I. Introduction

The object of this exercise is to verify the equation for centripetal acceleration,

for an object moving with a constant speed, v, around a circle of radius, r. We will also verify that the direction of the acceleration vector is toward the center of the circle. We will use the graphical method of vector arithmetic.

I. Introduction

Suppose an object initially has velocity vector v_A, and after a time Dt it has velocity vector v_B. Then its average acceleration during that period is given by

The direction of a_avg is the same as the direction of v_B - v_A. We will also call this velocity difference Dv.

To find the instantaneous acceleration for an object at a single point P, we would like to take two points very close in time. Dt will be very small, and the vector difference, Dv, will also be very small. It is, however, not practical to calculate the vector difference this way, because small errors in drawing the vectors will lead to large errors in the vector difference. Hence in this lab we will take fairly large time differences, and we will still see that the direction and magnitude of the average acceleration come out reasonably close to the expected values.

II. Dividing the Work

Students will work in groups of two, but each student will do his or her own constructions. You will use a compass to draw an arc of a circle on graph paper. We will want to find the acceleration of an object as it moves through an angle q subtended at the center of the circle. We will do constructions for q equal to 30, 40, 50, 60, 70, 80, 90, 100, 110, and 120 degrees. Each group will do only two of these angles. The instructor will ask some groups to do 30 and 80 degrees, other groups to do 40 and 90 degrees, etc., so that we have some work for all these angles. Later you will put your resuts on the blackboard and you will look at the calculated average accelerations together.

Within the group of two, both students must do the construction for both angles. When you have your results you can compare them. If there seems to be a significant difference, try to figure out why, and if you can't, ask the instructor.

III. Construction

Set the compass to a radius of exactly 10 cm. Hold the graph paper vertically, place the center of the circle on the left side, and draw an arc starting directly below the center and extending around somewhat beyond your value of q.

We will think of this arc as a scale drawing of part of the circle, with the scale being 1 cm equals 1 m in the real motion.

Record the radius of your circle in meters.

Point A, the starting position of the motion, should be directly below the center, so that the radius vector from the center to A is vertical. The velocity vector, v_A, is tangent to the curve at A, and so must be drawn horizontal. (An important theorem of plane geometry is that the tangent to a circle is perpendicular to the radius at the point of contact.) Draw the velocity vector 10 cm long. In the velocity diagram we will take 1 cm equal to 1 m/s.

Record the velocity of your motion in m/s.

Now find point B, the final position of the motion, using the protractor. It is an angle q around the arc from A. Draw the radius vector from the center to B, and draw the tangent to the circle at B.

Note: To draw the tangent, do not try to draw a line that intersects the circle at just one point. A better way is to use the protractor to draw a perpendicular to the radius.

Draw the velocity vector v_B along this tangent, 10 cm long. (Remember, we are considering uniform circle motion, so the speed is constant.) Now draw the vector -v_A with its tail on the head of v_B. The vector -v_A is, by definition, opposite in direction to v_A, and so it is horizontal on your paper, and 10 cm long. Now use the definition of vector summation to construct v_B + (-v_A), which in turn is equal to v_B - v_A, or Dv.

IV. Direction of the Acceleration

The direction of Dv is the direction of the acceleration. Is your vector pointing toward the center of the circle?

Remember that the velocity vector does not have a location in space. It only has a length and a direction. (50 mph north in Chicago is the same velocity as 50 mph north in NY.) The acceleration you have found is the average acceleration of the object as it moves from point A to point B. Therefore, the way to look at this vector is to move it to a point somewhere between A and B. Select a point halfway along the circle from A to B, and move your Dv, so that its tail is at that point. To move it, use the technique described in part VII of this write-up. Now does it point toward the center of the circle?

V. Magnitude of the Acceleration

Using the two scales (for distance and for velocity), you can calculate the magnitude of the acceleration:

1. First recall your values of r (in meters) and v (in m/s).
2. Next find the magnitude of Dv on your diagram: Measure the vector's length with a ruler, in cm, and convert to m/s. Now you have to find the time interval, Dt.
3. First find the period, T, for one complete circle, given that the circumference is 2pr, and your velocity is v. This time will be in seconds.
4. Next find the time, Dt, to go from A to B, for your construction. To do this use the fact that the length of the arc from A to B is a fraction q/360 of the full circumference (with q in degrees). Therefore, Dt is the same fraction of the period. Find the fraction.
5. Find Dt.
6. Calculate the magnitude of the acceleration,
   a_avg = Dv/Dt, This will be in m/s².

What you have found, using the definition of acceleration as rate of change of velocity, is the average acceleration from point A to point B. The formula for instantaneous acceleration is v²/r. Calculate this value also.

VI. Collecting Results

To reduce the amount of data being considered, the two students in a group should average their two results for each angle. Then place on the blackboard their values of q and a_avg. All students in the lab should then copy the table, and plot a graph of a_avg vs. q, using all the points. (There will be about 20 points.) Draw a smooth curve as close as possible to all the points (not necessarily a straight line, unless it really looks like a straight line). Then extrapolate the curve to q = 0. This limit, as the distance (and time) from point A to point B gets very small, is the instantaneous acceleration. Mark the theoretical value, v²/r, on the vertical axis. What is the percent difference between your value and this theoretical value?

VII. Note on moving a vector:

An important property of a vector is that it does not have a location in space. If you have a vector at point, P (as shown below), you can re-draw it at point Q, with the same magnitude and direction, and it is the same vector.

In practice, you can measure the angle, b, between the vector and a horizontal line at P. Then you use a protractor to construct the same angle at Q; then draw the vector at Q by making it the same length as at P.