* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Circular Motion Acceleration and Centripetal Force
Survey
Document related concepts
Hooke's law wikipedia , lookup
Derivations of the Lorentz transformations wikipedia , lookup
Bra–ket notation wikipedia , lookup
Classical mechanics wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Four-vector wikipedia , lookup
Equations of motion wikipedia , lookup
Mass versus weight wikipedia , lookup
Velocity-addition formula wikipedia , lookup
Jerk (physics) wikipedia , lookup
Coriolis force wikipedia , lookup
Laplace–Runge–Lenz vector wikipedia , lookup
Fictitious force wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Centrifugal force wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Transcript
OpenStax-CNX module: m38406 1 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/ http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 2 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 http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 3 3 Discussion I purposely published an earlier module titled Vector Subtraction for Blind Students this module. 12 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). 12 http://cnx.org/content/m38374/latest/?collection=col11294/latest http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 4 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. 13 http://cnx.org/content/m38374/latest/?collection=col11294/latest http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 5 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 http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 6 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. http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 7 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. http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 8 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 http://cnx.org/content/m38406/1.2/ OpenStax-CNX module: m38406 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- http://cnx.org/content/m38406/1.2/ 9