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b/2
Radius (m)
N/T
Radius Error (m) N rotations
Time (s)
Frequency (Hz)
Trial 1
0.18
0.011
10
5.6
1.79
Trial 2
0.18
0.011
10
5.68
1.76
Trial 3
0.18
0.011
10
5.8
1.72
Trial 4
0.18
0.011
10
5.74
1.74
Trial 5
0.18
0.011
10
5.75
1.74
Error
Bob’s Mass
Hanger Mass
Total Hanging Mass
0.374 kg
0.0511 kg
0.7011 kg
0.187 kg
0.35055 kg
Avg. frequency (Hz)
1.75
Standard Deviation of frequency (Hz)
0.026457513
Tape (cm)
2.2
Uncertainty in tape:
1.1
Florida Institute of Technology
© 2023 by J. Gering
Experiment 7
Centripetal Force
Questions
If an object moves in a circle and the centripetal force “turns off” what happens? How are
centripetal and centrifugal forces related? How do you propagate errors through an equation that
has three terms? How do you compare two numbers when each has an error margin?
Concepts
Newton’s Third Law is the action-reaction law. Whenever something exerts a force on an object,
the object exerts a reaction force on that something. The reaction force points in the opposite
direction to the action force and it is equal in magnitude to the action force. Remember, the
action and reaction forces act on different objects. A rocket goes up because of the upward thrust
of the expanding exhaust gasses acting on the engine’s nozzle. The nozzle exerts a force on the
exhaust gasses that points down. The forces have the same magnitude but they act on different
objects. Newton’s Third Law is written as:
!
!
Fon A by B = −Fon B by A
(1)
Suppose you tie a rock to the end of a string and swing it in a horizontal circle over your head. A
horizontal circle ensures that gravity does not change the angular speed of the rock. If you
release the string, the rock flies out along a straight-line path, tangent to the circle. This suggests
that the rock’s velocity vector was tangent to the circle just before release.
A constant rate of rotation is called uniform circular motion UCM. Here, the object revolves in a
constant period of time, T. The length of the velocity vector is constant. The direction of the
velocity vector changes with time. Since velocity is a vector, any change in that vector (even in
direction) represents an acceleration. This acceleration is always directed toward the center of
the circle. It is called a centripetal acceleration, which means center seeking.
v
a
r
a
v
a
v
Figure 1. The velocity and acceleration vectors for uniform circular motion.
7 – 1
Florida Institute of Technology
© 2023 by J. Gering
From Newton’s Second Law, whenever a net force acts on an object it accelerates. Not
surprisingly, the force that produces the object’s centripetal acceleration is called the centripetal
force. In the example of the rock rotating over your head, the centripetal force acts on the rock
by the string. Physically, it is the tension in the string. Another example is when you drive a car
around a circular curve. The centripetal force is a center-directed frictional force that acts on the
car tires by the road. Normally, this force is large enough to keep the car in its lane of traffic. If
this force is not large enough (or if the car’s speed is too great) the radius must be larger to
compensate and the car slides outward into the other lane.
The term centrifugal, which means away from center is sometimes confusing. The standard
notion is that another force acts on the object moving in a circle and this force is directed
outward from center. This notion is NOT correct. The centrifugal force is actually the Newton’s
Third Law reaction force to the centripetal force. Recall you obtain a reaction force by switching
the labels “A” and “B” if you use the Fon A by B convention for describing a force F. In the two
examples mentioned previously:
EXAMPLE
rock tied to string
car rounding a turn
CENTRIPETAL
CENTRIFUGAL
Fon rock by string
Fon tire by road
Fon string by rock
Fon road by tire
As you can see, the centrifugal force does NOT act on the object of interest that is moving in a
circle (the rock and the tire). We can instead shift our focus to the string as the object moving in
a circle. Your hand holds the string and this is why you feel a pull outward. The force on your
hand by the string is directed outward and the force on the string by your hand is directed
inward. Once again, the object moving in a circle (now the string) experiences a centripetal
force and NOT a centrifugal force.
