Download Circular Motion Acceleration and Centripetal Force

OpenStax-CNX module: m38406
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Circular Motion Acceleration and
∗
Centripetal Force
R.G. (Dick) Baldwin
This work is produced by OpenStax-CNX and licensed under the
†
Creative Commons Attribution License 3.0
Abstract
This module explains acceleration and centripetal force in a format that is accessible to blind students.
1 Table of Contents
• Preface (p. 2)
· General (p. 2)
· Prerequisites (p. 2)
· Supplemental material (p. 2)
• Discussion (p. 3)
·
·
·
·
·
·
·
·
Subtraction of vectors (p. 3)
An old-fashioned merry-go-round (p. 4)
A weight on a string (p. 5)
Water in a bucket (p. 5)
The moon and the Earth (p. 6)
The work-energy explanation (p. 6)
Centripetal, not centrifugal (p. 6)
Summary (p. 7)
• Example scenarios (p. 7)
· Scenario #1 (p. 7)
· Scenario #2 (p. 7)
· Scenario #3 (p. 8)
• Resources (p. 8)
• Miscellaneous (p. 8)
∗ Version
1.2: Oct 8, 2012 1:44 pm -0500
† http://creativecommons.org/licenses/by/3.0/
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2 Preface
2.1 General
This module is part of a collection (see http://cnx.org/content/col11294/latest/ 1 ) 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 acceleration and centripetal force in a format that is accessible to blind students.
2.2 Prerequisites
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:
• A graph board for plotting graphs and vector diagrams ( http://www.youtube.com/watch?v=c8plj9UsJbg
2
).
• A protractor for measuring angles ( http://www.youtube.com/watch?v=v-F06HgiUpw 3 ).
• An audio screen reader that is compatible with your operating system, such as the NonVisual Desktop
Access program (NVDA), which is freely available at http://www.nvda-project.org/ 4 .
• A refreshable Braille display capable of providing a line by line tactile output of information displayed
on the computer monitor ( http://www.userite.com/ecampus/lesson1/tools.php 5 ).
• A device to create Braille labels. Will be used to label graphs constructed on the graph board.
The minimum prerequisites for understanding the material in these modules include:
• A good understanding of algebra.
• An understanding of the use of a graph board for plotting graphs and vector diagrams ( http://www.youtube.com/watch?
6
).
• An understanding of the use of a protractor for measuring angles ( http://www.youtube.com/watch?v=vF06HgiUpw 7 ).
• A basic understanding of the use of sine, cosine, and tangent from trigonometry ( http://www.clarku.edu/∼djoyce/trig/
8
).
• An introductory understanding of JavaScript programming ( http://www.dickbaldwin.com/tocjscript1.htm
9
and http://www.w3schools.com/js/default.asp 10 ).
• An understanding of all of the material covered in the earlier modules in this collection.
2.3 Supplemental material
I recommend that you also study the other lessons in my extensive collection of online programming tutorials.
You will nd a consolidated index at www.DickBaldwin.com 11 .
1 http://cnx.org/content/col11294/latest/
2 http://www.youtube.com/watch?v=c8plj9UsJbg
3 http://www.youtube.com/watch?v=v-F06HgiUpw
4 http://www.nvda-project.org/
5 http://www.userite.com/ecampus/lesson1/tools.php
6 http://www.youtube.com/watch?v=c8plj9UsJbg
7 http://www.youtube.com/watch?v=v-F06HgiUpw
8 http://www.clarku.edu/∼djoyce/trig/
9 http://www.dickbaldwin.com/tocjscript1.htm
10 http://www.w3schools.com/js/default.asp
11 http://www.dickbaldwin.com/toc.htm
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3 Discussion
I purposely published an earlier module titled Vector Subtraction for Blind Students
this module.
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in preparation for
An unbalanced net force is required
As you learned in an earlier module, in order for an object to travel in a circle under uniform circular
motion, a force must act on that object to cause it to stay on the circular path. Otherwise, according to
Newton, the object will y o into space at a constant velocity (ignoring both air resistance and gravity).
Change of direction equals acceleration
Whenever the velocity vector that describes the motion of an object changes direction, that object has
undergone an acceleration, even if the magnitude of the velocity vector hasn't changed.
Continual change in direction
The velocity vector for an object that is traveling on a circular path at a constant speed is constantly
changing direction. Otherwise, the object would not stay on the circular path. As you are aware, velocity is
a vector. That is, it has both magnitude and direction.
As you are also aware, acceleration (the time rate of change of velocity) is also a vector. That is to say,
it also has magnitude and direction. The direction of the acceleration vector is the same as the direction of
the force that caused the acceleration.
