Dimensional analysis — checking equations, deriving relations: Exercises

Level 1 — Recognition

Goal: read the "kind" straight off a definition.

L1.1

State the dimension of density (mass per unit volume).

Recall Solution

WHAT we do: go back to the definition. Density $ρ = \frac{mass}{volume}$ . WHY: dimensions are always found from the defining relation, never guessed.

Mass has dimension $[M]$ .
Volume is length × length × length, so $[L^{3}]$ . $[ρ] = \frac{[ M ]}{[ L ^{3} ]} = M L^{- 3}$ Answer: $M L^{- 3}$ .

L1.2

State the dimension of momentum ( $p = m v$ , mass times velocity).

Recall Solution

Velocity is displacement over time, $[L T^{- 1}]$ . Multiply by mass $[M]$ : $[p] = [M] [L T^{- 1}] = M L T^{- 1} .$ Answer: $M L T^{- 1}$ .

L1.3

Which of these can be added to a force without breaking homogeneity? (a) pressure × area, (b) mass × velocity, (c) energy / length.

Recall Solution

Force has dimension $[M L T^{- 2}]$ . Check each:

(a) pressure $= M L^{- 1} T^{- 2}$ , area $= L^{2}$ , product $= M L T^{- 2}$ ✅ matches force.
(b) $M \cdot L T^{- 1} = M L T^{- 1}$ (that's momentum) ❌.
(c) energy $= M L^{2} T^{- 2}$ , divided by $L$ gives $M L T^{- 2}$ ✅ matches force. Answer: (a) and (c) share force's dimension; (b) does not.

Level 2 — Application

Goal: run the homogeneity check on real equations.

L2.1

Is $v = u + a t$ dimensionally consistent? ( $v, u$ velocities, $a$ acceleration, $t$ time.)

Recall Solution

Each added/equated term must match.

$[v] = [u] = L T^{- 1}$ .
$[a t] = (L T^{- 2}) (T) = L T^{- 1}$ ✅. All three are $L T^{- 1}$ . Consistent. ✅

L2.2

The equation $E = \frac{1}{2} m v^{2} + m g h$ . Check that the two terms on the right have equal dimensions.

Recall Solution

$[\frac{1}{2} m v^{2}] = M (L T^{- 1})^{2} = M L^{2} T^{- 2}$ (the $\frac{1}{2}$ is a pure number, dropped).
$[m g h] = M (L T^{- 2}) (L) = M L^{2} T^{- 2}$ . Both equal $M L^{2} T^{- 2}$ — the dimension of energy. ✅ Consistent (and matches $[E]$ ).

L2.3

Spot the error: a student writes $s = u t + \frac{1}{2} a t^{3}$ .

Recall Solution

$[u t] = (L T^{- 1}) (T) = L$ ✅.
$[a t^{3}] = (L T^{- 2}) (T^{3}) = L T^{1}$ ❌ — that is length × time, not a length. The two added terms disagree, so the equation is wrong. The correct second term is $\frac{1}{2} a t^{2}$ , which gives $(L T^{- 2}) (T^{2}) = L$ . ✅

Level 3 — Analysis

Goal: reason about arguments, exponents, and hidden dimensionless groups.

L3.1

In the equation $y = A sin (k x - ω t)$ , $x$ is a length and $t$ is a time. Find the dimensions of $k$ and $ω$ .

Recall Solution

WHY the constraint: the argument of $sin$ must be dimensionless — you cannot take the sine of "2 metres." So inside the bracket, $k x$ and $ω t$ must each be pure numbers, and they must match each other (they are subtracted).

$[k x] = M^{0} L^{0} T^{0} \Rightarrow [k] [L] = 1 \Rightarrow [k] = L^{- 1}$ .
$[ω t] = M^{0} L^{0} T^{0} \Rightarrow [ω] [T] = 1 \Rightarrow [ω] = T^{- 1}$ . Answer: $[k] = L^{- 1}$ (per metre), $[ω] = T^{- 1}$ (per second).

L3.2

A quantity is defined by $F = \frac{G m _{1} m _{2}}{r ^{2}}$ , where $F$ is a force, $m_{1}, m_{2}$ masses, $r$ a distance. Find the dimension of $G$ .