Another example is when you feel pressed against the interior of a car door when the car takes a
corner. Suppose you were traveling in a straight line. Newton’s First Law dictates you will tend
to continue in that path. However, the car turns and the car door intercepts your straight-line
path. If a force (directed toward the center of the rotation) acts on you, you will remain upright
in the car seat. This is the centripetal force provided by the car door.
Method
The centripetal force apparatus consists of a heavy metal bob connected by a spring to a central,
rotating shaft. See Fig. 2. The shaft is turned by hand. As the bob sweeps out a circular path, it
stretches the spring. Were it not for the spring, the bob would tend to fly off in a straight-line
tangent to the circle. Once stretched, the spring will supply a centripetal force, which is directed
toward the center of the circle.
7 – 2
Florida Institute of Technology
© 2023 by J. Gering
Rotating
Shaft
Counter
Weight
Strings
Spring
Rotating
“bob”
Pulley
Index
Figure 2. The Beck Centripetal Force Apparatus
If the bob moves at a uniform speed, v, in a circle of constant radius, r, this force will also be of
constant magnitude and equal to the mass of the object multiplied by its centripetal acceleration:
F=
mv 2
r
(4)
In this experiment, you calculate the centripetal force exerted by the spring on the bob. The
calculation is checked by directly measuring the spring’s tension.
Procedure
1)
Derive the equation you will use to calculate the centripetal force from the mass of the
bob, m, the number of revolutions counted, N, and the frequency of revolutions, f. Check
with your instructor to verify you have the correct equation before continuing. Include
this derivation in the data analysis section of your report.
2)
Adjust where the cross-bar is clamped to the rotating shaft so the un-stretched spring
pulls the bob about 2 – 3 cm away from vertical. Disconnect the spring and record the
radius of rotation from the index.
3)
Reconnect the spring and rotate the shaft fast enough so the bob continuously maintains
this constant radius of rotation. Continually exert a small torque on the top of the rotating
shaft to keep the bob’s radius as constant as possible.
4)
Examine the bob as it flies past the chosen index marker. Estimate the variation in the
radius of the bob’s circular path. If the bob swings back and forth through a small
distance of b centimeters, then the error in the radius is ± b/2. Record this error (in
meters) on your data sheet. Calculate the percent error in the radius.
7 – 3
Florida Institute of Technology
© 2023 by J. Gering
5)
Measure the time T it takes for the bob to make N complete revolutions (you decide on
N). Record N and T then calculate (in a spreadsheet) the frequency f . Make more
timings and calculate the average frequency and its sample standard deviation.
6)
The error in the radius is not easily controlled. More than likely, the percent error in the
radius will be around 2% or larger. You can control the percent error in the frequency
with additional timings. Therefore, make additional timings until the percent error in the
frequency is significantly less than the percent error in the radius. Significantly less can
be half since addition in quadrature shows
x 2 + (0.5x ) ≈ x
2
7)
Measure the bob’s mass and estimate an error in this value. Then, calculate the percent
error in the bob’s mass.
8)
Slide a paper clip into the small hole drilled in the bob. Tie a length of string to the paper
clip and run it over the pulley. Tie a loop in the other end of the string and hang a weight
hanger from it. Add mass to the hanger until the bob hangs vertically. Record the mass
and calculate the static equivalent centripetal force. Note: The angled weight hangers are
not exactly 50 grams, so measure their mass using a triple beam balance.
9)
Derive an expression for the propagated error in the centripetal force calculated from m, r
and f. See Appendix C. Check with your instructor to verify you have the correct
equation before continuing. Include this derivation in the data analysis section of your
report.
10)
In a spreadsheet, calculate the experimental value for the mass multiplied by the
acceleration, which is equal to the centripetal force. Also, use your error propagation
equation to find the percent error in this calculated centripetal force. Then, convert this
percent error into an absolute error.
11)
Calculate the difference (or discrepancy) d between your best calculated centripetal force
and the static measurement of the same quantity. Also calculate the propagated error in
this difference, σd (see Appendix C). Organize all calculations in a spreadsheet. Include
the spreadsheet in your report.
12)
Question: Do your two values agree within the limits of experimental error? If not,
explain the discrepancy. Hint: There is a systematic error overlooked in the theory. Cite
specific factors and analyze how a specific, overlooked error would affect the result.
7 – 4
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