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course phy 201
10/14 2:54 am
`q001: Lab investigation:You investigated the question of whether a fixed amount of energy imparted to a system has a consistent result.
The energy came from a rubber band chain.
The system was a rotating ramp with dominoes on its ends.
Give a description of the system and how it was used.
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There was a rubber band chain, stretched a set distance from the ramp. This was done with the rubber band point 30, 15, and 7.5 cm from the center.
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Give your data.
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Distance from center; revolutions (in degrees)
30 cm; 3045 degrees
30 cm; 3330 degrees
30 cm; 3105 degrees
30 cm; 3050 degrees
30 cm; 3205 degrees
15 cm; 2835 degrees
15 cm; 2745 degrees
15 cm; 2790 degrees
15 cm; 2790 degrees
15 cm; 2700 degrees
7.5 cm; 1720 degrees
7.5 cm; 1215 degrees
7.5 cm; 2250 degrees
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Show how you analyzed your data.
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It was rather simple. Take the mean/median of each data set, and find if there is any obvious differences.
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Give your conclusions.
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Applying the force from the edge of the ramp (30 cm difference) very clearly had more revolutions than the other two lengths. Each set had a somewhat significant drop in average revolutions turned
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If you have collateral observations and/or ideas for extending this investigation, give a synopsis of your observations and/or ideas.
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It might be helpful to try using different lengths of rubber band, to see if increasing or decreasing the force applied has the same results.
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`q002. Your ramp has mass 200 grams and length 60 cm. By itself its moment of inertia is 1/12 M L^2, where M is its mass and L its length. What is its moment of inertia?
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Moment of inertia = (1/12*200) * 60^2; moment of inertia = 60,000 grams. Or 60 kg;
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To get your result you multiplied grams by centimeters^2. That doesn't give you units of grams.
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Your dominoes each had mass about 16 grams. How far was the center of each domino from the axis of rotation (your best estimate, recalling that the dimensions of a domino are 5 cm long by 2.5 cm wide by about .8 or .9 cm thick) and what was the moment of inertia of each?
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Each domino was placed on its side with the end of it matching to the end of the ramp. As such, its center would have been 2.5 cm from the end of the ramp, or 27.5 cm from the center; 16/12 = 1.33 grams; 27.5^2 = 756.25 meters; 1.33 * 756.25 = 1005.81 grams, roughly 1 kg;
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1/12 M L^2 applies to a uniform rod rotating about its center. It doesn't apply to a mass concentrated at or near the 27.5 cm position. The moment of inertia of a mass m located at distance r from the axis of rotation is just m r^2.
(27.5 meters)^2 = 756.25 meters,
but you don't have anything of length 27.5 meters.
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What therefore was the moment of inertia of your system?
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moment of inertia; 60,000 + 1005.81 = 61,005.81 grams
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If you correct your results you will find that the moment of inertia of each domino is around 12 000 gram cm^2. There are two dominoes, so the total moment of inertia will be over 80 000 gram cm^2.
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`q003. In the experiment your rubber band had some average tension as it snapped back from release. We want to find the work done on the system as it snapped back.
Using a reasonable rough estimate, how many dominoes do you think the rubber band would have supported at the stretched length of your chain?
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nine dominoes gives the exact length
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At .16 Newtons per supported domino, roughly how much force did the chain therefore exert at this length?
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1.44 Newtons
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What approximate average force did the chain therefore exert between release and losing contact with the rotating system?
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1.44 Newtons
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Through what distance do you think this average force was exerted?
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40 cm
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How much work did the chain therefore do on the system?
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1.44 * 40 = 57.6 cmNewtons
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You have calculated the moment of inertia I for the system, and you have the definition KE = 1/2 I omega^2. If the work done by the chain on the system all went into kinetic energy, what then should omega have been at the instant the rubber band chain lost contact? Include units throughout your solution. You probably won't know how to interpret the units of your final solution, but don't worry about that.
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I = 61,005.81 grams; KE = 57.6 cmNewtons; convert I into kilograms for proper comparison with newtons (kgm/s^2); I = 61.006 kg; 57.6 cmNewtons = ½(61.006 kg)*omega^2; 57.6 = 30.503 * omega^2; 1.88 = omega^2; roughly 1.37 = omega;
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Your units aren't correct (see previous notes).
I believe that with correct units you'll get something in the very general vicinity of 10 radians / sec^2, but check it out by redoing the calculation with correct units.
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How much kinetic energy would the system therefore have lost as it coasted to rest?
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57.6 cmNewtons; It would have lost all the kinetic energy it started with.
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How much work was therefore done by the net force acting on the system?
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57.6 cmNewtons;
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Through how many revolutions did the system you observed rotate, and how much kinetic energy would therefore have been lost per rotation?
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At 30 cm: average number of rotations was around 8 rotations; This amounts to around 7.2 cmNewtons lost per rotation. At 15 cm: the average number of rotations was around 7.5 rotations; This amounts to around 7.68 cmNewtons lost per rotation
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`q004. Suppose that another system gains 5 Joules of energy from an ideal (and hence conservative) rubber band chain, while losing 2 Joules of energy to friction. How many Joules of kinetic energy would you expect it to gain?
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3 joules
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For this situation, identify each of the quantities `dW_noncons_ON, `dKE and `dW_cons_ON.
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dW_noncons_ON: 3; dKE: 5; dW_con_ON: 2
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Reconcile your conclusions with the equation `dW_noncons_ON = `dKE - `dW_cons_ON.
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3 = 5-2
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Identify `dPE for this situation.
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dPE = 5
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Write the above equation in terms of `dW_noncons_ON, `dKE and `dPE
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DPE = dW_noncon_ON
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"
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You're on the right track, but you need to modify details of some of your calculations.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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