Question

A large steel wheel is to be fitted onto a shaft of the same material. At \[27^\circ C\] 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 the constant over the required temperature range: \[{\alpha _{steel}} = 1.20 \times {10^{ - 5}}{K^{ - 1}}\]

Answer 1

Hint: We are asked to find the temperature at which the expansion of steel to the constant over the given temperature range. We can start to answer this question by writing down the data given in the question. Then we can move onto finding the asked temperature by using the formula to find the difference of the diameter of the wheel before and after expansion.

Formulas used:
The formula used to find the difference between the diameter of the wheel before and after expansion is given by,
\[\Delta d = {d_1}{\alpha _{steel}}\left( {{T_1} - {T_2}} \right)\]
Where \[{\alpha _{steel}}\] is the coefficient of linear expansion of steel, \[{d_1}\] is the initial diameter at which the temperature is to be found, \[{T_1}\] and \[{T_2}\] are the temperatures.

Complete step by step answer:
Let us start to attempt this question by writing down the data given in the question. The coefficient of linear expansion of steel is given as,
\[{\alpha _{steel}} = 1.20 \times {10^{ - 5}}{K^{ - 1}}\]
The initial temperature is given as \[{T_2} = 27 + 273 = 300\,K\].
The diameter of the shaft is \[{d_1} = 8.7\].
The difference in diameter of the shaft and central hole is,
\[\Delta d = 8.7 = 8.69 = 0.01\]

Now we have written down the values, we can substitute the values in the formula,
\[\Delta d = {d_1}{\alpha _{steel}}\left( {{T_1} - {T_2}} \right)\]
Substituting,
\[{T_1} - {T_2} = \dfrac{{\Delta d}}{{{d_1}{\alpha _{steel}}}} \\
\Rightarrow {T_1} - {T_2} = \dfrac{{0.01}}{{8.7 \times 1.20 \times {{10}^{ - 5}}}} \\
\Rightarrow {T_1} - {T_2} = - 95.78\]
We can now bring the known value of temperature to the right hand side and get
\[{T_1} = - 95.78 + {T_2} \\
\therefore {T_1} = 204.22\,K\]
We can convert this into the Celsius scale and get \[60^\circ C\].

Therefore, at \[60^\circ C\] the wheel slips on the shaft.

Note: We always convert the temperature into kelvin because the unit of coefficient of linear expansion is given as kelvin inverse. We did not convert the diameter into meters because when we multiple with ten powers negative two on both sides, it gets cancelled.That is \[\Delta d = {d_1}{\alpha _{steel}}\left( {{T_1} - {T_2}} \right)\] becomes \[\Delta d \times {10^{ - 2}} = {d_1} \times {10^{ - 2}}{\alpha _{steel}}\left( {{T_1} - {T_2}} \right)\]