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Summary: This module explains the mathematics of circular motion in a format that is accessible to blind students.
This module is part of a collection (see http://cnx.org/content/col11294/latest/ ) of modules designed to make physics concepts accessible to blind students. The collection is intended to supplement but not to replace the textbook in an introductory course in high school or college physics.
This module explains the mathematics of circular motion in a format that is accessible to blind students.
In addition to an Internet connection and a browser, you will need the following tools (as a minimum) to work through the exercises in these modules:
The minimum prerequisites for understanding the material in these modules include:
I recommend that you also study the other lessons in my extensive collection of online programming tutorials. You will find a consolidated index at www.DickBaldwin.com .
Now that you have an idea of how circular motion behaves from a physical viewpoint, let's take a look at the mathematics that describe circular motion.
We begin this module with two new terms: angular displacement and angular velocity .
Dealing with points can be awkward
Up until now in this series of modules on circular motion, we have dealt mainly with the motion of points involved in uniform circular motion. However, in some situations, that is awkward. Consider a wheel on a car, for example. There are an infinite number of points on the wheel, and when the wheel is spinning, every point is moving with a different velocity and/or acceleration. It would be difficult for us to describe that motion in terms of the motions of all the points.
This causes us to seek a more comprehensive description that will encompass the motion of all of the points on the wheel. Two terms that accomplish that purpose are angular displacement and angular velocity.
Approaching the situation from this viewpoint, we concentrate on angles instead of distances. If a wheel spins through one-fourth of a complete revolution, every point on the wheel moves through the same 90-degree angle. (However, as you learned in earlier modules, points at different radii move different distances.)
A set of new variables
We will define a set of variables involving angular motion that are analogous to displacement, velocity, and acceleration in the realm of linear motion. However, we will use angular measurements instead of linear distance measurements.
Angular displacement
Instead of linear displacement, for example, we will speak of angular displacement. Angular displacement is the angle through which a rotating body turns based on some starting and stopping criteria.
A pie-shaped wedge
As you learned in an earlier module, a point on a wheel moves along the circumference of a circle as the wheel rotates. Viewing the rotating wheel from a vantage point that is perpendicular to the wheel, during a given time interval, we see that the point sweeps out a pie-shaped wedge with its point at the center of the wheel.
An arc of a circle
The motion of the point describes an arc of a circle directly opposite the point at the center of the circle. This pie-shaped wedge describes an angle, which is the angular displacement during that episode of movement. (You should be able to simulate this on your graph board in order to get a better picture in your mind.)
Definition of angular displacement
dA = Af - Ai
where
Physics books typically use Greek letters such as phi and theta to represent angles. However, it is unlikely that your Braille display can handle Greek characters, so I will stick with standard qwerty keyboard characters.
Angular displacement is a signed quantity
The direction of rotation is indicated by the algebraic sign of the angular displacement. It is conventional to consider a counter clockwise rotation to result in a positive angular displacement.
Units of angular displacement
The units of angular displacement are typically degrees or radians.
That brings us to angular velocity, The average angular velocity is the average rate of change of angular displacement.
Definition of angular velocity
wAvg = dA/dT
where
It is customary in physics books to represent angular velocity with the Greek letter omega. In this module, I will use a lower-case "w" character to represent angular velocity where appropriate, simply because it looks a lot like a Greek omega character.
Reduce the time interval
If we allow the time interval dT to be come shorter and shorter, we are averaging over smaller and smaller time intervals. In the limit, as dT approaches zero, wAvg becomes w, which is the instantaneous angular velocity.
Angular velocity is a signed quantity
Angular velocity is also a signed quantity with the sign indicating the direction of rotation. By convention, counter clockwise rotation is viewed as positive rotation. The sign of angular velocity is the same as the sign of the angular displacement that forms the basis for the angular velocity.
Units of angular velocity
The units of angular velocity are typically degrees per second or radians per second. You will learn later that radians is a dimensionless quantity. Therefore, when angular velocity is measured in radians per second, it often appears simply as
w = 10/sec
The most familiar measurement of angles, in the U.S. is in degrees. However, in some situations, it is more convenient to measure angles in radians than in degrees.
This becomes most apparent when we need to relate angular displacement or angular velocity with the distance traveled or the tangential speed of a point on a rotating object.
Definition of a radian
One radian is an angular measurement, which is equal to an angle at the center of a circle whose arc is equal in length to the radius of the circle.
Simulate with a graph board
I recommend that you use your graph board to simulate an angle of one radian.
Using your graph board along with some string and pushpins, draw a Cartesian coordinate system. Then draw a circle with a convenient radius with its center at the origin of your coordinate system.
