Length — Question 8

Question 8

The ruler expands along with the object

A metal rule is calibrated at $20\,{}^\circ\mathrm{C}$. In the supplied linear model, every interval on the rule expands at temperature $T$ by a factor $1+\alpha_r\Delta T$, where $\Delta T=T-20\,{}^\circ\mathrm{C}$ expressed as a temperature interval in kelvins. Take $\alpha_r=12.0\times10^{-6}\ \mathrm{K^{-1}}$.

A rod has expansion coefficient $\alpha_o=23.0\times10^{-6}\ \mathrm{K^{-1}}$ and physical length $L(T)=L_{20}(1+\alpha_o\Delta T)$. The rule and rod are both uniformly at $40\,{}^\circ\mathrm{C}$, and the rule indicates $R=1.00000\ \mathrm{m}$. The symbol $R$ denotes the indicated length using the rule’s labelled units, not its expanded physical span. Assume no other measurement error.

1. What physical distance is spanned by a rule interval labelled $1.00000\ \mathrm{m}$ at $40\,{}^\circ\mathrm{C}$? Hence find the rod’s actual length at that temperature.

2. Find $L_{20}$. Explain why correcting only the rod’s expansion while ignoring the rule’s expansion gives the wrong reference-temperature length.

3. Derive a general expression for $R(T)$ in terms of $L_{20}$, $\alpha_o$, $\alpha_r$, and $\Delta T$.

4. If $\alpha_o=\alpha_r$, what happens to the indicated length as temperature changes? Does a constant indication then establish a constant physical length?

Treat the coefficients and readings as exact within this stipulated linear model, not as universal values for all metals.

Question 8 - Solution

The ruler expands along with the object

(a) The physical scale intervals are larger when warm.

Here $\Delta T=20\ \mathrm{K}$, so the rule’s expansion factor is $1+(12.0\times10^{-6})(20)=1.00024$. An interval labelled one metre therefore spans $\boxed{1.00024\ \mathrm{m}}$ physically.

The indication $R=1.00000\ \mathrm{m}$ represents the rod’s actual warm length $$\boxed{L(40)=1.00024\ \mathrm{m}}.$$ Expanded scale intervals require a positive correction to the indicated length.

(b) Then refer the rod to its specified temperature.

The rod’s own expansion factor is $1+(23.0\times10^{-6})(20)=1.00046$. Hence $$L_{20}=\frac{1.00024}{1.00046}\ \mathrm{m} \approx\boxed{0.999780101\ \mathrm{m}}.$$ Ignoring rule expansion would instead give $$\frac{1.00000}{1.00046}\ \mathrm{m}\approx0.999540212\ \mathrm{m},$$ about $0.239890\ \mathrm{mm}$ too small. Both the measuring scale and the object must be referred consistently to physical length.

(c) Derive the indicated-length relation.

$$\frac{R(T)}{L_{20}}-1=\frac{(\alpha_o-\alpha_r)\Delta T}{1+\alpha_r\Delta T}.$$

The physical distance corresponding to indication $R(T)$ is $R(T)(1+\alpha_r\Delta T)$. Equating that to the rod length gives $$R(T)(1+\alpha_r\Delta T)=L_{20}(1+\alpha_o\Delta T),$$ $$\boxed{R(T)=L_{20}\frac{1+\alpha_o\Delta T}{1+\alpha_r\Delta T}}.$$

(d) Equal fractional expansion can conceal a change.

When the coefficients are equal, the ratio is one and $R(T)=L_{20}$ throughout the model’s range. Nevertheless the physical length is $L_{20}(1+\alpha_o\Delta T)$ and generally changes with temperature. The object and scale grow together, keeping the number of scale intervals spanned constant. A stable reading is not, by itself, evidence that the object’s physical length stayed fixed.