course PHY 201
12:11 AMMonday, September 28, 2009
Sorry for the late submission, it's been a rough week." "Class 090923
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Thanks.
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In this assignment you will be asked to
do a little more with your rubber band calibration graph, which you constructed previously
find or confirm the correct accelerations for the previously done Atwood and ball-down-ramp experiments
do a little more with the rubberband-chain force diagrams you sketched in class, and learn a little terminology
plug a = F_net / m into the second and fourth equation of motion, and do your best with a little algebra
Trapezoidal Graph of Rubber Band Calibration Data
When you calibrated your rubber band chain you observed the position of the end of the chain with various numbers of dominoes.
Plot the information on a graph of y = number of dominoes vs. x = position of end. Your graph will consist of five points.
Give the coordinates of each of your five points; be sure to include units.
65,1
69.5,2
73,3
77,4
79,5
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Now use your points to make a series of trapezoids:
Connect your five points with line segments. There will be four line segments, each with a slope you calculated in response to previous class notes.
From each point sketch the 'vertical' line segment from that point to the horizontal axis. You will sketch five line segments, each representing the number of dominoes suspended in the corresponding trial. Having sketched these segments, you will have constructed a series of four trapezoids.
The graph you have constructed will be called a 'trapezoidal approximation graph'.
Briefly describe your graph in words
It looks like half of a suspension bridge. :)
I have 5 points, each with vertical ""legs"" that reach the x-axis, and line segments between each point, forming 4 graph trapezoids. The graph ""moves"" from the bottom left to the upper right.
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Find the area of each trapezoid.
Label the slope of each trapezoid by placing it in a rectangular 'box' just above the 'slope segment' at the top of the trapezoid.
Label the area of each trapezoid writing it inside the trapezoid and circling it.
What are the four slopes?
slope = rise / run
= (2 - 1) / (69.5 - 65)
= 1 / 4.5
= 0.222
= 1 (73 - 69.5)
= 1 / 3.5
= 0.286
= 1 / (77 - 73)
= 1 / 4
= 0.25
= 1 / (79 - 77)
= 1 / 2
= 0.5
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What are the four areas?
area = [(a1 + a2) / 2] * b
= [(2+ 1) / 2] * 4.5
= 6.75
= [(3 + 2) / 2] * 3.5
= 8.75
= [(4+ 3) / 2] * 4
= 14
= [(5 + 4) / 2] * 2
= 9
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What does each slope represent (explain what the rise represents, what the run represents, and what the slope therefore represents; be sure to include units with each quantity):
The rise is the change in the number of dominoes, which is in units of dominoes.
The run is the change in length of the rubber band, in units of centimeters.
The slope, therefore is rise / run, or the change in dominoes divided by the change in rubber band length, whose units would be in dominoes / cm.
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What does each area represent (explain what the altitude of the equal-area rectangle represents, what the base represents, and what the area therefore represents).
The altitude of the equal-area rectangle is the average number of dominoes for the interval.
The base represents the change in rubber band length.
The area is then rubber band length times average change in dominoes, which I think would be force in cm * domino.
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Your correct accelerations for lab situations:
In a number of lab situations acceleration is assumed constant, and we observe the time needed for an object to undergo a known change in position, either starting or ending at rest.
To solve the motion in such a situation we the following procedure will always work efficiently:
Apply the definition of average velocity.
Sketch the v vs. t trapezoid.
Use the trapezoid to determine the initial and final velocity of the object on the interval.
Apply the definition of average acceleration.
One very common error is to divide average velocity by time interval (vAve / `dt). This is not a calculation that comes up when following the above procedure, and it isn't a calculation that tells us anything important about the motion. Be sure you understand why this calculation doesn't happen in the given procedure.
Another common error is for students to double the average velocity to get the final velocity. For the case where initial velocity is zero and acceleration is constant, this is actually done. However this is a result of the fact that the v vs. t trapezoid is in this case a triangle, and is not something we generally do. You shouldn't even think about doubling the average velocity, and certainly shouldn't get into the habit of doing so. Draw the trapezoid and do as the drawing dictates.