Recall Solution

WHAT we do: solve the homogeneity equation for the unknown $[G]$ . $[F] = \frac{[ G ] [ m _{1} ] [ m _{2} ]}{[ r ] ^{2}} \Rightarrow M L T^{- 2} = \frac{[ G ] M \cdot M}{L ^{2}} .$ Rearrange: $[G] = \frac{M L T ^{- 2} \cdot L ^{2}}{M ^{2}} = M^{- 1} L^{3} T^{- 2} .$ Answer: $M^{- 1} L^{3} T^{- 2}$ .

L3.3

The drag force on a sphere at low speed is claimed to be $F = 6 π η r v$ , where $η$ (viscosity) has dimension $M L^{- 1} T^{- 1}$ , $r$ is a radius, $v$ a speed. Verify the equation is dimensionally consistent.

Recall Solution

The factor $6 π$ is a pure number — drop it. $[η r v] = (M L^{- 1} T^{- 1}) (L) (L T^{- 1}) = M L^{- 1 + 1 + 1} T^{- 1 - 1} = M L^{1} T^{- 2} .$ That is exactly $[F] = M L T^{- 2}$ . ✅ Consistent.

Level 4 — Synthesis

Goal: derive the form of a law by matching powers of $M, L, T$ .

L4.1

The speed $v$ of a transverse wave on a stretched string may depend on the tension $T$ (a force, $[M L T^{- 2}]$ ) and the mass per unit length $μ$ ( $[M L^{- 1}]$ ). Derive how $v$ scales with $T$ and $μ$ .

Figure — Dimensional analysis — checking equations, deriving relations

Recall Solution

Step 1 — guess a product of powers. Assume $v = k T^{a} μ^{b}$ , with $k$ a dimensionless constant. WHY a product: with only two ingredients, a monomial is the only form homogeneity can pin down. Step 2 — write dimensions of both sides. $L T^{- 1} = (M L T^{- 2})^{a} (M L^{- 1})^{b} = M^{a + b} L^{a - b} T^{- 2 a} .$ Step 3 — match each base separately (each of $M, L, T$ is an independent axis — see the three sliders in the figure):

$M : a + b = 0$
$L : a - b = 1$
$T : - 2 a = - 1 \Rightarrow a = \frac{1}{2}$ From $a = \frac{1}{2}$ and $a + b = 0$ : $b = - \frac{1}{2}$ . (Check $L$ : $\frac{1}{2} - (- \frac{1}{2}) = 1$ ✅.) Step 4 — assemble. $v = k T^{1/2} μ^{- 1/2} = k \frac{T}{μ} .$ Insight: the true constant is $k = 1$ , but dimensional analysis cannot supply it. The form $v \propto T / μ$ is fully determined.

L4.2

Convert a pressure of $1 Pa = 1 N/m^{2}$ into the CGS unit (g, cm, s). $[P] = M L^{- 1} T^{- 2}$ .

Recall Solution

WHY the exponents matter: each base unit is scaled by its conversion factor raised to its power in the dimension $M^{1} L^{- 1} T^{- 2}$ .

$M : 1 kg = 1 0^{3} g$ , power $+ 1$ → factor $1 0^{3}$ .
$L : 1 m = 1 0^{2} cm$ , power $- 1$ → factor $1 0^{- 2}$ .
$T : 1 s$ , power $- 2$ → factor $1$ . $1 Pa = 1 0^{3} \times 1 0^{- 2} \times 1 = 1 0^{1} g\cdotcm^{- 1} s^{- 2} = 10 (barye/dyne\cdotcm^{- 2}) .$ Answer: $1 Pa = 10 g\cdotcm^{- 1} s^{- 2}$ .

Level 5 — Mastery

Goal: recognise where the method under-determines or fails — and say precisely why.

L5.1

It's proposed that the frequency $f$ of a vibrating water drop depends on its radius $r$ , density $ρ$ ( $[M L^{- 3}]$ ), and surface tension $σ$ ( $[M T^{- 2}]$ , i.e. force per length). Derive the scaling $f = k r^{a} ρ^{b} σ^{c}$ .

