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course Phy: 201
Class 090916Note: 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:
1, 65
2, 69
3. 71.5
4, 75
5, 78.5
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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.
Length # Dominos Rise Run Slope
65 1
69 2 4 1 4
71.5 3 2.5 1 2.5
75 4 3.5 1 3.5
78.5 4 3.5 1 3.5
you switched the rise and the run
# dominoes is on the 'vertical' axis and length on the 'horizontal'
otherwise very good
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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, 33.5
4, 34
6, 36
8, 37.5
10, 38
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Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
Length # Dominos Rise Run Slope
33.5 2
34 4 0.5 2 ¼
36 6 2 2 1
37.5 8 1.5 2 ¾
38 10 0.5 2 ¼
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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, and 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:
14 rotations in 25 seconds
10 rotations in 16 ½ seconds
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Double the chain and repeat.
Give your raw data and your (supported) conclusions:
11 rotations in 32 seconds
10 rotations in 14 seconds
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How does period of the oscillation compare between the two systems?
The doubled chain has a greater period of oscillations. The oscillations are much quicker because the chain is shorter. However, when we used a constant oscillation number, the clock time was very close together, only 2 ½ seconds apart.
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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.
We found that with 10 dominos, there were 24 bounces and it required 15 seconds to complete. With 8 dominos, there were 18 bounces and it required 10 seconds to complete. The hypothesis is correct. It is imaginable that the greater the amount the more bounces and the larger time interval to complete the bounces. There is a natural frequency because by performing many trials we found that the chain had about the same data each time we performed the initial experiment. The frequency does depend on the number of dominos present in the bag. We performed many trials with many different numbers of dominos and found different frequencies for each trial. The more dominos there are present in the bag, the greater number of bounces and the greater time interval to complete the experiment.
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How would you design an experiment, or experiments, to further test your hypotheses?
One experiment you could do is to just perform many trials. Start with 10 dominos and perform many trials with this 10. Then go down to 8 and do the same thing. Then 6, 4, and so on. However, one trial is not enough. Many times you will have an off set, or some crazy data. So of course you would need to perform many trials.
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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?
The frequencies are doubled.
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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.
It took 48 bounces in 17 seconds whereas it took 24 in 15. With the 8 dominos it took 38 bounces in 12 seconds whereas it took 18 in 10. So therefore the number of bounces doubled in the same number of around the same number of seconds. Because we doubled the chain the bounces doubled also. This is expected because the chain is half the whole chain.
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How would you design an experiment, or experiments, to further test your hypotheses?
The experiment would be the same as before. You would just do many trials using many different numbers of dominos.
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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 length of the pendulum does change. As the pendulum first begins swinging, the length of the pendulum increases as the pendulum’s frequency increases. After some time, the pendulum length will begin to decrease as the pendulum’s frequency begins to slow down and eventually come to rest.
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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:
6, 27 ½, 180°
6, 31, 270°
10, 49, 380°
15, 115, 720°
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Describe what you think is happening in this system related to force and energy.
As the chain is pulled back force is being used. As the chain is released energy is transferred from the person to the chain through friction and sends the dominos down the table.
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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:
The data are as follows:
‘ds ‘dt
450° 3 half-cycles
225° 3 half-cycles
405° 2.5 half-cycles
550° 4 half-cycles
225° 3 half-cycles
So first, I divided 450 degrees by 3 half-cycles. That gave me 150 degrees/ half-cycle. This is the average rate of change with respect to clock time. Then I had to find the change in velocity. So I set up the equation 150 = (0 – I) cm/ 2 half cycles. I cross multiplied and got 300° for the initial velocity. Now for velocity I have to take (0 – 300)/3 = -100°/half-cycle^2 for the average rate of change of velocity with respect to clock time or acceleration.
Then, I divided 225 degrees by 3 half-cycles. That gave me 75 degrees/ half-cycle. This is the average rate of change with respect to clock time. Then I had to find the change in velocity. So I set up the equation 75 = (0 – I) cm/ 2 half cycles. I cross multiplied and got 150° for the initial velocity. Now for velocity I have to take (0 – 150)/3 = -50°/half-cycle^2 for the average rate of change of velocity with respect to clock time or acceleration.
Then, I divided 405 degrees by 2.5 half-cycles. That gave me 162 degrees/ half-cycle. This is the average rate of change with respect to clock time. Then I had to find the change in velocity. So I set up the equation 162 = (0 – I) cm/ 2 half cycles. I cross multiplied and got 324° for the initial velocity. Now for velocity I have to take (0 – 324)/2.5 = -129.6°/half-cycle^2 for the average rate of change of velocity with respect to clock time or acceleration.
Then, I divided 350 degrees by 4 half-cycles. That gave me 87.5 degrees/ half-cycle. This is the average rate of change with respect to clock time. Then I had to find the change in velocity. So I set up the equation 87.5 = (0 – I) cm/ 2 half cycles. I cross multiplied and got 175° for the initial velocity. Now for velocity I have to take (0 – 175)/4 = -43.75°/half-cycle^2 for the average rate of change of velocity with respect to clock time or acceleration.
Finally, I divided 225 degrees by 3 half-cycles. That gave me 75 degrees/ half-cycle. This is the average rate of change with respect to clock time. Then I had to find the change in velocity. So I set up the equation 75 = (0 – I) cm/ 2 half cycles. I cross multiplied and got 150° for the initial velocity. Now for velocity I have to take (0 – 150)/3 = -50°/half-cycle^2 for the average rate of change of velocity with respect to clock time or acceleration.
very well done
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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.
The displacement is 80 centimeters. It took 4 half-cycles for the apparatus to complete a cycle. Therefore the average rate of change of velocity with respect to clock time is as follows.
