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course phy 201
9/23 2:40 amSorry this assignment is 2 days late, Ive been having major computer trouble. I completed most of it LATE night sunday night/ monday morning. Then i meant to wake up early and finish monday before class but i overslept and then had some fan issues with my laptop. They appear to be fixed now thankfully.
Class 090916
Note: When answering these questions, give your answer to a question before the &&&&. This is different than my previous request to place your answer after the &&&&.
Thanks.
Calibrate Rubber Band Chains:
Calibrate a rubber band chain (i.e., find its length as a function of the force exerted to stretch it) using 1, 2, 3, 4 and 5 dominoes. Give your raw data below in five lines, with number of dominoes and length of chain separated by a comma, and an explanation following in subsequent lines:
0, 49 cm
1, 60 cm
2, 65 cm
3, 70 cm
4, 75 cm
5, 81 cm
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Graph chain length vs. number of dominoes, and calculate graph slope between each pair of points. Give your results below. Table form would be good, with columns for length and number of dominoes, rise, run and slope. However as long as you include an explanation, any format would be acceptable.
0-1 dominos - slope 11cm/1 domino
1-2 - slope 5cm/1 domino
2-3 - slope 5 cm/1 domino
3-4 - slope 5 cm/ 1 domino
4-5 - slope 6 cm/ 1 domino
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Double the chain and calibrate it using 2, 4, 6, 8 and 10 dominoes. Give your raw data below, in the same format as before:
2, 31 cm
4, 33 cm
6, 35 cm
8, 37 cm
10, 39 cm
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Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
The slope was 2cm/2 dominos or 1cm/domino in every case
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Rotate the strap using the chain
Suspend the strap from your domino chain, supporting the strap at its center so it will rotate in (or close to) a horizontal plane, sort of like a helicopter rotor. Rotate the strap through a few revolutions and then release it. It will rotate first in one direction, then in the other, then back in the original direction, etc., with amplitude decreasing as the energy of the system is dissipated. Make observations that allow you to determine the period of its motion, and determine whether its period changes significantly.
Give your raw data and your (supported) conclusions:
Rotated 1 direction for 85 seconds, then back the other direction for 60 seconds
The period definitely decreased a lot during the second rotation, Common sense tells me that it should decrease but im not sure if it should by this much, maybe the chain was not at equilibrium when I started.
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Double the chain and repeat.
Give your raw data and your (supported) conclusions:
I did 2 trials here
1st time it rotated 28 seconds 1 way and 20 seconds back the other way
2nd time it rotated 30 seconds 1 way and 25 back the other way.
These two readings seemed to be pretty close to each other. Once again the chain loses energy during each ""switchback"" but at a more reasonable rate than in the single chain experiment.
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How does period of the oscillation compare between the two systems?
Both lose energy after each switch back, however the double chain seems to have closer times in each direction. The added torque of the double band seems to help it maintain momentum better
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'Bounce' the dominoes on the end of the chain
'Bounce' a bag of dominoes on the chain. Is there a natural frequency? Does the natural frequency depend on the number of dominoes? If so how does it depend on the number of dominoes?
You might not be able to give complete answers to these questions based on your data from class. Give your data, your conclusions, and you hypotheses (i.e., the answers you expect to get) regarding these questions.
5 dominos bounced 8 times in 5 seconds before coming to rest
10 dominos bounced 5 times in 5 seconds before coming to rest
13 dominos bounced 12 times in 12 seconds before coming to rest.
The heavier the dominos the longer amount of time it took for each bounce. Also the heavier the domino bag was it seems as if it kept the momentum going longer as well
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How would you design an experiment, or experiments, to further test your hypotheses?
I would do many more tests, including 1,2,3,4,6,7,8,9 dominos. With all these different runs you could graph your info and see if there is a correlation between weight and bounce frequency
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Repeat for doubled chain. How are the frequencies of doubled chain related to those of single chain, for same number of dominoes?
Did not complete this trial with a double band
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You might not be able to give complete answers to these questions based on your data from class. Give your data, your conclusions, and you hypotheses (i.e., the answers you expect to get) regarding these questions.
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How would you design an experiment, or experiments, to further test your hypotheses?
