course Phy: 201
Class 090923The following conventions will allow your instructor to quickly locate your answers and separate them from the rest of any submitted document, which will significantly increase the quality of the instructor's feedback to you and to other students.
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 &&&&.
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If you don't follow these guidelines you may well be asked to edit your document to make the changes before I can respond to it.
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.
1 domino, 65 centimeters
2 dominos, 69 centimeters
3 dominos, 71.5 centimeters
4 dominos, 75 centimeters
5 dominos, 78.5 centimeters
On a table of y vs. x, the x column (generally the independent variable) is listed first.
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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
The graph looks like 4 trapezoids. The first slope starts off steep and become less steep by each interval. The trapezoids are higher on each interval; however, the width of each is different.
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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?
Ό, 2/5, 2/7, 2/7
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What are the four areas?
6, 6.25, 12.25, 15.75
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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 represents the number of dominos, the run represents the length of the rubber bands in centimeters, the slope represents length of rubber band chain with respect to number of dominos.
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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 represents the number of dominos, the base represents the length of the chain, the area represents change of the length of rubber band chain.
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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:
1 rubber band, 6 half-cycles, 4.45 cm/s^2
2 rubber bands, 3.3 half-cycles, 14.7 cm/s^2
3 rubber bands, 1.7 half-cycles, 55.4 cm/s^2
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The two slopes obtained when you graph acceleration vs. number of added rubber bands for the Atwood machine:
10.25, 40.7
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The coordinates of the three points you graphed:
1, 4.45
2, 14.7
3, 55.4
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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:
Flat = 4 half-cycles, 3.75 cm/half-cycles^2
Top = 2.5 half-cycles, 9.6 cm/half-cycles^2
Side = 3 half-cycles, 6.7 cm/half-cycle^2
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The two 'graph slopes' obtained when you graph acceleration vs. ramp slope:
60/59, 30/29
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The coordinates of the three points you graphed:
1, 4.45
2, 14.7
3, 55.4
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Give a detailed description of how you proceeded from raw data to acceleration for one of the Atwood Machine trials:
First I figured the average rate of change or average velocity. I took 80 cm and divided it by the number of half-cycles; for the first trial it was 6. Then I took this number and set up the equation to find out final velocity. For the first trial, 13.3cm/hc = (0 + Vf)/2. So the final velocity is 26.7 cm/hc. Then I can use the equation for average acceleration. A_ave = (Vf V0)/ dt. So for the first ramp, A_ave = (26.7cm/hc 0 cm/hc)/6 hc = 4.45 cm/hc^2. Then I repeated this process for each ramp.
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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:
First I figured the average rate of change or average velocity. I took 30 cm and divided it by the number of half-cycles; for the first ramp it was 4. Then I took this number and set up the equation to find out final velocity. For the first trial, 7.5cm/hc = (0 + Vf)/2. So the final velocity is 15 cm/hc. Then I can use the equation for average acceleration. A_ave = (Vf V0)/ dt. So for the first ramp, A_ave = (15cm/hc 0 cm/hc)/4 hc = 3.75 cm/hc^2. Then I repeated this process for each ramp.
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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 = Vo + (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:
1. [Vf] Vo = [Vo + (F_net/m) dt] Vo
2. m [Vf Vo = (F_net/m) dt]
3. mVf mVo = F_net * dt
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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 = Vo^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:
1. [Vf^2] Vo^2 = [Vo^2 + 2(F_net/m) ds] Vo^2
2. ½ m [Vf^2 Vo^2 = 2(F_net/m) * ds
3. mVf^2 mVo^2 = F_net * ds
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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 amount of contact put on an object. If you apply more force, the more the object will move. This is the same with the time interval. The longer you push an object, the farther it will move.
Momentum is the amount of potential energy an object has. So the faster you push it, the more momentum it has.
Impulse Momentum Theorem is the higher the impulse the higher momentum or potential energy an object has.
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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 It takes twice as much work to lift twice the weight to the same height.
Kinetic Energy Energy an object has when its in motion.
Work Kinetic Energy Theorem work is when an oject is in motion; so therefore, the work and kinetic energy equal each other.
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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?
Does force B (the one in the y direction) act in the positive or negative y direction?
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)
Which is greater, force A or force B?
Which is longer, the A arrow or the B arrow?
Which is the longest of the three arrows?
Positive
Negative
II
A
B
C
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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?
Is one of your projection lines parallel to the y axis? If so, which one is it?
X projection line
Y projection 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?
Which is longer, the x projection of C or the vector A?
Which is longer, the y projection of C or the vector B?
Place in order of length, from shortest to longest: C, Cx, Cy, A, B.
X projection of C
They are the same
They are the same
B = Cy < A = Cx < C
I've looked at submissions from well over half the class, including most the students who (like you) seem to be understanding things well, and this is the first good answer I've seen to this question.
Keep up the good work.
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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. See my notes.