An Iron Ring on a Wheel

solved ThermalExpansion

Q. A blacksmith fixes an iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim and the iron ring are 5.243 m $L_{2} = 5.243$ m and 5.231 m $L_{1} = 5.231$ m, respectively at 27 °C $T_{1} = 27$ °C. Taking the coefficient of linear expansion of iron as 1.20 × 10⁻⁵ K⁻¹ $α_{l} = 1.20 \times 10^{−5}$ K⁻¹, to what temperature should the ring be heated so as to fit the rim of the wheel, in °C?

Inputs

rim

L_{2}

5.24 m

ring

L_{1}

5.23 m

temperature

T_{1}

27 °C

expansion

α_{l}

1.2e-05 K⁻¹

Outputs

hot

T_{2}

How it's solved

Solved with ThermalExpansion

Given

$L_{1} = 5.231$ m
$L_{2} = 5.243$ m
$T_{1} = 27$ °C
$α_{l} = 1.20 \times 10^{−5}$ K⁻¹

Find $T_{2}$ .

Working

$T_{1} = 27$ °C $= 300.15$ K
linear expansion(ring) $L_{2} - L_{1} = α_{l} \cdot L_{1} \cdot (T_{2} - T_{1})$
$5.243 - 5.231 = 1.2 \times 10^{−5} \cdot 5.231 \cdot (T_{2} - 300.15)$
$T_{2} = 218.2$ °C
$T_{2} = 491.318037$ K $= 218.168037$ °C
$T_{2} \approx 218$ °Cto 3 significant figures

Conclusion

The ring has to grow by the small gap between it and the rim; iron grows by a few parts in a hundred thousand for each degree, so it takes about two hundred degrees.

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 :an_iron_ring_on_a_wheel do4  kind :numerical5  describes "A blacksmith fixes an iron ring on the rim of the wooden wheel of a horse cart. The diameter of the rim " \6            "and the iron ring are %{rim} and %{ring}, respectively at %{temperature}. Taking the coefficient of " \7            "linear expansion of iron as %{expansion}, %{hot}, in °C?"8  given rim: "5.243 m"9  given ring: "5.231 m"10  given temperature: "27 °C"11  given expansion: "1.20e-5 K-1"12  asks hot: "to what temperature should the ring be heated so as to fit the rim of the wheel"13end

solution.rb

1require_relative "../questions/an_iron_ring_on_a_wheel"2require_relative "../science/thermal_properties_of_matter/thermal_expansion"34solve :an_iron_ring_on_a_wheel, using: ThermalExpansion do5  objects :ring67  take ring: ring.length_at_first8  take rim: ring.length_after9  take temperature: ring.temperature_at_first10  take expansion: ring.coefficient_of_linear_expansion1112  hot = ring.temperature_after13  answer hot:, in: "°C", figures: 31415  conclusion "The ring has to grow by the small gap between it and the rim; iron grows by a few parts in " \16             "a hundred thousand for each degree, so it takes about two hundred degrees."17end

playground.rb

1require_relative "../solutions/an_iron_ring_on_a_wheel"23play_with :an_iron_ring_on_a_wheel do4  title "An Iron Ring on a Wheel"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 ring	$L_{2}$	m	—
length at first of ring	$L_{1}$	m	—
coefficient of linear expansion of ring	$α_{l}$	K⁻¹	—
temperature after of ring	$T_{2}$	K	—
temperature at first of ring	$T_{1}$	K	—
change in length of ground	$Δ l_{g}$	m	—
change in length of ring	$Δ l$	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(ring)

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(ring)