Make the arc match the radius
Cut a piece of string to the length of the radius of the circle. Then, beginning at the intersection of the circle and the horizontal axis, lay the string along the circumference of the circle moving in a counter clockwise direction. Put a pushpin at the point where the string ends. Then stretch a rubber band from that point back to the center of the circle.
Measure the angle
Using your protractor, measure the angle that the rubber band makes with the horizontal axis. That angle should be about 57.3 degrees, which is one radian.
Measure the number of radians in 360 degrees
Now, using the string whose length is equal to the radius of the circle as a measuring tool, determine how many strings of that length you can lay end-to-end around the circumference of the circle.
You should find that about 6.28 (2*pi) such strings are required to go all the way around the circumference of the circle.
An angle in radians is a ratio of lengths
An angle measured in radians is a ratio of two values, each of which have units of length. Therefore, such an angle has no dimensions.
Measurement of an angle in radians
angle = s/r
where
An example measurement
Assume that the arc length is equal to one-half the circumference of the circle. This arc represents a subtended angle of 180 degrees. Then,
s = circumference/2 = pi * radius
angle = (pi * radius)/radius = pi
From this we can see that pi radians is equal to 180 degrees.
Earlier we saw that 2*pi radians equal 360 degrees.
Facts worth remembering
Consider the case of a 1.5-radian angular displacement of a wheel in a given time interval. What is the corresponding displacement of a point on the circumference of the wheel? Assume that the radius of the wheel is 0.5 meters.
angle = s/r, or
s = angle * r, or
s = 1.5 * 0.5m = 0.75m
A simple solution
Thus, we see that the tangential displacement of a point on the circumference of a wheel due to a given angular displacement of the wheel in radians is simply the product of the displacement and the radius of the wheel.
Solving for the same result using angular displacement in degrees would be somewhat more complicated.
A similar simplification occurs when dealing with the angular velocity of a wheel and the tangential speed of a point on the circumference of the wheel.
As in linear measurements, the average angular velocity of a wheel is equal to the angular displacement of the wheel divided by the time interval during which the displacement takes place.
Measurement of angular velocity in radians
w = dA/dT
By substitution,
w = (s/r)/dT = s/(r*dT)
where
Another example
Consider the case of a 1.5-radian/second angular velocity of a wheel. What is the corresponding tangential speed of a point on the circumference of the wheel? Assume that the radius of the wheel is 0.5 meters.
The tangential speed is equal to the tangential displacement, s, divided by the time interval over which the displacement occurs. Given the above information, we can write:
w = s/(r*dT)
Given that
v = s/dT
Substitution yields
v; = w*r, or
v = (1.5/s)*(0.5m) = 0.75 m/s
Another simple relationship
Once again, if you keep your units straight, the tangential speed of a point on the circumference of the wheel is simply equal to the angular velocity in radians per second multiplied by the radius of the wheel.
Facts worth remembering
tangential displacement = dA * r
tangential speed = w * r
where
As you already know, when the speed of a point moving in a circle is constant, its motion is called uniform circular motion.
As you also already know, even though the speed of the point is constant, the velocity is not constant. The velocity is constantly changing because the direction of the velocity vector is constantly changing.
The period
The amount of time required for the point to travel completely around the circle is called the period of the motion.
The frequency
The frequency of the motion, which is the number of revolutions per unit time, is defined as the reciprocal of the period. That is,
frequency in rev per sec = 1/(period in sec per rev), or
f = 1/T
where
The speed of a point moving completely around the circle is equal to the distance traveled divided by the time.
sT = 2*pi*r/T, or
sT = 2*pi*r*f
where
We know from before that
sT = w * r, or
w = sT/r
Therefore, by substitution from above,
w = 2*pi*r*f/r = 2*pi*f, or
the angular velocity in radians per second is the product of 2*pi and the frequency in revolutions per second.
where
The SI unit for frequency
The SI unit for frequency is hertz (Hz) where 1 Hz is equal to one revolution per second or one cycle per second.
Facts worth remembering
w = 2*pi*f
where
The SI unit for frequency is hertz (Hz) where 1 Hz is equal to one revolution per second or one cycle per second
In an earlier module, you learned how to subtract vectors and; demonstrate that the acceleration vector of an object moving with uniform circular motion always points toward the center of the circle. However, in that lesson, we did not address the magnitude of the acceleration vector. We will do that here.
A very difficult derivation
Deriving the magnitude of the acceleration vector depends very heavily on the use of vector diagrams, complex assumptions, complicated equations. Unfortunately, this is one of those times that I won't be able to present that derivation in a format that is accessible for blind students. In this case, blind students will simply have to accept the final results in equation form and use those equations for the solution of problems in this area.
Facts worth remembering
Ar = v^2/r, or
Ar = (w^2)*r
where
In this section, I will work through some examples that illustrate what you learned in the earlier section along with what you have learned in earlier modules.
