Free Fall Lab
(determination of "g")
Purpose: To examine and determine if the following statement is valid: "In the absence of all other external forces except gravity (g), a falling body will accelerate at a constant 9.81 m/s^2. Also, this lab is intended to introduce the usage and usefulness of numerical processing software (in this case, Microsoft Excel).
Experiment Equipment: In order to perform this experiment, we will utilize a Spark Generator (which sparks every 1/60th of a second) held in place by an Electromagnet which is attached to the top of a Sturdy Column. This set-up creates a 1.5 m falling distance (with the total height of the assembled apparatus being 1.86 m). In place along the path of free-fall, is a strip of Spark-sensitive Tape. The assembled apparatus includes a heavy tripod base with screws for leveling purposes, a free-fall body (object), a weighted clip to anchor the spark paper, as well as a power supply powering the spark generator.
Procedure: Steps listed below
1. Turn the dial hooked up to the electromagnet up slightly
2. Hang the wooden cylinder with the metal ring around it on the electromagnet
3. Turn on the power of the sparker thingamugug.
4. Hold down the spark button on the sparker box. (This begins to spark at 60 Hz.)
5. Turn the electromagnet off so that the object descends in free-fall.
6. Turn off power to the sparker thingamugug.
7. Tear off the spark-sensitive tape; replace with new strip for next student.
The spark burns on the tape indicate the position of the falling mass in intervals of 1/60th s. This strip of tape essentially reveals the relationship between position vs. time of the free-fall object (with gravity being the ONLY force acting on it).
Data Collection: In order to gather and interpret this data more clearly, we placed the spark-sensitive tape on the top of our desk and secured it in place (fully extended) with the aid of some adhesive tape. We then acquired a 2 m stick in order to measure the displacement of the falling mass over continuous intervals of 1/60th s. We recorded the 19 values depicted below.
(recorded value of displacement)
Organizing Collected Data into Excel: We began with a new, empty spreadsheet. In the first column (A), we entered in the time interval of 1/60th s into 19 rows. In the following column (B) we entered our recorded displacement values at each corresponding intervals of time. In column (C) we utilized the formula capabilities of Excel to calculate change in position of the mass for each interval of time. In column (D) we then calculated the mid-interval time (that is, the half-way point between each interval). In turn, we used the same tool once again in column (E) to determine the speed of the free-falling mass at every mid-interval point in time.
(spreadsheet generated using recorded data from spark-sensitive tape)
Once all the data was organized neatly via Excel, we then selected the last two columns (D & E) and used Excel to visually graph the data in a Speed vs. Time Graph. We also used the same method in order to graph a Position vs. Time Graph using the values of distance and time from the spreadsheet. Both graphs have their corresponding equations displayed as well, these are useful in calculating our recorded value of "g". The graphs are depicted below.
(Speed vs. Time Graph)
(Position vs. Time Graph)
Questions/Analysis:
1. Show that, for constant acceleration, the velocity in the middle of a time interval is the same as the average velocity for that time interval
- To demonstrate this, we will use the time interval of: to = 0.1000 s and tf = 0.1167 s.
We know that the average change in velocity is equal to it's change in position divided by the change in time during it's position change. So:
Vavg = (xf - xo) / (tf - to)
= (12.40 cm - 9.90 cm) / (0.1167 s - 0.1000 s)
= 150 cm/s
If we look at the graph of speed vs. time, and examine the speed of the object at the mid-interval at time 0.1083 s, we see that the instantaneous velocity is 150 cm/s, which agrees with our average velocity calculation.
2. Describe how you can get the acceleration due to gravity from your velocity/time graph. Compare result with the accepted value of g.
- The graph displays the corresponding equation of the line. This equation is velocity at any given time, or V(t). We know that acceleration is the change in velocity over the change in time. So,
V(t) = 950.53t + 50.658
V(0.3000) = 335.8 cm/s
V(0.0167) = 66.53 cm/s
a = (Vf - Vo) / (tf - to)
a = 950 cm/s^2 = 9.50 m/s^2
Our calculated value (utilizing the information provided from the graph above) of g seems to be off by 0.3 m/s^2.
3. Describe how you can get the acceleration due to gravity from y our position/time graph. Compare result with accepted value of g.
- We can derive an equation for the acceleration of the free-fall body at any given time by taking the second derivative of the equation displayed in the position vs. time graph.
f(x) = 472.88x^2 + 50.937x + 0.0373
f ' (x) = 945.76x + 50.937
f ''(x) = 945.76
The second derivative of the position function shows that no matter the time, the acceleration is constant at 945.76 cm/s^2, or 9.50 m/s^2.
Experimental Uncertainty: In order to obtain an actual value for the uncertainty in this experiment(which tells the reader about the instruments used to makes measurements), we gathered the entire class' calculated values of g and inputted them into a new Excel spreadsheet. What we did next, was use Excel to determine the Standard Deviation of the Mean of the class' values of g. This spreadsheet is depicted below.
2. Describe how you can get the acceleration due to gravity from your velocity/time graph. Compare result with the accepted value of g.
- The graph displays the corresponding equation of the line. This equation is velocity at any given time, or V(t). We know that acceleration is the change in velocity over the change in time. So,
V(t) = 950.53t + 50.658
V(0.3000) = 335.8 cm/s
V(0.0167) = 66.53 cm/s
a = (Vf - Vo) / (tf - to)
a = 950 cm/s^2 = 9.50 m/s^2
Our calculated value (utilizing the information provided from the graph above) of g seems to be off by 0.3 m/s^2.
3. Describe how you can get the acceleration due to gravity from y our position/time graph. Compare result with accepted value of g.
- We can derive an equation for the acceleration of the free-fall body at any given time by taking the second derivative of the equation displayed in the position vs. time graph.
f(x) = 472.88x^2 + 50.937x + 0.0373
f ' (x) = 945.76x + 50.937
f ''(x) = 945.76
The second derivative of the position function shows that no matter the time, the acceleration is constant at 945.76 cm/s^2, or 9.50 m/s^2.
Experimental Uncertainty: In order to obtain an actual value for the uncertainty in this experiment(which tells the reader about the instruments used to makes measurements), we gathered the entire class' calculated values of g and inputted them into a new Excel spreadsheet. What we did next, was use Excel to determine the Standard Deviation of the Mean of the class' values of g. This spreadsheet is depicted below.
(Spreadsheet used to determine standard deviation from accepted value of g)
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