Experiment 6

Centripetal Force


The objective is to experimentally verify the formula for centripetal force.


A computer with internet connection, a calculator (The built-in calculator of the computer may be used.), paper, and pencil


An object performing uniform circular motion is constantly under the action of a force that acts toward its center of rotation and has a magnitude of

where m is the mass of the object, v its liner speed, and r its radius of rotation.

An object tends to move along a straight line if it is under the action of a force that has a constant direction, but if the direction of the force keeps changing such that it is always directed toward the center of a circle, the object is then forced to follow a circular path.  Fig. 1 shows a small object of mass m that is connected to a string of length r and is being spun in a horizontal plane at constant linear speed v.  Note that the velocity is not constant. This is because of the fact that, in circular motionthe direction of velocity keeps changing.  The velocity vector v is always perpendicular to the radius of rotation r.

Figure 1

If the radius of rotation r is very big, the object is not greatly forced to curve, and therefore, a small centripetal force is needed for its mild curved motion.  You will notice this in the following applet from the way the necessary hanging mass becomes smaller when greater radius of rotation is selected.


If you have not done so, you may need to add the following Website to your Java exception list: http://www.phy.ntnu.edu.tw/

To do this, follow the path (Windows operating system),

Start → All Programs → Java → Configure Java → Security (use High) → Edit Site List … → Add → Type in the site URL (http://www.phy.ntnu.edu.tw/).

Click on the following link: 

www.phy.ntnu.edu.tw/oldjava/circularMotion/circular3D e.html

In the applet, the red mass m attached to a cord of length r = 10.0m (as shown in the top left corner) is performing uniform circular motion.   The length of the cord can be changed by moving the hanging mass M up or down by mouse.  As you move the hanging mass up and down, the radius of rotation r changes as its value can be read and adjusted at the top left corner.  r may be set at any desired value.  For simplicity of calculations, we will assume that the mass of the rotating object is m = 1.00kg.

The accepted value for ac can be read from the space at the top where it says Mg/m.  Note that this ratio has units of acceleration, or simply m/s2.   Setting m = 1.00kg, the product Mg gives us the centripetal acceleration of the rotating mass m.  (The kg in m cancels the kg in M, and the result has unit of m/s2 or unit of acceleration).

The measured value for centripetal acceleration can be found in a different way.  Count he number of rotations in each trial of the experiment and read the corresponding elapsed time from the applet Since the length traveled on the circle per turn is 2p r, the length traveled in n turns is Dx = 2p r n.  The elapsed time is t or Dt The average speed is therefore, v = Dx/Dt as shown in Table 1.   Now, knowing v, the measured ac can be found from v2/r.

Accordingly, the measured and accepted values for centripetal force can also be found as shown Table 1.

Note:  When you clean any previous run and reset the applet, it is ready to go.  Holding and releasing the left click runs and stops the applet.  You need to practice a few times to become comfortable with it.

1) Let the initial data be: r = 15.0m, and w = 0.69rd/s, and m = 1.00kg as was mentioned before.

2) click on the page and put m into motion.  Let it run for 5 turns, for example, and at the end of the 5th turn release the mouse and it will stop.  Record the values of t and n, the number of turns, in the Table 1.

3) Calculate V V2/r , and read Mg/m and record them in Table 1.

4) Calculate the corresponding centripetal forces in Table 1 as well.

5) Calculate the related % error and record it.

6) Repeat the experiment for cases 2 through 6 by changing the radius of rotation according to Table 1.  Of course, the hanging mass M has to be changed to provide the necessary centripetal force.  For precision, You need more number of turns as mass m turns faster at smaller radii.  This will make the error smaller because of your reaction time and your judgment at start or stop instants.

Note:  If the applet starts acting strange, refresh the screen. 


Given:       m = 1.00kg,    = 9.8 m/s2.


Table 1

Trial r







# of





v =




 ac= v2/r









 Fc= Mg





1 15.0 1.00 5.0              
2 12.0 1.00 10.0              
3 10.0 1.00 20.0              
4 8.0 1.00 30.0              
5 6.0 1.00 40.0              
6 5.0 1.00 50.0              


Show typical calculations done in the chart.

Comparison of the results: 

Write down the percent error formula used.


State your conclusions of the experiment.


Provide a discussion if necessary.


1) Give two examples on objects performing circular motion and determine the source of centripetal force for each. 

2) Suppose you tie a rock to a string and spin it in a vertical plane.  In what direction will the rock move if the string breaks exactly when the rock is at the lowest point in its circular path?

3) Using V = 2πrf, derive a formula for Fc in terms of m, r, and frequency f Simplify the formula after you substitute for v in the centripetal force formula.  Note that f is the frequency of rotation or the number of turns per second.