Dimensional Analysis (FM1.2): Your Mechanics Superpower
Hello! Welcome to Dimensional Analysis. Don't worry if the name sounds intense—this chapter is actually about giving you a brilliant quality control tool for all your Further Mechanics problems.
Dimensional analysis is essentially a method of checking the physics (or mechanics) of an equation by looking only at the fundamental building blocks of the quantities involved, like Mass, Length, and Time. If an equation's 'ingredients' don't match up on both sides, it must be wrong!
Key Learning Objectives
- Define the three fundamental dimensions (M, L, T).
- Determine the dimensions of derived physical quantities (like Force, Energy, Momentum).
- Use the principle of dimensional consistency to check the validity of equations.
- Predict the form of simple physical formulae.
1. The Three Fundamental Dimensions
In Further Mechanics (and classical physics), every quantity can be broken down into combinations of just three basic, independent dimensions. Think of these as the alphabet of mechanics!
The Foundational Trio
We use uppercase letters to represent these dimensions:
- Mass (M): The dimension for how much 'stuff' something is made of.
- SI Unit: kilograms (kg)
- Length (L): The dimension for distance, displacement, or size.
- SI Unit: metres (m)
- Time (T): The dimension for duration.
- SI Unit: seconds (s)
We write the dimensions of a quantity \(Q\) using square brackets: \([Q]\).
Quick Review: Basic Quantities
- \([\text{Mass}] = M\)
- \([\text{Displacement / Distance}] = L\)
- \([\text{Area}] = L^2\)
- \([\text{Volume}] = L^3\)
- \([\text{Time}] = T\)
Key Takeaway: Everything in mechanics, from speed to power, is just a specific combination of M, L, and T raised to certain powers.
2. Finding Dimensions of Derived Quantities
To find the dimension of any quantity, you must start with a known formula involving that quantity and substitute in the dimensions of the known terms.
Step-by-Step Derivation of Key Quantities
This section is crucial for FM1. You should be able to derive these quickly, especially Force and Energy.
-
Velocity (\(v\))
Formula: Velocity = Distance / Time
\([\text{Velocity}] = \frac{[\text{Distance}]}{[\text{Time}]} = \frac{L}{T} = L T^{-1}\) -
Acceleration (\(a\))
Formula: Acceleration = Change in Velocity / Time
\([\text{Acceleration}] = \frac{[\text{Velocity}]}{[\text{Time}]} = \frac{L T^{-1}}{T} = L T^{-2}\) -
Force (\(F\)) - This is the most important one!
Formula: Newton's Second Law, \(F = ma\)
\([\text{Force}] = [\text{Mass}] \times [\text{Acceleration}] = M \times L T^{-2} = \mathbf{M L T^{-2}}\)Memory Aid: If you know Force, you can derive almost everything else in mechanics!
-
Energy (Work or Potential Energy) (\(E\) or \(W\))
Formula: Work Done = Force \(\times\) Distance
\([\text{Energy}] = [\text{Force}] \times [\text{Distance}] = (M L T^{-2}) \times L = \mathbf{M L^2 T^{-2}}\) -
Momentum (\(p\))
Formula: Momentum = Mass \(\times\) Velocity
\([\text{Momentum}] = M \times L T^{-1} = \mathbf{M L T^{-1}}\)
What about constants?
When you use dimensional analysis, pure numbers, mathematical constants (\(\pi\), 2, 5.7), or trigonometric functions always have no dimensions.
Key Takeaway: Always start with a definition or a physical formula to find the dimensions. The dimension of Force, \(M L T^{-2}\), is your starting point for most complex derivations.
3. Dimensional Consistency (Homogeneity)
The most powerful principle in dimensional analysis is the Principle of Dimensional Consistency (or homogeneity).
The Golden Rule
For any equation to be physically valid, the dimensions of the terms on the Left Hand Side (LHS) must be exactly the same as the dimensions of the terms on the Right Hand Side (RHS).
Crucially, if an equation involves the sum or difference of several terms (e.g., \(A = B + C\)), then each individual term (\(A\), \(B\), and \(C\)) must have the exact same dimensions.
Analogy for Struggling Students
Imagine you are writing a recipe. If the recipe says:
Total volume of cake batter = (Volume of flour) + (Mass of eggs)
This recipe is nonsense! You cannot add a volume (\(L^3\)) to a mass (\(M\)).
In physics, if an equation suggests adding Force (\(M L T^{-2}\)) and Energy (\(M L^2 T^{-2}\)), that equation is fundamentally inconsistent and wrong.
Application A: Checking Dimensional Consistency
Let's check the consistency of a simple SUVAT equation: \(v^2 = u^2 + 2as\).
