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//What is Gravity?// Finding Acceleration Due to Gravity

Andrew Miller, Ryan Listro

Due: 28/9/10

Purpose: In this experiment, we are trying to find the acceleration of a free falling object due to gravity and if any other forces besides gravity are acting on that object.

Hypothesis: When we find the distance that the free falling object traveled using spark time and spark tape, we can use the equation of the line to find the velocity and with that, find the acceleration using the derivatives (slope of the tangent line). The derivative of position is the velocity, and the derivative of velocity is acceleration.

Materials: Spark timer/ tape, mass, ruler, Microsoft Excel, Clip, Masking tape

Procedure:


 * 1) Mount spark timer on cabinet with clip.
 * 2) Thread spark tape through spark timer, making sure the bright side is facing up.
 * 3) Tape the weight to the spark tape.
 * 4) Set spark timer to 60 Hertz, turn on spark timer, and release weight.
 * 5) Take spark tape out (ignore first part) and mark the “0” dot.
 * 6) Measure each dot (six dots = 0.10 seconds) from 0.
 * 7) Put measurements into table, and graph with Excel.
 * 8) Find line of best fit (use polynomial line) and find R2 value.

Data:

Table of Time vs. Distance Values:


 * Dot # || Time (s) || Distance (cm) ||
 * 0 || 0.000 || 0 ||
 * 1 || 0.017 || 0.45 ||
 * 2 || 0.033 || 1.35 ||
 * 3 || 0.050 || 2.81 ||
 * 4 || 0.067 || 4.35 ||
 * 5 || 0.083 || 6.40 ||
 * 6 || 0.100 || 8.60 ||
 * 7 || 0.117 || 10.81 ||
 * 8 || 0.133 || 13.49 ||
 * 9 || 0.150 || 16.50 ||
 * 10 || 0.167 || 19.60 ||
 * 11 || 0.183 || 23.10 ||
 * 12 || 0.200 || 26.91 ||
 * 13 || 0.217 || 31.20 ||
 * 14 || 0.233 || 35.43 ||
 * 15 || 0.250 || 40.25 ||
 * 16 || 0.267 || 44.90 ||
 * 17 || 0.283 || 50.20 ||
 * 18 || 0.300 || 55.60 ||
 * 19 || 0.317 || 60.96 ||
 * 20 || 0.333 || 66.81 ||
 * 21 || 0.350 || 72.85 ||
 * 22 || 0.367 || 79.05 ||
 * 23 || 0.383 || 85.20 ||
 * 24 || 0.400 || 92.40 ||
 * 25 || 0.417 || 99.41 ||
 * 26 || 0.433 || 106.55 ||
 * 27 || 0.450 || 114.35 ||
 * 28 || 0.467 || 121.50 ||
 * 29 || 0.483 || 130.35 ||
 * 30 || 0.500 || 139.00 ||





Calculations to find Acceleration due to Gravity:

Using the derivative:

s(x) = 473.43x^2 + 41.222x (Position)

v(x) = 947.86x + 41.222 (Velocity)

a(x) = 947.86 cm/s^2 (Acceleration)

Using the equation encompassing distance, initial velocity, acceleration, and time:

d = v1t + .5at2

d = Bt + At2

B = v1

B = 41.222 cm/s^2

A = .5a

473.43 = .5a

a = 947.86 cm/s^2

Percent Error:

Theoretical Value of G: 9.81 m/s2 = 981 cm/s^2

(abs(Experimental – Theoretical)/ Theoretical) X 100

(abs(947.86 – 981)/981) X 100 = 3.38% error

Discussion Questions

1. Does the shape of your graph agree with the expected graph? Why or why not? Yes, our graph agrees with the shape of the expected graph. Our graph starts at 0 and has a very small slope in the beginning. As time increases, the slope of graph increases and the object's speed increases to that of gravity. The curve of our graph displays an increase in both distance and velocity; therefore, we can then find the acceleration of our free-falling object.

2. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.) In order to compare our results to that of the class, percent difference needs to be calculated. The equation would be the difference between our experimental value (947.86) and the class average (848.65). That difference would then be divided by the class average and then multiplied by 100. Compared to the class, our value was on the higher side, but it was closer to the theoretical value of the acceleration of gravity. After calculations, the percent difference between our data and the class's was 11.69%.

3. Did the object accelerate uniformly? How do you know?

Uniform acceleration is when the velocity of an object changes equal amounts in equal time periods. In this lab, the free falling object did accelerate uniformly--in fact, all free falling objects accelerate uniformly because the only force that is acting on them is gravity (friction in the air is ignored). The evidence of this for our lab is on our graph, for the distance is increasing exponentially over a period of time, meaning the velocity is also increasing in the same way over the same period of time.

4. What should the velocity-time graph of this object look like? The velocity-time graph of this object should have a diagonal line with a positive slope. This line should also have a starting point at (0,0).

5. Write down the expected equation of the line from this v-t graph (use specific information from your x-t graph).

The equation for our position graph is: s(x) = 473.43x^2 + 41.222x, meaning that the equation for velocity is :

v(x) =947.86x + 41.222 (values are in cm)

6. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be? Acceleration would be higher than it should be if a force (i.e. wind or human interaction) acted on the free-falling weight and pushed it faster than it was dropping. An example would have been if instead of letting go of the spark tape, one pulled the tape down even slightly. Acceleration would be lower than it should be if something was keeping the free falling weight from falling on its own and due to gravity. An example of this would be if someone held the ticker tape or if he let it run through his fingers.

Conclusion:

In this experimented, my partner and I found an approximate value of the acceleration due to gravity of a freefalling object (a weight). Theoretically, our hypothesis to find the position graph and then take two derivatives of it to obtain the acceleration was correct; however, that is not the only way to find the acceleration. We used the equation d = vit + 1/2at2 as a basis, and substituted the results we obtained on the Excel graph. We replaced the “B” value in Bx of our line with initial velocity, and then replaced the “A” in our At2 for the equation of the line with 0.5a. By setting our A value equal to 0.5a, we found that the acceleration from the lab matched our theoretical data that we found by taking the derivative of the position graph twice. We had a minor percent error our value of g was less than the accepted value. This error probably came from the spark timer, for threading a piece of tape through it probably slows the weight (freefalling object) down slightly, meaning that it would take a little longer for the object to reach the ground, which explains why we had a lower value of g. This is simply human error—there really is not a way to completely fix this problem. If we used thinner tape, we could help the problem, but not completely solve it. For our “B” (initial velocity) value, we believe that it was much higher than normal because we did not start with the first dots on our spark tape; rather, we started from the third set of dots. Because of this, our initial velocity was not 0, for the tape had increased its velocity to a number greater than 0 at the third dot point. In order to fix this error, we could have been more delicate when threading the spark tape—we waited about half a second before releasing the weight, making the first part of the tape incredibly difficult to interpret. By knowing the acceleration due to gravity of a freefalling object, we can figure out its velocity and theoretically how long an object would take to reach the ground. Newton discovered gravity, and the way that G fits into his equation is that it is a constant—no matter the objects used, gravity will act on them the same, despite their mass.