Changing Units — Question 9

Question 9

A formula coefficient has a unit convention too

A fitted resistance-force model is supplied as an equation between numerical values: $$\frac{F}{\mathrm{N}}=0.0200\left(\frac{v}{\mathrm{km/h}}\right)^2.$$ Here $F$ is force magnitude and $v$ is speed. The notation $v/(\mathrm{km/h})$ means the number obtained when speed is expressed in kilometres per hour. Take the coefficient as exact within this exercise. No claim about a universal force law is intended.

A new program accepts the speed in $\mathrm{m/s}$ but keeps the numerical coefficient 0.0200. Use exact $1\ \mathrm{km}=1000\ \mathrm{m}$, $1\ \mathrm{h}=3600\ \mathrm{s}$, and $1\ \mathrm{N}=1\ \mathrm{kg\,m/s^2}$.

1. Rewrite the given model as $F/\mathrm{N}=B[v/(\mathrm{m/s})]^2$ and determine $B$. Show why the conversion factor is squared.

2. For $v=10.0\ \mathrm{m/s}$, compare the correct force with the new program’s force. Give the factor by which the new program underestimates the result.

3. Write a quantity equation $F=kv^2$ valid without a fixed choice of units, and give $k$ in coherent SI units. Explain why $k$ is not dimensionless even though the displayed coefficient 0.0200 is a pure number.

4. A colleague changes only the coefficient’s label from $\mathrm{N/(km/h)^2}$ to $\mathrm{N/(m/s)^2}$. Explain why this relabeling changes the model.

Question 9 - Solution

A formula coefficient has a unit convention too

(a) Substitute the relation between numerical speed values.

Since $1\ \mathrm{m/s}=3.6\ \mathrm{km/h}$, $$\frac{v}{\mathrm{km/h}}=3.6\frac{v}{\mathrm{m/s}}.$$ Therefore $$\frac{F}{\mathrm{N}}=0.0200(3.6)^2 \left(\frac{v}{\mathrm{m/s}}\right)^2, \qquad\boxed{B=0.2592}.$$ The complete numerical speed is squared, so its conversion factor is squared as well.

(b) Evaluate the same physical speed in both implementations.

At $10.0\ \mathrm{m/s}=36.0\ \mathrm{km/h}$, $$F_{\mathrm{correct}}=0.0200(36.0)^2\ \mathrm{N} =0.2592(10.0)^2\ \mathrm{N}=\boxed{25.92\ \mathrm{N}}.$$ The faulty program gives $[0.0200(10.0)^2]\ \mathrm{N}=2.000\ \mathrm{N}$. The correct force is larger by the factor $$\boxed{25.92/2.000=12.96}.$$

(c) Restore the units carried by the coefficient.

$$[k]=\frac{[F]}{[v]^2}=\frac{\mathrm{kg\,m\,s^{-2}}}{\mathrm{m^2\,s^{-2}}}=\mathrm{kg\,m^{-1}}.$$

Multiplying the numerical-value equation by newton yields $$F=kv^2,\qquad k=0.2592\ \frac{\mathrm{N}}{(\mathrm{m/s})^2} =\boxed{0.2592\ \mathrm{kg/m}}.$$ The unit follows from force divided by speed squared. The original 0.0200 is a pure number because its equation divides both physical quantities by stated units. Its corresponding physical coefficient is $0.0200\ \mathrm{N/(km/h)^2}$, which equals the $k$ above.

(d) Convert the coefficient; do not merely rename it.

$\mathrm{km/h}$ and $\mathrm{m/s}$ are different-sized units. Replacing the label while retaining 0.0200 changes the physical coefficient to $0.0200\ \mathrm{kg/m}$, reducing it by a factor of 12.96. A correct conversion preserves $k$ and changes both its numerical value and its unit representation together.