IEMDaily - Video Lecture Notes

Definition

Density

Density is the characteristic property of a substance. It is defined as the ratio of mass of a substance to it's volume.

ρ = \frac{m}{V}

Example

Example of how to calculate density?

A Sample of ethanol has a volume of

240 c m^{3}

. Its mass is found to be 190g. What is the density of ethanol?
Step 1 : Given
mass M

= 190 g

volume V

= 240 c m^{3}

Density D = ?
Step 2 : Write equation for density and put all the values in it.

D = \frac{M}{V}

D = \frac{190}{240}

= 0.79 \frac{g}{c m^{3}}

Definition

Dimensional Formula

The dimensional formula is defined as the expression of the physical quantity in terms of its basic unit with proper dimensions. For example, dimensional force is

F = [M L T^{- 2}]

It's because the unit of Force is N(newton) or

k g \times m / s^{2}

Definition

Dimensional formula of function of quantities

Let dimensional formulas of two quantities be given by

[A] = [M^{a m} L^{a l} T^{a t}]

and

[B] = [M^{b m} L^{b l} T^{b t}]

Then dimensional formula of

A^{x} B^{y}

is given by

[A^{x} B^{y}] = [M^{x a m + y b m} L^{x a l + y b l} T^{x a t + y b t}]

Example:
Angular momentum of a physical quantity is given by

I = m v r

. Its dimensional formula can be found as:

[m] = [M^{1} L^{0} T^{0}]

[v] = [M^{0} L^{1} T^{- 1}]

[r] = [M^{0} L^{1} T^{0}]

Hence,

[I] = [m v r] = [M^{1} L^{2} T^{- 1}]

Shortcut

Conversion between Units using dimensional Analysis

If density of a material is

1

k g / m^{3}

Density has dimensional formula:

[M L^{- 3}]

To convert density of the object in CGS system.
Density=

1000 / 10^{6}

10^{- 3}

g / c c

Definition

Correctness of Physical Equation Using Dimensional Analysis

Checking the correctness of physical equation is based on the principle of homogeneity of dimensions. According to this principle, only physical quantities of the same nature having the same dimensions can be added, subtracted or can be equated. To check correctness of given physical equation, the physical quantities on two side of the equations are expressed in terms of fundamental units of mass, length and time. The powers of

M, L

T

are same on two sides of the equations, then the physical equation is correct otherwise not.

Definition

Establishment of relationship between physical quantities

If all the factors affecting a derived quantity is known, then the function relating it from the quantities can be established using dimensional analysis.
Example: Finding time-period of a simple pendulum (

T

) given it depends on length of the pendulum (

l

) and acceleration due to gravity (

g

).
Dimensional formulae of the quantities are:

[T] = [M^{0} L^{0} T^{1}]

[l] = [M^{0} L^{1} T^{0}]

[g] = [M^{0} L^{1} T^{- 2}]

Let

T = k l^{A} g^{B}

where

k, A, B

are constants.
Then,

[T] = [k] [l^{A}] [g^{B}]

[M^{0} L^{0} T^{1}] = [M^{0} L^{A + B} T^{- 2 B}]

Equating the powers on LHS and RHS,

A + B = 0

- 2 B = 1

Solving,

A = \frac{1}{2}, B = \frac{- 1}{2}

Hence, time-period is given by:

T = k l^{1 / 2} g^{- 1 / 2}

Note:
The established relation between the physical quantities is not unique and hence may or may not be absolutely correct.

Definition

Dimension

Dimension, an expression of the character of a derived quantity in relation to fundamental quantities, without regard for its numerical value. In any system of measurement, such as the metric system, certain quantities are considered fundamental, and all others are considered to be derived from them.

Definition

Uses of Dimensional Analysis

1. To check the correctness of a physical equation.
2. To derive the relation between different physical quantities involved in a physical phenomenon.
3. To change from one system of units to another.

Definition

Limitations of Dimensional Methods

Dimensional analysis has no information on dimensionless constants.

If a quantity is dependent on logarithmic, trigonometric or exponential functions, this method cannot be used. e.g. :
$y = a \cos (ω t - k x)$ can not be derived using this method.

In some cases, it is difficult to guess the factors while deriving the relation connecting two or more physical quantities.

This method cannot be used in an equation containing two or more variables with same dimensions. Therefore, a dimensionally correct relation may not always be the actual correct relation. e.g. :
dimension of $\frac{1}{T}$ and $\frac{2}{T}$ are same.

It cannot be used if the physical quantity is dependent on more than three unknown variables. e.g. $T = 2 π \sqrt{\frac{I}{m g L}}$ cannot be derived by using dimensions.
This method cannot be used if the physical quantity contains more than one term, say sum or difference of two terms i.e it does not always tell us the exact form of a relation. eg. $v = u + a t$ cannot be derived using this relation.
It does not tell whether a given physical quantity is a scalar or a vector.