Unit Conversion and Dimensional Analysis

Site: Saylor Academy
Course: PHYS101: Introduction to Mechanics
Book: Unit Conversion and Dimensional Analysis
Printed by: Guest user
Date: Wednesday, May 8, 2024, 11:24 AM

Description

Read this text for examples of how to calculate physical quantities and units of measurement.

Unit Conversion and Dimensional Analysis

It is often necessary to convert from one type of unit to another. For example, if you are reading a European cookbook, some quantities may be expressed in units of liters and you need to convert them to cups. Or, perhaps you are reading walking directions from one location to another and you are interested in how many miles you will be walking. In this case, you will need to convert units of feet to miles.

Let's consider a simple example of how to convert units. Let us say that we want to convert 80 meters (m) to kilometers (km).

The first thing to do is to list the units that you have and the units that you want to convert to. In this case, we have units in meters and we want to convert to kilometers.

Next, we need to determine a conversion factor relating meters to kilometers. A conversion factor is a ratio expressing how many of one unit are equal to another unit. For example, there are 12 inches in 1 foot, 100 centimeters in 1 meter, 60 seconds in 1 minute, and so on. In this case, we know that there are 1,000 meters in 1 kilometer.

Now we can set up our unit conversion. We will write the units that we have and then multiply them by the conversion factor so that the units cancel out, as shown:

80 \not x \times \frac{1 \mathrm{~km}}{1000 \not \mathrm{m}}=0.080 \mathrm{~km}

Note that the unwanted m unit cancels, leaving only the desired km unit. You can use this method to convert between any types of unit.

Lengths in meters Masses in kilograms (more precise values in parentheses) Times in seconds (more precise values in parentheses)
10^{-18} Present experimental limit to smallest observable detail
10^{-30}
Mass of an electron \left(9.11 \times 10^{-31} \mathrm{~kg}\right)
10^{-23}
Time for light to cross a proton
10^{-15} Diameter of a proton
10^{-27}
Mass of a hydrogen atom \left(1.67 \times 10^{-27} \mathrm{~kg}\right)
10^{-22}
Mean life of an extremely unstable nucleus
10^{-14} Diameter of a uranium nucleus
10^{-15}
Mass of a bacterium
10^{-15}
Time for one oscillation of visible light
10^{-10} Diameter of a hydrogen atom
10^{-5}
Mass of a mosquito
10^{-13}
Time for one vibration of an atom in a solid
10^{-8} Thickness of membranes in cells of living organisms
10^{-2}
Mass of a hummingbird
10^{-8}
Time for one oscillation of an FM radio wave
10^{-6} Wavelength of visible light
1
Mass of a liter of water (about a quart)
10^{-3}
Duration of a nerve impulse
10^{-3} Size of a grain of sand 10^{2}
 
Mass of a person
\quad 1
Time for one heartbeat
1 Height of a 4-year-old child
10^{3}
Mass of a car
10^{5}
One day \left(8.64 \times 10^{4} \mathrm{~s}\right)
10^{2} Length of a football field
10^{8}
Mass of a large ship
10^{7}
One year (y) \left(3.16 \times 10^{7} \mathrm{~s}\right)
10^{4} Greatest ocean depth
10^{12}
Mass of a large iceberg
10^{9}
About half the life expectancy of a human
10^{7} Diameter of the Earth
10^{15}
Mass of the nucleus of a comet
10^{11}
Recorded history
10^{11} Distance from the Earth to the Sun
10^{23}
Mass of the Moon \left(7.35 \times 10^{22} \mathrm{~kg}\right)
10^{17}
Age of the Earth
10^{16} Distance traveled by light in 1 year (a light year)
10^{25}
Mass of the Earth \left(5.97 \times 10^{24} \mathrm{~kg}\right)
10^{18}
Age of the universe
10^{21} Diameter of the Milky Way galaxy
10^{30}
Mass of the Sun \left(1.99 \times 10^{30} \mathrm{~kg}\right)    
10^{22} Distance from the Earth to the nearest large galaxy (Andromeda)
10^{42}
Mass of the Milky Way galaxy (current upper limit)    
 10^{26} Distance from the Earth to the edges of the known universe
10^{53}
Mass of the known universe (current upper limit)    


Table 1.3 Approximate Values of Length, Mass, and Time

Nonstandard Units

While there are numerous types of units that we are all familiar with, there are others that are much more obscure. For example, a firkin is a unit of volume that was once used to measure beer. One firkin equals about 34 liters.

To learn more about nonstandard units, use a dictionary or encyclopedia to research different "weights and measures". Take note of any unusual units, such as a barleycorn, that are not listed in the text. Think about how the unit is defined and state its relationship to SI units.

