9/2/2014 Free Fall Lab: determination of the gravity on Earth.

9/2/2014 Free Fall Lab: determination of the gravity on Earth.

Introduction: Gravity is everywhere, and physics has so much to do with it. We have always been told that gravity accelerate objects at 9.8 m/s^2 when they are falling since high school. We have done an experiment in class to examine this statement.

Apparatus:

We used the free fall apparatus in class to get precise records of motion of the falling object. The falling motion was precisely recorded by a spark generator on a spark- sensitive tape. The spark was generated at 60 Hz, which left a dot on the tape for every 1/60 second of the falling motion. we eventually got series of dots that corresponded the position of the falling object for the each entire falling motion. To be more precise in this lab, we dropped the mass for many times, and each team in class analyzed one of the different tapes.









Thesis:
We assumed the value of the acceleration gravity gives an object while falling down is 9.8 m/s^2 toward the center of the earth (toward the floor in this experiment.)
We measured and record the distance from each dot on the tape to another, and put the set of value under a column named X into Microsoft Excel. In Excel, we apply the equation:  ∆X= X-Xprevious to another column ∆X. We then got the another series of value, velocity, by applying equation: V = ∆X/∆t, where ∆t was 1/60 second. Finally, we got the series of value, acceleration, by applying a=∆V/∆t in Excel, and the average of the series of acceleration would be our experimental gravitational acceleration. Hopefully, the experimental gravitational acceleration should be close to our theoretical gravitational acceleration, 9.8 m/s^2.

Data and Analysis:
Here is the table that we have analyzed in Excel in class:

After trying to graph t vs. v with all the data we have, we found that our data is off by too much from cell F19 to F23, so we decided to only use the part of data that seems to be more reasonable, which is the data selected in the above picture, to graph:
 v(cm/s) vs. t(sec) (reasonable data)

In our data, the average value of acceleration is the average value of all the data include those we decided not to use, so we used the linear fit feature to get the slope of the best fit of the graph, which will still equal the average acceleration of our falling object.

The slope of  v(cm/s) vs. t(sec) is the average acceleration because we know that derivative of velocity function is acceleration:
v'(t)=a(t)

After acquired the gravitational acceleration of our sample falling object, we compare our value with the value other groups in the class got:
Comparison of all the gravitational acceleration for standard deviation from 10 groups in the class, and ours is number 2. 

The standard deviation for 10 different values our class got is 10.74

Conclusion:
Discussing assumptions you made in doing the lab
From the actual data we found, we ignored the data from cells F19 to F23 because we assume the acceleration is a constant value 9.8 m/s^2 such that the pattern in our data is that value of v was increasing as time pasts, and data from cells F19 to F23 was jumping up and down. 
Discussing the effect of experimental uncertainty
Since the ruler we used to measure the distance traveled by the falling object has precision up to 10th decimal place of a centimeter, all the length we got have uncertainty of                + or - 0.05 cm.
Differences between expected and experimental values
All the data and calculation was collected and calculated in centimeter, so the average gravitational acceleration according to our class was actually  9.51 m/s^2. According to standard deviation theory, 68% of 10 groups got the gravitational acceleration, g = 9.51 m/s^2 + or - 0.11 m/s^2. That is, 7 groups out of 10 groups have their value falls in one standard deviation from the mean value.
According to the statement: "In the absence of all other external forces except gravity, a falling body will accelerate at 9.8 m/s^2," our experimental value has % error equals to         (9.51-9.8)/9.8*100% = -3.0 %.