Being very sure you analyze the motion correctly, please report the following. If you've done some or all of these correctly, you can insert a copy (or copies) from your previous document(s):
The data and accelerations of the three trials for the Atwood machine:
80cm displacement for each
1 rubber band: 5 half-cycles
`ds / `dt = 80cm / 5 half-cycles = 16 cm/half-cycle
a_ave = (vf - v0) / `dt
= 32 cm / half-cycle / 5 half-cycles
= 6.4 cm/half-cycle^2
2 rubber bands: 2.5 half-cycles
`ds / `dt = 80cm / 2.5 half-cycles = 32 cm/half-cycle
a_ave = 64 cm / half-cycle / 2.5 half-cycles
= 25.6 cm/half-cycle^2
3 rubber bands: 2 half-cycles
`ds / `dt = 80cm / 2 half-cycles = 40 cm/half-cycle
a_ave = 80 cm / half-cycle / 2 half-cycles
= 40 cm/half-cycle^2
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The two slopes obtained when you graph acceleration vs. number of added rubber bands for the Atwood machine:
slope = rise / run
= (25.6 - 6.4) / (2 - 1)
= 19.2 / 1
= 19.2
= (40 - 25.6) / (3 - 2)
= 14.4 / 1
= 14.4
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The coordinates of the three points you graphed:
(1, 6.4)
(2, 25.6)
(3, 40)
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The data and accelerations for the three trials (domino flat, on long side, on short side) for the ball-down-ramp experiment:
Pendulum used for flat domino: 25cm
Flat domino time interval: 2.5 half-cycles = 1.25 cycles = 1.25 seconds
Flat domino ramp slope: 0.0333
Flat domino v_ave: 24 cm/s
Flat domino a_ave: 38.4 cm/s^2
Pendulum used for side domino: 20cm
Side domino time interval: 2.5 half-cycles = 1.25 cycles = 1.1175 s
Side domino ramp slope: 0.0833
Side domino v_ave: 26.846 cm/s
Side domino a_ave: 47.152 cm/s^2
Pendulum used for tall domino: 20cm
Tall domino time interval: 1.5 half-ycles = 0.75 cycles = 0.671 s
Tall domino ramp slope: 0.167
Tall domino v_ave: 44.709 cm/s
Tall domino a_ave: 133.261 cm/s^2
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The two 'graph slopes' obtained when you graph acceleration vs. ramp slope:
= (47.152 - 38.4) / (0.0833 - 0.0333)
= 8.752 / 0.05
= 175.04
= (133.261 - 47.152) / (0.167 - 0.0833)
= 86.109 / 0.0837
= 1,028.781
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The coordinates of the three points you graphed:
38.4, 0.0333
47.152, 0.0833
133.261, 0.167
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Give a detailed description of how you proceeded from raw data to acceleration for one of the Atwood Machine trials:
Given the raw data of a displacement, or `ds, and a clock time, I proceeded to calculuate average velocity with `ds / `dt. I then doubled this figure for final velocity (since it begins at rest, or v0 = 0), and used it to determine average acceleration with the formula a_ave = (vf - v0) / `dt.
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Give a detailed description of how you proceeded from raw data to acceleration for one of the ball-down-ramp trials:
I used the same process. Given a change in position of 30cm on the ramp, and a clock time, I divided `ds / `dt to arrive at the average velocity. Using this figure, I solved for a_ave = (vf - v0) / `dt, where vf was double the average velocity and v0 was 0.
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Newtons Second Law
Here are a few fairly obvious statements.
It takes more force to accelerate lots of mass than just a little mass.
The more acceleration you want the more force you need.
More force implies greater acceleration.
All these statements are a bit imprecise. More precise statements require more words, with a potential for confusion. So were going to work from the imprecise statements that first form our ideas, to the precise statements on which we base the actual science.
More precise and accurate versions of the above statements might look like the following:
To give two different masses the same acceleration, the greater mass requires the greater force.
To give a certain mass a greater acceleration, you need to apply a greater force.
