Post on 07-Apr-2015
Investigating Centripetal Force:
The Relationship between Frequency vs. Force
Data:
# of Washers Trial 1 Period
(s/10 cycle)
Trial 2 Period
(s/10 cycle)
Trial 3 Period
(s/10 cycle)
5 6.63 6.63 6.51
6 6.08 6.44 6.26
7 5.42 6.11 5.59
8 6.05 5.27 4.90
9 5.09 5.11 5.05
10 4.99 5.17 4.81
11 4.81 4.35 4.69
12 4.26 4.38 4.39
13 4.30 4.75 4.51
14 4.42 4.17 3.78
# of Washers Average Period (s/10 cycle)
5 6.59
6 6.26
7 5.71
8 5.41
9 5.08
10 4.99
11 4.62
12 4.34
13 4.52
14 4.12
Data Process:
I averaged each of my data sets for number of washers from 5-14. And for
my uncertainties, I found the range between the max and min of each data set
and divided it by 2.
6.63+6.63+6.44 = 6.59
3
(6.63-6.44)/2 ≈ 0.093
6.59±0.093 ≈ 6.59±0.01
Then I found the frequency by inversing my period. For my uncertainties
for frequency, I used the same amount of uncertainties as I did for my period.
T = cycle/s
f = s/cycle = T-1
Fin= (m4π2r) (1/T2)
Fin= (m4π 2r) f2
Force (# of Washers) Average Frequency (10 cycles/s) ±0.01
5 0.15
6 0.16
7 0.18
8 0.18
9 0.20
10 0.20
11 0.22
12 0.23
13 0.22
14 0.24
In order to straighten the graph I square rooted my force.
Fin= (m4π 2r) f2
Fin1/2= (m4π 2r) ½ f
(Fin/ m4π 2r) ½=f
Force1/2 (# of Washers) Average Frequency (10 cycles/s)
√5 0.15
√6 0.16
√7 0.18
√8 0.18
√9 0.20
√10 0.20
√11 0.22
√12 0.23
√13 0.22
√14 0.24
Conclusion:
In this lab, I was looking for the relationship of the inward force of an
object (rubber stopper) as it was spinning for a limited number of revolutions. I did
this by making the number of washers (each washer being 7 grams) my
independent variable and keeping the radius of the string, mass of stopper, mass of
washers, and number of revolution as my controlled variables.
My data shows me that the inward force of the rubber stopper is directly
proportional to the f2. In order to straighten my data on the graph so to figure out
what my slope represented, I square rooted both F and f2. For my slope I got
0.05860. My slope represents my controlled variables of this specific experiment
such as the radius and mass of the stopper. This slope can vary widely depending
on what the radius and mass of the stopper is.
My graph does fit through the error bars and it theoretically should go
through the origin because if there is no inward force, there would not be
frequency. But my graph shows that it does not go through the origin. I believe my
range of data was sufficient enough to conclude that the lighter inward force an
object has as it is spinning, the lower its frequency will be. And as the inward force
increases, the frequency will also increase as well but as the inward force reaches
infinity, the slope of frequency will reach 0 making frequency be a constant.
Evaluation:
I do not believe that my data has systematic error. My data does go through
all my error bars even though at √13 it barely goes through. My one major
uncertainty in this lab was at √13