course phy 201
20090921 0225
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, 71.4 cm
2, 75.2 cm
3, 79.5 cm
4, 84.05 cm
5, 90.3 cm
The measurements were obtained from placing a paperclip at the top of a meter stick and letting the paperclip pertrude beyond
in a
matter so that the rubber bands could hang from the paperclip freely about 4 cm from the meter stick. Once the dominoes
supported by
the rubber band chain quit bouncing, the paperclip supporting the bag was carefully pushed against the ruler to get a precise
measurement.
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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.
71.4cm, 1 dominoe, 71.4 cm/dominoe (slope)
75.2cm, 2 dom, 37.6 cm/dom
79.5cm, 3 dom, 26.5 cm/dom
84.05cm, 4 dom, 21.012 cm/dom
90.3cm, 5 dom, 18.06 cm/dom
you haven't calculated rise/run from point to point; you are calculating y coordinate / x coordinate, not change in y coordinate / change in x coordinate
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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, 35.3 cm
4, 37.5 cm
6, 39.7 cm
8, 41.9 cm
10, 44.4 cm
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Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
Graphed and calculated slopes for each pair of points.
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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:
101 seconds
146 seconds
172 seconds
191 seconds
216 seconds
237 seconds
255 seconds
288 seconds
The strap was first rotated five 360 degrees and then released. Each recording above represents the time in which the strap
changed
angular direction. Experiment was concluded once 8 intervals were obtained to get an idea of the rate in which the periods
were
changing.
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Double the chain and repeat.
Give your raw data and your (supported) conclusions:
35 seconds
65 seconds
97 seconds
126 seconds
153 seconds
180 seconds
204 seconds
225 seconds
As before, the strap was rotated five 360 degrees and then released. This experiment did not keep a parrallel plane as the
before
experiment did, was slanted maybe 15-20 degrees. Again 8 intervals were observed and recorded with a running time from
0'00''
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How does period of the oscillation compare between the two systems?
They both seemed to have run almost a similiar period relatively but the doubled chain would of ended earlier than the single
chain.
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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.
10 dominoes, 20 seconds
8 dom, 15 s
6 dom, 12 s
4 dom, 10 s
2 dom, 6 s
Would have to conclude, first, the inaccuracies of this experiment. While observing the chain bouncing, the time in which if
quit
bouncing is inaccurate .5 second to 1 second due to the fact of not being able to physically being able to see when the chain
truely
quit bouncing. However, with our data, it is safe to hypothesis that the more dmonoes that are added to the chain, the more
time it
will require the system to come to rest from a bounce. Also, a good control we maintained was the height of the drop (which
was
measured from the top of the chain to about half way in the chain - constant measure was from the bottom of the table)
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How would you design an experiment, or experiments, to further test your hypotheses?
Would continue to test multiple observations with not only the same number of dominoes but also a different set of numbers of
dominoes and compare.
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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?
:( This data was not recorded or observed.
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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.
n/a
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How would you design an experiment, or experiments, to further test your hypotheses?
same as before, continue to experiment with different numbers of dominoes.
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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.
I would expect the length of the chain to be greatest at it's midpoint (or center of it's half cycles) due to the face of
gravity
having it's greatest effect on the rubber bands at this point. Also, just like game of tetherball, the ball is applying more
force
on the string it's attached to at the beginning of it's swing. I can relate this to the rubber band pendulum's length even
though
cannot fully describe as much as I understood.
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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, 90 cm to 33 cm, 90 degrees
40 cm, 100 to -26 cm, 270 degrees
35 cm, 0 cm to 67 cm, 0 degrees
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Describe what you think is happening in this system related to force and energy.
Energy and force is being used to create potential (stored) energy in the rubber band system. This energy is then converted
to
kinetic energy and elastic energy which in return uses its force to push the block down the table. Some energy is lost from
the
system due mainly to thermal energy (friction).