The direction of the acceleration vector
The direction of the acceleration vector for an object under uniform circular motion always points to the
center of the circle.
That may be dicult for you to believe. I will try to convince you using several dierent approaches,
some graphically satisfying, and some more anecdotal.
3.1 Subtraction of vectors
For this approach, I would like for you to use your graph board and follow along by creating and subtracting
vectors. Use your graph board, some string, and a few pushpins to create an arc of a circle centered on the
origin of a Cartesian coordinate system. All you will need is about one-half of the circle, on and to the right
of the vertical axis.
Uniform circular motion
We will assume that on object is moving counter-clockwise around that circle with a uniform speed.
The initial velocity vector
Draw a vector with a length of ten units with it's tail at the intersection of the circle and the positive
horizontal axis. This vector should point straight up in the direction of the positive y-axis. This vector
represents the velocity of the object as it crosses the horizontal axis on its trip around the circle.
The nal velocity vector
Now go up to a point on the circle about 30 degrees relative to the positive horizontal axis (the exact
angle isn't critical). Draw a vector with a length of 10 units that is tangent to the circle at that point.
It isn't easy to draw a vector that is tangent to a circle. However, the direction of that vector should be
perpendicular to a line that extends from the point on the circle to the center of the circle. That will cause
the vector to be tangent to the circle at that point.
The initial and nal velocity vectors
We now have two vectors that represent the velocity vectors for the object at two dierent points along
its circular path separated by a time interval. (The exact time interval doesn't matter at this point in
the discussion.) Label the vector at the horizontal axis as Vi (for initial vector). Label the vector at the
30-degree point as Vf (for nal vector).
Although there is no way for you to label it on the graph, we will represent the time interval required for
the object to move from the rst point to the second point as dt (which is an abbreviation for delta-time,
or change in time).
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Average acceleration
We know that the average acceleration is the time rate of change of velocity. Therefore, we can write
Aavg = (Vf - Vi)/dt
where
•
•
•
•
•
Aavg represents the average acceleration
Vf represents the velocity at the second point
Vi represents the velocity at the rst point
(Vf - Vi) represents the change in velocity
dt represents the time interval over which that change took place
Subtracting vectors
In the earlier module titled Vector Subtraction for Blind Students 13 , you learned how to subtract the
vector named Vi from the vector named Vf using the parallelogram method.
Let the time interval get smaller and smaller
What we want to do in this module is to estimate what happens to (Vf - Vi) as the time interval, dt,
gets smaller and smaller.
Given that the speed of the object remains constant, the point on the circumference of the circle that
represents the tail of Vf will move closer and closer to the horizontal axis as the time interval gets smaller.
The angle between the vectors gets smaller and smaller
That, in turn, will cause the angle that Vf forms with the horizontal axis to move closer and closer to
90 degrees. That means that the dierence in the angles that the two vectors make with the horizontal axis
will grow smaller and smaller, and the angle between the two vectors will grow smaller and smaller.
The dierence vector
In the earlier module, you learned that for very small angles, the dierence vector is very close to being
perpendicular to both Vf and Vi. Being very close to perpendicular to both Vf and Vi means that the
acceleration vector is very close to pointing directly at the center of the circle.
Using calculus, it can be shown that in the limit, as the time interval, dt, approaches zero, the direction
of the acceleration vector points directly at the center of the circle.
The magnitude of the acceleration vector
In the earlier module, you also learned that the magnitude of the dierence vector approaches zero as the
angle between the two velocity vectors approaches zero. However, the acceleration is equal to the magnitude
of the dierence vector divided by the scalar value of the time interval, dt, which is also approaching zero.
One very small value divided by another very small value is not necessarily a very small value. It can be
shown using calculus that in the limit, the magnitude of the acceleration vector is not zero.
The conclusion
The conclusion is that an object undergoing uniform circular motion experiences an acceleration vector
that points directly at the center of the circle and it has a non-zero magnitude.
3.2 An old-fashioned merry-go-round
That was the graphical explanation of the acceleration vector. Now I will cite a few anecdotal explanations.
A long wooden plank
When I was around ten or eleven years old, the child across the street from my house had an old-fashioned
homemade merry-go-round in his back yard. This device consisted of a long wooden plank with a hole in
the center. A large bolt was threaded through the hold and pushed vertically into a hole in the top of a post
about two feet tall. The bolt formed a spindle and the plank was able to rotate around the spindle.
Two handles
Two short skinny boards were fastened to the plank about 18 inches from each end to form a sort of a
cross in each end. These boards extended about 9 inches on either side of the plank and were intended to
be handles.
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A three-child operation
It took three children to operate this merry-go-round. One child sat on each end and the third child
pushed it around and around as fast as possible.
