Phy 231
Your 'ball down ramp' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Your general comment **
** Will a steeper ramp give greater or lesser time? **
the steepest ramp will require the least amount of time for the ball to make it to the end because it will be rolling faster than the other two.
** As slopes increases will time intervals increase, decrease or show no pattern? **
i would expect the time intervals to be decreasing as you went from least to greatest slope.
** Your report of 5 trials each way for 1 domino **
1.828
1.875
1.672
1.734
1.750
1.453
1.500
1.359
1.422
1.484
the first 5 lines represent the 5 trials in one direction and the last 5 represent the ball rolling down in the opposite direction using the same parameters. as you can see i have a slightly unlevel table and the ball has more of a slope when facing in the opposite direction.
** Your report of 5 trials each way for 2 dominoes **
1.234
1.156
1.109
1.172
1.094
1.031
1.000
0.984
1.000
1.000
the first 5 is one direction and the last 5 is the ball rolling the opposite way showing a slope in my table top
** Your report of 5 trials each way for 3 dominoes **
0.891
0.906
0.922
0.891
0.859
0.844
0.781
0.859
0.875
0.750
the first 5 is the original direction and the last 5 is the ball rolling the opposite directiono showing a slope in my table top
** Do your results support or fail to support your hypothesis about increasing or decreasing times? **
no they agree that the bigger the slope the faster the time will be for the ball to reach the bottom.
** How do you think ave velocity is related to slope? **
the average velocity is related to the slope of the ramp by the greater the slope the greater the vAve while the smaller the slope the smaller the vAve. on a graph of distance vs. time, the slope represents the vAve.
** Speculate on why ave velocity changes with slope. **
the ball is rolling faster becasue there is a greater incline so the ball is picking up more speed in less time. so the slope of the ramp causes the ball to roll faster or with less trouble causing the vAve to increase.
** How could you test your speculations? **
you can find out the average velocity for the three trials of 1,2,and 3 dominoes and compare based on the number of dominoes which has the greater vAve by measuring the distance it traveled over the amount of time.
** **
40 min
** **
1.391
1.531
1.484
1.406
1.562
1.398
1.695
1.367
1.281
1.516
Each number represents the time it took for the ball to roll down a ramp that was raised by one domino.
** **
1.063
.969
.875
1.109
1.234
1.047
1.125
1.000
1.055
1.000
Each number represents the time it took for the ball to roll down a ramp that was raised by two dominoes.
** **
0.953
0.953
0.984
0.938
0.984
0.923
0.906
0.984
1.000
0.984
Each number represents the time it took for the ball to roll down a ramp that was raised by three dominoes.
** **
1.475, .07517
1.451, .1601
1.05, .1366
1.045, .05136
.9624, .02065
.9594, .04194
The first column contains the mean average of the time it took for the ball to roll down the ramp. The second column contains the standard deviation of the set of data that the mean averages are made up of.
** **
18.98 cm/s
19.3 cm/s
26.67 cm/s
26.8 cm/s
29.1 cm/s
29.18 cm/s
The results were obtained by dividing 28 by the time interval. This gives average velocity since distance/time = vAve. For example, the first time interval was 1.475 seconds. We assume that the ball traveled 28 cm, so 28/1.475 = 18.98 cm/s.
** **
25.74
26.60
50.80
51.29
60.47
60.83
The average rate of change of velocity with respect to time is known as the acceleration. The acceleration is found by dividing the change in velocity by the time. Since we have the average velocity of each interval, and we know that each ball started at rest (v0 = 0cm/s) we know that the change in velocity is vAve*2. So to find each acceleration, we use the equation (vAve*2)/ time. For example, in the first trial, we found the vAve to be 18.98 cm/s. Multiplying this by 2 to find dv, we get 37.96. Now, we dividide this by its time interval which we found to be 1.475, and we get its acceleration at 25.74 cm/s^2.
** **
26.17
51.045
60.65
The results were found by adding the two values for each trial and dividing by 2. Each result is the average acceleration for each trial.
** **
26.17
51.05
60.65
These results were found by first average the time intervals for the left to right and the right to left runs. Then, the the average change in velocities were found and these values were used to find the new accelerations.
** **
No, I would not expect the accelerations to be the same for each slope. The results I calculated to be the average of the two times are also the same of the average of the two rates. I would expect them to be the same because we are still finding the average each time. No matter if you average the times before or after the calculations, they will be the same, especially since the accelerations are uniform.
** **
.03, 26.17
.06, 51.05
.09, 60.65
The first column is ramp slope and is a unitless quantity. The second line is acceleration and is recorded in cm/s^2. The information above indicates that acceleration increases as slope increases.
** **
-.018
10
The slope is a unitless number. The number represents the point on the graph in which the best-fit line crosses the x-axis.
The unit for this number is cm/s^2 and is the point on the graph where the best-fit lines crosses the y-axis.
** **
.1, 79
The estimate was made by first drawing a best-fit line with the points, then locating .1 on the graph and estimating where the best-fit line would cross this point. I expect this to be relatively accurate since the best-fit line seems to be accurate.
** **
.118
79
669.5
The slope indicates the change in acceleration in the system.
** **
1.475, 1.451, 1.463
.07517, .1601, .117
The mean time and the average of the standard deviations were found.
** **
1.346, 1.58
These numbers are the left and right boundaries of the time interval based on the standard deviation. No numbers found during the experiment should be below 1.346 or above 1.58 based on the standard deviation.
** **
20.80, 30.91
17.72, 22.43
22.43, 30.91
These are the min/max values of velocity and acceleration based on the standard deviation. It is likely that the acceleration would be between these two numbers because it would lie between the boundaries determined by the standard deviation.
** **
29.36, 61.59
24.52, 42.97
42.97, 61.59
These are the min/max values of velocity and acceleration based on the standard deviation.
** **
30.12, 64.80
28.22, 56.88
56.88, 64.80
These are the min/max values of velocity and acceleration based on the standard deviation.
** **
.03, 22.43
.06, 42.97
.09, 56.88
.03, 30.91
.06, 61.59
.09, 64.80
These numbers are the slopes/ upper and lower limites of acceleration. They show how the acceleration changes with respsect to slope.
** **
My graph does fit the description. They do pass through the vertical segments. Since the max/min accelerations were more or less close to the best-fit line before. The line didn't change much at all.
** **
42.37, .06, 706.17
These are the rise/run/ and slope based on two points from the graph.
** **
25.97, .06, 432.83
These are the rise/run/ and slope based on two points from the grpah.
** **
2 hours 30 minutes
** **
&#Very good work. Let me know if you have questions. &#