The radius of the Earth at the equator is equal to approximately 6378km. What is the circumference of the earth at the equator.
Solution:
If you were to travel around the Earth at the equator, you would travel along a circular arc that subtends an angle of 2*pi radians. We know how to compute the length of the circular arc given the radius and the subtended angle:
arc length = (subtended angle) * radius, or
circumference = 2*pi*6378km = 40074 km
The Earth rotates around its axis once each 24 hours. Therefore, a point on the equator makes one full trip around a circle with the circumference of the Earth each 24 hours.
Assume you are standing at a point on the equator. Ignoring all of the other motions of the universe, what is the speed with which you are traveling around that circle?
What is the angular velocity of the earth in radians per second.
Does the angular velocity of the Earth change when you drive North from the equator?
Speed
Since we already know the circumference of the Earth, we know that you will travel 40074 km each 24 hours. Therefore,
speed = 40074km/24hr = 463.8 meters/second, or
speed = 1037 miles/hour
Did you know that you are constantly moving through space at a speed slightly greater than 1000 miles per hour?
Angular velocity
We also know that the earth rotates around its axis by 2*pi radians each 24 hours. Therefore, the angular velocity of the earth is
w = 2*pi radians/24 hours = 7.27 *10^(-5) radians / second
Differences in angular velocity
For purposes of this discussion, the Earth does not distort as it rotates. Therefore, the angular velocity of every point on the surface of the Earth rotates around the Earth's axis with the same angular velocity; namely 2*pi radians every 24 hours.
However I did see on TV the other day that the angular velocity of the Earth's core may be different from the angular velocity of the outer crust of the Earth. Apparently this can happen because the core is not firmly attached to the crust, but rather is suspended in a bath of molten rock.
You learned earlier that the speed is equal to the product of the angular velocity and the radius. Therefore,
speed = w * r, or
speed = (7.27 *10^(-5) radians / second) * 6378km, or
speed = 463.7 m/s
which agrees with Solution A above.
Note that w represents angular velocity and r represents radius in the above calculations.
A child's toy contains a round disk that rotates with a period of 0.628 seconds.
What is the frequency with which a spot on the edge of the disk passes a fixed mark on the body of the toy.
What is the angular velocity of the toy?
Solution:
f = 1/T = 1/(0.628) = 1.59 Hz
w = 2*pi*f = 10 radian/second
where
A student ties a 10 kg mass onto a fishing line with a breaking strength of 5 newtons, and then starts swinging the mass around over his head. The student tries very hard to cause the path to be circular. The distance from the center of the circle to the mass is 3 meters.
As time goes on, the student swings the mass faster and faster until the fishing line breaks. What is the tangential velocity of the mass when the fishing line breaks.
Solution:
Centripetal force = mass * (centripetal acceleration), or
f = m * Ar, or
f = m * v^2/r, or
v^2 = f*r/m, or
v = (f*r/m)^(1/2), or
v = (5 newtons * 3 meters/10 kg)^(1/2), or
v = 1.22 m/s
The magnitude of the tangential velocity of the mass when the fishing line breaks is 1.22 meters/second
Check the solution using angular velocity, w
w = v / r, or
w = (1.22 m/s) / 3m = 0.407 radians/second
The angular velocity for a tangential velocity of 1.22 m/s is 0.407 radians/second
Ar = (w^2)*r, or
Ar = (( 0.407 /second)^2)*3m, or
Ar = 0.5 m/s^2
The magnitude of the radial acceleration is 0.5 m/s^2
f = m * Ar, or
f = 10kg*(0.5 m/s^2), or
f = 5 newtons, or
the force equals 5 newtons, which matches the breaking strength of the fishing line.
I encourage you to work through the examples that I have presented in this module to confirm that you get the same results. Experiment with the scenarios, making changes, and observing the results of your changes. Make certain that you can explain why your changes behave as they do.
I will publish a module containing consolidated links to resources on my Connexions web page and will update and add to the list as additional modules in this collection are published.
This section contains a variety of miscellaneous information.
Financial : Although the Connexions site makes it possible for you to download a PDF file for this module at no charge, and also makes it possible for you to purchase a pre-printed version of the PDF file, you should be aware that some of the HTML elements in this module may not translate well into PDF.
I also want you to know that I receive no financial compensation from the Connexions website even if you purchase the PDF version of the module.
Affiliation : I am a professor of Computer Information Technology at Austin Community College in Austin, TX.
-end-
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This work is licensed by Richard Baldwin under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.
Last edited by Richard Baldwin on Oct 8, 2012 1:55 pm -0500.
"Blind students should not be excluded from physics courses because of inaccessible textbooks. The modules in this collection present physics concepts in a format that blind students can read […]"