Module 5  Uniform Circular Motion
From PER wiki
Up until now in this Unit we have been specifying details of the acceleration experienced by an object (e.g. constant acceleration, zero acceleration) and then using the concepts of mechanics to learn about the motion. In this Module, we will work the other way: we will specify a motion we are interested in and then use the concepts of mechanics to learn about the acceleration that is being experienced. The motion we are interested in is motion in a circle of constant radius at constant speed, which is referred to as uniform circular motion.
Learning Goals
After working through this module, you should be able to:
 Explain how it is possible that an object moving with constant speed can be undergoing an acceleration.
 Define centripetal acceleration.
 Draw free body diagrams for objects executing uniform circular motion.
Observing the Motion using the Physics Teaching Technology Resource
[What is the Physics Teaching Technology Resource?]
In order to describe a motion, we have to first observe it carefully. One of the great advantages of mechanics over other branches of physics is the fact that we can use our eyes to make the observation (this is not true when studying electricity or quantum mechanics, for example). The Physics and Astronomy Education Research Group at Rutgers University have created a library of simple video experiments to illustrate many kinds of motion. Below, we present three of their videos that involve circular motion. Observe them carefully. Particularly, think about the forces acting on the person or object that is moving in a circle in each video.
These videos are part of the Physics Teaching Technology Resource © Rutgers University, used with permission. Please wait while the movies load up. (This may take a while.) You will need QuickTime to view the movies.  


Centripetal Acceleration
We are now ready to discuss the properties of the acceleration associated with uniform circular motion. There are three important claims we will make about this acceleration.
Claim 1: An object executing uniform circular motion must be experiencing an interaction.
This claim is based upon Newton's First Law, which says that an object that is not experiencing any interaction will necessarily move in a straight line. You can see the evidence for this claim in the video segments above. The bowling ball in the first video moves in a straight line unless acted upon by the hammer. In the second video segment, Professor Etkina skates in a straight line unless acted upon by the rope (first circular path) or by friction from the ground (second circular path). In the third video segment, the ball only moves in a circular path when constrained to do so by the interaction from the wall. When the wall is removed, the ball travels a straight path.
Claim 2: The net interaction on an object in uniform circular motion must produce a force pointing toward the center of the circular path.
The evidence for this claim can be seen by careful observation of the videos using what we have already learned about the properties of interactions. In the first video segment, the hammer is always applied to the bowling ball on the side directly opposite the center of the circle. If we assume that the ball and hammer are fairly smooth (so there is little friction) then the contact interaction points toward the center of the circle. In the second video, the rope turning Professor Etkina clearly pulls her toward the center of the circle when she first makes a circular path. In her second circular path, she kicks out away from the center, producing a reaction force that (according to Newton's Third Law) points inward toward the center. In the third video segment, the normal force applied by the wall is perpendicular to the wall and therefore points toward the center of the circle (again, we assume that friction is small).
Claim 3: The size of the object's acceleration is given by v^{2}/r, where v is the speed of the object and r is the radius of the circular path.
Evidence for our final claim cannot be observed directly in the video, but must be derived using what we have assumed about the motion (uniform speed and uniform radius). Below, we present two possible derivations. One uses trigonometry plus limits, and the other uses calculus.
Derivation, Option 1: Trigonometry plus Limits 

