Worked examples · Sheet 2

Dimensional analysis problems

Theory used: Dimensions of a quantity Principle of dimensional homogeneity Base and derived quantities International System of Units

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine which of the following relations are dimensionally correct, where $x$ and $Δ x$ are lengths, $t$ is a time, $v$ and $v_{0}$ are velocities and $a$ is an acceleration:

(i) v^{2} = v_{0}^{2} + 2 a Δ x, (ii) x = v_{0} t + a t .

[v] = [L T^{- 1}], [a] = [L T^{- 2}], [Δ x] = [L], [t] = [T]

(i) [v^{2}] = [v_{0}^{2}] = [L^{2} T^{- 2}], [a Δ x] = [L T^{- 2}] [L] = [L^{2} T^{- 2}]

(ii) [x] = [L], [v_{0} t] = [L T^{- 1}] [T] = [L], [a t] = [L T^{- 2}] [T] = [L T^{- 1}]

By the principle of dimensional homogeneity, all the terms added in a relation and the two sides of the equality must have the same dimensions. The factor $2$ is a pure number and does not change the dimensions. In relation (i) the three terms all have dimension $[L^{2} T^{- 2}]$ , so the relation is dimensionally correct. In relation (ii) the term $a t$ has the dimension of a velocity, whereas $x$ and $v_{0} t$ are lengths; therefore relation (ii) is wrong. Relation (i) is a formula of uniformly accelerated motion; its dimensional correctness is a necessary condition, and establishing that it is exact requires its derivation, carried out in the topic on one-dimensional kinematics.

Exercise 2

Determine the dimensions and the SI unit of the constants $k$ , $b$ and $G$ that appear in the laws

F = k x, F = b v, F = G \frac{m _{1} m _{2}}{r ^{2}},

where $F$ is a force, $x$ and $r$ are lengths, $v$ is a velocity, $m_{1}$ and $m_{2}$ are masses.

[F] = [M L T^{- 2}]

[k] = \frac{[ F ]}{[ x ]} = \frac{[ M L T ^{- 2} ]}{[ L ]} = [M T^{- 2}] ⟹ kg/ s^{2} = N/m

[b] = \frac{[ F ]}{[ v ]} = \frac{[ M L T ^{- 2} ]}{[ L T ^{- 1} ]} = [M T^{- 1}] ⟹ kg/s

[G] = \frac{[ F ] [ r ^{2} ]}{[ m _{1} m _{2} ]} = \frac{[ M L T ^{- 2} ] [ L ^{2} ]}{[ M ^{2} ]} = [M^{- 1} L^{3} T^{- 2}] ⟹ m^{3} / (kg \cdot s^{2}) = N \cdot m^{2} /k g^{2}

Each constant is isolated in its law, and its dimensions are those of the resulting ratio, because the two sides of a physical relation have the same dimensions. Force has the dimensions of mass times acceleration, that is, $[M L T^{- 2}]$ . The spring constant $k$ is a force divided by a length, so it is measured in $N/m$ , which in base units is $kg/ s^{2}$ . The coefficient $b$ is a force divided by a velocity and is measured in $kg/s$ . For the gravitational constant the force is multiplied by the square of the distance and divided by the product of the masses; the unit is $N \cdot m^{2} /k g^{2}$ , which in base units becomes $m^{3} / (kg \cdot s^{2})$ .

Exercise 3

The speed $v$ of waves on a stretched string depends on the tension $F$ of the string, which is a force, and on its linear density $μ$ , that is, on the mass per unit length. Assuming $v = C F^{α} μ^{β}$ with $C$ dimensionless, determine $α$ and $β$ .

[v] = [L T^{- 1}], [F] = [M L T^{- 2}], [μ] = [M L^{- 1}]

[L T^{- 1}] = [M L T^{- 2}]^{α} [M L^{- 1}]^{β} = [M^{α + β} L^{α - β} T^{- 2 α}]

⎩ ⎨ ⎧ α + β = 0 α - β = 1 - 2 α = - 1 ⟹ α = \frac{1}{2}, β = - \frac{1}{2}

v = C \frac{F}{μ}

The dimensions of the two sides are written down: the linear density is a mass divided by a length, so it has dimensions $[M L^{- 1}]$ . The powers of a product distribute over the factors, therefore the right-hand side has exponent $α + β$ for mass, $α - β$ for length and $- 2 α$ for time. By dimensional homogeneity these exponents must coincide with those of the left-hand side, namely $0$ , $1$ and $- 1$ , and the three conditions must hold simultaneously. The third equation gives $α = \frac{1}{2}$ , the first $β = - \frac{1}{2}$ , and the second is satisfied, since $\frac{1}{2} + \frac{1}{2} = 1$ . Hence the speed is proportional to the square root of the tension and inversely proportional to $μ$ , and quadrupling the tension doubles the speed.

Exercise 4

The drag force $F$ that the air exerts on a fast car depends on the density $ρ$ of the air, on the speed $v$ of the car and on the area $A$ of its frontal cross section. Assuming $F = C ρ^{a} v^{b} A^{c}$ with $C$ dimensionless, determine the exponents and establish by how much the drag increases if the speed goes from $90 km/h$ to $126 km/h$ .

[F] = [M L T^{- 2}], [ρ] = [M L^{- 3}], [v] = [L T^{- 1}], [A] = [L^{2}]

[M L T^{- 2}] = [M^{a} L^{- 3 a + b + 2 c} T^{- b}]

⎩ ⎨ ⎧ a = 1 - 3 a + b + 2 c = 1 - b = - 2 ⟹ a = 1, b = 2, c = 1

F = C ρ v^{2} A, \frac{F _{2}}{F _{1}} = (\frac{v _{2}}{v _{1}})^{2} = (\frac{126}{90})^{2} = 1. 4^{2} = 1.96

The dimensions of the right-hand side are obtained by multiplying those of the three factors raised to their respective exponents: mass appears only in the density, time only in the speed, length in all three. Equating the exponents of $[M]$ , $[L]$ and $[T]$ to those of the force gives a system of three equations in three unknowns. The first and the third immediately give $a = 1$ and $b = 2$ ; substituting into the second gives $- 3 + 2 + 2 c = 1$ , hence $c = 1$ . The drag is therefore proportional to the square of the speed. The ratio of the two drag forces does not depend on the units, since the speeds appear as a ratio, and a $40%$ increase in speed almost doubles the drag, which becomes $1.96$ times the initial one.