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If you swing the chain like a pendulum, does its length change? Describe how the length of the pendulum might be expected to change as it swings back and forth.
The band gets longer during some points of its swing. This was proven by the experiment you showed in class when you put dominos under the bag and then had me pull the pendulum and release it. The swinging of the weight on the end of the bands seem to exert different amounts of force on the band at different points in its swing.
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Slingshot a domino block across the tabletop
Use your chain like a slingshot to 'shoot' a domino block so that it slides along the tabletop. Observe the translational and rotational displacements of the block between release and coming to rest, vs. pullback distance.
Give your results, in a series of lines. Each line should have pullback distance, translational displacement and rotational displacement, separated by commas:
30 cm, 27 cm, 90 degrees
40 cm, 41 cm, 180 degrees
50 cm, 67 cm, 0 degrees
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Describe what you think is happening in this system related to force and energy.
The rotation seems to be pretty random due to variations in how the block of dominos were held and the friction between the block and the table.
As the band was stretched further it exerted more force on the block causing it to travel exponentially further each time. This is because more energy was conveyed each time.
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Complete analysis of systems observed in previous class
Rotating Strap:
For last time you calculated the average rate of change of position with respect to clock time for each of five trials on the rotating strap. This average rate of change of position is an average velocity. Find the average rate of change of velocity with respect to clock time for each trial. As always, include a detailed explanation:
My 5 averages were
57.5 degrees/tick
72.5 degrees/tick
61 degrees/ tick
73.33 degrees/ tick
70.66 degrees/tick
To get the avg A you would take these numbers and divide by the number of ticks
57.5/6 = 9.58 degrees/tick^2
72.5/4= 18.13 degrees/tick^2
61/5= 12.2 degrees / tick^2
73.33/9 = 8.15 degrees/tick^2
70.66/7.5= 9.42 degrees/tick^2
it's not average velocity / change in clock time, but change in velocity / change in clock time
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(Note: Since the system is rotating its positions, velocities and accelerations are actually rotational positions, rotational velocities and rotational accelerations. They are technically called angular positions, angular velocity and angular accelerations, because the position of the system is measured in units of angle (e.g., for this experiment, the position is measured in degrees). These quantities even use different symbols, to avoid confusion between rotational motion and translational motion (motion from one place to another). So technically the question above doesn't use the terms 'position', 'velocity', etc. quite correctly. However the reasoning and the analysis are identical to the reasoning we've been using to analyze motion, and for the moment we're not going to worry about the technical terms and symbols.)
Atwood Machine:
Find the average rate of change of velocity with respect to clock time for each trial of the Atwood machine.
My three trials were
80/ 8 = 10 cm / tick
80 / 7 = 11.4 cm / tick
80/5.5 = 14.5 cm / tick
You then take these avg Velocities and divide by ticks
10/8 = 1.25 cm/ticks^2
11.4/ 7 = 1.63 cm/ticks^2
14.5/5.5 = 2.64 cm/ ticks^2
it's not average velocity / change in clock time, but change in velocity / change in clock time
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Hotwheels car:
For the Hotwheels car observed in the last class, double-check to be sure you have your signs right:
• You pushed the car in two different directions on your two trials, one in the direction you chose as positive, and one in the direction you chose as negative.
• You will therefore have one trial in which your displacement was positive and one in which it was negative.
• Your final velocity in each case was zero. In one case your initial velocity was positive, in the other it was negative. Be careful that your change in velocity for each trial has the correct sign, and that the corresponding acceleration therefore has the correct sign.
Signs were correct
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New Exercises
Exercise 1:
A ball rolls from rest down each of 3 ramps, the first supported by 1 domino at one end, the second by 2 dominoes, the third by 3 dominoes. The ramp is 60 cm long, and a domino is 1 cm thick. The motion is in every case measured by the same simple pendulum.
It requires 6 half-cycles to roll down the first, 4 half-cycles to roll down the second and 3 half-cycles to roll down the third.
Assuming constant acceleration on each ramp, find the average acceleration on each. Explain the details of your calculation:
1st ramp: (im assuming each of the 3 parts of the ramp are 20 cm each, since you said the ramp was 60 cm. If each of the 3 are 60 each then these calculations are off.)
