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The Atwood system consists of the paperclips suspended over the pulley. A total of six large clips connected by a thread were suspended, three from each side of the pulley. The system was released and, one side being slightly more massive than the other due to inconsistencies in the masses of the clips, accelerated from rest, with one side descending and the other ascending. The system accelerated through 50 cm in a time interval between 4 and 6 seconds; everyone used their 8-count to more accurately estimate the interval. Then a small clip was attached to the side that had previously ascended. This side now descended and the system was observed to now descend is an interval that probably lasted between 1 and 2 seconds.
If you weren't in class you can assume time intervals of 5 seconds and 1.5 seconds. Alternatively you can wait until tomorrow and observe the system yourself; the initial observation requires only a couple of minutes.
`qx001. What were your counts for the 50 cm descent of the Atwood system?
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3.4 and 1.2 seconds
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`qx002. What were the two accelerations?
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Average Velocities are:
50cm / 3.4s = 14.7cm/s
50cm/1.2s= 41.7 cm/s
The accelerations would respectively be:
14.7cm/s / 3.4s = 4.3cm/s^2
41.7cm/s / 1.2s = 34.8cm/s^2
Good, except that acceleration is change in velocity / change in clock time, not average velocity / change in clock time.
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`qx003. Why did the systems accelerate?
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Gravity was working on the paper clips at 9.8 m/s^2 giving it a uniform acceleration towards the core of the earth.
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`qx004. Suppose the large paperclips all had mass 10 grams, the small clip a mass of 1 gram. What then was the net force accelerating the system on the first trial, and what was the net force on the second?
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1 large paper clip and 1 small paperclip would amount to 11 grams in total for the first trial.
3 large paper clips and 1 small paperclip would amount to 31 grams in total for the second trial.
11 grams * 4.3cm/s^2 =47.3 g*cm/s^2
31 grams * 34.8cm/s^2 = 1078.8 g*cm/s^2
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.. if uncertainty +-1%
`qx005. Given the masses assumed in the preceding, what is the force acting on each side of the system? What therefore is the net force on the system?
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Since trial one had a force of 47.3 g*cm/s^2 descending, the other side of the system would have the same net force ascending presuming the acceleration remained constant, same for trial 2.
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`qx006. Based on your counts and the resulting accelerations, do you think the ratio of the masses of the large to small paperclips is greater than, or less than, the 10-to-1 ratio assumed in the preceding two questions?
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It is less than a 10-to-1 ratio, it is more of a 3-to-1 ratio, divide 31 grams / 11 grams = about 3 to 1
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`qx007. If the mass of each larger clip is M and the mass of a smaller clip is m, what would be the expressions for the net force accelerating the system? What would be the expression for the acceleration of the system?
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Netforce = M+m * a
Acceleration = (v0 + vf) / 2
this would not give you acceleration. THis is the average of initial and final velocities.
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`qx008. If the mass of the each of the larger clips is considered accurate to within +-1%:, would this be sufficient to explain the acceleration observed when 3 large clips were hung from each side?
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There would be a +-6% error, it would be relatively sufficient being under 10% error.
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... sample the accelerations for random divisions of the six large clips ... predict what the distribution of masses would look like ...
Magnet and Balance
Everyone was given a small magnet and asked to achieve a state where the balance was in an equilibrium position significantly different from that observed without the magnet. It was suggested that the length of the suspended clip beneath the surface of the water should differ by at least a centimeter.
... assuming 1 mm diam ...
`qx009. Describe in a few lines your efforts to achieve the desired result. What worked, what didn't, what difficulties presented themselves, etc.?
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To actually get the paperclip to stay submerged while balancing the beam presented to take a while. After the beam was perfectly balanced however, the paperclip easily stayed submerged because of the surface tension of the water acting as a grip, it slightly pulled the beam down on that side.
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`qx010. How much difference was there in the length of clip suspended in the water? If you didn't actually measure this, give a reasonable estimate.
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The length of the clip suspended was about 2 centimeters
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`qx011. How did you adjust the magnet? If you wanted to quickly increase or decrease the length of the suspended paper clip beneath the surface by 1 millimeter, using only what you had in front of you during the experiment, how would you go about it?
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To shift the system downward towards the paperclip side you would move the magnet more toward the outside of the beam, and move it to the inside of the system to move the beam upward on the paperclip side.
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`qx012. Assuming the diameter of the suspended clip to be 1 millimeter, by how much did the buoyant force on the suspended clip change? How much force do you therefore infer the magnet exerted? If you have accurate measurements, then use them. Otherwise use estimates of the positions of various components as a basis for your responses.
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The magnet would have to have exerted just a little bit less force than the paperclip’s buoyant force, otherwise the paperclip would not have stayed partially submerged.
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&#Good responses. Let me know if you have questions. &#
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