Uniform Circular Motion
PES 115 Report
Objective
The purpose of this experiment is to determine the relationships between radiuses, mass,
velocity and centripetal force of a spinning body. We used logger pro to accurately
measure the orbital period of the spinning mass and used these measurements to
determine the interrelated interactions of the specified properties and viewed the results
graphically.
Data and Calculations

The black markings on the string are about 10 cm apart in length, measured from the
center of the spinning mass.
Part A: Factors that influence Circular Motion
Velocity versus Centripetal Force

Fill out the table holding the Spinning mass (M) and the radius (R) constant.
Figure 1: Experimental setup for the lab
Which Spinning Mass did you select _hook with foam wrapping_ (Tennis ball, etc..)
UCM - 1
What is the mass of the Spinning mass _0.0283_ kg.
What Radius did you select _0.30_ m (around 20 cm is a good choice).

Fill out the tables for five different hanging mass values.
Revolution Number and
Time per Revolution (T)
[sec]
Hanging Mass (m) [kg]
1
2
3
4
5
6
7
8
9
10
Average Time per Revolution
(T) [sec]
Calculate the Speed (v) [m/s]
0.1001 kg
0.1992 kg
0.2992 kg
0.4000 kg
0.4997 kg
0.61337 s
0.613087 s
0.613727 s
0.611319 s
0.618954 s
0.589000 s
0.572337 s
0.567165 s
0.553799 s
0.549274 s
0.413210 s
0.403737 s
0.393689 s
0.39364 s
0.388600 s
0.386300 s
0.393600 s
0.409900 s
0.417200 s
0.418100 s
0.367288 s
0.370600 s
0.374100 s
0.368047 s
0.365853 s
0.367541 s
0.375694 s
0.376971 s
0.372394 s
0.363807 s
0.316510 s
0.310189 s
0.316308 s
0.309619 s
0.300742 s
0.294794 s
0.297564 s
0.299070 s
0.307155 s
0.299924 s
0.271455 s
0.274200 s
0.273700 s
0.279400 s
0.282000 s
0.286041 s
0.286679 s
0.285880 s
0.288500 s
0.285612 s
0.590204 s
0.41083 s
0.370230 s
0.304845 s
0.281347 s
3.19 m/s
4.69 m/s
5.09 m/s
6.18 m/s
6.70 m/s
Show an Example Calculation Here
Determining the speed:
v
2R
T
v
2 0.3 m 
m
 3.19
0.590204 s
s
By re-examining the figure above, we can see that the forces are equal due to the
connection of the tensile forces though the string.
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F1  F2  centripetal force
F1  mg
m

F1  0.1001 kg   9.81 2   0.981981 N
s 

F1  F2  centripetal force  0.981981 N

Consolidate all your previous data in one nice and neat table.
Radius (m)

Spinning
mass (kg)
Hanging Mass
Centripetal
(kg)  x 9.8 Force (N)
Speed
v (m/s)
Speed2
v (m2/s2)
2
0.30 m
0.0283 kg
0.1001 kg
- 0.981981 N
3.19 m/s
10.1761 m2/s2
0.30 m
0.0283 kg
0.1992 kg
- 1.954152 N
4.69 m/s
21.9961 m2/s2
0.30 m
0.0283 kg
0.2992 kg
- 2.935152 N
5.09 m/s
25.9081 m2/s2
0.30 m
0.0283 kg
0.4000 kg
- 3.92400 N
6.18 m/s
38.1924 m2/s2
0.30 m
0.0283 kg
0.4997 kg
- 4.902057 N
6.70 m/s
44.8900 m2/s2
What can you conclude about the relationship between the speed and the centripetal
force? A graph would be very helpful here to see any trends.

People can usually recognize a linear relationship, we don’t often think in terms of
quadratic relationships. For that reason, we usually try to plot graphs in such a way
that the graph should appear linear. This may mean changing the form of what we are
plotting from lets say v to v2. From your data table, is it more likely that a straightline graph will be achieved by plotting Fcent against v or v2? Plot both on the same
graph and discuss their implications.
Using the table above, we shall plot the centripetal force versus the velocity and the
centripetal force versus the velocity-squared.
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Force versus Velocity
0
0
1
2
3
4
5
6
7
8
-1
Force [N]
-2
-3
-4
y = -0.1231x2 + 0.1041x
R2 = 0.982
-5
-6
Velocity [m/s]
Force versus Velocity^2
0
0
5
10
15
20
25
30
-1
35
40
45
50
y = -0.1054x
R2 = 0.9769
Force [N]
-2
-3
-4
-5
-6
Velocity^2 [m^2/s^2]
The graph of Force versus Velocity-squared produces the most linear result.
By analyzing both the graphs above, we can see that as velocity increases, so does the
centripetal force. This means that velocity and force are directly related. This is further
reinforced by the equation of centripetal force:
UCM - 4
 Mv 2
 rˆ 
F
R
Notice that if we plot Force on the y-axis and v-squared on the x-axis, this is effectively a
straight line:
y  mx  b
F  slope v 2   y  i ntercept 
compared to
The variance in the data is most likely due to a variable tilt on the system due to the
acceleration due to gravity.
Velocity versus Spinning Mass

