Video Version Experiment #2

I'm going to ask you to submit a modification of your results here.
Noting that the pendulum requires 1/4 cycle between release and the first tap, then 1/2 cycle for each subsequent tap, and also noting that the period is T = .2 sqrt(L) with T in sec when L is in cm, find the average velocity and acceleration on each run and please send me your analysis.

Physics 201

In this experiment, I am going to measure the velocity of a cart traveling down a slope at different inclines. The goal of the experiment is to determine if the velocity of the cart is directly proportional to the slope of the ramp. I will determine this by measuring the velocity of the cart at four different slopes, keeping track of how long it takes for the cart to reach the end of the track. My measure of time will be the pearl pendulum, so my time will be counted in taps against the bracket. I will record the velocity of the cart at four separate slopes, and then graph my results in an average velocity vs. slope graph. The increase in velocity will be proportional to the slope of ramp if the graph forms a straight line. With my timer, I used the same setup as the one I constructed for the first mini-experiment. During the first segment of the video clip, I was instructed to set my pendulum to where the first run would yield a time of 2 cycles. My pendulum, however, was set up to count much smaller cycles. The length of my pendulum is 8 cm. In the time that my pendulum should have made two cycles, it made exactly six. I did not want to alter the pendulum since I know it to be accurate. Also, the count is exactly one third of the count asked for in the video clip. During the first run, the rise of the ramp was 1.26 cm, and the run was 44 cm. This gave me a slope of .03. The car traveled 46.2 cm in 6 taps, giving it an average velocity of 7.7 cm/tap. On the second run, the slope was increased to .046. This resulted in the car traveling 10.27 cm/tap. On the third run, the slope was decreased to .038, which is larger than the first slope, but smaller than the second. The velocity of that run was 8.4 cm/tap. On the final run, the slope was decreased even further to .022, resulting in an average velocity of 6.6 cm/tap. This is what my table and graph look like.

Pendulum (cm) Slope Vave (cm/tap) 8 .029 7.7 8 .046 10.27 8 .038 8.4 8 .022 6.6

As the graph shows, the plotted points appear to be almost completely linear. This would suggest that the velocity of the object increases proportionally to the slope of the ramp. The slope of the graph is 144.13, or 1.44 cm/sec for every increment of .01 slope. To find the final velocity of the cart, I would just have to double the average velocity from the table above. A table of slope vs. final velocity would like the following table.

Pendulum (cm) Slope Vfinal (cm/tap) 8 .029 15.4 8 .046 20.54 8 .038 16.8 8 .022 13.2 As you can see, the table reflects the same tendancies as the first, just with higher velocities. A graph of this table would also look similar to the previous graph, with the only difference being the actual values of velocity, which would change the slope of the line passing through the points. From this table, we can produce another table, which compares the change in velocity with the time it took to make that change. This would in essence be a table that would be used to determine acceleration.

Pendulum (cm) taps Slope `dV (cm/tap) accel. cm/tap^2 8 6 .029 15.4 2.57 8 4.5 .046 20.54 4.56 8 5.5 .038 16.8 3.05 8 7 .022 13.2 1.89 In this table, I included the number of taps, so it could be shown how I obtained the acceleration of each run. These results give me this graph.

The slope of this graph is also linear, increasing at a constant rate. What this tells me is that the acceleration of the cart increases at a proportional rate to the slope of the ramp. The slope of the line is 288.26, or 2.88 cm/tap for each increment of .01 slope. From the results of the experiment, it would appear that the velocity of an object accelerating down a slope increases directly proportional to the slope of the ramp. Also, since the velocity is proportional, the change in velocity would also proportional, meaning that the acceleration increases proportionally as the slope increases.

Mini-experiment 3 Rubber Bands

For Phy 241 I'm going to ask you to do a little more with this experiment. Before I can do that, please tell me if you have already completed your Calculus sequence or are in the process of completing it this year? What I ask you to do--or more correctly the way I ask it--depends on your status with that course.

