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course phy 201
10/8 1230 am
120924The experiment Error Analysis I is assigned. It involves timing a pendulum using the TIMER program, obtaining 30 time intervals corresponding to 30 cycles, calculating the mean time interval for a cycle, calculating by how much each cycle deviates frrm the mean, and using the deviations to calculate the mean and standard deviations. This experiment generally requires about an hour. A spreadsheet isn't necessary, but if you know how to use it the spreadsheet would be appropriate for calculating deviations, squaring them and adding them up. This would probably save you about 20-30 minutes of tedious calculation. Instructions for this experiment are detailed and you should have little trouble following them.
You are invited to submit the Practice Major Quiz (qa) for my scrutiny and feedback.
You were asked last time to read Chapter 2 of the text. You should now work out some prohlems:
Text assignment 10:
`gGeneral College Physics (Giancoli text)
Text Problems Chapter 2, Problems 1-5, 6, 9, 11, 16, 22, 27
`uUniversity Physics (Young text)
Text Chapter 2, Problems 54, 57, 60, 63, 66
Optional Openstax: (vectors quantities defined section 2, graphs in section 3, acceleration section 4, eqns section 5, prob solving section 6, free fall section 7, graphical analysis section 8, )
General College Physics: Problems 2, 3, 6, 7, 9, 10, 13, 14, 16, 19
qa documents 9 and 10 introduce forces and units of force. The questions for today's class will make more sense if you do these qa's first.
qa document 11 isn't actually assigned until 120926. It's listed here in case you're on a roll after #'s 9 and 10 and want to go ahead and knock out 11 as well.
qa_09 Forces exerted by gravity; nature of force; units of force
qa_10 Force and Acceleration
qa_11 Situations involving forces and accelerations.
Hypothesis Testing with Data from Pearl Pendulum and Ball Down Ramp
'Seed' questions corresponding to the Major Quiz are listed below. You should at least read through them, and if you think it would be beneficial and have time you're welcome to submit some (or all) of them. However they aren't required; they are simply available if you have occasion to use them.
1.1
1.2
2.1
2.2
3.1
3.2
4.1
5.1
6.1
7.1
7.2
8.1
8.2
9.1
Introductory Problem Sets: Work through Set 3, #'s 1-11, which are related to force, work and energy. As always with Introductory Problem Sets you should be sure you know how to work all of these problems, since about 30% of each test consists of Introductory Problem Set problems. These problems should reinforce and give an additional perspective on the rest of this assignment.
synopsis: You might find the synopsis useful through Idea #8 (Newton's Laws). The synopsis is for your reference.
You should also take a good look at the Linked Outline, which many students have found very beneficial in preparing for the Major Quiz.
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Information useful for today's questions:
The work done by a force on an object, as the object is displaced, is equal to the average force multiplied by the displacement. Average force is found with respect to position rather than time.
As stated in class, the net force on an object can accelerate and/or deform the object. If the object is deformed, it may be deformed in such a way that when the force is removed it can return to its original shape, or it can remain wholly or partially deformed.
If the object is not deformed then the work done by the net force goes into changing the object's kinetic energy. If the object is deformed the work that goes into deforming it will not go into changing its kinetic energy. The energy that goes into deforming the object will go into some combination of potential energy and thermal energy. The potential energy can be recovered in the form of work; some portion but not all of the thermal energy can also be recovered in the form of work.
There's a whole lot in those statements, and to tell the whole truth they need just a little more information. We aren't going to try to get to the whole truth at this point. We're just going to worry about kinetic energy and work for right now, and soon we'll worry a little bit about potential energy. We'll keep in mind that thermal energy can also be involved, but it will be quite awhile before we attempt to do much with it except acknowledge that when work is done, changes in thermal energy are usually involved.
If the work done by the net force all goes into accelerating the mass, with none going into rotating it, then the net force, mass and acceleration are related by Newton's Second Law, which states that under these conditions
F_net = mass * acceleration.