The average velocity must be figured out first. So 80 cm/ 4 half-cycles = 20 cm/half-cycle. We take this to find the change in velocity. We know the final velocity is 0 because the system comes to rest. So 20cm/half-cycle = (0 – I)/2= 40 cm/ half-cycle^2. So now we can find average acceleration. So 40 cm/ half-cycle^2 / 4 half-cycles = 10 cm/ half-cycle.
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Hot wheels car:
For the Hot wheels 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.
I chose south for the positive direction. For this direction the average rate of change of position with respect to clock time was: (0+22.5) cm/3 half-cycles = 7.5 cm/half-cycle. Then the average rate of change of velocity would be figured out next; however we have to find initial velocity. So, 7.5 cm/half-cycle = (0-I)/2. I cross multiply and get 15 cm/half-cycles as the initial velocity. Now the average acceleration can be figured out, (0+15)/3half-cycles = 5 cm/half-cycle^2.
I chose north for the negative direction. For this direction the average rate of change of position with respect to clock time was: (0 – 25) cm/4 half-cycles = -6.25 cm/half-cycle. Then the average rate of change of velocity would be figured out next; however we have to find initial velocity. So, -6.25 cm/half-cycle = (0-I)/2. I cross multiply and get -12.5 cm/half-cycles as the initial velocity. Now the average acceleration can be figured out, (0 – 12.5)/4half-cycles = -3.125 cm/half-cycle^2.
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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, and 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:
First I have to find the average rate of change with respect to clock time, or average velocity. So, (60cm)/ 6 half-cycles = 10 cm/half-cycle. Then I have to find the final velocity. S (vf + 0)/2, cross-multiply and dived to get 20 cm/half. Finally average acceleration is configured. (20 – 0) cm/ 6 half-cycles = 3.33 cm/half-cycles ^2.
First I have to find the average rate of change with respect to clock time, or average velocity. So, (60cm)/ 4 half-cycles = 15 cm/half-cycle. Then I have to find the final velocity. S (vf + 0)/2, cross-multiply and divide to get 30 cm/half. Finally average acceleration is configured. (30 – 0) cm/ 4 half-cycles = 7.5 cm/half-cycles ^2.
First I have to find the average rate of change with respect to clock time, or average velocity. So, (60cm)/ 3 half-cycles = 20 cm/half-cycle. Then I have to find the final velocity. S (vf + 0)/2, cross-multiply and divide to get 40 cm/half. Finally average acceleration is configured. (40 – 0) cm/ 3 half-cycles = 13.33 cm/half-cycles ^2.
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Find the slope of each ramp.
60cm/ 1 cm = 60 cm
60cm/ 2 cm = 30 cm
60cm/ 3 cm = 20 cm
the rise is much smaller than the run; you reversed the two
easy to fix
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Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
3.33, 60
7.5, 30
13.33, 20
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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.
The slopes are (-50/7) and (-50/29). Quantity A is slope. So the change in A is the change in slope. Quantity B is average acceleration. So the change in B is the change in average acceleration. Therefore the average rate of change of A with respect to B is (60-30)/ (3.3 – 7.5) and (30-20)/ (7.5 – 13.33). This slope tells us the average rate of change of the difference between the acceleration of the different slopes on the different ramps.
acceleration vs ramp slope has ramp slope in the first column
your ramp slopes need to be corrected
however except for these details you are following the process and the reasoning very nicely
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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.
For motion down the first ramp:
What event begins the interval and what even ends the interval?
The ball is released from rest; the ball hits the end of the first ramp.
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What are the initial velocity, acceleration and displacement?
Initial velocity is 0, acceleration is 15cm/s^2, and displacement is 30 cm.
Using the equations of motion find the final velocity for this interval.
I used the equation vf^2= v0^2 + 2a ‘ds. So I plug in the numbers. Vf^2 = (0cm) + 2(15cm/s^2) (30 cm) = 900cm^2/s^2. Then you take the square root of each side to get 30 cm/s as the final velocity.
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
So I used the equation of average acceleration. It is a_ave= (vf-v0)/ ‘dt. So I plug in the numbers and 15cm/s^2 = (30cm/s – 0cm/s)/ ‘dt = 2 seconds on the first ramp.
For motion down the second ramp:
What event begins the interval and what even ends the interval?
The beginning event is the moment it hits the second ramp and the ending event is when it leaves the second ramp.
What are the initial velocity, acceleration and displacement?
Initial velocity is 30cm/s, acceleration is 30 cm/s^2, displacement is 30 cm.
Using the equations of motion find the final velocity for this interval.
I used the equation vf^2= v0^2 + 2a ‘ds. So I plug in the numbers. Vf^2 = (30cm/s) + 2(30cm/s^2) (30 cm) = 1830cm^2/s^2. Then you take the square root of each side to get 42.77 cm/s as the final velocity.
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
So I used the equation of average acceleration. It is a_ave= (vf-v0)/ ‘dt. So I plug in the numbers and 30cm/s^2 = (42.77cm/s – 0cm/s)/ ‘dt = about 1.5 seconds on the second ramp.
your equation is correct but you weren't careful enough in your solution
you don't specify your steps but I'm pretty sure that as you applied them they would lead you to
1.5 / s = 1 / `dt.
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?
22.5 cm/s^2; I just took the two accelerations of the two 30cm ramps and averaged them. (15+30)/2 = 22.5 cm/s^2.
not a bad attempt, but the accelerations last for different time intervals so a simple average doesn't work
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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?
5 cm, because the ramp is increased by 5cm/s^2, I think it will have travelled 5 cm when it passes the other ball.
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.
you've got a few errors but most of your work looks very good
see my notes