Recall Solution

Step 1 — dimensions of both sides. $[f] = T^{- 1}$ . $T^{- 1} = L^{a} (M L^{- 3})^{b} (M T^{- 2})^{c} = M^{b + c} L^{a - 3 b} T^{- 2 c} .$ Step 2 — match bases.

$M : b + c = 0$
$T : - 2 c = - 1 \Rightarrow c = \frac{1}{2} \Rightarrow b = - \frac{1}{2}$
$L : a - 3 b = 0 \Rightarrow a = 3 b = - \frac{3}{2}$ Step 3 — assemble. $f = k r^{- 3/2} ρ^{- 1/2} σ^{1/2} = k \frac{σ}{ρ r ^{3}} .$ Insight: three ingredients, three base dimensions → a unique solution. The problem is exactly determined.

L5.2

A student tries to derive the full range formula of a projectile, $R = \frac{v ^{2} s i n ( 2 θ )}{g},$ purely from dimensional analysis, guessing $R = k v^{a} g^{b}$ . Explain what dimensional analysis can and cannot recover here.

Recall Solution

What it CAN do. Match $R = k v^{a} g^{b}$ : $L = (L T^{- 1})^{a} (L T^{- 2})^{b} = L^{a + b} T^{- a - 2 b} .$

$L : a + b = 1$
$T : - a - 2 b = 0 \Rightarrow a = - 2 b$ Substitute: $- 2 b + b = 1 \Rightarrow b = - 1, a = 2$ . So $R = k \frac{v ^{2}}{g} .$ The form $R \propto v^{2} / g$ is correct. What it CANNOT do.

The dimensionless factor $sin (2 θ)$ is invisible — $θ$ is an angle (dimensionless), so it never enters the base-matching. Dimensional analysis can't tell you $R$ depends on launch angle at all.
The pure number $k$ is unknown (here it would be $sin 2 θ$ , capped at 1). Conclusion: dimensional analysis gives $v^{2} / g$ scaling but is blind to the angular factor and any numeric constant. See Equations of motion (kinematics) for the full derivation.

L5.3

A drag force $F$ on a fast-moving body is thought to depend on speed $v$ , cross-sectional area $A$ ( $[L^{2}]$ ), fluid density $ρ$ ( $[M L^{- 3}]$ ), and viscosity $η$ ( $[M L^{- 1} T^{- 1}]$ ). Explain why a single monomial $F = k v^{a} A^{b} ρ^{c} η^{d}$ is under-determined, and state the number of free parameters.

Recall Solution

Count the equations vs unknowns. We have 4 unknown exponents ( $a, b, c, d$ ) but only 3 base dimensions ( $M, L, T$ ) → only 3 equations. $M L T^{- 2} = (L T^{- 1})^{a} (L^{2})^{b} (M L^{- 3})^{c} (M L^{- 1} T^{- 1})^{d} .$

$M : c + d = 1$
$L : a + 2 b - 3 c - d = 1$
$T : - a - d = - 2$ Three equations, four unknowns → one free parameter ( $4 - 3 = 1$ ). The solution is a family, not a single answer: physically this free parameter is the Reynolds number $Re = ρ v A / η$ , a dimensionless group. Drag becomes $F = ρ v^{2} A \cdot ϕ (Re),$ where $ϕ$ is an unknown dimensionless function the method cannot pin down. Answer: under-determined by exactly 1 parameter; you get a dimensionless group, not a formula. This is precisely the content of the Buckingham Pi theorem.

Score yourself

Recall Mastery checklist

Read the dimension off any definition (L1) ::: $M, L, T$ from the defining ratio. Run a homogeneity check and read the verdict correctly (L2) ::: failed = wrong; passed = not disproved. Force a $sin / ln / e^{x}$ argument to be dimensionless (L3) ::: gives the dimension of the coefficient. Derive a two-ingredient law by matching powers (L4) ::: monomial → match $M, L, T$ → solve. Recognise under-determination when variables > base dimensions (L5) ::: leftover count = dimensionless groups.

Connections

Units and the SI system — the unit conversions behind L1 and L4.
Significant figures and error propagation — the numeric side of these answers.
Equations of motion (kinematics) — full derivation of the projectile range (L5.2).
Buckingham Pi theorem — the general rule governing L5.3.
Vectors — components and addition — dimensions of vector quantities.