Measurement, units and dimensional analysis

The first week of Physics I looks light, but on the midterm the fastest points come from here. A unit conversion or a “units of b?” question is over in a minute. The unit checking you learn here also helps you eliminate wrong options in every question later: a formula that gives m/s in an acceleration question is wrong, done.

The five rules you should take from this page:

1. Every number has a unit. Carry the unit through the calculation, cancel it, and check it at the end.
2. Two added terms must have the same dimension. You cannot add apples and oranges; nor meters and seconds.
3. Conversion = multiplying by 1. $\frac{1000\,\text{m}}{1\,\text{km}}$ is a fraction, and its value is 1.
4. For squared and cubed units, the conversion factor is squared/cubed too. $1\,\text{m}^3 = 10^6\,\text{cm}^3$, not $10^2$.
5. Significant figures: in multiplication the fewest significant figures wins, in addition the fewest decimal places.

SI units and base quantities

In mechanics there are three base quantities. Everything else is derived from them.

Derived quantities are products of these three: velocity $[v] = \text{L/T}$, acceleration $[a] = \text{L/T}^2$, force $[F] = \text{M L T}^{-2}$, density $[\rho] = \text{M/L}^3$.

$1\,\text{N} = 1\,\text{kg}\cdot\text{m/s}^2 \qquad 1\,\text{J} = 1\,\text{N}\cdot\text{m} = 1\,\text{kg}\cdot\text{m}^2/\text{s}^2$

The square bracket $[\;]$ means “the dimension of”. $[F]$ = the dimension of force.

Prefixes

On the exam the data usually comes with prefixes (5.0 mm, 250 g, 3.0 μs). Convert to SI before putting it in a formula. Forgetting to convert g to kg is the most common lost point.

Conversion of units

Method: multiply by fractions whose value is 1 and which cancel the unit you do not want.

Example 1. What is 90 km/h in m/s?

$90\,\frac{\text{km}}{\text{h}}\cdot\frac{1000\,\text{m}}{1\,\text{km}}\cdot\frac{1\,\text{h}}{3600\,\text{s}} = \frac{90\,000}{3600}\,\frac{\text{m}}{\text{s}} = 25\,\text{m/s}$

Shortcut: for km/h → m/s divide by 3.6, for m/s → km/h multiply by 3.6. Which one? Common sense: the number in m/s is always smaller than the number in km/h.

Conversion for area and volume

When a unit is squared, the conversion fraction is squared too:

$1\,\text{m}^3 = (100\,\text{cm})^3 = 10^6\,\text{cm}^3 \qquad 1\,\text{cm}^2 = (10^{-2}\,\text{m})^2 = 10^{-4}\,\text{m}^2$

Also $1\,\text{L} = 1000\,\text{cm}^3 = 10^{-3}\,\text{m}^3$.

PITFALL · forgetting the cube. Converting 5000 cm³ to m³ by dividing by 100 and writing 50 m³. The right way is $5000/10^6 = 5.0\times10^{-3}\,\text{m}^3$. A bucket of water cannot be 50 m³; a common-sense check catches this mistake.

Density and atomic mass

Density is mass per unit volume:

$\rho = \frac{m}{V} \qquad [\rho] = \text{kg/m}^3$

The practical unit is g/cm³: $1\,\text{g/cm}^3 = \dfrac{10^{-3}\,\text{kg}}{10^{-6}\,\text{m}^3} = 1000\,\text{kg/m}^3$. Water is about 1.00 g/cm³, aluminum 2.70, iron 7.86, copper 8.92.

Example 2. A solid cylinder of radius 1.50 cm and height 4.00 cm has a mass of 76.3 g. What is its density?

V = π r² h = π (1.50 cm)² (4.00 cm) = 28.3 cm³
ρ = m / V = 76.3 g / 28.3 cm³ = 2.70 g/cm³
SI: 2.70 × 1000 = 2.70 × 10³ kg/m³   (aluminum)

Atomic mass and the mole: the molar mass $M$ (g/mol) of an element is the mass in grams of 1 mol of atoms. 1 mol = $N_A = 6.022\times10^{23}$ items.

$n = \frac{m}{M} \qquad N = n\,N_A$

The $M$ of aluminum is 27.0 g/mol. In 27.0 g of aluminum there are $6.022\times10^{23}$ atoms. The mass of one atom: $27.0\,\text{g} / 6.022\times10^{23} = 4.48\times10^{-23}\,\text{g}$.

Dimensional analysis

The two sides of an equation, and every added term, must have the same dimension. This simple rule produces three types of questions.

Type 1: the unit of a constant

$x = pt^2 + qt^4$ ($x$ in meters, $t$ in seconds). Every term must be in meters:

$[p]\,\text{s}^2 = \text{m} \Rightarrow [p] = \text{m/s}^2 \qquad [q]\,\text{s}^4 = \text{m} \Rightarrow [q] = \text{m/s}^4$

Type 2: is the equation dimensionally correct?

$x = v_0 t + \tfrac12 a t^2$: $[v_0 t] = \tfrac{\text{L}}{\text{T}}\cdot\text{T} = \text{L}$, $[at^2] = \tfrac{\text{L}}{\text{T}^2}\cdot\text{T}^2 = \text{L}$. All are L, so the equation is dimensionally consistent. Since $\tfrac12$ is dimensionless, this analysis cannot see it.

PITFALL · dimensionally correct ≠ correct. $x = v_0 t + 2at^2$ is also dimensionally correct but physically wrong. Dimensional analysis cannot find numerical coefficients (½, 2, π); it can only rule out wrong equations.

Type 3: finding the exponents

Example 3. Let the period of a pendulum be of the form $T = k\,L^a g^b$ ($k$ dimensionless). What are $a$ and $b$?

Step 1. Expand everything into base dimensions:

$\text{T}^1 = \text{L}^a\,(\text{L}\,\text{T}^{-2})^b = \text{L}^{a+b}\,\text{T}^{-2b}$

Step 2. Equate the exponent of each base dimension:

L:  0 = a + b
T:  1 = -2b      →  b = -1/2
                  →  a = +1/2

Result: $T = k\sqrt{L/g}$. (In the real formula $k = 2\pi$; dimensional analysis cannot give this.) M does not appear in this equation at all; this also tells you that the period does not depend on the mass.

Significant figures

A measurement result cannot claim more than the precision of the measuring instrument.

Counting rules:

Operation rules:

• Multiplication and division: the result has as many significant figures as the data with the fewest significant figures. $3.47 \times 2.1 = 7.287 \to 7.3$.
• Addition and subtraction: the result has as many decimal places as the data with the fewest decimal places. $12.4 + 0.356 + 101 = 113.756 \to 114$.

TIP · rounding in intermediate steps. Keep intermediate results in full on the calculator and round only the final answer. If you round at every step, the last digit drifts and may not match any of the options.

Order of magnitude

The nearest power of 10 to a number. $8.3\times10^4 \to 10^5$, $2.1\times10^{-3} \to 10^{-3}$. It is useful in Fermi-type estimates and in eliminating options: the mass of a car is $\sim10^3$ kg, the height of a person is $\sim10^0$ m.

Common mistakes