course 201
8/21 1am
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.
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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:
5, 71
4, 68.7
3, 66
2, 63
1, 60.5
The length of the rubber band chain was calculated in centimeters
&&&&
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.
Number of dominoes, band length in cm, rise (`dband length), run (`dtdominos), slope
5,71
4, 68.7, 2.3, 1, 2.3
3, 66, 2.7, 1, 2.7
2, 63, 3, 1, 3
1, 60.5, 2.5, 1, 2.5
&&&&
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:
10, 35.5
8, 34
6, 32
4, 31.5
2, 30
Length of chain was measured in centimeters
&&&&
Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
Number of dominoes, band length in cm, rise (`dband length), run (`dtdominos), slope
10, 35.5
8, 34, 1.5, 2, .75
6, 32, 2, 2, 1
4, 31.5, .5, .25
2, 30, 1.5, .75
&&&&
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:
6.75 rotations (360 degrees) forward (clockwise) at 25 seconds on stopwatch
9.75 rotations backward at 98 seconds
1.75 rotations forward at 135 seconds
.75 rotations backward at 166 seconds
The period was not constant because as the system spun in one direction, the rubber bands twist together and get to a point
at which they want to unwind. This pushes the system to slow and eventually spin in the opposite direction, but it has lost a
lot of momentum. Each back spin is slower than the previous. In the first rotation forward, the system had about 1 rotation
every 3.7 seconds.(25 seconds/6.75 rotations=3.7 seconds/rotation) In the first rotation backward, the system had slowed to
one rotation every 7.5 seconds. (98 seconds-25 seconds=73 seconds for this direction, 73 seconds/9.75 rotations=7.5
seconds/rotation) The second rotation forward slows even further to 21.1 seconds for one rotation. (135 seconds-98 seconds=37
seconds for this direction, 37 seconds/1.75 rotations=21.1 seconds/rotation) The last rotation backward, which was not even a
full rotation, was even slower at 41.3 seconds per rotation. (166 seconds-135 seconds=31 seconds, 31 seconds/.75
rotations=41.3 seconds/rotation) The system operated with the following velocities: forward 3.7 sec/rot, backward 7.5
sec/rot, forward 21.1 sec/rot, backward to stop at 41.3 sec/rot.
Double the chain and repeat.
Give your raw data and your (supported) conclusions:
4 rotations forward at 14 seconds on stopwatch
5.75 rotations backward at 38 seconds
3 rotations forward at 58 seconds
1.75 rotations backward at 84 seconds
.5 rotations forward at 100 seconds
.5 rotations backward at 122 seconds
.25 rotations forward at 139 seconds
Again, the period was not uniform. The difference in this situation, the shortened rubber band chain, lessened both the
maximum amount of rotations in each direction the strap turned and the total time for the system to come to a stop. It did,
however, increase the number of cycles of forwards and backwards spins. I would attribute this to the shorter amount of space
for the rubber band chain to twist, giving less time before it was too tightly wound and forced the system to spin in the
opposite direction. Using the same calculations as above, I found the individual velocities of the trials to be forward 3.5
sec/rotation, backward 4.2 sec/rot, forward 6.7 sec/rot, backward 14.9 sec/rot, forward 32sec/rot, backward 44 sec/rot,
forward 68 sec/rot.
&&&&
How does period of the oscillation compare between the two systems?
The two systems seem to operate similarly, but the second system was able to increase its number of periods forward and
backward overall in a less amount of time to a stop than the first. I would attribute this to the shorter rubber band chain.
The strap was not able to rotate as many times before the tension on the twisted chain was too much, forcing the strap to
stop in spin in the opposite direction. But since I gave both systems about the same amount of power initially, the second
system was able to make it forward and backward more times.
&&&&
'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 rubber band chain supporting 2 dominos- 9 bounces,
4 dominos- 10 bounces
8 dominos- 20 bounces
The rubber band chain hung from a ruler on the table, with the plastic bag of dominoes about 10 inches above floor, and the
bag was pulled to floor level and released to start the bouncing.
