7-Sep-2016: Free Fall Lab

PURPOSE:

"To examine the validity of this statement:
In the absence of all other external forces except gravity, a falling body will accelerate at 9.81 m/s/s."

In other words, we will try to determine the value of g while simultaneously learning about Excel and some statistics for analyzing data.

INRODUCTION:

The goal of this lab was to derive a value for the acceleration due to gravity on planet Earth. Using the Spark Generator Free Fall Apparatus 9000, our group was able to directly measure and record the displacement of an object in free fall. With this information -- and a few other calculated variables -- our group made a spreadsheet on Excel and plotted a position vs. time graph, as well as a mid-interval speed vs. time graph.

APPARATUS:

This is the Spark Generator Free Fall Apparatus 9000, with corresponding electromagnet and variable power supply. Its main purpose is to provide students with a permanent record of a free falling body. The free falling body (blue thing) is held in place at the top by an electromagnet. When released, a spark generator precisely records the fall of this object onto a thin strip of spark-sensitive paper, which can be used to create a position vs. time graph and a velocity vs. time graph to calculate acceleration.

While it is always a wise convention to know how to use your equipment before doing lab, today I will make an exception in the interest of time. In my particular class, our professor decided to just hand us the strips of "pre-sparked" paper. For more information on how to use this device, refer to your lab module.

Adult supervision required

PROCEDURE:

Equipment needed: Spark tape from teacher; computer with Microsoft Excel; meter-stick

1) Lay your spark tape on a long, flat surface (such as a table). Notice the dots and their relative distances. Make sure that the distance between the dots becomes larger from left to right. These dots correspond to the position of the falling mass every 1/60th of a second:

spark tape with series of dots

2) Place a meter stick next to the tape. Line up the 0-cm mark with the first dot and record the position of each dot from the 0-cm mark. In other words, record the displacement of each dot.

3) Create an Excel Spreadsheet with the following columns: time, distance, delta x, mid-interval time,  and mid-interval speed. Like so:

The time column is in increments of 1/60th of a second. The distance column shows the relative displacement of each dot from the origin -- the 0-cm mark. Although unmarked, it should be noted that each distance has an uncertainty of +/- 0.1 cm. The delta x column shows the distance between two consecutive dots on the spark tape. For example, the value in cell C2 is obtained by =(B3-B2). The mid-interval time column gives the time for the middle of the each 1/60th s interval. The mid-interval speed is essentially (delta x)/time, where time = 1/60th of a second.

4) Create a mid-interval speed vs. time graph using your data from columns D and E. Make sure to do a linear fit afterwards to obtain an equation with a correlation value of R². 

5) Create a position vs. time graph using columns A and B. Do a polynomial fit of order 2 in order to obtain a polynomial equation with a correlation value of R² as well.

Here is a picture of our graphs:

Distance v. Time (left) and Mid-interval speed v. Time (right)

QUESTIONS:

1) Consider an object in free fall with an acceleration constant, a = 10 m/s/s. The object was initially at rest and fell a certain distance for three seconds. If I arbitrarily choose my interval to be three seconds, then my mid-interval would be defined as:

V_mid-interval, @t_1.5 = V₀ + at 

= 0 m/s + (10 m/s/s)(1.5 s) = 15 m/s

If I wanted to know the average velocity for my interval, then I would simply use the equation: 

 V_avg = (V₀ +V_f) / 2

Where

V₀, @t₀ = 0 m/s,

V_f, @t₃ = 30 m/s,

Plugging everything in, we find:

V_avg = (0 m/s + 30 m/s) / 2 = 15 m/s

Therefore,

V_avg = 15 m/s = V_mid-interval

2) The slope of the velocity vs. time graph for the free falling body yields the acceleration due to gravity. This means that our graph gave us a value for gravity, g ≈ 9.54 m/s/s. This corresponds to roughly a 3% error compared to the accepted value of 9.81 m/s/s.

3) Since the equation of the position vs. time graph is a polynomial of the form y = ax² + bx + c, we can assume that it corresponds to the kinematic equation: Δy = V_0yt + (1/2)gt². Therefore, we can conclude that:

(1/2)g = 4.7591 m/s/s

g = 9.5182 m/s/s

This means that our graph gave us a value for gravity, g ≈ 9.52 m/s/s. This corresponds to roughly a 3% error compared to the accepted value of 9.81 m/s/s.

CONCLUSION: 

After carefully analyzing our results, I think it is safe to say that our results for the theoretical value of g were fairly accurate. This is because both of our derived values of g were within 3% of the accepted value of 9.8 m/s/s. However, there is no doubt that there are bound to be sources of uncertainty inherent to even our most precise calculations. One possible source of error in our calculations was the variation in significant figures between the distances we measured. For example, in cells B2 through B7, we only used one to two significant figures, while in cells B8 through B11, we used three significant figures. Furthermore, it is entirely possible that we could have misread numbers in our measurements of distance, leading to inaccuracies in our derivation of g. 

GROUP MEMBERS: Matthew I., Xavier L., Billy J.