Oops
We quickly learned that if we sat on the plank between the handle and the end of the plank and tried
to hold onto the handle, we would slide o the end of the plank when the rotation of the plank reached a
certain speed. Therefore, we learned to sit with our legs across the handle, with the handle pressing against
the back insides of our knees. By doing this, we could survive without sliding o the plank no matter how
fast the third child caused the plank to rotate.
Why am I telling you this?
Although I didn't understand it at the time, when the plank was rotating around its spindle at a relatively
high rate of speed, the handle exerted a force on the inside of both knees. In other words, the handle exerted
a force on each leg in the direction of the center of the plank. This is a force that we will call a centripetal
force .
note:
Centripetal force
According to The Physics Classroom 14 , any object moving in a circle (or along a circular path)
experiences a centripetal force. That is, there is some physical force pushing or pulling the object
towards the center of the circle. This is the centripetal force requirement.
The word centripetal is merely an adjective used to describe the direction of the force. We are not
introducing a new type of force but rather describing the direction of the net force acting upon the
object that moves in the circle. Whatever the object, if it moves in a circle, there is some force
acting upon it to cause it to deviate from its straight-line path, accelerate inwards and move along
a circular path.
A centripetal force on my body
The eect of this centripetal force was to cause my body to accelerate in the direction of the force, which
was directly toward the spindle at the center of the plank. The spindle was at the center of the circle around
which I was experiencing circular motion (although probably not uniform circular motion).
3.3 A weight on a string
Consider tying a weight, such as a full bottle of soda, to the end of a string. Then swing the bottle around
and around above your head in an approximate circle. You will feel a force pulling your hand toward the
periphery of the circle, so you will exert a force on the string to keep the bottle from ying away. The force
that you exert on the string will be exerted on the bottle at the other end of the string and that force will be
directed along the string towards the center of the circle. That is the force that we will refer to as centripetal
(center seeking) force, and that force will act on the mass of the bottle to cause the bottle to accelerate in
the direction of the center of the circle.
What happens if the string breaks?
If the string breaks, there will no longer be a centripetal force acting on the bottle and it will y o into
space along a line that is tangential to its location on the circle at the instant the string breaks. It won't
continue moving in a circle. Instead, it will move in a straight line until the resistance of the air and force
of gravity bring it to a sudden stop on the ground. (Actually, it will move along a parabolic arc under the
force of gravity until it strikes the ground.)
3.4 Water in a bucket
If you don't mind taking a chance on getting wet, ll a small bucket about half full of water. Then swing
the bucket rapidly in a circle in a vertical plane.
14 http://www.physicsclassroom.com/Class/circles/u6l1c.cfm
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If you can swing the bucket fast enough, the water will stay in the bucket even when the bucket is upside
down. Why is that? The water molecules want to move in a straight line. However, the inside surface of
the bucket exerts a centripetal force on the water molecules causing them to accelerate toward the center of
the circle. The centripetal force increases with the speed of the bucket. As long as the centripetal force is
greater than the weight of the water, the water won't fall out of the bucket onto your head. However, if you
allow the speed of the bucket to decrease, you will reach the point where gravity will overcome, and you will
probably get wet.
3.5 The moon and the Earth
Don't ask me how it got started in the rst place, but somehow the moon got started circling the Earth.
The speed of the moon and the radius of its orbit is just exactly right so that the gravitational force that
the Earth exerts on the moon causes the moon to accelerate towards the Earth. The amount of acceleration
toward the earth, when combined with the speed of the moon, causes the moon to move in a uniform circular
orbit around the Earth instead of either ying o into space or crashing into the Earth.
3.6 The work-energy explanation
It is probably time to start explaining this phenomena in a more technical and less anecdotal manner.
The application of a centripetal force for uniform circular motion causes the direction of the object to be
changed without changing its speed. Let's see if we can explain this from a work-energy viewpoint.
Work
Recall from an earlier module that work is a force acting upon an object to cause a
Also recall that the amount of work done on an object, expressed in Joules, is given by
Work = F * D * cosine(theta)
where
displacement
.
• F represents the force that causes the displacement expressed in Newtons.
• D represents the amount of the displacement expressed in meters.
• theta is the angle between the direction of the displacement and the line of action of the force.
Centripetal force points toward center of circle
As we showed (or claimed to show) earlier, the centripetal force for an object in uniform circular motion
always points in the direction of the center of the circle. At the same time, the velocity vector, which
represents the direction of the displacement is tangential to the circle. Therefore, the angle between the
centripetal force and the direction of displacement is 90 degrees. This means that the centripetal force does
no work on the object, because the cosine of 90 degrees is zero.
