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Check the dimensional homogeneity of the following equations: (a) \(F=m a\), (b) \(W=F s\), and (c) \(\dot{W}=T \omega\), where \(m=\) mass, \(a=\) acceleration, \(F=\) force, \(W=\) work, \(s=\) distance, \(\omega=\) angular velocity, \(T=\) torque, and \(\dot{W}=\) power.

Short Answer

Expert verified
The equations \(F = ma\), \(W = Fs\), and \(\dot{W} = T\omega\) are all dimensionally homogeneous.

Step by step solution

01

Substitute the dimensions of physical quantities in the first equation

In the first equation \(F = ma\), we substitute the dimensions of Force (F) [M L T^-2], mass (m) [M], and acceleration (a) [L T^-2]. So we get [M L T^-2] = [M] [L T^-2]. Both sides of the equation have the same dimensions, so this equation is dimensionally homogeneous.
02

Substitute the dimensions of physical quantities in the second equation

In the second equation \(W = Fs\), we substitute the dimensions of Work (W) [M L^2 T^-2], Force (F) [M L T^-2] and distance (s) [L]. We get [M L^2 T^-2] = [M L T^-2] [L]. Both sides of the equation have the same dimensions, so this equation is also dimensionally homogeneous.
03

Substitute the dimensions of physical quantities in the third equation

In the third equation \(\dot{W} = T\omega\), we substitute the dimensions of Power (P or \(\dot{W}\)) [M L^2 T^-3], Torque (T) [M L^2 T^-2] and angular velocity (\(\omega\)) [T^-1]. We get [M L^2 T^-3] = [M L^2 T^-2] [T^-1]. Both sides of the equation have the same dimensions so this equation is also dimensionally homogeneous.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Dimensional Analysis
Understanding dimensional analysis is critical for ensuring equations in physics make sense. It's a method used to convert one set of units to another, to check the validity of equations, and to deduce relationships between physical quantities. By comparing the dimensions on both sides of an equation, like in the exercise given, one can determine if an equation is dimensionally homogeneous, which means it has the same dimensional expression on both sides.

For example, looking at the equation for force, which is given by Newton's second law as force (F) equals mass (m) multiplied by acceleration (a), or in formula form, F=ma. Using dimensional analysis, we assess the homogeneity by substituting the dimensional units for mass ([M]) and acceleration ([L T^-2]). If both sides of the equation balance dimensionally, as they do here with [M L T^-2] being equivalent on both sides, the equation passes this test of dimensional analysis.
Physical Quantities
Physical quantities such as mass, length, time, and force are the building blocks of physics, used to describe the universe in a measurable way. Each physical quantity is expressed with a specific unit of measurement, which provides a standard for comparison. The equations in the exercise—F=ma, W=Fs, and \(\dot{W}=T\omega\)—are all expressions of these physical quantities interacting.

When we substitute dimensions into the equations during a dimensional analysis, we're actually probing the consistency of these physical quantities. As we assess an equation like work (W) equals force (F) times distance (s), or W=Fs, we have to ensure that the dimensions of work ([M L^2 T^-2]) correctly correspond to the multiplication of the dimensions for force and distance. The dimensional homogeneity verifies the consistent interplay between these quantities, a foundation without which our understanding of physics would crumble.
Units of Measurement
Units of measurement act as the language through which we can communicate and understand the physical quantities. They serve as references that allow us to measure and compare quantities effectively. Some of the fundamental units include meters for distance (m), kilograms for mass (kg), seconds for time (s), and newtons for force (N).

Every physical equation must make sense in terms of these units, which is what we check using the principle of dimensional homogeneity. For instance, power is measured in watts, which can be broken down into the units of measurement for the involved quantities, in this case, work done over time. In the exercise \(\dot{W}=T\omega\), power (usually in watts, [M L^2 T^-3]) can be derived from torque (in newton-meters, [M L^2 T^-2]) and angular velocity (in radians per second, [T^-1]), reaffirming the importance of units of measurement in understanding and validating physical laws.

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