Fill out the table for four different spinning mass values holding the hanging mass
(Centripetal force) and the radius constant.
Hanging Mass (m) [kg]
Revolution Number and
Time per Revolution (T)
[sec]
1
2
3
4
5
6
7
8
9
10
Average Time per Revolution
(T) [sec]
Calculate the Speed (v) [m/s]
0.0283 kg
0.0338 kg
0.0425 kg
0.0597 kg
0.335436s
0.345177 s
0.345790 s
0.341797 s
0.344900 s
0.345528 s
0.338092 s
0.341980 s
0.343600 s
0.340333 s
0.352764 s
0.356300 s
0.355410 s
0.351290 s
0.348822 s
0.346520 s
0.341358 s
0.334109 s
0.329102 s
0.320019 s
0.359099s
0.347119 s
0.345707 s
0.343074 s
0.351000 s
0.351700 s
0.353498 s
0.371901 s
0.327260 s
0.366474 s
0.418771 s
0.417716 s
0.414485 s
0.416000 s
0.420500 s
0.421500 s
0.413144 s
0.419256 s
0.408700 s
0.397500 s
0.342263 s
0.345570 s
0.356230 s
0.414757 s
5.51 m/s
5.45 m/s
5.29 m/s
4.54 m/s
Show an Example Calculation Here
Determining the speed:
v
2R
T
v
2 0.3 m 
m
 5.51
0.342263 s
s
UCM - 5

Consolidate all your previous data in one nice and neat table.
Hanging
Radius (m)
mass (kg)
Centripetal

Force (N)
Spinning
Mass (kg)
Speed
v (m/s)
Speed2
v (m2/s2)
1/Speed2
v-2 (s2/m2)
2
0.30 m
0.2989 kg
- 2.9332209 N
0.0283 kg
5.51 m/s
30.3601 m2/s2
0.032938 s2/m2
0.30 m
0.2989 kg
- 2.9332209 N
0.0338 kg
5.45 m/s
29.7025 m2/s2
0.033667 s2/m2
0.30 m
0.2989 kg
- 2.9332209 N
0.0425 kg
5.29 m/s
27.9841 m2/s2
0.035735 s2/m2
0.30 m
0.2989 kg
- 2.9332209 N
0.0597 kg
4.54 m/s
20.6116 m2/s2
0.048516 s2/m2

What can you conclude about the relationship between the speed and the size of the
spinning mass? A graph would be very helpful here to see any trends.
From your data table, is it more likely that a straight-line graph will be achieved by
plotting Spinning Mass against v or v2 or v-2? Plot all three graphs and discuss their
implications.
Spinning Mass versus Velocity
0.07
0.06
Spinning Mass [kg]

0.05
0.04
0.03
0.02
0.01
0
4
4.2
4.4
4.6
4.8
Velocity [m/s]
UCM - 6
5
5.2
5.4
5.6
Spinning Mass versus Velocity^2
0.07
Spinning Mass [kg]
0.06
0.05
0.04
0.03
0.02
0.01
0
16
18
20
22
24
26
28
30
32
Velocity^2 [m^2/s^2]
Spinning Mass versus 1/Velocity^2
0.06
Spinning Mass [kg]
0.05
y = 0.5112x + 0.0167
R2 = 0.9239
0.04
0.03
0.02
0.01
0
0
0.01
0.02
0.03
0.04
0.05
0.06
0.07
1/Velocity^2 [s^2/m^2]
The graph of Spinning Mass versus 1/Velocity-squared provides the most linear graph.
By analyzing the graphs above, we can see that as spinning mass increases, the velocity
decreases (or the inverse velocity increases). This means that the mass and velocity are
inversely related. This is further reinforced by the equation of centripetal force solved for
the spinning mass:
UCM - 7
M 
FR const ant

v2
v2
The variance in the data is most likely due to small changes in the radius during spinning
(i.e., the radius is not perfectly held constant during the experiment).
Velocity versus Radius

Fill out the table holding the hanging mass and the spinning mass constant and vary
the radius.
Spinning Mass: 0.0283 kg
Hanging Mass: 0.3029 kg
0.1 m
0.2 m
0.3 m
0.4 m
0.231275 s
0.228528 s
0.231670 s
0.229426 s
0.227274 s
0.229200 s
0.224000 s
0.229408 s
0.224927 s
0.230484 s
0.258021 s
0.251310 s
0.255169 s
0.259700 s
0.254200 s
0.250618 s
0.255110 s
0.255589 s
0.257383 s
0.256600 s
0.323926 s
0.318991 s
0.322382 s
0.317700 s
0316900 s
0.319419 s
0.325483 s
0.329034 s
0.329204 s
0.316460 s
0.427919 s
0.421983 s
0.424371 s
0.422900 s
0.416631 s
0.412697 s
0.409972 s
0.393100 s
0.387500 s
0.396598 s
0.228629 s
0.25504 s
0.321950 s
0.411367 s
2.75 m/s
4.92 m/s
5.85 m/s
6.11 m/s
Radius (R) [m]
Revolution Number and
Time per Revolution (T)
[sec]
1
2
3
4
5
6
7
8
9
10
Average Time per Revolution
(T) [sec]
Calculate the Speed (v) [m/s]
Show an Example Calculation Here
Determining the speed:
v

