A Wheel Slipped on a Shaft

solved ThermalExpansion

Q. A large steel wheel is to be fitted on to a shaft of the same material. At 27 °C $T_{1} = 27$ °C, the outer diameter of the shaft is 8.70 cm $L_{1} = 8.70$ cm and the diameter of the central hole in the wheel is 8.69 cm $L_{2} = 8.69$ cm. The shaft is cooled using 'dry ice'. Assume coefficient of linear expansion of the steel to be constant over the required temperature range: 1.20 × 10⁻⁵ K⁻¹ $α_{l} = 1.20 \times 10^{−5}$ K⁻¹. At what temperature of the shaft does the wheel slip on the shaft, in °C?

Inputs

warm

T_{1}

27 °C

shaft

L_{1}

8.7 cm

hole

L_{2}

8.69 cm

expansion

α_{l}

1.2e-05 K⁻¹

Outputs

cold

T_{2}

How it's solved

Solved with ThermalExpansion

Given

$T_{1} = 27$ °C
$L_{1} = 8.70$ cm
$L_{2} = 8.69$ cm
$α_{l} = 1.20 \times 10^{−5}$ K⁻¹

Find $T_{2}$ .

Working

$T_{1} = 27$ °C $= 300.15$ K
$L_{1} = 8.70$ cm $= 0.087$ m
$L_{2} = 8.69$ cm $= 0.0869$ m
linear expansion(shaft) $L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})$
$0.0869 - 0.087 = 1.2 \times 10^{−5} \cdot 0.087 \cdot (T_{2} - 300.15)$
$T_{2} = −68.79$ °C
$T_{2} = 204.364559$ K $= −68.7854406$ °C
$T_{2} \approx −69$ °Cto 2 significant figures

Conclusion

The shaft has to shrink to the size of the hole; steel shrinks by about a hundred-thousandth of itself for each degree, so it must be cooled well below freezing.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_wheel_slipped_on_a_shaft do4  kind :numerical5  describes "A large steel wheel is to be fitted on to a shaft of the same material. At %{warm}, the outer diameter " \6            "of the shaft is %{shaft} and the diameter of the central hole in the wheel is %{hole}. The shaft is " \7            "cooled using 'dry ice'. Assume coefficient of linear expansion of the steel to be constant over the " \8            "required temperature range: %{expansion}. %{cold}, in °C?"9  given warm: "27 °C"10  given shaft: "8.70 cm"11  given hole: "8.69 cm"12  given expansion: "1.20e-5 K-1"13  asks cold: "At what temperature of the shaft does the wheel slip on the shaft"14end

solution.rb

1require_relative "../questions/a_wheel_slipped_on_a_shaft"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"34solve :a_wheel_slipped_on_a_shaft, using: ThermalExpansion do5  objects :shaft67  take warm: shaft.temperature_at_first8  take shaft: shaft.length_at_first9  take hole: shaft.length_after10  take expansion: shaft.coefficient_of_linear_expansion1112  cold = shaft.temperature_after13  answer cold:, in: "°C", figures: 21415  conclusion "The shaft has to shrink to the size of the hole; steel shrinks by about a hundred-thousandth " \16             "of itself for each degree, so it must be cooled well below freezing."17end

playground.rb

1require_relative "../solutions/a_wheel_slipped_on_a_shaft"23play_with :a_wheel_slipped_on_a_shaft do4  title "A Wheel Slipped on a Shaft"5end

The laws it drew on

quantity	written	units	within	from
length at first	$L_{1}$	m	—	ThermalExpansion
length after	$L_{2}$	m	—	ThermalExpansion
change in length	$Δ l$	m	—	ThermalExpansion
temperature at first	$T_{1}$	K	—	ThermalExpansion
temperature after	$T_{2}$	K	—	ThermalExpansion
coefficient of linear expansion	$α_{l}$	K⁻¹	—	ThermalExpansion
length after of ground	$L_{2,g}$	m	—
length at first of ground	$L_{1,g}$	m	—
coefficient of linear expansion of ground	$α_{l,g}$	K⁻¹	—
temperature after of ground	$T_{2,g}$	K	—
temperature at first of ground	$T_{1,g}$	K	—
length after of shaft	$L_{2}$	m	—
length at first of shaft	$L_{1}$	m	—
coefficient of linear expansion of shaft	$α_{l}$	K⁻¹	—
temperature after of shaft	$T_{2}$	K	—
temperature at first of shaft	$T_{1}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of shaft	$Δ l$	m	—

L_{2,b} - L_{1,b} = α_{l,b} \cdot L_{1,b} \cdot (T_{2,b} - T_{1,b})

linear expansion

Δ l_{b} = L_{2,b} - L_{1,b}

the change is the length after less the length at first

L_{2,g} - L_{1,g} = α_{l,g} \cdot L_{1,g} \cdot (T_{2,g} - T_{1,g})

linear expansion(ground)

L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})

linear expansion(shaft)

Δ l_{g} = L_{2,g} - L_{1,g}

the change is the length after less the length at first(ground)

Δ l = L_{2} - L_{1}

the change is the length after less the length at first(shaft)