As with any mathematical description of motion, we begin by setting up a coordinate system. We will choose our coordinates such that the circular motion takes place in the xy plane with its center at the origin. Further, we will arrange things so that the object is moving counterclockwise, and we will take the initial time t_{i} to occur at an instant when the object is touching the positive xaxis. With these choices in mind, we can draw the velocity of the object at two different times: t_{i}, when the object is on the xaxis moving counterclockwise, and t_{i}+Δt, a time Δt later. The velocity of the object at each of these times is shown in the figure below.
Note that the velocity is always tangential to the path of the object. It is also important to note that the object's movement along the circle has swept out an angle Δθ in the time Δt. Because acceleration is defined as the time rate of change of the velocity, we want to look at the change in the velocity during the time Δt. To do this, we set up a vector subtraction, taking:
Graphically, this subtraction looks like: Now, we have to use a trick. We know that the velocity has a constant size v because of our assumption of uniform circular motion. Thus, the changing velocity vector will itself sweep out a circle of radius v, as indicated by the dotted line in the figure below: The trick is this: as the time difference between the two velocities we are subtracting becomes very small, so does the angle Δθ. In the limit of a very small Δθ, the length of a small portion of the circle swept out by the velocity vector as it turns becomes essentially equal to the straightline difference Δv that we are trying to find. The important thing about this is that we know the length of the curved portion of the circle. That length is simply Δθ times v. Thus, in the limit of a very short time interval, we know:
Now, we can go back to the original circle in position space (the green circle in the first figure) and we can find an expression for Δθ. We can do this by using the fact that in a time Δt the object will cover an arc length of vΔt along the edge of the circle, as shown in the figure below. Thus, as long as we are measuring θ in radians, we can write:
Substituting this into our expression for the change in velocity gives:
Now, we simply use the definition of acceleration as the time rate of change of velocity:
where the Δt's have canceled. 
Derivation, Option 2: Calculus 

We can use the definition of acceleration as the second time derivative of position, provided that we can write a mathematical expression for the position of an object executing uniform circular motion. Before we can define the position, we must set up a coordinate system. We will choose our coordinates such that the circular motion takes place in the xy plane with its center at the origin. Further, we will arrange things so that the object is moving counterclockwise, and we will take the initial time t = 0 to occur at an instant when the object is touching the positive xaxis. The situation is shown below:
The components of the position vector can then be written using trigonometry as:
and
where we have used the fact that we can find the angle in radians by dividing the arc length traveled by the object by the radius of the circle. The arc length traveled around the circle is simply vt because the object is moving with constant speed v. With our position functions in hand, we can find the velocity of the object by taking derivatives:
and
By plotting this vector and the position vector for an arbitrary time t, you can show that the velocity of the object is perpendicular to the position vector and tangent to the circle as shown in the figure below. With the velocity determined, we can now take another derivative to find the acceleration:
and
We can now find the size of the acceleration by using the Pythagorean theorem plus the trig identity that sine squared plus cosine squared equals 1:
and we have demonstrated Fact 3, which completes our goal. We can actually go farther and use our work to demonstrate Fact 2 as well. Looking carefully at the form of the acceleration, you can see that:
in other words, the acceleration points directly opposite to the position vector, as shown in the figure below. Thus, the acceleration points to the center of the circle, and we have demonstrated Fact 2. 
The acceleration associated with circular motion is generally called centripetal acceleration because of fact 2. "Centripetal" means "centerseeking" and is meant to remind us that the acceleration of an object moving in a circle constantly changes direction so as to always point at the center of the circular path.
The Uniform Circular Motion (Centripetal Acceleration) Model
We are now ready to introduce the Uniform Circular Motion (Centripetal Acceleration) Model. In principle, this Model involves a full description of the position, velocity and acceleration of an object moving in a circle of constant radius with constant speed. In practice, the most important thing to remember about Uniform Circular Motion is that any object executing this motion must be experiencing a centripetal acceleration that obeys the three facts outlined above. By adding in what we have learned about Newton's Laws, we can see that if we are given a problem involving an object executing Uniform Circular Motion we have important information about the magnitude and direction of the net force on this object (since the net force is equal to ma, and we know a = v^{2}/r and that it points toward the center of the circle).
Illustrative Example: The International Space Station
By navigating to the site http://spaceflight.nasa.gov/realdata/tracking/ you can obtain tracking information giving the altitude and speed of the International Space Station (ISS). Suppose that the information displayed on the site was as shown in the screen capture above. By making the assumption that the space station's orbit is a circle with its center at the center of the earth, find the approximate magnitude of the acceleration experienced by the space station as a result of the gravitational pull of the earth.
Solution 

System: [show system] Interactions: [show interactions] Model:[show Model] Quantitative Approach:[show approach] 