Step 1: Find the dimension of the LHS.
LHS is \(v^2\). Since \([v] = L T^{-1}\),
\([LHS] = [v^2] = (L T^{-1})^2 = L^2 T^{-2}\)
Step 2: Find the dimension of each term on the RHS.
Term 1: \(u^2\)
\([u^2] = (L T^{-1})^2 = L^2 T^{-2}\)
Term 2: \(2as\)
The number 2 has no dimension.
\([as] = [\text{Acceleration}] \times [\text{Displacement}]\)
\([as] = (L T^{-2}) \times L = L^2 T^{-2}\)
Step 3: Compare.
Since \([LHS] = L^2 T^{-2}\), \([u^2] = L^2 T^{-2}\), and \([2as] = L^2 T^{-2}\), the dimensions match. The equation is dimensionally consistent.
Did you know? Dimensional consistency confirms an equation *might* be correct, but it doesn't prove it. An equation could be dimensionally consistent but still have the wrong constant (e.g., \(v^2 = 5 u^2 + 2as\)).
Key Takeaway: Every term added or subtracted in a physically meaningful equation must have identical dimensions.
4. Application B: Predicting Formulae
Dimensional analysis is powerful because it allows us to predict the exponents (or indices) of quantities in an unknown formula. This is a common exam skill.
Step-by-Step Example: The Simple Pendulum
We want to find a formula for the Period (\(T\)) of a simple pendulum. We hypothesize that the period depends on:
- Mass of the bob (\(m\))
- Length of the string (\(l\))
- Acceleration due to gravity (\(g\))
We assume the relationship is a product of these quantities raised to unknown powers \(a, b, c\):
$$T \propto m^a l^b g^c$$
Or, including an unknown constant \(k\) (which is dimensionless):
$$T = k m^a l^b g^c$$
Step 1: Write down the dimensions of all quantities.
- \([T] = T^1\)
- \([m] = M^1\)
- \([l] = L^1\)
- \([g] = L T^{-2}\) (Gravity is an acceleration)
Step 2: Substitute the dimensions into the assumed formula.
$$[T] = [m]^a [l]^b [g]^c$$
$$M^0 L^0 T^1 = (M^1)^a (L^1)^b (L T^{-2})^c$$
Don't worry about the \(M^0 L^0\) on the LHS; we just write the dimensions explicitly to make the comparison easier.
Step 3: Equate the powers of M, L, and T on both sides.
We rearrange the RHS by combining the exponents:
$$M^0 L^0 T^1 = M^a L^{b+c} T^{-2c}$$
Now we equate the powers:
For M (Mass):
$$0 = a \implies \mathbf{a=0}$$
For T (Time):
$$1 = -2c \implies c = -\frac{1}{2}$$
For L (Length):
$$0 = b + c$$
Substitute \(c = -\frac{1}{2}\):
$$0 = b + (-\frac{1}{2}) \implies \mathbf{b=\frac{1}{2}}$$
Step 4: Substitute the indices back into the original formula.
$$T = k m^a l^b g^c$$
$$T = k m^0 l^{1/2} g^{-1/2}$$
$$T = k \sqrt{\frac{l}{g}}$$
This dimensional analysis correctly predicts that the period is independent of the mass (\(m^0\)), and proportional to the square root of the length divided by gravity. (In actual physics, the constant \(k = 2\pi\), but dimensional analysis cannot determine this constant).
Key Takeaway: Formula prediction involves setting up simultaneous equations based on the exponents of M, L, and T.
Common Mistakes and Handy Tricks
1. Forgetting the Dimensions of Acceleration (\(g\))
In mechanics problems, the acceleration due to gravity, \(g\), is often mistakenly treated as having units of Force or Mass. Remember:
\([g] = [\text{Acceleration}] = L T^{-2}\)
2. Trying to add Different Dimensions
If a question asks you to check the dimensional consistency of \(E = 2F + 3t^2\), stop immediately! \(F\) is Force and \(t^2\) is Time squared. There is no need to proceed; the equation is inconsistent unless \(F\) and \(t^2\) somehow produce the dimension of \(E\), which is highly unlikely in simple expressions.
3. Units vs. Dimensions
While they are related, they are not the same!
- Dimension: The nature of the quantity (e.g., Length, L).
- Unit: The specific scale we use to measure it (e.g., metres, m).
Quick Review Box: Essential Dimensions to Memorise
| Quantity | Dimension |
| Force (\(F\)) | \(M L T^{-2}\) |
| Energy/Work (\(E/W\)) | \(M L^2 T^{-2}\) |
| Momentum (\(p\)) | \(M L T^{-1}\) |