Class 11 > Unit # 01: Physical Quantities & Measurements > Deriving Equations Using Dimensions

Derivation of Physics Equations Using Dimensional Analysis

By applying the Principle of Homogeneity of Dimensions, we can construct relationships between different physical quantities if we know the factors upon which a specific quantity depends. By setting up a proportionality expression with unknown exponents ($a, b, c$), we can solve a system of linear equations to deduce the exact formula up to a dimensionless constant ($k$). Below are five common examples illustrating this process.

Example 1: Centripetal Force ($F_c$)

Problem Statement: Determine the force ($F$) acting on a body moving in a circle, assuming it depends on its mass ($m$), velocity ($v$), and radius ($r$).

Step 1: Set up the proportionality relation:
$$F \propto m^a v^b r^c \implies F = k \cdot m^a v^b r^c \quad \text{--- (Eq. 1)}$$

Step 2: Substitute dimensional formulas:
Dimension of Force $F = $ [MLT⁻²]
$$\text{[MLT⁻²]} = \text{[M]}^a \cdot \text{[LT⁻¹]}^b \cdot \text{[L]}^c$$ $$\text{[M¹L¹T⁻²]} = \text{[M]}^a \cdot \text{[L]}^{b+c} \cdot \text{[T]}^{-b}$$

Step 3: Equate the exponents of M, L, and T:
For $\text{M}$: $a = 1$
For $\text{T}$: $-b = -2 \implies b = 2$
For $\text{L}$: $b + c = 1 \implies 2 + c = 1 \implies c = -1$

Step 4: Substitute back into Eq. 1:

$$F = k \cdot m^1 v^2 r^{-1}$$

Result: $F = k \frac{m v^2}{r}$  (Experimentally, $k = 1$, giving $F = \frac{mv^2}{r}$)

Example 2: Time Period of a Simple Pendulum ($T$)

Problem Statement: Derive the expression for the time period ($T$) of a simple pendulum, assuming it depends on the mass of the bob ($m$), length of the string ($l$), and acceleration due to gravity ($g$).

Step 1: Set up the proportionality relation:
$$T = k \cdot m^a l^b g^c \quad \text{--- (Eq. 1)}$$

Step 2: Substitute dimensional formulas:
Dimension of Time Period $T = $ [T]
$$\text{[M⁰L⁰T¹]} = \text{[M]}^a \cdot \text{[L]}^b \cdot \text{[LT⁻²]}^c$$ $$\text{[M⁰L⁰T¹]} = \text{[M]}^a \cdot \text{[L]}^{b+c} \cdot \text{[T]}^{-2c}$$

Step 3: Equate the exponents:
For $\text{M}$: $a = 0$ (Time period does not depend on mass)
For $\text{T}$: $-2c = 1 \implies c = -\frac{1}{2}$
For $\text{L}$: $b + c = 0 \implies b - \frac{1}{2} = 0 \implies b = \frac{1}{2}$

Step 4: Substitute back into Eq. 1:

$$T = k \cdot m^0 l^{1/2} g^{-1/2} = k \cdot \sqrt{\frac{l}{g}}$$

Result: $T = k \sqrt{\frac{l}{g}}$  (Experimentally, $k = 2\pi$, giving $T = 2\pi\sqrt{\frac{l}{g}}$)

Example 3: Kinetic Energy ($K.E.$)

Problem Statement: Establish a formula for the kinetic energy ($E$) of a moving body, assuming it depends on its mass ($m$) and its velocity ($v$).

Step 1: Set up the proportionality relation:
$$E = k \cdot m^a v^b \quad \text{--- (Eq. 1)}$$

Step 2: Substitute dimensional formulas:
Dimension of Energy $E = $ [ML²T⁻²]
$$\text{[ML²T⁻²]} = \text{[M]}^a \cdot \text{[LT⁻¹]}^b$$ $$\text{[M¹L²T⁻²]} = \text{[M]}^a \cdot \text{[L]}^b \cdot \text{[T]}^{-b}$$

Step 3: Equate the exponents:
For $\text{M}$: $a = 1$
For $\text{L}$: $b = 2$
For $\text{T}$: $-b = -2 \implies b = 2$ (Consistent with the line above)

Step 4: Substitute back into Eq. 1:

$$E = k \cdot m^1 v^2$$

Result: $E = k m v^2$  (Experimentally, $k = \frac{1}{2}$, giving $K.E. = \frac{1}{2}mv^2$)

Example 4: Velocity of Sound Waves in a Medium ($v$)

Problem Statement: Derive an expression for the velocity ($v$) of sound waves traveling through a medium, depending on the modulus of elasticity ($E$) and the density ($\rho$) of the medium.

Step 1: Set up the proportionality relation:
$$v = k \cdot E^a \rho^b \quad \text{--- (Eq. 1)}$$

Step 2: Substitute dimensional formulas:
Velocity $v = $ [LT⁻¹], Elasticity $E = \text{[ML⁻¹T⁻²]}$, Density $\rho = \text{[ML⁻³]}$
$$\text{[M⁰L¹T⁻¹]} = \text{[ML⁻¹T⁻²]}^a \cdot \text{[ML⁻³]}^b$$ $$\text{[M⁰L¹T⁻¹]} = \text{[M]}^{a+b} \cdot \text{[L]}^{-a-3b} \cdot \text{[T]}^{-2a}$$

Step 3: Equate the exponents:
For $\text{T}$: $-2a = -1 \implies a = \frac{1}{2}$
For $\text{M}$: $a + b = 0 \implies \frac{1}{2} + b = 0 \implies b = -\frac{1}{2}$
For $\text{L}$: $-a - 3b = -\frac{1}{2} - 3(-\frac{1}{2}) = -\frac{1}{2} + \frac{3}{2} = 1$ (Consistent)

Step 4: Substitute back into Eq. 1:

$$v = k \cdot E^{1/2} \rho^{-1/2} = k \cdot \sqrt{\frac{E}{\rho}}$$

Result: $v = k \sqrt{\frac{E}{\rho}}$  (Newton-Laplace formula style configuration)

Example 5: Stokes' Law for Viscous Drag Force ($F$)

Problem Statement: Denuce the viscous drag force ($F$) operating on a small spherical ball falling through a fluid, assuming it depends on the coefficient of viscosity ($\eta$), radius of the ball ($r$), and terminal velocity ($v$).

Step 1: Set up the proportionality relation:
$$F = k \cdot \eta^a r^b v^c \quad \text{--- (Eq. 1)}$$

Step 2: Substitute dimensional formulas:
Force $F = \text{[MLT⁻²]}$, Coefficient of Viscosity $\eta = \text{[ML⁻¹T⁻¹]}$, Radius $r = \text{[L]}$, Velocity $v = \text{[LT⁻¹]}$
$$\text{[M¹L¹T⁻²]} = \text{[ML⁻¹T⁻¹]}^a \cdot \text{[L]}^b \cdot \text{[LT⁻¹]}^c$$ $$\text{[M¹L¹T⁻²]} = \text{[M]}^a \cdot \text{[L]}^{-a+b+c} \cdot \text{[T]}^{-a-c}$$

Step 3: Equate the exponents:
For $\text{M}$: $a = 1$
For $\text{T}$: $-a - c = -2 \implies -1 - c = -2 \implies c = 1$
For $\text{L}$: $-a + b + c = 1 \implies -1 + b + 1 = 1 \implies b = 1$

Step 4: Substitute back into Eq. 1:

$$F = k \cdot \eta^1 r^1 v^1$$

Result: $F = k \eta r v$  (According to Stokes' Law, $k = 6\pi$, yielding $F = 6\pi\eta rv$)