course PHY 201
8:33 PMSunday, September 27, 2009
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:
1,65
2,69.5
3,73
4,77
5,79
column 1 is # of dominoes, column 2 is length of rubber band in 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.
x y
1 65
2 69.5
3 73
4 77
5 79
rise run slope
4.5 1 4.5
3.5 1 3.5
4 1 4
2 1 2
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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:
2,32.5
4,34.5
6,36
8,38
10,40.5
column 1 is # of dominoes, column 2 is length of rubber band in cm
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Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
x y
2 32.5
4 34.5
6 36
4 38
8 40.5
rise run slope
2 2 1
1.5 2 0.75
2 2 1
2.5 2 1.25
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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:
I'm not sure I recorded proper information for this one, but the strap rotated 360 degrees in 3 seconds as the period stabilized somewhat in the middle.
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Double the chain and repeat.
Give your raw data and your (supported) conclusions:
Again, I think I didn't record enough data to properly respond to this one, but doubling the chain cause the rotation of 360 degrees to last approx 2.5 sec at the ""stabilized"" speed.
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How does period of the oscillation compare between the two systems?
The doubled strap rotates a bit more quickly than the single chain setup.
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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.
A bag with more dominoes had a longer period as the chain is stretched further with added weight.
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How would you design an experiment, or experiments, to further test your hypotheses?
I would repeat the experiment with an accurate timing device and a helper to bounce the bag down a meter stick as I recorded the chain length at each bounce.
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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 doubled chain did not stretch as far, but more dominioes lengthened the chain at approximately the same ratio.
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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.
The more dominoes in the bag, the more the chain stretches, the longer the period of the system.
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How would you design an experiment, or experiments, to further test your hypotheses?
Repeat with a wider variety of ""chains"" and time against an accurate clock.
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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.
Centrifugal force causes the rubber band to stretch as it swings like a pendulum.
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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:
40,17,0
45,55,20
50,84,10
55,114.5,45
column 1 is pullback in cm, column 2 is translational displacement in cm, and column 3 is rotational displacement in degrees
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Describe what you think is happening in this system related to force and energy.
The rubber band being pulled back stores potential energy, which when released exerts its force onto the domino block, which is propelled forwards.
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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 average velocities from last time were as follows:
= 90 degrees / half-cycle
= 22.5 degrees / half-cycle
= 35.714 degrees / half-cycle
To find average change in velocity with respect to clock time, we divide the change in velocity by the change in clock time. To do this we need an initial and final velocity as well as a time interval:
a_ave = (vf - v0) / `dt
a_ave = (180 degrees/half-cycle - 0 degrees/half-cycle) / 9 half-cycles
= 20 degrees/half-cycle^2
a_ave = (45 degrees/half-cycle - 0 degrees/half-cycle) / 4 half-cycles
= 11.25 degrees/half-cycle^2
a_ave = (71.428 degrees/half-cycle - 0 degrees/half-cycle) / 7 half-cycles
= 10.204 degrees/half-cycle^2
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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.
I found the velocities:
= 80 cm / half-cyles
= 14.545 cm / half-cycle
Using the time intervals:
8 half-cycles
5.5 half-cycles
Using the same process outlined above, we divide change in velocity by change in clock time
a_ave = (vf - v0) / `dt
= (160 cm/half-cycle - 0 half-cycle) / 8 half-cycles
= 20 cm /half-cycle^2
= (29.09 cm/half-cycle - 0 half-cycle) / 5.5 half-cycles
= 5.289 cm /half-cycle^2
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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.
Doh! My signs were backwards. The car is actually decreasing in speed so the positive direciton should have the negative sign.
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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:
If each ramp is 60cm long, the change in position, or `ds = 60cm.
The clock times are given to us at 6, 4, and 3 half-cycles, these are the `dt values.
