# Can't solve? Try dimensional analysis! British Physics Olympiad, BpHo Solution

TLDRThis script introduces dimensional analysis as a powerful technique in physics, often overlooked in courses. It demonstrates how to use this method to solve a differential equation involving mass change over time, density, area, and gravitational field strength. By setting up a system of equations based on unit equivalence, the video derives the values of the exponents alpha, beta, and gamma. The process is applied to a problem from the 2018 British Physics Olympiad, showcasing the technique's utility in determining coefficients in physical equations.

###### Takeaways

- π Dimensional analysis is an important but often overlooked technique in physics.
- π§© The video demonstrates how to use dimensional analysis to solve a differential equation involving mass change over time.
- π The differential equation presented is dm/dt = k, where k is a dimensionless constant.
- π The units on the left-hand side of the equation (kg/s) must match the units on the right-hand side after applying dimensional analysis.
- π Density (rho) is raised to the power of alpha, and its units are kg/m^3, which will be adjusted by the power of alpha.
- π The area (a) is raised to the power of beta, with units of m^2.
- π Gravitational field strength (g) is raised to the power of gamma, with units of m/s^2.
- βοΈ By equating the units on both sides of the equation, we can set up a system of equations to solve for alpha, beta, and gamma.
- π Alpha is determined to be 1, as the units of mass on both sides of the equation must match.
- π Beta is calculated to be 5/4 by solving the equation -3*alpha + 2*beta + gamma = 0, with known values of alpha and gamma.
- π Gamma is found to be 1/2, by equating the powers of time units on both sides of the equation.
- π The technique demonstrated was part of a question from the 2018 British Physics Olympiad, highlighting its significance in problem-solving.

###### Q & A

### What is dimensional analysis, and why is it important in physics?

-Dimensional analysis is a technique used in physics to determine the relationships between different physical quantities by analyzing their units. It is important because it allows physicists to verify equations, derive formulas, and understand the dependencies between variables.

### What is the purpose of using dimensional analysis in the given differential equation?

-The purpose of using dimensional analysis in the given differential equation is to determine the values of the exponents alpha, beta, and gamma, which correspond to the powers of density, area, and gravitational field strength, respectively.

### What are the SI units for density (rho), area (A), and gravitational field strength (g)?

-In SI units, the density (rho) is measured in kilograms per cubic meter (kg/m^3), the area (A) is measured in square meters (m^2), and the gravitational field strength (g) is measured in meters per second squared (m/s^2).

### How do you determine the units on both sides of the differential equation?

-To determine the units on both sides of the differential equation, you analyze the units for each term. On the left side, dm/dt represents mass per time, so the units are kilograms per second (kg/s). On the right side, you combine the units of density, area, and gravitational field strength, each raised to their respective powers.

### What is the significance of the dimensionless constant 'k' in the equation?

-The dimensionless constant 'k' in the equation has no units and does not affect the dimensional analysis. It serves as a scaling factor that does not alter the relationship between the physical quantities involved.

### How is the exponent alpha determined from the equation?

-The exponent alpha is determined by comparing the units of mass on both sides of the equation. Since the left side has kilograms raised to the power of 1, alpha must also be equal to 1 to maintain dimensional consistency.

### What equation is formed to determine the values of beta and gamma?

-The equation formed to determine the values of beta and gamma is: -3*alpha + 2*beta + gamma = 0. This equation is derived from combining the units of meters from density, area, and gravitational field strength and setting them equal to zero since there are no meters on the left side of the equation.

### What is the value of gamma, and how is it calculated?

-The value of gamma is 1/2. It is calculated by equating the power of seconds on both sides of the equation. Since the left side has seconds raised to the power of -1, and the right side has seconds raised to the power of -2*gamma, setting -1 = -2*gamma gives gamma = 1/2.

### How is the exponent beta calculated?

-The exponent beta is calculated by solving the equation -3*alpha + 2*beta + gamma = 0. With alpha = 1 and gamma = 1/2, the equation simplifies to -3 + 2*beta + 1/2 = 0. Solving for beta gives beta = 5/4.

### What are the final values of alpha, beta, and gamma, and how do they affect the rewritten equation?

-The final values are alpha = 1, beta = 5/4, and gamma = 1/2. These values allow you to rewrite the differential equation with the correct exponents, providing a deeper understanding of the relationship between the variables in the context of the problem.

###### Outlines

##### π Dimensional Analysis in Physics

This paragraph introduces the concept of dimensional analysis, a powerful technique often overlooked in physics education. The speaker demonstrates how to apply this method to a differential equation involving mass change over time (dm/dt), a dimensionless constant (k), density (rho), area (a), and gravitational field strength (g). The goal is to determine the exponents alpha, beta, and gamma for these variables. The explanation involves equating the units on both sides of the equation and setting up a system of equations to solve for the exponents. The process concludes with the identification of alpha as 1, beta as 5/4, and gamma as 1/2, showcasing the utility of dimensional analysis in deriving physical relationships.

##### π Solving the 2018 British Physics Olympiad Problem

The second paragraph reveals that the equation discussed is from the 2018 British Physics Olympiad, known for its challenging questions. The speaker emphasizes the importance of learning dimensional analysis as a key technique in physics. They also invite viewers to check out the Olympiad questions via a link provided in the description. The speaker expresses hope that the audience has enjoyed learning about this technique and teases curiosity by mentioning the temperature of an average black hole, suggesting further exploration of physics concepts.

###### Mindmap

###### Keywords

##### π‘Dimensional Analysis

##### π‘Differential Equation

##### π‘Dimensionless Constant

##### π‘Density

##### π‘Gravitational Field Strength

##### π‘Power

##### π‘Units

##### π‘Rate of Change

##### π‘System of Equations

##### π‘Coefficients

##### π‘British Physics Olympiad

###### Highlights

Dimensional analysis is introduced as an underappreciated technique in physics.

A differential equation is presented involving mass change over time, density, area, and gravitational field strength.

The process of using dimensional analysis to determine the exponents alpha, beta, and gamma is explained.

The units of the rate of change of mass are identified as kilograms per second.

The dimensional constant k is noted as dimensionless, thus having no units.

Density is defined with its units, kilograms per cubic meter, and its relation to the exponent alpha.

The area A is described with its units and relation to the exponent beta.

Gravitational field strength is detailed with its units and connection to the exponent gamma.

The expansion of the right-hand side of the equation is shown, leading to a comparison of units.

A system of equations is formed to solve for the exponents based on unit equality.

Alpha is determined to be one based on the units of mass on both sides of the equation.

A method to find beta and gamma using the remaining units is outlined.

Gamma is solved to be one-half by equating the powers of time units.

Beta is calculated to be five quarters by solving the remaining unit equation.

The final coefficients for alpha, beta, and gamma are summarized.

The technique's application is demonstrated with a question from the 2018 British Physics Olympiad.

The importance of dimensional analysis in physics is emphasized for its problem-solving capabilities.

An invitation to explore the temperature of an average black hole is extended as a curiosity pique.

###### Transcripts

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