Uses Of Dimensional Analysis (Board Level)

Application of Dimensional Analysis

 There are mainly three applications of dimensional analysis.

1.   To check the correctness of a equation,
2.   To convert one system of units into other
3.   To find relation between different physical quantities

To check the dimensional correctness of a given physical relation :

This is based on the ‘principle of homogeneity’. According to this principle the dimensions of each term on both sides of an equation must be the same. It is based on the fact that only similar quantities can be added or subtracted.

For example consider an equation If Y=P\pm {{(Q)}^{{1/2}}}\pm \sqrt{{(D+E)F}},

Where Y, P, Q, D, E and F are some physical quantities. Then according to principle of homogeneity all terms on RHS like P, Q^1/2, \sqrt{{(D+E)F}} must have same dimensions, and also the dimensions of Y must also be same.

Hence   [Y] = [P] = [Q^1/2] = [\sqrt{{(D+E)F}}]

If the dimensions of each term on both sides are same, the equation is dimensionally correct, otherwise not. But Note that a dimensionally correct equation may or may not be physically correct.

Example :

(i) F=m{{v}^{2}}/{{r}^{2}}

By substituting dimension of the physical quantities in the above relation, [ML{{T}^{{-2}}}]=[M]{{[L{{T}^{{-1}}}]}^{2}}/{{[L]}^{2}}

i.e. [ML{{T}^{{-2}}}]=[M{{T}^{{-2}}}]

As in the above equation dimensions of both sides are not same; this formula is not correct dimensionally, so can never be physically correct

(ii) S=ut+(1/2)a{{t}^{2}}

By substituting dimension of the physical quantities in the above relation

 [L] = [LT^–1][T] + [LT^–2][T²]

i.e.  [L] = [L] + [L]

As in the above equation dimensions of each term on both sides are same, so this equation is dimensionally correct.

(iii) Look at the equation : \displaystyle {{v}^{2}}={{u}^{2}}+2as        

Dimensions of \displaystyle {{v}^{2}}:[{{L}^{2}}{{T}^{{-2}}}]

Dimensions of \displaystyle {{u}^{2}}:[{{L}^{2}}{{T}^{{-2}}}]

Dimensions of \displaystyle 2as:[L{{T}^{{-2}}}][L]=[{{L}^{2}}{{T}^{{-2}}}]

The equation \displaystyle {{v}^{2}}={{u}^{2}}+2as is dimensionally consistent, or dimensionally correct.

 To convert one system of units to other :

The measure of a physical quantity is  `nu’  where n is the numerical value and u be the units of the physical quantity. Because we are finding the equivalent value in second system means nu remains constant in the two systems and hence nu = constant

If a physical quantity X has dimensional formula [M^aL^bT^c] and if (derived) units of that physical quantity in two systems are [M_{1}^{a}L_{1}^{b}T_{1}^{c}] and [M_{2}^{a}L_{2}^{b}T_{2}^{c}] respectively and n₁ and n₂ be the numerical values in the two systems respectively, then {{n}_{1}}[{{u}_{1}}]={{n}_{2}}[{{u}_{2}}]

Þ {{n}_{1}}[M_{1}^{a}L_{1}^{b}T_{1}^{c}]={{n}_{2}}[M_{2}^{a}L_{2}^{b}T_{2}^{c}]

Þ {{n}_{2}}={{n}_{1}}{{\left[ {\frac{{{{M}_{1}}}}{{{{M}_{2}}}}} \right]}^{a}}{{\left[ {\frac{{{{L}_{1}}}}{{{{L}_{2}}}}} \right]}^{b}}{{\left[ {\frac{{{{T}_{1}}}}{{{{T}_{2}}}}} \right]}^{c}}

where M₁, L₁ and T₁­ are fundamental units of mass, length and time in the first (known) system and M₂, L₂ and T₂ are fundamental units of mass, length and time in the second (unknown) system. Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated.

Example :

(i) conversion of 1newton into dyne.

The newton is the S.I. unit of force and has dimensional formula [MLT^–2].

So 1 N = 1 kg-m/ sec²

By using   {{n}_{2}}={{n}_{1}}{{\left[ {\frac{{{{M}_{1}}}}{{{{M}_{2}}}}} \right]}^{a}}{{\left[ {\frac{{{{L}_{1}}}}{{{{L}_{2}}}}} \right]}^{b}}{{\left[ {\frac{{{{T}_{1}}}}{{{{T}_{2}}}}} \right]}^{c}}

=1\,{{\left[ {\frac{{kg}}{{gm}}} \right]}^{1}}\,{{\left[ {\frac{m}{{cm}}} \right]}^{1}}{{\left[ {\frac{{sec}}{{sec}}} \right]}^{{-2}}}

=1\,{{\left[ {\frac{{{{{10}}^{3}}gm}}{{gm}}} \right]}^{1}}\,{{\left[ {\frac{{{{{10}}^{2}}cm}}{{cm}}} \right]}^{1}}{{\left[ {\frac{{sec}}{{sec}}} \right]}^{{-2}}}

