CHAPTER 1.3: DIMENSIONS

CHAPTER 1.3

DIMENSIONAL ANALYSIS

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Measurements Quantities

Units Instruments

Vector QuantitiesScalar Quantities

Accuracy & UncertaintyDimension Analysis Significant Figures

DIMENSIONAL ANALYSIS

DIMENSIONSDIMENSIONS

What is “Dimension” ?

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

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.

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.

Derived an Equation (Quantities)Derived an Equation (Quantities)

Example 1

Velocity = displacement / time

[velocity] = [displacement] / [time]

= L / T

= LT-1

v = s / t

Example 2 The period of a pendulum

The periodThe period PP of a swinging pendulum depends only on the length of of 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

T = kma ℓbgc

Write a general equation:Write a general equation:

T α ma ℓ bgc

By using the dimension method, an expression could By 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

Write out the dimensions of the variablesWrite out the dimensions of the variables

T = MaLb(LT-2)c

T = MaLbLcT-2c

T1 = MaLb+cT-2c

[T] = [ma][ℓ b][gc]

Using indicesUsing indices

a = 0 -2c = 1 → c = -½

b + c = 0 b = -c = ½

T = kma ℓbgc

T = km0 ℓ½g-½

g

lkT

Whereby, the value of k is known by experimentWhereby, the value of k is known by experiment

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

To check whether a specific formula or To 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] = LThus, the left hand side = right hand side, rendering the equation as homogenous

m

FC

][

][][ 2

m

FC

Example 2

Given that the speed for the wave of a rope is ,

Check its homogenity by using the dimensional analysis

m

FC 2

L.H.S

[C] = (LT-1)2

[C] = L2T-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)

ExercisesExercises

Show that the equations below are

either homogenous or otherwise

v = u + 2as

s = ut + ½ at2

Find out dimension or units of derived Find out dimension or units of derived quantitiesquantities

k

mT 22

k

mT 2

Example

Consider the equation ,

where m is the mass and T is a time, therefore dimension of k can be describe as

k

mT 2

2

2

T

mk

][

][][

2T

mk

2T

M

2MT → unit: kgs-2

thus, the units of k is in kgs-2

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

SCALAR AND VECTOR

Dimensional AnalysisDimensional Analysis

Example:Example:

The periodThe period PP of a swinging pendulum depends only on of a swinging pendulum depends only on the length of the pendulumthe length of the pendulum ll and the acceleration of and the acceleration of gravitygravity gg.. Which of the following formulas forWhich of the following formulas for PP couldcould be correct ?be correct ?

g

lP 2

g

lP 2(a)(a) (b)(b) (c)(c)

Given: d has units of length (L) and g has units of (L / T 2).

P = 2 (lg)2

Dimensional AnalysisDimensional Analysis

Example continue…Example continue…Realize that the left hand side P has units of time (TT )

Try the first equation

P dg2 2(a)(a) (b)(b) (c)(c)

(a)(a) LL

T

L

TT

2

2 4

4 Not Right !!Not Right !!

Pdg

2Pdg

2

LL

T

T T

2

2

P dg2 2(a)(a) (b)(b) (c)(c)

(b)(b) Not Right !!Not Right !!

Example continue…Example continue…Try the second equation

Dimensional AnalysisDimensional Analysis

Pdg

2Pdg

2

TT

TLL 2

2

P dg2 2(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

Pdg

2Pdg

2