystyle -69{}^{\circ }C}$