20 cm/6 ticks = 3.33 cm/tick (avg Vel)
3.33 cm/tick / 6 ticks = .555 cm/tick^2 (this is Avg A)
2nd ramp:
20 cm/ 4 ticks = 5 cm/tick (avg V)
5cm/tick / 4 ticks = 1.25 cm/tick^2 (avg A)
3rd ramp:
20 cm/ 3 ticks = 6.66 cm/tick (avg v)
6.66/ 3 = 2.22 cm/tick^2 (Avg A)
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Find the slope of each ramp.
Starting ramp:
3 dominos x 1 cm = 3 cm rise, 20 cm run
3cm/20cm
or 1cm/6.66cm
Mid ramp:
2cm/20cm
of 1cm/10cm
Ending ramp:
1cm/20cm
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Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
1cm/6.66cm , .55cm/ ticks^2
1 cm/ 10 cm, 1.25 cm / ticks^2
1 cm/ 20 cm, 2.22 cm / ticks^2
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Connect the three points with straight line segments, and find the slope of each line segment. Each slope represents a average rate of change of A with respect to B. Identify the A quantity and the B quantity, and explain as best you can what this rate of change tells you.
Slope between 1st and second point:
(.7cm/ticks^2)/ - .0515cm
bt 2nd and 3rd
(.97cm/ticks^2) / - .05cm
Quantity A is the acceleration
Quantity B is the ramp slope.
This rate of change tells us how the acceleration is affected by changes in the slope of each ramp.
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Exercise 2: A ball rolls down two consecutive ramps, starting at the top of the first and rolling without interruption onto and down the second. Each ramp is 30 cm long.
The acceleration on the first ramp is 15 cm/s^2, and the acceleration on the second is 30 cm/s^2.
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For motion down the first ramp:
What event begins the interval and what even ends the interval?
When the ball is released to when it touches the 2nd ramp
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What are the initial velocity, acceleration and displacement?
0, 15 cm/s^2, 30 cm
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Using the equations of motion find the final velocity for this interval.
Vf^2 = 0 + (2* 15cm/s^2) *30 cm
Vf^2 = 30 cm/s^2 * 30 cm
Vf^2 = 900 cm^2/s^2
Take square root, VF=30 cm/s
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Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
2 seconds
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For motion down the second ramp:
What event begins the interval and what even ends the interval?
Hits the 2nd ramp to leaves the ramp
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What are the initial velocity, acceleration and displacement?
30 cm/s, 30 cm/s^2, 30 cm
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Using the equations of motion find the final velocity for this interval.
Skipped the first few steps
Vf= 30 cm/s + 1800 cm^2/s^2
Im guessing this is where you take the square root, because I don’t think the 2 quantities on the right can be added together since they are different
So
Vf= 5.48 cm/s + 42.43 cm/s =
47.91 cm/s
sqrt(v0^2 + 2 a `ds) is not the same as v0 + sqrt(2 a `ds).
For example sqrt(25) = 5, but sqrt(25) = sqrt(16 + 9), which is not 4 + 3.
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Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
I think you mean time spent on second ramp.
.77 seconds
Challenge Exercise:
The first part of this exercise is no more challenging than the preceding problem. It uses the result of that problem:
A ball accelerates uniformly down a ramp of length 60 cm, right next to the two 30-cm ramps of the preceding exercise. The ball is released from rest at the same instant as the ball in the preceding exercise.
What is its acceleration if it reaches the end of its ramp at the same instant the other ball reaches the end of the second ramp?
60 cm/ 2.77 seconds = 21.66 cm/s
this is avg v
21.66cm/s / 2.77 seconds =
7.82 cm/s^2 = A
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The second part is pretty challenging:
The 60 cm ramp is made a bit steeper, so that its acceleration is increased by 5 cm/s^2. The experiment is repeated. How far will the ball on this ramp have traveled when it passes the other ball?
?????No real clue here, please explain
Homework:
Your label for this assignment:
ic_class_090916
Copy and paste this label into the form.
Report your results from today's class using the Submit Work Form. Answer the questions posed above.
Note also that you will receive a subsequent document with some alternative materials, and that you will be asked to complete a short portion of that document.
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&#Good responses. See my notes and let me know if you have questions. &#