course Phy 201
ic_class_090916Calibrate 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:
Trial 1: 1 domino, 65cm
Trial 2: 2 dominos, 69cm
Trial 3: 3 dominos, 71.5cm
Trial 4: 4 dominos, 75cm
Trial 5: 5 dominos, 78.5cm
&&&&
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.
P1(1,65) & P2(2,69): m=(69-65)/(2-1)=4
P2(2,69) & P3(3,71.5): m=(71.5-69)/(3-2)=2.5
P3(3,71.5) & P4(4,75): m=(75-71.5)/(4-3)=3.5
P4(4,75) & P5(5,78.5): m=(78.5-75)/(5-4)=3.5
&&&&
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:
Trial 1: 2 dominos, 33.5cm
Trial 2: 4 dominos, 34cm
Trial 3: 6 dominos, 36cm
Trial 4: 8 dominos, 37.5cm
Trial 5: 10 dominos, 38cm
&&&&
Graph length of doubled chain vs. number of dominoes, and calculate graph slope between each pair of points.
P1(2,33.5) & P2(4,34): m=(34-33.5)/(4-2)= .5/2= 1/4
P2(4,34) & P3(6,36): m=(36-34)/(6-4)= 2/2= 1
P3(6,36) & P4(8,37.5): m=(37.5-36)/(8-6)= 1.5/2= 3/4
P4(8,37.5) & P5(10,38): m=(38-37.5)/(10-8)= .5/2= 1/4
&&&&
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:
Single Chain:
Trial 1: 14 full rotations, 25 seconds,
Trial 2: 10 full rotations, 16 seconds
&&&&
Double the chain and repeat.
Give your raw data and your (supported) conclusions:
Trial 1: 11 full rotations, 32 seconds
Trial 2: 10 full rotations, 14 seconds
&&&&
How does period of the oscillation compare between the two systems?
In Both of these experiments we made 2 observations,through these observations we discovered that the tighter you twist the
rubberbands the faster the strap will rotate. Through these experiments we have concluded that the period changes are very
significant and the period is not constant due to the fact you could twist it to much and it rotates faster than any of the
other data.
&&&&
'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.
Trial 1: 12 bounces(from start to stop) in 10 seconds, 5 dominos
Trial 2: 24 bounces in 15 seconds, 5 dominos
Trial 3: 18 bounces in 10 seconds, 5 dominos
Based upon what I observed I feel that there isn't a natural frequency, however the length of the bounces do depend on how
many dominoes are in the bag which suggest that there is a frequency. If there was a frequency it would get larger( or
longer) based on how many Dominoes the rubberband is carrying.
&&&&
How would you design an experiment, or experiments, to further test your hypotheses?
I wouldn't really change much from what we did in class, however I would make more observations in a single trial so that I
could truly determine if there is a frequency or not. What I mean is that I would repeat the experiment agian without making
any changes so that I could confirm my observations.
&&&&
Repeat for doubled chain. How are the frequencies of doubled chain related to those of single chain, for same 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.
The number of times the dominoes bounced is close to the same, give or take a few bounces.
&&&&
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.
Trial 1: 15 bounces(from start to stop) in 10 seconds, 5 dominos
Trial 2: 20 bounces in 15 seconds, 5 dominos
Trial 3: 17 bounces in 10 seconds, 5 dominos
Based upon what I observed I feel that there isn't a natural frequency, however the length of the bounces do depend on how
many dominoes are in the bag which suggest that there is a frequency. If there was a frequency it would get larger( or
longer) based on how many Dominoes the rubberband is carrying.
&&&&
How would you design an experiment, or experiments, to further test your hypotheses?
I would make a set of experiments that changed the number of dominoes in the bag in each trial.
&&&&
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 would definately change because the weight of the dominos would tug on the rubberband as they were falling
which would cause the rubberband to stretch and would therefore increase its length.
&&&&
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:
Trial 1: Pull back distance: 6cm, rotational displacement: 180 degrees, Translational displacement: 27cm.
Trial 2: Pull back distance: 6cm, rotational displacement: 270 degrees, Translational displacement: 31cm.
Trial 3: Pull back distance: 10cm, rotational displacement: 380 degrees, Translational displacement: 49cm.
Trial 4: Pull back distance: 15cm, rotational displacement: 720 degrees, Translational displacement: 115cm.
We used 3 dominos tied together for each trial.
&&&&
Describe what you think is happening in this system related to force and energy.
Your body is using energy to create the force necessary to pull the rubberband back into place. The rubbeband is then
building tension in its lines which is causing it to pull back, thus there is a transfer of energy. When we released the
rubberband the energy that was being created by the tension is released at once pulling the dominos then lauching them across
the table with force created by the energy, which is created by the tension.