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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:
Trial 1 - ave acceleration = ((0(vf)-(-270degrees/(12*.667)(v0))*2)/(12*.667) ---> a = -8.429 degrees/sec^2
T2 - ave accel. = -10.36 deg/s^2
T3 - ave accel. = -5.396 deg/s^2
T4 - ave accel. = -7.645 deg/s^2
T5 - ave accel. = -8.966 deg/s^2
1/2 osc - .667 sec --- seems as if the differences in these shows either human errors in the experiment or errors with the
experiment
itself (eg the strap sitting proportionately even on the di). The acceleration is negative due to the system slowing down,
not
speeding up. And I would of said deacceleration but you having cleared/informed of it's impossibility in a previous semester
changed
my views and ideas in that matter.
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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.
Trial 1 - ave acceleration = (80 cm/(4cycles*1.333s)*2(find ave vel so multiply by 2 to find vf, so I believe but do not
agree on my
reasoning)/(4cycles*1.333s) ---> ave acceleration = 7.260 cm/s^2
T2 - ave accel = 11.91 cm/s^s
T3 - ave accel = 22.51 cm/s^s
1 cycle = 1.333 seconds
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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 are 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.
_____
3 -----
2 ``````
1 0
Lr = 60 cm, Hd = 1 cm, 1t = 6(half-cycles), 2t = 4(half-cycles), 3t = 3(half-cycles)
Assuming constant acceleration on each ramp, find the average acceleration on each. Explain the details of your calculation:
Ramp 1 - ave accel = (20cm/(6hc)*2/6hc = 1.11 cm/half-cycles
Ramp 2 - ave accel = (20cm/(4hc)*2/4hc = 2.50 cm/half-cycles
Ramp 3 - ave accel = (20cm/(3hc)*2/3hc = 4.44 cm/half-cycles
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Find the slope of each ramp.
Each ramp should have the same slope if setup evenly (correctly). Slope = 1cm(change in height)/20cm(change in run) = 1/20
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Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
(.05,1.11)(.05,2.50)(.05,4.44)
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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.
With these points, a slope cannot be determined. The connecting line only shows the variation in accelerations on the y axis
(going
straight up) and so does not show any relationship between acceleration and the constant ramp slope.
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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? when the ball is realeased about the left end(or top) of the
ramp and
then once it reaches the right end(bottom) of the 1st ramp/begins the 2nd ramp.
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What are the initial velocity, acceleration and displacement?
v0 = 0, a = 15cm/s^2, `ds = 30cm
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Using the equations of motion find the final velocity for this interval.
vf^2 = v0^2 + 2s`ds
vf^2 = 0cm/s + 2(15cm/s^2)30cm
vf^2 = 900cm^2/s^s
vf = 30cm/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
2`ds/(vf+vo) = `dt
`dt = 2(30cm)/(30cm/s+0cm/s)
`dt = 60cm/30cm/s
`dt = 2 seconds
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For motion down the second ramp:
What event begins the interval and what even ends the interval? beginning - point when the balls leaves the first ramp and
enters the
second ramp ends - point in which the ball leave the second ramp
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What are the initial velocity, acceleration and displacement?
v0 = 30cm/s, acceleration = 30cm/s^2, `ds = 30cm
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Using the equations of motion find the final velocity for this interval.
vf^2 = v0^2 + 2a`ds
vf^2 = (30cm/s)^2 + 2(30cm/s^2)30cm
vf^2 = 900cm^2/s^2 + 1800cm^2/s^2
vf = 51.96 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.
Will assume this if for the second ramp.
`dt = 2(30cm)/(51.96cm/s+30cm/s)
`dt = 60cm/(81.96cm/s)
`dt = .732 seconds
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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?
a = (vf-v0)/`dt
a = (51.96 cm/s - 0 cm/s)/2.732 seconds
a = 19.02 cm/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?
acceleration before 19.02 cm/s^2 and increases 5 cm/s^2 so new acceleration will be 24.02 cm/s^2. We still know that v0 is 0
and
that `ds is 60 cm.
vf^2 = v0^2+2a`ds (2(24.02cm/s^s)60cm)
vf = 53.69 cm/s
Not sure of how you are asking the question, for the ball will almost instantly pass the 2-ramped ball.
`dt = 2`ds/(vf+v0)
`dt = 2(60cm)/(56.69cm/s)
`dt = 2.117 s
Therefore, the ball on this 60 cm ramp will reach the end (2.732s-2.117s) .615 seconds before the two 30-cm ramped ball will.
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You're in good shape. You did make an error or two so see my notes.