A large steel wheel is to be fitted on to a shaft of the same mat… - Vidyadip

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 $8.70cm$ and the diameter of the central hole in the wheel is $8.69cm$. 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: asteel = $1.20 \times 10^{–5} K^{–1}$.

✓

Answer

The given temperature, T = 27°C can be written in Kelvin as: 27 + 273 = 300K
Outer diameter of the steel shaft at T, $d_1 = 8.70cm$
Diameter of the central hole in the wheel at T, $d_2 = 8.69cm$
Coefficient of linear expansion of steel, $\alpha_{\text{steel}}=1.20\times10^{-5}\text{K}^{-1}$
After the shaft is cooled using ‘dry ice’, its temperature becomes $T_1$.
The wheel will slip on the shaft, if the change in diameter, $\triangle\text{d}=8.69-8.70$
Temperature $T_1$, can be calculated from the relation:
$\triangle\text{d}=\text{d}_1\alpha_{\text{steel}}(\text{T}_1-\text{T})$
$0.01 = 8.70 \times 1.20 \times 10^{–5} (T_1 – 300)$
$(T_1 – 300) = 95.78$
$\therefore T_1= 204.21K$
$= 204.21 – 273.16$
$= –68.95^\circ C$
Therefore, the wheel will slip on the shaft when the temperature of the shaft is –69°C.

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