Mass could be in dominoMasses, grams, kilograms or other units. Acceleration could be in cm / second, meters / second or other units. Any unit of mass multiplied by any unit of acceleration will give you a unit of force. Examples would include grams * meters / second^2, kilograms * cm / second^2, dominoMasses * miles/hour. These units are all interconvertible (i.e., by using basic conversion factors any of these units can be expressed as a certain number of any of the others).
Another possible unit of force is the kilogram * meter / second^2. This is the standard SI unit of force, and is given the special name Newton. A Newton is a kilogram * meter / second^2.
In some questions you will want to express your units of force in Newtons, but unless you have a good reason to do so it isn't at this point necessary, as long as you use valid units of force.
Use these ideas as necessary to help answer the questions below.
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`q001. The work done on a certain rubber band, as it stretches from length 64 cm to 68 cm, is represented by the area of a force vs. length trapezoid with 'graph altitudes' 0.4 Newtons and 0.8 Newtons, and trapezoid 'width' 4 cm.
What is the area of this trapezoid?
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.6 * 4 = 2.4n cm
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What is the average force exerted over this 4 cm interval?
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.06 newtons
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What is the product of the average force and the displacement?
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.6* 4 = 2.4n cm
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Why is the answer to the last question equal to the area of the trapezoid?
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Because the average force is the average distance between .4n and .8 n, when you use the average force it is like cutting the top off a triangle and making it into a rectangle then multipy it by the base width to get the area.
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How much work is done on the rubber band?
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.6 * 4 = 2.4ncm
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Where did the energy to do that work come from?
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the energy came from the recoil of the rubberband as potential energy that was stretched using the power of cheerios.
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`q002. If the rubber band in the preceding exerts a force of 0 Newtons when its length is 60 cm, then assuming that the force vs. length graph is a straight line what is the area of the corresponding trapezoid?
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.4n * 8cm = 3.2ncm
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What is the average force?
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(.8+0)/ 2 = .4n
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Why should the work done on the rubber band for this interval be equal to the area under the trapezoid?
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because the rubberband didn't start stretching until it reached 60cm.
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`q003. The kinetic energy of a mass as it moves from point to point is 1/2 m v^2, where m is its mass and v its velocity.
A typical domino has mass around 0.015 kilograms.
When a domino hits the floor after being dropped from a height of 1 meter its velocity is about 4.4 meters / second.
What is its kinetic energy at that velocity?
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.5 * .015 *4.4^2 = .1452kgm/s
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When it counts (e.g., on a test), be sure you use units throughout your calculations.
.5 * .015 kg * (4.4 m/s)^2 = .1452 kg * m^2 / s^2, not kg m/s.
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Gravity exerts a force of about 0.15 Newtons on this domino. How much work is done on it by gravity as the domino falls that 1 meter to the floor?
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A Newton is a kilogram * meter / second^2. How does the kinetic energy gained by the domino as it falls compare to the work done on it by gravity?
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`q004. Give your data from today's experiments, along with a brief explanation.
**** We attached a magnet to the back of a toy car and used another magnet to move the car
Distance from each other distance car traveled
7cm 17.1cm
7cm 17cm
6cm 26cm
6cm 25cm
5cm 33cm
5cm 34.5cm
4cm 44cm
4cm 45cm
3cm 69cm
3cm 71cm
2cm 104cm
2cm 101cm
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Sketch a graph of coasting distance vs. proximity of the magnets and give a brief description of the graph.
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It is decreasing to the right, not a straight line
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What is your best result for the acceleration of the car as it moves across the tabletop?
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31cm/s^2
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Let's assume that the object consisting of the car plus the magnet and domino(es) has a mass of 0.100 kilograms. How much force does it take to accelerate it at the rate you just gave?
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31cm/s^2 * .1 = 3.1 newtons
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A Newton is a kg * m / s^2.
The unit of your calculation would be kg cm / s^2.
Replacing the cm by .01 m you would find that
31 kg cm/s^2 = 31 kg (.01 m) / s^2 = .31 kg m/s^2.
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.31 kg m^2 = .31 Joule.