I would say that there is a natural frequency, but that there has to be enough weight to keep the system going back and
forth. The initial pull will propel the smaller bags of dominos to bounce for a short amount of time, but as there is not
much weight to pull the rubber bands to the floor on the down fall, the bounces start to even out and stop. In the trial
with more dominos, the initial pull makes the rubber bands stretch out, and when you let go they contract in again. However
gravity the pulls the bag of rubber bands back closer to the floor, and the rubber bands have to stretch back out. It is a
much more even situation.
&&&&
How would you design an experiment, or experiments, to further test your hypotheses?
I would experiment with a system that had longer, stronger rubber band chains, and use a more substantial amount of weight,
which I would vary with each trial. I would then try to find out if there were patterns in the results.
&&&&
Repeat for doubled chain. How are the frequencies of doubled chain related to those of single chain, for same number of
dominoes?
Time did not allow for us to get these calculations
&&&&
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.
&&&&
How would you design an experiment, or experiments, to further test your hypotheses?
&&&&
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.
Yes, the length does change. This was illustrated to us in class using a stack of dominos that they system did not touch at
equilibrium when it was still, but knocked over when it was set in motion and passed through equilibrium. I would attribute
this to the weight of the rubber band bag on the end, and the force of gravity. When the rubber band bag is pulled back to
one extreme of the pendulum cycle, and then let go, it free falls and gravity pulls it closer to the ground faster than it
would if the bag was not moving. The pendulum length is then closest to the ground as it passes through equilibrium, and
farthest when it reaches the outer extremes and almost gets a bounce higher because of the rubber bands contracting back. It
then is pulled back down as it swings towards equilibrium.
&&&&
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 cm, 140 cm, 90 degrees
30 cm, 43 cm, none
35 cm, 70 cm, 30 degrees
45 cm, 120 degrees, 180 cm
&&&&
Describe what you think is happening in this system related to force and energy.
We held the rubber band ends lab table width apart and measured the length of the pull back from the middle of the rubber
band chain using a meter stick. We placed a stack of dominos in front of the point we pulled back, and when we let go they
were pushed forward, like it was in a slingshot. There was a large difference in the distance the domino stack was able to
cover corresponding to how many centimeters the rubber band was pulled back. Any less than thirty centimeters was barely
readable, while any more than 45 centimeters shot the dominos further than our lab table. The system worked by putting
tension in the rubber bands, stretching them to the point that they contracted quickly when we let go. This quick snap back
projected the dominos forward. The force is given to the system on the pullback, and the energy comes when the system of
rubber bands propels the dominos.
&&&&
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:
V0= 2*average velocity because the system comes to rest at vf=0
#1- `ds/`dt=(sf-s0)/(tf-t0)= (160 degrees- 0 degrees)/(4 half cycles-0 half cycles)=160 degrees/4 half cycles=40 degrees/half
cycle
`dv=vf-v0=0 degrees/halfcycle- 80 degrees/halfcycle=-80 degrees
`dv/`dt=-80 degrees/halfcycle / 4 half cycles = -20 degrees/halfcycle^2
#2-`ds/`dt=185 degrees/8 half cycles=23.1 degrees/half cycle
`dv/`dt=-46.2 deg/hc /8 hc= -5.8 deg/hc ^2
#3-`ds/`dt=85 degrees/4 half cycles=21.3 degrees/half cycle
`dv/`dt= -42.6 deg/hc / 4 hc= -10.7 deg/hc^2
¬#4-`ds/`dt=380 degrees/12 half cycles=31.7 degrees/half cycle
`dv/`dt= -63.4 deg/hc / 12 hc= -5.3 deg/hc ^2
#5- `ds/`dt=60 degrees/3 half cycles=20 degrees/half cycle
`dv/`dt= -40 deg/hc / 3 hc= -13.3 deg/hc^2
#6-`ds/`dt=110 degrees/5 half cycles=22 degrees/half cycle
`dv/`dt= -44 deg/hc / 5 hc = -8.8 deg/ hc^2
&&&&
(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.