If you apply the same force to two different masses, the lesser mass will have the greater acceleration.
The last statement 'If you apply the same force to two different masses, the lesser mass will have the greater acceleration' is not really so.
For example you can apply all sorts of force to a merry-go-round and it never goes anywhere, it just rotates, with the parts near the rim moving faster than the parts near the center.
So for now we're going to remember that real objects can rotate, but we're going to confine our attention to situations that dont involve rotation.
It an object doesn't rotate when a force is applied to it, we say that it is acting like a particle.
Particles dont rotate. They only translate.
We can (and will) verify by experiment that the following statement holds. This is Newton's Second Law:
The net force exerted on a particle is the product of its mass and its acceleration:
F_net = m a.
You need to memorize this entire statement, along with the definitions of average rate, average velocity, average acceleration, the equations of uniformly accelerated motion and the interpretation of a v vs. t trapezoid.
The net force on a particle is what you get if you combine all the forces acting on it.
A couple of important consequences of Newton's Second Law:
If its acceleration is zero then the net force on it is zero.
If its sitting still then its acceleration is zero.
If its moving at constant speed in a constant direction then its acceleration is zero.
Otherwise its acceleration isnt zero and the net force on it is not zero.
Substituting a = F_net / m into the second and fourth equations of motion:
If we solve Newtons Second Law for acceleration we get
a = F_net / m
If substitute F_net / m for a in the second equation of uniformly accelerated motion, vf = v0 + a `dt, what equation do we get?
vf = v0 + (F_net / m) `dt
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We can rearrange your equation to get
F_net * `dt = m vf m v0
Show, as best you can at this point, the steps needed to get from your previous answer to this form:
vf = v0 + (F_net / m) `dt
vf - v0 = (F_net / m) `dt
(vf - v0) * `dt = F_net / m
m * [(vf - v0) / `dt) = F_net
F_net = m * [(vf - v0) / `dt)
F_net = (m vf - m v0) / `dt
F_net * `dt = (m vf - m v0)
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If substitute F_net / m for a in the fourth equation of uniformly accelerated motion, vf^2 = v0^2 + 2 a `ds, what equation do we get?
vf^2 = v0^2 + 2 (F_net / m) `ds
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We can rearrange your equation to get
F_net * `ds = ½ m vf^2 ½ m v0^2.
Show, as best you can at this point, the steps needed to get from your previous answer to this form:
vf^2 = v0^2 + 2 (F_net / m) `ds
vf^2 - v0^2 = 2 (F_net / m) `ds
(vf^2 - v0^2) / 2`ds = (F_net / m)
m [(vf^2 - v0^2) / 2`ds] = F_net
F_net = m [(vf^2 - v0^2) / 2`ds]
F_net = (m vf^2 - m v0^2) / 2`ds
F_net * `ds = (m vf^2 - m v0^2) / 2
F_net * `ds = 1/2 m vf^2 - 1/2 m v0^2
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Impulse, momentum, energy, work
The quantities F_net * `ds, F_net * `dt, 1/2 m v^2 and m v arise naturally when a = F_net / m is substituted into the second and fourth equations of motion.
We give these quantities names, which you should learn immediately:
F_net * `ds is the work done by the net force F_net on an interval for which the displacement is `ds
F_net * `dt is the impulse of the net force F_net on an interval of duration `dt.
1/2 m v^2 is the kinetic energy of a particle of mass m moving at velocity v.
m v is the momentum of a particle of mass m moving at velocity v.
Our substitutions give us the two equations
F_net * `dt = m vf - m v0
and
F_net * `ds = ½ m vf^2 ½ m v0^2.
Both of these equations apply to the behavior of a particle of mass m on an interval.
F_net * `dt is the impulse of F_net on the interval.
m vf is the momentum of the particle at the end of the interval
m v0 is the momentum of the particle at the beginning of the interval
so m vf - m v0 is the change in the particle's momentum on that interval
F_net * `dt = m vf - m v0 therefore states that, on the interval in question,
the impulse of the net force acting on the particle is equal to the change in its momentum.