&&&&
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:
&&&&
(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: (11.4cm/ halfcycle)/ 7 halfcycles= 1.63cm/halfcycle^2
Trial 2: (17.8cm/ halfcycle)/ 4.5 halfcycles= 3.95cm/halfcycle^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:
Ramp 1: vAve= 60cm/ 6 halfcycle= 10cm/halfcycle, aAve= (10cm/halfcycle -0cm/halfcycle)/6 halfcycles= 1.6cm/halfcycle^2
Ramp 2: vAve= 60cm/ 4 halfcycle= 15cm/halfcycle, aAVe= (15cm/halfcycle -0cm/halfcycle)/6 halfcycles= 3.75cm/halfcycle^2
Ramp 3: vAve= 60cm/ 3 halfcycle= 20cm/halfcycle, aAve= (20cm/halfcycle -0cm/halfcycle)/6 halfcycles= 6.66cm/halfcycle^2
First I figured out the average Velocity of each ramp. Then I subtracted the initial velocity from the final velocity and
divided that by the clock time to find the average acceleration.
good, except that you used vAve for vf
&&&&
Find the slope of each ramp.
Ramp 1: m=1/60
Ramp 2: m=1/30
Ramp 3: m=1/10
&&&&
Graph acceleration vs. ramp slope. Your graph will consist of three points. Give the coordinates of these points.
P1: (1.6, 1/60)
P2: (3.75, 1/30)
P3: (6.66, 1/10)
acceleration vs. ramp slope would have ramp slope as the first coordinate, acceleration as the second
&&&&
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.
P1 & P2: [(1/30)-(1/60)]/(3.75-1.6)= 0.00775
The slope represents the average rate of change of slope by the change in acceleration
P2 & P3: [(1/10)-(1/30)]/(6.66-3.75)= .023
The slope represents the average rate of change of slope by the change in acceleration
&&&&
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?
I think that it begins when the ball is made to role down the ramp by a force and it ends when the ball roles on to the next
ramp at an acceleration of 15cm/s^2. (If I'm wrong please explain this too me.)
&&&&
What are the initial velocity, acceleration and displacement?
The initial velocity is 0cm/s, the acceleration is 15cm/s^2, and I guess the displacement is the length of one ramp, 30cm.
&&&&
Using the equations of motion find the final velocity for this interval.
First we need to find time: using the 3rd equation 'dp= Vo*'dt +(a*'dt^2)/2 , 30cm=(0cm/s * 'dt) +(15cm/s^2 * 'dt^2)/2
(since Vo is being multiplied by 'dt, 'dt is canceled because Vo is 0cm/s), 2* 30cm=(15cm/s^2 * 'dt^2)/2 *2, 60cm= 15cm/s^2 * 'dt^2, 60cm/(15cm/s^2)='dt^2, 4s^2='dt^2, 4seconds='dt
the square root of 4 s^2 is 2 s, not 4 s
Second we can now find the final velocity: By using the 2nd equation, aAve=(Vf-Vo)/'dt, 15cm/s^2=(Vf- 0cm/s)/4seconds,
15cm/s^2 *(4 seconds)= Vf, Vf=60cm/s
&&&&
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
using the 3rd equation 'dp= Vo*'dt +(a*'dt^2)/2 , 30cm=(0cm/s * 'dt) +(15cm/s^2 * 'dt^2)/2 (since Vo is being multiplied
by 'dt, 'dt is canceled because Vo is 0cm/s), 2* 30cm=(15cm/s^2 * 'dt^2)/2 *2, 60cm= 15cm/s^2 * 'dt^2,
60cm/(15cm/s^2)='dt^2, 4s^2='dt^2, 4seconds
The time spent on the first ramp was 4 seconds.
&&&&
For motion down the second ramp:
What event begins the interval and what even ends the interval?
It begins when it roles off the first ramp at 15cm/s^2, and it ends when it hits the end of the second ramp at 30cm/s^2.
&&&&
What are the initial velocity, acceleration and displacement?
The initial velocity is 60cm/s, the acceleration is 30cm/s^2, and the dispacement is 30cm.
&&&&
Using the equations of motion find the final velocity for this interval.
using the fourth equation: Vf^2= Vo^2 +2*aAve * 'ds, Vf^2= (60cm/s)^2 + 2*(30cm/s^2)(30cm), Vf= 73.48 cm/s
&&&&
Using the final velocity with the other information about this interval, reason out the time spent on the first ramp.
using the second equation: 30cm/s^2= (73.48cm/s -60cm/s)/'dt, 'dt= (73.48cm/s -60cm/s)/(30cm/s^2), 'dt= approximately .5
seconds
well done; the 4 s vs. 2 s led to some wrong conclusions but they could easily be corrected.
&&&&
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?
&&&&
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?
Very good work; see my notes on a couple of easily-corrected errors.