Evan+and+Aaron

9/27/10

What is the acceleration of a falling body?

The purpose of this lab is to find the acceleration of a body of mass that is falling.
 * Purpose**

If the object attached to the end of the ticker tape is dropped and is in free fall, then its acceleration will be 9.8 m/s 2. why do you think so?
 * Hypothesis**

1. Take a clamp and clamp a ticker tape timer onto a cabinet door. 2. Place a long piece of ticker tape through the timer and tape a mass onto the front end of the tape. 3. Turn on the ticker tape timer and allow the mass to drop down to the floor while guiding the tape through the timer. 4. Tape the ticker tape onto the table and use a meter-stick to measure the distances of each dot from the beginning of the tape. 5. Put the data into Excel and make a graph.
 * Procedure**

__Our Data Collected__
 * Data/Graphs**
 * Dot # || Time (s) || position Length (cm) ||
 * 0 || 0.0000 || 0 ||
 * 1 || 0.0167 || 0.5 ||
 * 2 || 0.0333 || 1.04 ||
 * 3 || 0.0500 || 1.66 ||
 * 4 || 0.0667 || 2.55 ||
 * 5 || 0.0833 || 3.67 ||
 * 6 || 0.1000 || 5.4 ||
 * 7 || 0.1167 || 7.11 ||
 * 8 || 0.1333 || 9.03 ||
 * 9 || 0.1500 || 11.29 ||
 * 10 || 0.1667 || 13.65 ||
 * 11 || 0.1833 || 16.5 ||
 * 12 || 0.2000 || 19.34 ||
 * 13 || 0.2167 || 22.68 ||
 * 14 || 0.2333 || 26.02 ||
 * 15 || 0.2500 || 29.98 ||
 * 16 || 0.2667 || 34.05 ||
 * 17 || 0.2833 || 38.27 ||
 * 18 || 0.3000 || 43.16 ||
 * 19 || 0.3167 || 48.01 ||
 * 20 || 0.3333 || 53.23 ||
 * 21 || 0.3500 || 58.77 ||
 * 22 || 0.3667 || 64.55 ||
 * 23 || 0.3833 || 70.6 ||
 * 24 || 0.4000 || 76.69 ||
 * 25 || 0.4167 || 83.45 ||
 * 26 || 0.4333 || 90.14 ||
 * 27 || 0.4500 || 97.04 ||
 * 28 || 0.4667 || 104.57 ||
 * 29 || 0.4833 || 112.28 ||
 * 30 || 0.5000 || 120.05 ||

great graph!

__Class Averages for A, B, and G values__ (**Bold values** = our calculated values)
 * **Group** || **Coefficient A** || **Coefficient B** || **Experimental Value of g** ||
 * 1 || 432.82 || 5.77 || 865.64 ||
 * 2 || 389.94 || 1.81 || 779.88 ||
 * 3 || 434.47 || 4.68 || 868.94 ||
 * 4 || 375.28 || 13.46 || 750.56 ||
 * 5 || 446.4 || -33.1 || 892.8 ||
 * 6 || 405.01 || 0.75 || 810 ||
 * 7 || 473.43 || 41.22 || 947.86 ||
 * 8 || 421.08 || 7.2 || 842.16 ||
 * The Real 4 || 433.97 || 53.41 || 867.94 ||
 * 9 || 548.45 || 22.97 || 1096.9 ||
 * **6** || **475.74** || **1.78** || **951.48** ||
 * 10 || 252.48 || 9.39 || 504.96 ||
 * 11 || 426.66 || 37.06 || 853.32 ||

__Our Calculation__ y= Ax 2 + Bx y = 475.74x 2 + 1.7773x y = 1.7773x + 475.74x 2 d = Bt + At 2 d = Vi(t) + (1/2)at 2 A = 1/2a 475.74 = 1/2a 951.48 cm/s 2 = a
 * Calculations**

fantastic results!

__ Sample Calculation (Ideal Calculation) __ A = 1/2a 490.50 = 1/2a
 * 981.00 cm/s ** 2 ** = a **

__ ** Percent Error ** __ % Error = [(981 - 951.48) / 981] x 100 % Error = [29.52 / 981] x 100 % Error = 3.01%


 * Discussion Questions**

The shape of the graph reflects the expected graph. The model equation is a quadratic, so the graph would presumably be a parabola. Sure enough, the graph that is showing is one half of a parabola because the values increase exponentially.
 * 1. Does the shape of your graph agree with the expected graph? Why or why not?**

Our results tend to be much better than that of the class average. When calculated, our percent error was 3.01%. The class average was 20.10%. Our results were a lot more accurate than the class'. It would be good to do percent difference to compare
 * 2. How do your results compare to that of the class? (Use Percent difference to discuss quantitatively.)**

The object did accelerate uniformly because the acceleration is increasing exponentially, and our graph shows this. As the time increases, the distance between each pair of dots increases because acceleration is increasing exponentially.
 * 3. Did the object accelerate uniformly? How do you know?**

The velocity-time graph of this object should look like a straight line that has a positive slope, and it should start at zero since the mass starts at rest and its initial velocity is zero. slope =?
 * 4. What should the velocity-time graph of this object look like?**

y = 951.48x + 1.7773
 * 5. Write down the expected equation of the line from this v-t graph (use specific information from your x-t graph).**

﻿ ﻿ ﻿If someone was pulling the mass down or pushing the object down while it were falling, the acceleration would be higher than it should be. If the ticker tape was not guided through the timer, it could rub against the top of the cabinet door which would cause friction and lower the acceleration.
 * 6. What factor(s) would cause acceleration due to gravity to be higher than it should be? Lower than it should be?**

From this lab, we discovered that the experimental value of our object's acceleration is 951.48 cm/s 2. This tells us that the object in question was not in a free-fall during the experiment; otherwise the acceleration value would have been 981 cm/s 2. Our percent error was only 3.01%, so the answer was not very far off, especially compared to the rest of the class. A reason for this could be because of other forces acting against gravity, which is pulling the object down. One of these could be friction, which is created by the ticker tape rubbing against either the cabinet or through the machine.
 * Conclusion**