Video: Dimensional Analysis | Nagwa Video: Dimensional Analysis | Nagwa

Video: Dimensional Analysis

In this video we learn about dimensional analysis, and how to use the units of the elements in our calculations to verify the dimensional consistency of our calculations, as well as to determine the appropriate units and dimensionality of our results.


Video Transcript

In this video, we’re going to learn about dimensional analysis, what it is, how to use it, and why it’s important. To begin, imagine that you’re in the middle of a physics examination. You’re working on a problem with all sorts of different kinds of information provided. And the problem asks, ultimately, for the acceleration, 𝑎, of the object described in the problem. You take some of the information given in the problem statement and combine those values. Subtracting, adding, dividing, and finally coming up with an answer on your calculator screen for the acceleration.

Of course, on an exam, we all want to get the questions right as much as possible. And the question is, is there a way that we could check our result for acceleration, 𝑎, to give us some confidence that we’re on the right track? The answer is yes. And we would do it through a method called dimensional analysis.

Dimensional analysis is analyzing relationships between physical quantities using units. There are two main applications for dimensional analysis. The first application is to verify dimensional consistency. Every physically measurable quantity has units that go along with it that describe that quality. For example, a measurement of force has units of newtons, which can be simplified in the basic SI terms: kilograms, meters, and seconds.

There’s a statement known as Newton’s second law, which says that the force on an object is equal to that object’s mass times its acceleration. This is a statement we could evaluate for dimensional consistency. A measured value of force has units of kilograms meters per second squared. Kilograms has a dimension of mass, meters has a dimension of length, and inverse second squared has a dimension of 𝑇 to the negative two.

Evaluating the right side of our equation, mass times acceleration, the dimensions of that product equals dimensions of mass, 𝑀, multiplied by a length per time squared. So we see that, dimensionally, Newton’s second law is consistent. So that’s one application of dimensional analysis.

A second application is to convert between units. Say you’re given a time in hours, but you want to report that time in seconds. That would involve converting the unit of time from hours to seconds. Or if you had a value of length in meters and you wanted to express that length in units of inches. That also would involve unit conversion, which would rely on dimensional analysis. Let’s get some practice trying dimensional analysis out with a couple example problems.

A student is trying to remember a formula from geometry. Assuming that 𝐴 corresponds to area, 𝑉 corresponds to volume, and all other variables are lengths. What missing dimension in the formula 𝑉 equals 𝐴 must the right-hand side of the equation be multiplied by to make the formula dimensionally consistent?

Considering these two terms, 𝐴 corresponding to area and 𝑉 to volume, if we write out their dimensions, the dimensions of 𝐴 are length squared, while the dimensions of 𝑉 are length cubed. Considering the formula 𝑉 equals 𝐴 multiplied by something, we’re asked what dimension needs to fit in to that blank spot. If we look at the equation from a dimension perspective, we can see that, on the left-hand side, we have 𝐿 cubed. And currently, on the right-hand side, we have 𝐿 squared. So we’re missing a dimension of 𝐿.

This means we would want to multiply the right-hand side of our equation by a length 𝐿 in order to make it dimensionally consistent. And with this 𝐿 in place, our equation is.

Now, let’s try a second practice problem involving dimensional analysis.

Consider the physical quantities 𝑚, 𝑠, 𝑣, 𝑎, and 𝑡 with dimensions. Dimensions of 𝑚 equals capital 𝑀. Dimensions of 𝑠 equals capital 𝐿. Dimensions of 𝑣 equals 𝐿𝑇 to the negative one. Dimensions of 𝑎 equals 𝐿 times 𝑇 to the negative two. And dimensions of 𝑡 equals capital 𝑇. The equation 𝑇 equals 𝑚𝑠 over 𝑎 is dimensionally consistent. Find the dimension of the quantity on the left-hand side of the equation.

We want to solve for the dimension of the term capital 𝑇 that appears on the left-hand side of the given equation. Since we’re given an equation that we’re told is dimensionally consistent, we can write that the dimensions of 𝑇 is equal to the dimension of 𝑚 times the dimension of 𝑠 divided by the dimension of 𝑎. If we enter these dimensions in. Where the dimension of 𝑚 is capital 𝑀. The dimension of 𝑠 is capital 𝐿. And the dimension of 𝑎 is 𝐿 times 𝑇 to the negative two. We see that, in this fraction, the factors of length 𝐿 cancel out. And we can write it in a simplified form as capital 𝑀 times 𝑇 squared. This is the net dimension of the right-hand side of the equation. And since the equation is dimensionally consistent, it’s also the dimension of capital 𝑇. So our result is capital 𝑀 times capital 𝑇 squared.

In summary of dimensional analysis, dimensional analysis involves analyzing relationships between physical quantities using units. The symbol of brackets indicates dimensions of some quantity. For example, the dimensions of seconds, 𝑠, is equal to time, capital 𝑇. And finally, dimensional analysis enables answer-checking.

Remember that hypothetical exam question that asked us to solve for acceleration, 𝑎. We know the units of acceleration are meters per second squared. That is, its dimension is length per time squared. And we can check to see if the way we’ve combined given variables yields those units. If it does, we can be confident that our numerical answer is likely on the right track. So dimensional analysis is very helpful for letting us compare physical quantities.

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