2R
T
v
2 0.1 m 
m
 2.75
0.228629 s
s
Consolidate all your previous data in one nice and neat table.
Spinning
Spinning
Hanging
Mass type
Mass (kg)
mass (kg)
Washers
0.0283 kg
0.3029 kg

Centripetal
Force (N)
-2.97 N
UCM - 8
Radius (m)
0.10 m
Speed
Speed2
v (m/s)
v2 (m2/s2)
2.75 m/s
7.5625 m2/s2
Washers
0.0283 kg
0.3029 kg
-2.97 N
0.20 m
4.92 m/s
24.2064 m2/s2
Washers
0.0283 kg
0.3029 kg
-2.97 N
0.30 m
5.85 m/s
34.2225 m2/s2
Washers
0.0283 kg
0.3029 kg
-2.97 N
0.40 m
6.11 m/s
37.3321 m2/s2

What can you conclude about the relationship between the speed and the radius of the
circle? A graph would be very helpful here to see any trends.
From your data table, is it more likely that a straight-line graph will be achieved by
plotting R against v or v2? Plot both graphs and discuss their implications.
Radius versus Velocity
0.45
0.4
y = 0.0094x2 + 0.0012x
R2 = 0.9132
0.35
Radius [m]

0.3
0.25
0.2
0.15
0.1
0.05
0
0
1
2
3
4
Velocity [m/s]
UCM - 9
5
6
7
Radius versus Velocity^2
0.45
0.4
y = 0.0096x
R2 = 0.9131
Radius [m]
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
0
5
10
15
20
25
30
35
40
Velocity^2 [m^2/s^2]
The graph of Radius versus Velocity-squared provides the most linear graph.
By analyzing the graphs above, we can see that as radius increases, the velocity also
increases. This means that the radius and velocity are directly related. This is further
reinforced by the equation of centripetal force solved for the spinning mass:
R
 
Mv 2
 consta nt v 2
F
Part B: Experiment vs. Theory

Pick one set of data from the pervious section. Fill out the table below. Explain
why you pick one set over the other two.

Calculate the expected velocity according to Newton’s laws based on m, M and R
(Equation 1).

Compare this theoretical velocity to the one you measured.

Perform a %difference calculation. Explain the discrepancy and any sources that
would have caused them.
UCM - 10
Hanging
mass (kg)
Spinning
Mass (kg)
Radius (m)
Calculated v
(m/s)
Measured
v (m/s)
% Diff.
0.3029 kg
0.0283 kg
0.10 m
3.24 m/s
2.75 m/s
15%
0.3029 kg
0.0283 kg
0.20 m
4.58 m/s
4.92 m/s
7.4%
0.3029 kg
0.0283 kg
0.30 m
5.61 m/s
5.85 m/s
4.3%
0.3029 kg
0.0283 kg
0.40 m
6.48 m/s
6.11 m/s
5.7%
Show at least one example of each type of calculation.
v
mgR

M
0.3029 kg  9.81 m2 0.10 m 
s 

0.0283 kg 
v  3.24
m
s
Results and Questions

Explain any discrepancy and the sources that would have caused them.

Does the theory hold-up (Part B)?

What effect would friction between the tubing and string have on this lab?

In the discussion of the background physics we ignored the force due to
gravity on the spinning mass (M). Discuss how would this affect our results if
accounted for this force.

Name three common physical situations where you would feel a centripetal
force.

Name three common physical situations where you would feel a centrifugal
force.
UCM - 11
Conclusion

This closing paragraph is where it is appropriate to conclude and express your
opinions about the results of the experiment and all its parts. Only the final
result(s) needs to be restated.
You are intelligent scientists. Follow the guidelines provided and write an appropriate
conclusion section based on your results and deductive reasoning. See if you can think of
any possible causes of error.
** NOTE: There are several components of error which could significantly modify the
results of this experiment. Some of these are listed below:
 Ignoring acceleration due to gravity on the spinning mass
 Tilt on the measured object
 Drag and air resistance
 Variable lengths of the radius
 Measurement of the markings on the string
 Snagging and catching
 Calibration
 Sensor limitation parameters
 Computer processor speed and reading registration
 Sensor Alignment
 Other …
A few of the potential errors listed above may be applicable to YOUR experiment
UCM - 12