Experiment E2025 Uniform Circular Motion

Figure 1. A puck travels in a horizontal circle on an air table. Discussion The Model Marking the Video Graphical Analysis

Discussion

Figure 1 shows a puck traveling in a horizontal circle on an air table. One end of a string is attached to a fixed point shown with the red angle in Figure 1. The other end of the string is attached to the puck. When set into motion, the puck travels in a circle with a radius equal to the length of the string. After marking the video you will see that the puck travels equal distances in equal times. This motion is called Uniform Circular Motion.

Uniform Circular Motion

An object undergoing uniform circular motion has these two observable characteristics:

1. The object is traveling in a circular path.
2. The speed is constant, that is the object travels equal distances in equal times.

Now, the puck travels equal distances in equal times, however, the direction of motion is continually changing. This means there is an acceleration. Recall the definition of net force:

F_net = ma

Examining this equation, we see that an observed acceleration indicates there must be at least one force causing it. Also, the directions of the acceleration and the net force are the same since the mass is positive.

What force or forces can cause the acceleration in Figure 1? Remember that the motion is a horizontal circle, so both x and y are horizontal. The weight and normal are both perpendicular to the page and can not cause the observed acceleration (the weight is into the page and the normal is out of the page). Also, the air table reduces friction, so we will ignore it in this experiment. The only force that is left is tension. Recall that the direction of the tension is away from the object and along the string. Can you see that its direction will always be toward the center of the circle in this experiment?

We now know the direction of the acceleration. It must be in the same direction as the net force. In this experiment, the net force is the tension and its direction is always toward the center of the circle. So, the acceleration is also always toward the center of the circle. Now, what is the magnitude of the acceleration? How does it depend on the speed, v, and the radius, r?

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The Model

Let's examine the relationship between acceleration and speed first. Figure 2 shows two diagrams of an object traveling in uniform circular motion. In the right diagram, the radius is the same but the speed has been doubled.

Look at the left diagram in Figure 2 first. The object is traveling counterclockwise and at point a its velocity is v₁ up. At point b its velocity is v₂ to the left. The time it takes to travel from a to b is t. We can write this acceleration as:

a₁ = Dv/Dt = (v₂ - v₁)/t

Here v₁ and v₂ have the same speed, but their directions are different. So, their difference is not zero.

Looking at the right diagram, the speed has been doubled. At point a the velocity is 2v₁ up and at point b the velocity is 2v₂ to the left. The time it takes to travel from a to b is now t/2. This is true because the distance from a to b is the same, however the speed is doubled so the time is halved. We can write this acceleration as:

a₂ = (2v₂ - 2v₁)/(t/2) = 2(v₂ - v₁)/(t/2) = 4(v₂ - v₁)/(t)

Or:

a₂ = 4a₁

So, doubling the speed did not double the acceleration, it increased it by four times. What would happen if we tripled the speed? Then at point a the velocity is 3v₁ up and at point b the velocity is 3v₂ to the left. The time it takes to travel from a to b is now t/3. Calling the acceleration for this case a₃, see if you can show that:

a₃ = 9a₁

Try this example when the speed is quadrupled. You should see this pattern:

• Increasing the speed by a factor of two increases the acceleration by a factor of four.
• Increasing the speed by a factor of three increases the acceleration by a factor of nine.
• Increasing the speed by a factor of four increases the acceleration by a factor of sixteen.

Therefore, in uniform circular motion, the acceleration depends on the speed squared.

Now lets examine the relationship between the acceleration and the radius. Figure 3 shows two diagrams. In the right diagram, the speed remains the same but the radius is doubled.

Look at the left diagram in Figure 3 first. The object is traveling counterclockwise and at point a, its velocity is v₁ up. At point b, its velocity is v₂ to the left. The time it takes to travel from a to b is t. We can write this acceleration as:

a₁ = (v₂ - v₁)/t

Looking at the right diagram, the radius has been doubled. At point a the velocity is v₁ up and at point b the velocity is v₂ to the left. The time it takes to travel from a to b is now 2t. This is true because the distance from a to b is now twice as large, however the speed is the same so the time is doubled. We can write this acceleration as:

a₂ = (v₂ - v₁)/(2t) = a₁/2

So, doubling the radius halved the acceleration. What would happen if we tripled the radius? Then at point a the velocity is still v₁ up and at point b the velocity is still v₂ to the left. However, the time it takes to travel from a to b is now 3t since the distance is tripled. Calling the acceleration for this case a₃, see if you can show that:

a₃ = a₁/3

Try this example when the speed is quadrupled. You should see this pattern:

• Increasing the radius by a factor of two decreases the acceleration by a factor of two.
• Increasing the radius by a factor of three decreases the acceleration by a factor of three.
• Increasing the radius by a factor of four decreases the acceleration by a factor of four.

Therefore, in uniform circular motion, the acceleration depends on the inverse of the radius.

This analysis shows that the acceleration for uniform circular motion depends on the speed squared and on the inverse of the radius. Therefore we can write:

a_c = v²/r

Here the subscript, c, means centripetal (center seeking). Now we know that acceleration has dimensions of L/T² so the above equation must have these dimensions. Does it?

Summarizing, we can say these things for uniform circular motion:

• The object is traveling in a circular path.
• The speed is constant, that is the object travels equal distances in equal times.
• The direction of the net force is toward the center of the circle.
• The direction of the acceleration is toward the center of the circle.
• The magnitude of the acceleration is v²/r.

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Marking the Video

1. Play the video and observe the motion.
2. Select the Video Scale menu item, then the New Origin sub item. Place the new origin at the center of the circular path.
3. Mark the object until the video does not step any more.
4. Use the Circles - Vectors menu item to observe first the velocity and acceleration and then the momentum and net force. What are your observations?
5. Does the motion have the characteristics of uniform circular motion?
6. When you move the mouse on the video, notice the x and y positions changing below the video. Move the mouse to the new origin and you will see that both x and y are zero. Move the mouse to a point on the circular path and find the radius of the motion.

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Graphical Analysis

1. Press the GRAPHS Button and the Data Analysis Choices will appear. Pick Option 1. Click the Next button.
2. Direction of Motion? Select x and y.
3. From the plots options, select velocity and acceleration. Click the Next button.
4. Observe the velocity versus time graph. There are three plots, one for x, one for y and one for the magnitude. Is the magnitude (speed) fairly constant? Record the magnitude. Small lines are drawn from each data point on the magnitude plot. These lines show the direction of the resultant velocity. How does the direction of the velocity correlate with the object's motion in the video?
5. Observe the acceleration versus time graph. There are three plots, one for x, one for y and one for the magnitude. Is the magnitude fairly constant? Is the magnitude equal to v²/r as the model predicts? Small lines are drawn from each data point on the magnitude plot. These lines show the direction of the acceleration. Is the direction always toward the center of the circular path? Is the direction of the acceleration in the same direction as the net force?
6. Does this experiment support the model?

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