\ 1 N = 10⁵ dyne

(ii) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system

The value of G in C.G.S. system is 6.67 ´ 10^–8 C.G.S. units while its dimensional formula is [M^–1L³T^–2]

So G = 6.67 ´ 10^–8 cm³/g s² 

By using {{n}_{2}}={{n}_{1}}{{\left[ {\frac{{{{M}_{1}}}}{{{{M}_{2}}}}} \right]}^{a}}{{\left[ {\frac{{{{L}_{1}}}}{{{{L}_{2}}}}} \right]}^{b}}{{\left[ {\frac{{{{T}_{1}}}}{{{{T}_{2}}}}} \right]}^{c}}

=6.67\times {{10}^{{-8}}}{{\left[ {\frac{{gm}}{{kg}}} \right]}^{{-1}}}{{\left[ {\frac{{cm}}{m}} \right]}^{3}}{{\left[ {\frac{{sec}}{{sec}}} \right]}^{{-2}}}

=6.67\times {{10}^{{-8}}}{{\left[ {\frac{{gm}}{{{{{10}}^{3}}gm}}} \right]}^{{-1}}}{{\left[ {\frac{{cm}}{{{{{10}}^{2}}cm}}} \right]}^{3}}{{\left[ {\frac{{sec}}{{sec}}} \right]}^{{-2}}}

=6.67\times {{10}^{{-11}}}

\  G =  6.67 ´ 10^–11 M.K.S. units

To find relation between different physical quantities:

If one knows the dependency of a physical quantity on other quantities and if the dependency is of the product type, then using the method of dimensional analysis, relation between the quantities can be derived.

Example :

(i) Let’s find the relation for the time period of a simple pendulum.

Let time period of a simple pendulum is a function of mass of the bob (m), effective length (l), acceleration due to gravity (g) then assuming the function to be product of power function of m, l and g 

Hence \displaystyle T\propto {{m}^{x}}{{l}^{y}}{{g}^{z}}

i.e., T=K{{m}^{x}}{{l}^{y}}{{g}^{z}}  ….(1)

where K = dimensionless constant

Now lets substitute the dimensions of each quantity. 

[T] = [M]^x [L]^y [LT^–2]^z       

or     [M⁰ L⁰ T¹]  = [ M^x L^y+z T^–2z ]

For equation to be correct the dimensions must be same on LHS and RHS . Hence we equate the dimensions (powers) of similar quantities in LHS and RHS

On equating powers of M we get x = 0,

On equating powers of L  we get y+z = 0

and on equating powers of T we get  -2z = 1   or z= -1/2,

since y+z = 0 on putting z=-1/2 we get y= 1/2.

Now we put back these values of x, y, z in equation (1)

So the required physical relation becomes T=K\sqrt{{\frac{l}{g}}}

The value of dimensionless constant is found (2p ) through experiments so T=2\pi \sqrt{{\frac{l}{g}}}

(ii) Find a relation for the viscous force acting on a spherical body of radius `r’, moving with a speed  `v’  through a liquid of coefficient of viscosity `h‘

So  F = f (h, r, v) 

hence \displaystyle F\propto {{\eta }^{x}}{{r}^{y}}{{v}^{z}}

or    F=K{{\eta }^{x}}{{r}^{y}}{{v}^{z}} …. (2)

where K is dimensionless constant.

If the above relation is dimensionally correct

[ML{{T}^{{-2}}}]={{[M{{L}^{{-1}}}{{T}^{{-1}}}]}^{x}}{{[L]}^{y}}{{[L{{T}^{{-1}}}]}^{z}}

or [ML{{T}^{{-2}}}]=[{{M}^{x}}{{L}^{{-x+y+z}}}{{T}^{{-x-z}}}]

Equating the powers of similar quantities

x = 1;   – x + y + z = 1  and  – x – z = – 2

Solving these for x, y and z, we get x = y = z = 1, Now put back these values in equation(2)

we get    F = Khr v

On experimental grounds, K = 6p; so  F = 6phrv

This is the famous Stoke’s law.

Limitations of Dimensional Analysis

Although dimensional analysis is very useful but it has some limitations as given below

(1) If dimensions are given, physical quantity may not be unique as many physical quantities have same dimensions. For example if the dimensional formula of a physical quantity is [M{{L}^{2}}{{T}^{{-2}}}]it may be work or energy or torque.

(2) Numerical constant having no dimensions [K] such as (1/2), 1 or 2p etc. cannot be deduced by the methods of dimensions.

(3) The method of dimensions can not be used to derive relations other than product of power functions. For example,

s=u\,t+\,(1/2)\,a\,{{t}^{2}} or y=a\sin \omega \,t

cannot be derived by using this theory (try if you can). However, the dimensional correctness of these can be checked.

(4) The method of dimensions cannot be applied to derive formula if in mechanics a physical quantity depends on more than 3 physical quantities as then there will be less number (= 3) of equations than the unknowns (>3). However still we can check correctness of the given equation dimensionally. For example T=2\pi \sqrt{{{I}/{{mgl}}\;}}can not be derived by theory of dimensions but its dimensional correctness can be checked.

