Dimensional Analysis Problems With Answers

Mastering Dimensional Analysis: Problems and Solutions

Dimensional analysis is a powerful technique used in physics and engineering to check the validity of equations and to derive relationships between physical quantities. It's based on the principle that any equation describing a physical phenomenon must be dimensionally consistent; that is, the dimensions on both sides of the equation must be the same. This seemingly simple principle can solve complex problems and prevent costly errors. This article provides a comprehensive guide to dimensional analysis, covering fundamental concepts, worked-out problems, and frequently asked questions to solidify your understanding. Mastering dimensional analysis will significantly enhance your problem-solving skills in various scientific and engineering fields.

Understanding the Fundamentals of Dimensional Analysis

Before diving into problems, let's establish a solid foundation. At its core, dimensional analysis revolves around the fundamental dimensions: mass (M), length (L), time (T), and electric current (I). Other dimensions, like temperature (Θ), luminous intensity (J), and amount of substance (N), are also used depending on the context but these four are the most common. Every physical quantity can be expressed as a combination of these fundamental dimensions.

For instance:

• Speed: Has dimensions of L/T (length per time)
• Acceleration: Has dimensions of L/T² (length per time squared)
• Force: Has dimensions of MLT⁻² (mass times length per time squared - derived from Newton's second law, F=ma)
• Energy: Has dimensions of ML²T⁻² (derived from the work-energy theorem, W=Fd)

We represent dimensions using square brackets [ ]. For example, [speed] = LT⁻¹. This notation allows us to systematically analyze the dimensions of any equation.

Step-by-Step Approach to Solving Dimensional Analysis Problems

The process of solving dimensional analysis problems typically involves these steps:

1. Identify the variables involved: List all the physical quantities relevant to the problem.

2. Assign dimensions to each variable: Express each variable in terms of its fundamental dimensions (M, L, T, I, etc.).

3. Formulate a possible equation: Based on your understanding of the physical phenomenon, write a general equation relating the variables. You might use proportionality constants (dimensionless numbers) as needed.

4. Check dimensional homogeneity: Ensure that the dimensions on both sides of the equation are identical. If not, there's an error in the equation. This step is crucial for validating the equation.

5. Determine the unknown exponents: If the equation contains unknown exponents, solve for them by equating the exponents of each fundamental dimension on both sides.

6. Write the final equation: Substitute the determined exponents back into the general equation to obtain the final equation relating the variables.

Worked-Out Dimensional Analysis Problems

Let's illustrate the process with a series of examples, progressively increasing in complexity.

Problem 1: Period of a Simple Pendulum

The period (T) of a simple pendulum is believed to depend on its length (l), mass (m), and the acceleration due to gravity (g). Find the relationship between these quantities using dimensional analysis.

Solution:

1. Variables: T, l, m, g

2. Dimensions:

   • [T] = T
   • [l] = L
   • [m] = M
   • [g] = LT⁻²

3. Possible equation: We assume a relationship of the form T ∝ l^am^bg^c, where a, b, and c are unknown exponents.

4. Dimensional homogeneity: [T] = [l^am^bg^c] => T = L^aM^b(LT⁻²)^c = L^a+cM^bT^-2c

5. Equating exponents:

   • For T: 1 = -2c => c = -1/2
   • For M: 0 = b => b = 0
   • For L: 0 = a + c => a = -c = 1/2

6. Final equation: T ∝ l^1/2m⁰g^-1/2 => T ∝ √(l/g). This indicates that the period is proportional to the square root of the length divided by the acceleration due to gravity. Note that the mass is irrelevant in this simplified analysis.

Problem 2: Determining the Drag Force

The drag force (F_d) on a sphere moving through a fluid is believed to depend on the sphere's radius (r), its velocity (v), and the fluid's density (ρ) and viscosity (η). Find the relationship using dimensional analysis.

Solution:

1. Variables: F_d, r, v, ρ, η

2. Dimensions:

   • [F_d] = MLT⁻²
   • [r] = L
   • [v] = LT⁻¹
   • [ρ] = ML⁻³
   • [η] = ML⁻¹T⁻¹

3. Possible equation: F_d ∝ r^av^bρ^cη^d

4. Dimensional homogeneity: MLT⁻² = L^a(LT⁻¹)^b(ML⁻³)^c(ML⁻¹T⁻¹)^d = L^a+b-3c-dM^c+dT^-b-d

5. Equating exponents:

   • For M: 1 = c + d
   • For L: 1 = a + b - 3c - d
   • For T: -2 = -b - d

Solving this system of equations (there are multiple solutions, reflecting the possibility of multiple contributing effects) is more complex than in Problem 1. A common approach is to solve for some variables in terms of others. One solution is a = 1, b = 1, c = 1, d = 0, yielding F_d ∝ r v ρ (Stokes' law for low Reynolds numbers). Another solution, relevant at high Reynolds numbers, might involve different exponents.

Problem 3: Energy of a satellite in orbit

The total energy (E) of a satellite in a circular orbit around a planet depends on its mass (m), the planet’s mass (M), and the orbital radius (r). Find the relationship.

Solution:

1. Variables: E, m, M, r

2. Dimensions:

   • [E] = ML²T⁻²
   • [m] = M
   • [M] = M
   • [r] = L

3. Possible equation: E ∝ m^aM^br^c

4. Dimensional homogeneity: ML²T⁻² = M^a+bL^cT⁰

5. Equating exponents:

   • For M: 1 = a + b
   • For L: 2 = c
   • For T: 0 = 0 (this equation provides no additional information).

Solving this system requires additional physical insight. We know the energy should be proportional to both masses. A simple assumption (a=1, b=1) leads to E ∝ mMr⁻². The minus sign is crucial for showing that the energy is negative (representing a bound state). A more rigorous derivation using Newtonian mechanics would confirm this result, including a proportionality constant related to the gravitational constant.

Frequently Asked Questions (FAQ)

Q: What if the dimensions don't match?

A: If the dimensions on both sides of an equation don't match, it means the equation is incorrect. You've made a mistake either in the equation itself or in assigning dimensions to the variables. Review your work carefully.

Q: Can dimensional analysis determine the exact value of constants?

A: No, dimensional analysis cannot determine the exact value of dimensionless constants. It only provides information about the relationship between the variables' dimensions. It provides the form of the equation, but not the numerical coefficients.

Q: Are there any limitations to dimensional analysis?

A: Yes, dimensional analysis has limitations. It cannot determine equations involving dimensionless constants, it cannot handle equations with trigonometric functions or exponentials directly, and it doesn't provide information about the physical constants. Also, it may yield multiple possible solutions.

Q: How is dimensional analysis used in real-world applications?

A: Dimensional analysis has wide applications. In physics, it's used to verify equations and to derive relationships between quantities. In engineering, it's used to design experiments, check the consistency of formulas, and scale up or down experimental results.

Conclusion

Dimensional analysis is a valuable tool for anyone working with physical quantities. By systematically analyzing the dimensions of variables, you can verify the consistency of equations, uncover relationships between variables, and even derive simplified equations. While it has limitations, its power in checking for errors and guiding problem-solving makes it an indispensable technique in physics and engineering. Mastering dimensional analysis, as illustrated through the worked-out problems, will significantly improve your problem-solving capabilities and deepen your understanding of physical phenomena. Remember to practice regularly to fully grasp this powerful tool.