This is called the impulse momentum theorem.
F_net * `ds is the work done by the force F_net on the interval.
1/2 m vf^2 is the kinetic energy of the particle at the end of the interval
1/2 m v0^2 is the kinetic energy of the particle at the beginning of the interval
so m vf - m v0 is the change in the particle's kinetic energy on that interval
F_net * `ds = ½ m vf^2 ½ m v0^2 therefore states that, on the interval in question,
the work done by the net force acting on the particle is equal to the change in its kinetic energy.
This is called the work-kinetic energy theorem.
Briefly explain in your own words how we get the definitions of impulse, momentum and the impulse-momentum theorem.
Impulse is the net force times a time interval `dt. Momentum is mass m times times velocity v. The impulse-momentum theorem is the idea that the net force over a time interval (impulse) is equal to the momentem of a particle moving at a velocity (momentum) -- or rather, the change in momentem over an interval.
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Briefly explain in your own words how we get the definitions of work, kinetic energy and the work-kinetic energy theorem.
Work is net force times a displacment interval, `ds. Kinetic energy is the energy of motion, determined by a particle's mass and velocity. The work-kinetic energy theorem sets these two values equal to one another.
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Force vectors
The three arrows you drew to depict the forces exerted by the rubber bands on the common paperclip are called force vectors.
If you put the arrows end-to-end they should ideally form a closed triangle (i.e., the last arrow should end where the first one began). This would indicate that when the forces are combined, they add up to zero.
If two of the rubber bands are perpendicular, then you can define an x-y coordinate plane so that one force acts along the x axis and the other along the y axis.
The force in the direction you chose as the x direction can point either in the positive or negative x direction. The same is so for the force in the y direction.
The third force will not be directed along either of these axes; its direction will be partially in the x and partially in the y direction.
Label the force in the x direction A, the force in the y direction B and the third force C.
Each arrow represents the direction of one of the forces, and the relative lengths of the arrows represent the relative 'strengths' of the three forces (i.e., the longest vector represents the greatest of the three forces, the shortest vector represents the least of the three forces, with the lengths in proportion to the forces).
We will refer to the arrows as the force vectors A, B and C.
Look at the sketch you made of the three forces.
Does force A (the one in the x direction) act in the positive or negative x direction?
negative
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Does force B (the one in the y direction) act in the positive or negative y direction?
positive
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Into what quadrant does force C (the third force) point? (if the x-y plane is in the standard vertical vs. horizontal orientation, the first quadrant corresponds to the upper right, the second to upper left, the third to lower left, the fourth to lower right)
fourth
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Which is greater, force A or force B?
force B
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Which is longer, the A arrow or the B arrow?
B arrow
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Which is the longest of the three arrows?
C arrow
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Now draw the projection lines from the tip of arrow C to the x and y axes:
The x projection line is a dotted line from the tip of the C arrow to the x axis, and is perpendicular to the x axis.
The y projection line is a dotted line from the tip of the C arrow to the y axis, and is perpendicular to the x axis.
Is one of your projection lines parallel to the x axis? If so, which one is it?
The y projection line
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Is one of your projection lines parallel to the y axis? If so, which one is it?
The x-projectin line
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Finally draw arrows from the origin to the ends of your projection lines:
Draw an arrow starting at the origin and running along the x axis ,ending at the point where the x projection line meets the x axis.
This arrow represents what we call the x projection of the vector C.
Label this arrow Cx.
Draw an arrow starting at the origin and running along the y axis to the point where the y projection line meets the y axis.
This arrow represents what we call the y projection of the vector C.
Label this arrow Cy.
Which is longer, the x projection of C or the y projection of C?
Cx
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Which is longer, the x projection of C or the vector A?
Cx, but in theory they should be equal?
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Which is longer, the y projection of C or the vector B?
They appear approxmiately equal.
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Place in order of length, from shortest to longest: C, Cx, Cy, A, B.
A, B, Cy, Cx, C
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Homework:
Your label for this assignment:
ic_class_090923
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.
Very good work.