Numbers, Units & Dimensional Analysis
π― What You'll Learn
- Understand SI units and perform unit conversions confidently
- Apply dimensional analysis to check equation validity
- Express numbers in scientific notation and use significant figures correctly
- Distinguish between base and derived quantities
π Prerequisites
Basic algebra (powers, fractions, solving simple equations). No prior physics knowledge needed.
1. The Concept
Why Units Matter
In physics, every measurement is meaningless without a unit. Saying "the car traveled 50" tells us nothingβ50 what? Meters? Miles? Light-years?
Units transform raw numbers into physical information. They let us communicate precisely, convert between different measurement systems, and verify that our equations make physical sense.
The SI System
The International System (SI) defines seven base units from which all other units are derived:
| Quantity | Unit Name | Symbol |
|---|---|---|
| Length | meter | m |
| Mass | kilogram | kg |
| Time | second | s |
| Electric Current | ampere | A |
| Temperature | kelvin | K |
| Amount of Substance | mole | mol |
| Luminous Intensity | candela | cd |
All other units are derived units, built from combinations of base units. For example:
- Speed: meters per second (m/s)
- Force: kilogram-meter per second squared (kgΒ·m/sΒ²) = Newton (N)
- Energy: kilogram-meter squared per second squared (kgΒ·mΒ²/sΒ²) = Joule (J)
Dimensional Analysis
Every physical quantity has a dimension β a description of what type of measurement it is. We write dimensions using square brackets:
- Length: $[L]$
- Mass: $[M]$
- Time: $[T]$
- Speed: $[L T^{-1}]$ (length divided by time)
- Acceleration: $[L T^{-2}]$
- Force: $[M L T^{-2}]$
Key Principle: In any valid physics equation, both sides must have the same dimensions. This is called dimensional homogeneity.
Dimensional analysis is a powerful tool for:
- Checking if an equation is dimensionally correct
- Deriving relationships between physical quantities
- Converting between unit systems
2. Key Equations
Speed Formula
Where $v$ = speed (m/s), $d$ = distance (m), $t$ = time (s)
Dimensions: $[L T^{-1}] = \frac{[L]}{[T]}$ β
Dimensional Formula for Force
From Newton's second law: $F = ma$, where $[a] = [L T^{-2}]$
Unit Conversion Formula
Example: Convert km to m by multiplying by $\frac{1000 \text{ m}}{1 \text{ km}}$
3. Worked Examples
Example 1: Unit Conversion
Question: Convert 72 km/h to m/s.
Solution:
We need to convert kilometers to meters and hours to seconds:
Tip: To convert km/h to m/s, divide by 3.6
Example 2: Dimensional Analysis Check
Question: A student claims that velocity can be calculated using $v = at^2$, where $a$ is acceleration and $t$ is time. Is this dimensionally correct?
Solution:
Let's check the dimensions on both sides:
Left side: $[v] = [L T^{-1}]$
Right side: $[a t^2] = [L T^{-2}] \cdot [T^2] = [L]$
Conclusion: $[L T^{-1}] \neq [L]$ β The equation is dimensionally incorrect!
The correct relationship is $v = at$ (velocity equals acceleration times time).
Example 3: Deriving Relationships Using Dimensions
Question: The period $T$ of a simple pendulum depends on its length $L$ and gravitational acceleration $g$. Use dimensional analysis to find the form of the relationship.
Solution:
Assume: $T \propto L^a g^b$ for some powers $a$ and $b$.
Writing dimensions:
For dimensional consistency:
- Power of $L$: $a + b = 0$ β $a = -b$
- Power of $T$: $-2b = 1$ β $b = -\frac{1}{2}$
- Therefore: $a = \frac{1}{2}$
Result: $T \propto L^{1/2} g^{-1/2}$ β $T \propto \sqrt{\frac{L}{g}}$
The complete formula is $T = 2\pi\sqrt{\frac{L}{g}}$ (dimensional analysis can't find the $2\pi$ constant).
4. Practice Problems
Problem 1 (Easy): Convert 5 m/s to km/h.
Problem 2 (Medium): The kinetic energy $E$ of an object depends on its mass $m$ and velocity $v$. Use dimensional analysis to determine if $E \propto m v^2$ is dimensionally correct. (Energy has dimensions $[M L^2 T^{-2}]$)
Problem 3 (Medium): Check if the equation $s = ut + \frac{1}{2}at^2$ is dimensionally consistent, where $s$ is displacement, $u$ is initial velocity, $a$ is acceleration, and $t$ is time.
Problem 4 (Hard): The drag force $F_d$ on a sphere moving through a fluid depends on the sphere's radius $r$, velocity $v$, and the fluid's viscosity $\eta$ (dimensions $[M L^{-1} T^{-1}]$). Find the form of $F_d$ using dimensional analysis.
Show Answers & Solutions
Answer 1: 18 km/h
Solution: $5 \text{ m/s} = 5 \times 3.6 = 18 \text{ km/h}$
Answer 2: Yes, dimensionally correct β
Solution:
$[E] = [M L^2 T^{-2}]$
$[m v^2] = [M] \cdot [L T^{-1}]^2 = [M L^2 T^{-2}]$ β
Answer 3: Yes, both terms are dimensionally consistent β
Solution:
$[s] = [L]$
$[ut] = [L T^{-1}] \cdot [T] = [L]$ β
$[\frac{1}{2}at^2] = [L T^{-2}] \cdot [T^2] = [L]$ β
All terms have dimension $[L]$, so the equation is valid.
Answer 4: $F_d \propto \eta r v$
Solution:
Assume $F_d = k \eta^a r^b v^c$ where $k$ is dimensionless.
$[M L T^{-2}] = [M L^{-1} T^{-1}]^a [L]^b [L T^{-1}]^c$
$[M L T^{-2}] = [M^a L^{-a+b+c} T^{-a-c}]$
Matching powers: $a=1$, $-a+b+c=1$ β $b+c=2$, $-a-c=-2$ β $c=1$
Therefore: $b=1$, giving us $F_d \propto \eta r v$ (Stokes' Law)
5. Summary & Common Mistakes
π Key Takeaways
- Always write units with every measurement β numbers without units are meaningless in physics
- The SI system has 7 base units; all others are derived from these
- Use dimensional analysis to verify equations and derive relationships
- Both sides of any valid equation must have the same dimensions
- Unit conversion: multiply by conversion factors equal to 1
β οΈ Common Mistakes
- Forgetting to convert all units: Converting km to m but forgetting to convert hours to seconds (results in being off by factor of 3.6!)
- Adding quantities with different dimensions: You can't add 5 meters + 3 seconds β it's physically meaningless
- Confusing dimensional analysis with complete solutions: Dimensional analysis gives you the form of relationships but can't determine dimensionless constants like $2\pi$
- Mixing up base and derived units: Remember that force (N), energy (J), and power (W) are derived units, not base units
- Using wrong conversion factors: Double-check your conversion factors β is 1 km = 1000 m or 100 m?
