course 201
Updated with the new information you requested.
I chose experiment number nine, which asks how the horizontal range of a projectile which leaves the end of a straight ramp at a constant height above the floor, after accelerating the full length of the ramp, depends on the slope of the ramp. Assuming slope and acceleration are proportional; how then would the horizontal range depend on the velocity at the end of the ramp?The materials used in conducting this experiment, up to this point, includes; a steel metal ramp (length 57.7cm) attached to plywood by a screw at the lower end of its slope to ensure its position stays the same, a screw adjustment mechanism placed at the other end of the ramp to be used for slope adjustment, one steel ball with directional/positioning marks placed on its top and lateral side to ensure proper ball placement on the ramp for each trial, a piece of white paper taped to the floor with its end corresponding to the end of the ramp from which the ball leaves 94.2cm above that spot on the floor, a piece of carbon paper laying on top of the white paper to ensure the exact landing spot of the ball can be marked, a metal ball, dominoes (used for ramp elevation above 1.5in or 3.81cm), and a meter stick for measuring.
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Twenty trials have been conducted up to this point. The first trial began with a ramp slope of .006cm. The ball was placed on the ramp in the position corresponding to the markings placed on it and the ramp and allowed to accelerate down the ramp from rest through a distance of 57.7cm before reaching the end of the ramp. The ball was then allowed to land on the carbon paper on the floor (94.2 cm below), which caused an ink dot to be left on the white paper. The dot was then marked with its corresponding slope and measured to obtain its horizontal range of travel after leaving the ramp. Each of the 20 trials has been conducted in this manner and the data obtained is listed below. (Note that each trial was conducted at an increased slope from the previous trial. This was done in the first 12 trials by adjusting the screw mechanism (4 turns which corresponds to 1/8 of an inch in rise of the ramp after each trial). Each rise was converted to centimeters in order to determine what percent of a domino with the thickness of .95cm it was. The remaining 8 trials were conducted by removing the screw mechanism and using whole dominos to adjust the ramp slope.
The following page contains a table consisting of the data collected.
# Dominos Rise (cm) Run (cm) Slope Ramp Length (cm) Vertical Drop (cm) Horiz. Range
(cm) Final Velocity (cm/s) Change
Velocity
(cm/s) Ave.
Velocity
(cm/s) Ramp
Time
(s) Accel.
(ramp)
(cm/s2)
.3342 .3175 57.7 .006 57.7 94.2 10.85 24.71 24.71 12.36 4.67 2.65
.6684 .635 57.7 .011 57.7 94.2 13.4 30.53 30.53 15.27 3.78 4.04
1.0026 .9525 57.7 .017 57.7 94.2 15.7 35.81 35.81 17.91 3.22 5.56
1.3368 1.27 57.7 .022 57.7 94.2 17.75 40.55 40.55 20.28 2.85 7.12
1.6711 1.5875 57.7 .028 57.7 94.2 19.4 44.40 44.40 22.20 2.6 8.54
2.0053 1.905 57.7 .033 57.7 94.2 21.1 48.42 48.42 24.21 2.38 10.17
2.3395 2.2225 57.7 .039 57.7 94.2 22.5 51.78 51.78 25.89 2.23 11.61
2.6737 2.54 57.6 .044 57.7 94.2 24.25 56.01 56.01 28.01 2.06 13.6
3.0079 2.8575 57.6 .050 57.7 94.2 25.5 59.12 59.12 29.56 1.95 15.16
3.3421 3.175 57.6 .055 57.7 94.2 26.9 62.64 62.64 31.32 1.84 17.02
3.6763 3.4925 57.6 .061 57.7 94.2 28.0 65.60 65.60 32.8 1.76 18.64
4.0105 3.81 57.6 .066 57.7 94.2 29.1 68.42 68.42 34.21 1.69 20.24
5.0 4.75 57.5 .083 57.7 94.2 31.6 75.55 75.55 37.78 1.53 24.7
6.0 5.7 57.4 .099 57.7 94.2 34.4 83.94 83.94 41.97 1.37 30.64
7.0 6.65 57.3 .116 57.7 94.2 37.0 92.45 92.45 46.23 1.25 37
8.0 7.6 57.2 .133 57.7 94.2 39.3 100.8 100.8 50.4 1.14 44.21
9.0 8.55 57.1 .150 57.7 94.2 41.4 109.4 109.4 54.7 1.05 52.1
10.0 9.5 56.9 .167 57.7 94.2 43.45 118.6 118.6 59.3 0.97 61.13
11.0 10.45 56.7 .184 57.7 94.2 44.85 126.5 126.5 63.25 0.91 69.51
12.0 11.4 56.6 .201 57.7 94.2 46.75 136.9 136.9 68.45 0.84 81.49
Graph 1: Slope vs. Horizontal Range
Graph 2: Final Velocity vs. Horizontal Range
Graph 3: Acceleration vs. Slope
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Good work.
If there is no slipping, the acceleration of the ball on the ramp will be the result of the parallel component of the ball's weight, which for a ramp at angle theta has magnitude weight * sin(theta).
Since the weight of the ball is constant, a graph of acceleration vs. sin(theta) will therefore be linear, again provided there is no slipping.
If a 'slope triangle' is constructed with run = 1 and rise = slope = m, then its hypotenuse will be sqrt(1 + m^2).
Regarding this hypotenuse as a vector whose initial point is the origin, its magnitude is sqrt(1 + m^2) and its x and y components are respectively 1 and m.
Since its y component is equal to magnitude * sin(theta) = sqrt(1 + m^2) sin(theta), we have sqrt(1 + m^2) sin(theta) = m so that sin(theta) = m / sqrt(1 + m^2).
For each slope you can therefore determine acceleration and sin(theta), and you will be able to construct the desired graph and determine with reasonable precision the point at which the graph diverges from linear, from which you can determine the slope at which slipping begins.
The expression
weight * sin(theta)
for the magnitude of the parallel component comes from the fact that With respect to an x-y coordinate system with the x axis along the incline, theta will be the angle of the weight vector with the negative y axis. This is explained in the text.
This also comes your free-body diagram as follows:
With respect to an x-y coordinate system with the x axis along the incline, the angle of the weight vector will be either 270 deg + theta or 270 deg - theta (depending on whether you picture the incline moving down and to the right or down and to the left). So the parallel component of the weight, which is along the x axis, will be weight * cos(270 deg +- theta) and its magnitude will be | weight * cos(270 deg + - theta) | = | weight * sin(theta) |. This uses the fact that the sine and cosine functions are 90 deg out of phase, which since their periods are both 360 deg also makes them 270 deg out of phase.