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How much work is therefore done by the accelerating force as the car travels 10 centimeters?
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.1kg * 10cm = 1 newton
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That would be 1 kg * cm, which is .01 kg * m = .01 N.
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`q005. We will define rubber band stretch as its length, minus the shortest length observed. So for example if you observed chain lengths 21.5, 23.0, 24.0 and 24.5 centimeters, you would subtract the 21.5 centimeters to get stretches of 0, 1.5, 2.5 and 3.0 centimeters.
Sketch a graph of rubber band stretch vs. magnet proximity, including an approximate curve following the trend of your data, and briefly describe your graph.
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Plots are (.6,7.5) (1.7 , 3.7) (1 , 5.3) (2.5 , 3.5)
Decreasing to the right, big curve
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Good.
Generally you want to present your data in an ordered fashion; easier on the reader.
e.g.,
(.6,7.5) , (1 , 5.3), (1.7 , 3.7), (2.5 , 3.5)
In this order the eye can quickly discern the trend: Stretch increasing from left to right, proximity decreasing.
Also stretch vs. proximity would have proximity first.
No problem, but note for future reference.
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Estimate the area under your graph to the right of length 4 centimeters.
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3.5prox * .8 stretch = 2.8
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Estimate the area under your graph to the right of length 3 centimeters.
****.95* 4.5 = 4.28
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If every centimeter of stretch corresponds to a force of 0.1 Newton, how many Newton * cm are represented by each of your graph areas?
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2.5*.1n = .25ncm
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`q006. University Physics:
The graph of coasting distance vs. proximity is given by y = 7800 cm^3.32 * x^(-2.32), where y is coasting distance and x is proximity.
Coasting distance is assumed to be proportional to the work done by the magnetic forces. If A is the conversion factor from centimeters of coasting to Newton * cm of work, then, W = 7800 cm^3.32 * A x^(-2.32) gives the work done by the magnetic force as a function of proximity.
Evaluate your function to find the force at x = 3, 5 and 7 cm.
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What is the average rate of change of W with respect to x on each of the two x intervals corresponding to your previous results?
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What is the meaning of this rate of change? (hint, if needed: what do you get if you multiply this rate of change on an interval by `dx?)
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What therefore is the relationship between the work vs. x function and the force vs. x function?
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Based on W = 7800 cm^3.32 * A x^(-2.32), what is the function for force vs. x?
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The conversion factor A is the number of Newton * cm of work done by the accelerating force, per centimeter of coasting. Assuming (as in an earlier question) that your car plus magnets and dominoes has a mass of .1 kilogram, you can use the acceleration of your car to figure out the frictional force which brings it to rest. Then you can figure out how many Newton * cm of work are done as the car moves 1 centimeter. This result will be your value of A.
So what is the value of A?
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What therefore is the force when proximity x is 4 centimeters?
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According to your force vs. stretch graph, if each centimeter of stretch corresponds to .16 Newtons of force, how much force is exerted when proximity x is 4 centimeters?
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Your two results for force should theoretically be the same. However experimental uncertainties, not to mention the law of natural cussedness than infests actual physics systems causing them to misbehave in the most inconvenient possible manner, dictate that there will be a difference. How well do your two results compare and to what do you attribute any differences?
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Very good.
Do check my notes, but you're in good shape here.
Note on today's class:
The moment of inertia of your ramp is 1/12 * .2 kg * (.6 m)^2 = 1/12 * .072 kg m^2 = .006 kg m^2.
At 5.4 rad / second, its KE would be
KE = .5 * .006 kg m^2 * (5.4 rad/s)^2 = .09 Joules.
You guys were getting .9 Joules and I couldn't figure out what was wrong, since this is about 6 times as great as the energy you put into the rubber band.
The problem was that you were using .06 kg m^2 rather than .006 kg m^2, and that was my fault for writing as .06.
If you include the moments of inertia of the dominoes, it looks like you're going to end up with at least 70% of the energy you put into the dominoes represented as the KE of the rotataing system.
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