V0= 2*average velocity because the system comes to rest at vf=0
First trial with system as is: `ds/`dt= 80 cm/10 half cycles= 8 cm/half cycle
`dv= vf-v0=0-2*8cm/halfcycle= -16 cm/hc
`dv/`dt=-16 cm/hc/10 hc= -1.6 cm/hc^2
First(and second) trial with rubber band added to system: `ds/`dt= 80 cm/7 half cycles=11.4 cm/half cycle
`dv/`dt= -22.8 cm/hc / 7 hc= 3.4 cm/hc^2
&&&&
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.
&&&&
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:
First ramp:
First we have to find average velocity, which is `ds/`dt. 60cm/3ramps=20 cm for each ramp, so `ds/dt=20 cm/ 6 half cycles=
3.33 cm/hc
Then to find average acceleration, we need `dv, which is vf-v0. Since v0=0, it can be assumed that vf=2*average velocity. So
`dv/`dt= 2*3.33 cm/hc / 6 half cycles = 1.11 cm/hc^2
Second ramp:
`ds/`dt=20 cm/4 hc=5 cm/hc
`dv/`dt= 10 cm/hc / 4 hc= 2.5 cm/hc
Third ramp:
`ds/`dt= 20 cm/ 3 hc= 6.67 cm/hc
`dv/`dt= 13.33 cm/hc / 3 hc= 4.44 cm/hc^2
&&&&
Find the slope of each ramp.
First ramp:
Slope =rise/run=`dy/`dx=3cm/20 cm=.15
Second ramp:
`dy/`dx=2cm /20 cm=.1
Third ramp:
`dy/`dx=1cm/20 cm=.05
&&&&
Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
(1.11, .15), (2.5, .1) (4.44, .05)
are your slopes in the same order as your accelerations?
&&&&
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 = `dy/`dx= .1-.15/2.5-1.11= -.05/1.39= -.04
Slope= `dy/`dx= .05-.1/4.44-2.5=-.05/1.94= -.03
The situation we have calculated would be the average rate of change of acceleration with respect to ramp slope. This would
seem to show the amount the acceleration differs based on each unit of ramp slope.
&&&&
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?
S0=0cm, sf=30 cm
What are the initial velocity, acceleration and displacement?
V0=0, a= 15 cm/s^2, `ds=30 cm
Using the equations of motion find the final velocity for this interval.
Vf^2= v0^2 +2a`ds
Vf^2=0^2 + 2(15cm/s^2)(30 cm)
Vf^2=0+900 cm^2/s^2
Vf^2=900 cm^2/s^2
Vf=900 cm/s
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
If the final velocity of the ball was 900 cm/s , its average velocity would be half that, or 450 cm/s. Since average velocity
can also be expressed as `ds/`dt, 450 cm/s=30 cm/`dt, 450 cm/s *`dt=30 cm, `dt=30 cm/450 cm/s, `dt=.067 seconds
For motion down the second ramp:
What event begins the interval and what even ends the interval?
S0=30 cm, sf=60 cm
What are the initial velocity, acceleration and displacement?
Initial velocity=900 cm/sec, a=30 cm/sec^2, `ds=30 cm
Using the equations of motion find the final velocity for this interval.
Vf^2= v0^2 +2a`ds
Vf^2=900 cm/sec^2 +2*30 cm/sec^2*30 cm
Vf^2=810000 cm^2/sec^2 + 1800 cm^2/sec^2
Vf^2=811800 cm^2/sec^2
Vf=901 cm/sec
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
If the final velocity of the ball was 901 cm/s , its average velocity would be half that, or 450.5 cm/s.
The initial velocity on the second ramp isn't zero, so its average velocity wouldn't be 450 cm/s.
Since average
velocity can also be expressed as `ds/`dt, 450.5 cm/s=30 cm/`dt, 450.5 cm/s *`dt=30 cm, `dt=30 cm/450.5 cm/s, `dt=.067
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?
30 cm/sec^2
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?
Homework:
Your label for this assignment:
ic_class_090916
You're in great shape. There are a couple of minor errors, so do see my notes.