For this experiment, I had four rubber bands, marked 1 through 4. Upon measuring my rubber bands, I determined that each one was 9 cm long. This would mean that to reach the critical 35% increase in length, the rubber bands would each have to stretch to 12.15 cm. I filled the putty tubs with water, one at a time. Each time I measured the bands, they got slightly longer. Each band was increasing the same distance, so it will only be necessary to make one graph of force vs. distance. At 12.5 tubs of water, all four of my rubber bands reached a length of 12.15 cm, which is 35% past its un-stretched length.

With this information, and the information provided in the assignment that gravity exerts .21 N of force on each tub, I determined that gravity is exerting 2.625 N on the baggie of water (ignoring the slight weight of the baggie itself). Number of tubs length of band (cm) Force exerted (N) 1 9 .21 2 9.1 .42 3 9.2 .63 4 9.4 .84 5 9.6 1.05 6 9.9 1.26 7 10.3 1.47 8 10.4 1.68 9 10.6 1.89 10 11.1 2.1 11 11.5 2.31 12 12 2.52 12.5 12.15 2.625

-Estimate the force corresponding to a length of 9.8 cm. With the graph that I constructed, it would appear that at a length of 9.8 cm, the force exerted is 1.2 N.

-Estimate the length of a rubber band that gives a force of 2.4 Newtons. The length of the rubber band when there is 2.4 Newtons applied is approximately 11.5 cm. -From the curve estimate the force corresponding to each of the lengths 8.5, 9, 9.5, 10, 10.5, 11 and 11.5 cm. Give your estimates. Since my rubber band was 9 cm long, I did not have a measurement at 8.5 cm. at 9 cm, I would estimate .20 N. at 9.5 cm, I would estimate .85 N. at 10 cm, I would estimate 1.25 N. At 10.5 N, I would estimate 1.65 N. At 11 cm, I would estimate 2 N. at 11.5 cm, I would estimate 2.25 N.

-By how much did each of the forces you just estimated differ from the data you obtained from the calibration graph? Each estimation was fairly close to the actual data I obtained from the calibration graph. -Which do you have more faith in, the values from the curve you just created or the values you read from the original calibration graph, and why? I have more faith in the curve because it can average out where I might have made minor mistakes. Since each of my rubber bands measured the same length, and each stretched at the same rate, all of my measurements should be the same. The only difference at all is that as the rubber bands are higher in the chain, they have to support more weight in rubber band and paper clip, but that weight is almost negligible.

Video Version Experiment 3

My followup on this experiment will also depend on your status with respect to calculus.

In the first part of this experiment, I used video clip #7 to check the velocities of the two parts of the ramp. My timing turned out to show that the average velocity on the second ramp is twice that as the first. After completing this part, I proceeded with video clip #8.

In this part of the experiment, I observed the same system as before, with a ramp at a slight angle transferring to a rail that would allow a ball to travel at a constant velocity. I recorded the time it took for the ball to travel on the first ramp, and then from the first ramp to the second ramp at various distances. The following table shows the raw data of the experiment. The length is the distance of the first part of the ramp followed by the distance of the second part of the ramp. The first time is for the first part of the ramp, followed by the average velocity. The third column is for the second part of the ramp, followed by the second average velocity. Length (1st/2nd) 1st time 1st V 2nd time 2nd V 10 cm/20 cm 1.25 sec 8 cm/s .82 sec 24.39 cm/s 15 cm/30 cm 1.375 sec 10.91 cm/s 1.375 sec 21.82 cm/s 20 cm/40 cm 1.70 sec 11.76 cm/s 1.60 sec 25 cm/s 25 cm/40 cm 2.00 sec 12.5 cm/s 1.4 sec 28.57 cm/s 30 cm/40 cm 2.14 sec 14.02 cm/s 1.25 sec 32 cm/s 35 cm/40 cm 2.23 sec 15.70 cm/s 1.20 sec 33.33 cm/s 40 cm/40 cm 2.30 sec 17.39 cm/s 1.14 sec 35.09 cm/s 45cm/40 cm 2.56 sec 17.58 cm/s 1.07 sec 37.38 cm/s 50 cm/40 cm 2.81 sec 17.80 cm/s 1.04 sec 38.46 cm/s

