This document discusses dimensional analysis and dimensions. It defines dimension as fundamental quantities like length, time, mass, etc. that physical quantities can be expressed as combinations of. Dimensional analysis can be used to derive equations, check if equations are dimensionally correct, and determine the dimensions or units of derived quantities. An example shows using dimensional analysis to derive an equation for the period of a pendulum based on length and acceleration due to gravity. Exercises are provided to practice these concepts.

1. CHAPTER 1.3: DIMENSIONS
CHAPTER 1.3
DIMENSIONAL ANALYSIS

2. Grab the whole picture !
Measurements Quantities
Units Instruments
Vector Quantities
Scalar Quantities
Accuracy & Uncertainty
Dimension
Analysis Significant Figures
DIMENSIONAL ANALYSIS

3. DIMENSIONSDIMENSIONS
What is “Dimension” ?

4. Many physical quantities can be expressed in terms of aMany physical quantities can be expressed in terms of a
combination ofcombination of fundamental dimensions such asfundamental dimensions such as
[Length] L
[Time] T
[Mass] M
[Current] A
[Temperature] θ
[Amount] N
The symbol [ ] means dimension or stands for dimensionThe symbol [ ] means dimension or stands for dimension

5. There are physical quantities which are
dimensionless:
numerical value
ratio between the same quantity angle
some of the known constants like ln,
log and etc.

6. Dimensional AnalysisDimensional Analysis
Dimension analysis can be used to:
Derive an equation.
Check whether an equation is dimensionally
correct. However, dimensionally correct doesn’t
necessarily mean the equation is correct
Find out dimension or units of derived quantities.

7. Derived an Equation (Quantities)Derived an Equation (Quantities)
Example 1
Velocity = displacement / time
[velocity] = [displacement] / [time]
= L / T
= LT-1
v = s / t

8. Example 2 The period of a pendulum
The periodThe period PP of a swinging pendulum depends only on the length ofof a swinging pendulum depends only on the length of
the pendulumthe pendulum ll and the acceleration of gravityand the acceleration of gravity gg..
What are the dimensions of the variables?What are the dimensions of the variables?
t → T
m → M
ℓ → L
g → LT-2

9. T = kma
ℓb
gc
Write a general equation:Write a general equation:
T α ma
ℓ b
gc
By using the dimension method, an expression couldBy using the dimension method, an expression could
be derived that relates T, l and gbe derived that relates T, l and g
whereby a, b and c are dimensionless constantwhereby a, b and c are dimensionless constant
thusthus

10. Write out the dimensions of the variablesWrite out the dimensions of the variables
T = Ma
Lb
(LT-2
)c
T = Ma
Lb
Lc
T-2c
T1
= Ma
Lb+c
T-2c
[T] = [ma
][ℓ b
][gc
]
Using indicesUsing indices
a = 0
-2c = 1 → c = -½
b + c = 0
b = -c = ½

11. T = kma
ℓb
gc
T = km0
ℓ½
g-½
g
l
kT =
Whereby, the value of k is known by experimentWhereby, the value of k is known by experiment

12. ExercisesExercises
The viscosity force, F going against the movement of
a sphere immersed in a fluid depends on the radius of
the sphere, a the speed of the sphere, v and the
viscosity of the fluid, η. By using the dimension
method, derive an equation that relates F with a, v
and η.
(given that )
Av
Fl
=η

13. To check whether a specific formula orTo check whether a specific formula or
an equation is homogenousan equation is homogenous
Example 1
S = vt
[s] = [v] [t]
L.H.S
[s] = L
R.H.S
[v] [t] = LT-1
(T)
[v] [t] = L
Thus, the left hand side = right hand side, rendering the
equation as homogenous

14. m
F
C =
][
][
][ 2
m
F
C =
Example 2
Given that the speed for the wave of a rope is ,
Check its homogenity by using the dimensional analysis
m
F
C =2

15. L.H.S
[C] = (LT-1
)2
[C] = L2
T-2
R.H.S
[F] = MLT-2
,
= LT-2
M
MLT
M
F 2
][
][ −
=
[M] = M
Conclusion: The above equation is not homogenous
(L.H.S ≠ R.H.S)

16. ExercisesExercises
Show that the equations below are
either homogenous or otherwise
v = u + 2as
s = ut + ½ at2

17. Find out dimension or units of derivedFind out dimension or units of derived
quantitiesquantities
k
m
T π22
=
k
m
T π2=
Example
Consider the equation ,
where m is the mass and T is a time, therefore dimension of
k can be describe as
k
m
T π2=

18. 2
2
T
m
k
π
=
][
][
][ 2
T
m
k =
2
T
M
=
2−
= MT → unit: kgs-2
thus, the units of k is in kgs-2

19. ExerciseExercise
The speed of a sound wave, v going through an
elastic matter depends on the density of the
elastic matter, ρ and a constant E given as
equation
V = E½
- ρ-½
Determine the dimension for E in its SI units

20. SCALAR AND VECTOR

21. Dimensional AnalysisDimensional Analysis
Example:Example:
The periodThe period PP of a swinging pendulum depends only onof a swinging pendulum depends only on
the length of the pendulumthe length of the pendulum ll and the acceleration ofand the acceleration of
gravitygravity gg.. Which of the following formulas forWhich of the following formulas for PP couldcould
be correct ?be correct ?
g
l
P π2=
g
l
P π2=(a)(a) (b)(b) (c)(c)
Given: d has units of length (L) and g has units of (L / T 2
).
P = 2π (lg)2

22. Dimensional AnalysisDimensional Analysis
Example continue…Example continue…
Realize that the left hand side P has units of time (TT )
Try the first equation
( )P dg= 2
2
π(a)(a) (b)(b) (c)(c)
(a)(a) L
L
T
L
T
T⋅





 = ≠2
2 4
4 Not Right !!Not Right !!
P
d
g
= 2πP
d
g
= 2π

23. L
L
T
T T
2
2
= ≠
( )P dg= 2
2
π(a)(a) (b)(b) (c)(c)
(b)(b) Not Right !!Not Right !!
Example continue…Example continue…
Try the second equation
Dimensional AnalysisDimensional Analysis
P
d
g
= 2πP
d
g
= 2π

24. TT
T
L
L 2
2
==
( )P dg= 2
2
π(a)(a) (b)(b) (c)(c)
(c)(c) This has the correct units!!This has the correct units!!
This must be the answer!!This must be the answer!!
Example continue…Example continue…
Try the third equation
Dimensional AnalysisDimensional Analysis
P
d
g
= 2πP
d
g
= 2π