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Uniform Circular Motion
Uniform circular motion is the motion of an object in a circular path with a
velocity that has a constant magnitude and a direction that is constantly changing. This is
due to a force exerted on the object along the radius of the circle. This force is called the
centripetal force. The acceleration of the object is the rate of change of the velocity and
is given by
a = v2/r
(l)
where v is the speed of the body and r is the radius of the circle. The direction of the
acceleration is along the radius toward the center of the circle. This is the centripetal or
radial acceleration.
Newton’s 2nd law for circular motion then is
F = m v2 / r
(2)
The force F is called the centripetal force.
It is convenient in this experiment to write the velocity in quantities that can be
measured. The angular velocity ω is related to velocity by the radius of the circle. That
is,
v = rω
(3)
The angular velocity can be written in terms of frequency f, the number of revolutions per
second. That is
ω = 2πf
(4)
Substituting Eqn. (3) and (4) into (2), the centripetal force is equal to
F = (4 π2 m r ) f2
(5)
In this experiment, a mass is rotated at the end of a string. The mass is located on
a rod that can rotate and the mass is free to move along the rod. The string is fed through
a pulley and connected to a force probe. The force probe measures the centripetal force.
A photogate is attached to the apparatus such that it can read the period of one revolution.
There is a plastic rod that extends below the fixed mass that breaks the light beam of the
photogate once each revolution. The inverse of the period is the frequency of the motion.
The apparatus is illustrated in Fig. 1.
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Examining Eqn (5), if the force is plotted versus the square of the frequency, then
the data should produce a straight line that should go through the origin with a slope
equal to 4π2mr. In this experiment, part 1 is to hold the mass and radius constant and
vary the frequency and measure the force. In part 2, the radius is constant and different
masses are added. However, the frequency will change due to the different load on the
electric motor. Therefore, the force, mass, and frequency are measured at a fixed radius
and the force is calculated using Eqn (5).
APPARATUS:
1.Centripetal force apparatus
2. Photogate
3. Supporting stand
4. Force Probe
5. Logger Pro
6. 12 V Power Supply
Procedures:
1. The Centripetal Force Apparatus should already be assembled. There is a switch on
the force probe. Make sure the switch is set to the 10 N setting. Check to make sure the
force probe and the photogate are plugged into Logger Pro and Logger Pro is plugged
into an electrical outlet. Connect your laptop to Logger Pro and open the program. Go to
File/Open/Probe & Sensors/Photogates/strobe. The program should sense that a force
probe is also connected and you should click the box accordingly. The table should
display column heading of Time, Gate Status, and Frequency. Next, add the Force
column by navigating to Experiement/Set up Sensors/Show all Interfaces. Click on
Channel 1 and add the Dual Range Force Sensor. We will be using the Frequency and
Force columns.
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Part 1- Force versus frequency
2. The rotating mass is the mass that is attached to the string. The mass is the sum of the
mass of the weights plus the mass of the thumbscrew (4 g). The fixed mass is placed at
the same position on the other side. It is important that the rod be balanced so that it will
not wobble when rotated. The radius can be read on the scale below the mass. The string
may be slack so pull the mass out to read the radius. Add a 20-gram mass to both sides of
the apparatus and record the mass and the radius in the data table.
3. Set the voltage to 2 V. When the apparatus starts rotating, push collect. Collect data
for 3 seconds and push stop. The tables should be filled with data. The force column is
not so constant and does oscillate some. To find the average value, go to the Data menu
and open new calculated column. Name the column “Newtons” and go to
function/statistics/mean and then go to variable and choose force. The average value
should appear in next column. Do the same for average frequency and name it “Hertz”.
Record the frequency and force in data table. Repeat for voltages 3 thru 10 V at 1 V
intervals.
4. Plot the Force versus the frequency squared and fit the curve to a straight line. Record
the slope in data table. Compare the experimental slope with the theoretical value
(4π2mr).
Part 2 - Force versus mass
5. Set the power supply to 5 V and keep it constant throughout this part. Place 10 g on
both sides of the apparatus. The rotating mass now totals 14 g. Start the apparatus
rotating and measure the frequency and force. Record the mass and the radius in the data
table.
6. Increase mass on each side of rod by 5 g and repeat the measurements. Continue in 5
g increments to a maximum of 34 g. Be sure the radii of both masses remain equal
throughout experiment.
7. Using an Excel spreadsheet, calculate the force using Eqn (5) and find a percent error.
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Data Table:
Part 1 - Force versus Frequency
Mass =____________________
Radius= _____________
Experimental Slope = _______________
Theoretical Slope _____________
Percent error = __________________
Frequency
Frequency squared
Force
2V
3V
4V
5V
6V
7V
8V
9V
10V
Part 2 - Force Versus Mass
1.
Mass (kg)
0.014
2.
0.019
3.
0.024
4.
0.029
5.
0.034
Measured Force
Radius= _____________
Frequency
Show a sample calculation below.
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Calculated Force
Error
Questions:
1) How would the motion of the mass be affected if the apparatus was rotated such that
the plane of the circle was vertical?
2) Because the Earth rotates about its axis, you will have an apparent weight that is
slightly less at the equator than at the poles. Compare the normal force of an 80 kg
person at the equator and at the north pole.
npole = _________
nequator = _________
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