 


Source: Rice University, https://openstax.org/books/college-physics/pages/1-2-physical-quantities-and-units
Creative Commons License This work is licensed under a Creative Commons Attribution 4.0 License.


Example 1.1 Unit Conversions: A Short Drive Home

Example 1.1 Unit Conversions: A Short Drive Home

Suppose that you drive the 10.0 km from your university to home in 20.0 min. Calculate your average speed (a) in kilometers per hour (km/h) and (b) in meters per second (m/s). (Note: Average speed is distance traveled divided by time of travel).

Strategy

First we calculate the average speed using the given units. Then we can get the average speed into the desired units by picking the correct conversion factor and multiplying by it. The correct conversion factor is the one that cancels the unwanted unit and leaves the desired unit in its place.

Solution for (a)

(1) Calculate average speed. Average speed is distance traveled divided by time of travel. (Take this definition as a given for now - average speed and other motion concepts will be covered in a later module). In equation form,

\text{average speed =} \, \dfrac{\text { distance }}{\text { time }}.

(2) Substitute the given values for distance and time.

\text{average speed =} \, \dfrac{10.0 \mathrm{~km}}{20.0 \mathrm{~min}}=0.500 \dfrac{\mathrm{km}}{\mathrm{min}}.

(3) Convert \mathrm{km} / \mathrm{min} to \mathrm{km} / \mathrm{h} : multiply by the conversion factor that will cancel minutes and leave hours. That conversion factor is 60 \mathrm{~min} / \mathrm{hr}. Thus,

\text{average speed =} \, 0.500 \dfrac{\mathrm{km}}{\min } \times \dfrac{60 \mathrm{~min}}{1 \mathrm{~h}}=30.0 \dfrac{\mathrm{km}}{\mathrm{h}}.

Discussion for (a)

To check your answer, consider the following:

(1) Be sure that you have properly cancelled the units in the unit conversion. If you have written the unit conversion factor upside down, the units will not cancel properly in the equation. If you accidentally get the ratio upside down, then the units will not cancel; rather, they will give you the wrong units as follows:

\dfrac{\mathrm{km}}{\min } \times \dfrac{1 \mathrm{hr}}{60 \mathrm{~min}}=\dfrac{1}{60} \dfrac{\mathrm{km} \cdot \mathrm{hr}}{\min ^{2}},

which are obviously not the desired units of km/h.

(2) Check that the units of the final answer are the desired units. The problem asked us to solve for average speed in units of km/h and we have indeed obtained these units.

(3) Check the significant figures. Because each of the values given in the problem has three significant figures, the answer should also have three significant figures. The answer 30.0 km/hr does indeed have three significant figures, so this is appropriate. Note that the significant figures in the conversion factor are not relevant because an hour is defined to be 60 minutes, so the precision of the conversion factor is perfect.

(4) Next, check whether the answer is reasonable. Let us consider some information from the problem - if you travel 10 km in a third of an hour (20 min), you would travel three times that far in an hour. The answer does seem reasonable.

Solution for (b)

There are several ways to convert the average speed into meters per second.

(1) Start with the answer to (a) and convert km/h to m/s. Two conversion factors are needed - one to convert hours to seconds, and another to convert kilometers to meters.

(2) Multiplying by these yields

\text { Average speed }=30.0 \dfrac{\mathrm{km}}{\mathrm{h}} \times \dfrac{1 \mathrm{~h}}{3,600 \mathrm{~s}} \times \dfrac{1,000 \mathrm{~m}}{1 \mathrm{~km}},

\text { Average speed }=8.33 \dfrac{\mathrm{m}}{\mathrm{s}} \text {. }

Discussion for (b)

If we had started with 0.500 km/min, we would have needed different conversion factors, but the answer would have been the same: 8.33 m/s.

You may have noted that the answers in the worked example just covered were given to three digits. Why? When do you need to be concerned about the number of digits in something you calculate? Why not write down all the digits your calculator produces? The module Accuracy, Precision, and Significant Figures will help you answer these questions.

Check Your Understanding

1. Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird to beat its wings once. Which fundamental unit should the scientist use to describe the measurement? Which factor of 10 is the scientist likely to use to describe the motion precisely? Identify the metric prefix that corresponds to this factor of 10.

Solution

The scientist will measure the time between each movement using the fundamental unit of seconds. Because the wings beat so fast, the scientist will probably need to measure in milliseconds, or 10^{-3} seconds. (50 beats per second corresponds to 20 milliseconds per beat).


2. One cubic centimeter is equal to one milliliter. What does this tell you about the different units in the SI metric system?

Solution

The fundamental unit of length (meter) is probably used to create the derived unit of volume (liter). The measure of a milliliter is dependent on the measure of a centimeter.