No work is done
When no work is done upon an object by external forces, the total mechanical energy, consisting of
potential energy plus kinetic energy, of the object remains constant. If an object is moving in a circle in a
plane that is parallel to the surface of the Earth, the eect of gravity may pull the entire plane toward the
surface of the Earth, but it won't eect the circular motion unequally with respect to the position of the
object in the circle.
Therefore, if an object is moving in a horizontal circle at constant speed, the centripetal force does no
work on the object and cannot change the total mechanical energy of the object. The kinetic energy will
remain constant and therefore, the speed of the object will remain constant. The centripetal force will
accelerate the object by changing its direction but will not change its speed.
3.7 Centripetal, not centrifugal
Don't confuse centripetal force with the word centrifugal . For some reason, many people mistakenly use
the word centrifugal (meaning outward) when they should use the word centripetal. In this case, centripetal
is the correct word.
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Maybe this is like the use of the words nuclear and nucular. Which is correct? I will leave that as an
exercise for the student to determine.
3.8 Summary
An object in uniform circular motion must experience an unbalanced force pointing towards the center of
the circle. This force is referred to as a centripetal force, where centripetal means inward seeking .
This force is required to cause the object to continually change its direction in order to move along a
circular path without changing its speed.
Because the centripetal force is directed perpendicular to the tangential velocity vector, the centripetal
force cannot change the total mechanical energy possessed by the object (the cosine of 90 degrees is 0).
Because the centripetal force has no impact on the potential energy possessed by the object, and because
it cannot change the total mechanical energy, it cannot change the kinetic energy possessed by the object.
Since it can't change the kinetic energy, it can't change the object's speed.
However, it can change the object's direction without changing its speed.
4 Example scenarios
4.1 Scenario #1
An object is moving with uniform circular motion in a counter-clockwise direction around a circle whose
center is at the origin in a Cartesian coordinate system. When the object passes through the intersection
of the circle with the positive horizontal axis, what is the direction of the velocity vector relative to the
horizontal axis?
1.
2.
3.
4.
5.
0 degrees
180 degrees
90 degrees
270 degrees
30 degrees
The correct answer is #3, 90 degrees. At that point, the line tangent to the circle touches the circle where
it intersects the positive horizontal axis and is perpendicular to the horizontal axis. Because the motion is
counter-clockwise, the vector points up at 90 degrees (instead of down at 270 degrees) relative to the positive
horizontal axis.
4.2 Scenario #2
An object is moving with uniform circular motion in a counter-clockwise direction around a circle whose
center is at the origin in a Cartesian coordinate system. When the object passes through the intersection
of the circle with the positive horizontal axis, what is the direction of the acceleration vector relative to the
positive horizontal axis?
1.
2.
3.
4.
5.
0 degrees
180 degrees
90 degrees
270 degrees
30 degrees
The correct answer is #2, 180 degrees. The acceleration vector for an object under uniform circular motion
always points in the direction of the center of the circle. At the intersection of the circle with the positive
horizontal axis, the direction of the center of the circle is 180 degrees relative to the positive horizontal axis.
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4.3 Scenario #3
A heavy object is being swung on a string in a counter-clockwise direction around a circle whose center is
at the origin in a Cartesian coordinate system. The plane of the circle is parallel to the ground so that we
can ignore the eects of gravity.
Just as the object passes through the intersection of the circle with the positive horizontal axis, the string
breaks allowing the object to y o into the nearby playground space. What is the direction of motion of
the object relative to the positive horizontal axis?
1.
2.
3.
4.
5.
6.
0 degrees
180 degrees
90 degrees
270 degrees
30 degrees
The object will continue to move in a circle.
The correct answer is #3, 90 degrees, which is the direction of the velocity vector at the instant that the
string breaks. Because there will no longer be a centripetal force to cause the object to change direction and
stay on the circular path, it will continue moving in the direction that it is moving when the string breaks.
5 Resources
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.
6 Miscellaneous
This section contains a variety of miscellaneous information.
note:
Housekeeping material
• Module name: Circular Motion Acceleration and Centripetal Force for Blind Students
• File: Phy1250.htm
• Keywords:
· physics
· accessible
· accessibility
· blind
· graph board
· protractor
· screen reader
· refreshable Braille display
· JavaScript
· trigonometry
· uniform circular motion
· tangential
· velocity vector
· centripetal force
· work-energy
note:
Disclaimers: Financial : Although the Connexions site makes it possible for you to
download a PDF le for this module at no charge, and also makes it possible for you to purchase a
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pre-printed version of the PDF le, 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 nancial compensation from the Connexions website even
if you purchase the PDF version of the module.
Aliation : I am a professor of Computer Information Technology at Austin Community College
in Austin, TX.
-end-
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