We begin by finding the average velocities of the balls on each ramp:
v_ave1 = `ds / dt
= 60cm / 6 half-cycles
= 10cm / half-cycle
v_ave2 = `ds / dt
= 60cm / 4 half-cycles
= 15cm / half-cycle
v_ave3 = `ds / dt
= 60cm / 3 half-cycles
= 20cm / half-cycle
Using the average velocity of the first ramp we can determine the final velocity easily since the ball starts at rest, or v0 = 0 cm / half-cycle, thus the final velocity is twice the average velocity.
a_ave1 = (vf - v0) / 6 half-cycles
= (20 cm / half-cycle - 0 cm/half-cycle) / 6 half-cycles
= 20 cm / half-cycle / 6 half-cycles
= 3.333 cm / half-cycle^2
Using the formula vf = [(`ds / `dt) * 2] - v0, we find the final velocities of the next two ramps. We use the final velocity of the first ramp as the initial velocity of the second, then the second for the third:
vf2 = [(`ds / `dt) * 2] - v0
= 30cm / half-cycle - 20 cm / half-cycle
= 10 cm /half-cycle
a_ave2 = (vf - v0) / 6 half-cycles
= (10 cm / half-cycle - 20 cm/half-cycle) / 4 half-cycles
= -10 cm / half-cycle / 4 half-cycles
= -2.5cm / half-cycle^2
I'm not sure a negative acceleration is appropriate here, and I'm not sure where I might have went wrong.
vf3 = [(`ds / `dt) * 2] - v0
= 40cm / half-cycle - 10 cm / half-cycle
= 30 cm /half-cycle
a_ave3 = (vf - v0) / 6 half-cycles
= (30 cm / half-cycle - 10 cm/half-cycle) / 4 half-cycles
= 20 cm / half-cycle / 3 half-cycles
= 6.667 cm / half-cycle^2
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Find the slope of each ramp.
The slope of each ramp is approx 60cm / # of dominoes, which is equal to 1 cm:
1cm / 60 cm = 0.0167
2cm / 60 cm = 0.0333
3cm / 60 cm = 0.05
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Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
3.333, 0.0167
-2.5, 0.0333
6.667, 0.05
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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 A quantity is acceleration and the B quantity is ramp slope. The slope of the line segments defines the change in acceleration with respect to ramp slope. I think my response is incorrect for the second acceleration so the ramp slopes here would be way off.
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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 initial event is the ball starting at the top of the first ramp (which I will assume is starting from rest). The final event is the ball leaing the first ramp and entering the second.
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What are the initial velocity, acceleration and displacement?
These values are all 0, in the respective units.
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Using the equations of motion find the final velocity for this interval.
vf = sqrt(v0^2 + 2 a_ave `ds)
= sqrt((0cm/s)^2 + 2 (15 cm/s^2) * 30cm)
= sqrt(2 * 15 cm/s^2 * 30cm)
= sqrt(30 cm/s^2 * 30cm)
= sqrt(900 cm^2 / s^2)
= 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.
`ds = [(vf + v0) / 2] `dt
`dt = `ds / [(vf + v0) / 2]
= 30cm / [(30cm/s + 0cm/s) / 2]
= 30cm / 15cm/s
= 2 s
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For motion down the second ramp:
What event begins the interval and what even ends the interval?
The ball leaving the first ramp and entering the second ramp begins the interval.
The interval ends with the ball reaching the end of the second ramp.
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What are the initial velocity, acceleration and displacement?
Initial velocity is the final velocity of the first ramp, 30cm/s.
Initial acceleration could be 0 or could be the average acceleration given in the preceding problem.
Initial displacement is 0 cm.
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Using the equations of motion find the final velocity for this interval.
vf = sqrt((30 cm/s)^2 + 2 * 30 cm/s^2 * 30cm)
= sqrt(900 cm^2/s^2 + 2 * 1800cm^2/s^2)
= sqrt(2700cm^2/s^2)
= 51.962 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.
`dt = `ds / [(vf + v0) / 2]
= 30cm / [(51.962 cm/s + 30 cm/s) / 2]
= 30cm / (81.962 cm/s / 2)
= 30cm / 40.981cm/s
= 0.732 s
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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?
I think it's accelaration would be the average of the first two ramps:
(15cm/s^2 + 30cm/s^2) / 2
(45cm/s^2) / 2
22.5cm/s^2
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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?
If the second ramp's acceleratoin is 22.5cm/s^2 + 5cm/s^2 = 27.5 cm/s^2, it will reach the end of the ramp in less time:
vf = sqrt(v0^2 + 2 a_ave `ds)
= sqrt((0cm/s)^2 + 2 * 27.5 * 60cm)
= sqrt(2 * 27.5 * 60cm)
= sqrt(3,300 cm^2 / s^2)
= 57.446 cm/s
`dt = `ds / [(vf + v0) / 2]
= 60cm / [(57.446cms/s) / 2]
= 60cm / 28.723 cm /s
= 2.0889 s
I'm not sure where to go from there, but I think they it would be about the midpoint, 30cm.
Homework:
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