As the table and graph shows, my points are somewhat sporadic. This does however show that the relationship between the two velocities is linear. The slope of my line appears to be approximately 1.65, where in reality, it should be 2. My line does not pass through the origin of the graph. I suppose that technically it could, since if the velocity on the first ramp is zero, than the velocity on the second ramp is zero. If this were a perfect experiment, the line would travel through the origin since for every 1 cm/sec increase on the x axis, there is a 2 cm/sec increase on the y axis. This graph obviously does not demonstrate that fact. The ball has the greater average velocity on the second ramp since the average velocity of this ramp is equal to the final velocity of the first ramp. The initial velocity of the ball on the first ramp is the smallest figure, followed by the average velocity of the ball on the first ramp, then the final velocity of the ball on the first ramp. The velocity of the ball on the ramps are as follows, from slowest to fastest. 1. Initial velocity of the ball on the first ramp. 2. Average velocity of the ball on the first ramp. 3. Final velocity of the ball on the first ramp is equal to the initial velocity of the ball on the second ramp. 4. The average velocity of the ball on the second ramp is equal to the initial velocity on the second ramp. 5. The final velocity of the ball on the second ramp is equal to its average velocity on the second ramp.

In order to graph a velocity vs. time graph, I chose to examine the run where the first part of the ramp was 30 cm, and the second part was 40 cm. As the graph shows, the velocity is increasing at a constant rate throughout the travel on ramp 1. As soon as the ball hits ramp 2 at 2.14 seconds (30 cm), the velocity becomes constant.

The average velocity of the first part of the ramp is indeed half the average velocity of the second part of the ramp. Though the first graph does not portray this idea, it is not difficult to see where some more accurate timing would result in a graph that is much closer to the actual behavior of the ball. The final velocity of the ball on the first ramp is the initial velocity and final velocity on the second ramp. Since average velocity is the average of initial and final velocity, the average velocity of the second ramp is twice that of the first ramp.

I believe that the level of uncertainty for my measurements on the 30 cm incline is .05 sec. I came to this conclusion after studying my findings, and realizing that almost all of my times were within .05 seconds of each other. Based on the observed time and the level of uncertainty, the minimum and maximum values for the actual time of the 30 cm run is 2.09 sec to 2.19 sec for the ball to reach the end of the first ramp. Had the actual time been equal to the minimum value, than the average velocity of the 30 cm run would be 14.35 cm/sec, as opposed to 14.02 cm/sec for my actual time. Had the actual time been equal to the maximum value, than the average velocity of the 30 cm run would be 13.70 cm/sec, as opposed to 14.02 cm/sec for my actual time. I believe that the actual minimum and maximum average velocities are 14.35 cm/sec and 13.70 cm/sec. The range of velocities will be on the y axis at 2.14 seconds, extending up and down .65 cm/sec.

I believe that the level of uncertainty for my measurements on the 40 cm incline is .05 sec as well. I came to this conclusion after studying my findings, and realizing that almost all of my times were within .05 seconds of each other. Based on the observed time and the level of uncertainty, the minimum and maximum values for the actual time of the 40 cm run is 1.20 sec to 1.30 sec for the ball to reach the end of the second ramp. Had the actual time been equal to the minimum value, than the average velocity of the 40 cm run would be 33.33 cm/sec, as opposed to 32.00 cm/sec for my actual time. Had the actual time been equal to the maximum value, than the average velocity of the 40 cm run would be 30.80 cm/sec, as opposed to 32.00 cm/sec for my actual time. I believe that the actual minimum and maximum average velocities are 30.80 cm/sec and 33.33 cm/sec. The range of velocities will be on the y axis at 3.39 seconds, extending up and down 1.27 cm/sec.

When these areas are blocked in on my graph, the line actually goes through the point. After drawing rectangle blocks for each point on my first graph, the line travels through every rectangle except for the first two. A line could be drawn through all rectangles, but it would be further away from the origin.

It is apparent that the only reason that this experiment was not completely accurate was my timing skills. Had the times been as accurate as the could have been, the graph would show that the average velocity of the first ramp is half that of the second ramp. Any doubt about this fact could be solved by using the four velocity and acceleration formulas that we work with.