(5) Even if a physical quantity depends on 3 physical quantities, out of which two have same dimensions, the formula cannot be derived by theory of dimensions, e.g., formula for the frequency of a tuning fork f=(d/{{L}^{2}})\,v cannot be derived by theory of dimensions but can be checked.

Illustration

Using the method of dimensions, find the acceleration of a particle moving with a constant speed v in a circle of radius r.

Solution

Assuming that the acceleration of a particle depends on v and r

                        a a v^x r^y

                Þ      a = k v^x r^y      ….. (i)  

                        Now as we know dimensions of acceleration (a) = M°LT^–2

                        and dimensions of velocity (v) = M°LT^–1

                        dimension of radius (r) = M°LT°

                        Putting all three dimensions  in (1), we get

                        M°LT^–2 = (M°LT^–1)^x (M°LT°)^y

                        M°LT^–2 = M° L^{x + y} T^–x

                        Comparing the powers, we get

                        x + y = 1

                        x = 2

                        \ y = 1-2 = -1

                        \ from equation (i) a = k v²r^–1

                        \displaystyle a=\frac{{k{{v}^{2}}}}{r}

Illustration

In the expression \displaystyle \left( {P+\frac{a}{{{{v}^{2}}}}} \right)(v-b)=RT

 P is pressure and V is the volume. Calculate the dimensions of a and b.

Solution         

Only physical quantities having same dimensions are added or subtracted. So \displaystyle \frac{a}{{{{v}^{2}}}} has the same dimensions as that of pressure.

                        As pressure = \displaystyle \frac{{Force}}{{Area}}

                        Dimensions of pressure \displaystyle =\frac{{ML{{T}^{{-2}}}}}{{{{L}^{2}}}}=M{{L}^{{-1}}}{{T}^{{-2}}}

                        \ Dimensions of \displaystyle \frac{a}{{{{v}^{2}}}}=M{{L}^{{-1}}}{{T}^{{-2}}}

                        Dimensions of a \displaystyle =M{{L}^{{-1}}}{{T}^{{-2}}}{{({{V}^{3}})}^{2}}

                        \displaystyle =(M{{L}^{{-1}}}{{T}^{{-2}}}){{({{L}^{3}})}^{2}}

                        \displaystyle =M{{L}^{{-1}}}{{T}^{{-2}}}{{L}^{6}}=M{{L}^{5}}{{T}^{{-2}}}

                        Similarly dimensions of b is same as that of volume.

                        Dimensions of  \displaystyle b={{M}^{0}}{{L}^{3}}{{T}^{0}}.

Illustration   Does \displaystyle {{S}_{{nth}}}=u+\frac{a}{2}(n-1) dimensionally correct?

Solution        

Yes, this expression is dimensionally correct, yet it appears to be incorrect. As we are taking it to be for n^th second. Here one second is divided through the equation. 

Illustration

A displacement of a particle is given by equation y = A sin wt, where y is in metres and A is also in metres, t is in seconds. What are the dimensions of w.

Solution       

As the angles are always dimensionless, so

                        wt = dimensionless quantity

                        Dimensions of wt = [M°L°T°]    

                        Hence  [w].[ T ] = [M°L°T°]

                       therefore dimensions of w = {M°L°T^–1]

Illustration

If density r, acceleration due to gravity g and frequency f are the basic quantities, find the dimensions of force.

Solution:          We know that            [r] = ML^–3,         [g] = LT^–2           [ f ] = T^–1

\displaystyle F\propto {{\rho }^{x}}{{g}^{y}}{{f}^{z}}            

hence   \displaystyle F=K{{\rho }^{x}}{{g}^{y}}{{f}^{z}}        …(A)

Now we put dimensions of each quantity

we get  \displaystyle [ML{{T}^{{-2}}}]={{[M{{L}^{{-3}}}]}^{x}}\text{ }{{[L{{T}^{{-2}}}]}^{y}}\text{ }{{[{{T}^{{-1}}}]}^{z}}

Now as usual  equating powers of M,L,T on LHS and RHS we can find the values of x,y,z. Which are

x=1 , y=4 , z=-6. On substituting these values in equation (A) we find the dimensions of force in terms of r,g,f.

Hence  \displaystyle F=K\rho {{g}^{4}}{{f}^{{-6}}}

therefore the dimensions of F=[rg⁴f^-6]

Illustration 

If force, length and time would have been the fundamental units what would have been the dimensional formula for mass.

Solution          Let   M = K F^x L^y T^z                                   ….(B) 

                      Put the dimensions in both the sides

                               [M]  =  [MLT^–2]^x [L]^y [T]^z

                or            [ML⁰T⁰] = [M]^x [L]^(x+y) [T]^(–2x+z)]

                     On equating the powers of similar quantities 

                     we get   x = 1, 

                                   x + y = 0    

                     &       – 2x + z = 0

                     Hence  x = 1, y = – 1,  z = 2

                using equation (B) we get  M = K FL^-1T².

                therefore dimensions of mass M = [ FL^-1T²]