Examples on Dimensional Analysis

Every physical quantity is measured by comparing it to a fundamental, arbitrarily chosen, internationally agreed on reference standard known as a unit. A physical quantity measurement is stated as a number (or numerical measure) followed by a unit that is further calculated by dimensional formula. Even though the number of physical quantities appears to be huge, we only use a small number of units to describe them since they are all connected. Fundamental units are the units for fundamental or base quantities. The system of units refers to the whole collection of these units, including both basic and derived units.

Errors in Measurement, Accuracy and Instrument Precision

All experimental science and technology are built on the foundation of measurement. Every measurement made using any measuring equipment yields some amount of uncertainty. This uncertainty is termed ‘error.’ There is an inaccuracy in every computed amount based on measurable quantities. We will distinguish between these two terms: precision and accuracy. A measurement’s accuracy measures how near the measured value is to the quantity’s real value. While precision specifies the resolution or limit to which a quantity may be measured.

Examples of Dimensional Analysis

1. S = ut + 0.5at2 dimensional analysis

We will use the concept of homogeneity to examine the accuracy of the following equation:

Dimensional formula:  S=ut+1/2at2 

Where, S represents distance with a dimension of [L], which is on the left side of the equation. On the right side, we have u, which is velocity with the dimension [LT-1] and t, which is time with the dimension [T]. So, the net dimension of the product of velocity u and time t will be LT-1T=[L]. Another part of the right side equation is 0.5at2, where a is acceleration with the dimension [LT-2], and the dimension of the square of t is [T2], so the net dimension of term 0.5at2 will be L. As a result, the dimension of the net right-sided equation by the dimensional formula is L + L = [L] because the sum of two dimensions equals the same dimension relation. As a result, the left and right sides of the equation have the same [L] dimension and the equation S=ut+0.5at2 is accurate according to the concept of homogeneity.

2. Verify that the equation v2 – u2 = 2aS is accurate.

We’ll use the dimensional formula and the idea of homogeneity as a guide. The equation will be dimensionally valid if the dimensions of quantities on both sides are equal. In quantity v2 – u2, v and u represent velocity and have a dimension of [LT-1], and in the dimension of quantity 2aS where a represents acceleration and has a dimension of [LT-2] and S represents distance and has a dimension of [L]. Now, let us find the dimension of amount v2 – u2 as [LT-1]2=[L2T-2], since u and v are velocities and their difference is also a velocity.

Let us now calculate the dimension of the quantity 2aS by dimensional formula. Now, 2 is a dimensionless constant and the multiplication of a and S will have the dimension of LT-2L=[L2T-2]. As a result, we can observe that both components (right and left-hand side quantities) have the same [L2T-2] dimension relation. As a result, the principle of homogeneity dictates that both components have the same dimension, indicating that the supplied equation is accurate.

3. Verify that the equation F = mv2/r is accurate.

Suppose the quantity ‘f’ indicates force and the quantity mv2/r, where m represents mass, v represents velocity and r represents radius are the same, then the above equation will be dimensionally valid if the dimensions on both sides of the equation are the same.

The dimension of ‘F’ may be represented by the dimensional formula M[LT-2] or dimension of force F = [MLT-2] for a force with a mass acceleration dimension. The radius dimension, which is just distance, will be r = [L] and the mass dimension will be m=[M]. The velocity dimension v = [LT-1]. Now, when these dimensions are applied to the right-hand side by the dimensional formula, mv2/r, we obtain [M][LT-1]2/[L], which yields [MLT-2].

As a result, we see that the dimension of quantity F, the dimension of force f=[MLT-2] and the dimension of quantity mv2/r is [MLT-2] are the same, indicating that this physical quantity equation is the same as the same dimensions on both sides, indicating that it is valid.

Conclusion 

Understanding dimension ideas that drive the description of physical behaviour is critical since only physical quantities with the same dimensions may be added or removed. A good grasp of dimensional analysis helps us in deducing specific relationships between distinct physical quantities and assessing the derivation, correctness and dimensional consistency or homogeneity of various mathematical expressions. When multiplying the magnitudes of two or more physical quantities, their units should be ordinary algebraic symbols. Identical units in the numerator and denominator can be cancelled. The same may be said about a physical quantity’s dimensions. Similarly, the size of physical quantities represented by symbols on both sides of a mathematical equation must be the same.