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MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics

Physics 8.01

Supplementary Notes 1: Dimensional Analysis

1.1 International System of System of Units

There are only four fundamental quantities (measurements) necessary to

specify all physical phenomena: length, time, mass and charge. All other quantities are

expressible in terms of these, constructed as a matter of convenience. The basic system of units used throughout science and technology today is the

internationally accepted Système International (SI) (Table 1). It consists of four base

quantities and their corresponding base units: length (meter), mass (kilogram), time

(second), and electric current (ampere). The unit for electric charge, the coulomb, is

defined in terms of the ampere, and hence is referred to as a derived unit. In addition,

three other quantities, temperature, amount of substance, and luminous intensity are part

of the SI base quantities with corresponding units shown in Table 1.

Mechanics is based on just the first three of these quantities, the MKS or meter-

kilogram-second system. An alternative metric system to this, still widely used, is the so-

called CGS system (centimeter-gram-second). For distance and time measurements,

British Imperial units (especially in the USA) based on the foot (ft), the yard (yd), the

mile (mi), etc., as units of length, and also the minute, hour, day and year as units of time.

Table 1 Système International (SI) System of Units

Base Quantity Base Unit

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)

We shall refer to the dimension of the base quantity by the quantity itself, for example

dim length [length] = L, dim mass [mass] M, dim time [time] T. (1)

1.2 Dimensions of Commonly Encountered Quantities

Many physical quantities are derived from the base quantities by a set of algebraic

relations defining the physical relation between these quantities. The dimension of the

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derived quantity can always be written as a product of the powers of the dimensions of

the base quantities.

Example 1 Derived Dimensions of Mechanical Quantities

(i) The dimensions of velocity are given by the relationship

[velocity] = [length]/[time] = L T-1 . (2)

(ii) Force is also a derived quantity and using the definition of force F = ma and

acceleration a = dv / dt , force has dimensions

[force] =[mass][velocity]

[time]. (3)

We could express force in terms of mass, length, and time by the relationship

[force] =[mass][length]

[time]2= M L T-2 . (4)

(iii) The derived dimension of kinetic energy follows from the definition that K = 1

2mv

2 ,

thus

[kineticenergy] = [mass][velocity]2 , (5)

which in terms of mass, length, and time is

[kineticenergy] =[mass][length]2

[time]2= M L2 T-2 (6)

(iv) The derived dimension of work is

[work] = [force][length] , (7)

which in terms of our fundamental dimensions is

[work] =[mass][length]2

[time]2= M L2 T-2 (8)

So work and kinetic energy have the same dimensions.

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(v) Power is defined to be the rate of change in time of work so the dimensions are

[power] =[work]

[time]=

[force][length]

[time]=

[mass][length]2

[time]3= M L2 T-3 (9)

In Table 2 we list the derived dimensions of some common mechanical quantities

in terms of mass, length, and time.

Table 2 Dimensions of Some Common Mechanical Quantities

M mass , L length , T time

Quantity Dimension MKS unit

Angle dimensionless1 Dimensionless = radian

Steradian dimensionless Dimensionless = radian2

Area L

2 m

2

Volume L

3 m

3

Frequency T

-1 s

1= hertz = Hz

Velocity L T

-1 m s

1

Acceleration L T

-2 m s

2

Angular Velocity T

-1 rad s

1

Angular Acceleration T

-2 rad s

2

Density M L

-3

kg m 3

Momentum M L T

-1

kg m s 1

Angular Momentum M L

2T

-1

kg m2 s 1

Force M L T

-2

kg m s 2= newton= N

Work, Energy M L

2T

-2

kg m2 s 2= joule=J

Torque M L

2T

-2

kg m2 s 2

Power M L

2T

-3

kg m2 s 3= watt = W

Pressure M L

-1T

-2 kg m 1 s 2

= pascal= Pa

1 Even though angle and steradian are dimensionless quantities, it is often helpful to carry around a “unit” associated with them, like the radian, to understand their role in an expression or to determine if a result makes sense.

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1.3 Dimensional Analysis

There are many phenomena in nature that can be explained by simple relationships

between the observed phenomena. When trying to find a dimensional correct formula for

a quantity from a set of given quantities, an answer that is dimensionally correct will

scale properly and is generally off by a constant of order unity.

Consider a simple pendulum consisting of a massive bob suspended from a fixed

point by a string. Let T denote the time (period of the pendulum) that it takes the bob to

complete one cycle of oscillation. How does the period of the simple pendulum depend

on the quantities that define the pendulum and the quantities that determine the motion?

What possible quantities are involved? The length of the pendulum l , the mass of

the pendulum bob m , the gravitational acceleration g

, and the initial angular amplitude

of the bob 0

are all possible quantities that may enter into the formula for the period of

the swing. Have we included every possible quantity? We can never be sure but let’s first

work with this set and if we need more than we will have to think harder!

Our problem is then to find a function f such that

T = f l,m,g,

0( ) (10)

We first make a list of the dimensions of our quantities as shown in Table 3. Choose the

set: mass, length, and time, to use as the base dimensions.

Table 3 Dimensions of quantities that may describe the period of pendulum

Name of Quantity Symbol Dimensional Formula

Time of swing t T

Length of pendulum l L

Mass of pendulum m M

Gravitational acceleration g

L T-2

Angular amplitude of swing 0

No dimension

We begin by writing the period as a product of these given quantities that have

dimensions, (thus the initial angular amplitude of swing cannot enter into our expression),

with each given quantity raised to a rational power,

T = bl X mY g Z (11)

where b is a dimensionless constant. Our first observation is that since the period has

only dimensions of time, the mass of the bob cannot enter into our relationship since

neither length of the gravitational acceleration can remove the dimension of the

pendulum mass. Therefore the power Y = 0 , hence

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T = bl X g Z (12)

A dimensional analysis of Eq. (12) yields

[T ] = [L]X[L T

2]

Z= [L]

X 2Z[T ]

2Z (13)

Let’s focus on the length. In order to eliminate length, we see that the condition

0 = X + Z (14)

must be satisfied in Eq.(13). In a similar fashion, when we consider time, Eq.(13)

requires that

1= 2Z . (15)

It is a rather straightforward exercise to solve these two equations to find that

X = Z = 1 / 2 (16)

Thus a dimensionally correct formula for the period of the pendulum is

T = b l / g . (17)

Since the angular amplitude 0

is dimensionless, it may or may not appear. We

can account for this by introducing some function y(

0) into our relationship, which is

beyond the limits of this type of analysis. Then the period for a complete swing is

T = y(

0) l / g (18)

We shall discover later on by solving the problem exactly that y(

0) is independent of

the angular amplitude 0

for very small amplitudes and is equal to y(

0) = 2 ,

T = 2 l / g (19)

Example 2 Work and Kinetic Energy

A constant force of magnitude F acts over a distance d on an object of mass m . Find the

final speed v of the body.

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Solution: Here the list of quantities that might be involved is given for us. We write the

speed as

v = bFXd

Ym

Z (20)

where X , Y , and Z are rational numbers. A dimensional analysis yields

[L T1] = [M L T

2]

X[L]

Y[M ]

Z (21)

Therefore we have the following set of algebraic relations

1 X Y= + (22)

1 2X= (23)

0 X Z= + (24)

Solving this set of equations we have that

1/ 2, , 1/ 2, 1/ 21/ 2X X YZ= = == . (25)

So the final speed is

/v b Fd m= . (26)

Later on that using the work–kinetic energy theorem we will discover that the constant

2b = , and therefore

v = 2Fd / m . (27)