Dimensional Analysis

This lesson covers dimensional analysis — a powerful technique for checking equations, deriving possible forms of relationships, and converting between unit systems.

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1 Fundamental Dimensions

All mechanical quantities can be expressed in terms of three fundamental dimensions:

Dimension	Symbol
Mass	$M$M
Length	$L$L
Time	$T$T

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2 Dimensions of Common Quantities

Quantity	Formula	Dimensions
Displacement	$s$s	$L$L
Velocity	$s/t$s/t	$LT^{-1}$LT−1
Acceleration	$v/t$v/t	$LT^{-2}$LT−2
Force	$ma$ma	$MLT^{-2}$MLT−2
Momentum	$mv$mv	$MLT^{-1}$MLT−1
Energy / Work	$Fs$Fs	$ML^2T^{-2}$ML2T−2
Power	$W/t$W/t	$ML^2T^{-3}$ML2T−3
Pressure	$F/A$F/A	$ML^{-1}T^{-2}$ML−1T−2
Angular velocity	$\theta/t$θ/t	$T^{-1}$T−1
Moment of inertia	$mr^2$mr2	$ML^2$ML2
Torque	$Fr$Fr	$ML^2T^{-2}$ML2T−2
Spring constant	$F/x$F/x	$MT^{-2}$MT−2
Gravitational const $G$G	$Fr^2/(m_1 m_2)$Fr2/(m1​m2​)	$M^{-1}L^3T^{-2}$M−1L3T−2

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3 Principle of Dimensional Homogeneity

A physically valid equation must be dimensionally homogeneous: every term must have the same dimensions.

Example 1 — Checking an Equation

Check whether $v^2 = u^2 + 2as$v2=u2+2as is dimensionally consistent.

Solution

$[v^2] = (LT^{-1})^2 = L^2 T^{-2}$[v2]=(LT−1)2=L2T−2

$[u^2] = L^2 T^{-2}$[u2]=L2T−2 $\checkmark$✓

$[2as] = LT^{-2} \times L = L^2 T^{-2}$[2as]=LT−2×L=L2T−2 $\checkmark$✓

All terms have dimensions $L^2 T^{-2}$L2T−2. The equation is dimensionally consistent.

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Example 2 — Spotting an Error

Is the formula $E = mgh + \frac{1}{2}mv$E=mgh+21​mv dimensionally correct?

Solution

$[mgh] = M \times LT^{-2} \times L = ML^2T^{-2}$[mgh]=M×LT−2×L=ML2T−2

$[\frac{1}{2}mv] = M \times LT^{-1} = MLT^{-1}$[21​mv]=M×LT−1=MLT−1

These have different dimensions, so the formula is wrong. It should be $\frac{1}{2}mv^2$21​mv2.

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4 Finding a Formula by Dimensional Analysis

If a quantity $Q$Q depends on several variables, dimensional analysis can determine the form of the relationship (up to a dimensionless constant).

Method:

1. List the variables on which $Q$Q depends.
2. Assume $Q = k \cdot x_1^a \cdot x_2^b \cdot x_3^c \cdots$Q=k⋅x1a​⋅x2b​⋅x3c​⋯ where $k$k is dimensionless.
3. Equate dimensions on both sides.
4. Solve for the exponents $a, b, c, \ldots$a,b,c,….

Example 3 — Period of a Simple Pendulum

Assume the period $T$T depends on the length $l$l, mass $m$m, and gravitational acceleration $g$g.

$T = k \cdot l^a \cdot m^b \cdot g^c$T=k⋅la⋅mb⋅gc

Dimensions:

$[T] = T$[T]=T

$[l^a] = L^a$[la]=La, $[m^b] = M^b$[mb]=Mb, $[g^c] = L^c T^{-2c}$[gc]=LcT−2c

Equating:

$M$M: $0 = b$0=b, so $b = 0$b=0.

$L$L: $0 = a + c$0=a+c, so $a = -c$a=−c.

$T$T: $1 = -2c$1=−2c, so $c = -1/2$c=−1/2 and $a = 1/2$a=1/2.

Therefore: $T = k\sqrt{l/g}$T=kl/g​.